Worn-tile Scrabble












35












$begingroup$


Problem



You're stuck in a cabin in the middle of the woods, with only an old scrabble set to entertain yourselves. Upon inspection you see that the scrabble letters are so worn, that only the points for each letter are visible.



Nonetheless you decide to play a game. You pull seven letters from the bag and place them on your tray, your challenge is to determine what those letters could be.



So generally, given a list of points convert it into any possible string or list of letters.





Scrabble Tiles and Distributions




  • 2 blank tiles (scoring 0 points)

  • 1 point: E ×12, A ×9, I ×9, O ×8, N ×6, R ×6, T ×6, L ×4, S ×4, U ×4

  • 2 points: D ×4, G ×3

  • 3 points: B ×2, C ×2, M ×2, P ×2

  • 4 points: F ×2, H ×2, V ×2, W ×2, Y ×2

  • 5 points: K ×1

  • 8 points: J ×1, X ×1

  • 10 points: Q ×1, Z ×1


So if you have a list of points [10,10,8,5,1,1,1] then "QZJKEEE" would be valid but "QQJKEEE" would not be valid (since there is only 1 Q tile in the bag)





Problem Specific Rules




  • You may assume all inputs are valid and that there will always be 7 tiles (i.e it wont be a list of seven 10 point tiles and won't be 9 tiles)

  • You can assume no tiles have been previously pulled from the bag (so the distribution is the standard distribution of english tiles as defined above)

  • You do not have to generate a valid word, only a valid string of letters.

  • The order of your string is irrelevant as long as for each tile there is a corresponding letter.

  • Points are based on the standard english scrabble tile points as defined above.

  • You may output in upper or lower case, for a blank tile you may output either a space character or an underscore '_'

  • Your answer may output as any reasonable representation of the tiles such as a List, String, Array or Sequence





General rules:




  • This is code-golf, so shortest answer in bytes wins.

    Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.


  • Standard rules apply for your answer with default I/O rules, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.


  • Default Loopholes are forbidden.

  • If possible, please add a link with a test for your code (i.e. TIO).

  • Also, adding an explanation for your answer is highly recommended.




Test Cases



Obviously since you can output any possible value it's difficult to define strict test cases.



Some cases with a possible valid return value:



[10,0,10,5,8,8,0] -> "Q ZKJX "
[1,1,1,1,1,1,1] -> "EEEEEEE"
[1,2,3,4,5,8,0] -> "NDBHKJ "
[2,2,2,2,2,2,2] -> "DGDGDGD"


Some cases with an invalid return value:



[10,0,10,5,8,8,0] -> "Q QKJX "  - Too many Qs 
[1,1,1,1,1,1,1] -> "EEEEEE " - Space is 0 points not 1
[1,2,3,4,5,8,0] -> "NDBH" - Too short
[1,2,3,4,5,8,0] -> "NDBHKJ I" - Too long
[1,2,3,4,5,8,0] -> "ÉDBHKJ1" - Contains none scrabble characters
[2,2,2,2,2,2,2] -> "GDGDGDG" - Contains too many Gs (case for invalid cycling)









share|improve this question











$endgroup$












  • $begingroup$
    Do I need to output a string or is a list ok?
    $endgroup$
    – Maltysen
    2 days ago










  • $begingroup$
    You can output a list, I'll update the question
    $endgroup$
    – Expired Data
    2 days ago






  • 1




    $begingroup$
    What can I output for a blank?
    $endgroup$
    – Maltysen
    2 days ago






  • 3




    $begingroup$
    Suggested test case: [2,2,2,2,2,2,2] (the only case where it's important to start with a D rather than a G if a cycling method is used)
    $endgroup$
    – Arnauld
    2 days ago






  • 1




    $begingroup$
    Notifications are @ then the person's name without spaces. I.e. Expired Data would become @ExpiredData.
    $endgroup$
    – Tau
    2 days ago
















35












$begingroup$


Problem



You're stuck in a cabin in the middle of the woods, with only an old scrabble set to entertain yourselves. Upon inspection you see that the scrabble letters are so worn, that only the points for each letter are visible.



Nonetheless you decide to play a game. You pull seven letters from the bag and place them on your tray, your challenge is to determine what those letters could be.



So generally, given a list of points convert it into any possible string or list of letters.





Scrabble Tiles and Distributions




  • 2 blank tiles (scoring 0 points)

  • 1 point: E ×12, A ×9, I ×9, O ×8, N ×6, R ×6, T ×6, L ×4, S ×4, U ×4

  • 2 points: D ×4, G ×3

  • 3 points: B ×2, C ×2, M ×2, P ×2

  • 4 points: F ×2, H ×2, V ×2, W ×2, Y ×2

  • 5 points: K ×1

  • 8 points: J ×1, X ×1

  • 10 points: Q ×1, Z ×1


So if you have a list of points [10,10,8,5,1,1,1] then "QZJKEEE" would be valid but "QQJKEEE" would not be valid (since there is only 1 Q tile in the bag)





Problem Specific Rules




  • You may assume all inputs are valid and that there will always be 7 tiles (i.e it wont be a list of seven 10 point tiles and won't be 9 tiles)

  • You can assume no tiles have been previously pulled from the bag (so the distribution is the standard distribution of english tiles as defined above)

  • You do not have to generate a valid word, only a valid string of letters.

  • The order of your string is irrelevant as long as for each tile there is a corresponding letter.

  • Points are based on the standard english scrabble tile points as defined above.

  • You may output in upper or lower case, for a blank tile you may output either a space character or an underscore '_'

  • Your answer may output as any reasonable representation of the tiles such as a List, String, Array or Sequence





General rules:




  • This is code-golf, so shortest answer in bytes wins.

    Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.


  • Standard rules apply for your answer with default I/O rules, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.


  • Default Loopholes are forbidden.

  • If possible, please add a link with a test for your code (i.e. TIO).

  • Also, adding an explanation for your answer is highly recommended.




Test Cases



Obviously since you can output any possible value it's difficult to define strict test cases.



Some cases with a possible valid return value:



[10,0,10,5,8,8,0] -> "Q ZKJX "
[1,1,1,1,1,1,1] -> "EEEEEEE"
[1,2,3,4,5,8,0] -> "NDBHKJ "
[2,2,2,2,2,2,2] -> "DGDGDGD"


Some cases with an invalid return value:



[10,0,10,5,8,8,0] -> "Q QKJX "  - Too many Qs 
[1,1,1,1,1,1,1] -> "EEEEEE " - Space is 0 points not 1
[1,2,3,4,5,8,0] -> "NDBH" - Too short
[1,2,3,4,5,8,0] -> "NDBHKJ I" - Too long
[1,2,3,4,5,8,0] -> "ÉDBHKJ1" - Contains none scrabble characters
[2,2,2,2,2,2,2] -> "GDGDGDG" - Contains too many Gs (case for invalid cycling)









share|improve this question











$endgroup$












  • $begingroup$
    Do I need to output a string or is a list ok?
    $endgroup$
    – Maltysen
    2 days ago










  • $begingroup$
    You can output a list, I'll update the question
    $endgroup$
    – Expired Data
    2 days ago






  • 1




    $begingroup$
    What can I output for a blank?
    $endgroup$
    – Maltysen
    2 days ago






  • 3




    $begingroup$
    Suggested test case: [2,2,2,2,2,2,2] (the only case where it's important to start with a D rather than a G if a cycling method is used)
    $endgroup$
    – Arnauld
    2 days ago






  • 1




    $begingroup$
    Notifications are @ then the person's name without spaces. I.e. Expired Data would become @ExpiredData.
    $endgroup$
    – Tau
    2 days ago














35












35








35


1



$begingroup$


Problem



You're stuck in a cabin in the middle of the woods, with only an old scrabble set to entertain yourselves. Upon inspection you see that the scrabble letters are so worn, that only the points for each letter are visible.



Nonetheless you decide to play a game. You pull seven letters from the bag and place them on your tray, your challenge is to determine what those letters could be.



So generally, given a list of points convert it into any possible string or list of letters.





Scrabble Tiles and Distributions




  • 2 blank tiles (scoring 0 points)

  • 1 point: E ×12, A ×9, I ×9, O ×8, N ×6, R ×6, T ×6, L ×4, S ×4, U ×4

  • 2 points: D ×4, G ×3

  • 3 points: B ×2, C ×2, M ×2, P ×2

  • 4 points: F ×2, H ×2, V ×2, W ×2, Y ×2

  • 5 points: K ×1

  • 8 points: J ×1, X ×1

  • 10 points: Q ×1, Z ×1


So if you have a list of points [10,10,8,5,1,1,1] then "QZJKEEE" would be valid but "QQJKEEE" would not be valid (since there is only 1 Q tile in the bag)





Problem Specific Rules




  • You may assume all inputs are valid and that there will always be 7 tiles (i.e it wont be a list of seven 10 point tiles and won't be 9 tiles)

  • You can assume no tiles have been previously pulled from the bag (so the distribution is the standard distribution of english tiles as defined above)

  • You do not have to generate a valid word, only a valid string of letters.

  • The order of your string is irrelevant as long as for each tile there is a corresponding letter.

  • Points are based on the standard english scrabble tile points as defined above.

  • You may output in upper or lower case, for a blank tile you may output either a space character or an underscore '_'

  • Your answer may output as any reasonable representation of the tiles such as a List, String, Array or Sequence





General rules:




  • This is code-golf, so shortest answer in bytes wins.

    Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.


  • Standard rules apply for your answer with default I/O rules, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.


  • Default Loopholes are forbidden.

  • If possible, please add a link with a test for your code (i.e. TIO).

  • Also, adding an explanation for your answer is highly recommended.




Test Cases



Obviously since you can output any possible value it's difficult to define strict test cases.



Some cases with a possible valid return value:



[10,0,10,5,8,8,0] -> "Q ZKJX "
[1,1,1,1,1,1,1] -> "EEEEEEE"
[1,2,3,4,5,8,0] -> "NDBHKJ "
[2,2,2,2,2,2,2] -> "DGDGDGD"


Some cases with an invalid return value:



[10,0,10,5,8,8,0] -> "Q QKJX "  - Too many Qs 
[1,1,1,1,1,1,1] -> "EEEEEE " - Space is 0 points not 1
[1,2,3,4,5,8,0] -> "NDBH" - Too short
[1,2,3,4,5,8,0] -> "NDBHKJ I" - Too long
[1,2,3,4,5,8,0] -> "ÉDBHKJ1" - Contains none scrabble characters
[2,2,2,2,2,2,2] -> "GDGDGDG" - Contains too many Gs (case for invalid cycling)









share|improve this question











$endgroup$




Problem



You're stuck in a cabin in the middle of the woods, with only an old scrabble set to entertain yourselves. Upon inspection you see that the scrabble letters are so worn, that only the points for each letter are visible.



Nonetheless you decide to play a game. You pull seven letters from the bag and place them on your tray, your challenge is to determine what those letters could be.



So generally, given a list of points convert it into any possible string or list of letters.





Scrabble Tiles and Distributions




  • 2 blank tiles (scoring 0 points)

  • 1 point: E ×12, A ×9, I ×9, O ×8, N ×6, R ×6, T ×6, L ×4, S ×4, U ×4

  • 2 points: D ×4, G ×3

  • 3 points: B ×2, C ×2, M ×2, P ×2

  • 4 points: F ×2, H ×2, V ×2, W ×2, Y ×2

  • 5 points: K ×1

  • 8 points: J ×1, X ×1

  • 10 points: Q ×1, Z ×1


So if you have a list of points [10,10,8,5,1,1,1] then "QZJKEEE" would be valid but "QQJKEEE" would not be valid (since there is only 1 Q tile in the bag)





Problem Specific Rules




  • You may assume all inputs are valid and that there will always be 7 tiles (i.e it wont be a list of seven 10 point tiles and won't be 9 tiles)

  • You can assume no tiles have been previously pulled from the bag (so the distribution is the standard distribution of english tiles as defined above)

  • You do not have to generate a valid word, only a valid string of letters.

  • The order of your string is irrelevant as long as for each tile there is a corresponding letter.

  • Points are based on the standard english scrabble tile points as defined above.

  • You may output in upper or lower case, for a blank tile you may output either a space character or an underscore '_'

  • Your answer may output as any reasonable representation of the tiles such as a List, String, Array or Sequence





General rules:




  • This is code-golf, so shortest answer in bytes wins.

    Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.


  • Standard rules apply for your answer with default I/O rules, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.


  • Default Loopholes are forbidden.

  • If possible, please add a link with a test for your code (i.e. TIO).

  • Also, adding an explanation for your answer is highly recommended.




Test Cases



Obviously since you can output any possible value it's difficult to define strict test cases.



Some cases with a possible valid return value:



[10,0,10,5,8,8,0] -> "Q ZKJX "
[1,1,1,1,1,1,1] -> "EEEEEEE"
[1,2,3,4,5,8,0] -> "NDBHKJ "
[2,2,2,2,2,2,2] -> "DGDGDGD"


Some cases with an invalid return value:



[10,0,10,5,8,8,0] -> "Q QKJX "  - Too many Qs 
[1,1,1,1,1,1,1] -> "EEEEEE " - Space is 0 points not 1
[1,2,3,4,5,8,0] -> "NDBH" - Too short
[1,2,3,4,5,8,0] -> "NDBHKJ I" - Too long
[1,2,3,4,5,8,0] -> "ÉDBHKJ1" - Contains none scrabble characters
[2,2,2,2,2,2,2] -> "GDGDGDG" - Contains too many Gs (case for invalid cycling)






code-golf string scrabble






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edited 2 days ago







Expired Data

















asked 2 days ago









Expired DataExpired Data

898216




898216












  • $begingroup$
    Do I need to output a string or is a list ok?
    $endgroup$
    – Maltysen
    2 days ago










  • $begingroup$
    You can output a list, I'll update the question
    $endgroup$
    – Expired Data
    2 days ago






  • 1




    $begingroup$
    What can I output for a blank?
    $endgroup$
    – Maltysen
    2 days ago






  • 3




    $begingroup$
    Suggested test case: [2,2,2,2,2,2,2] (the only case where it's important to start with a D rather than a G if a cycling method is used)
    $endgroup$
    – Arnauld
    2 days ago






  • 1




    $begingroup$
    Notifications are @ then the person's name without spaces. I.e. Expired Data would become @ExpiredData.
    $endgroup$
    – Tau
    2 days ago


















  • $begingroup$
    Do I need to output a string or is a list ok?
    $endgroup$
    – Maltysen
    2 days ago










  • $begingroup$
    You can output a list, I'll update the question
    $endgroup$
    – Expired Data
    2 days ago






  • 1




    $begingroup$
    What can I output for a blank?
    $endgroup$
    – Maltysen
    2 days ago






  • 3




    $begingroup$
    Suggested test case: [2,2,2,2,2,2,2] (the only case where it's important to start with a D rather than a G if a cycling method is used)
    $endgroup$
    – Arnauld
    2 days ago






  • 1




    $begingroup$
    Notifications are @ then the person's name without spaces. I.e. Expired Data would become @ExpiredData.
    $endgroup$
    – Tau
    2 days ago
















$begingroup$
Do I need to output a string or is a list ok?
$endgroup$
– Maltysen
2 days ago




$begingroup$
Do I need to output a string or is a list ok?
$endgroup$
– Maltysen
2 days ago












$begingroup$
You can output a list, I'll update the question
$endgroup$
– Expired Data
2 days ago




$begingroup$
You can output a list, I'll update the question
$endgroup$
– Expired Data
2 days ago




1




1




$begingroup$
What can I output for a blank?
$endgroup$
– Maltysen
2 days ago




$begingroup$
What can I output for a blank?
$endgroup$
– Maltysen
2 days ago




3




3




$begingroup$
Suggested test case: [2,2,2,2,2,2,2] (the only case where it's important to start with a D rather than a G if a cycling method is used)
$endgroup$
– Arnauld
2 days ago




$begingroup$
Suggested test case: [2,2,2,2,2,2,2] (the only case where it's important to start with a D rather than a G if a cycling method is used)
$endgroup$
– Arnauld
2 days ago




1




1




$begingroup$
Notifications are @ then the person's name without spaces. I.e. Expired Data would become @ExpiredData.
$endgroup$
– Tau
2 days ago




$begingroup$
Notifications are @ then the person's name without spaces. I.e. Expired Data would become @ExpiredData.
$endgroup$
– Tau
2 days ago










14 Answers
14






active

oldest

votes


















8












$begingroup$

JavaScript (ES6), 72 bytes



A shorter variant suggested by @supercat



a=>a.map(o=n=>'?ED?BWQ?_EG?CFZ?_EDJMH?K?EGXPV'[n*9.4+(o[n]=7-~o[n])&31])


Try it online!





JavaScript (ES6),  137 ... 84 78 77  76 bytes



Saved 10 bytes by using Neil's cycling method



Returns a list of tiles. Uses _ for blank tiles.



a=>a.map(o=n=>"____FHVWGDGD_K__BCMPEEEE_ZQ__XJ"[n*20%44%32+(o[n]=-~o[n])%4])


Try it online!



How?



For each number of points, we cycle through a group of exactly 4 tiles, starting with the second tile of each group (this is important for G vs D):



 points | group | max. sequence
--------+-------+---------------
0 | ____ | __
1 | EEEE | EEEEEEE
2 | GDGD | DGDGDGD
3 | BCMP | CMPBCMP
4 | FHVW | HVWFHVW
5 | _K__ | K
8 | _XJ_ | XJ }--- these letters may only appear once each
10 | _ZQ_ | ZQ /


All these groups are stored as a single string of 31 characters:



____FHVWGDGD_K__BCMPEEEE_ZQ__XJ
^ ^ ^ ^ ^ ^ ^ ^
0 4 8 12 16 20 24 28


NB: We don't need to store the final "_" in "_XJ_", as it will never be accessed anyway.



The number of points $n$ is converted to the correct index $i_n$ into this string with:



$$i_n=((20times n)bmod 44)bmod 32$$



  n | *20 | mod 44 | mod 32 | group
----+-----+--------+--------+-------
0 | 0 | 0 | 0 | ____
1 | 20 | 20 | 20 | EEEE
2 | 40 | 40 | 8 | GDGD
3 | 60 | 16 | 16 | BCMP
4 | 80 | 36 | 4 | FHVW
5 | 100 | 12 | 12 | _K__
8 | 160 | 28 | 28 | _XJ_
10 | 200 | 24 | 24 | _ZQ_


The current position in each group is stored in the object $o$.






share|improve this answer











$endgroup$













  • $begingroup$
    Advancing o[n] by 8 each time would cost one extra character for the advancement, but allow one to replace %4 and %32 both with &31 for a net win. My best, based on yours, would be a=>a.map(o=n=>('?ED?BWQ?_EG?CFZ?_EDJMH?K?EGXPV'[n*9.4+(o[n]=7-~o[n])&31])). A shorter, "almost" version is a=>a.map(o=n=>("_EDBFK_EDCHJQEGMVXZEGPW"[n+(o[n]=5-~o[n])%24])) but that approach would needs a compact way to map the values 8 and 10 into 11 and 12, plus a slight adjustment to the string to fix an off-by-one problem.
    $endgroup$
    – supercat
    2 days ago










  • $begingroup$
    @supercat Sounds good! I'll have a closer look at it tomorrow.
    $endgroup$
    – Arnauld
    2 days ago










  • $begingroup$
    @supercat Another interesting formula is '_??VKWZHQFP?M?CGBGXDJD'[(n*96+(o[n]=32-~o[n]))%68%33]||'E', with a lookup string of only 22 characters. The full code is still 2 bytes longer than your solution, though.
    $endgroup$
    – Arnauld
    yesterday



















7












$begingroup$


Charcoal, 33 bytes



⭆觧⪪”&↖“Vh_z↶∕¡⌈∨₂χ¹‖◨⌊″”¶ι№…θκι


Try it online! Link is to verbose version of code. Explanation:



 θ                  Input array
⭆ Map over elements and join
”...” Literal string " nEnDGnBCMPnFHVWnKnnnJXnnQZ"
⪪ ¶ Split on newlines
§ ι Indexed by current element
§ Cyclically indexed by
№…θκι Number of times current element has already appeared
Implcitly print





share|improve this answer









$endgroup$





















    5












    $begingroup$


    Jelly,  31 30 27  26 bytes



    “ñẒẎYñ(“Nut¦hß’ṃØA;€⁶ɓṢĖœị


    A monadic Link accepting a list of integers which yields a list of characters.

    - a mishmash of my previous, below, and my improvement of Nick Kennedy's



    Try it online!



    The output is not given in the same order as the input (this is allowed).



    Using 2 of my own additions to the language in an answer does not happen often! ( and ɓ here).



    How?



    “...“...’ṃØA;€⁶ɓṢĖœị - Link: list of integers, V     e.g. [10,1,0,3,2,1,10]
    “...“...’ - list of base 250 integers [28089224382041, 77611203526272]
    ØA - 'ABC...XYZ'
    ṃ - base decompress (vectorises) ["EDMFKZZJZQ", "NGPYKZZXZZ"]
    ;€ - for €ach: concatenate:
    ⁶ - a space ["EDMFKZZJZQ ", "NGPYKZZXZZ "]
    ɓ - start a new dyadic chain with swapped arguments - i.e. f(V,that)
    Ṣ - sort [0,1,1,2,3,10,10]
    Ė - enumerate [[1,0],[2,1],[3,1],[4,2],[5,3],[6,10],[7,10]]
    œị - multi-dimensional index into " NEGMZQ"
    (1-based and modular)




    previous @ 30



    “²rṛʂṂø5=Ɓṇ^N¥Y»⁾tky;⁶s2ɓṢĖUœị


    A monadic Link accepting a list of integers which yields a list of characters.



    Try it online!



    This one's output is also mixed-case (this is allowed).



    How?



    “...»⁾tky;⁶s2ɓṢĖUœị - Link: list of integers, V          e.g. [10,1,0,3,2,1,10]
    “...» - compression of dictionary entries:
    - "end", "GMP", "fyttes", "adj", and "xci" and the string "qz"
    - "endGMPfyttesadjxciqz"
    y - translate with:
    ⁾tk - ['t', 'k'] "endGMPfykkesadjxciqz"
    ;⁶s2ɓṢĖUœị - ...
    - ...then like the above method (except U reverses each pair of indices)
    " neGMzq"





    share|improve this answer











    $endgroup$













    • $begingroup$
      I think you made a typo in your first explanation. ' NWGMZQ' after the multi-dimensional index into would be quite a feat without any W in the string. ;)
      $endgroup$
      – Kevin Cruijssen
      yesterday






    • 1




      $begingroup$
      @KevinCruijssen - yws, typo fixwd; thanks!
      $endgroup$
      – Jonathan Allan
      yesterday



















    4












    $begingroup$

    Pyth - 92 86 83 81 80 75 60 52 49 42 36 bytes



    Loops through input, popping off the available letters. I just have one of each letter that together give 7 for that point category. Now using packed string encoding.



    K[M*L7c."B_êº çÑOÒ
    7âCkÑ"Lm.)@K

    K Assign to K
    [M Map list(for popping). Uses a quirk of M to splat each first
    *L7 Map repeating each string by 7
    c L Split on occurrences of 'L'
    ."..." Packed string encoding of the needed letters
    m (Q) Map on input (input is taken implicitly)
    .) Pop. This returns the first element after removing it
    @K Index into K
    (d) The loop variable is given implicitly


    Btw, this is the original letter string before encoding: "_ E DG BCMP FHVW K JX QZ".



    Try it online.






    share|improve this answer











    $endgroup$





















      3












      $begingroup$


      Perl 5, 71 bytes





      @a=(__,'E'x7,DDDDGGG,BBCCMMP,FFHHVVW,K,1,1,JX,1,QZ);say chop$a[$_]for<>


      Try it online!






      share|improve this answer









      $endgroup$





















        3












        $begingroup$


        05AB1E, 70 52 39 38 29 26 25 bytes



        {ε.•3Oû}α›ηö‡.ÝŽ{•2ôÁyèNè?


        -18 bytes thanks to @ExpiredData.

        -13 bytes by using the same extend to size 7 from @Maltysen's Pyth answer.

        -9 bytes by creating a port of @JonathanAllan's Jelly answer, so make sure to upvote him!

        -3 bytes thanks to @Emigna.



        Results in a list of characters, and uses lowercase letters and a space for blanks.



        Try it online or verify some more test cases.



        Explanation:





        {                      # Sort the (implicit) input-list
        ε # Map each character `y` in this list to:
        .•3Oû}α›ηö‡.ÝŽ{• # Push compressed string "endgmpfykkzzzzjxzzqz "
        2ô # Split into parts of size 2
        Á # Rotate it once towards the left so the space is leading
        yè # Use integer `y` to index into the string-pairs
        Nè # Then get the `N`'th character of the string-pair (with automatic
        # wraparound), where `N` is the index of the loop


        See this 05AB1E tip of mine (section How to compress strings not part of the dictionary?) to understand why .•3Oû}α›ηö‡.ÝŽ{• is "endgmpfykkzzzzjxzzqz ".





        Previous 38 bytes answer:



        .•Mñ&Àû«ì{₆v*Å+µ-•#ðšε7∍}IvDyèн©?ε®õ.;


        Try it online or verify some more test cases.



        Explanation:





        .•Mñ&Àû«ì{₆v*Å+µ-•    # Push compressed string "e dg bcmp fhvw k   jx  qz"
        # # Split on spaces: ["e","dg","bcmp","fhvw","k","","","jx","","qz"]
        ðš # Prepend a space to this list
        ε7∍} # Extend each string to size 7:
        # [" ","eeeeeee","dgdgdgd","bcmpbcm","fhvwfhv","kkkkkkk","","","jxjxjxj","","qzqzqzq"]
        Iv # Loop `y` over the input-list:
        Dyè # Get the `y`'th string from a copy of the list
        н # Get it's first character
        ©? # Store it in the register, and print it without trailing newline
        ε # Then map each string in the list to:
        ®õ.; # Remove the first occurrence of the character from the register


        See this 05AB1E tip of mine (section How to compress strings not part of the dictionary?) to understand why .•Mñ&Àû«ì{₆v*Å+µ-• is "e dg bcmp fhvw k jx qz".






        share|improve this answer











        $endgroup$













        • $begingroup$
          Can't you use " 0eeeeeee0ddddggg0bbccmmp0ffhhvvw0k000jx00qz"?
          $endgroup$
          – Expired Data
          2 days ago












        • $begingroup$
          @ExpiredData Ah, of course. You only draw 7 letter.. Thanks! Will change it.
          $endgroup$
          – Kevin Cruijssen
          2 days ago






        • 1




          $begingroup$
          You can save 3 bytes using {v instead of 7F and y instead of I{Nè.
          $endgroup$
          – Emigna
          yesterday










        • $begingroup$
          @Emigna Ah, of course.. Thanks!
          $endgroup$
          – Kevin Cruijssen
          yesterday



















        2












        $begingroup$


        C (gcc), 110 bytes





        _={0,7,14,21,0,0,22,0,24};f(char*s){for(;*s+1;s++)*s=*s?*s-1?"DDDDGGGBBCCMMPFFHHVVWKJXQZ"[_[*s-2]++]:69:32;}


        Try it online!



        Uses the _ array as an index into the static string "DDDDGGGBBCCMMPFFHHVVWKJXQZ" dynamically with exceptions for 0 and 1.



        Argument is a -1-terminated array of scores which is transformed in-place into a -1-terminated string.






        share|improve this answer









        $endgroup$





















          1












          $begingroup$


          C# (Visual C# Interactive Compiler), 104 90 bytes





          a=>a.OrderBy(x=>x).Select((x,i)=>(x="_ E DG BCMP FHVW K   JX  QZ".Split()[x])[i%x.Length])


          Try it online!






          share|improve this answer











          $endgroup$





















            1












            $begingroup$


            Jelly, 34 32 bytes



            “¿RÇĊƈ⁸⁾%ỵṆþœsṀṂ’ṃØAṣ”A;⁶ẋ€7⁸ịḢ€


            Try it online!



            I hadn’t seen there was a shorter Jelly answer when I wrote this, and this uses a different approach so I thought was worth posting as well.



            Thanks to @JonathanAllan for saving 2 bytes!






            share|improve this answer











            $endgroup$













            • $begingroup$
              By using base-decompression, , you can save 2 bytes
              $endgroup$
              – Jonathan Allan
              2 days ago



















            1












            $begingroup$


            Python 3, 178 142 135 127 112 117 bytes





            def f(l):
            d=list(map(list," _EEEEEEE_DDDDGGG_BBCCMMP_FFHHVVW_K___JX__QZ".split('_')))
            return[d[i].pop()for i in l]


            Try it online!



            -1 byte thanks to cdlane



            correct thanks to mathmandan






            share|improve this answer











            $endgroup$













            • $begingroup$
              in " -> in" for 111
              $endgroup$
              – cdlane
              2 days ago










            • $begingroup$
              d=list(map(list,"...".split('_'))) to save another byte
              $endgroup$
              – cdlane
              2 days ago












            • $begingroup$
              This function f probably doesn't need to be named, so you can save 2 bytes. However, f consumes the entries of d, so I'm not sure that it fits the consensus requirement that "the function has to be reusable arbitrarily often, without...restating...any other code accompanying the submission." (For example, running f([10,0,10,5,8,8,0]) more than once would result in an error.) Please see meta discussion here: codegolf.meta.stackexchange.com/a/7615/36885
              $endgroup$
              – mathmandan
              2 days ago



















            0












            $begingroup$


            Python 2, 102 bytes (or maybe 95?)



            (Also fine for Python 3.)





            lambda a:''.join([r*7for r in'_ E DG BCMP FHVW K * * JX * QZ'.split()][x][:a.count(x)]for x in set(a))


            Try it online!



            I don't think the following would be acceptable:



            lambda a:[[r*7for r in'_ E DG BCMP FHVW K * * JX * QZ'.split()][x][:a.count(x)]for x in set(a)]


            This second version would give output like ['__', 'JX', 'QZ', 'K']. So the letters would be correct, but they would be collected by point value. (If this were acceptable, it would save 7 bytes.)






            share|improve this answer









            $endgroup$





















              0












              $begingroup$


              PHP, 101 bytes





              $b=[_,E,DG,BCMP,FHVW,K,8=>JX,0,QZ];foreach($argv as$t){echo$c=($d=$b[$t])[0];$b[$t]=substr($d,1).$c;}


              As a standalone program, input via command line:



              $ php s.php 10 0 10 5 8 8 0
              "Q_ZKJX_"


              Try it online!



              Or 112 bytes as a function



              function($a){$b=[_,E,DG,BCMP,FHVW,K,8=>JX,0,QZ];foreach($a as$t)$b[$t]=substr($d=$b[$t],1).$c=$d[0];return$c;}


              Try it online!



              Output



              [10,0,10,5,8,8,0]   "Q_ZKJX_"
              [1,1,1,1,1,1,1] "EEEEEEE"
              [1,2,3,4,5,8,0] "EDBFKJ_"
              [2,2,2,2,2,2,2] "DGDGDGD"





              share|improve this answer











              $endgroup$





















                0












                $begingroup$


                Ruby, 77 76 bytes





                ->a{r=%w{_ E DG BCMP FHVW K . . JX . QZ};a.map{|i|(r[i]<<r[i][0]).slice! 0}}


                Try it online!






                share|improve this answer











                $endgroup$





















                  0












                  $begingroup$


                  Perl 6, 63 bytes





                  *>>.&{(<_ E DG BCMP FHVW K _ _ JX _ QZ>[$_]x 7).comb[%.{$_}++]}


                  Try it online!



                  <_ E DG BCMP FHVW K _ _ JX _ QZ> # array indexed on tile value
                  (<...>[$_] x 7) # pull letters for this value, repeat 7 times to catch E
                  % # anonymous stateful hash
                  .{$_} # element for this tile value
                  ++ # post increment value to move position
                  .comb[...] # characters to array, pull this incrementing index


                  So essentially it keeps a lookup of offsets for each tile value and increments them as needed, using the offset to pull a character from the available set.






                  share|improve this answer









                  $endgroup$














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                    14 Answers
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                    8












                    $begingroup$

                    JavaScript (ES6), 72 bytes



                    A shorter variant suggested by @supercat



                    a=>a.map(o=n=>'?ED?BWQ?_EG?CFZ?_EDJMH?K?EGXPV'[n*9.4+(o[n]=7-~o[n])&31])


                    Try it online!





                    JavaScript (ES6),  137 ... 84 78 77  76 bytes



                    Saved 10 bytes by using Neil's cycling method



                    Returns a list of tiles. Uses _ for blank tiles.



                    a=>a.map(o=n=>"____FHVWGDGD_K__BCMPEEEE_ZQ__XJ"[n*20%44%32+(o[n]=-~o[n])%4])


                    Try it online!



                    How?



                    For each number of points, we cycle through a group of exactly 4 tiles, starting with the second tile of each group (this is important for G vs D):



                     points | group | max. sequence
                    --------+-------+---------------
                    0 | ____ | __
                    1 | EEEE | EEEEEEE
                    2 | GDGD | DGDGDGD
                    3 | BCMP | CMPBCMP
                    4 | FHVW | HVWFHVW
                    5 | _K__ | K
                    8 | _XJ_ | XJ }--- these letters may only appear once each
                    10 | _ZQ_ | ZQ /


                    All these groups are stored as a single string of 31 characters:



                    ____FHVWGDGD_K__BCMPEEEE_ZQ__XJ
                    ^ ^ ^ ^ ^ ^ ^ ^
                    0 4 8 12 16 20 24 28


                    NB: We don't need to store the final "_" in "_XJ_", as it will never be accessed anyway.



                    The number of points $n$ is converted to the correct index $i_n$ into this string with:



                    $$i_n=((20times n)bmod 44)bmod 32$$



                      n | *20 | mod 44 | mod 32 | group
                    ----+-----+--------+--------+-------
                    0 | 0 | 0 | 0 | ____
                    1 | 20 | 20 | 20 | EEEE
                    2 | 40 | 40 | 8 | GDGD
                    3 | 60 | 16 | 16 | BCMP
                    4 | 80 | 36 | 4 | FHVW
                    5 | 100 | 12 | 12 | _K__
                    8 | 160 | 28 | 28 | _XJ_
                    10 | 200 | 24 | 24 | _ZQ_


                    The current position in each group is stored in the object $o$.






                    share|improve this answer











                    $endgroup$













                    • $begingroup$
                      Advancing o[n] by 8 each time would cost one extra character for the advancement, but allow one to replace %4 and %32 both with &31 for a net win. My best, based on yours, would be a=>a.map(o=n=>('?ED?BWQ?_EG?CFZ?_EDJMH?K?EGXPV'[n*9.4+(o[n]=7-~o[n])&31])). A shorter, "almost" version is a=>a.map(o=n=>("_EDBFK_EDCHJQEGMVXZEGPW"[n+(o[n]=5-~o[n])%24])) but that approach would needs a compact way to map the values 8 and 10 into 11 and 12, plus a slight adjustment to the string to fix an off-by-one problem.
                      $endgroup$
                      – supercat
                      2 days ago










                    • $begingroup$
                      @supercat Sounds good! I'll have a closer look at it tomorrow.
                      $endgroup$
                      – Arnauld
                      2 days ago










                    • $begingroup$
                      @supercat Another interesting formula is '_??VKWZHQFP?M?CGBGXDJD'[(n*96+(o[n]=32-~o[n]))%68%33]||'E', with a lookup string of only 22 characters. The full code is still 2 bytes longer than your solution, though.
                      $endgroup$
                      – Arnauld
                      yesterday
















                    8












                    $begingroup$

                    JavaScript (ES6), 72 bytes



                    A shorter variant suggested by @supercat



                    a=>a.map(o=n=>'?ED?BWQ?_EG?CFZ?_EDJMH?K?EGXPV'[n*9.4+(o[n]=7-~o[n])&31])


                    Try it online!





                    JavaScript (ES6),  137 ... 84 78 77  76 bytes



                    Saved 10 bytes by using Neil's cycling method



                    Returns a list of tiles. Uses _ for blank tiles.



                    a=>a.map(o=n=>"____FHVWGDGD_K__BCMPEEEE_ZQ__XJ"[n*20%44%32+(o[n]=-~o[n])%4])


                    Try it online!



                    How?



                    For each number of points, we cycle through a group of exactly 4 tiles, starting with the second tile of each group (this is important for G vs D):



                     points | group | max. sequence
                    --------+-------+---------------
                    0 | ____ | __
                    1 | EEEE | EEEEEEE
                    2 | GDGD | DGDGDGD
                    3 | BCMP | CMPBCMP
                    4 | FHVW | HVWFHVW
                    5 | _K__ | K
                    8 | _XJ_ | XJ }--- these letters may only appear once each
                    10 | _ZQ_ | ZQ /


                    All these groups are stored as a single string of 31 characters:



                    ____FHVWGDGD_K__BCMPEEEE_ZQ__XJ
                    ^ ^ ^ ^ ^ ^ ^ ^
                    0 4 8 12 16 20 24 28


                    NB: We don't need to store the final "_" in "_XJ_", as it will never be accessed anyway.



                    The number of points $n$ is converted to the correct index $i_n$ into this string with:



                    $$i_n=((20times n)bmod 44)bmod 32$$



                      n | *20 | mod 44 | mod 32 | group
                    ----+-----+--------+--------+-------
                    0 | 0 | 0 | 0 | ____
                    1 | 20 | 20 | 20 | EEEE
                    2 | 40 | 40 | 8 | GDGD
                    3 | 60 | 16 | 16 | BCMP
                    4 | 80 | 36 | 4 | FHVW
                    5 | 100 | 12 | 12 | _K__
                    8 | 160 | 28 | 28 | _XJ_
                    10 | 200 | 24 | 24 | _ZQ_


                    The current position in each group is stored in the object $o$.






                    share|improve this answer











                    $endgroup$













                    • $begingroup$
                      Advancing o[n] by 8 each time would cost one extra character for the advancement, but allow one to replace %4 and %32 both with &31 for a net win. My best, based on yours, would be a=>a.map(o=n=>('?ED?BWQ?_EG?CFZ?_EDJMH?K?EGXPV'[n*9.4+(o[n]=7-~o[n])&31])). A shorter, "almost" version is a=>a.map(o=n=>("_EDBFK_EDCHJQEGMVXZEGPW"[n+(o[n]=5-~o[n])%24])) but that approach would needs a compact way to map the values 8 and 10 into 11 and 12, plus a slight adjustment to the string to fix an off-by-one problem.
                      $endgroup$
                      – supercat
                      2 days ago










                    • $begingroup$
                      @supercat Sounds good! I'll have a closer look at it tomorrow.
                      $endgroup$
                      – Arnauld
                      2 days ago










                    • $begingroup$
                      @supercat Another interesting formula is '_??VKWZHQFP?M?CGBGXDJD'[(n*96+(o[n]=32-~o[n]))%68%33]||'E', with a lookup string of only 22 characters. The full code is still 2 bytes longer than your solution, though.
                      $endgroup$
                      – Arnauld
                      yesterday














                    8












                    8








                    8





                    $begingroup$

                    JavaScript (ES6), 72 bytes



                    A shorter variant suggested by @supercat



                    a=>a.map(o=n=>'?ED?BWQ?_EG?CFZ?_EDJMH?K?EGXPV'[n*9.4+(o[n]=7-~o[n])&31])


                    Try it online!





                    JavaScript (ES6),  137 ... 84 78 77  76 bytes



                    Saved 10 bytes by using Neil's cycling method



                    Returns a list of tiles. Uses _ for blank tiles.



                    a=>a.map(o=n=>"____FHVWGDGD_K__BCMPEEEE_ZQ__XJ"[n*20%44%32+(o[n]=-~o[n])%4])


                    Try it online!



                    How?



                    For each number of points, we cycle through a group of exactly 4 tiles, starting with the second tile of each group (this is important for G vs D):



                     points | group | max. sequence
                    --------+-------+---------------
                    0 | ____ | __
                    1 | EEEE | EEEEEEE
                    2 | GDGD | DGDGDGD
                    3 | BCMP | CMPBCMP
                    4 | FHVW | HVWFHVW
                    5 | _K__ | K
                    8 | _XJ_ | XJ }--- these letters may only appear once each
                    10 | _ZQ_ | ZQ /


                    All these groups are stored as a single string of 31 characters:



                    ____FHVWGDGD_K__BCMPEEEE_ZQ__XJ
                    ^ ^ ^ ^ ^ ^ ^ ^
                    0 4 8 12 16 20 24 28


                    NB: We don't need to store the final "_" in "_XJ_", as it will never be accessed anyway.



                    The number of points $n$ is converted to the correct index $i_n$ into this string with:



                    $$i_n=((20times n)bmod 44)bmod 32$$



                      n | *20 | mod 44 | mod 32 | group
                    ----+-----+--------+--------+-------
                    0 | 0 | 0 | 0 | ____
                    1 | 20 | 20 | 20 | EEEE
                    2 | 40 | 40 | 8 | GDGD
                    3 | 60 | 16 | 16 | BCMP
                    4 | 80 | 36 | 4 | FHVW
                    5 | 100 | 12 | 12 | _K__
                    8 | 160 | 28 | 28 | _XJ_
                    10 | 200 | 24 | 24 | _ZQ_


                    The current position in each group is stored in the object $o$.






                    share|improve this answer











                    $endgroup$



                    JavaScript (ES6), 72 bytes



                    A shorter variant suggested by @supercat



                    a=>a.map(o=n=>'?ED?BWQ?_EG?CFZ?_EDJMH?K?EGXPV'[n*9.4+(o[n]=7-~o[n])&31])


                    Try it online!





                    JavaScript (ES6),  137 ... 84 78 77  76 bytes



                    Saved 10 bytes by using Neil's cycling method



                    Returns a list of tiles. Uses _ for blank tiles.



                    a=>a.map(o=n=>"____FHVWGDGD_K__BCMPEEEE_ZQ__XJ"[n*20%44%32+(o[n]=-~o[n])%4])


                    Try it online!



                    How?



                    For each number of points, we cycle through a group of exactly 4 tiles, starting with the second tile of each group (this is important for G vs D):



                     points | group | max. sequence
                    --------+-------+---------------
                    0 | ____ | __
                    1 | EEEE | EEEEEEE
                    2 | GDGD | DGDGDGD
                    3 | BCMP | CMPBCMP
                    4 | FHVW | HVWFHVW
                    5 | _K__ | K
                    8 | _XJ_ | XJ }--- these letters may only appear once each
                    10 | _ZQ_ | ZQ /


                    All these groups are stored as a single string of 31 characters:



                    ____FHVWGDGD_K__BCMPEEEE_ZQ__XJ
                    ^ ^ ^ ^ ^ ^ ^ ^
                    0 4 8 12 16 20 24 28


                    NB: We don't need to store the final "_" in "_XJ_", as it will never be accessed anyway.



                    The number of points $n$ is converted to the correct index $i_n$ into this string with:



                    $$i_n=((20times n)bmod 44)bmod 32$$



                      n | *20 | mod 44 | mod 32 | group
                    ----+-----+--------+--------+-------
                    0 | 0 | 0 | 0 | ____
                    1 | 20 | 20 | 20 | EEEE
                    2 | 40 | 40 | 8 | GDGD
                    3 | 60 | 16 | 16 | BCMP
                    4 | 80 | 36 | 4 | FHVW
                    5 | 100 | 12 | 12 | _K__
                    8 | 160 | 28 | 28 | _XJ_
                    10 | 200 | 24 | 24 | _ZQ_


                    The current position in each group is stored in the object $o$.







                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited yesterday

























                    answered 2 days ago









                    ArnauldArnauld

                    80.9k797334




                    80.9k797334












                    • $begingroup$
                      Advancing o[n] by 8 each time would cost one extra character for the advancement, but allow one to replace %4 and %32 both with &31 for a net win. My best, based on yours, would be a=>a.map(o=n=>('?ED?BWQ?_EG?CFZ?_EDJMH?K?EGXPV'[n*9.4+(o[n]=7-~o[n])&31])). A shorter, "almost" version is a=>a.map(o=n=>("_EDBFK_EDCHJQEGMVXZEGPW"[n+(o[n]=5-~o[n])%24])) but that approach would needs a compact way to map the values 8 and 10 into 11 and 12, plus a slight adjustment to the string to fix an off-by-one problem.
                      $endgroup$
                      – supercat
                      2 days ago










                    • $begingroup$
                      @supercat Sounds good! I'll have a closer look at it tomorrow.
                      $endgroup$
                      – Arnauld
                      2 days ago










                    • $begingroup$
                      @supercat Another interesting formula is '_??VKWZHQFP?M?CGBGXDJD'[(n*96+(o[n]=32-~o[n]))%68%33]||'E', with a lookup string of only 22 characters. The full code is still 2 bytes longer than your solution, though.
                      $endgroup$
                      – Arnauld
                      yesterday


















                    • $begingroup$
                      Advancing o[n] by 8 each time would cost one extra character for the advancement, but allow one to replace %4 and %32 both with &31 for a net win. My best, based on yours, would be a=>a.map(o=n=>('?ED?BWQ?_EG?CFZ?_EDJMH?K?EGXPV'[n*9.4+(o[n]=7-~o[n])&31])). A shorter, "almost" version is a=>a.map(o=n=>("_EDBFK_EDCHJQEGMVXZEGPW"[n+(o[n]=5-~o[n])%24])) but that approach would needs a compact way to map the values 8 and 10 into 11 and 12, plus a slight adjustment to the string to fix an off-by-one problem.
                      $endgroup$
                      – supercat
                      2 days ago










                    • $begingroup$
                      @supercat Sounds good! I'll have a closer look at it tomorrow.
                      $endgroup$
                      – Arnauld
                      2 days ago










                    • $begingroup$
                      @supercat Another interesting formula is '_??VKWZHQFP?M?CGBGXDJD'[(n*96+(o[n]=32-~o[n]))%68%33]||'E', with a lookup string of only 22 characters. The full code is still 2 bytes longer than your solution, though.
                      $endgroup$
                      – Arnauld
                      yesterday
















                    $begingroup$
                    Advancing o[n] by 8 each time would cost one extra character for the advancement, but allow one to replace %4 and %32 both with &31 for a net win. My best, based on yours, would be a=>a.map(o=n=>('?ED?BWQ?_EG?CFZ?_EDJMH?K?EGXPV'[n*9.4+(o[n]=7-~o[n])&31])). A shorter, "almost" version is a=>a.map(o=n=>("_EDBFK_EDCHJQEGMVXZEGPW"[n+(o[n]=5-~o[n])%24])) but that approach would needs a compact way to map the values 8 and 10 into 11 and 12, plus a slight adjustment to the string to fix an off-by-one problem.
                    $endgroup$
                    – supercat
                    2 days ago




                    $begingroup$
                    Advancing o[n] by 8 each time would cost one extra character for the advancement, but allow one to replace %4 and %32 both with &31 for a net win. My best, based on yours, would be a=>a.map(o=n=>('?ED?BWQ?_EG?CFZ?_EDJMH?K?EGXPV'[n*9.4+(o[n]=7-~o[n])&31])). A shorter, "almost" version is a=>a.map(o=n=>("_EDBFK_EDCHJQEGMVXZEGPW"[n+(o[n]=5-~o[n])%24])) but that approach would needs a compact way to map the values 8 and 10 into 11 and 12, plus a slight adjustment to the string to fix an off-by-one problem.
                    $endgroup$
                    – supercat
                    2 days ago












                    $begingroup$
                    @supercat Sounds good! I'll have a closer look at it tomorrow.
                    $endgroup$
                    – Arnauld
                    2 days ago




                    $begingroup$
                    @supercat Sounds good! I'll have a closer look at it tomorrow.
                    $endgroup$
                    – Arnauld
                    2 days ago












                    $begingroup$
                    @supercat Another interesting formula is '_??VKWZHQFP?M?CGBGXDJD'[(n*96+(o[n]=32-~o[n]))%68%33]||'E', with a lookup string of only 22 characters. The full code is still 2 bytes longer than your solution, though.
                    $endgroup$
                    – Arnauld
                    yesterday




                    $begingroup$
                    @supercat Another interesting formula is '_??VKWZHQFP?M?CGBGXDJD'[(n*96+(o[n]=32-~o[n]))%68%33]||'E', with a lookup string of only 22 characters. The full code is still 2 bytes longer than your solution, though.
                    $endgroup$
                    – Arnauld
                    yesterday











                    7












                    $begingroup$


                    Charcoal, 33 bytes



                    ⭆觧⪪”&↖“Vh_z↶∕¡⌈∨₂χ¹‖◨⌊″”¶ι№…θκι


                    Try it online! Link is to verbose version of code. Explanation:



                     θ                  Input array
                    ⭆ Map over elements and join
                    ”...” Literal string " nEnDGnBCMPnFHVWnKnnnJXnnQZ"
                    ⪪ ¶ Split on newlines
                    § ι Indexed by current element
                    § Cyclically indexed by
                    №…θκι Number of times current element has already appeared
                    Implcitly print





                    share|improve this answer









                    $endgroup$


















                      7












                      $begingroup$


                      Charcoal, 33 bytes



                      ⭆觧⪪”&↖“Vh_z↶∕¡⌈∨₂χ¹‖◨⌊″”¶ι№…θκι


                      Try it online! Link is to verbose version of code. Explanation:



                       θ                  Input array
                      ⭆ Map over elements and join
                      ”...” Literal string " nEnDGnBCMPnFHVWnKnnnJXnnQZ"
                      ⪪ ¶ Split on newlines
                      § ι Indexed by current element
                      § Cyclically indexed by
                      №…θκι Number of times current element has already appeared
                      Implcitly print





                      share|improve this answer









                      $endgroup$
















                        7












                        7








                        7





                        $begingroup$


                        Charcoal, 33 bytes



                        ⭆觧⪪”&↖“Vh_z↶∕¡⌈∨₂χ¹‖◨⌊″”¶ι№…θκι


                        Try it online! Link is to verbose version of code. Explanation:



                         θ                  Input array
                        ⭆ Map over elements and join
                        ”...” Literal string " nEnDGnBCMPnFHVWnKnnnJXnnQZ"
                        ⪪ ¶ Split on newlines
                        § ι Indexed by current element
                        § Cyclically indexed by
                        №…θκι Number of times current element has already appeared
                        Implcitly print





                        share|improve this answer









                        $endgroup$




                        Charcoal, 33 bytes



                        ⭆觧⪪”&↖“Vh_z↶∕¡⌈∨₂χ¹‖◨⌊″”¶ι№…θκι


                        Try it online! Link is to verbose version of code. Explanation:



                         θ                  Input array
                        ⭆ Map over elements and join
                        ”...” Literal string " nEnDGnBCMPnFHVWnKnnnJXnnQZ"
                        ⪪ ¶ Split on newlines
                        § ι Indexed by current element
                        § Cyclically indexed by
                        №…θκι Number of times current element has already appeared
                        Implcitly print






                        share|improve this answer












                        share|improve this answer



                        share|improve this answer










                        answered 2 days ago









                        NeilNeil

                        82.8k745179




                        82.8k745179























                            5












                            $begingroup$


                            Jelly,  31 30 27  26 bytes



                            “ñẒẎYñ(“Nut¦hß’ṃØA;€⁶ɓṢĖœị


                            A monadic Link accepting a list of integers which yields a list of characters.

                            - a mishmash of my previous, below, and my improvement of Nick Kennedy's



                            Try it online!



                            The output is not given in the same order as the input (this is allowed).



                            Using 2 of my own additions to the language in an answer does not happen often! ( and ɓ here).



                            How?



                            “...“...’ṃØA;€⁶ɓṢĖœị - Link: list of integers, V     e.g. [10,1,0,3,2,1,10]
                            “...“...’ - list of base 250 integers [28089224382041, 77611203526272]
                            ØA - 'ABC...XYZ'
                            ṃ - base decompress (vectorises) ["EDMFKZZJZQ", "NGPYKZZXZZ"]
                            ;€ - for €ach: concatenate:
                            ⁶ - a space ["EDMFKZZJZQ ", "NGPYKZZXZZ "]
                            ɓ - start a new dyadic chain with swapped arguments - i.e. f(V,that)
                            Ṣ - sort [0,1,1,2,3,10,10]
                            Ė - enumerate [[1,0],[2,1],[3,1],[4,2],[5,3],[6,10],[7,10]]
                            œị - multi-dimensional index into " NEGMZQ"
                            (1-based and modular)




                            previous @ 30



                            “²rṛʂṂø5=Ɓṇ^N¥Y»⁾tky;⁶s2ɓṢĖUœị


                            A monadic Link accepting a list of integers which yields a list of characters.



                            Try it online!



                            This one's output is also mixed-case (this is allowed).



                            How?



                            “...»⁾tky;⁶s2ɓṢĖUœị - Link: list of integers, V          e.g. [10,1,0,3,2,1,10]
                            “...» - compression of dictionary entries:
                            - "end", "GMP", "fyttes", "adj", and "xci" and the string "qz"
                            - "endGMPfyttesadjxciqz"
                            y - translate with:
                            ⁾tk - ['t', 'k'] "endGMPfykkesadjxciqz"
                            ;⁶s2ɓṢĖUœị - ...
                            - ...then like the above method (except U reverses each pair of indices)
                            " neGMzq"





                            share|improve this answer











                            $endgroup$













                            • $begingroup$
                              I think you made a typo in your first explanation. ' NWGMZQ' after the multi-dimensional index into would be quite a feat without any W in the string. ;)
                              $endgroup$
                              – Kevin Cruijssen
                              yesterday






                            • 1




                              $begingroup$
                              @KevinCruijssen - yws, typo fixwd; thanks!
                              $endgroup$
                              – Jonathan Allan
                              yesterday
















                            5












                            $begingroup$


                            Jelly,  31 30 27  26 bytes



                            “ñẒẎYñ(“Nut¦hß’ṃØA;€⁶ɓṢĖœị


                            A monadic Link accepting a list of integers which yields a list of characters.

                            - a mishmash of my previous, below, and my improvement of Nick Kennedy's



                            Try it online!



                            The output is not given in the same order as the input (this is allowed).



                            Using 2 of my own additions to the language in an answer does not happen often! ( and ɓ here).



                            How?



                            “...“...’ṃØA;€⁶ɓṢĖœị - Link: list of integers, V     e.g. [10,1,0,3,2,1,10]
                            “...“...’ - list of base 250 integers [28089224382041, 77611203526272]
                            ØA - 'ABC...XYZ'
                            ṃ - base decompress (vectorises) ["EDMFKZZJZQ", "NGPYKZZXZZ"]
                            ;€ - for €ach: concatenate:
                            ⁶ - a space ["EDMFKZZJZQ ", "NGPYKZZXZZ "]
                            ɓ - start a new dyadic chain with swapped arguments - i.e. f(V,that)
                            Ṣ - sort [0,1,1,2,3,10,10]
                            Ė - enumerate [[1,0],[2,1],[3,1],[4,2],[5,3],[6,10],[7,10]]
                            œị - multi-dimensional index into " NEGMZQ"
                            (1-based and modular)




                            previous @ 30



                            “²rṛʂṂø5=Ɓṇ^N¥Y»⁾tky;⁶s2ɓṢĖUœị


                            A monadic Link accepting a list of integers which yields a list of characters.



                            Try it online!



                            This one's output is also mixed-case (this is allowed).



                            How?



                            “...»⁾tky;⁶s2ɓṢĖUœị - Link: list of integers, V          e.g. [10,1,0,3,2,1,10]
                            “...» - compression of dictionary entries:
                            - "end", "GMP", "fyttes", "adj", and "xci" and the string "qz"
                            - "endGMPfyttesadjxciqz"
                            y - translate with:
                            ⁾tk - ['t', 'k'] "endGMPfykkesadjxciqz"
                            ;⁶s2ɓṢĖUœị - ...
                            - ...then like the above method (except U reverses each pair of indices)
                            " neGMzq"





                            share|improve this answer











                            $endgroup$













                            • $begingroup$
                              I think you made a typo in your first explanation. ' NWGMZQ' after the multi-dimensional index into would be quite a feat without any W in the string. ;)
                              $endgroup$
                              – Kevin Cruijssen
                              yesterday






                            • 1




                              $begingroup$
                              @KevinCruijssen - yws, typo fixwd; thanks!
                              $endgroup$
                              – Jonathan Allan
                              yesterday














                            5












                            5








                            5





                            $begingroup$


                            Jelly,  31 30 27  26 bytes



                            “ñẒẎYñ(“Nut¦hß’ṃØA;€⁶ɓṢĖœị


                            A monadic Link accepting a list of integers which yields a list of characters.

                            - a mishmash of my previous, below, and my improvement of Nick Kennedy's



                            Try it online!



                            The output is not given in the same order as the input (this is allowed).



                            Using 2 of my own additions to the language in an answer does not happen often! ( and ɓ here).



                            How?



                            “...“...’ṃØA;€⁶ɓṢĖœị - Link: list of integers, V     e.g. [10,1,0,3,2,1,10]
                            “...“...’ - list of base 250 integers [28089224382041, 77611203526272]
                            ØA - 'ABC...XYZ'
                            ṃ - base decompress (vectorises) ["EDMFKZZJZQ", "NGPYKZZXZZ"]
                            ;€ - for €ach: concatenate:
                            ⁶ - a space ["EDMFKZZJZQ ", "NGPYKZZXZZ "]
                            ɓ - start a new dyadic chain with swapped arguments - i.e. f(V,that)
                            Ṣ - sort [0,1,1,2,3,10,10]
                            Ė - enumerate [[1,0],[2,1],[3,1],[4,2],[5,3],[6,10],[7,10]]
                            œị - multi-dimensional index into " NEGMZQ"
                            (1-based and modular)




                            previous @ 30



                            “²rṛʂṂø5=Ɓṇ^N¥Y»⁾tky;⁶s2ɓṢĖUœị


                            A monadic Link accepting a list of integers which yields a list of characters.



                            Try it online!



                            This one's output is also mixed-case (this is allowed).



                            How?



                            “...»⁾tky;⁶s2ɓṢĖUœị - Link: list of integers, V          e.g. [10,1,0,3,2,1,10]
                            “...» - compression of dictionary entries:
                            - "end", "GMP", "fyttes", "adj", and "xci" and the string "qz"
                            - "endGMPfyttesadjxciqz"
                            y - translate with:
                            ⁾tk - ['t', 'k'] "endGMPfykkesadjxciqz"
                            ;⁶s2ɓṢĖUœị - ...
                            - ...then like the above method (except U reverses each pair of indices)
                            " neGMzq"





                            share|improve this answer











                            $endgroup$




                            Jelly,  31 30 27  26 bytes



                            “ñẒẎYñ(“Nut¦hß’ṃØA;€⁶ɓṢĖœị


                            A monadic Link accepting a list of integers which yields a list of characters.

                            - a mishmash of my previous, below, and my improvement of Nick Kennedy's



                            Try it online!



                            The output is not given in the same order as the input (this is allowed).



                            Using 2 of my own additions to the language in an answer does not happen often! ( and ɓ here).



                            How?



                            “...“...’ṃØA;€⁶ɓṢĖœị - Link: list of integers, V     e.g. [10,1,0,3,2,1,10]
                            “...“...’ - list of base 250 integers [28089224382041, 77611203526272]
                            ØA - 'ABC...XYZ'
                            ṃ - base decompress (vectorises) ["EDMFKZZJZQ", "NGPYKZZXZZ"]
                            ;€ - for €ach: concatenate:
                            ⁶ - a space ["EDMFKZZJZQ ", "NGPYKZZXZZ "]
                            ɓ - start a new dyadic chain with swapped arguments - i.e. f(V,that)
                            Ṣ - sort [0,1,1,2,3,10,10]
                            Ė - enumerate [[1,0],[2,1],[3,1],[4,2],[5,3],[6,10],[7,10]]
                            œị - multi-dimensional index into " NEGMZQ"
                            (1-based and modular)




                            previous @ 30



                            “²rṛʂṂø5=Ɓṇ^N¥Y»⁾tky;⁶s2ɓṢĖUœị


                            A monadic Link accepting a list of integers which yields a list of characters.



                            Try it online!



                            This one's output is also mixed-case (this is allowed).



                            How?



                            “...»⁾tky;⁶s2ɓṢĖUœị - Link: list of integers, V          e.g. [10,1,0,3,2,1,10]
                            “...» - compression of dictionary entries:
                            - "end", "GMP", "fyttes", "adj", and "xci" and the string "qz"
                            - "endGMPfyttesadjxciqz"
                            y - translate with:
                            ⁾tk - ['t', 'k'] "endGMPfykkesadjxciqz"
                            ;⁶s2ɓṢĖUœị - ...
                            - ...then like the above method (except U reverses each pair of indices)
                            " neGMzq"






                            share|improve this answer














                            share|improve this answer



                            share|improve this answer








                            edited yesterday

























                            answered 2 days ago









                            Jonathan AllanJonathan Allan

                            54.2k537174




                            54.2k537174












                            • $begingroup$
                              I think you made a typo in your first explanation. ' NWGMZQ' after the multi-dimensional index into would be quite a feat without any W in the string. ;)
                              $endgroup$
                              – Kevin Cruijssen
                              yesterday






                            • 1




                              $begingroup$
                              @KevinCruijssen - yws, typo fixwd; thanks!
                              $endgroup$
                              – Jonathan Allan
                              yesterday


















                            • $begingroup$
                              I think you made a typo in your first explanation. ' NWGMZQ' after the multi-dimensional index into would be quite a feat without any W in the string. ;)
                              $endgroup$
                              – Kevin Cruijssen
                              yesterday






                            • 1




                              $begingroup$
                              @KevinCruijssen - yws, typo fixwd; thanks!
                              $endgroup$
                              – Jonathan Allan
                              yesterday
















                            $begingroup$
                            I think you made a typo in your first explanation. ' NWGMZQ' after the multi-dimensional index into would be quite a feat without any W in the string. ;)
                            $endgroup$
                            – Kevin Cruijssen
                            yesterday




                            $begingroup$
                            I think you made a typo in your first explanation. ' NWGMZQ' after the multi-dimensional index into would be quite a feat without any W in the string. ;)
                            $endgroup$
                            – Kevin Cruijssen
                            yesterday




                            1




                            1




                            $begingroup$
                            @KevinCruijssen - yws, typo fixwd; thanks!
                            $endgroup$
                            – Jonathan Allan
                            yesterday




                            $begingroup$
                            @KevinCruijssen - yws, typo fixwd; thanks!
                            $endgroup$
                            – Jonathan Allan
                            yesterday











                            4












                            $begingroup$

                            Pyth - 92 86 83 81 80 75 60 52 49 42 36 bytes



                            Loops through input, popping off the available letters. I just have one of each letter that together give 7 for that point category. Now using packed string encoding.



                            K[M*L7c."B_êº çÑOÒ
                            7âCkÑ"Lm.)@K

                            K Assign to K
                            [M Map list(for popping). Uses a quirk of M to splat each first
                            *L7 Map repeating each string by 7
                            c L Split on occurrences of 'L'
                            ."..." Packed string encoding of the needed letters
                            m (Q) Map on input (input is taken implicitly)
                            .) Pop. This returns the first element after removing it
                            @K Index into K
                            (d) The loop variable is given implicitly


                            Btw, this is the original letter string before encoding: "_ E DG BCMP FHVW K JX QZ".



                            Try it online.






                            share|improve this answer











                            $endgroup$


















                              4












                              $begingroup$

                              Pyth - 92 86 83 81 80 75 60 52 49 42 36 bytes



                              Loops through input, popping off the available letters. I just have one of each letter that together give 7 for that point category. Now using packed string encoding.



                              K[M*L7c."B_êº çÑOÒ
                              7âCkÑ"Lm.)@K

                              K Assign to K
                              [M Map list(for popping). Uses a quirk of M to splat each first
                              *L7 Map repeating each string by 7
                              c L Split on occurrences of 'L'
                              ."..." Packed string encoding of the needed letters
                              m (Q) Map on input (input is taken implicitly)
                              .) Pop. This returns the first element after removing it
                              @K Index into K
                              (d) The loop variable is given implicitly


                              Btw, this is the original letter string before encoding: "_ E DG BCMP FHVW K JX QZ".



                              Try it online.






                              share|improve this answer











                              $endgroup$
















                                4












                                4








                                4





                                $begingroup$

                                Pyth - 92 86 83 81 80 75 60 52 49 42 36 bytes



                                Loops through input, popping off the available letters. I just have one of each letter that together give 7 for that point category. Now using packed string encoding.



                                K[M*L7c."B_êº çÑOÒ
                                7âCkÑ"Lm.)@K

                                K Assign to K
                                [M Map list(for popping). Uses a quirk of M to splat each first
                                *L7 Map repeating each string by 7
                                c L Split on occurrences of 'L'
                                ."..." Packed string encoding of the needed letters
                                m (Q) Map on input (input is taken implicitly)
                                .) Pop. This returns the first element after removing it
                                @K Index into K
                                (d) The loop variable is given implicitly


                                Btw, this is the original letter string before encoding: "_ E DG BCMP FHVW K JX QZ".



                                Try it online.






                                share|improve this answer











                                $endgroup$



                                Pyth - 92 86 83 81 80 75 60 52 49 42 36 bytes



                                Loops through input, popping off the available letters. I just have one of each letter that together give 7 for that point category. Now using packed string encoding.



                                K[M*L7c."B_êº çÑOÒ
                                7âCkÑ"Lm.)@K

                                K Assign to K
                                [M Map list(for popping). Uses a quirk of M to splat each first
                                *L7 Map repeating each string by 7
                                c L Split on occurrences of 'L'
                                ."..." Packed string encoding of the needed letters
                                m (Q) Map on input (input is taken implicitly)
                                .) Pop. This returns the first element after removing it
                                @K Index into K
                                (d) The loop variable is given implicitly


                                Btw, this is the original letter string before encoding: "_ E DG BCMP FHVW K JX QZ".



                                Try it online.







                                share|improve this answer














                                share|improve this answer



                                share|improve this answer








                                edited 2 days ago

























                                answered 2 days ago









                                MaltysenMaltysen

                                21.4k445116




                                21.4k445116























                                    3












                                    $begingroup$


                                    Perl 5, 71 bytes





                                    @a=(__,'E'x7,DDDDGGG,BBCCMMP,FFHHVVW,K,1,1,JX,1,QZ);say chop$a[$_]for<>


                                    Try it online!






                                    share|improve this answer









                                    $endgroup$


















                                      3












                                      $begingroup$


                                      Perl 5, 71 bytes





                                      @a=(__,'E'x7,DDDDGGG,BBCCMMP,FFHHVVW,K,1,1,JX,1,QZ);say chop$a[$_]for<>


                                      Try it online!






                                      share|improve this answer









                                      $endgroup$
















                                        3












                                        3








                                        3





                                        $begingroup$


                                        Perl 5, 71 bytes





                                        @a=(__,'E'x7,DDDDGGG,BBCCMMP,FFHHVVW,K,1,1,JX,1,QZ);say chop$a[$_]for<>


                                        Try it online!






                                        share|improve this answer









                                        $endgroup$




                                        Perl 5, 71 bytes





                                        @a=(__,'E'x7,DDDDGGG,BBCCMMP,FFHHVVW,K,1,1,JX,1,QZ);say chop$a[$_]for<>


                                        Try it online!







                                        share|improve this answer












                                        share|improve this answer



                                        share|improve this answer










                                        answered 2 days ago









                                        XcaliXcali

                                        5,505520




                                        5,505520























                                            3












                                            $begingroup$


                                            05AB1E, 70 52 39 38 29 26 25 bytes



                                            {ε.•3Oû}α›ηö‡.ÝŽ{•2ôÁyèNè?


                                            -18 bytes thanks to @ExpiredData.

                                            -13 bytes by using the same extend to size 7 from @Maltysen's Pyth answer.

                                            -9 bytes by creating a port of @JonathanAllan's Jelly answer, so make sure to upvote him!

                                            -3 bytes thanks to @Emigna.



                                            Results in a list of characters, and uses lowercase letters and a space for blanks.



                                            Try it online or verify some more test cases.



                                            Explanation:





                                            {                      # Sort the (implicit) input-list
                                            ε # Map each character `y` in this list to:
                                            .•3Oû}α›ηö‡.ÝŽ{• # Push compressed string "endgmpfykkzzzzjxzzqz "
                                            2ô # Split into parts of size 2
                                            Á # Rotate it once towards the left so the space is leading
                                            yè # Use integer `y` to index into the string-pairs
                                            Nè # Then get the `N`'th character of the string-pair (with automatic
                                            # wraparound), where `N` is the index of the loop


                                            See this 05AB1E tip of mine (section How to compress strings not part of the dictionary?) to understand why .•3Oû}α›ηö‡.ÝŽ{• is "endgmpfykkzzzzjxzzqz ".





                                            Previous 38 bytes answer:



                                            .•Mñ&Àû«ì{₆v*Å+µ-•#ðšε7∍}IvDyèн©?ε®õ.;


                                            Try it online or verify some more test cases.



                                            Explanation:





                                            .•Mñ&Àû«ì{₆v*Å+µ-•    # Push compressed string "e dg bcmp fhvw k   jx  qz"
                                            # # Split on spaces: ["e","dg","bcmp","fhvw","k","","","jx","","qz"]
                                            ðš # Prepend a space to this list
                                            ε7∍} # Extend each string to size 7:
                                            # [" ","eeeeeee","dgdgdgd","bcmpbcm","fhvwfhv","kkkkkkk","","","jxjxjxj","","qzqzqzq"]
                                            Iv # Loop `y` over the input-list:
                                            Dyè # Get the `y`'th string from a copy of the list
                                            н # Get it's first character
                                            ©? # Store it in the register, and print it without trailing newline
                                            ε # Then map each string in the list to:
                                            ®õ.; # Remove the first occurrence of the character from the register


                                            See this 05AB1E tip of mine (section How to compress strings not part of the dictionary?) to understand why .•Mñ&Àû«ì{₆v*Å+µ-• is "e dg bcmp fhvw k jx qz".






                                            share|improve this answer











                                            $endgroup$













                                            • $begingroup$
                                              Can't you use " 0eeeeeee0ddddggg0bbccmmp0ffhhvvw0k000jx00qz"?
                                              $endgroup$
                                              – Expired Data
                                              2 days ago












                                            • $begingroup$
                                              @ExpiredData Ah, of course. You only draw 7 letter.. Thanks! Will change it.
                                              $endgroup$
                                              – Kevin Cruijssen
                                              2 days ago






                                            • 1




                                              $begingroup$
                                              You can save 3 bytes using {v instead of 7F and y instead of I{Nè.
                                              $endgroup$
                                              – Emigna
                                              yesterday










                                            • $begingroup$
                                              @Emigna Ah, of course.. Thanks!
                                              $endgroup$
                                              – Kevin Cruijssen
                                              yesterday
















                                            3












                                            $begingroup$


                                            05AB1E, 70 52 39 38 29 26 25 bytes



                                            {ε.•3Oû}α›ηö‡.ÝŽ{•2ôÁyèNè?


                                            -18 bytes thanks to @ExpiredData.

                                            -13 bytes by using the same extend to size 7 from @Maltysen's Pyth answer.

                                            -9 bytes by creating a port of @JonathanAllan's Jelly answer, so make sure to upvote him!

                                            -3 bytes thanks to @Emigna.



                                            Results in a list of characters, and uses lowercase letters and a space for blanks.



                                            Try it online or verify some more test cases.



                                            Explanation:





                                            {                      # Sort the (implicit) input-list
                                            ε # Map each character `y` in this list to:
                                            .•3Oû}α›ηö‡.ÝŽ{• # Push compressed string "endgmpfykkzzzzjxzzqz "
                                            2ô # Split into parts of size 2
                                            Á # Rotate it once towards the left so the space is leading
                                            yè # Use integer `y` to index into the string-pairs
                                            Nè # Then get the `N`'th character of the string-pair (with automatic
                                            # wraparound), where `N` is the index of the loop


                                            See this 05AB1E tip of mine (section How to compress strings not part of the dictionary?) to understand why .•3Oû}α›ηö‡.ÝŽ{• is "endgmpfykkzzzzjxzzqz ".





                                            Previous 38 bytes answer:



                                            .•Mñ&Àû«ì{₆v*Å+µ-•#ðšε7∍}IvDyèн©?ε®õ.;


                                            Try it online or verify some more test cases.



                                            Explanation:





                                            .•Mñ&Àû«ì{₆v*Å+µ-•    # Push compressed string "e dg bcmp fhvw k   jx  qz"
                                            # # Split on spaces: ["e","dg","bcmp","fhvw","k","","","jx","","qz"]
                                            ðš # Prepend a space to this list
                                            ε7∍} # Extend each string to size 7:
                                            # [" ","eeeeeee","dgdgdgd","bcmpbcm","fhvwfhv","kkkkkkk","","","jxjxjxj","","qzqzqzq"]
                                            Iv # Loop `y` over the input-list:
                                            Dyè # Get the `y`'th string from a copy of the list
                                            н # Get it's first character
                                            ©? # Store it in the register, and print it without trailing newline
                                            ε # Then map each string in the list to:
                                            ®õ.; # Remove the first occurrence of the character from the register


                                            See this 05AB1E tip of mine (section How to compress strings not part of the dictionary?) to understand why .•Mñ&Àû«ì{₆v*Å+µ-• is "e dg bcmp fhvw k jx qz".






                                            share|improve this answer











                                            $endgroup$













                                            • $begingroup$
                                              Can't you use " 0eeeeeee0ddddggg0bbccmmp0ffhhvvw0k000jx00qz"?
                                              $endgroup$
                                              – Expired Data
                                              2 days ago












                                            • $begingroup$
                                              @ExpiredData Ah, of course. You only draw 7 letter.. Thanks! Will change it.
                                              $endgroup$
                                              – Kevin Cruijssen
                                              2 days ago






                                            • 1




                                              $begingroup$
                                              You can save 3 bytes using {v instead of 7F and y instead of I{Nè.
                                              $endgroup$
                                              – Emigna
                                              yesterday










                                            • $begingroup$
                                              @Emigna Ah, of course.. Thanks!
                                              $endgroup$
                                              – Kevin Cruijssen
                                              yesterday














                                            3












                                            3








                                            3





                                            $begingroup$


                                            05AB1E, 70 52 39 38 29 26 25 bytes



                                            {ε.•3Oû}α›ηö‡.ÝŽ{•2ôÁyèNè?


                                            -18 bytes thanks to @ExpiredData.

                                            -13 bytes by using the same extend to size 7 from @Maltysen's Pyth answer.

                                            -9 bytes by creating a port of @JonathanAllan's Jelly answer, so make sure to upvote him!

                                            -3 bytes thanks to @Emigna.



                                            Results in a list of characters, and uses lowercase letters and a space for blanks.



                                            Try it online or verify some more test cases.



                                            Explanation:





                                            {                      # Sort the (implicit) input-list
                                            ε # Map each character `y` in this list to:
                                            .•3Oû}α›ηö‡.ÝŽ{• # Push compressed string "endgmpfykkzzzzjxzzqz "
                                            2ô # Split into parts of size 2
                                            Á # Rotate it once towards the left so the space is leading
                                            yè # Use integer `y` to index into the string-pairs
                                            Nè # Then get the `N`'th character of the string-pair (with automatic
                                            # wraparound), where `N` is the index of the loop


                                            See this 05AB1E tip of mine (section How to compress strings not part of the dictionary?) to understand why .•3Oû}α›ηö‡.ÝŽ{• is "endgmpfykkzzzzjxzzqz ".





                                            Previous 38 bytes answer:



                                            .•Mñ&Àû«ì{₆v*Å+µ-•#ðšε7∍}IvDyèн©?ε®õ.;


                                            Try it online or verify some more test cases.



                                            Explanation:





                                            .•Mñ&Àû«ì{₆v*Å+µ-•    # Push compressed string "e dg bcmp fhvw k   jx  qz"
                                            # # Split on spaces: ["e","dg","bcmp","fhvw","k","","","jx","","qz"]
                                            ðš # Prepend a space to this list
                                            ε7∍} # Extend each string to size 7:
                                            # [" ","eeeeeee","dgdgdgd","bcmpbcm","fhvwfhv","kkkkkkk","","","jxjxjxj","","qzqzqzq"]
                                            Iv # Loop `y` over the input-list:
                                            Dyè # Get the `y`'th string from a copy of the list
                                            н # Get it's first character
                                            ©? # Store it in the register, and print it without trailing newline
                                            ε # Then map each string in the list to:
                                            ®õ.; # Remove the first occurrence of the character from the register


                                            See this 05AB1E tip of mine (section How to compress strings not part of the dictionary?) to understand why .•Mñ&Àû«ì{₆v*Å+µ-• is "e dg bcmp fhvw k jx qz".






                                            share|improve this answer











                                            $endgroup$




                                            05AB1E, 70 52 39 38 29 26 25 bytes



                                            {ε.•3Oû}α›ηö‡.ÝŽ{•2ôÁyèNè?


                                            -18 bytes thanks to @ExpiredData.

                                            -13 bytes by using the same extend to size 7 from @Maltysen's Pyth answer.

                                            -9 bytes by creating a port of @JonathanAllan's Jelly answer, so make sure to upvote him!

                                            -3 bytes thanks to @Emigna.



                                            Results in a list of characters, and uses lowercase letters and a space for blanks.



                                            Try it online or verify some more test cases.



                                            Explanation:





                                            {                      # Sort the (implicit) input-list
                                            ε # Map each character `y` in this list to:
                                            .•3Oû}α›ηö‡.ÝŽ{• # Push compressed string "endgmpfykkzzzzjxzzqz "
                                            2ô # Split into parts of size 2
                                            Á # Rotate it once towards the left so the space is leading
                                            yè # Use integer `y` to index into the string-pairs
                                            Nè # Then get the `N`'th character of the string-pair (with automatic
                                            # wraparound), where `N` is the index of the loop


                                            See this 05AB1E tip of mine (section How to compress strings not part of the dictionary?) to understand why .•3Oû}α›ηö‡.ÝŽ{• is "endgmpfykkzzzzjxzzqz ".





                                            Previous 38 bytes answer:



                                            .•Mñ&Àû«ì{₆v*Å+µ-•#ðšε7∍}IvDyèн©?ε®õ.;


                                            Try it online or verify some more test cases.



                                            Explanation:





                                            .•Mñ&Àû«ì{₆v*Å+µ-•    # Push compressed string "e dg bcmp fhvw k   jx  qz"
                                            # # Split on spaces: ["e","dg","bcmp","fhvw","k","","","jx","","qz"]
                                            ðš # Prepend a space to this list
                                            ε7∍} # Extend each string to size 7:
                                            # [" ","eeeeeee","dgdgdgd","bcmpbcm","fhvwfhv","kkkkkkk","","","jxjxjxj","","qzqzqzq"]
                                            Iv # Loop `y` over the input-list:
                                            Dyè # Get the `y`'th string from a copy of the list
                                            н # Get it's first character
                                            ©? # Store it in the register, and print it without trailing newline
                                            ε # Then map each string in the list to:
                                            ®õ.; # Remove the first occurrence of the character from the register


                                            See this 05AB1E tip of mine (section How to compress strings not part of the dictionary?) to understand why .•Mñ&Àû«ì{₆v*Å+µ-• is "e dg bcmp fhvw k jx qz".







                                            share|improve this answer














                                            share|improve this answer



                                            share|improve this answer








                                            edited yesterday

























                                            answered 2 days ago









                                            Kevin CruijssenKevin Cruijssen

                                            42.7k571217




                                            42.7k571217












                                            • $begingroup$
                                              Can't you use " 0eeeeeee0ddddggg0bbccmmp0ffhhvvw0k000jx00qz"?
                                              $endgroup$
                                              – Expired Data
                                              2 days ago












                                            • $begingroup$
                                              @ExpiredData Ah, of course. You only draw 7 letter.. Thanks! Will change it.
                                              $endgroup$
                                              – Kevin Cruijssen
                                              2 days ago






                                            • 1




                                              $begingroup$
                                              You can save 3 bytes using {v instead of 7F and y instead of I{Nè.
                                              $endgroup$
                                              – Emigna
                                              yesterday










                                            • $begingroup$
                                              @Emigna Ah, of course.. Thanks!
                                              $endgroup$
                                              – Kevin Cruijssen
                                              yesterday


















                                            • $begingroup$
                                              Can't you use " 0eeeeeee0ddddggg0bbccmmp0ffhhvvw0k000jx00qz"?
                                              $endgroup$
                                              – Expired Data
                                              2 days ago












                                            • $begingroup$
                                              @ExpiredData Ah, of course. You only draw 7 letter.. Thanks! Will change it.
                                              $endgroup$
                                              – Kevin Cruijssen
                                              2 days ago






                                            • 1




                                              $begingroup$
                                              You can save 3 bytes using {v instead of 7F and y instead of I{Nè.
                                              $endgroup$
                                              – Emigna
                                              yesterday










                                            • $begingroup$
                                              @Emigna Ah, of course.. Thanks!
                                              $endgroup$
                                              – Kevin Cruijssen
                                              yesterday
















                                            $begingroup$
                                            Can't you use " 0eeeeeee0ddddggg0bbccmmp0ffhhvvw0k000jx00qz"?
                                            $endgroup$
                                            – Expired Data
                                            2 days ago






                                            $begingroup$
                                            Can't you use " 0eeeeeee0ddddggg0bbccmmp0ffhhvvw0k000jx00qz"?
                                            $endgroup$
                                            – Expired Data
                                            2 days ago














                                            $begingroup$
                                            @ExpiredData Ah, of course. You only draw 7 letter.. Thanks! Will change it.
                                            $endgroup$
                                            – Kevin Cruijssen
                                            2 days ago




                                            $begingroup$
                                            @ExpiredData Ah, of course. You only draw 7 letter.. Thanks! Will change it.
                                            $endgroup$
                                            – Kevin Cruijssen
                                            2 days ago




                                            1




                                            1




                                            $begingroup$
                                            You can save 3 bytes using {v instead of 7F and y instead of I{Nè.
                                            $endgroup$
                                            – Emigna
                                            yesterday




                                            $begingroup$
                                            You can save 3 bytes using {v instead of 7F and y instead of I{Nè.
                                            $endgroup$
                                            – Emigna
                                            yesterday












                                            $begingroup$
                                            @Emigna Ah, of course.. Thanks!
                                            $endgroup$
                                            – Kevin Cruijssen
                                            yesterday




                                            $begingroup$
                                            @Emigna Ah, of course.. Thanks!
                                            $endgroup$
                                            – Kevin Cruijssen
                                            yesterday











                                            2












                                            $begingroup$


                                            C (gcc), 110 bytes





                                            _={0,7,14,21,0,0,22,0,24};f(char*s){for(;*s+1;s++)*s=*s?*s-1?"DDDDGGGBBCCMMPFFHHVVWKJXQZ"[_[*s-2]++]:69:32;}


                                            Try it online!



                                            Uses the _ array as an index into the static string "DDDDGGGBBCCMMPFFHHVVWKJXQZ" dynamically with exceptions for 0 and 1.



                                            Argument is a -1-terminated array of scores which is transformed in-place into a -1-terminated string.






                                            share|improve this answer









                                            $endgroup$


















                                              2












                                              $begingroup$


                                              C (gcc), 110 bytes





                                              _={0,7,14,21,0,0,22,0,24};f(char*s){for(;*s+1;s++)*s=*s?*s-1?"DDDDGGGBBCCMMPFFHHVVWKJXQZ"[_[*s-2]++]:69:32;}


                                              Try it online!



                                              Uses the _ array as an index into the static string "DDDDGGGBBCCMMPFFHHVVWKJXQZ" dynamically with exceptions for 0 and 1.



                                              Argument is a -1-terminated array of scores which is transformed in-place into a -1-terminated string.






                                              share|improve this answer









                                              $endgroup$
















                                                2












                                                2








                                                2





                                                $begingroup$


                                                C (gcc), 110 bytes





                                                _={0,7,14,21,0,0,22,0,24};f(char*s){for(;*s+1;s++)*s=*s?*s-1?"DDDDGGGBBCCMMPFFHHVVWKJXQZ"[_[*s-2]++]:69:32;}


                                                Try it online!



                                                Uses the _ array as an index into the static string "DDDDGGGBBCCMMPFFHHVVWKJXQZ" dynamically with exceptions for 0 and 1.



                                                Argument is a -1-terminated array of scores which is transformed in-place into a -1-terminated string.






                                                share|improve this answer









                                                $endgroup$




                                                C (gcc), 110 bytes





                                                _={0,7,14,21,0,0,22,0,24};f(char*s){for(;*s+1;s++)*s=*s?*s-1?"DDDDGGGBBCCMMPFFHHVVWKJXQZ"[_[*s-2]++]:69:32;}


                                                Try it online!



                                                Uses the _ array as an index into the static string "DDDDGGGBBCCMMPFFHHVVWKJXQZ" dynamically with exceptions for 0 and 1.



                                                Argument is a -1-terminated array of scores which is transformed in-place into a -1-terminated string.







                                                share|improve this answer












                                                share|improve this answer



                                                share|improve this answer










                                                answered 2 days ago









                                                LambdaBetaLambdaBeta

                                                2,239418




                                                2,239418























                                                    1












                                                    $begingroup$


                                                    C# (Visual C# Interactive Compiler), 104 90 bytes





                                                    a=>a.OrderBy(x=>x).Select((x,i)=>(x="_ E DG BCMP FHVW K   JX  QZ".Split()[x])[i%x.Length])


                                                    Try it online!






                                                    share|improve this answer











                                                    $endgroup$


















                                                      1












                                                      $begingroup$


                                                      C# (Visual C# Interactive Compiler), 104 90 bytes





                                                      a=>a.OrderBy(x=>x).Select((x,i)=>(x="_ E DG BCMP FHVW K   JX  QZ".Split()[x])[i%x.Length])


                                                      Try it online!






                                                      share|improve this answer











                                                      $endgroup$
















                                                        1












                                                        1








                                                        1





                                                        $begingroup$


                                                        C# (Visual C# Interactive Compiler), 104 90 bytes





                                                        a=>a.OrderBy(x=>x).Select((x,i)=>(x="_ E DG BCMP FHVW K   JX  QZ".Split()[x])[i%x.Length])


                                                        Try it online!






                                                        share|improve this answer











                                                        $endgroup$




                                                        C# (Visual C# Interactive Compiler), 104 90 bytes





                                                        a=>a.OrderBy(x=>x).Select((x,i)=>(x="_ E DG BCMP FHVW K   JX  QZ".Split()[x])[i%x.Length])


                                                        Try it online!







                                                        share|improve this answer














                                                        share|improve this answer



                                                        share|improve this answer








                                                        edited 2 days ago

























                                                        answered 2 days ago









                                                        Expired DataExpired Data

                                                        898216




                                                        898216























                                                            1












                                                            $begingroup$


                                                            Jelly, 34 32 bytes



                                                            “¿RÇĊƈ⁸⁾%ỵṆþœsṀṂ’ṃØAṣ”A;⁶ẋ€7⁸ịḢ€


                                                            Try it online!



                                                            I hadn’t seen there was a shorter Jelly answer when I wrote this, and this uses a different approach so I thought was worth posting as well.



                                                            Thanks to @JonathanAllan for saving 2 bytes!






                                                            share|improve this answer











                                                            $endgroup$













                                                            • $begingroup$
                                                              By using base-decompression, , you can save 2 bytes
                                                              $endgroup$
                                                              – Jonathan Allan
                                                              2 days ago
















                                                            1












                                                            $begingroup$


                                                            Jelly, 34 32 bytes



                                                            “¿RÇĊƈ⁸⁾%ỵṆþœsṀṂ’ṃØAṣ”A;⁶ẋ€7⁸ịḢ€


                                                            Try it online!



                                                            I hadn’t seen there was a shorter Jelly answer when I wrote this, and this uses a different approach so I thought was worth posting as well.



                                                            Thanks to @JonathanAllan for saving 2 bytes!






                                                            share|improve this answer











                                                            $endgroup$













                                                            • $begingroup$
                                                              By using base-decompression, , you can save 2 bytes
                                                              $endgroup$
                                                              – Jonathan Allan
                                                              2 days ago














                                                            1












                                                            1








                                                            1





                                                            $begingroup$


                                                            Jelly, 34 32 bytes



                                                            “¿RÇĊƈ⁸⁾%ỵṆþœsṀṂ’ṃØAṣ”A;⁶ẋ€7⁸ịḢ€


                                                            Try it online!



                                                            I hadn’t seen there was a shorter Jelly answer when I wrote this, and this uses a different approach so I thought was worth posting as well.



                                                            Thanks to @JonathanAllan for saving 2 bytes!






                                                            share|improve this answer











                                                            $endgroup$




                                                            Jelly, 34 32 bytes



                                                            “¿RÇĊƈ⁸⁾%ỵṆþœsṀṂ’ṃØAṣ”A;⁶ẋ€7⁸ịḢ€


                                                            Try it online!



                                                            I hadn’t seen there was a shorter Jelly answer when I wrote this, and this uses a different approach so I thought was worth posting as well.



                                                            Thanks to @JonathanAllan for saving 2 bytes!







                                                            share|improve this answer














                                                            share|improve this answer



                                                            share|improve this answer








                                                            edited 2 days ago

























                                                            answered 2 days ago









                                                            Nick KennedyNick Kennedy

                                                            1,53649




                                                            1,53649












                                                            • $begingroup$
                                                              By using base-decompression, , you can save 2 bytes
                                                              $endgroup$
                                                              – Jonathan Allan
                                                              2 days ago


















                                                            • $begingroup$
                                                              By using base-decompression, , you can save 2 bytes
                                                              $endgroup$
                                                              – Jonathan Allan
                                                              2 days ago
















                                                            $begingroup$
                                                            By using base-decompression, , you can save 2 bytes
                                                            $endgroup$
                                                            – Jonathan Allan
                                                            2 days ago




                                                            $begingroup$
                                                            By using base-decompression, , you can save 2 bytes
                                                            $endgroup$
                                                            – Jonathan Allan
                                                            2 days ago











                                                            1












                                                            $begingroup$


                                                            Python 3, 178 142 135 127 112 117 bytes





                                                            def f(l):
                                                            d=list(map(list," _EEEEEEE_DDDDGGG_BBCCMMP_FFHHVVW_K___JX__QZ".split('_')))
                                                            return[d[i].pop()for i in l]


                                                            Try it online!



                                                            -1 byte thanks to cdlane



                                                            correct thanks to mathmandan






                                                            share|improve this answer











                                                            $endgroup$













                                                            • $begingroup$
                                                              in " -> in" for 111
                                                              $endgroup$
                                                              – cdlane
                                                              2 days ago










                                                            • $begingroup$
                                                              d=list(map(list,"...".split('_'))) to save another byte
                                                              $endgroup$
                                                              – cdlane
                                                              2 days ago












                                                            • $begingroup$
                                                              This function f probably doesn't need to be named, so you can save 2 bytes. However, f consumes the entries of d, so I'm not sure that it fits the consensus requirement that "the function has to be reusable arbitrarily often, without...restating...any other code accompanying the submission." (For example, running f([10,0,10,5,8,8,0]) more than once would result in an error.) Please see meta discussion here: codegolf.meta.stackexchange.com/a/7615/36885
                                                              $endgroup$
                                                              – mathmandan
                                                              2 days ago
















                                                            1












                                                            $begingroup$


                                                            Python 3, 178 142 135 127 112 117 bytes





                                                            def f(l):
                                                            d=list(map(list," _EEEEEEE_DDDDGGG_BBCCMMP_FFHHVVW_K___JX__QZ".split('_')))
                                                            return[d[i].pop()for i in l]


                                                            Try it online!



                                                            -1 byte thanks to cdlane



                                                            correct thanks to mathmandan






                                                            share|improve this answer











                                                            $endgroup$













                                                            • $begingroup$
                                                              in " -> in" for 111
                                                              $endgroup$
                                                              – cdlane
                                                              2 days ago










                                                            • $begingroup$
                                                              d=list(map(list,"...".split('_'))) to save another byte
                                                              $endgroup$
                                                              – cdlane
                                                              2 days ago












                                                            • $begingroup$
                                                              This function f probably doesn't need to be named, so you can save 2 bytes. However, f consumes the entries of d, so I'm not sure that it fits the consensus requirement that "the function has to be reusable arbitrarily often, without...restating...any other code accompanying the submission." (For example, running f([10,0,10,5,8,8,0]) more than once would result in an error.) Please see meta discussion here: codegolf.meta.stackexchange.com/a/7615/36885
                                                              $endgroup$
                                                              – mathmandan
                                                              2 days ago














                                                            1












                                                            1








                                                            1





                                                            $begingroup$


                                                            Python 3, 178 142 135 127 112 117 bytes





                                                            def f(l):
                                                            d=list(map(list," _EEEEEEE_DDDDGGG_BBCCMMP_FFHHVVW_K___JX__QZ".split('_')))
                                                            return[d[i].pop()for i in l]


                                                            Try it online!



                                                            -1 byte thanks to cdlane



                                                            correct thanks to mathmandan






                                                            share|improve this answer











                                                            $endgroup$




                                                            Python 3, 178 142 135 127 112 117 bytes





                                                            def f(l):
                                                            d=list(map(list," _EEEEEEE_DDDDGGG_BBCCMMP_FFHHVVW_K___JX__QZ".split('_')))
                                                            return[d[i].pop()for i in l]


                                                            Try it online!



                                                            -1 byte thanks to cdlane



                                                            correct thanks to mathmandan







                                                            share|improve this answer














                                                            share|improve this answer



                                                            share|improve this answer








                                                            edited 2 days ago

























                                                            answered 2 days ago









                                                            Noodle9Noodle9

                                                            30137




                                                            30137












                                                            • $begingroup$
                                                              in " -> in" for 111
                                                              $endgroup$
                                                              – cdlane
                                                              2 days ago










                                                            • $begingroup$
                                                              d=list(map(list,"...".split('_'))) to save another byte
                                                              $endgroup$
                                                              – cdlane
                                                              2 days ago












                                                            • $begingroup$
                                                              This function f probably doesn't need to be named, so you can save 2 bytes. However, f consumes the entries of d, so I'm not sure that it fits the consensus requirement that "the function has to be reusable arbitrarily often, without...restating...any other code accompanying the submission." (For example, running f([10,0,10,5,8,8,0]) more than once would result in an error.) Please see meta discussion here: codegolf.meta.stackexchange.com/a/7615/36885
                                                              $endgroup$
                                                              – mathmandan
                                                              2 days ago


















                                                            • $begingroup$
                                                              in " -> in" for 111
                                                              $endgroup$
                                                              – cdlane
                                                              2 days ago










                                                            • $begingroup$
                                                              d=list(map(list,"...".split('_'))) to save another byte
                                                              $endgroup$
                                                              – cdlane
                                                              2 days ago












                                                            • $begingroup$
                                                              This function f probably doesn't need to be named, so you can save 2 bytes. However, f consumes the entries of d, so I'm not sure that it fits the consensus requirement that "the function has to be reusable arbitrarily often, without...restating...any other code accompanying the submission." (For example, running f([10,0,10,5,8,8,0]) more than once would result in an error.) Please see meta discussion here: codegolf.meta.stackexchange.com/a/7615/36885
                                                              $endgroup$
                                                              – mathmandan
                                                              2 days ago
















                                                            $begingroup$
                                                            in " -> in" for 111
                                                            $endgroup$
                                                            – cdlane
                                                            2 days ago




                                                            $begingroup$
                                                            in " -> in" for 111
                                                            $endgroup$
                                                            – cdlane
                                                            2 days ago












                                                            $begingroup$
                                                            d=list(map(list,"...".split('_'))) to save another byte
                                                            $endgroup$
                                                            – cdlane
                                                            2 days ago






                                                            $begingroup$
                                                            d=list(map(list,"...".split('_'))) to save another byte
                                                            $endgroup$
                                                            – cdlane
                                                            2 days ago














                                                            $begingroup$
                                                            This function f probably doesn't need to be named, so you can save 2 bytes. However, f consumes the entries of d, so I'm not sure that it fits the consensus requirement that "the function has to be reusable arbitrarily often, without...restating...any other code accompanying the submission." (For example, running f([10,0,10,5,8,8,0]) more than once would result in an error.) Please see meta discussion here: codegolf.meta.stackexchange.com/a/7615/36885
                                                            $endgroup$
                                                            – mathmandan
                                                            2 days ago




                                                            $begingroup$
                                                            This function f probably doesn't need to be named, so you can save 2 bytes. However, f consumes the entries of d, so I'm not sure that it fits the consensus requirement that "the function has to be reusable arbitrarily often, without...restating...any other code accompanying the submission." (For example, running f([10,0,10,5,8,8,0]) more than once would result in an error.) Please see meta discussion here: codegolf.meta.stackexchange.com/a/7615/36885
                                                            $endgroup$
                                                            – mathmandan
                                                            2 days ago











                                                            0












                                                            $begingroup$


                                                            Python 2, 102 bytes (or maybe 95?)



                                                            (Also fine for Python 3.)





                                                            lambda a:''.join([r*7for r in'_ E DG BCMP FHVW K * * JX * QZ'.split()][x][:a.count(x)]for x in set(a))


                                                            Try it online!



                                                            I don't think the following would be acceptable:



                                                            lambda a:[[r*7for r in'_ E DG BCMP FHVW K * * JX * QZ'.split()][x][:a.count(x)]for x in set(a)]


                                                            This second version would give output like ['__', 'JX', 'QZ', 'K']. So the letters would be correct, but they would be collected by point value. (If this were acceptable, it would save 7 bytes.)






                                                            share|improve this answer









                                                            $endgroup$


















                                                              0












                                                              $begingroup$


                                                              Python 2, 102 bytes (or maybe 95?)



                                                              (Also fine for Python 3.)





                                                              lambda a:''.join([r*7for r in'_ E DG BCMP FHVW K * * JX * QZ'.split()][x][:a.count(x)]for x in set(a))


                                                              Try it online!



                                                              I don't think the following would be acceptable:



                                                              lambda a:[[r*7for r in'_ E DG BCMP FHVW K * * JX * QZ'.split()][x][:a.count(x)]for x in set(a)]


                                                              This second version would give output like ['__', 'JX', 'QZ', 'K']. So the letters would be correct, but they would be collected by point value. (If this were acceptable, it would save 7 bytes.)






                                                              share|improve this answer









                                                              $endgroup$
















                                                                0












                                                                0








                                                                0





                                                                $begingroup$


                                                                Python 2, 102 bytes (or maybe 95?)



                                                                (Also fine for Python 3.)





                                                                lambda a:''.join([r*7for r in'_ E DG BCMP FHVW K * * JX * QZ'.split()][x][:a.count(x)]for x in set(a))


                                                                Try it online!



                                                                I don't think the following would be acceptable:



                                                                lambda a:[[r*7for r in'_ E DG BCMP FHVW K * * JX * QZ'.split()][x][:a.count(x)]for x in set(a)]


                                                                This second version would give output like ['__', 'JX', 'QZ', 'K']. So the letters would be correct, but they would be collected by point value. (If this were acceptable, it would save 7 bytes.)






                                                                share|improve this answer









                                                                $endgroup$




                                                                Python 2, 102 bytes (or maybe 95?)



                                                                (Also fine for Python 3.)





                                                                lambda a:''.join([r*7for r in'_ E DG BCMP FHVW K * * JX * QZ'.split()][x][:a.count(x)]for x in set(a))


                                                                Try it online!



                                                                I don't think the following would be acceptable:



                                                                lambda a:[[r*7for r in'_ E DG BCMP FHVW K * * JX * QZ'.split()][x][:a.count(x)]for x in set(a)]


                                                                This second version would give output like ['__', 'JX', 'QZ', 'K']. So the letters would be correct, but they would be collected by point value. (If this were acceptable, it would save 7 bytes.)







                                                                share|improve this answer












                                                                share|improve this answer



                                                                share|improve this answer










                                                                answered yesterday









                                                                mathmandanmathmandan

                                                                92367




                                                                92367























                                                                    0












                                                                    $begingroup$


                                                                    PHP, 101 bytes





                                                                    $b=[_,E,DG,BCMP,FHVW,K,8=>JX,0,QZ];foreach($argv as$t){echo$c=($d=$b[$t])[0];$b[$t]=substr($d,1).$c;}


                                                                    As a standalone program, input via command line:



                                                                    $ php s.php 10 0 10 5 8 8 0
                                                                    "Q_ZKJX_"


                                                                    Try it online!



                                                                    Or 112 bytes as a function



                                                                    function($a){$b=[_,E,DG,BCMP,FHVW,K,8=>JX,0,QZ];foreach($a as$t)$b[$t]=substr($d=$b[$t],1).$c=$d[0];return$c;}


                                                                    Try it online!



                                                                    Output



                                                                    [10,0,10,5,8,8,0]   "Q_ZKJX_"
                                                                    [1,1,1,1,1,1,1] "EEEEEEE"
                                                                    [1,2,3,4,5,8,0] "EDBFKJ_"
                                                                    [2,2,2,2,2,2,2] "DGDGDGD"





                                                                    share|improve this answer











                                                                    $endgroup$


















                                                                      0












                                                                      $begingroup$


                                                                      PHP, 101 bytes





                                                                      $b=[_,E,DG,BCMP,FHVW,K,8=>JX,0,QZ];foreach($argv as$t){echo$c=($d=$b[$t])[0];$b[$t]=substr($d,1).$c;}


                                                                      As a standalone program, input via command line:



                                                                      $ php s.php 10 0 10 5 8 8 0
                                                                      "Q_ZKJX_"


                                                                      Try it online!



                                                                      Or 112 bytes as a function



                                                                      function($a){$b=[_,E,DG,BCMP,FHVW,K,8=>JX,0,QZ];foreach($a as$t)$b[$t]=substr($d=$b[$t],1).$c=$d[0];return$c;}


                                                                      Try it online!



                                                                      Output



                                                                      [10,0,10,5,8,8,0]   "Q_ZKJX_"
                                                                      [1,1,1,1,1,1,1] "EEEEEEE"
                                                                      [1,2,3,4,5,8,0] "EDBFKJ_"
                                                                      [2,2,2,2,2,2,2] "DGDGDGD"





                                                                      share|improve this answer











                                                                      $endgroup$
















                                                                        0












                                                                        0








                                                                        0





                                                                        $begingroup$


                                                                        PHP, 101 bytes





                                                                        $b=[_,E,DG,BCMP,FHVW,K,8=>JX,0,QZ];foreach($argv as$t){echo$c=($d=$b[$t])[0];$b[$t]=substr($d,1).$c;}


                                                                        As a standalone program, input via command line:



                                                                        $ php s.php 10 0 10 5 8 8 0
                                                                        "Q_ZKJX_"


                                                                        Try it online!



                                                                        Or 112 bytes as a function



                                                                        function($a){$b=[_,E,DG,BCMP,FHVW,K,8=>JX,0,QZ];foreach($a as$t)$b[$t]=substr($d=$b[$t],1).$c=$d[0];return$c;}


                                                                        Try it online!



                                                                        Output



                                                                        [10,0,10,5,8,8,0]   "Q_ZKJX_"
                                                                        [1,1,1,1,1,1,1] "EEEEEEE"
                                                                        [1,2,3,4,5,8,0] "EDBFKJ_"
                                                                        [2,2,2,2,2,2,2] "DGDGDGD"





                                                                        share|improve this answer











                                                                        $endgroup$




                                                                        PHP, 101 bytes





                                                                        $b=[_,E,DG,BCMP,FHVW,K,8=>JX,0,QZ];foreach($argv as$t){echo$c=($d=$b[$t])[0];$b[$t]=substr($d,1).$c;}


                                                                        As a standalone program, input via command line:



                                                                        $ php s.php 10 0 10 5 8 8 0
                                                                        "Q_ZKJX_"


                                                                        Try it online!



                                                                        Or 112 bytes as a function



                                                                        function($a){$b=[_,E,DG,BCMP,FHVW,K,8=>JX,0,QZ];foreach($a as$t)$b[$t]=substr($d=$b[$t],1).$c=$d[0];return$c;}


                                                                        Try it online!



                                                                        Output



                                                                        [10,0,10,5,8,8,0]   "Q_ZKJX_"
                                                                        [1,1,1,1,1,1,1] "EEEEEEE"
                                                                        [1,2,3,4,5,8,0] "EDBFKJ_"
                                                                        [2,2,2,2,2,2,2] "DGDGDGD"






                                                                        share|improve this answer














                                                                        share|improve this answer



                                                                        share|improve this answer








                                                                        edited yesterday

























                                                                        answered 2 days ago









                                                                        gwaughgwaugh

                                                                        2,1581518




                                                                        2,1581518























                                                                            0












                                                                            $begingroup$


                                                                            Ruby, 77 76 bytes





                                                                            ->a{r=%w{_ E DG BCMP FHVW K . . JX . QZ};a.map{|i|(r[i]<<r[i][0]).slice! 0}}


                                                                            Try it online!






                                                                            share|improve this answer











                                                                            $endgroup$


















                                                                              0












                                                                              $begingroup$


                                                                              Ruby, 77 76 bytes





                                                                              ->a{r=%w{_ E DG BCMP FHVW K . . JX . QZ};a.map{|i|(r[i]<<r[i][0]).slice! 0}}


                                                                              Try it online!






                                                                              share|improve this answer











                                                                              $endgroup$
















                                                                                0












                                                                                0








                                                                                0





                                                                                $begingroup$


                                                                                Ruby, 77 76 bytes





                                                                                ->a{r=%w{_ E DG BCMP FHVW K . . JX . QZ};a.map{|i|(r[i]<<r[i][0]).slice! 0}}


                                                                                Try it online!






                                                                                share|improve this answer











                                                                                $endgroup$




                                                                                Ruby, 77 76 bytes





                                                                                ->a{r=%w{_ E DG BCMP FHVW K . . JX . QZ};a.map{|i|(r[i]<<r[i][0]).slice! 0}}


                                                                                Try it online!







                                                                                share|improve this answer














                                                                                share|improve this answer



                                                                                share|improve this answer








                                                                                edited yesterday

























                                                                                answered yesterday









                                                                                iamnotmaynardiamnotmaynard

                                                                                95349




                                                                                95349























                                                                                    0












                                                                                    $begingroup$


                                                                                    Perl 6, 63 bytes





                                                                                    *>>.&{(<_ E DG BCMP FHVW K _ _ JX _ QZ>[$_]x 7).comb[%.{$_}++]}


                                                                                    Try it online!



                                                                                    <_ E DG BCMP FHVW K _ _ JX _ QZ> # array indexed on tile value
                                                                                    (<...>[$_] x 7) # pull letters for this value, repeat 7 times to catch E
                                                                                    % # anonymous stateful hash
                                                                                    .{$_} # element for this tile value
                                                                                    ++ # post increment value to move position
                                                                                    .comb[...] # characters to array, pull this incrementing index


                                                                                    So essentially it keeps a lookup of offsets for each tile value and increments them as needed, using the offset to pull a character from the available set.






                                                                                    share|improve this answer









                                                                                    $endgroup$


















                                                                                      0












                                                                                      $begingroup$


                                                                                      Perl 6, 63 bytes





                                                                                      *>>.&{(<_ E DG BCMP FHVW K _ _ JX _ QZ>[$_]x 7).comb[%.{$_}++]}


                                                                                      Try it online!



                                                                                      <_ E DG BCMP FHVW K _ _ JX _ QZ> # array indexed on tile value
                                                                                      (<...>[$_] x 7) # pull letters for this value, repeat 7 times to catch E
                                                                                      % # anonymous stateful hash
                                                                                      .{$_} # element for this tile value
                                                                                      ++ # post increment value to move position
                                                                                      .comb[...] # characters to array, pull this incrementing index


                                                                                      So essentially it keeps a lookup of offsets for each tile value and increments them as needed, using the offset to pull a character from the available set.






                                                                                      share|improve this answer









                                                                                      $endgroup$
















                                                                                        0












                                                                                        0








                                                                                        0





                                                                                        $begingroup$


                                                                                        Perl 6, 63 bytes





                                                                                        *>>.&{(<_ E DG BCMP FHVW K _ _ JX _ QZ>[$_]x 7).comb[%.{$_}++]}


                                                                                        Try it online!



                                                                                        <_ E DG BCMP FHVW K _ _ JX _ QZ> # array indexed on tile value
                                                                                        (<...>[$_] x 7) # pull letters for this value, repeat 7 times to catch E
                                                                                        % # anonymous stateful hash
                                                                                        .{$_} # element for this tile value
                                                                                        ++ # post increment value to move position
                                                                                        .comb[...] # characters to array, pull this incrementing index


                                                                                        So essentially it keeps a lookup of offsets for each tile value and increments them as needed, using the offset to pull a character from the available set.






                                                                                        share|improve this answer









                                                                                        $endgroup$




                                                                                        Perl 6, 63 bytes





                                                                                        *>>.&{(<_ E DG BCMP FHVW K _ _ JX _ QZ>[$_]x 7).comb[%.{$_}++]}


                                                                                        Try it online!



                                                                                        <_ E DG BCMP FHVW K _ _ JX _ QZ> # array indexed on tile value
                                                                                        (<...>[$_] x 7) # pull letters for this value, repeat 7 times to catch E
                                                                                        % # anonymous stateful hash
                                                                                        .{$_} # element for this tile value
                                                                                        ++ # post increment value to move position
                                                                                        .comb[...] # characters to array, pull this incrementing index


                                                                                        So essentially it keeps a lookup of offsets for each tile value and increments them as needed, using the offset to pull a character from the available set.







                                                                                        share|improve this answer












                                                                                        share|improve this answer



                                                                                        share|improve this answer










                                                                                        answered 18 hours ago









                                                                                        Phil HPhil H

                                                                                        1,112817




                                                                                        1,112817






























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                                                                                            If this is an answer to a challenge…




                                                                                            • …Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.


                                                                                            • …Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
                                                                                              Explanations of your answer make it more interesting to read and are very much encouraged.


                                                                                            • …Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.



                                                                                            More generally…




                                                                                            • …Please make sure to answer the question and provide sufficient detail.


                                                                                            • …Avoid asking for help, clarification or responding to other answers (use comments instead).





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