In the RSA DES challenges, how did the contestants know they had found the right key considering they weren't...
If the contestants were given both the plaintext and ciphertext, it's straightforward. Just bruteforce all 56-bit keys until you find one that maps the given plaintext to the given ciphertext.
But from what I understand, the contestants were given only the ciphertext and the initialization vector. I'm confused as to how they cracked the challenge without any plaintext.
What I imagine they did was:
- Ok, we know the plaintext is less than or equal to the size of the ciphertext.
- Calculate all possible plaintext from 1 bit up to the bit size of the ciphertext.
- Calculate all possible 56-bit keys.
- Run each plaintext through all 56-bit keys until they found a mapping.
But this doesn't make sense given the sheer size of the possibilities.
So since they weren't given any plaintext, how did they know they'd found the right plaintext/key combination?
rsa des brute-force-attack history
edited Dec 15 at 14:26
kelalaka
4,97321939
asked Dec 14 at 8:54
Bastien
333
New contributor
Bastien is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
If the contestants were given both the plaintext and ciphertext, it's straightforward. Just bruteforce all 56-bit keys until you find one that maps the given plaintext to the given ciphertext.
But from what I understand, the contestants were given only the ciphertext and the initialization vector. I'm confused as to how they cracked the challenge without any plaintext.
What I imagine they did was:
- Ok, we know the plaintext is less than or equal to the size of the ciphertext.
- Calculate all possible plaintext from 1 bit up to the bit size of the ciphertext.
- Calculate all possible 56-bit keys.
- Run each plaintext through all 56-bit keys until they found a mapping.
But this doesn't make sense given the sheer size of the possibilities.
So since they weren't given any plaintext, how did they know they'd found the right plaintext/key combination?
rsa des brute-force-attack history
edited Dec 15 at 14:26
kelalaka
4,97321939
asked Dec 14 at 8:54
Bastien
333
New contributor
Bastien is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
If the contestants were given both the plaintext and ciphertext, it's straightforward. Just bruteforce all 56-bit keys until you find one that maps the given plaintext to the given ciphertext.
But from what I understand, the contestants were given only the ciphertext and the initialization vector. I'm confused as to how they cracked the challenge without any plaintext.
What I imagine they did was:
- Ok, we know the plaintext is less than or equal to the size of the ciphertext.
- Calculate all possible plaintext from 1 bit up to the bit size of the ciphertext.
- Calculate all possible 56-bit keys.
- Run each plaintext through all 56-bit keys until they found a mapping.
But this doesn't make sense given the sheer size of the possibilities.
So since they weren't given any plaintext, how did they know they'd found the right plaintext/key combination?
rsa des brute-force-attack history
edited Dec 15 at 14:26
kelalaka
4,97321939
asked Dec 14 at 8:54
Bastien
333
New contributor
Bastien is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
If the contestants were given both the plaintext and ciphertext, it's straightforward. Just bruteforce all 56-bit keys until you find one that maps the given plaintext to the given ciphertext.
But from what I understand, the contestants were given only the ciphertext and the initialization vector. I'm confused as to how they cracked the challenge without any plaintext.
What I imagine they did was:
- Ok, we know the plaintext is less than or equal to the size of the ciphertext.
- Calculate all possible plaintext from 1 bit up to the bit size of the ciphertext.
- Calculate all possible 56-bit keys.
- Run each plaintext through all 56-bit keys until they found a mapping.
But this doesn't make sense given the sheer size of the possibilities.
So since they weren't given any plaintext, how did they know they'd found the right plaintext/key combination?
rsa des brute-force-attack history
rsa des brute-force-attack history
edited Dec 15 at 14:26
kelalaka
4,97321939
asked Dec 14 at 8:54
Bastien
333
New contributor
Bastien is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 15 at 14:26
kelalaka
4,97321939
asked Dec 14 at 8:54
Bastien
333
New contributor
Bastien is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 15 at 14:26
kelalaka
4,97321939
edited Dec 15 at 14:26
kelalaka
4,97321939
edited Dec 15 at 14:26
kelalaka
4,97321939
4,97321939
asked Dec 14 at 8:54
Bastien
333
New contributor
Bastien is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 14 at 8:54
Bastien
333
asked Dec 14 at 8:54
Bastien
333
333
New contributor
Bastien is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Bastien is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Bastien is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
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One can still access the challenge rules from the archive.org
Each contest is based on a specified cipher. A brief piece of printable ASCII text (containing byte values in hexadecimal notation from 0x20 to 0x7e) will be appended to the fixed 24-character string "The unknown message is:". The result will be padded and then encrypted with the associated cipher under a randomly-generated key.
The message is also padded by PKCS#5 padding. Thus, the attacker has three options to test their results;
- Check the beginning of the message; it must start with "The unknown message is:"
- A valid PKCS#5 padding at he the end.
- And, one general rule when only ciphertexts are given; one can check the result is valid (English) language. For this, one block may not be enough, since there are many valid words in the 64-bit block. One block will result in many keys. To narrow the keys, additional ciphertexts are required. This approach can be work even 1. and 2. cases are not known. For this specific challenge the range of the characters given between
0x20and0x7e, i.e. betweenand~
distributed.net found this message in DES Challenge II-1;
The secret message is: Many hands make light work.
Note : It is interesting that the Wikipedia says; The secret message is: where the source is lists.distributed.net, however, the RSA archives says The unknown message is:.
answered Dec 14 at 10:53
kelalaka
4,97321939
Super clear explanation. Thank you.
– Bastien
Dec 17 at 7:12
add a comment |
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One can still access the challenge rules from the archive.org
Each contest is based on a specified cipher. A brief piece of printable ASCII text (containing byte values in hexadecimal notation from 0x20 to 0x7e) will be appended to the fixed 24-character string "The unknown message is:". The result will be padded and then encrypted with the associated cipher under a randomly-generated key.
The message is also padded by PKCS#5 padding. Thus, the attacker has three options to test their results;
- Check the beginning of the message; it must start with "The unknown message is:"
- A valid PKCS#5 padding at he the end.
- And, one general rule when only ciphertexts are given; one can check the result is valid (English) language. For this, one block may not be enough, since there are many valid words in the 64-bit block. One block will result in many keys. To narrow the keys, additional ciphertexts are required. This approach can be work even 1. and 2. cases are not known. For this specific challenge the range of the characters given between
0x20and0x7e, i.e. betweenand~
distributed.net found this message in DES Challenge II-1;
The secret message is: Many hands make light work.
Note : It is interesting that the Wikipedia says; The secret message is: where the source is lists.distributed.net, however, the RSA archives says The unknown message is:.
answered Dec 14 at 10:53
kelalaka
4,97321939
Super clear explanation. Thank you.
– Bastien
Dec 17 at 7:12
add a comment |
One can still access the challenge rules from the archive.org
Each contest is based on a specified cipher. A brief piece of printable ASCII text (containing byte values in hexadecimal notation from 0x20 to 0x7e) will be appended to the fixed 24-character string "The unknown message is:". The result will be padded and then encrypted with the associated cipher under a randomly-generated key.
The message is also padded by PKCS#5 padding. Thus, the attacker has three options to test their results;
- Check the beginning of the message; it must start with "The unknown message is:"
- A valid PKCS#5 padding at he the end.
- And, one general rule when only ciphertexts are given; one can check the result is valid (English) language. For this, one block may not be enough, since there are many valid words in the 64-bit block. One block will result in many keys. To narrow the keys, additional ciphertexts are required. This approach can be work even 1. and 2. cases are not known. For this specific challenge the range of the characters given between
0x20and0x7e, i.e. betweenand~
distributed.net found this message in DES Challenge II-1;
The secret message is: Many hands make light work.
Note : It is interesting that the Wikipedia says; The secret message is: where the source is lists.distributed.net, however, the RSA archives says The unknown message is:.
answered Dec 14 at 10:53
kelalaka
4,97321939
Super clear explanation. Thank you.
– Bastien
Dec 17 at 7:12
add a comment |
One can still access the challenge rules from the archive.org
Each contest is based on a specified cipher. A brief piece of printable ASCII text (containing byte values in hexadecimal notation from 0x20 to 0x7e) will be appended to the fixed 24-character string "The unknown message is:". The result will be padded and then encrypted with the associated cipher under a randomly-generated key.
The message is also padded by PKCS#5 padding. Thus, the attacker has three options to test their results;
- Check the beginning of the message; it must start with "The unknown message is:"
- A valid PKCS#5 padding at he the end.
- And, one general rule when only ciphertexts are given; one can check the result is valid (English) language. For this, one block may not be enough, since there are many valid words in the 64-bit block. One block will result in many keys. To narrow the keys, additional ciphertexts are required. This approach can be work even 1. and 2. cases are not known. For this specific challenge the range of the characters given between
0x20and0x7e, i.e. betweenand~
distributed.net found this message in DES Challenge II-1;
The secret message is: Many hands make light work.
Note : It is interesting that the Wikipedia says; The secret message is: where the source is lists.distributed.net, however, the RSA archives says The unknown message is:.
answered Dec 14 at 10:53
kelalaka
4,97321939
One can still access the challenge rules from the archive.org
Each contest is based on a specified cipher. A brief piece of printable ASCII text (containing byte values in hexadecimal notation from 0x20 to 0x7e) will be appended to the fixed 24-character string "The unknown message is:". The result will be padded and then encrypted with the associated cipher under a randomly-generated key.
The message is also padded by PKCS#5 padding. Thus, the attacker has three options to test their results;
- Check the beginning of the message; it must start with "The unknown message is:"
- A valid PKCS#5 padding at he the end.
- And, one general rule when only ciphertexts are given; one can check the result is valid (English) language. For this, one block may not be enough, since there are many valid words in the 64-bit block. One block will result in many keys. To narrow the keys, additional ciphertexts are required. This approach can be work even 1. and 2. cases are not known. For this specific challenge the range of the characters given between
0x20and0x7e, i.e. betweenand~
distributed.net found this message in DES Challenge II-1;
The secret message is: Many hands make light work.
Note : It is interesting that the Wikipedia says; The secret message is: where the source is lists.distributed.net, however, the RSA archives says The unknown message is:.
answered Dec 14 at 10:53
kelalaka
4,97321939
edited Dec 14 at 19:10
answered Dec 14 at 10:53
kelalaka
4,97321939
answered Dec 14 at 10:53
kelalaka
4,97321939
answered Dec 14 at 10:53
kelalaka
4,97321939
4,97321939
Super clear explanation. Thank you.
– Bastien
Dec 17 at 7:12
add a comment |
Super clear explanation. Thank you.
– Bastien
Dec 17 at 7:12
Super clear explanation. Thank you.
– Bastien
Dec 17 at 7:12
Super clear explanation. Thank you.
– Bastien
Dec 17 at 7:12
add a comment |
Bastien is a new contributor. Be nice, and check out our Code of Conduct.
Bastien is a new contributor. Be nice, and check out our Code of Conduct.
Bastien is a new contributor. Be nice, and check out our Code of Conduct.
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