If a non-deterministic Turing machine runs in f(n) space, then why does it run in 2^O(f(n)) time?
Assuming that f(n) >= n.
If possible, I'd like a proof in terms of Turing machines. I understand the reason why with machines that run on binary, because each "tape cell" is a bit with either 0 or 1, but in Turing machines a tape cell could hold any number of symbols. I'm having trouble why the base is '2' and not something like 'b' where 'b' is the number of types of symbols of the Turing machine tape.
complexity-theory turing-machines
asked 5 hours ago
Taking1n1
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Assuming that f(n) >= n.
If possible, I'd like a proof in terms of Turing machines. I understand the reason why with machines that run on binary, because each "tape cell" is a bit with either 0 or 1, but in Turing machines a tape cell could hold any number of symbols. I'm having trouble why the base is '2' and not something like 'b' where 'b' is the number of types of symbols of the Turing machine tape.
complexity-theory turing-machines
asked 5 hours ago
Taking1n1
62
New contributor
Taking1n1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I think your reasoning is correct, but we measure space in the number of bits required to represent the space. So even if a single symbol could be 0,1,..7, we count that as 3 bits.
– Albert Hendriks
4 hours ago
That is the worst case. As is also shown by the use of O notation. It is commonly referred to the encoding in the CLRS. Since every character can be encoded in a minimum of 2 characters. If you encode using only one than you will have a linear complexity in input length rather than the logarithmic complexity the binary encoding offers. Which when applied to a linear problem will give a theoretical exponential complexity. Please refer to CLRS, Introduction to Algorithms part on NP completeness for better explanation.
– anon
3 hours ago
Is it necessary or helpful in any way to assume $f(n)ge n$?
– Apass.Jack
3 hours ago
add a comment |
Assuming that f(n) >= n.
If possible, I'd like a proof in terms of Turing machines. I understand the reason why with machines that run on binary, because each "tape cell" is a bit with either 0 or 1, but in Turing machines a tape cell could hold any number of symbols. I'm having trouble why the base is '2' and not something like 'b' where 'b' is the number of types of symbols of the Turing machine tape.
complexity-theory turing-machines
asked 5 hours ago
Taking1n1
62
New contributor
Taking1n1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Assuming that f(n) >= n.
If possible, I'd like a proof in terms of Turing machines. I understand the reason why with machines that run on binary, because each "tape cell" is a bit with either 0 or 1, but in Turing machines a tape cell could hold any number of symbols. I'm having trouble why the base is '2' and not something like 'b' where 'b' is the number of types of symbols of the Turing machine tape.
complexity-theory turing-machines
complexity-theory turing-machines
asked 5 hours ago
Taking1n1
62
New contributor
Taking1n1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
Taking1n1
62
New contributor
Taking1n1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
Taking1n1
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Taking1n1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 5 hours ago
Taking1n1
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asked 5 hours ago
Taking1n1
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Taking1n1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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I think your reasoning is correct, but we measure space in the number of bits required to represent the space. So even if a single symbol could be 0,1,..7, we count that as 3 bits.
– Albert Hendriks
4 hours ago
That is the worst case. As is also shown by the use of O notation. It is commonly referred to the encoding in the CLRS. Since every character can be encoded in a minimum of 2 characters. If you encode using only one than you will have a linear complexity in input length rather than the logarithmic complexity the binary encoding offers. Which when applied to a linear problem will give a theoretical exponential complexity. Please refer to CLRS, Introduction to Algorithms part on NP completeness for better explanation.
– anon
3 hours ago
Is it necessary or helpful in any way to assume $f(n)ge n$?
– Apass.Jack
3 hours ago
add a comment |
I think your reasoning is correct, but we measure space in the number of bits required to represent the space. So even if a single symbol could be 0,1,..7, we count that as 3 bits.
– Albert Hendriks
4 hours ago
That is the worst case. As is also shown by the use of O notation. It is commonly referred to the encoding in the CLRS. Since every character can be encoded in a minimum of 2 characters. If you encode using only one than you will have a linear complexity in input length rather than the logarithmic complexity the binary encoding offers. Which when applied to a linear problem will give a theoretical exponential complexity. Please refer to CLRS, Introduction to Algorithms part on NP completeness for better explanation.
– anon
3 hours ago
Is it necessary or helpful in any way to assume $f(n)ge n$?
– Apass.Jack
3 hours ago
I think your reasoning is correct, but we measure space in the number of bits required to represent the space. So even if a single symbol could be 0,1,..7, we count that as 3 bits.
– Albert Hendriks
4 hours ago
I think your reasoning is correct, but we measure space in the number of bits required to represent the space. So even if a single symbol could be 0,1,..7, we count that as 3 bits.
– Albert Hendriks
4 hours ago
That is the worst case. As is also shown by the use of O notation. It is commonly referred to the encoding in the CLRS. Since every character can be encoded in a minimum of 2 characters. If you encode using only one than you will have a linear complexity in input length rather than the logarithmic complexity the binary encoding offers. Which when applied to a linear problem will give a theoretical exponential complexity. Please refer to CLRS, Introduction to Algorithms part on NP completeness for better explanation.
– anon
3 hours ago
That is the worst case. As is also shown by the use of O notation. It is commonly referred to the encoding in the CLRS. Since every character can be encoded in a minimum of 2 characters. If you encode using only one than you will have a linear complexity in input length rather than the logarithmic complexity the binary encoding offers. Which when applied to a linear problem will give a theoretical exponential complexity. Please refer to CLRS, Introduction to Algorithms part on NP completeness for better explanation.
– anon
3 hours ago
Is it necessary or helpful in any way to assume $f(n)ge n$?
– Apass.Jack
3 hours ago
Is it necessary or helpful in any way to assume $f(n)ge n$?
– Apass.Jack
3 hours ago
add a comment |
1 Answer
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why the base is '2' and not something like 'b' where 'b' is the number of types of symbols of the Turing machine tape?
Because it does not matter in the sense that $2^{O(f(n))} = b^{O(f(n))}$ for any positive $b > 1$.
According to the popular set-theoretic understanding of the big $O$-notation,
$$begin{align}
2^{O(f(n))}&={2^{g(n)}mid g(n)in O(f(n))} \
&={left(b^{log_b2}right)^{g(n)}mid g(n)in O(f(n))} \
&={b^{h(n)}mid exists g(n), h(n)= log_b2g(n), g(n)in O(f(n))} \
&={b^{h(n)}mid h(n)in O(f(n))} \
&=b^{O(f(n))}end{align}$$
answered 2 hours ago
Apass.Jack
7,1841533
add a comment |
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why the base is '2' and not something like 'b' where 'b' is the number of types of symbols of the Turing machine tape?
Because it does not matter in the sense that $2^{O(f(n))} = b^{O(f(n))}$ for any positive $b > 1$.
According to the popular set-theoretic understanding of the big $O$-notation,
$$begin{align}
2^{O(f(n))}&={2^{g(n)}mid g(n)in O(f(n))} \
&={left(b^{log_b2}right)^{g(n)}mid g(n)in O(f(n))} \
&={b^{h(n)}mid exists g(n), h(n)= log_b2g(n), g(n)in O(f(n))} \
&={b^{h(n)}mid h(n)in O(f(n))} \
&=b^{O(f(n))}end{align}$$
answered 2 hours ago
Apass.Jack
7,1841533
add a comment |
why the base is '2' and not something like 'b' where 'b' is the number of types of symbols of the Turing machine tape?
Because it does not matter in the sense that $2^{O(f(n))} = b^{O(f(n))}$ for any positive $b > 1$.
According to the popular set-theoretic understanding of the big $O$-notation,
$$begin{align}
2^{O(f(n))}&={2^{g(n)}mid g(n)in O(f(n))} \
&={left(b^{log_b2}right)^{g(n)}mid g(n)in O(f(n))} \
&={b^{h(n)}mid exists g(n), h(n)= log_b2g(n), g(n)in O(f(n))} \
&={b^{h(n)}mid h(n)in O(f(n))} \
&=b^{O(f(n))}end{align}$$
answered 2 hours ago
Apass.Jack
7,1841533
add a comment |
why the base is '2' and not something like 'b' where 'b' is the number of types of symbols of the Turing machine tape?
Because it does not matter in the sense that $2^{O(f(n))} = b^{O(f(n))}$ for any positive $b > 1$.
According to the popular set-theoretic understanding of the big $O$-notation,
$$begin{align}
2^{O(f(n))}&={2^{g(n)}mid g(n)in O(f(n))} \
&={left(b^{log_b2}right)^{g(n)}mid g(n)in O(f(n))} \
&={b^{h(n)}mid exists g(n), h(n)= log_b2g(n), g(n)in O(f(n))} \
&={b^{h(n)}mid h(n)in O(f(n))} \
&=b^{O(f(n))}end{align}$$
answered 2 hours ago
Apass.Jack
7,1841533
why the base is '2' and not something like 'b' where 'b' is the number of types of symbols of the Turing machine tape?
Because it does not matter in the sense that $2^{O(f(n))} = b^{O(f(n))}$ for any positive $b > 1$.
According to the popular set-theoretic understanding of the big $O$-notation,
$$begin{align}
2^{O(f(n))}&={2^{g(n)}mid g(n)in O(f(n))} \
&={left(b^{log_b2}right)^{g(n)}mid g(n)in O(f(n))} \
&={b^{h(n)}mid exists g(n), h(n)= log_b2g(n), g(n)in O(f(n))} \
&={b^{h(n)}mid h(n)in O(f(n))} \
&=b^{O(f(n))}end{align}$$
answered 2 hours ago
Apass.Jack
7,1841533
answered 2 hours ago
Apass.Jack
7,1841533
answered 2 hours ago
Apass.Jack
7,1841533
answered 2 hours ago
Apass.Jack
7,1841533
7,1841533
add a comment |
add a comment |
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I think your reasoning is correct, but we measure space in the number of bits required to represent the space. So even if a single symbol could be 0,1,..7, we count that as 3 bits.
– Albert Hendriks
4 hours ago
That is the worst case. As is also shown by the use of O notation. It is commonly referred to the encoding in the CLRS. Since every character can be encoded in a minimum of 2 characters. If you encode using only one than you will have a linear complexity in input length rather than the logarithmic complexity the binary encoding offers. Which when applied to a linear problem will give a theoretical exponential complexity. Please refer to CLRS, Introduction to Algorithms part on NP completeness for better explanation.
– anon
3 hours ago
Is it necessary or helpful in any way to assume $f(n)ge n$?
– Apass.Jack
3 hours ago