If a non-deterministic Turing machine runs in f(n) space, then why does it run in 2^O(f(n)) time?
Clash Royale CLAN TAG#URR8PPP
$begingroup$
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 Jan 1 at 22:02
Taking1n1Taking1n1
62
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add a comment |
$begingroup$
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 Jan 1 at 22:02
Taking1n1Taking1n1
62
$endgroup$
$begingroup$
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.
$endgroup$
– anon
Jan 1 at 23:16
add a comment |
$begingroup$
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 Jan 1 at 22:02
Taking1n1Taking1n1
62
$endgroup$
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 Jan 1 at 22:02
Taking1n1Taking1n1
62
asked Jan 1 at 22:02
Taking1n1Taking1n1
62
asked Jan 1 at 22:02
Taking1n1Taking1n1
62
asked Jan 1 at 22:02
Taking1n1Taking1n1
62
asked Jan 1 at 22:02
Taking1n1Taking1n1
62
62
$begingroup$
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.
$endgroup$
– anon
Jan 1 at 23:16
add a comment |
$begingroup$
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.
$endgroup$
– anon
Jan 1 at 23:16
$begingroup$
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.
$endgroup$
– anon
Jan 1 at 23:16
$begingroup$
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.
$endgroup$
– anon
Jan 1 at 23:16
add a comment |
2 Answers
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$begingroup$
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 $b > 1$.
Using the popular set-theoretic understanding of the big $O$-notation, we have
$$beginalign
2^O(f(n))&=2^g(n)mid g(n)in O(f(n)) \
&=left(b^log_b2right)^g(n)mid g(n)in O(f(n)) \
&=b^h(n)mid exists g(n), h(n)= (log_b2)g(n), g(n)in O(f(n)) \
&=b^h(n)mid h(n)in O(f(n)) \
&=b^O(f(n))endalign$$
By the way, there is no need to assume $f(n)ge n$. For example, all above hold when $f(n) = lceilsqrt nrceil$ or $f(n) = lceillog_2nrceil$.
answered Jan 2 at 0:19
Apass.JackApass.Jack
8,1521633
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Because an exponential function to any base can be expressed in terms of any exponential function of another base, in a way which also happens to make this possible. Namely, for any real number $a > 0$, $a^x = 2^lg(a) x$. Thus if you have $a^O(f(n))$ it is the same as $2^lg(a) O(f(n))$. But a constant multiplying a big-O simply gets absorbed into it, since the whole point of the notation is that it's the shape, not the scale, of the function wrapped by it that counts.
answered Jan 2 at 5:14
The_SympathizerThe_Sympathizer
1011
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2 Answers
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2 Answers
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$begingroup$
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 $b > 1$.
Using the popular set-theoretic understanding of the big $O$-notation, we have
$$beginalign
2^O(f(n))&=2^g(n)mid g(n)in O(f(n)) \
&=left(b^log_b2right)^g(n)mid g(n)in O(f(n)) \
&=b^h(n)mid exists g(n), h(n)= (log_b2)g(n), g(n)in O(f(n)) \
&=b^h(n)mid h(n)in O(f(n)) \
&=b^O(f(n))endalign$$
By the way, there is no need to assume $f(n)ge n$. For example, all above hold when $f(n) = lceilsqrt nrceil$ or $f(n) = lceillog_2nrceil$.
answered Jan 2 at 0:19
Apass.JackApass.Jack
8,1521633
$endgroup$
add a comment |
$begingroup$
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 $b > 1$.
Using the popular set-theoretic understanding of the big $O$-notation, we have
$$beginalign
2^O(f(n))&=2^g(n)mid g(n)in O(f(n)) \
&=left(b^log_b2right)^g(n)mid g(n)in O(f(n)) \
&=b^h(n)mid exists g(n), h(n)= (log_b2)g(n), g(n)in O(f(n)) \
&=b^h(n)mid h(n)in O(f(n)) \
&=b^O(f(n))endalign$$
By the way, there is no need to assume $f(n)ge n$. For example, all above hold when $f(n) = lceilsqrt nrceil$ or $f(n) = lceillog_2nrceil$.
answered Jan 2 at 0:19
Apass.JackApass.Jack
8,1521633
$endgroup$
add a comment |
$begingroup$
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 $b > 1$.
Using the popular set-theoretic understanding of the big $O$-notation, we have
$$beginalign
2^O(f(n))&=2^g(n)mid g(n)in O(f(n)) \
&=left(b^log_b2right)^g(n)mid g(n)in O(f(n)) \
&=b^h(n)mid exists g(n), h(n)= (log_b2)g(n), g(n)in O(f(n)) \
&=b^h(n)mid h(n)in O(f(n)) \
&=b^O(f(n))endalign$$
By the way, there is no need to assume $f(n)ge n$. For example, all above hold when $f(n) = lceilsqrt nrceil$ or $f(n) = lceillog_2nrceil$.
answered Jan 2 at 0:19
Apass.JackApass.Jack
8,1521633
$endgroup$
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 $b > 1$.
Using the popular set-theoretic understanding of the big $O$-notation, we have
$$beginalign
2^O(f(n))&=2^g(n)mid g(n)in O(f(n)) \
&=left(b^log_b2right)^g(n)mid g(n)in O(f(n)) \
&=b^h(n)mid exists g(n), h(n)= (log_b2)g(n), g(n)in O(f(n)) \
&=b^h(n)mid h(n)in O(f(n)) \
&=b^O(f(n))endalign$$
By the way, there is no need to assume $f(n)ge n$. For example, all above hold when $f(n) = lceilsqrt nrceil$ or $f(n) = lceillog_2nrceil$.
answered Jan 2 at 0:19
Apass.JackApass.Jack
8,1521633
edited Jan 2 at 5:49
answered Jan 2 at 0:19
Apass.JackApass.Jack
8,1521633
answered Jan 2 at 0:19
Apass.JackApass.Jack
8,1521633
answered Jan 2 at 0:19
Apass.JackApass.Jack
8,1521633
8,1521633
add a comment |
add a comment |
$begingroup$
Because an exponential function to any base can be expressed in terms of any exponential function of another base, in a way which also happens to make this possible. Namely, for any real number $a > 0$, $a^x = 2^lg(a) x$. Thus if you have $a^O(f(n))$ it is the same as $2^lg(a) O(f(n))$. But a constant multiplying a big-O simply gets absorbed into it, since the whole point of the notation is that it's the shape, not the scale, of the function wrapped by it that counts.
answered Jan 2 at 5:14
The_SympathizerThe_Sympathizer
1011
$endgroup$
add a comment |
$begingroup$
Because an exponential function to any base can be expressed in terms of any exponential function of another base, in a way which also happens to make this possible. Namely, for any real number $a > 0$, $a^x = 2^lg(a) x$. Thus if you have $a^O(f(n))$ it is the same as $2^lg(a) O(f(n))$. But a constant multiplying a big-O simply gets absorbed into it, since the whole point of the notation is that it's the shape, not the scale, of the function wrapped by it that counts.
answered Jan 2 at 5:14
The_SympathizerThe_Sympathizer
1011
$endgroup$
add a comment |
$begingroup$
Because an exponential function to any base can be expressed in terms of any exponential function of another base, in a way which also happens to make this possible. Namely, for any real number $a > 0$, $a^x = 2^lg(a) x$. Thus if you have $a^O(f(n))$ it is the same as $2^lg(a) O(f(n))$. But a constant multiplying a big-O simply gets absorbed into it, since the whole point of the notation is that it's the shape, not the scale, of the function wrapped by it that counts.
answered Jan 2 at 5:14
The_SympathizerThe_Sympathizer
1011
$endgroup$
Because an exponential function to any base can be expressed in terms of any exponential function of another base, in a way which also happens to make this possible. Namely, for any real number $a > 0$, $a^x = 2^lg(a) x$. Thus if you have $a^O(f(n))$ it is the same as $2^lg(a) O(f(n))$. But a constant multiplying a big-O simply gets absorbed into it, since the whole point of the notation is that it's the shape, not the scale, of the function wrapped by it that counts.
answered Jan 2 at 5:14
The_SympathizerThe_Sympathizer
1011
answered Jan 2 at 5:14
The_SympathizerThe_Sympathizer
1011
answered Jan 2 at 5:14
The_SympathizerThe_Sympathizer
1011
answered Jan 2 at 5:14
The_SympathizerThe_Sympathizer
1011
1011
add a comment |
add a comment |
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$begingroup$
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.
$endgroup$
– anon
Jan 1 at 23:16