Product of differences of circular permutation
$begingroup$
Numbers $1,2,ldots,n$ are arranged into a circle. What is the maximum product of the differences $|x_1-x_2|times|x_2-x_3|timescdotstimes|x_{n-1}-x_n|times|x_n-x_1|$?
I think the maximum should occur when the numbers are arranged $n,1,n-1,2,n-2,3,ldots$. The sum for this arrangement is $(n-1)(n-2)cdots1cdotlfloor n/2rfloor = (n-1)!cdotlfloor n/2rfloor$.
This question was inspired by the question asking for the maximum sum. There, it is possible to prove optimality by noting that we have $2n$ terms ($n$ with $+$ and $n$ with $-$), and each number occurs twice.
Here, it is still true that we have $2n$ terms ($n$ with $+$ and $n$ with $-$), and each number occurs twice. But since we're taking the product instead of the sum, optimality is no longer clear.
combinatorics permutations
edited Dec 8 '18 at 18:36
Jam
5,00821431
asked Nov 9 '14 at 18:00
DexterDexter
891417
$endgroup$
add a comment |
$begingroup$
Numbers $1,2,ldots,n$ are arranged into a circle. What is the maximum product of the differences $|x_1-x_2|times|x_2-x_3|timescdotstimes|x_{n-1}-x_n|times|x_n-x_1|$?
I think the maximum should occur when the numbers are arranged $n,1,n-1,2,n-2,3,ldots$. The sum for this arrangement is $(n-1)(n-2)cdots1cdotlfloor n/2rfloor = (n-1)!cdotlfloor n/2rfloor$.
This question was inspired by the question asking for the maximum sum. There, it is possible to prove optimality by noting that we have $2n$ terms ($n$ with $+$ and $n$ with $-$), and each number occurs twice.
Here, it is still true that we have $2n$ terms ($n$ with $+$ and $n$ with $-$), and each number occurs twice. But since we're taking the product instead of the sum, optimality is no longer clear.
combinatorics permutations
edited Dec 8 '18 at 18:36
Jam
5,00821431
asked Nov 9 '14 at 18:00
DexterDexter
891417
$endgroup$
add a comment |
$begingroup$
Numbers $1,2,ldots,n$ are arranged into a circle. What is the maximum product of the differences $|x_1-x_2|times|x_2-x_3|timescdotstimes|x_{n-1}-x_n|times|x_n-x_1|$?
I think the maximum should occur when the numbers are arranged $n,1,n-1,2,n-2,3,ldots$. The sum for this arrangement is $(n-1)(n-2)cdots1cdotlfloor n/2rfloor = (n-1)!cdotlfloor n/2rfloor$.
This question was inspired by the question asking for the maximum sum. There, it is possible to prove optimality by noting that we have $2n$ terms ($n$ with $+$ and $n$ with $-$), and each number occurs twice.
Here, it is still true that we have $2n$ terms ($n$ with $+$ and $n$ with $-$), and each number occurs twice. But since we're taking the product instead of the sum, optimality is no longer clear.
combinatorics permutations
edited Dec 8 '18 at 18:36
Jam
5,00821431
asked Nov 9 '14 at 18:00
DexterDexter
891417
$endgroup$
Numbers $1,2,ldots,n$ are arranged into a circle. What is the maximum product of the differences $|x_1-x_2|times|x_2-x_3|timescdotstimes|x_{n-1}-x_n|times|x_n-x_1|$?
I think the maximum should occur when the numbers are arranged $n,1,n-1,2,n-2,3,ldots$. The sum for this arrangement is $(n-1)(n-2)cdots1cdotlfloor n/2rfloor = (n-1)!cdotlfloor n/2rfloor$.
This question was inspired by the question asking for the maximum sum. There, it is possible to prove optimality by noting that we have $2n$ terms ($n$ with $+$ and $n$ with $-$), and each number occurs twice.
Here, it is still true that we have $2n$ terms ($n$ with $+$ and $n$ with $-$), and each number occurs twice. But since we're taking the product instead of the sum, optimality is no longer clear.
combinatorics permutations
combinatorics permutations
edited Dec 8 '18 at 18:36
Jam
5,00821431
asked Nov 9 '14 at 18:00
DexterDexter
891417
edited Dec 8 '18 at 18:36
Jam
5,00821431
asked Nov 9 '14 at 18:00
DexterDexter
891417
edited Dec 8 '18 at 18:36
Jam
5,00821431
edited Dec 8 '18 at 18:36
Jam
5,00821431
edited Dec 8 '18 at 18:36
Jam
5,00821431
5,00821431
asked Nov 9 '14 at 18:00
DexterDexter
891417
asked Nov 9 '14 at 18:00
DexterDexter
891417
asked Nov 9 '14 at 18:00
DexterDexter
891417
891417
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your conjecture is false. Just generating some random permutations, for example, I find a permutation of 10, $2, 6, 1, 7, 3, 9, 5, 10, 4, 8$, for which the product of differences is $8294400$; compare $9! times 5 = 1814400$ for yours. Note that all the differences in that permutation are 4, 5, or 6 (that is, near 10/2); I think the small factors in your product towards the end hurt you more than the large ones help.
As a result it makes sense to have as many factors near $n/2$ as possible, and I think something like $1, (n/2+1), 2, (n/2)+2, cdots, n/2, n$ (for $n$ even) will do quite well - in this case you get a product of $(n/2)^{n/2} (n/2-1)^{n/2-1} (n-1)$.
answered Nov 9 '14 at 18:26
Michael LugoMichael Lugo
18.2k33576
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add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
Your conjecture is false. Just generating some random permutations, for example, I find a permutation of 10, $2, 6, 1, 7, 3, 9, 5, 10, 4, 8$, for which the product of differences is $8294400$; compare $9! times 5 = 1814400$ for yours. Note that all the differences in that permutation are 4, 5, or 6 (that is, near 10/2); I think the small factors in your product towards the end hurt you more than the large ones help.
As a result it makes sense to have as many factors near $n/2$ as possible, and I think something like $1, (n/2+1), 2, (n/2)+2, cdots, n/2, n$ (for $n$ even) will do quite well - in this case you get a product of $(n/2)^{n/2} (n/2-1)^{n/2-1} (n-1)$.
answered Nov 9 '14 at 18:26
Michael LugoMichael Lugo
18.2k33576
$endgroup$
add a comment |
$begingroup$
Your conjecture is false. Just generating some random permutations, for example, I find a permutation of 10, $2, 6, 1, 7, 3, 9, 5, 10, 4, 8$, for which the product of differences is $8294400$; compare $9! times 5 = 1814400$ for yours. Note that all the differences in that permutation are 4, 5, or 6 (that is, near 10/2); I think the small factors in your product towards the end hurt you more than the large ones help.
As a result it makes sense to have as many factors near $n/2$ as possible, and I think something like $1, (n/2+1), 2, (n/2)+2, cdots, n/2, n$ (for $n$ even) will do quite well - in this case you get a product of $(n/2)^{n/2} (n/2-1)^{n/2-1} (n-1)$.
answered Nov 9 '14 at 18:26
Michael LugoMichael Lugo
18.2k33576
$endgroup$
add a comment |
$begingroup$
Your conjecture is false. Just generating some random permutations, for example, I find a permutation of 10, $2, 6, 1, 7, 3, 9, 5, 10, 4, 8$, for which the product of differences is $8294400$; compare $9! times 5 = 1814400$ for yours. Note that all the differences in that permutation are 4, 5, or 6 (that is, near 10/2); I think the small factors in your product towards the end hurt you more than the large ones help.
As a result it makes sense to have as many factors near $n/2$ as possible, and I think something like $1, (n/2+1), 2, (n/2)+2, cdots, n/2, n$ (for $n$ even) will do quite well - in this case you get a product of $(n/2)^{n/2} (n/2-1)^{n/2-1} (n-1)$.
answered Nov 9 '14 at 18:26
Michael LugoMichael Lugo
18.2k33576
$endgroup$
Your conjecture is false. Just generating some random permutations, for example, I find a permutation of 10, $2, 6, 1, 7, 3, 9, 5, 10, 4, 8$, for which the product of differences is $8294400$; compare $9! times 5 = 1814400$ for yours. Note that all the differences in that permutation are 4, 5, or 6 (that is, near 10/2); I think the small factors in your product towards the end hurt you more than the large ones help.
As a result it makes sense to have as many factors near $n/2$ as possible, and I think something like $1, (n/2+1), 2, (n/2)+2, cdots, n/2, n$ (for $n$ even) will do quite well - in this case you get a product of $(n/2)^{n/2} (n/2-1)^{n/2-1} (n-1)$.
answered Nov 9 '14 at 18:26
Michael LugoMichael Lugo
18.2k33576
answered Nov 9 '14 at 18:26
Michael LugoMichael Lugo
18.2k33576
answered Nov 9 '14 at 18:26
Michael LugoMichael Lugo
18.2k33576
answered Nov 9 '14 at 18:26
Michael LugoMichael Lugo
18.2k33576
18.2k33576
add a comment |
add a comment |
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