Runtime of computing the coefficient of a product of multinomials?
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Suppose I have $k$ variables, $ x_1, x_2, ... x_k $ and $m$ expressions in the form $ (1 +$ the product of some subset of $x_1 ... x_k)$ – for instance, $(1 + x_1)$ or $(1 + x_1x_2x_5)$ could be one of the $m$ equations, but not something like $(1 + x_1^2)$. What is the runtime of calculating a coefficient of a term such as $x_1^{10}x_2^{20}$ with the above constraints?
My intuition is that for $m$ expressions, you can end up with $(m+1)^k$ potential combinations of variables, and you can create counters for them. Then, for each expression, you increment the appropriate counters after running the calculations, arriving at a runtime of $ O(m * (m + 1)^k)$ which can be simplified in big-O notation as $O(m^{k+1})$. Does this make sense? Is there a quicker solution for it?
combinatorics algorithms computational-complexity multinomial-coefficients
edited Dec 15 '18 at 12:22
Alex Ravsky
41.9k32383
asked Dec 12 '18 at 17:11
mjkaufermjkaufer
10614
$endgroup$
add a comment |
$begingroup$
Suppose I have $k$ variables, $ x_1, x_2, ... x_k $ and $m$ expressions in the form $ (1 +$ the product of some subset of $x_1 ... x_k)$ – for instance, $(1 + x_1)$ or $(1 + x_1x_2x_5)$ could be one of the $m$ equations, but not something like $(1 + x_1^2)$. What is the runtime of calculating a coefficient of a term such as $x_1^{10}x_2^{20}$ with the above constraints?
My intuition is that for $m$ expressions, you can end up with $(m+1)^k$ potential combinations of variables, and you can create counters for them. Then, for each expression, you increment the appropriate counters after running the calculations, arriving at a runtime of $ O(m * (m + 1)^k)$ which can be simplified in big-O notation as $O(m^{k+1})$. Does this make sense? Is there a quicker solution for it?
combinatorics algorithms computational-complexity multinomial-coefficients
edited Dec 15 '18 at 12:22
Alex Ravsky
41.9k32383
asked Dec 12 '18 at 17:11
mjkaufermjkaufer
10614
$endgroup$
$begingroup$
When you expand out the product, there are $2^m$ summands. You can count the number of summands equal to your term in something like $2^mcdot k$ time. I feel like your $m^{k+1}$ algorithm is not actually achievable.
$endgroup$
– Mike Earnest
Dec 12 '18 at 21:40
$begingroup$
This smart algorithm avoids full product expanding into $2^m$ summands. Assuming it makes arithmetic operations in $O(1)$ time, the algorithm and bound are correct.
$endgroup$
– Alex Ravsky
Dec 15 '18 at 12:15
add a comment |
$begingroup$
Suppose I have $k$ variables, $ x_1, x_2, ... x_k $ and $m$ expressions in the form $ (1 +$ the product of some subset of $x_1 ... x_k)$ – for instance, $(1 + x_1)$ or $(1 + x_1x_2x_5)$ could be one of the $m$ equations, but not something like $(1 + x_1^2)$. What is the runtime of calculating a coefficient of a term such as $x_1^{10}x_2^{20}$ with the above constraints?
My intuition is that for $m$ expressions, you can end up with $(m+1)^k$ potential combinations of variables, and you can create counters for them. Then, for each expression, you increment the appropriate counters after running the calculations, arriving at a runtime of $ O(m * (m + 1)^k)$ which can be simplified in big-O notation as $O(m^{k+1})$. Does this make sense? Is there a quicker solution for it?
combinatorics algorithms computational-complexity multinomial-coefficients
edited Dec 15 '18 at 12:22
Alex Ravsky
41.9k32383
asked Dec 12 '18 at 17:11
mjkaufermjkaufer
10614
$endgroup$
Suppose I have $k$ variables, $ x_1, x_2, ... x_k $ and $m$ expressions in the form $ (1 +$ the product of some subset of $x_1 ... x_k)$ – for instance, $(1 + x_1)$ or $(1 + x_1x_2x_5)$ could be one of the $m$ equations, but not something like $(1 + x_1^2)$. What is the runtime of calculating a coefficient of a term such as $x_1^{10}x_2^{20}$ with the above constraints?
My intuition is that for $m$ expressions, you can end up with $(m+1)^k$ potential combinations of variables, and you can create counters for them. Then, for each expression, you increment the appropriate counters after running the calculations, arriving at a runtime of $ O(m * (m + 1)^k)$ which can be simplified in big-O notation as $O(m^{k+1})$. Does this make sense? Is there a quicker solution for it?
combinatorics algorithms computational-complexity multinomial-coefficients
combinatorics algorithms computational-complexity multinomial-coefficients
edited Dec 15 '18 at 12:22
Alex Ravsky
41.9k32383
asked Dec 12 '18 at 17:11
mjkaufermjkaufer
10614
edited Dec 15 '18 at 12:22
Alex Ravsky
41.9k32383
asked Dec 12 '18 at 17:11
mjkaufermjkaufer
10614
edited Dec 15 '18 at 12:22
Alex Ravsky
41.9k32383
edited Dec 15 '18 at 12:22
Alex Ravsky
41.9k32383
edited Dec 15 '18 at 12:22
Alex Ravsky
41.9k32383
41.9k32383
asked Dec 12 '18 at 17:11
mjkaufermjkaufer
10614
asked Dec 12 '18 at 17:11
mjkaufermjkaufer
10614
asked Dec 12 '18 at 17:11
mjkaufermjkaufer
10614
10614
$begingroup$
When you expand out the product, there are $2^m$ summands. You can count the number of summands equal to your term in something like $2^mcdot k$ time. I feel like your $m^{k+1}$ algorithm is not actually achievable.
$endgroup$
– Mike Earnest
Dec 12 '18 at 21:40
$begingroup$
This smart algorithm avoids full product expanding into $2^m$ summands. Assuming it makes arithmetic operations in $O(1)$ time, the algorithm and bound are correct.
$endgroup$
– Alex Ravsky
Dec 15 '18 at 12:15
add a comment |
$begingroup$
When you expand out the product, there are $2^m$ summands. You can count the number of summands equal to your term in something like $2^mcdot k$ time. I feel like your $m^{k+1}$ algorithm is not actually achievable.
$endgroup$
– Mike Earnest
Dec 12 '18 at 21:40
$begingroup$
This smart algorithm avoids full product expanding into $2^m$ summands. Assuming it makes arithmetic operations in $O(1)$ time, the algorithm and bound are correct.
$endgroup$
– Alex Ravsky
Dec 15 '18 at 12:15
$begingroup$
When you expand out the product, there are $2^m$ summands. You can count the number of summands equal to your term in something like $2^mcdot k$ time. I feel like your $m^{k+1}$ algorithm is not actually achievable.
$endgroup$
– Mike Earnest
Dec 12 '18 at 21:40
$begingroup$
When you expand out the product, there are $2^m$ summands. You can count the number of summands equal to your term in something like $2^mcdot k$ time. I feel like your $m^{k+1}$ algorithm is not actually achievable.
$endgroup$
– Mike Earnest
Dec 12 '18 at 21:40
$begingroup$
This smart algorithm avoids full product expanding into $2^m$ summands. Assuming it makes arithmetic operations in $O(1)$ time, the algorithm and bound are correct.
$endgroup$
– Alex Ravsky
Dec 15 '18 at 12:15
$begingroup$
This smart algorithm avoids full product expanding into $2^m$ summands. Assuming it makes arithmetic operations in $O(1)$ time, the algorithm and bound are correct.
$endgroup$
– Alex Ravsky
Dec 15 '18 at 12:15
add a comment |
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$begingroup$
When you expand out the product, there are $2^m$ summands. You can count the number of summands equal to your term in something like $2^mcdot k$ time. I feel like your $m^{k+1}$ algorithm is not actually achievable.
$endgroup$
– Mike Earnest
Dec 12 '18 at 21:40
$begingroup$
This smart algorithm avoids full product expanding into $2^m$ summands. Assuming it makes arithmetic operations in $O(1)$ time, the algorithm and bound are correct.
$endgroup$
– Alex Ravsky
Dec 15 '18 at 12:15