Counting $k$-sets with sum $n$ and xor-sum $0$.
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Given $k, n > 0$, how many ordered lists $a_1, a_2, dots, a_k$ are there such that $a_i geq 0$ for all $i$, such that $sum_i a_i = n$ and $oplus_i a_i = 0$, where the latter operation denotes bitwise xor (i.e. $15 oplus 3 = 12$).
It is likely there is no closed-form expression, in that case a reasonably fast algorithm would also be interesting.
EDIT: Context: this is the number of losing starting positions in a game of Nim with $k$ initial piles containing a total of $n$ stones.
combinatorics combinatorial-game-theory
asked Jan 10 at 19:49
Timon KniggeTimon Knigge
370110
$endgroup$
add a comment |
$begingroup$
Given $k, n > 0$, how many ordered lists $a_1, a_2, dots, a_k$ are there such that $a_i geq 0$ for all $i$, such that $sum_i a_i = n$ and $oplus_i a_i = 0$, where the latter operation denotes bitwise xor (i.e. $15 oplus 3 = 12$).
It is likely there is no closed-form expression, in that case a reasonably fast algorithm would also be interesting.
EDIT: Context: this is the number of losing starting positions in a game of Nim with $k$ initial piles containing a total of $n$ stones.
combinatorics combinatorial-game-theory
asked Jan 10 at 19:49
Timon KniggeTimon Knigge
370110
$endgroup$
2
$begingroup$
See section 5 of Tanya Khovanova and Joshua Xiong, "Nim Fractals", arxiv.org/abs/1405.5942, in particular Theorem 27.
$endgroup$
– Michael Lugo
Jan 10 at 20:22
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@MichaelLugo Thanks, that's pretty good. It only gives $n$ even though.
$endgroup$
– Timon Knigge
Jan 10 at 21:32
1
$begingroup$
Aha, but if $n equiv 1 mod 2$ then there is an odd number of odd sized piles and the nim sum is never $0$.
$endgroup$
– Timon Knigge
Jan 10 at 21:36
add a comment |
$begingroup$
Given $k, n > 0$, how many ordered lists $a_1, a_2, dots, a_k$ are there such that $a_i geq 0$ for all $i$, such that $sum_i a_i = n$ and $oplus_i a_i = 0$, where the latter operation denotes bitwise xor (i.e. $15 oplus 3 = 12$).
It is likely there is no closed-form expression, in that case a reasonably fast algorithm would also be interesting.
EDIT: Context: this is the number of losing starting positions in a game of Nim with $k$ initial piles containing a total of $n$ stones.
combinatorics combinatorial-game-theory
asked Jan 10 at 19:49
Timon KniggeTimon Knigge
370110
$endgroup$
Given $k, n > 0$, how many ordered lists $a_1, a_2, dots, a_k$ are there such that $a_i geq 0$ for all $i$, such that $sum_i a_i = n$ and $oplus_i a_i = 0$, where the latter operation denotes bitwise xor (i.e. $15 oplus 3 = 12$).
It is likely there is no closed-form expression, in that case a reasonably fast algorithm would also be interesting.
EDIT: Context: this is the number of losing starting positions in a game of Nim with $k$ initial piles containing a total of $n$ stones.
combinatorics combinatorial-game-theory
combinatorics combinatorial-game-theory
asked Jan 10 at 19:49
Timon KniggeTimon Knigge
370110
asked Jan 10 at 19:49
Timon KniggeTimon Knigge
370110
edited Jan 10 at 20:13
Timon Knigge
asked Jan 10 at 19:49
Timon KniggeTimon Knigge
370110
asked Jan 10 at 19:49
Timon KniggeTimon Knigge
370110
asked Jan 10 at 19:49
Timon KniggeTimon Knigge
370110
370110
2
$begingroup$
See section 5 of Tanya Khovanova and Joshua Xiong, "Nim Fractals", arxiv.org/abs/1405.5942, in particular Theorem 27.
$endgroup$
– Michael Lugo
Jan 10 at 20:22
$begingroup$
@MichaelLugo Thanks, that's pretty good. It only gives $n$ even though.
$endgroup$
– Timon Knigge
Jan 10 at 21:32
1
$begingroup$
Aha, but if $n equiv 1 mod 2$ then there is an odd number of odd sized piles and the nim sum is never $0$.
$endgroup$
– Timon Knigge
Jan 10 at 21:36
add a comment |
2
$begingroup$
See section 5 of Tanya Khovanova and Joshua Xiong, "Nim Fractals", arxiv.org/abs/1405.5942, in particular Theorem 27.
$endgroup$
– Michael Lugo
Jan 10 at 20:22
$begingroup$
@MichaelLugo Thanks, that's pretty good. It only gives $n$ even though.
$endgroup$
– Timon Knigge
Jan 10 at 21:32
1
$begingroup$
Aha, but if $n equiv 1 mod 2$ then there is an odd number of odd sized piles and the nim sum is never $0$.
$endgroup$
– Timon Knigge
Jan 10 at 21:36
2
2
$begingroup$
See section 5 of Tanya Khovanova and Joshua Xiong, "Nim Fractals", arxiv.org/abs/1405.5942, in particular Theorem 27.
$endgroup$
– Michael Lugo
Jan 10 at 20:22
$begingroup$
See section 5 of Tanya Khovanova and Joshua Xiong, "Nim Fractals", arxiv.org/abs/1405.5942, in particular Theorem 27.
$endgroup$
– Michael Lugo
Jan 10 at 20:22
$begingroup$
@MichaelLugo Thanks, that's pretty good. It only gives $n$ even though.
$endgroup$
– Timon Knigge
Jan 10 at 21:32
$begingroup$
@MichaelLugo Thanks, that's pretty good. It only gives $n$ even though.
$endgroup$
– Timon Knigge
Jan 10 at 21:32
1
1
$begingroup$
Aha, but if $n equiv 1 mod 2$ then there is an odd number of odd sized piles and the nim sum is never $0$.
$endgroup$
– Timon Knigge
Jan 10 at 21:36
$begingroup$
Aha, but if $n equiv 1 mod 2$ then there is an odd number of odd sized piles and the nim sum is never $0$.
$endgroup$
– Timon Knigge
Jan 10 at 21:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $F(n,k,x)$ denote the amount of ordered lists $a_1,...,a_k$ such that $a_igeq0$, $sum a_i=n$ and $oplus a_i=x$. Then considering all possible values of $a_k$ we find the recursive formula $$F(n,k,x)=sum_{i=0}^nF(n-i,k-1,xoplus i).$$ This gives us an $mathcal{O}(n^3k)$ time algorithm. Not very fast, but at least polynomial in $n$ and $k$.
answered Jan 10 at 20:12
SmileyCraftSmileyCraft
3,571518
$endgroup$
1
$begingroup$
One can use binary doubling (as in squaring by exponentiation) to get this to $O(n^3 log k)$. Still very slow though.
$endgroup$
– Timon Knigge
Jan 10 at 20:15
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let $F(n,k,x)$ denote the amount of ordered lists $a_1,...,a_k$ such that $a_igeq0$, $sum a_i=n$ and $oplus a_i=x$. Then considering all possible values of $a_k$ we find the recursive formula $$F(n,k,x)=sum_{i=0}^nF(n-i,k-1,xoplus i).$$ This gives us an $mathcal{O}(n^3k)$ time algorithm. Not very fast, but at least polynomial in $n$ and $k$.
answered Jan 10 at 20:12
SmileyCraftSmileyCraft
3,571518
$endgroup$
1
$begingroup$
One can use binary doubling (as in squaring by exponentiation) to get this to $O(n^3 log k)$. Still very slow though.
$endgroup$
– Timon Knigge
Jan 10 at 20:15
add a comment |
$begingroup$
Let $F(n,k,x)$ denote the amount of ordered lists $a_1,...,a_k$ such that $a_igeq0$, $sum a_i=n$ and $oplus a_i=x$. Then considering all possible values of $a_k$ we find the recursive formula $$F(n,k,x)=sum_{i=0}^nF(n-i,k-1,xoplus i).$$ This gives us an $mathcal{O}(n^3k)$ time algorithm. Not very fast, but at least polynomial in $n$ and $k$.
answered Jan 10 at 20:12
SmileyCraftSmileyCraft
3,571518
$endgroup$
1
$begingroup$
One can use binary doubling (as in squaring by exponentiation) to get this to $O(n^3 log k)$. Still very slow though.
$endgroup$
– Timon Knigge
Jan 10 at 20:15
add a comment |
$begingroup$
Let $F(n,k,x)$ denote the amount of ordered lists $a_1,...,a_k$ such that $a_igeq0$, $sum a_i=n$ and $oplus a_i=x$. Then considering all possible values of $a_k$ we find the recursive formula $$F(n,k,x)=sum_{i=0}^nF(n-i,k-1,xoplus i).$$ This gives us an $mathcal{O}(n^3k)$ time algorithm. Not very fast, but at least polynomial in $n$ and $k$.
answered Jan 10 at 20:12
SmileyCraftSmileyCraft
3,571518
$endgroup$
Let $F(n,k,x)$ denote the amount of ordered lists $a_1,...,a_k$ such that $a_igeq0$, $sum a_i=n$ and $oplus a_i=x$. Then considering all possible values of $a_k$ we find the recursive formula $$F(n,k,x)=sum_{i=0}^nF(n-i,k-1,xoplus i).$$ This gives us an $mathcal{O}(n^3k)$ time algorithm. Not very fast, but at least polynomial in $n$ and $k$.
answered Jan 10 at 20:12
SmileyCraftSmileyCraft
3,571518
answered Jan 10 at 20:12
SmileyCraftSmileyCraft
3,571518
answered Jan 10 at 20:12
SmileyCraftSmileyCraft
3,571518
answered Jan 10 at 20:12
SmileyCraftSmileyCraft
3,571518
3,571518
1
$begingroup$
One can use binary doubling (as in squaring by exponentiation) to get this to $O(n^3 log k)$. Still very slow though.
$endgroup$
– Timon Knigge
Jan 10 at 20:15
add a comment |
1
$begingroup$
One can use binary doubling (as in squaring by exponentiation) to get this to $O(n^3 log k)$. Still very slow though.
$endgroup$
– Timon Knigge
Jan 10 at 20:15
1
1
$begingroup$
One can use binary doubling (as in squaring by exponentiation) to get this to $O(n^3 log k)$. Still very slow though.
$endgroup$
– Timon Knigge
Jan 10 at 20:15
$begingroup$
One can use binary doubling (as in squaring by exponentiation) to get this to $O(n^3 log k)$. Still very slow though.
$endgroup$
– Timon Knigge
Jan 10 at 20:15
add a comment |
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See section 5 of Tanya Khovanova and Joshua Xiong, "Nim Fractals", arxiv.org/abs/1405.5942, in particular Theorem 27.
$endgroup$
– Michael Lugo
Jan 10 at 20:22
$begingroup$
@MichaelLugo Thanks, that's pretty good. It only gives $n$ even though.
$endgroup$
– Timon Knigge
Jan 10 at 21:32
1
$begingroup$
Aha, but if $n equiv 1 mod 2$ then there is an odd number of odd sized piles and the nim sum is never $0$.
$endgroup$
– Timon Knigge
Jan 10 at 21:36