Mixing
Proving that the Nim-sum cannot be zero for two turns in a row
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In proving certain results about Nim, I found a lemma that is causing me trouble:
Lemma 1 If the Nim-sum is $0$ after a player’s turn, then the next move must change it.
To prove this, let the number of stones in the heaps be $x_1, x_2, ... x_n$, and $s$ be the nim-sum $$s=x_1⊕x_2⊕x_3⊕ . . . ⊕x_n$$ Let $t$ be the sum of the heaps after the move, $$t=y_1⊕y_2⊕y_3⊕ . . . ⊕y_n$$ Then if $s= 0$, the next move causes some $x_k=y_k$ and the rest of the $x_i=y_i$ for $i neq k$, since only one pile of stones is changed.
Then:
$$t=0⊕t$$
$$=s⊕s⊕t$$
$$=s⊕(x_1⊕x_2⊕...⊕x_n)⊕(y_1⊕y_2⊕...⊕y_n)$$
$$=s⊕(x_1⊕y_1)⊕(x_2⊕y_2)⊕...⊕(x_k⊕y_k)$$
$$=s⊕x_k⊕y_k$$
If $s$ is 0, then $t$ must be nonzero, since $x_k⊕y_k$ will never be 0. Therefore, if you make the nim-sum $0$ on your turn, your opponent must make it nonzero.
I understand why $x_k ⊕ y_k$ can't be $0$, but why can't it be equal to $s$? This would make the nimsum $0$ still, and I don't see why this wouldn't be possible.
game-theory combinatorial-game-theory
edited Jan 30 at 15:30
Jaap Scherphuis
4,232717
asked Jan 30 at 13:35
Tiwa AinaTiwa Aina
2,720421
$endgroup$
add a comment |
$begingroup$
In proving certain results about Nim, I found a lemma that is causing me trouble:
Lemma 1 If the Nim-sum is $0$ after a player’s turn, then the next move must change it.
To prove this, let the number of stones in the heaps be $x_1, x_2, ... x_n$, and $s$ be the nim-sum $$s=x_1⊕x_2⊕x_3⊕ . . . ⊕x_n$$ Let $t$ be the sum of the heaps after the move, $$t=y_1⊕y_2⊕y_3⊕ . . . ⊕y_n$$ Then if $s= 0$, the next move causes some $x_k=y_k$ and the rest of the $x_i=y_i$ for $i neq k$, since only one pile of stones is changed.
Then:
$$t=0⊕t$$
$$=s⊕s⊕t$$
$$=s⊕(x_1⊕x_2⊕...⊕x_n)⊕(y_1⊕y_2⊕...⊕y_n)$$
$$=s⊕(x_1⊕y_1)⊕(x_2⊕y_2)⊕...⊕(x_k⊕y_k)$$
$$=s⊕x_k⊕y_k$$
If $s$ is 0, then $t$ must be nonzero, since $x_k⊕y_k$ will never be 0. Therefore, if you make the nim-sum $0$ on your turn, your opponent must make it nonzero.
I understand why $x_k ⊕ y_k$ can't be $0$, but why can't it be equal to $s$? This would make the nimsum $0$ still, and I don't see why this wouldn't be possible.
game-theory combinatorial-game-theory
edited Jan 30 at 15:30
Jaap Scherphuis
4,232717
asked Jan 30 at 13:35
Tiwa AinaTiwa Aina
2,720421
$endgroup$
3
$begingroup$
$s$ is the state before the move, and is assumed to be zero. So you are effectively asking: I understand $x_k⊕y_k$ can't be zero, but why can't it be equal to zero?
$endgroup$
– Jaap Scherphuis
Jan 30 at 15:26
add a comment |
$begingroup$
In proving certain results about Nim, I found a lemma that is causing me trouble:
Lemma 1 If the Nim-sum is $0$ after a player’s turn, then the next move must change it.
To prove this, let the number of stones in the heaps be $x_1, x_2, ... x_n$, and $s$ be the nim-sum $$s=x_1⊕x_2⊕x_3⊕ . . . ⊕x_n$$ Let $t$ be the sum of the heaps after the move, $$t=y_1⊕y_2⊕y_3⊕ . . . ⊕y_n$$ Then if $s= 0$, the next move causes some $x_k=y_k$ and the rest of the $x_i=y_i$ for $i neq k$, since only one pile of stones is changed.
Then:
$$t=0⊕t$$
$$=s⊕s⊕t$$
$$=s⊕(x_1⊕x_2⊕...⊕x_n)⊕(y_1⊕y_2⊕...⊕y_n)$$
$$=s⊕(x_1⊕y_1)⊕(x_2⊕y_2)⊕...⊕(x_k⊕y_k)$$
$$=s⊕x_k⊕y_k$$
If $s$ is 0, then $t$ must be nonzero, since $x_k⊕y_k$ will never be 0. Therefore, if you make the nim-sum $0$ on your turn, your opponent must make it nonzero.
I understand why $x_k ⊕ y_k$ can't be $0$, but why can't it be equal to $s$? This would make the nimsum $0$ still, and I don't see why this wouldn't be possible.
game-theory combinatorial-game-theory
edited Jan 30 at 15:30
Jaap Scherphuis
4,232717
asked Jan 30 at 13:35
Tiwa AinaTiwa Aina
2,720421
$endgroup$
In proving certain results about Nim, I found a lemma that is causing me trouble:
Lemma 1 If the Nim-sum is $0$ after a player’s turn, then the next move must change it.
To prove this, let the number of stones in the heaps be $x_1, x_2, ... x_n$, and $s$ be the nim-sum $$s=x_1⊕x_2⊕x_3⊕ . . . ⊕x_n$$ Let $t$ be the sum of the heaps after the move, $$t=y_1⊕y_2⊕y_3⊕ . . . ⊕y_n$$ Then if $s= 0$, the next move causes some $x_k=y_k$ and the rest of the $x_i=y_i$ for $i neq k$, since only one pile of stones is changed.
Then:
$$t=0⊕t$$
$$=s⊕s⊕t$$
$$=s⊕(x_1⊕x_2⊕...⊕x_n)⊕(y_1⊕y_2⊕...⊕y_n)$$
$$=s⊕(x_1⊕y_1)⊕(x_2⊕y_2)⊕...⊕(x_k⊕y_k)$$
$$=s⊕x_k⊕y_k$$
If $s$ is 0, then $t$ must be nonzero, since $x_k⊕y_k$ will never be 0. Therefore, if you make the nim-sum $0$ on your turn, your opponent must make it nonzero.
I understand why $x_k ⊕ y_k$ can't be $0$, but why can't it be equal to $s$? This would make the nimsum $0$ still, and I don't see why this wouldn't be possible.
game-theory combinatorial-game-theory
game-theory combinatorial-game-theory
edited Jan 30 at 15:30
Jaap Scherphuis
4,232717
asked Jan 30 at 13:35
Tiwa AinaTiwa Aina
2,720421
edited Jan 30 at 15:30
Jaap Scherphuis
4,232717
asked Jan 30 at 13:35
Tiwa AinaTiwa Aina
2,720421
edited Jan 30 at 15:30
Jaap Scherphuis
4,232717
edited Jan 30 at 15:30
Jaap Scherphuis
4,232717
edited Jan 30 at 15:30
Jaap Scherphuis
4,232717
4,232717
asked Jan 30 at 13:35
Tiwa AinaTiwa Aina
2,720421
asked Jan 30 at 13:35
Tiwa AinaTiwa Aina
2,720421
asked Jan 30 at 13:35
Tiwa AinaTiwa Aina
2,720421
2,720421
3
$begingroup$
$s$ is the state before the move, and is assumed to be zero. So you are effectively asking: I understand $x_k⊕y_k$ can't be zero, but why can't it be equal to zero?
$endgroup$
– Jaap Scherphuis
Jan 30 at 15:26
add a comment |
3
$begingroup$
$s$ is the state before the move, and is assumed to be zero. So you are effectively asking: I understand $x_k⊕y_k$ can't be zero, but why can't it be equal to zero?
$endgroup$
– Jaap Scherphuis
Jan 30 at 15:26
3
3
$begingroup$
$s$ is the state before the move, and is assumed to be zero. So you are effectively asking: I understand $x_k⊕y_k$ can't be zero, but why can't it be equal to zero?
$endgroup$
– Jaap Scherphuis
Jan 30 at 15:26
$begingroup$
$s$ is the state before the move, and is assumed to be zero. So you are effectively asking: I understand $x_k⊕y_k$ can't be zero, but why can't it be equal to zero?
$endgroup$
– Jaap Scherphuis
Jan 30 at 15:26
add a comment |
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$begingroup$
$s$ is the state before the move, and is assumed to be zero. So you are effectively asking: I understand $x_k⊕y_k$ can't be zero, but why can't it be equal to zero?
$endgroup$
– Jaap Scherphuis
Jan 30 at 15:26