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Incubator Q&A Delta and Epsilon play a game with marbles

Solution Yes, by Sprague-Grundy. The same comment as I made on the related question applies. Answer to what I suppose to be the intended question $146$ is an N-position (the next player to mov...

posted 4mo ago by Peter Taylor‭ · edited 4mo ago by Peter Taylor‭

Answer

#3: Post edited by

Peter Taylor‭ · 2026-05-29T09:04:17Z (4 months ago)
Apparently Markdown link syntax doesn't work inside <summary> blocks

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<details><summary>Solution</summary>
~~Yes, by [Sprague-Grundy](https://en.wikipedia.org/wiki/Sprague%E2%80%93Grundy_theorem). The same comment as I made on the related question applies.~~
</details>
<details><summary>Answer to what I suppose to be the intended question</summary>
$146$ is an N-position (the next player to move wins), so Delta can win.
</details>
<details><summary>Details</summary>
The P-positions (previous player wins) are $5k$ and $5k+2$, and the N-positions are $5k+1$, $5k+3$, $5k+4$, where we assume throughout that $k$ is an integer.
From $5k$ marbles either $k=0$ and the previous player has won, or we go to one of $5k'+1$ or $5k'+4$. From $5k+2$ marbles we go to $5k'+1$ or $5k'+3$.
From $5k+1$ or $5k+4$ we can choose to go to $5k$, and from $5k+3$ we can choose to go to $5k+2$.
---
For those who are not familiar with this style of argument, the key points are:
* The possible game states are partitioned into two: those where the remainder after dividing the number of marbles by $5$ is $0$ or $2$, and those where the remainer is $1$, $3$, or $4$. Every game state is in exactly one of these groups.
* The game state where the game has terminated and the next player has no move ($0$ marbles left, with remainder $0$) is in the first group.
* From the first group, every legal move goes to the second group.
* From the second group, there is always a legal move to the first group.
* Therefore when it is your turn, if the game state is in the second group you should force it to the first group, and then your next turn will also be from the second group. Since you never play a state from the first group, you will never be the player to (fail to) move from the $0$ marble state.
</details>

<details><summary>Solution</summary>
Yes, by <a href="https://en.wikipedia.org/wiki/Sprague%E2%80%93Grundy_theorem">Sprague-Grundy</a>. The same comment as I made on the related question applies.
</details>
<details><summary>Answer to what I suppose to be the intended question</summary>
$146$ is an N-position (the next player to move wins), so Delta can win.
</details>
<details><summary>Details</summary>
The P-positions (previous player wins) are $5k$ and $5k+2$, and the N-positions are $5k+1$, $5k+3$, $5k+4$, where we assume throughout that $k$ is an integer.
From $5k$ marbles either $k=0$ and the previous player has won, or we go to one of $5k'+1$ or $5k'+4$. From $5k+2$ marbles we go to $5k'+1$ or $5k'+3$.
From $5k+1$ or $5k+4$ we can choose to go to $5k$, and from $5k+3$ we can choose to go to $5k+2$.
---
For those who are not familiar with this style of argument, the key points are:
* The possible game states are partitioned into two: those where the remainder after dividing the number of marbles by $5$ is $0$ or $2$, and those where the remainer is $1$, $3$, or $4$. Every game state is in exactly one of these groups.
* The game state where the game has terminated and the next player has no move ($0$ marbles left, with remainder $0$) is in the first group.
* From the first group, every legal move goes to the second group.
* From the second group, there is always a legal move to the first group.
* Therefore when it is your turn, if the game state is in the second group you should force it to the first group, and then your next turn will also be from the second group. Since you never play a state from the first group, you will never be the player to (fail to) move from the $0$ marble state.
</details>

#2: Post edited by

Peter Taylor‭ · 2026-05-29T09:01:53Z (4 months ago)
Link Sprague-Grundy and explain N- and P-positions

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<details><summary>Solution</summary>
~~Yes, by Sprague-Grundy. The same comment as I made on the related question applies.~~
</details>
<details><summary>Answer to what I suppose to be the intended question</summary>
~~$146$ is an N-position, so Delta can win.~~
</details>
<details><summary>Details</summary>
~~The P-positions are $5k$ and $5k+2$, and the N-positions are $5k+1$, $5k+3$, $5k+4$, where we assume throughout that $k$ is an integer.~~
From $5k$ marbles either $k=0$ and the previous player has won, or we go to one of $5k'+1$ or $5k'+4$. From $5k+2$ marbles we go to $5k'+1$ or $5k'+3$.
From $5k+1$ or $5k+4$ we can choose to go to $5k$, and from $5k+3$ we can choose to go to $5k+2$.
</details>

<details><summary>Solution</summary>
Yes, by [Sprague-Grundy](https://en.wikipedia.org/wiki/Sprague%E2%80%93Grundy_theorem). The same comment as I made on the related question applies.
</details>
<details><summary>Answer to what I suppose to be the intended question</summary>
$146$ is an N-position (the next player to move wins), so Delta can win.
</details>
<details><summary>Details</summary>
The P-positions (previous player wins) are $5k$ and $5k+2$, and the N-positions are $5k+1$, $5k+3$, $5k+4$, where we assume throughout that $k$ is an integer.
From $5k$ marbles either $k=0$ and the previous player has won, or we go to one of $5k'+1$ or $5k'+4$. From $5k+2$ marbles we go to $5k'+1$ or $5k'+3$.
From $5k+1$ or $5k+4$ we can choose to go to $5k$, and from $5k+3$ we can choose to go to $5k+2$.
---
For those who are not familiar with this style of argument, the key points are:
* The possible game states are partitioned into two: those where the remainder after dividing the number of marbles by $5$ is $0$ or $2$, and those where the remainer is $1$, $3$, or $4$. Every game state is in exactly one of these groups.
* The game state where the game has terminated and the next player has no move ($0$ marbles left, with remainder $0$) is in the first group.
* From the first group, every legal move goes to the second group.
* From the second group, there is always a legal move to the first group.
* Therefore when it is your turn, if the game state is in the second group you should force it to the first group, and then your next turn will also be from the second group. Since you never play a state from the first group, you will never be the player to (fail to) move from the $0$ marble state.
</details>

#1: Initial revision by

Peter Taylor‭ · 2026-05-27T16:10:46Z (4 months ago)

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<details><summary>Solution</summary>
Yes, by Sprague-Grundy. The same comment as I made on the related question applies.
</details>

<details><summary>Answer to what I suppose to be the intended question</summary>
$146$ is an N-position, so Delta can win.
</details>

<details><summary>Details</summary>
The P-positions are $5k$ and $5k+2$, and the N-positions are $5k+1$, $5k+3$, $5k+4$, where we assume throughout that $k$ is an integer.

From $5k$ marbles either $k=0$ and the previous player has won, or we go to one of $5k'+1$ or $5k'+4$. From $5k+2$ marbles we go to $5k'+1$ or $5k'+3$.

From $5k+1$ or $5k+4$ we can choose to go to $5k$, and from $5k+3$ we can choose to go to $5k+2$.
</details>

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