# Last Biscuit: Getting Started Suppose it is your turn to play Last Biscuit. Ideally, you would like to leave your opponent with a configuration of biscuits from which they cannot possibly win. We will call this a losing configuration: when faced with a losing configuration the player to go next will definitely lose.

Can you find such a losing configuration for Last Biscuit? (Hint: start with as few biscuits as possible.)

If one stack contains one biscuit and the other stack contains two biscuits, the player to go next cannot possibly win. (We'll write this configuration as $(1,2)$.)

• If the first player takes away the biscuit from the stack containing only one biscuit, then only the 2-biscuit stack is left, which the second player can take away to win.
• If the first player takes away a biscuit from the stack containing two biscuits, then two stacks with one biscuit each are left, which the second player can take away to win.
• If the first player takes away both biscuits from the stack containing two biscuits, then then only the 1-biscuit stack is left, which the second player can take away to win.

Therefore, $(1,2)$ is a losing configuration for the player to go next.

If a configuration of biscuits allows you to reach a losing position in one move, then you can definitely win the game from that configuration: make the appropriate move, and your opponent faces the losing position. We call such a configuration a winning configuration.

Can you find a winning configuration for Last Biscuit? (Hint: work backwards from the losing configuration you found above.)

One winning configuration is $(2,3)$. If you take one biscuit from each stack, your opponent is faced with the losing configuration $(1,2)$. Another is $(1,3)$. Taking one biscuit from the 3-stack also leaves your opponent with the losing position $(1,2)$.

Still working backwards from the losing configuration, can you find more winning configurations? How many are there?

There are infinitely many winning configurations from which you can reach $(1,2)$ in one move.

• Any configuration of the form $(1,n)$ where $n>2$. You simply take $n-2$ biscuits from the $n$-stack to leave your opponent with the losing configuration. (Note that a configuration $(n,m)$ offers the same play options as $(m,n)$ so we will treat these as the same.)
• Any configuration of the form $(2,n)$ where $n>1$. You simply take $n-1$ biscuits from the $n$-stack to leave your opponent with the losing configuration.
• Any configuration of the form $(n,n+1)$, where $n > 1$. You simply take $n-1$ biscuits from both stacks to leave your opponent with the losing configuration.

Using what you have just found out, can you find other losing configurations and the corresponding winning configurations?

As we have just seen, configurations $(1,n)$, $(2,n)$ and $(n,n+1)$ for $n>1$ are ruled out because they are winning positions. The next configuration to consider is $(3,5)$. You can quite easily convince yourself that whatever the player to go first does will leave their opponent with a winning position. Therefore $(3,5)$ is a losing position for the player who goes first.

Can you continue this line of thought to find all the losing and winning configurations?