Alice and Bob play a game by taking turns removing 1 or 2 stones from a pile that initially has n stones. The person that removes the last stone wins the game. Alice plays always first.
(a) Prove by induction that if n is a multiple of 3 then Bob has a wining strategy.
(b) Prove that if n is not a multiple of 3 then Alice has a wining strategy.

Answer:

Step-by-step explanation:

(a) We will prove by induction that if n is a multiple of 3 then Bob has a winning strategy.

Let n=3

It is given that Alice always plays first.

Then, for the first move, Alice has a choice of removing 1 or 2 stones.

Case 1:

Alice removes 1 stone. Then it is now Bob's turn. There are 2 stones left. Bob has the choice of removing 1 or 2 stones. Then Bob's winning strategy should be to remove 2 stones. Then Bob removes the last stone and wins.

Case 2:

Alice removes 2 stones. Then it is Bob's turn. There is exactly 1 stone left, which Bob removes in his turn. Since he removes the last stone, he wins.

Thus, in either case, for n=3, Bob has a winning strategy. ____ (A)

Now, let us assume that Bob has a winning strategy for n=3p.

We will now check if Bob has a winning strategy for n=3p+3

It is given that Alice plays first.

Case 1:

Alice starts the game by removing 1 stone. Then, Bob has a choice of removing 1 or 2 stones. If he chooses to remove 2 stones, then the number of stones left is 3p+3-3=3p and it is now Alice's turn to play. This is exactly the game when n=3p and we have already assumed that Bob has a winning strategy for n=3p. Thus, in this case Bob has a winning strategy.

Case 2:

Alice starts the game by removing 2 stones. Then, Bob has a choice of removing 1 or 2 stones. If he chooses to remove 1 stone, then the number of stones left is 3p+3-3=3p and it is now Alice's turn to play. This is again exactly the game when n=3p and we have already assumed that Bob has a winning strategy for n=3p. Thus, in this case also, Bob has a winning strategy.

Thus, in either case Bob has a winning strategy for n=3p+3 if he has a winning strategy for n=3p. _______ (B)

Thus, from (A) & (B), using induction, we can say that Bob has a winning strategy if n is a multiple of 3.

(b) We will now prove that if n is not a multiple of 3, then Alice has a winning strategy.

If n is not a multiple of 3, then n can have either of the forms of 3m+1 or 3m+2, m ∈ W.

We will prove the given fact for both the forms simultaneously

Let m=0, i.e., n=1 or n=2

Since Alice starts first, she removes 1 stone, if n=1 or 2 stones if n=2 and thus wins. Thus Alice has a winning strategy if m=0. ______ (A)

Let us assume that Alice has a winning strategy for m=k, i.e., for n=3k+1 & n=3k+2

Now, we will check if Alice has a winning strategy for m=k+1, i.e., for n=3(k+1)+1=3k+4 or n=3(k+1)+2=3k+5

Let n=3k+4

Since Alice plays first, she has a choice to remove 1 or 2 stones.

Note that, if Alice removes 2 stones and in turn Bob removes 2 stones, then the number of stones becomes a multiple of 3 such that it is Alice's turn to play. In that case, Bob will have a winning strategy as shown in the previous part.

Then, Alice should remove 1 stone. Then, it is now Bob's turn and he has a choice of removing 1 or 2 stones.

Case 1:

Bob removes 1 stone. Then there are 3k+4-1-1=3k+2 stones remaining and it is Alice's turn. This is identical to the game where n=3k+2. We have already assumed that Alice has a winning strategy in this case.

Case 2:

Bob removes 2 stones. Then there are 3k+4-1-2=3k+1 stones remaining and it is Alice's turn. This is identical to the game where n=3k+1. We have already assumed that Alice has a winning strategy in this case.

Thus, in either case, Alice has a winning strategy.

Let n=3k+5

Since Alice plays first, she has a choice to remove 1 or 2 stones.

Note that, if Alice removes 1 stone and in turn Bob removes 1 stone, then the number of stones becomes a multiple of 3 such that it is Alice's turn to play. In that case, Bob will have a winning strategy as shown in the previous part.

Then, Alice should remove 2 stones. Then, it is now Bob's turn and he has a choice of removing 1 or 2 stones.

Case 1:

Bob removes 1 stone. Then there are 3k+5-2-1=3k+2 stones remaining and it is Alice's turn. This is identical to the game where n=3k+2. We have already assumed that Alice has a winning strategy in this case.

Case 2:

Bob removes 2 stones. Then there are 3k+5-2-2=3k+1 stones remaining and it is Alice's turn. This is identical to the game where n=3k+1. We have already assumed that Alice has a winning strategy in this case.

Thus, in either case, Alice has a winning strategy.

Then, we can say that Alice has a winning strategy for m=k+1 if she has a winning strategy for m=k. _____ (B)

Then, by induction, from (A) & (B), we can say that if n is not a multiple of 3, then Alice has a winning strategy.

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