Alice and Bob play the following game: Alice begins by picking a natural number \( n \geqslant 2 \). Then, with Bob starting first, they alternately choose one number from the set \( A = \{1, 2, \ldots, n\} \) according to the following condition: The number chosen at each step should be distinct from all the already chosen numbers, and should differ by 1 from an already chosen number. (At the very first step Bob can choose any number he wants.) The game ends when all numbers from the set \( A \) are chosen.

For example, if Alice picks \( n = 4 \), then a valid game would be for Bob to choose 2, then Alice to choose 3, then Bob to choose 1, and then Alice to choose 4.

Alice wins if the sum \( S \) of all of the numbers that she has chosen is composite. Otherwise Bob wins. (In the above example \( S = 7 \), so Bob wins.)

Decide which player has a winning strategy.
Alice has a winning strategy. She initially picks \( n=8 \). We will give a strategy so that she can end up with \( S \) even, or \( S=15 \), or \( S=21 \), so she wins.

Case 1: If Bob chooses 1, then the game ends with Alice choosing 2, 4, 6, 8 so \( S \) is even (larger than 2) and Alice wins.

Case 2: If Bob chooses 2, then Alice chooses 3. Bob can now choose either 1 or 3.

Case 2A: If Bob chooses 1, then Alice's numbers are 3, 4, 6, 8. So \( S=21 \) and Alice wins.

Case 2B: If Bob chooses 4, then Alice chooses 1 and ends with the numbers 1, 3, 6, 8. So \( S \) is even and Alice wins.

Case 3: If Bob chooses 3, then Alice chooses 2. Bob can now choose either 1 or 4.

Case 3A: If Bob chooses 1, then Alice's numbers are 2, 4, 6, 8. So \( S \) is even and Alice wins.

Case 3B: If Bob chooses 4, then Alice chooses 5. Bob can now choose either 1 or 6.

Case 3Bi: If Bob chooses 1, then Alice's numbers are 2, 5, 6, 8. So \( S=21 \) and Alice wins.

Case 3Bii: If Bob chooses 6, then Alice chooses 1. Then Alice's numbers are 2, 5, 1, 8. So \( S \) is even and Alice wins.

Case 4: If Bob chooses 4, then Alice chooses 5. Bob can now choose either 3 or 6.

Case 4A: If Bob chooses 3, then Alice chooses 6. Bob can now choose either 2 or 7.

Case 4Ai: If Bob chooses 2, then Alice chooses 1 and ends up with 5, 6, 1, 8. So \( S \) is even and Alice wins.

Case 4Aii: If Bob chooses 7, then Alice chooses 8 and ends up with 5, 6, 8, 1. So \( S \) is even and Alice wins.

Case 4B: If Bob chooses 6, then Alice chooses 7. Bob can now choose either 3 or 8.

Case 4Bi: If Bob chooses 3, then Alice chooses 2 and ends up with 5, 7, 2 and either 1 or 8. So \( S=15 \) or \( S=22 \) and Alice wins.

Case 4Bii: If Bob chooses 8, then Alice's numbers are 5, 7, 3, 1. So \( S \) is even and Alice wins.

Cases 5-8: If Bob chooses \( k \in \{5, 6, 7, 8\} \) then Alice follows the strategy in case \( 9-k \) but whenever she had to choose \( \ell \), she instead chooses \( 9-\ell \). If at the end of that strategy she ended up with \( S \), she will now end up with \( S' = 4 \cdot 9 - S = 36 - S \). Then \( S' \) is even or \( S' = 15 \) or \( S' = 21 \) so again she wins.