再，再改進一下。

都結巴了。
Let me define losing hand as whenever you create this hand to your opponent, you win.

1. 任何時候，不能留給對手有兩堆數目相等的情況。
That’s right-
I will summarize your conclusion one step further – 0-X-X is a losing hand.

2. 如果有一堆變成1的時候，留給對手讓兩堆數目相差是1.
That’s incorrect(practically correct though). 1-X-(X+1) is NOT necessarily a losing hand. For example 1-3-4. You are just one step close to the common solution.
-- 1, x,x+1 (x>4, otherwise, pick to 1,4,5 ; 1,3,4;1,2,3
)

3. 如果任何一堆是2的時候。保持另兩堆的數目相差等於2.
That’s incorrect (practically correct though). 2-X-X+2 is NOT necessary a losing hand. For example 2-6-8. You are just one step close to the common solution.
2,x,x+2 (x>6, otherwise, pick to 2,6,4; 2,5,3;

4. 任何一堆都大於2的時候，保持另兩堆的數目相差大於2.
That’s incorrect for obvious reasons (3-4-7; 3-5-6). It might be more difficult than you think.
-- 3,x,x+3 (x>7，）
此題無解，或者說無窮多解