suppose there are x stones left for first player (A), he can take 1,2,3 stones away, so second player B will have three cases to deal with (x -1), (x-2), (x-3). after he pick the stones, there will be 9 cases left for A.

```
B (x-1) -> A: (x-2), (x-3), (x-4)
B (x-2) -> A: (x-3), (x-4), (x-5)
B (x-3) -> A: (x-4), (x-5), (x-6)
```

Now, if A can guarantee he win at either of three groups, then he can force B to into that one of the three states and A can end up in that particular group after B's move.

```
f(x) = (f(x-2)&&f(x-3)&&f(x-4)) || (f(x-3)&&f(x-4)&&f(x-5)) || (f(x-4)&&f(x-5)&&f(x-6))
```

if we examine the equation a little closer, we can find f(x - 4) is a critical point, if f(x-4) is false, then f(x) will be always false.

we can also find out the initial conditions, f(1), f(2), f(3) will be true (A always win), and f(4) will be false. so

based on previous equation and initial conditions f(5) = f(6) = f(7) = true, f(8) = false;

obviously, f(1), f(2), f(3) can make all f(4n + 1), f(4n + 2), f(4n + 3) to be true, only f(4n) will be false then.

so here we go our one line solution:

return (n % 4 != 0);