# 292 Nim Game

You are playing the following Nim Game with your friend: There is a heap of stones on the table, each time one of you take turns to remove 1 to 3 stones. The one who removes the last stone will be the winner. You will take the first turn to remove the stones.
Both of you are very clever and have optimal strategies for the game. Write a function to determine whether you can win the game given the number of stones in the heap.
For example, if there are 4 stones in the heap, then you will never win the game: no matter 1, 2, or 3 stones you remove, the last stone will always be removed by your friend.
Hint:
If there are 5 stones in the heap, could you figure out a way to remove the stones such that you will always be the winner?
class Solution {
public:
bool canWinNim(int n) {
return n % 4 != 0;
}
};
Another way of approaching this problem is through dynammic programming. The idea goes as follows: First consider the first three cases: 0, 1, 2, 3. In each of these scenarios, we can just take the maximum number of stones, and win the game. I can use this to initialize my map first the first 3 cases.
Next, we consider 4. Each of the case we take 3, 2, or 1 - this will lead your opponent is taking the remainder. We can model this to say that if for all cache[n-i] in 1:3, then this leads to a lose. However, if in any case, we can find at least a single path that leads us to victory, then we can select this path.
Using this method, we can then familiarize ourselfs with the pattern, and the first solution mentioned.
def canWinNim(self, n):
"""
:type n: int
:rtype: bool
"""
# start us off
cache = {0: True, 1: True, 2: True, 3: True}
if cache.__contains__(n):
return cache[n]
result = False
for n_stones in range(4, n+1):
if all(cache[n_stones-pick] for pick in range(1,4)):
cache[n_stones] = False
else:
cache[n_stones] = True
return cache[n]