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  1. Prepare
  2. Algorithms
  3. Game Theory
  4. Misère Nim
  5. Discussions

Misère Nim

  • rishabh_ags
     + 1 comment

    my python code:

    g = int(input().strip())

for _ in range(g):
    n = int(input().strip())
    s = [int(x) for x in input().strip().split(' ')]
    x = 0
    for i in s:
        x ^= i
    if len(set(s))==1 and 1 in s:
        #here x=0 or 1
        if x:#odd no. of ones
            print("Second")
        else:#even no. of ones
            print("First")
    else:
        if x:
            print("First")
        else:
            print("Second")

    I don't understand that when the s(array containing the number of stones in piles) is not having only one, how can it be same as the normal nim game. I mean see this part:

       else:
        if x:
            print("First")
        else:
            print("Second")

    this is same for nim as well as misere nim, how is this possible?? when in nim last player to pick the stone wins but in misere nim he/she loses!!