Nim With a Twist
A Nim game with a twist: you can reject your opponent's move.
You and AI take turns moving. Each person can take away any number of stones from one of the piles. Whoever takes the last stone wins. The catch is that each turn when a move is made, that move can be rejected by the other person, as long as there is another move. Rejection can only be used once per turn. There are 3 levels in total. In each level, you have a winning strategy, but it's up to you to find it. Level 1 is not that difficult, since it has only a single pile of stones.
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#Here's my code to calculate all must-win or must-lose
#forms in 3d.
set_lose={(0,0,0)}
set_win={(0,1,0),(1,0,0),(0,0,1)}
limit=5
def add_win(i,j,k):
set_win.add((i,j,k))
set_win.add((i,k,j))
set_win.add((k,i,j))
set_win.add((k,j,i))
set_win.add((j,i,k))
set_win.add((j,k,i))
def have_common(a,b):
if a[0]==b[0]:
if a[1]==b[1]:
for i in range(max(a[2],b[2])+1,limit+1):
add_win(a[0],a[1],i)
elif a[2]==b[2]:
for i in range(max(a[1],b[1])+1,limit+1):
add_win(a[0],i,a[2])
else:
if (a[1]>b[1]) and (a[2]<b[2]):
add_win(a[0],a[1],b[2])
if(a[1]<b[1]) and (a[2]>b[2]):
add_win(a[0],b[1],a[2])
elif a[1]==b[1]:
if a[2]==b[2]:
for i in range(max(a[0],b[0])+1,limit+1):
add_win(i,a[1],a[2])
else:
if(a[0]>b[0]) and (a[2]<b[2]):
add_win(a[0],a[1],b[2])
if(a[0]<b[0]) and(a[2]>b[2]):
add_win(b[0],a[1],a[2])
elif a[2]==b[2]:
if(a[0]>b[0]) and (a[1]<b[1]):
add_win(a[0],b[1],a[2])
if(a[0]<b[0]) and (a[1]>b[1]):
add_win(b[0],a[1],a[2])
def xinzeng(dot):
for i in set_lose:
have_common(i,dot)
for i in range(0,limit+1):
for j in range(0,limit+1):
for k in range(0,limit+1):
dot=(i,j,k)
if (dot not in set_win)and(dot not in set_lose):
# print(i,j,k)
xinzeng(dot)
set_lose.add(dot)
print(len(set_lose))
for i in set_lose:
print(i)