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Program to find winner of a rower breaking game in Python
Suppose we have an array height. There are n different towers with different height. Amal and Bimal are playing a game. The game rules are like below
Amal always plays first
During each move, the current player selects a tower of height X and break it down into Y different towers of height Z each. [Y*Z = X; X and Y > 1]
Whoever has no move will lose the game
We have to find the winner's name.
So, if the input is like height = [3,1,2], then the output will be Bimal, because the initial heights are {3,1,2}. If Amal breaks the height of tower 2 to two towers of height 1, then new height array will be {3,1,1,1}, Bimal can break tower with height 3 and make three towers of height 1, so Amal has no move hence Bimal wins.
To solve this, we will follow these steps −
- Define a function util() . This will take limit, initial limit value is 10^3+5
- result := an array of size limit and fill with 0
- for i in range 2 to limit - 1, do
- s := a new set
- for j in range 1 to floor of square root of i, do
- d := quotient of i/j, r := remainder of i/j
- if r is same as 0, then
- if j is odd, then
- insert result[d]) into s
- if d is odd, then
- insert result[j] into s
- if j is odd, then
- j := 0
- while j is present in s, do
- j := j + 1
- result[i] := j
- return result
- g := util()
- From the main method, do the following −
- r := 0
- for each i in height, do
- r := r XOR g[i]
- if r is non-zero, then
- return "Amal"
- otherwise,
- return "Bimal"
Example
Let us see the following implementation to get better understanding −
def util(limit=10**3+5):
result = [0] * limit
for i in range(2, limit):
s = set()
for j in range(1, int(i**0.5)+1):
d, r = divmod(i, j)
if r == 0:
if j & 1:
s.add(result[d])
if d & 1:
s.add(result[j])
j = 0
while j in s: j += 1
result[i] = j
return result
g = util()
def solve(height):
r = 0
for i in height:
r ^= g[i]
if r:
return "Amal"
else:
return "Bimal"
height = [3,1,2]
print(solve(height))Input
[3,1,2]
Output
Bimal
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