We are given an infinite two dimensional plane made of hexagons connected together. We can visualize this plane as a honeycomb. Element X is present on one of the cells / hexagon.
We are given N steps, the task is to calculate number of such hexagonal paths possible in which element X has to perform a walk of N steps and return back to the original hexagon, where
Input : 1 Output : Number of walks possible is/are 0 Explanation : 0 because using just one step we can move to any of the adjacent cells but we cannot trace back to the original hexagon. Input : 2 Output : Number of walks possible is/are 6 Input : 4 Output : Number of walks possible is/are 90
- A hexagonal walk can be defined as walking through adjacent hexagons and returning to the original cell. We know the fact that a hexagon contains six sides i.e. a hexagon is surrounded by six hexagons. Now, we have to count number of ways we take N steps and come back to the original hexagon.
- Now, let us suppose the original hexagon (where element X was initially present) to be the origin. We need all possible ways we can take (N-k) steps such that we have some steps which would trace back to our original hexagon. We can visualize this hexagon and it’s related coordinate system from the picture below.
- Now, let’s assume, our element X was present at 0:0 of the given picture. Thus, we can travel in six possible directions from a hexagon. Now, using the directions above we memorize all possible movements such that we trace back to the original 0:0 index. For memorizing we use a 3D array and we preprocess our answer for a given number of steps and then query accordingly.
Below is the implementation of above approach :
# Python 3 implementation of counting
# number of possible hexagonal walks
depth = 16
ways = [[[0 for i in range(17)]
for i in range(17)]
for i in range(17)]
def preprocess(list, steps):
# We initialize our origin with 1
ways = 1
# For each N = 1 to 14, we traverse in
# all possible direction. Using this 3D
# array we calculate the number of ways
# at each step and the total ways for a
# given step shall be found at ways[step
# number] because all the steps after
# that will be used to trace back to the
# original point index 0:0 according to the image.
for N in range(1, 16, 1):
for i in range(1, depth, 1):
for j in range(1, depth, 1):
ways[N][i][j] = (ways[N – 1][i][j + 1] +
ways[N – 1][i][j – 1] +
ways[N – 1][i + 1][j] +
ways[N – 1][i – 1][j] +
ways[N – 1][i + 1][j – 1] +
ways[N – 1][i – 1][j + 1])
# This array stores the number of ways
# possible for a given step
list[N] = ways[N]
print(“Number of walks possible is/are”,
# Driver Code
if __name__ == ‘__main__’:
list = [0 for i in range(16)]
steps = 4
# Preprocessing all possible ways
# This code is contributed by
Number of walks possible is/are 90
The time complexity of the above code is and the space complexity is also similar due to the 3D array used.
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