Knight Solution using Numpy's linalg.matrix_power

NPKnightSolution.py

import numpy as np

def getPossibilities(startingPosition, turns):
    """
    Calculates the total number of uniques paths from the starting position in N turns for the transition matrix
    """
    
    # edgecase: 0 turns, return 1
    if turns == 0:
        return 1
        
    # our transition matrix
    transitionMatrix = np.matrix([
        [0, 0, 0, 0, 1, 0, 1, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 1, 0, 1, 0],
        [0, 0, 0, 0, 0, 0, 0, 1, 0, 1],
        [0, 0, 0, 0, 1, 0, 0, 0, 1, 0],
        [1, 0, 0, 1, 0, 0, 0, 0, 0, 1],
        [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [1, 1, 0, 0, 0, 0, 0, 1, 0, 0],
        [0, 0, 1, 0, 0, 0, 1, 0, 0, 0],
        [0, 1, 0, 1, 0, 0, 0, 0, 0, 0],
        [0, 0, 1, 0, 1, 0, 0, 0, 0, 0]], dtype=object)
    
    # get our new matrix N turns ahead
    newMatrix = np.linalg.matrix_power(transitionMatrix, turns)
    
    # sum the relevant row
    return np.sum(newMatrix[startingPosition])