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@CamDavidsonPilon
Last active September 1, 2020 11:13
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import numpy as np
from numpy.linalg import matrix_power
from matplotlib import pyplot as plt
import seaborn as sns
SIZE = 100
M = np.zeros((SIZE, SIZE))
# encoding rolls of die
for y in xrange(SIZE):
if y <= (SIZE - 6 - 1):
M[y, np.arange(y+1, y+7)] = 1
elif y == SIZE - 1: # end, absorbing state
M[y, SIZE-1] = 6
else:
# this should be checked more carefully. This is the "bounce-off" behaviour.
M[y, (y+1):] = 1
M[y, (SIZE-(6 - (SIZE-y-2))):SIZE-1] += 1
def slip_effect(M, start_node, end_node):
"""
Ex: if am in position 1, and then roll a 2, I should temporarily be on position 3 before
slipping to position 13. Thus everywhere that _should_ have landed me on position 3
should instead be moved to position 13, and all position 3s should be erased.
"""
ix, = np.where(M[:, start_node])
M[ix, end_node] += M[ix, start_node]
M[:, start_node] = 0
return M
# encoding of snakes
M = slip_effect(M, 16, 6)
M = slip_effect(M, 86, 23)
M = slip_effect(M, 98, 77)
M = slip_effect(M, 63, 59)
M = slip_effect(M, 61, 18)
M = slip_effect(M, 94, 74)
M = slip_effect(M, 92, 72)
M = slip_effect(M, 53, 33)
# encoding of ladders
M = slip_effect(M, 3, 13)
M = slip_effect(M, 8, 30)
M = slip_effect(M, 19, 37)
M = slip_effect(M, 27, 83)
M = slip_effect(M, 39, 59)
M = slip_effect(M, 50, 66)
M = slip_effect(M, 62, 80)
M = slip_effect(M, 70, 90)
M = M/6.0
# check for first "win" in 7 moves - note my initial state is technically move 2, hence the off-by-one.
for n in xrange(1, 6):
assert matrix_power(M, n)[0, SIZE-1] == 0
M_6 = matrix_power(M, 6)
assert M_6[0, SIZE-1] > 0
# what is the probability of this event?
M[0,SIZE-1] # 0.0135%
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