これが誰かを助ける場合に備えて、私はここにPythonで完全なソリューションを書きました：https ：//github.com/goblinhack/towers-of-hanoi
#!/usr/bin/env python
#
# The way to solve this is quite simple but does differ slightly for N = odd or even numbers of rings.
#
# At each move you do either a) or b):
#
# a) move the "1" value to the peg to the right, wrapping around to the first peg if needed
#
# b) make the only other legal move
#
# And then repeat either a) or b) for (2 ^ numrings) - 1.
#
# So for N=3, you would do the above steps 7 times.
#
# The catch that I alluded to earlier is that for N == odd (3,5,...), you will need to repeat this
# entire algorithm one more time as the above will only move the rings one peg to the right.
#
import sys
#
# Print the tower so we can check our progress
#
def print_tower(pegs, nrings):
npegs = len(pegs)
for y in range(0, nrings):
h = nrings - y
for x in range(0, npegs):
if len(pegs[x]) >= h:
sys.stdout.write(str(pegs[x][len(pegs[x]) - h]) + " ")
else:
sys.stdout.write("| ")
print("")
print("-----")
def solve_tower(nrings, npegs):
pegs = []
for peg in range(0, npegs):
pegs.append([])
#
# push the nrings on
#
for i in range(0, nrings):
pegs[0].append(i + 1)
#
# For N == odd numbers we will need to repeat this twice
#
for tries in range(0, 1 + nrings % 2):
print_tower(pegs, nrings)
move_peg_one_right = True
#
# Repeat the steps a) or b) for 2^N-1 times
#
for moves in range(0, (1 << nrings) - 1):
#
# step a)
#
if move_peg_one_right:
for peg in range(0, npegs):
if len(pegs[peg]):
if pegs[peg][0] == 1:
next_peg = (peg + 1) % npegs
pegs[next_peg].insert(0, pegs[peg].pop(0))
print("Moving value 1 from peg {} to peg {}\n".format(peg + 1, next_peg + 1))
break
else:
#
# step b)
#
moved_a_ring = False
for peg in range(0, npegs):
#
# Look for a ring on a peg to move
#
if len(pegs[peg]):
value = pegs[peg][0]
#
# Don't move the ring value "1" as we move that in a)
#
if value != 1:
for next_peg in range(0, npegs):
#
# The next peg is the one to the right of this peg. If we reach the last peg then we
# need to move to the first peg.
#
next_peg = (peg + next_peg) % npegs
#
# Don't move to the same peg; that would be silly
#
if next_peg == peg:
continue
#
# If the destination peg is empty, move there
#
if not len(pegs[next_peg]):
pegs[peg].pop(0)
pegs[next_peg].insert(0, value)
moved_a_ring = True
print("Moving value {} from peg {} to empty peg {}\n".format(value, peg + 1, next_peg + 1))
break
elif value < pegs[next_peg][0]:
#
# Else if the destination peg has a lower value, move there
#
pegs[peg].pop(0)
pegs[next_peg].insert(0, value)
moved_a_ring = True
print("Moving < value {} from peg {} to peg {} dest {}\n".format(value, peg + 1, next_peg + 1, pegs[next_peg][0]))
break
if moved_a_ring:
break
if not moved_a_ring:
print("Error, failed to move")
sys.exit(1)
print_tower(pegs, nrings)
#
# Alternate between a) and b)
#
move_peg_one_right = not move_peg_one_right
print("Finished pass\n")
nrings = 3
npegs = 3
solve_tower(nrings, npegs)