# Using python to implement the recursive version of the hannotta example of hannotta recursive algorithm

• 2020-04-02 13:33:49
• OfStack

Hannotta implemented in python. With graphic demonstration

from time import sleep
def disp_sym(num, sym):
print(sym*num, end='')
#recusion
def hanoi(a, b, c, n, tray_num):
if n == 1:
move_tray(a, c)
disp(tray_num)
sleep(0.7)
else:
hanoi(a, c, b, n-1, tray_num)
move_tray(a, c)
disp(tray_num)
sleep(0.7)
hanoi(b, a, c, n-1, tray_num)
def move_tray(a, b):
for i in a:
if i != 0:
for j in b:
if j != 0:
b[b.index(j) - 1] = i
a[a.index(i)] = 0
return
b.append(i)
b.pop(0)
a[a.index(i)] = 0
return

def disp(tray_num):
global a, b, c
for i in range(tray_num):
for j in ['a', 'b', 'c']:
disp_sym(5, ' ')
eval('disp_sym(tray_num - ' + j + "[i], ' ')")
eval('disp_sym(' + j + "[i], '=')")
disp_sym(1, '|')
eval('disp_sym(' + j + "[i], '=')")
eval('disp_sym(tray_num - ' + j + "[i], ' ')")
print()
print('---------------------------------------------------------------------------')
tray_num=int(input("Please input the number of trays:"))
tray=[]
for i in range(tray_num):
tray.append(i + 1)
a=[0]*tray_num
b=a[:]
c=a[:]
a = tray[:]
disp(tray_num)
hanoi(a, b, c, tray_num, tray_num)

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