# Answer to Question #58039 in Python for locky

Question #58039

How to Permute the positions of the last two nibbles of a number. In other words, a block

Expert's answer

x=int(input())

x=list(bin(x)[2:])

x_change=[]

x_finish=[]

if len(x)<8 :

x_change = [ '0' for i in range (8-len(x))]

for i in range(len(x)):

x_change.append(x[i])

else :

x_change=x

if len(x_change)==8 :

for i in range(4,8): x_finish.append(x_change[i])

for i in range(4): x_finish.append(x_change[i])

else:

for i in range(len(x_change)-8): x_finish.append(x_change[i])

for i in range(len(x_change)-4,len(x_change)): x_finish.append(x_change[i])

for i in range(len(x_change)-8,len(x_change)-4): x_finish.append(x_change[i])

x_finish=int(''.join(x_finish),2)

print(x_finish)

x=list(bin(x)[2:])

x_change=[]

x_finish=[]

if len(x)<8 :

x_change = [ '0' for i in range (8-len(x))]

for i in range(len(x)):

x_change.append(x[i])

else :

x_change=x

if len(x_change)==8 :

for i in range(4,8): x_finish.append(x_change[i])

for i in range(4): x_finish.append(x_change[i])

else:

for i in range(len(x_change)-8): x_finish.append(x_change[i])

for i in range(len(x_change)-4,len(x_change)): x_finish.append(x_change[i])

for i in range(len(x_change)-8,len(x_change)-4): x_finish.append(x_change[i])

x_finish=int(''.join(x_finish),2)

print(x_finish)

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