Understanding Cryptography by Christof Paar and Jan Pelzl - Chapter 3 Solutions - Ex3.13 Friday. 22 December 2017 - 6 mins Exercise 3.13 This problem deals with the lightweight cipher PRESENT.

Calculate the state of PRESENT-80 after the execution of one round. You can use the following table to solve this problem with paper and pencil. Use the following values (in hexadecimal notation): plaintext = 0000 0000 0000 0000, key = BBBB 5555 5555 EEEE FFFF. Now calculate the round key for the second round using the following table. Solution This solution is verified as correct by the official Solutions for Odd-Numbered Questions manual.

1. After round one, the state is as follows:

2. The generation of the round key for the second round is as follows (Round 1 does not modify the key register):

I wrote a python script which can perform PRESENT-80 encryption:

Note : This is not a remotely efficient implementation of PRESENT-80.

from bitarray import bitarray
def generate_round_keys ( key , verbose = False ):
yield key [: 64 ]
for i in range ( 1 , 32 ):
key = key [ - 19 :] + key [: - 19 ]
if verbose :
print "KeyState After Rotation: \t " , bit_array_to_hex ( key )
key = sbox ( key [: 4 ]) + key [ 4 :]
if verbose :
print "KeyState After S-box: \t\t " , bit_array_to_hex ( key )
round_counter = make_bit_array ( 5 , i )
key = key [: - 20 ] + ( key [ - 20 : - 15 ] ^ round_counter ) + key [ - 15 :]
if verbose :
print "KeyState After Counter Add: \t " , bit_array_to_hex ( key )
yield key [: 64 ]
def sbox ( n ):
sbox = {
'0000' : bitarray ( '1100' ),
'0001' : bitarray ( '0101' ),
'0010' : bitarray ( '0110' ),
'0011' : bitarray ( '1011' ),
'0100' : bitarray ( '1001' ),
'0101' : bitarray ( '0000' ),
'0110' : bitarray ( '1010' ),
'0111' : bitarray ( '1101' ),
'1000' : bitarray ( '0011' ),
'1001' : bitarray ( '1110' ),
'1010' : bitarray ( '1111' ),
'1011' : bitarray ( '1000' ),
'1100' : bitarray ( '0100' ),
'1101' : bitarray ( '0111' ),
'1110' : bitarray ( '0001' ),
'1111' : bitarray ( '0010' ),
}
return sbox [ n . to01 ()][:]
def apply_sbox ( ciphertext ):
return \
sbox ( ciphertext [ 0 : 4 ]) + \
sbox ( ciphertext [ 4 : 8 ]) + \
sbox ( ciphertext [ 8 : 12 ]) + \
sbox ( ciphertext [ 12 : 16 ]) + \
sbox ( ciphertext [ 16 : 20 ]) + \
sbox ( ciphertext [ 20 : 24 ]) + \
sbox ( ciphertext [ 24 : 28 ]) + \
sbox ( ciphertext [ 28 : 32 ]) + \
sbox ( ciphertext [ 32 : 36 ]) + \
sbox ( ciphertext [ 36 : 40 ]) + \
sbox ( ciphertext [ 40 : 44 ]) + \
sbox ( ciphertext [ 44 : 48 ]) + \
sbox ( ciphertext [ 48 : 52 ]) + \
sbox ( ciphertext [ 52 : 56 ]) + \
sbox ( ciphertext [ 56 : 60 ]) + \
sbox ( ciphertext [ 60 : 64 ])
def apply_pLayer ( ciphertext ):
temp = ciphertext [:]
temp . setall ( 0 )
for i in range ( 0 , 64 ):
if i == 63 :
temp [ 63 ] = ciphertext [ 63 ]
else :
pi = ( i * 16 ) % 63
temp [ pi ] = ciphertext [ i ]
return temp
def present_80_encrypt ( plaintext , key , verbose = False ):
ciphertext = make_bit_array ( 64 , plaintext )
key = make_bit_array ( 80 , key )
i = 0
for ki in generate_round_keys ( key , verbose ):
i += 1
ciphertext ^= ki
if verbose :
print "Round {i:02d} Key: \t\t\t " . format ( i = i ), bit_array_to_hex ( ki )
print "State After KeyAdd: \t\t " , bit_array_to_hex ( ciphertext )
if i == 32 :
break
ciphertext = apply_sbox ( ciphertext )
if verbose :
print "State After SBox: \t\t " , bit_array_to_hex ( ciphertext )
ciphertext = apply_pLayer ( ciphertext )
if verbose :
print "State After pLayer: \t\t " , bit_array_to_hex ( ciphertext )
if verbose :
print "--------"
def make_bit_array ( length , n ):
return bitarray ( bin ( n ). lstrip ( '-0b' ). zfill ( length ))
def bit_array_to_hex ( b ):
return hex ( int ( b . to01 (), 2 )). lstrip ( '-0x' ). rstrip ( 'L-' ). zfill ( 16 )
if __name__ == "__main__" :
plaintext = int ( '0000000000000000' , 16 )
key = int ( 'BBBB55555555EEEEFFFF' , 16 )
present_80_encrypt ( plaintext , key , verbose = True )

Thomas Busby I write about computing stuff