variant. Define a function, x ⊕ K
1
= f(x), then, we
can arrive at equations similar to those above where
( f ◦ E
K
2
◦ f ◦E
−1
K
3
)
3
(P) = C for the 3-Key Variant and
( f ◦ E
K
2
◦ f ◦ E
−1
K
3
)
2
(P) = C for the 4-Key Variant.
Notice then that we need to change the value of n
(number of pairs of K, φ(K) we consider) since key-
length has been increased, in 4-Key DES-EXEXEX.
We therefore have the following calculations (adapted
from (Vaudenay, 2011), Section 3.1, pg 5-6):
First, we calculate the expected f n
∗
, the number
of lists with an odd number of fixed points. Let n = 6,
E(n
∗
) = 1+ (n− 1)
1− e
−
3
2
2
≈ 2.94 (1)
Then, we have that there are 2
2(56)+(64)
possible
combinations of keys but an equation to satisfy on
(2.94)(64) bits. This gives us the respective value
of R
n
(expected number of wrong keys in R given n
Related-Key pairs) as:
R
n
≈ 2
2(56)+(64)−(2.94)(64)
= 2
−12.16
(2)
Notice that the value of n does not impact the
choice to repeat the entire attack 2 times (i.e. N
n
is
unrelated to n so long as a > 0). Therefore, with an
identical success rate, we require 6(2
64+1
) ≈ 2
68
BKP.
The only difference in the time complexity which
sees a 2
64
increase in the calculations to arrive at a
fixed point, since each K
1
must be guessed separately.
Note that the XOR functions to derive each set of
(P,C⊕K
1
) or (P⊕ K
1
,C) values are assumed to be of
negligible complexity. This has a time complexity of
2(2· 2
56+64
+ 2
56
· 2
−12.16
) + 2
56
≈ 2
122
encryptions.
Memory can be reused for each guess of K
1
, there-
fore, we have that the memory complexity is 2
63
.
For the Three-Key DES-XEXEXE, we can adopt
the same method of finding fixed points, however, as
in the original attack on Two-Key Triple-DES, take
n = 1. This has time complexity of 4(2
56+64
) = 2
122
encryptions and a memory complexity of 2
63
.
Notice that in both these cases, the time complex-
ities are comparable to that of the attack we con-
sidered on DES-XEXEXEX. This, we realized, is
because Vaudenay’s attacks on DES-EXEXEX and
DES-XEXEXE involve guessing each K
1
in turn, re-
turning to an attack very reminiscent of that of DES-
XEXEXEX. This makes the memory space 2
64
times
less, and reduces the required value of n, other in-
dicators that the attack is identical in nature. There-
fore, we have shown the robustness of our method of
strengthening Triple-DES, in that, even if RK attacks
such as Vaudenay’sattack could be more than trivially
applied, we still achieve a 2
64
complexity increase for
the 64-bits of added keylength.
3.4 Phan RK Attack
Phan’s RK slide-attack can be applied to both the 2-
Key and the 3-Key Triple Encryption effectively, as
discussed in his paper (Phan, 2004). We refer the
reader to his paper for the exact details of each attack.
With 2
32
PT-CT pairs each for the original key
and the Related-Key, we can expect 1 pair with the
desired relation by the Birthday Paradox. The first
set of encryptions (for all possible K
1
on all values
of P) dominates the time complexity, meaning that
2
56
· 2
32
= 2
88
single-DES encryptions are required
for the attack. The memory complexity is also domi-
nated by this step, 2
88
· (56+ 64+ 64) ≈ 2
96
.
In the 3-Key Triple DES, we consider PT-
CT pairs encrypted under the keys (K
1
, K
2
, K
3
)
and (K
2
, K
3
, K
1
). We then search for (P,C), en-
crypted under (K
1
, K
2
, K
3
) and (P
′
,C
′
) encrypted un-
der (K
2
, K
3
, K
1
) such that C
′
= E
K
1
(P) and C =
E
K
1
(P
′
). Once again, we obtain 2
32
PT-CT pairs for
each set of keys and create a list of candidates for K
1
by encrypting each P and decrypting each C accord-
ing to each K
1
. Those satisfying the collision con-
ditions give candidate values for K
1
. This, as he re-
ports, requires 2
88
DES encryptions and a memory
complexity of 2
32
· (64+ 64) = 2
39
.
Notice that an exhaustive search for K
2
, K
3
via a
traditional MITM attack applies once K
1
has been
determined. This can be achieved with 2
39
memory
complexity by portioning the 2
56
candidates for K
1
into sets of 2
32
values, a separate MITM attack is then
performed using an exhaustive key-search for K
2
and
matching against possible ciphertext values given for
each group. The total time complexity of this search
should be2
(56−32)
· (2
56
) = 2
80
, negligible in compar-
ison to the time-complexity of the main attack.
Similar to our analysis in other sections of this pa-
per, to achievea time-complexity lower than 2
64
times
that of the original attack, the attack must segment
the keys into two mutually exclusive groups. No-
tice that should be attempt a slide attack on either
DES-XEXEXEX variant, to isolate one or more en-
cryptions, XOR functions both inside and outside the
shared segment of the encryption scheme encryption
under the pair of Related-Keys must follow, making
them not mutually exclusive. Therefore, our focus
turns to the search for a method for the Phan attack
to achieve this complexity.
For this, we obtain 2 sets of 2
32
PT-CT pairs, en-
crypted under (K
1
, K
2
, K
3
, K
4
), denoted (P
∗
,C
∗
), and
(K
1
, K
3
, K
2
, K
4
), denoted (P
′∗
,C
′∗
). We guess a par-
ticular K
1
and XOR all 2
33
PT-CT pairs by it, to arrive
at (P,C) = (P
∗
⊕ K
1
,C
∗
⊕ K
1
) and (P
′
,C
′
) = (P
′∗
⊕
K
1
,C
′∗
⊕ K
1
), which is of negligible time-complexity.
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