Scribed by Siu-Man Chan

Summary

Given one way permutations (of which discrete logarithm is a candidate), we know how to construct pseudorandom functions. Today, we are going to construct pseudorandom permutations (block ciphers) from pseudorandom functions.

1. Pseudorandom Permutations

Recall that a pseudorandom function {F} is an efficient function {\colon\{0,1\}^k\times\{0,1\}^n\rightarrow\{0,1\}^n}, such that every efficient algorithm {A} cannot distinguish well {F_K(\cdot)} from {R(\cdot)}, for a randomly chosen key {K\in\{0,1\}^k} and a random function {R\colon\{0,1\}^n\rightarrow \{0,1\}^n}. That is, we want that {A^{F_K(\cdot)}} behaves like {A^{R(\cdot)}}.

A pseudorandom permutation {P} is an efficient function {\colon\{0,1\}^k\times \{0,1\}^n\rightarrow\{0,1\}^n}, such that for every key {K}, the function {P_K} mapping {x\mapsto P_K(x)} is a bijection. Moreover, we assume that given {K}, the mapping {x\mapsto P_K(x)} is efficiently invertible (i.e. {P_K^{-1}} is efficient). The security of {P} states that every efficient algorithm {A} cannot distinguish well {\langle P_K(\cdot), P_K^{-1}(\cdot)\rangle} from {\langle \Pi(\cdot), \Pi^{-1}(\cdot)\rangle}, for a randomly chosen key {K\in\{0,1\}^k} and a random permutation {\Pi\colon\{0,1\}^n\rightarrow\{0,1\}^n}. That is, we want that {A^{P_K(\cdot),P_K^{-1}(\cdot)}} behaves like {A^{\Pi(\cdot),\Pi^{-1}(\cdot)}}.

We note that the algorithm {A} is given access to both an oracle and its (supposed) inverse.

2. Feistel Permutations

Given any function {F\colon\{0,1\}^m\rightarrow\{0,1\}^m}, we can construct a permutation {D_F\colon\{0,1\}^{2m}\rightarrow\{0,1\}^{2m}} using a technique named after Horst Feistel. The definition of {D_F} is given by

\displaystyle  D_F(x,y)\mathrel{:}= y,F(y)\oplus x, \ \ \ \ \ (1)

where {x} and {y} are {m}-bit strings. Note that this is an injective (and hence bijective) function, because its inverse is given by

\displaystyle  D_F^{-1}(z,w)\mathrel{:}= F(z)\oplus w, z. \ \ \ \ \ (2)

Also, note that {D_F} and {D^{-1}_F} are efficiently computable given {F}.

However, {D_F} needs not be a pseudorandom permutation even if {F} is a pseudorandom function, because the output of {D_F(x,y)} must contain {y}, which is extremely unlikely for a truly random permutation.

To avoid the above pitfall, we may want to repeat the construction twice. We pick two independent random keys {K_1} and {K_2}, and compose the permutations {P(\cdot)\mathrel{:}= D_{F_{K_2}}(D_{F_{K_1}}(\cdot))}.

Indeed, the output does not always contain part of the input. However, this construction is still insecure, no matter whether {F} is pseudorandom or not, as the following example shows.

Here, {\overline0} denotes the all-zero string of length {m}, {\overline1} denotes the all-one string of length {m}, and {F(\cdot)} is {F_{K_1}(\cdot)}. This shows that, restricting to the first half, {P(\overline0\overline0)} is the complement of {P(\overline1\overline0)}, regardless of {F}.

What happens if we repeat the construction three times? We still do not get a pseudorandom permutation.

Exercise 1 (Not Easy) Show that there is an efficient oracle algorithm {A} such that

\displaystyle  \mathop{\mathbb P}_{\Pi : \{ 0,1 \}^{2m} \rightarrow \{ 0,1 \}^{2m} } [ A^{\Pi,\Pi^{-1} } = 1] = 2^{-\Omega(m) }

where {\Pi} is a random permutation, but for every three functions {F_1,F_2,F_3}, if we define {P(x) := D_{F_3} (D_{F_2} ( D_{F_1} (x )))} we have

\displaystyle  A^{P,P^{-1} } = 1

Finally, however, if we repeat the construction four times, with four independent pseudorandom functions, we get a pseudorandom permutation.

3. The Luby-Rackoff Construction

Let {F: \{ 0,1 \}^k \times \{ 0,1 \}^m \rightarrow \{ 0,1 \}^m} be a pseudorandom function, we define the following function {P: \{ 0,1 \}^{4k} \times \{ 0,1 \}^{2m} \rightarrow \{ 0,1 \}^{2m}}: given a key {\overline K=\langle K_1,\dotsc,K_4\rangle} and an input {x},

\displaystyle   P_{\overline K}(x)\mathrel{:}= D_{F_{K_4}}(D_{F_{K_3}}(D_{F_{K_2}}(D_{F_{K_1}}(x)))). \ \ \ \ \ (3)

It is easy to construct the inverse permutation by composing their inverses backwards.

Theorem 1 (Pseudorandom Permutations from Pseudorandom Functions) If {F} is a {(O(tr),\epsilon)}-secure pseudorandom function computable in time {r}, then {P} is a {(t, 4\epsilon + t^2 \cdot 2^{-m} + t^2 \cdot 2^{-2m} )} secure pseudorandom permutation.

4. Analysis of the Luby-Rackoff Construction

Given four random functions {\overline R = \langle R_1,\ldots,R_4\rangle}, {R_i: \{ 0,1 \}^m \rightarrow \{ 0,1 \}^m}, let {P_{\overline R}} be the analog of Construction (3) using the random function {R_i} instead of the pseudorandom functions {F_{K_i}},

\displaystyle  P_{\overline{R}} (x) = D_{{R_4}} ( D_{{R_3}} ( D_{{R_2}} (D_{{R_1}} (x )))) \ \ \ \ \ (4)

We prove Theorem 1 by showing that

  1. {P_{\overline K}} is indistinguishable from {P_{\overline R}} or else we can break the pseudorandom function

  2. {P_{\overline R}} is indistinguishable from a random permutation

The first part is given by the following lemma, which we prove via a standard hybrid argument.

Lemma 2 If {F} is a {(O(tr),\epsilon)}-secure pseudorandom function computable in time {r}, then for every algorithm {A} of complexity {\leq t} we have

\displaystyle   \left | \mathop{\mathbb P}_{\overline {K} } [ A^{P_{\overline{K}}, P^{-1} _{\overline{K}} } () =1 ] \right. \ \ \ \ \ (5)

And the second part is given by the following lemma:

Lemma 3 For every algorithm {A} of complexity {\leq t} we have

\displaystyle  \left| \mathop{\mathbb P}_{\overline {R} } [ A^{P_{\overline{R}}, P^{-1} _{\overline{R}} } () =1 ] - \mathop{\mathbb P}_{\Pi} [ A^{\Pi, \Pi^{-1} } () = 1] \right| \leq \frac{t^2}{ 2^{2m}} + \frac {t^2}{ 2^{m}}

where {\Pi : \{ 0,1 \}^{2m} \rightarrow \{ 0,1 \}^{2m}} is a random permutation.

We now prove Lemma 2 using a hybrid argument.

Proof: Consider the following five algorithms from {\{0,1\}^{2m}} to {\{0,1\}^{2m}}:

  • {H_0}: pick random keys {K_1}, {K_2}, {K_3}, {K_4},
    {H_0(\cdot)\mathrel{:}= D_{F_{K_4}}(D_{F_{K_3}}(D_{F_{K_2}}(D_{F_{K_1}}(\cdot))));}
  • {H_1}: pick random keys {K_2}, {K_3}, {K_4} and a random function {F_1\colon\{0,1\}^m\rightarrow\{0,1\}^m},
    {H_1(\cdot)\mathrel{:}= D_{F_{K_4}}(D_{F_{K_3}}(D_{F_{K_2}}(D_{F_1}(\cdot))));}
  • {H_2}: pick random keys {K_3}, {K_4} and random functions {F_1,F_2\colon\{0,1\}^m\rightarrow\{0,1\}^m},
    {H_2(\cdot)\mathrel{:}= D_{F_{K_4}}(D_{F_{K_3}}(D_{F_2}(D_{F_1}(\cdot))));}
  • {H_3}: pick a random key {K_4} and random functions {F_1,F_2,F_3\colon\{0,1\}^m\rightarrow\{0,1\}^m},
    {H_3(\cdot)\mathrel{:}= D_{F_{K_4}}(D_{F_3}(D_{F_2}(D_{F_1}(\cdot))));}
  • {H_4}: pick random functions {F_1,F_2,F_3,F_4 \colon\{0,1\}^m\rightarrow\{0,1\}^m},
    {H_4(\cdot)\mathrel{:}= D_{F_4}(D_{F_3}(D_{F_2}(D_{F_1}(\cdot)))).}

Clearly, referring to (5), {H_0} gives the first probability of using all pseudorandom functions in the construction, and {H_4} gives the second probability of using all completely random functions. By triangle inequality, we know that

\displaystyle   \exists i\quad\Bigl\lvert\mathop{\mathbb P}[A^{H_i,H_i^{-1}}=1]- \mathop{\mathbb P}[A^{H_{i+1},H_{i+1}^{-1}}=1]\Bigr\rvert>\epsilon. \ \ \ \ \ (6)

We now construct an algorithm {A'^{G(\cdot)}} of complexity {O(tr)} that distinguishes whether the oracle {G(\cdot)} is {F_K(\cdot)} or a random function {R(\cdot)}.

  • The algorithm {A'} picks {i} keys {K_1,K_2,\dotsc,K_i} and initialize {4-i-1} data structures {S_{i+2},\dotsc,S_4} to {\emptyset} to store pairs.
  • The algorithm {A'} simulates {A^{O,O^{-1}}}. Whenever {A} queries {O} (or {O^{-1}}), the simulating algorithm {A'} uses the four compositions of Feistel permutations, where

    • On the first {i} layers, run the pseudorandom function {F} using the {i} keys {K_1,K_2,\dotsc,K_i};
    • On the {i}-th layer, run an oracle {G};
    • On the remaining {4-i-1} layers, simulate a random function: when a new value for {x} is needed, use fresh randomness to generate the random function at {x} and store the key-value pair into the appropriate data structure; otherwise, simply return the value stored in the data structure.

When {G} is {F_K}, the algorithm {A'^{G}} behaves like {A^{H_i,H_i^{-1}}}; when {G} is a random function {R}, the algorithm {A'^{G}} behaves like {A^{H_{i+1}, H_{i+1}^{-1}}}. Rewriting (6),

\displaystyle \Bigl\lvert\mathop{\mathbb P}_K[A'^{F_K(\cdot)}=1]-\mathop{\mathbb P}_R[A'^{R(\cdot)}=1] \Bigr\rvert>\epsilon,

and {F} is not {(O(tr),\epsilon)}-secure. \Box

We say that an algorithm {A} is non-repeating if it never makes an oracle query to which it knows the answer. (That is, if {A} is interacting with oracles {g,g^{-1}} for some permutation {g}, then {A} will not ask twice for {g(x)} for the same {x}, and it will not ask twice for {g^{-1}(y)} for the same {y}; also, after getting the value {y=g(x)} in an earlier query, it will not ask for {g^{-1} (y)} later, and after getting {w=g^{-1} (z)} it will not ask for {g(w)} later. )

We shall prove Lemma 3 for non-repeating algorithms. The proof can be extended to arbitrary algorithms with some small changes. Alternatively, we can argue that an arbitrary algorithm can be simulated by a non-repeating algorithm of almost the same complexity in such a way that the algorithm and the simulation have the same output given any oracle permutation.

In order to prove Lemma 3 we introduce one more probabilistic experiment: we consider the probabilistic algorithm {S(A)} that simulates {A()} and simulates every oracle query by providing a random answer. (Note that the simulated answers in the computation of {SA} may be incompatible with any permutation.)

We first prove the simple fact that {S(A)} is close to simulating what really happen when {A} interacts with a truly random permutation.

Lemma 4 Let {A} be a non-repeating algorithm of complexity at most {t}. Then

\displaystyle  \left| \mathop{\mathbb P} [ S(A) =1 ] - \mathop{\mathbb P}_{\Pi} [ A^{\Pi, \Pi^{-1} } () = 1] \right| \leq \frac{t^2}{ 2 \cdot 2^{2m}} \ \ \ \ \ (7)

where {\Pi : \{ 0,1 \}^{2m} \rightarrow \{ 0,1 \}^{2m}} is a random permutation.

Finally, it remains to prove:

Lemma 5 For every non-repating algorithm {A} of complexity {\leq t} we have

\displaystyle  \left| \mathop{\mathbb P}_{\overline {R} } [ A^{P_{\overline{R}}, P^{-1} _{\overline{R}} } () =1 ] - \mathop{\mathbb P} [ S(A) = 1] \right| \leq \frac{t^2}{ 2\cdot 2^{2m}} + \frac {t^2}{ 2^{m}}

It is clear that Lemma 3 follows Lemma 4 and Lemma 5.

We now prove Lemma 4.

\displaystyle  \begin{array}{rcl}  && \left| \mathop{\mathbb P} [ S(A) =1 ] - \mathop{\mathbb P}_{\Pi} [ A^{\Pi, \Pi^{-1} } () = 1] \right| \\ &\leqslant & \mathop{\mathbb P}[\mbox{when simulating }S \mbox{, get answers inconsistent with any permutation}]\\ &\leqslant & \frac1{2^{2m}}(1+2+\cdots+t-1)\\ &= & \binom t2\frac1{2^{2m}}\\ &\leqslant & \frac{t^2}{2\cdot2^{2m}}. \end{array}

We shall prove Lemma 5 next time.

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