** Summary **

Today we begin a tour of the theory of one-way functions and pseudorandomness.

The highlight of the theory is a proof that if *one-way functions* exist (with good asymptotic security) then pseudorandom permutations exist (with good asymptotic security). We have seen that pseudorandom permutations suffice to do encryption and authentication with extravagantly high levels of security (respectively, CCA security and existential unforgeability under chosen message attack), and it is easy to see that if one-way functions do not exist, then every encryption and authentication scheme suffers from a total break.

Thus the conclusion is a strong “dichotomy” result, saying that either cryptography is fundamentally impossible, or extravagantly high security is possible.

Unfortunately the proof of this result involves a rather inefficient reduction, so the concrete parameters for which the dichotomy holds are rather unrealistic. (One would probably end up with a system requiring gigabyte-long keys and days of processing time for each encryption, with the guarantee that if it is not CCA secure then every 128-bit key scheme suffers a total break.) Nonetheless it is one of the great unifying achievements of the asymptotic theory, and it remains possible that a more effective proof will be found.

In this lecture and the next few ones we shall prove the weaker statement that if *one-way permutations* exist then pseudorandom permutations exist. This will be done in a series of four steps each involving reasonable concrete bounds. A number of combinatorial and number-theoretic problems which are believed to be intractable give us highly plausible candidate one-way permutations. Overall, we can show that if any of those well-defined and well-understood problems are hard, then we can get secure encryption and authentication with schemes that are slow but not entirely impractical. If, for example, solving discrete log with a modulus of the order of is hard, then there is a CCA-secure encryption scheme requiring a -bit key and fast enough to carry email, instant messages and probably voice communication. (Though probably too slow to encrypt disk access or video playback.)

**1. One-way Functions and One-way Permutations **

A one-way function is a function such that, for a random , it is hard to find a pre-image of .

Definition 1 (One-way Function)A function is -one way if for every algorithm of complexity we have

In the asymptotic theory, one is interested in one-way functions that are defined for all input lengths and are efficiently computable. Recall that a function is called *negligible* if for every polynomial we have .

Definition 2 (One-way Function — Asymptotic Definition)A function is one-way if

- is polynomial time computable and
- for every polynomial there is a negligible function such that for all large enough the function is -one way.

Example 1 (Subset Sum)On input , where , parses as a sequence of integers, each -bit long, plus a subset .The output is

Some variants of subset-sum have been broken, but it is plausible that is a -one way function with and super-polynomial in , maybe even as large as .

Exercise 1Let be a -secure one-way function. Show that

Definition 3 (One-way Permutation)If is a bijective -one way function, then we call a –one-way permutation.

If is an (asymptotic) one-way function, and for every is a bijection from into , then we say that is an (asymptotic) one-way permutation.

There is a non-trivial general attack against one-way permutations.

Exercise 2Let be a -secure one-way permutation. Show that

This means that we should generally expect the input length of a secure one-way permutation to be at least 200 bits or so. (Stronger attacks than the generic one are known for the candidates that we shall consider, and their input length is usually 1000 bits or more.)

Example 2 (Modular Exponentiation)Let be a prime, and be the group whose elements are and whose operation is multiplication . It is a fact (which we shall not prove) that is cyclic, meaning that there is an element such that the mappingis a permutation on . Such an element is called a generator, and in fact most elements of are generators. is conjectured to be one-way for most choices of and .

The problem of inverting is called the discrete logarithm problem.

The best known algorithm for the discrete logarithm is conjectured to run in time . It is plausible that for most and most the discrete logarithm is a one way permutation with and of the order of .

Problems like exponentiation do not fit well in the asymptotic definition, because of the extra parameters . (Technically, they do not fit our definitions at all because the input is an element of instead of a bit string, but this is a fairly trivial issue of data representation.) This leads to the definition of *family of one-way functions (and permutations)*.

**2. A Preview of What is Ahead **

Our proof that a pseudorandom permutation can be constructed from any one-way permutation will proceed via the following steps:

- We shall prove that for any one-way permutation we can construct a
*hard-core predicate*, that is a predicate such that is easy to compute given , but it is hard to compute given . - From a one-way function with a hard-core predicate, we shall show how to construct a pseudorandom generator with
*one-bit expansion*, mapping bits into . - From a pseudorandom generator with one-bit expansion, we shall show how to get generators with essentially
*arbitrary expansion*. - From a length-doubling generator mapping bits into , we shall show how to get
*pseudorandom functions*. - For a pseudorandom function, we shall show how to get
*pseudorandom permutations*.

**3. Hard-Core Predicate **

Definition 4 (Hard-Core Predicate)A boolean function is -hard core for a permutation if for every algorithm of complexity

Note that only one-way permutations can have efficiently computable hard-core predicates.

Exercise 3Suppose that is a -hard core predicate for a permutation , and is computable in time . Show that is -one way.

It is known that if is one-way, then every bit of is hard-core.

Our first theorem will be that a random XOR is hard-core for every one-way permutation.

We will use the following notation for “inner product” modulo 2:

Theorem 5 (Goldreich and Levin)Suppose that is an algorithm of complexity such that

We begin by establishing the following weaker result.

Theorem 6 (Goldreich and Levin — Weak Version)Suppose that is an algorithm of complexity such thatThen there is an algorithm of complexity at most such that

Before getting into the proof of Theorem 6, it is useful to think of the “super-weak” version of the Goldreich-Levin theorem, in which the right-hand-side in (3) is 1. Then inverting is very easy. Call the vector that has in the -th position and zeroes everywhere else, thus . Now, given and an algorithm for which the right-hand-side of (3) is 1, we have for every , and so we can compute given via invocations of . In order to prove the Goldreich-Levin theorem we will do something similar, but we will have to deal with the fact that we only have an algorithm that approximately computes inner products.

We derive the Weak Goldreich-Levin Theorem from the following reconstruction algorithm.

Lemma 7 (Goldreich-Levin Algorithm — Weak Version)There is an algorithm that given oracle access to a function such that, for some ,

runs in time , makes queries into , and with probability outputs .

Before proving Lemma 7, we need to state the following version of the *Chernoff Bound*.

Lemma 8 (Chernoff Bound)Let be mutually independent random variables. Then, for every , we have

*Proof:* We only give a sketch. Let . Then we want to prove that

For every fixed , Markov’s inequality gives us

We can use independence to write

and some calculus shows that for every we have

So we get

Equation (5) holds for every , and in particular for giving us

as desired.

We can proceed with the design and the analysis of the algorithm of Lemma 7.

*Proof:* [Of Lemma 7] The idea of the algorithm is that we would like to compute for , but we cannot do so by simply evaluating , because it is entirely possible that is incorrect on those inputs. If, however, we were just interested in computing for a random , then we would be in good shape, because would be correct with resonably large probability. We thus want to *reduce the task of computing on a specific , to the task of computing for a random .* We can do so by observing the following identity: for every and every , we have

where all operations are mod 2. (And bit-wise, when involving vectors.) So, in order to compute we can pick a random , and then compute . If is uniformly distributed, then and are uniformly distributed, and we have

Suppose now that we pick independently several random vectors , and that we compute for and we take the majority value of the as our estimate for . By the above analysis, each equals with probability at least ; furthermore, the events are mutually independent. We can then invoke the Chernoff bound to deduce that the probability that the majority value is wrong is at most . (If the majority vote of the is wrong, it means that at least or the are wrong, even though the expected number of wrong ones is at most , implying a deviation of from the expectation; we can invoke the Chernoff bound with .) The algorithm GLW is thus as follows:

- Algorithm GLW
- for to
- for to
- pick a random

- for to
- return

For every , the probability fails to compute is at most . So the probability that the algorithm fails to return is at most . The algorithm takes time and makes oracle queries into .

In order to derive Theorem 6 from Lemma 7 we will need the following variant of Markov’s inequality.

Lemma 9Let be a discrete bounded non-negative random variable ranging over . Then for every ,

*Proof:* Let be the set of values taken by with non-zero probability. Then

So we have .

We can now prove Theorem 6.

*Proof:* [Of Theorem 6] The assumption of the Theorem can be rewritten as

From Lemma 9 we have

Call an “good” if it satisfies .

The inverter , on input , runs the algorithm GLW using the oracle . If is good, then the algorithm finds with probability at least . At least half of the choices of are good, so overall the algorithm inverts on at least a fraction of inputs. The running time of the algorithm if plus the cost of calls to , each costing time .

one way function

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Lemma 9 looks like a geometric series