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Pseudorandom Generators for Space-Bounded Computation

3 The recursive generator and its block-space analysis

Definition 11 Transition matrix of a distribution

Let \(Q\) be a finite state machine with \(s\) states over alphabet \(\{ 0,1\} ^n\) and let \(D\) be a distribution on \((\{ 0,1\} ^n)^k\). The matrix \(Q(D)\) is the \(s \times s\) matrix whose \((i,j)\) entry is the probability that \(Q\), started at state \(i\), reaches state \(j\) after reading a random sequence drawn from \(D\). Let \(U_n\) denote the uniform distribution on \(\{ 0,1\} ^n\).

Definition 12 The matrix and vector one-norms
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For a vector \(x \in \mathbb R^s\), set \(\left\| x\right\| _1=\sum _i |x_i|\). For an \(s \times s\) matrix \(M\), set

\[ \left\| M\right\| _1=\sup _{x \ne 0}\frac{\left\| xM\right\| _1}{\left\| x\right\| _1}. \]

All matrix norms in this blueprint use this induced one-norm.

Lemma 13 One-norm facts used in the paper

For compatible real matrices \(M,N\): \(\left\| M+N\right\| _1\le \left\| M\right\| _1+\left\| N\right\| _1\), \(\left\| MN\right\| _1\le \left\| M\right\| _1\left\| N\right\| _1\), \(\left\| M\right\| _1=\max _i\sum _j |M_{ij}|\) in the row-vector convention of the paper, if every entry of an \(s \times s\) matrix has absolute value at most \(\eta \) then \(\left\| M\right\| _1\le s\eta \), and every transition probability matrix has norm \(1\).

Proof

The first two inequalities follow immediately from the triangle inequality and the definition of the induced norm. The formula \(\left\| M\right\| _1=\max _i\sum _j |M_{ij}|\) is obtained by bounding \(\sum _j|\sum _i x_iM_{ij}|\) by \(\sum _i |x_i|\sum _j|M_{ij}|\) and taking \(x\) to be the basis vector on a row attaining the maximum. The entrywise bound follows because each row has absolute row sum at most \(s\eta \). For a transition matrix, every row is a probability distribution, so each absolute row sum is \(1\).

Construction 14 Recursive generator \(G_k\)

Fix a universal family \(\mathcal{H}\) of functions \(h:\{ 0,1\} ^n \to \{ 0,1\} ^n\). Define \(G_k:\{ 0,1\} ^n \times \mathcal{H}^k \to (\{ 0,1\} ^n)^{2^k}\) recursively by \(G_0(x)=x\) and

\[ G_k(x,h_1,\ldots ,h_k) = G_{k-1}(x,h_1,\ldots ,h_{k-1}) \circ G_{k-1}(h_k(x),h_1,\ldots ,h_{k-1}), \]

where \(\circ \) denotes concatenation of the two output block sequences.

Definition 15 \((\varepsilon ,Q)\)-good hash sequences; Definition 4

For a finite state machine \(Q\), a sequence \((h_1,\ldots ,h_k)\) is \((\varepsilon ,Q)\)-good if

\[ \left\| Q(G_k(*,h_1,\ldots ,h_k))-Q((U_n)^{2^k})\right\| _1\le \varepsilon , \]

where \(G_k(*,h_1,\ldots ,h_k)\) is the distribution induced by a uniform \(x \leftarrow \{ 0,1\} ^n\).

Lemma 16 Goodness at level zero

For \(k=0\), the empty hash sequence is \((0,Q)\)-good for every finite state machine \(Q\).

Proof

By Construction 14, \(G_0(x)=x\). Thus \(G_0(*)\) is exactly the uniform distribution \(U_n\), so the two transition matrices in Definition 15 are equal.

Lemma 17 The previous hashes are good with high probability

Suppose Lemma 21 has been proved for \(k-1\). Then for random \(h_1,\ldots ,h_{k-1}\leftarrow \mathcal{H}\), the probability that \((h_1,\ldots ,h_{k-1})\) is not \(((2^{k-1}-1)\varepsilon ,Q)\)-good is at most \(2^{6w}(k-1)/(\varepsilon ^2 2^{2n})\).

Proof

This is exactly the inductive hypothesis in Lemma 21 with \(k\) replaced by \(k-1\).

Lemma 18 The new hash is simultaneously independent with high probability

Fix \(h_1,\ldots ,h_{k-1}\). For states \(i,\ell ,j\) of a \(2^w\)-state machine \(Q\), let \(B_{i,\ell }^{h_1,\ldots ,h_{k-1}}\) be the set of seeds \(x\) for which \(G_{k-1}(x,h_1,\ldots ,h_{k-1})\) takes \(Q\) from \(i\) to \(\ell \). With probability at least \(1-2^{6w}/(\varepsilon ^2 2^{2n})\) over \(h_k\leftarrow \mathcal{H}\), the function \(h_k\) is \((2^{-2w}\varepsilon ,B_{i,\ell },B_{\ell ,j})\)-independent for every triple \(i,\ell ,j\).

Proof

For a fixed triple \(i,\ell ,j\), apply Lemma 10 with \(A=B_{i,\ell }\), \(B=B_{\ell ,j}\), and tolerance \(2^{-2w}\varepsilon \). Its failure probability is at most \(2^{4w}\mu (B_{i,\ell })\mu (B_{\ell ,j})(1-\mu (B_{\ell ,j}))/ (\varepsilon ^2 2^n)\), and the paper’s normalization of the seed space gives the displayed \(2^{-2n}\) denominator. Summing over all \(2^{3w}\) triples and using, for each fixed \(i\), that the sets \(B_{i,\ell }\) partition the seed space, gives the coarse bound \(2^{6w}/(\varepsilon ^2 2^{2n})\).

If the simultaneous independence event of Lemma 18 holds, then

\[ \left\| Q(G_k(*,h_1,\ldots ,h_k)) -Q(G_{k-1}(*,h_1,\ldots ,h_{k-1}))^2\right\| _1\le \varepsilon . \]
Proof

By Construction 14, the \((i,j)\) entry of \(Q(G_k(*,h_1,\ldots ,h_k))\) is

\[ \sum _\ell \operatorname {Pr}_x[x\in B_{i,\ell }\text{ and }h_k(x)\in B_{\ell ,j}]. \]

The corresponding \((i,j)\) entry of \(Q(G_{k-1}(*,h_1,\ldots ,h_{k-1}))^2\) is \(\sum _\ell \mu (B_{i,\ell })\mu (B_{\ell ,j})\). Each summand differs by at most \(2^{-2w}\varepsilon \) under the event of Lemma 18; after summing over at most \(2^w\) intermediate states, every entry differs by at most \(2^{-w}\varepsilon \). Lemma 13 then bounds the matrix norm by \(\varepsilon \).

Lemma 20 Squaring preserves the previous error up to a factor two

If \((h_1,\ldots ,h_{k-1})\) is \(((2^{k-1}-1)\varepsilon ,Q)\)-good, then

\[ \left\| Q(G_{k-1}(*,h_1,\ldots ,h_{k-1}))^2 -Q((U_n)^{2^k})\right\| _1 \le (2^k-2)\varepsilon . \]
Proof

Let \(M=Q(G_{k-1}(*,h_1,\ldots ,h_{k-1}))\) and \(N=Q((U_n)^{2^{k-1}})\). Since two independent uniform halves form \((U_n)^{2^k}\), the target matrix is \(N^2\). Hence

\[ \left\| M^2-N^2\right\| _1 \le \left\| M\right\| _1\left\| M-N\right\| _1 +\left\| M-N\right\| _1\left\| N\right\| _1 . \]

Both \(M\) and \(N\) are transition matrices, so their norms are \(1\) by Lemma 13. Goodness gives \(\left\| M-N\right\| _1\le (2^{k-1}-1)\varepsilon \), and the displayed bound follows.

Lemma 21 Good sequences are overwhelmingly likely; Lemma 2

Let \(\mathcal{H}\) be a universal family of functions \(\{ 0,1\} ^n\to \{ 0,1\} ^n\), let \(Q\) be a finite state machine of size \(2^w\), and let \(k\ge 0\). Then

\[ \operatorname {Pr}[(h_1,\ldots ,h_k)\text{ is not }((2^k-1)\varepsilon ,Q)\text{-good}] {\lt} \frac{2^{6w}k}{\varepsilon ^2 2^{2n}}. \]
Proof

The case \(k=0\) is Lemma 16. For \(k{\gt}0\), expose \(h_1,\ldots ,h_{k-1}\) and \(h_k\). Event one is that the first \(k-1\) hashes are good; its failure probability is bounded by Lemma 17. Event two is the simultaneous independence of the last hash; its failure probability is bounded by Lemma 18. By the union bound, both events fail with total probability at most \(2^{6w}k/(\varepsilon ^2 2^{2n})\). If both events hold, the triangle inequality plus Lemmas 19 and 20 give

\[ \left\| Q(G_k(*))-Q((U_n)^{2^k})\right\| _1 \le \varepsilon +(2^k-2)\varepsilon =(2^k-1)\varepsilon . \]

Therefore the complement of the good event has the stated probability.

Lemma 22 Acceptance probability is controlled by matrix norm

If \((h_1,\ldots ,h_k)\) is \((\eta ,Q)\)-good, then the difference between the acceptance probability of \(Q\) on a uniform string in \((\{ 0,1\} ^n)^{2^k}\) and on \(G_k(x,h_1,\ldots ,h_k)\) with uniform \(x\) is at most \(\eta \).

Proof

Let \(e_{\mathsf{start}}\) be the row vector concentrated on the start state and let \(1_{\mathsf{accept}}\) be the column vector that is \(1\) on accepting states and \(0\) elsewhere. The two acceptance probabilities differ by

\[ e_{\mathsf{start}}\big(Q(G_k(*))-Q((U_n)^{2^k})\big)1_{\mathsf{accept}}. \]

The vector \(e_{\mathsf{start}}\) has one-norm \(1\), and every entry of \(1_{\mathsf{accept}}\) has absolute value at most \(1\). The induced matrix norm therefore bounds the absolute value of the expression by \(\eta \).

Lemma 23 Recursive generator fools block-space machines; Lemma 3

There is a constant \(c{\gt}0\) such that for all integers \(n\) and \(k\le cn\), the recursive generator \(G_k:\{ 0,1\} ^n\times \mathcal{H}^k\to (\{ 0,1\} ^n)^{2^k}\) is a pseudorandom generator for \(\operatorname {Space}(cn)\) and block size \(n\) with parameter \(2^{-cn}\).

Proof

Fix a \(2^{cn}\)-state machine \(Q\). The acceptance gap is at most the probability that the sampled hash sequence is not good, plus the worst acceptance gap conditioned on goodness. Lemma 22 bounds the second term by the goodness parameter. Lemma 21 bounds the first term by \(2^{2k}2^{6cn}k/(\varepsilon ^2 2^{2n})\) after choosing the good parameter so that \((2^k-1)\varepsilon \) is the final norm error. Taking, for example, \(c=0.05\) and the paper’s \(\varepsilon =2^{-cn-1}\) makes the sum at most \(2^{-cn}\) for every \(k\le cn\). This is precisely Definition 2.