A Fast Recursive Algorithm and Architecture for Pruned Bit-reversal Interleavers

Abstract

In this paper, pruned bit-reversal permutations employed in variable-length interleavers and their associated fast pruning algorithms and architectures are considered. Pruning permutations is mathematically formulated as a counting problem in a set of k integers and any subset of $\alpha$ consecutive integers under some permutation, where integers from this subset that map into indices less than some $\beta <k$ are to be counted. A solution to this problem using sums involving integer floors and related functions is proposed. It is shown that these sums can be evaluated recursively using integer operations. Specifically, a mathematical treatment for bit-reversal permutations (BRPs) and their permutation statistics are presented. These permutations have been mainly addressed using numerical techniques in the literature to speed up in-place computations of fast Fourier and related transforms. Closed-form expressions for BRP statistics including inversions, serial correlations, and a new statistic called permutation inliers that characterizes the pruning gap of pruned interleavers, are derived. Using the inliers statistic, a recursive algorithm that computes the minimum number of inliers in a pruned BR interleaver (PBRI) in logarithmic time complexity is presented. This algorithm enables parallelizing a serial PBRI algorithm by any desired parallelism factor by computing the pruning gap in lookahead rather than a serial fashion, resulting in significant reduction in interleaving latency and memory overhead. Extensions to 2-D block and stream interleavers are also presented. Moreover, efficient hardware architectures for the proposed algorithms employing simple logic gates are presented. Simulation results of interleavers employed in modern communication standards demonstrate 3 to 4 orders of magnitude improvement in interleaving time compared to existing approaches.

Appendices

Appendix

We start by proving the following lemma that will be used in subsequent proofs:

Lemma 13

Let k be even and $-k< c<k$ . Then for $x=0,\cdots ,k/2-1$ , we have

$$ \begin{array}{rll}\label{eq50} \Lambda&\triangleq&\left(\left(\frac{2x+c}{k}\right)\right) - \left(\left(\frac{2x+c+1}{k}\right)\right)\\ &=& \left\{ \begin{array}{ll} -\frac{1}{k} + {\frac{1}{2}}\delta\left( \frac{2x+c+1}{k} \right), & c\; odd; \\ -\frac{1}{k} + {\frac{1}{2}}\delta\left( \frac{2x+c}{k} \right), & c\; even. \end{array} \right.\end{array} $$

(37)

Proof

First write $\Lambda =-\frac {1}{k}-\left \lfloor \frac {2x+c}{k} \right \rfloor +\left \lfloor \frac {2x+c+1}{k} \right \rfloor +{\frac {1}{2}}\delta \left ( \frac {2x+c}{k} \right )-{\frac {1}{2}}\delta \left ( \frac {2x+c+1}{k} \right )$. Case c odd: Then $2x+c\neq 0$ (odd) and $2x+c+1\neq 0$ (even). If $2x+c+1<k$ or $2x+c+1>k$, then $\left \lfloor \frac {2x+c}{k} \right \rfloor =\left \lfloor \frac {2x+c+1}{k} \right \rfloor$, so $\Lambda =-1/k$. Otherwise if $2x+c+1=k$, then $\Lambda =-1/k+1/2$. Therefore, $\Lambda =-\frac {1}{k} + {\frac {1}{2}}\delta \left ( \frac {2x+c+1}{k} \right )$. Case 2 c even: Then $2x+c+1\neq 0$. If $2x+c= 0$ or $2x+c=k$, then $\Lambda =-1/k+1/2$. Otherwise, if $2x+c<k$ and $2x+c\neq 0$, or $2x+c>k$, then $\left \lfloor \frac {2x+c}{k} \right \rfloor =\left \lfloor \frac {2x+c+1}{k} \right \rfloor$, so $\Lambda =-1/k$. Therefore, $\Lambda =-\frac {1}{k} + {\frac {1}{2}}\delta \left ( \frac {2x+c}{k} \right )$. □

Proof of Lemma 6

From Eq. 17, it is obvious that if $c=0$ or $b=0$, then $T(k,b,c)=0$. Hence in the following we assume $c\neq 0, b\neq 0$. Using the following property of bit-reversal permutations

$$ \pi_n(j) = \left\{ \begin{array}{ll}2\pi_{n-1}(j), & j=0,1,\cdots,k/2-1; \\2\pi_{n-1}(j)+1, & j=k/2,\cdots,k-1, \end{array} \right.$$

(38)

we can split $T(k,b,c)$ into two sums and then apply Eq. 37 with $c=0$ to obtain

$$ \begin{array}{rll}T(k,b,c) &=& 4k\sum\limits_{j=0}^{k/2-1}\left[\left(\left(\frac{j-b}{k}\right)\right)-\left(\left(\frac{j}{k}\right)\right) \right] \left[\left(\left(\frac{2\pi_{n-1}(j)-c}{k}\right)\right)-\left(\left(\frac{2\pi_{n-1}(j)-c+1}{k}\right)\right) -\left\{ -\frac{1}{k} + {\frac{1}{2}}\delta\left( \frac{2\pi_{n-1}(j)}{k} \right) \right\} \right]\\ && + 4k\sum\limits_{j=0}^{k/2-1} \left[\left(\left(\frac{2(j-b)}{k}\right)\right)- \left(\left(\frac{2j}{k}\right)\right)\right] \left[\left(\left(\frac{2\pi_{n-1}(j)-c+1}{k}\right)\right) - \left\{ \left(\left(\frac{2\pi_{n-1}(j)}{k}\right)\right) +\frac{1}{k}-{\frac{1}{2}}\delta\left( \frac{2\pi_{n-1}(j)}{k} \right)\right\} \right]\end{array} $$

Denote by $T_1$ the first sum, and by $T_2$ the second.

Case 1: When c is odd, applying Eq. 37 again, $T_1$ and $T_2$ reduce to $T_1=2k\left [\left (\left (\frac {c^*-b}{k}\right )\right ) - \left (\left (\frac {c^*}{k}\right )\right ) +\left (\left (\frac {\vphantom {c^*}b}{k}\right )\right ) \right ]$ and $T_2=2T(k/2,b,(c-1)/2) - 2k\left (\left (\frac {\vphantom {c^*}2b}{k}\right )\right )$, where $c^*=\pi _{n-1}((c-1)/2)$. Adding the two and using Eq. 14, we obtain $T(k,b,c) = 2T(k/2,b,(c-1)/2) + 2k\left [ \left (\left (\frac {c^*-b}{k}\right )\right ) - \left (\left (\frac {c^*}{k}\right )\right ) -\left (\left (\frac {\vphantom {c^*}b}{k}+{\frac {1}{2}}\right )\right ) \right ]$. Simplifying the saw functions using Eq. 11, the first equation in Eq. 18 follows. Moreover, it is easy to show that these saw functions evaluate to either $-4b,-4b+k,-4b+2k,-4b+3k,-4b+4k$ depending on $c^*$ and b. Case 2: When c is even, $T_1$ still simplifies as shown above but with $c^*=\pi _{n-1}(c/2)$, while $T_2$ becomes $T_2=-2k\left [\left (\left (\frac {2(c^*-b)}{k}\right )\right )-\left (\left (\frac {2c^*}{k}\right )\right )+\left (\left (\frac {\vphantom {c^*}2b}{k}\right )\right ) \right ] + 2T(k/2,b,c/2)$. Hence $T(k,b,c) = 2T(k/2,b,c/2) -2k\left [\left (\left (\frac {c^*-b}{k}+{\frac {1}{2}}\right )\right ) - \left (\left (\frac {c^*}{k}+{\frac {1}{2}}\right )\right ) + \left (\left (\frac {\vphantom {c^*}b}{k}+{\frac {1}{2}}\right )\right ) \right ]$. Again, simplifying the saw functions, the second equation in Eq. 18 follows. Moreover, it is easy to show that these saw functions evaluate to either $0,k,2k$ depending on $c^*$ and b. □

Proof of Lemma 7

We split $U(k)$ using Eq. 38 with respect to both j and b, and then apply Eq. 37 to get

$$ \begin{array}{rll} U(k) &=& \sum\limits_{b=0}^{k/2-1}\left\{ \sum\limits_{j=0}^{k/2-1}\left[\left(\left(\frac{j-b}{k}\right)\right)-\left(\left(\frac{j}{k}\right)\right) \right] \left[\left(\left(\frac{2\pi_{n-1}(j)-\pi_n(b)}{k}\right)\right)-\left(\left(\frac{2\pi_{n-1}(j)-\pi_n(b)+1}{k}\right)\right) -\left\{ -\frac{1}{k} + {\frac{1}{2}}\delta\left( \frac{2\pi_{n-1}(j)}{k} \right) \right\} \right]\right.\\ && +\left. \sum\limits_{j=0}^{k/2-1} \left[\left(\left(\frac{2(j-b)}{k}\right)\right)- \left(\left(\frac{2j}{k}\right)\right)\right] \left[\left(\left(\frac{2\pi_{n-1}(j)-\pi_n(b)+1}{k}\right)\right) - \left\{ \left(\left(\frac{2\pi_{n-1}(j)}{k}\right)\right) +\frac{1}{k}-{\frac{1}{2}}\delta\left( \frac{2\pi_{n-1}(j)}{k} \right)\right\} \right]\right\} \\ &&+\sum\limits_{b=k/2}^{k-1}\left\{ \sum\limits_{j=0}^{k/2-1}\left[\left(\left(\frac{j-b}{k}\right)\right)-\left(\left(\frac{j}{k}\right)\right) \right] \left[\left(\left(\frac{2\pi_{n-1}(j)-\pi_n(b)}{k}\right)\right)-\left(\left(\frac{2\pi_{n-1}(j)-\pi_n(b)+1}{k}\right)\right) -\left\{ -\frac{1}{k} + {\frac{1}{2}}\delta\left( \frac{2\pi_{n-1}(j)}{k} \right) \right\} \right]\right.\\ &&\left. + \sum\limits_{j=0}^{k/2-1} \left[\left(\left(\frac{2(j-b)}{k}\right)\right)- \left(\left(\frac{2j}{k}\right)\right)\right] \left[\left(\left(\frac{2\pi_{n-1}(j)-\pi_n(b)+1}{k}\right)\right) - \left\{ \left(\left(\frac{2\pi_{n-1}(j)}{k}\right)\right) +\frac{1}{k}-{\frac{1}{2}}\delta\left( \frac{2\pi_{n-1}(j)}{k} \right)\right\} \right]\right\}\\ &= &\sum\limits_{b=0}^{k/2-1}\left\{ U_1 + U_2 \right\} + \sum\limits_{b=k/2}^{k-1}\left\{ U_3 + U_4 \right\}\end{array} $$

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For $b=0,\cdots ,k/2-1$, then $\pi _n(b)$ is even. Therefore $c_{\text {even}}^* \triangleq \pi _{n-1}\left ( \pi _n(b)/2 \right )= b$. On the other hand, for $b=k/2,\cdots ,k-1$, then $\pi _n(b)$ is odd. Therefore $c_{\text {odd}}^* \triangleq \pi _{n-1}\left ( (\pi _n(b)-1)/2 \right )= b - k/2$. Applying Eq. 37 when $\pi _n(b)$ is even for $U_1$ and $U_2$, we get $U_1=0$ and $U_2=\sum _{j=0}^{k/2-1}\left [\left (\left (\frac {j-b}{k/2}\right )\right )- \left (\left (\frac {j}{k/2}\right )\right )\right ] \left [\left (\left (\frac {\pi _{n-1}(j)-\pi _n(b)/2}{k/2}\right )\right ) - \left (\left (\frac {\pi _{n-1}(j)}{k/2}\right )\right ) \right ]$. Next, applying Eq. 37 when $\pi _n(b)$ odd for $U_3$, and rearranging terms in $U_4$, we get $U_3=\frac {1}{4} - \frac {1}{4}\delta \left ( \frac {b-k/2}{k} \right )$ and $U_4=\sum _{j=0}^{k/2-1} \left [\left (\left (\frac {j-b}{k/2}\right )\right )- \left (\left (\frac {j}{k/2}\right )\right )\right ] \left [\left (\left (\frac {\pi _{n-1}(j)-(\pi _n(b)-1)/2}{k/2}\right )\right ) -\left (\left (\frac {\pi _{n-1}(j)}{k/2}\right )\right )\right ] - {\frac {1}{2}}\left (\left (\frac {\vphantom {\pi _{n-1}(j)}2b}{k}\right )\right )$. After substituting for $U_1,U_2,U_3,U_4$ in Eq. 39, applying Eq. 38 on $\pi _n{(b)}$, and simplifying terms, $U(k)$ reduces to $U(k)=2U(k/2) + \frac {k}{8} - \frac {1}{4}$. Solving this recurrence with initial condition $U(1)=0$, the result in Eq. 19 follows. □

Proof of Lemma 8

When $p=0$, we have $C(k,0) = k^2\sum _{j=0}^{k-1}\left (\left (\frac {\pi _{n}(j)}{k}\right )\right )\left (\left (\frac {\pi _n(j)}{k}\right )\right ) =k^2\sum _{j=0}^{k-1}\left (\left (\frac {\vphantom {\pi _{n}(j)}j}{k}\right )\right )\left (\left (\frac {\vphantom {\pi _{n}(j)}j}{k}\right )\right ) = \frac {k(k-1)(k-2)}{12}$, where the second equality follows the first since the two sum the same elements but in a different order. When $p=k/2$, we split $C(k,k/2)$ and apply Eq. 38 to obtain $C(k,k/2)=8(k/2)^2\sum _{j=0}^{k/2-1}\left (\left (\frac {\pi _{n-1}(j)}{k/2}\right )\right )\left (\left (\frac {\pi _{n-1}(j)}{k/2}\right )\right )=8C(k/2,0)=\frac {k(k-2)(k-4)}{12}$. When $1\leq p<k/2$, we split $C(k,p)$ and apply Eq. 38 to $\pi _n(j)$ and $\pi _n(j+p)$. For simplicity, let $q=j+p$. Then

$$ C(k,p) = k^2\sum\limits_{j=0}^{k/2-1}\left(\left(\frac{2\pi_{n-1}(j)}{k}\right)\right)\left(\left(\frac{2\pi_{n-1}(q)}{k}\right)\right)+ k^2\sum\limits_{j=k/2-p}^{k/2-1}\left(\left(\frac{2\pi_{n-1}(j)}{k}\right)\right) \left[\left(\left(\frac{2\pi_{n-1}(q)+1}{k}\right)\right)- \left(\left(\frac{2\pi_{n-1}(q)}{k}\right)\right)\right] $$

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$$ + k^2\sum\limits_{j=0}^{k/2-1}\left(\left(\frac{2\pi_{n-1}(j)+1}{k}\right)\right)\left(\left(\frac{2\pi_{n-1}(q)+1}{k}\right)\right)+ k^2\sum\limits_{j=k/2-p}^{k/2-1}\left(\left(\frac{2\pi_{n-1}(j)+1}{k}\right)\right) \left[\left(\left(\frac{2\pi_{n-1}(q)}{k}\right)\right)- \left(\left(\frac{2\pi_{n-1}(q)+1}{k}\right)\right)\right] $$

(41)

In Eq. 40, the terms of the first sum when $\pi _n(j+p)=2\pi _{n-1}(j+p)$ are subtracted and replaced by $\pi _n(j+p)=2\pi _{n-1}(j+p)+1$ for $j=k/2-p,\cdots ,k/2-1$ in the second sum. Similarly in Eq. 41. Combining Eqs. 40 and 41, applying Eq. 37 four times, multiplying out terms, and simplifying the resulting expressions, $C(k,p)$ reduces to $C(k,p) = 8C(k/2,p) + k^2/2 - p - k\left [ \pi _{n-1}(k/2-p) + \pi _{n-1}(p) \right ]$. Let $v$ denote the position of the least significant one bit in the binary representation of $p$. Then

$$ \begin{array}{rll}\label{eq53} \pi_{n-1}(k/2-p) + \pi_{n-1}(p) &=& 2^{n-v-1} + 2^{n-v-2} - 1\\ &=& 3k/2^{v+2} - 1 \end{array}$$

(42)

Substituting back in $C(k,p)$ we get $C(k,p) = 8C(k/2,p) +\left (1 - \frac {3}{2^{v+1}}\right )\frac {k^2}{2} +k - p$. Finally, when $k/2<p<k-1$, following a similar derivation as above, we obtain $C(k,p) = 8C(k/2,p) +\left (1 - \frac {3}{2^{v+1}}\right )\frac {k^2}{2} + p$. □

Proof of Lemma 9

Assume $a,b$ are both even. The proof for other cases is similar. Splitting the summation, applying Eq. 38 to split $\pi _n(j)$ and $\pi _n(j+1)$, then adjusting missing terms for $j=k/2-1$ and $j=k-1$, we obtain

$$ \begin{array}{rll} V(k,a,b) &=& k^2\sum\limits_{j=0}^{k/2-1}\left(\left(\frac{2\pi_{n-1}(j)-a}{k}\right)\right)\left(\left(\frac{2\pi_{n-1}(j+1)-b}{k}\right)\right) + k^2\left[\left(\left(\frac{a+1}{k}\right)\right) - \left(\left(\frac{a+2}{k}\right)\right) \right] \left[\left(\left(\frac{b}{k}\right)\right) - \left(\left(\frac{b-1}{k}\right)\right) \right]\\ && + k^2\sum\limits_{j=0}^{k/2-1}\left[\left(\left(\frac{2\pi_{n-1}(j)-a}{k}\right)\right)+\frac{1}{k}-{\frac{1}{2}}\delta\left( \frac{2\pi_{n-1}(j)-a}{k} \right)\right]\left[ \left(\left(\frac{2\pi_{n-1}(j+1)-b}{k}\right)\right)+\frac{1}{k}-{\frac{1}{2}}\delta\left( \frac{2\pi_{n-1}(j+1)-b}{k} \right) \right],\end{array}$$

where Lemma (Eq. 13) is applied twice. Next, multiplying out terms and using property 16, the above expression simplifies to

$$\begin{array}{rll}\hspace*{-1pt} V(k,a,b) &=& 2k^2\sum\limits_{j=0}^{k/2-1} \left(\left(\frac{2\pi_{n-1}(j)-a}{k}\right)\right)\left(\left(\frac{2\pi_{n-1}(j+1)-b}{k}\right)\right) + k^2\left[\left(\left(\frac{a+1}{k}\right)\right) - \left(\left(\frac{a+2}{k}\right)\right) \right] \left[\left(\left(\frac{b}{k}\right)\right) - \left(\left(\frac{b-1}{k}\right)\right) \right]\\&&- \frac{k^2}{2}\left(\left(\frac{2\pi_{n-1}(\pi_{n-1}(b/2)-1)-a}{k}\right)\right) -\frac{k^2}{2}\left(\left(\frac{2\pi_{n-1}(\pi_{n-1}(a/2)+1)-b}{k}\right)\right) - \frac{k}{2} +\frac{k^2}{4}\delta\left( \frac{2\pi_{n-1}(\pi_{n-1}(a/2)+1)-b}{k} \right) \end{array}$$

which is equivalent to Eq. 23 with $a'=a,b'=b,a'' = 2\pi _{n-1}(\pi _{n-1}(a/2)+1)-b, b'' = 2\pi _{n-1}(\pi _{n-1}(b/2)-1)-a$, and $e=1$. □

Proof of Lemma 10

Assume $a,b$ are both even. The proof for other cases is similar. Proceeding similar to Lemma 9, we obtain

$$\begin{array}{rll}W(k,a,b) &=& 8W(k/2,a/2,b/2) + \frac{k^2}{4}\left[\delta\left( \frac{2\pi_{n-1}(\pi_{n-1}(a/2)+1)-b}{k} \right) - \delta\left( \frac{2\pi_{n-1}(\pi_{n-1}(a/2)+1)}{k} \right) - \delta\left( \frac{k/2-b}{k} \right)\right]\\ &&-\frac{k^2}{2} \left[\left(\left(\frac{2\pi_{n-1}(\pi_{n-1}(b/2)-1)-a}{k}\right)\right)-\left(\left(\frac{2\pi_{n-1}(\pi_{n-1}(b/2)-1)}{k}\right)\right) \right.\\&&\left. +\, \left(\left(\frac{2\pi_{n-1}(\pi_{n-1}(a/2)+1)-b}{k}\right)\right) -\left(\left(\frac{2\pi_{n-1}(\pi_{n-1}(a/2)+1)}{k}\right)\right) \right]\\ &&+ \frac{k^2}{2}\left[\left(\left(\frac{k/2-b}{k}\right)\right)-\left(\left(\frac{a+2}{k}\right)\right) +\frac{2}{k}-{\frac{1}{2}}\right] + k^2\left[\left(\left(\frac{a+2}{k}\right)\right)-\left(\left(\frac{a+1}{k}\right)\right)-\frac{1}{k} \right] \left[\left(\left(\frac{b-1}{k}\right)\right)-\left(\left(\frac{b}{k}\right)\right)+\frac{1}{k}-\frac{1}{2} \right]\end{array} $$

Simplifying the saw-fractions using Eq. 11, and noting that $\delta \left ( \frac {2\pi _{n-1}(\pi _{n-1}(a/2)+1)}{k} \right )= \delta \left ( \frac {a+2}{k} \right )$ since both give 1 when $a=k-2$, expression 25 follows with $e_a=e_b=0,e=1,a'=a,b'=b,a''= 2\pi _{n-1}\left (\pi _{n-1}\left (a/2\right )+1\right ), b'' = 2\pi _{n-1}\left (\pi _{n-1}\left (b/2\right )-1\right )$. □

Proof of Theorem 1

First write Eq. 3 as $\#\text {INL}_{\alpha ,\beta } = \sum _{j=0}^{k-1} \left (\left \lfloor \frac {j-\alpha }{k}\right \rfloor -\right . \left . \left \lfloor \frac {j}{k}\right \rfloor \right )$ $\left ( \left \lfloor \frac {\pi _n(j)-\beta }{k} \right \rfloor - \left \lfloor \frac {\pi (j)}{k} \right \rfloor \right )$, then replace the floor functions with $\left (\left (\cdot \right )\right )$. Multiplying out terms in the summation and using Eq. 15, Eq. 16 we obtain $\#\text {INL}_{\alpha ,\beta } = \frac {\alpha \beta }{k} + \sum _{j=0}^{k-1}\left (\left (\frac {\vphantom {\pi (j)}j-\alpha }{k}\right )\right ) \left (\left (\frac {\pi (j)-\beta }{k}\right )\right )-\sum _{j=0}^{k-1}\left (\left (\frac {\vphantom {\pi (j)}j-\alpha }{k}\right )\right ) \left (\left (\frac {\pi (j)}{k}\right )\right ) - \sum _{j=0}^{k-1}\left (\left (\frac {\vphantom {\pi (j)}j}{k}\right )\right ) \left (\left (\frac {\pi (j)-\beta }{k}\right )\right )+ \sum _{j=0}^{k-1}\left (\left (\frac {\vphantom {\pi (j)}j}{k}\right )\right ) \left (\left (\frac {\pi (j)}{k}\right )\right ) + K_{\text {INL}}(\alpha ,\beta )$, where

$$\begin{array}{rll}K_{\text{INL}}(\alpha,\beta) &\triangleq& - {\frac{1}{2}}\left(\left(\frac{\beta'-\alpha}{k}\right)\right)-{\frac{1}{2}}\left(\left(\frac{\vphantom{\beta}\alpha}{k}\right)\right) + {\frac{1}{2}}\left(\left(\frac{\beta'}{k}\right)\right) \\ &&- {\frac{1}{2}}\left(\left(\frac{\pi(\alpha) -\beta}{k}\right)\right)+ {\frac{1}{2}}\left(\left(\frac{\pi(\alpha)}{k}\right)\right) \\&&+\, \frac{1}{4}\delta\left( \frac{\pi(\alpha) - \beta}{k} \right) -\frac{1}{4}\delta\left( \frac{\pi(\alpha)}{k} \right)\\ &&- {\frac{1}{2}}\left(\left(\frac{\beta}{k}\right)\right) - \frac{1}{4}\delta\left( \frac{-\beta}{k} \right) + \frac{1}{4}\end{array}$$

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and $\beta '=\pi ^{-1}(\beta )$. Equation 43 can be further simplified by expanding $\left (\left (\cdot \right )\right )$ in terms of floor functions, resulting in Eq. 27. The condition $\pi (0)=0$ slightly simplifies the expression for the constant $K_{\text {INL}}$ but does not make the result less general. □

Proof of Corollary 2

We have $\sum _{j=0}^{k-1}\left \lfloor \frac {j-\alpha }{k} \right \rfloor \left \lfloor \frac {\pi _n(j+1)-\beta }{k} \right \rfloor = \sum _{j=1}^{k}\left \lfloor \frac {j-(\alpha +1)}{k} \right \rfloor$ $\left \lfloor \frac {\pi _n(j)-\beta }{k} \right \rfloor$ after change of variables. The kth term of the second sum is 0 since $\alpha \neq 0$. Adding and subtracting the 0th term, $\left \lfloor \frac {-\beta }{k} \right \rfloor =-1$ (since $\beta \neq 0$), the result follows. □

Proof of Theorem 5

$$\begin{array}{rll} \text{Cov}(X_i,X_{i+p}) &=& \frac{1}{k}\sum\limits_{j=0}^{k-1}\left(\pi_n(j)-\text{E}\left[ X_{j} \right]\right)\\ &&\times \left(\pi_n(j+p)-\text{E}\left[ X_{j+p} \right]\right) \\ &=& k\sum_{j=0}^{k-1}\left[\frac{\pi_n(j)}{k}-{\frac{1}{2}}+\frac{1}{2k}\right]\\&&\times\left[\frac{\pi_n(j+p)}{k}-{\frac{1}{2}}+\frac{1}{2k}\right]\end{array} $$

for $0<p<k$. Writing the summand terms using $\left (\left (\cdot \right )\right )$, multiplying out terms, and using $C(k,p)$ in Eq. 20, the above sum reduces to $\text {Cov}(X_i,X_{i+p})=\frac {1}{k}C(k,p)-\frac {1}{4} +\frac {k}{2} -\frac {1}{2}\left [\pi _n(k-p) + \pi _n(p)\right ]$. Let $0\leq v<n$ be the position of the least-significant one-bit in the binary representation of $p$ (starting from 0). Substituting $\pi _n(k-p) + \pi _n(p)=3k/2^{v+1}-1$ similar to Eqs. 42, 35 follows. □

Proof of Theorem 6

We first derive bounds on $T(k,\alpha ,\beta )$ in Eq. 17. Table 2 lists the first few terms of the minimum ($T_{\text {min}}(k)$) and maximum values ($T_{\text {max}}(k)$) of $T(k,\alpha ,\beta )$ empirically. It is easy to show by induction that $T_{\text {min}}(k)$ and $T_{\text {max}}(k)$ satisfy the recursions $T_{\text {min}}(k) = 2T_{\text {min}}(k/2) - ((2+\sqrt {2})\pm (2-\sqrt {2}))\sqrt {k} + 4$ and $T_{\text {max}}(k) = 2T_{\text {max}}(k/2) + 4(k\mp 4)/3$ for $k>2$ with initial conditions $T_{\text {min}}(2)=2$ and $T_{\text {max}}(2)=2$, where $\pm$ and $\mp$ represent cases when n is even/odd. Solving the recursions, we obtain the bound:

$$ \begin{array}{rll}&&\label{eq:Tmin_max_bound}-3k - 4 + ((4+3\sqrt{2})\pm(4-3\sqrt{2}))\sqrt{k} ~\leq~ T(k,\alpha,\beta)\\&&\quad\leq (12n-11)k/9 \mp 16/9\end{array} $$

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Table 2 Minimum and maximum values of $T(k,{\alpha },{\beta })$.

Let $\Delta ^{(t)}$ be the minimum integer added to $\alpha$ in Algorithm 3 at iteration $t$. Then at iteration $t+1$, $\Delta ^{(t+1)} = \#\text {OUL}_{\alpha +\Delta ^{(t)},\beta } = (\alpha +\Delta ^{(t)})-\#\text {INL}_{\alpha +\Delta ^{(t)},\beta } = (\alpha +\Delta ^{(t)})-(\alpha +\Delta ^{(t)})\beta /k-T(k,\alpha +\Delta ^{(t)},\beta )/4k - K_{\text {INL}}$from Eqs. 2, 28. Substituting the maximum and minimum values from Eq. 44 in this equation, and using the maximum and minimum values of $K_{\text {INL}}$ in Eq. 29, we obtain

$$\begin{array}{rll}\label{eq:delta_t+1_bound} && (\alpha+\Delta^{(t)})(1-\beta/k) + W_l ~\leq~ \Delta^{(t+1)} \\ &&\leq (\alpha+\Delta^{(t)})(1-\beta/k) + W_u,\end{array} $$

(45)

where $W_l=(11-12n)/36 \pm 4/9k -3/4$ and $W_u =1-$ $((1+3\sqrt {2}/4)\pm (1-3\sqrt {2}/4))/\sqrt {k}+1/k$. To determine the convergence rate, we study the convergence of the bounds in Eq. 45. The solution of the lower-bound recursion $\Delta _l^{(t+1)} =(\alpha +\Delta _{l}^{(t)})(1-\beta /k) + W_l$ is the sum of a geometric series $\Delta _{l}^{(t)} = \alpha \left (\left (1-\left (1-\beta /k\right )^{t+1}\right )k / \beta -1\right ) +\left (1-\left (1-\beta /k\right )^{t}\right )W_l k / \beta$ which converges to $\Delta _l^* \triangleq \lim _{t\rightarrow \infty }\Delta _{l}^{(t)} = \alpha (k/\beta -1)+W_lk/\beta$ at a rate $\left | \Delta _l^{(t+1)} - \Delta _l^* \right |/\left | \Delta _{l}^{(t)} - \Delta _l^* \right | = 1-\beta /k$. Similar equations hold for the upper bound recursion $\Delta _{u}^{(t)}$ and $\Delta _u^*$ with all subscripts l replaced by $u$. Hence $\Delta ^{(t)}$ converges at a rate $1-\beta /k$. □