**Vol:** 2 **Issue:** 1

**Published In: January 2014**

**Article No: **3 **Page:** 41-52 doi: https://doi.org/10.13052/jsame2245-4551.213

**Very Fast Algorithms for Eliminating the Diffraction Effects in Protein-Based Volumetric Memories**

Received 12 December 2014; Accepted 17 February 2015; Publication 31 March 2015

*Journal of Self-Assembly and Molecular Electronics, Vol. 2*, 41ȁ52.

doi: 10.13052/jsame2245-4551.213

© 2015 *River Publishers. All rights reserved.*

Dragos Trinca^{1} and Sanguthevar Rajasekaran^{2}

^{1}*Sc Piretus Prod Srl, Osoi, Jud. Iasi, 707110, Romania dntrinca@gmail.com*^{2}*Dep. of Computer Science and Engineering, University of Connecticut, Storrs, USA rajasek@engr.uconn.edu*

One of the current research directions in biological nanotechnology is the use of bacteriorhodopsin in the fabrication of protein-based volumetric memories. Bacteriorhodopsin, with its unique light-activated photocycle, nanoscale size, cyclicity (>10^{7}), and natural resistance to harsh environmental conditions, provides for protein-based volumetric memories that have a comparative advantage over magnetic and optical data storage devices. The construction of protein-based volumetric memories has been, however, severely limited by fundamental issues that exist with such devices, such as unwanted diffraction effects. In this paper, we propose some optimizations that can be applied to one of the previously proposed algorithms for eliminating the diffraction effects.

- compression
- biological nanotechnology
- optimization

Much of the current research effort in biological nanotechnology is directed toward self-assembled monolayers and thin films, biosensors, and protein-based photonic devices [1–6]. Although a number of proteins have been explored for device applications, bacteriorhodopsin [3] has received, it seems, the most attention. Bacteriorhodopsin protein memory devices exhibit increased thermal, chemical and photochromic stability, and have the advantage of being radiation-hardened, waterproof, and EMP-resistant. Such devices are capable of storing large amounts of data in a small volume.

Although there are a number of prototype systems and preliminary effort to apply them, the potential of this promising technology is relatively unexplored. Research on protein-based memories started in the late 1980s with considerable anticipation, but enthusiasm decreased shortly afterwards for several reasons. Commercial development of spatial light modulators (SLMs), that are an integral component of protein-based memories, was slow and there were several issues, such as unwanted diffraction effects, that limited performance in three-dimensional memory applications. More recently, however, the development of high-definition television projection equipment has resulted in the commercial availability of high-resolution, high-performance and relatively inexpensive SLMs. But, fundamental problems have remained. Two such problems are diffraction effects and scaling.

The branched-photocycle three-dimensional memory, as discussed in [2, 7], stores data by using a sequential two-photon process to convert bacteriorhodopsin (bR) in the activated region from the bR resting state to the Q state. This process involves using a paging beam to select a thin page of memory and a writing beam that is pixilated at those positions where data are to be written. The transition from the bR resting state to the Q state occurs via the intermediate states K, L, M, N, and O. Only the bR (bit 0) and Q (bit 1) states are stable for long periods of time. By using 32-level grey-scaling and two polarizations, each voxel can store 64 bits. Attempts to use higher levels of grey-scaling have failed, mainly due to the problems of diffraction introduced by having pages with significant differences in the average refractive indices. To understand this problem, we note that protein corresponding to bit 0 has a high refractive index and protein corresponding to bit 1 has a low refractive index with reference to the red laser beam at 633 *nm* that is used to read out the data. Prototypes made of the three-dimensional memory fail to work at high storage densities when individual pages of memory have a preponderance of bits of a given state. Consider the worst case – each page is either all 0’s or all 1’s. Then, a refractive index grating is created, that diffracts the laser beams quite efficiently because individual pages are stored with separations of 6-20 *µm*. While these separations are significantly larger than the diffraction limit would dictate, closer spacing is not possible, due to beam steering inside the data cuvettes. The unwanted beam steering is due to refractive index gradients. Algorithms that can store data at high resolution and that maintain the number of 0’s equal to the number of 1’s would solve this problem [7–12].

One way of ensuring an equal number of 0’s and 1’s is to replace a 0 with 01 and a 1 with 10. However, this procedure would reduce the available memory by a factor of 2 (which means that the *utility factor* would be 50% in this case). In [7] and [9] the authors have proposed some methods that provide utility factors of more than 50%. For completely random data, however, the best utility factor provided by the previously proposed algorithms is about 99.9%, and can be obtained by using the APPROXv1 algorithm proposed in [9], which works as follows. Let *I* be the input data, and *bI* the binary representation of *I*. If *I* is of length *L*, then *bI* is of length 8*L*, as each byte of data is represented binary on 8 bits. Scan *bI* and count the number of 0’s and 1’s. Suppose that *bI* doesn’t have an equal number of 0’s and 1’s. Scan *bI* from left to right, bit by bit. For each position *i* in the scanning, 1 ≤ *i* ≤ 8*L*, do as follows. Let *bI*_{1:i} be the portion already processed, and *bI*_{i+1:8L} the portion that has remained to be processed. Let $\overline{b{I}_{1:i}}$ be the complement of *bI*_{1:i}. If the binary string $\overline{b{I}_{1:i}}b{I}_{i+1:8L}$ has an equal number of 0’s and 1’s, then stop the scanning, let *q* be this first position *i* that leads to a $\overline{b{I}_{1:i}}b{I}_{i+1:8L}$ with an equal number of 0’s and 1’s, and the output of APPROXv1 is $x\overline{b{I}_{1:q}}b{I}_{q+1:8L}$, where *x* is a binary string of length 256 such that the first 120 bits of *x* store *q*, the next 3 bits store the version of the algorithm (in this case 001), the next 5 bits are 00000 (not used in APPROXv1) and the remaining 128 bits are used to make *x* a binary string with an equal number of 0’s and 1’s. So, in conclusion, the length of the output is 8*L*+256. (There is no particular reason for choosing the length of *x* to be 256, just that this value seems to be sufficient in practice.)

In [11], some optimizations that can be applied to APPROXv1 have been proposed. Precisely, the optimizations proposed in [11] apply to the second scanning step (when the position *q* is found), and work as follows. Suppose that *bI* has more 0’s than 1’s. Let *NofZs* be the number of 0’s in *bI*, and *NofOs* the number of 1’s in *bI*. Observe that every position *i* that leads to a $\overline{b{I}_{1:i}}b{I}_{i+1:8L}$ with an equal number of 0’s and 1’s has the property that *bI*_{1:i} has exactly (*i*-((*NofZs* – *NofOs*)/2))/2+(*NofZs* – *NofOs*)/2 0’s and (*i* – ((*NofZs* – *NofOs*)/2))/2 1’s. So, during the scanning, we can skip those positions that for sure don’t have this property, and examine only those positions that may have this property, until the desired position is found. Precisely, the modifications are the following.

- First, we don’t have to start the scanning from position 1. We can start directly with position (
*NofZs*–*NofOs*)/2, the reason being that we have to complement exactly (*NofZs*–*NofOs*)/2 0’s in order to transform*bI*into a binary string with an equal number of 0’s and 1’s. - Second, let
*i*be the current position in the scanning, and let*Zs*[*i*] be the number of 0’s in*bI*_{1:i}. The number of 1’s in*bI*_{1:i}is, clearly,*i*–*Zs*[*i*]. We have two cases, either*Zs*[*i*] ≥ (*i*–*Zs*[*i*]) or*Zs*[*i*] < (*i*–*Zs*[*i*]).- If $Zs[i]\ge (i-Zs[i])$, we check to see if $Zs[i]-(i-Zs[i])=(NofZs-NofOs)/2$. If yes, then the scanning stops, and let
*q*be this first position*i*that leads to a $\overline{b{I}_{1:i}}b{I}_{i+1:8L}$ with an equal number of 0’s and 1’s. Otherwise, if $Zs[i]-(i-Zs[i])\ne (NofZs-NofOs)/2$, then the next position in the scanning is $i+(NofZs-NofOs)/2-(Zs[i]-(i-Zs[i]))$. - If $Zs[i]<(i-Zs[i])$, the next position in the scanning is $i+(NofZs-NofOs)/2+((i-Zs[i])-Zs[i])$ .

- If $Zs[i]\ge (i-Zs[i])$, we check to see if $Zs[i]-(i-Zs[i])=(NofZs-NofOs)/2$. If yes, then the scanning stops, and let

This is the APPROXv2 algorithm recently proposed in [11]. The output in APPROXv2 is $x\overline{b{I}_{1:q}}b{I}_{q+1:8L}$, where the first 120 bits of *x* store *q*, the next 3 bits are 010 (the version of the algorithm, which is 2), the next 5 bits are 00000 (also, not used in APPROXv2), and the remaining 128 bits are used to make *x* a binary string with an equal number of 0’s and 1’s, as usual.

**Example 1** Suppose that

$$bI=01011010110000011010000100001111.$$

*bI has 18 0’s and 14 1’s. APPROXv1 works as follows.*

*Since*$\overline{b{I}_{1:1}}b{I}_{2:32}$*has 17 0’s and 15 1’s, the scanning goes on to the next position.**Since*$\overline{b{I}_{1:2}}b{I}_{3:32}$*has 18 0’s and 14 1’s, the scanning goes on to the next position.**Since*$\overline{b{I}_{1:3}}b{I}_{4:32}$*has 17 0’s and 15 1’s, the scanning goes on to the next position.**Since*$\overline{b{I}_{1:4}}b{I}_{5:32}$*has 18 0’s and 14 1’s, the scanning goes on to the next position.**Since*$\overline{b{I}_{1:5}}b{I}_{6:32}$*has 19 0’s and 13 1’s, the scanning goes on to the next position.**Since*$\overline{b{I}_{1:6}}b{I}_{7:32}$*has 18 0’s and 14 1’s, the scanning goes on to the next position.**Since*$\overline{b{I}_{1:7}}b{I}_{8:32}$*has 19 0’s and 13 1’s, the scanning goes on to the next position.**Since*$\overline{b{I}_{1:8}}b{I}_{9:32}$*has 18 0’s and 14 1’s, the scanning goes on to the next position.**Since*$\overline{b{I}_{1:9}}b{I}_{10:32}$*has 19 0’s and 13 1’s, the scanning goes on to the next position.**Since*$\overline{b{I}_{1:10}}b{I}_{11:32}$*has 20 0’s and 12 1’s, the scanning goes on to the next position.**Since*$\overline{b{I}_{1:11}}b{I}_{12:32}$*has 19 0’s and 13 1’s, the scanning goes on to the next position.**Since*$\overline{b{I}_{1:12}}b{I}_{13:32}$*has 18 0’s and 14 1’s, the scanning goes on to the next position.**Since*$\overline{b{I}_{1:13}}b{I}_{14:32}$*has 17 0’s and 15 1’s, the scanning goes on to the next position.**Since*$\overline{b{I}_{1:14}}b{I}_{15:32}$*has 16 0’s and 16 1’s, we conclude that $q=14$, and the scanning stops.*

**Example 2** *For the same bI, APPROXv2 works as follows.*

*Since we have Zs*[2] = 1*and*2 –*Zs*[2] = 1,*the next position in the scanning is 4.**Since we have Zs*[4] = 2*and*4 –*Zs*[4] = 2,*the next position in the scanning is 6.**Since we have Zs*[6] = 3*and*6 –*Zs*[6] = 3,*the next position in the scanning is 8.**Since we have Zs*[8] = 4*and*8 –*Zs*[8] = 4,*the next position in the scanning is 10.**Since Zs*[10] = 4*and*10 –*Zs*[10] = 6,*the next position in the scanning is 14.**Since Zs*[14] = 8*and*14 –*Zs*[14] = 6*and Zs*[14] – (14 –*Zs*[14]) = 2,*we conclude that q=14.*-
*So, significantly less positions are checked in APPROXv2 as compared to APPROXv1.*

APPROXv2 can be optimized as follows. Assume that, besides the input data, we have a parameter *sl*, such that 1 ≤ *sl* ≤ 8*L*, and such that 8*L* is a multiple of *sl*. Then, we consider the following substrings of *bI*:

$$b{I}_{1:sl},b{I}_{sl+1:2sl},\mathrm{...},b{I}_{8L-sl+1:8L}.$$

So, we are splitting *bI* into $\frac{8L}{sl}$ substrings, each of length *sl*. For each $b{I}_{i\xb7sl-sl+1:i\xb7sl},\text{\hspace{0.17em}}1\le i\le \frac{8L}{sl}$, consider the pairs (*a _{i}*,

$$\begin{array}{l}{a}_{i}=Zs[i\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}sl-sl+1]-(i\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}sl-sl+1-Zs[i\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}sl-sl+1]),\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{b}_{i}=Zs[i\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}sl]-(i\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}sl-Zs[i\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}sl]),\\ {c}_{i}=(i\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}sl-sl+1-Zs[i\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}sl-sl+1])-Zs[i\text{\hspace{0.17em}}\xb7\text{\hspace{0.05em}}\text{\hspace{0.17em}}sl-sl+1],\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{d}_{i}=(i\text{\hspace{0.17em}}\xb7\text{\hspace{0.05em}}\text{\hspace{0.17em}}sl-Zs[i\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}sl])-Zs[i\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}sl].\end{array}$$

So, for each $b{I}_{i\xb7sl-sl+1:i\xb7sl},\text{\hspace{0.17em}}{a}_{i}$ and s*b _{i}* are the differences between the number of 0’s and 1’s at the two ends, while

**Fact 1** *If NofZs* > *NofOs, then there exists at least one i*, $1\le i\le \frac{8L}{sl}$, *such that*

$${a}_{i}\le (NofZs-NofOs)/2\le {b}_{i}.$$

*Otherwise, if NofOs* > *NofZs, then there exists at least one i*, $1\le i\le \frac{8L}{sl}$, *such that*

$${c}_{i}\le (NofOs-NofZs)/2\le {d}_{i}.$$

Given this, APPROXv2 can be optimized as follows. Suppose that *NofZs* > *NofOs*. Let *j* be the first index that satisfies ${a}_{j}\le (NofZs-NofOs)/2\le {b}_{j}$. Then, we can search for the position *q* only in the substring $b{I}_{j\xb7sl-sl+1:j\xb7sl}$, because, at this point, we know that the position *q* we are looking for satisfies $j\xb7sl-sl+1\le q\le j\xb7sl$. This is the APPROXv3 algorithm, proposed in [12]. The output in APPROXv3 is as in APPROXv2, the only difference being *x*_{121:123} = 011 (the version of the algorithm, which is 3).

**Example 3** *For the same bI, and assumming sl* = 8, *APPROXv3 works as follows.*

*First, we compute the pairs:*(*a*_{1}= 1,*b*_{1}= 0), (*a*_{2}= -1,*b*_{2}= 2), (*a*_{3}= 1,*b*_{3}= 4), (*a*_{4}= 5,*b*_{4}= 4).*The first pair*(*a*,_{i}*b*)_{i}*that satisfies*$${a}_{i}\le (NofZs-NofOs)/2\le {b}_{i}$$

is (

*a*_{2},*b*_{2}).*So, we search for the position q only in the substring bI*_{9:16},*and we find q*= 14.

Some results when comparing APPROXv1 with APPROXv2 are given in Table 1. APPROXv3 takes less than one millisecond, in general.

In this paper, we propose some optimizations that can be applied to APPROXv3, and that could lead to faster practical algorithms for eliminating the diffraction effects. Suppose that *NofZs* > *NofOs*, and let *j* be the first index that satisfies

$${a}_{j}\le (NofZs-NofOs)/2\le {b}_{j}.$$

Then, we search for the position *q* in the substring $b{I}_{j\xb7sl-sl+1:j\xb7sl}$, and we find it.

If $q\le \frac{8L}{2}$, then the output is $x\overline{b{I}_{1:q}}b{I}_{q+1:8L}$, where *x* is a binary string of length 256 such that the first 120 bits of *x* store *q*, the next 3 bits are 100, the next 5 bits are 00000, and the remaining 128 bits are used to make *x* a binary string with an equal number of 0’s and 1’s. If $q\ge \text{\hspace{0.17em}}\left(\frac{8L}{2}+1\right)$, then the output is $xb{I}_{1:q}\overline{b{I}_{q+1:8L}}$, where *x* is a binary string of length 256 such that the first 120 bits of *x* store *q*, the next 3 bits are 100, the next 5 bits are 00000, and the remaining 128 bits are used to make *x* a binary string with an equal number of 0’s and 1’s.

Denote this version of APPROXv3 by APPROXv4. Clearly, in the case $q\le \frac{8L}{2}$, APPROXv4 is basically APPROXv3. But, when $q\ge \text{\hspace{0.17em}}\left(\frac{8L}{2}+1\right)$, we compute in the output of APPROXv4 the complement of *bI*_{q+1:8L} instead of the complement of *bI*_{1:q}, the reason being that in this case the length of *bI*_{q+1:8L} is strictly smaller than the length of *bI*_{1:q}. In practice, in the case $q\ge \text{\hspace{0.17em}}\left(\frac{8L}{2}+1\right)$, this means that the larger the difference $q-\frac{8L}{2}$ is, the faster APPROXv4 is as compared to APPROXv3.

**Example 4** *Suppose that*

$$bI=01011010011010010110000101100010.$$

*bI has 18 0’s and 14 1’s. Assumming sl = 8, APPROXv4 works as follows.*

*First, we compute the pairs:*(*a*_{1}= 1,*b*_{1}= 0), (*a*_{2}= 1,*b*_{2}= 0), (*a*_{3}= 1,*b*_{3}= 2), (*a*_{4}= 3,*b*_{4}= 4).*The first pair*(*a*,_{i}*b*)_{i}*that satisfies*$${a}_{i}\le (NofZs-NofOs)/2\le {b}_{i}$$

is (

*a*_{3},*b*_{3}).*Thus, we search for the position q only in the substring bI*_{17:24},*and we find q*= 22.*So, we are in the second case, when*$q\ge \text{\hspace{0.17em}}\left(\frac{8L}{2}+1\right)$,*that is, q is in the second half of bI.**The output is*$xb{I}_{1:22}\overline{b{I}_{23:32}}$,*where x is a binary string of length 256 such that the first 120 bits of x store q*= 22,*the next 3 bits are 100, the next 5 bits are 00000, and the remaining 128 bits are used to make x a binary string with an equal number of 0’s and 1’s.*

APPROXv4 can be further optimized as follows. Suppose that *NofZs* > *NofOs*, and let *j* be the first index that satisfies
$${a}_{j}\le (NofZs-NofOs)/2\le {b}_{j}.$$

Then, we search for the position *q* in the substring $b{I}_{j\xb7sl-sl+1:j\xb7sl}$, and find it.

If $q\le \frac{8L}{2}$, then the output is $x\overline{b{I}_{1:q}}b{I}_{q+1:8L}$, where *x* is a binary string of length 256 such that the first 120 bits of *x* store *q*, the next 3 bits are 101, the next 5 bits are 00000, and the remaining 128 bits are used to make *x* a binary string with an equal number of 0’s and 1’s. If $q\ge \text{\hspace{0.17em}}\left(\frac{8L}{2}+1\right)$, then consider the input *rI*, where *rI*_{i} = *bI*_{8L-i+1}, that is, *rI* is the reverse of *bI*. For *rI*, we know that *NofZs* > *NofOs*. From the pairs (*a _{i}*,

$${a}_{j\text{'}}^{\text{'}}\le (NofZs-NofOs)/2\le {b}_{j\text{'}}^{\text{'}}.$$

Then, we search for the position *q*′ in the substring $r{I}_{{j}^{\prime}\xb7sl-sl+1:{j}^{\prime}\xb7sl}$, and find it. That is, *q*′ is the first position *i* in *rI* such that $\overline{r{I}_{1:i}}r{I}_{i+1:8L}$ has an equal number of 0’s and 1’s. Then, the output is $x\overline{r{I}_{1:{q}^{\prime}}}r{I}_{{q}^{\prime}+1:8L}$, where *x* is a binary string of length 256 such that the first 120 bits of *x* store *q*′, the next 3 bits are 101, the next 5 bits are 10000, and the remaining 128 bits are used to make *x* a binary string with an equal number of 0’s and 1’s. So, in the case $q\ge \text{\hspace{0.17em}}\left(\frac{8L}{2}+1\right)$, the first 120 bits of *x* store the position *q*′ instead of *q* and since ${q}^{\prime}<\frac{8L}{2}$ then the 124-th bit of *x*, which is 1, will tell us, at decompression, that at compression the input *rI* has been considered instead of *bI*.

Denote this version of APPROXv4 by APPROXv5. In the case $q\le \frac{8L}{2}$, APPROXv5 is basically APPROXv4. When $q\ge \text{\hspace{0.17em}}\left(\frac{8L}{2}+1\right)$, then in APPROXv5 we consider the input *rI* instead of *bI*. In practice, in the case $q\ge \text{\hspace{0.17em}}\left(\frac{8L}{2}+1\right)$, if *q*′ is significantly smaller than 8*L* – *q* (the length of $\overline{b{I}_{q+1:8L}})$ then complementing *rI*_{1:q'} instead of *bI*_{q+1:8L} could lead to a significant speed-up.

We note that, in the case $q\ge \text{\hspace{0.17em}}\left(\frac{8L}{2}+1\right)$, APPROXv5 does additional work as compared to APPROXv4, by finding the position *q*′. However, this additional work would be negligible in general. In practice, if the decompression time is more important than the compression time, that is, if the time taken when decompressing the output $x\overline{r{I}_{1:{q}^{\prime}}}r{I}_{{q}^{\prime}+1:8L}$ is more important than the time taken when constructing it, then if *q*′ is significantly smaller than 8*L* – *q*, this means that decomplementing $\overline{r{I}_{1:{q}^{\prime}}}$ instead of $\overline{b{I}_{q+1:8L}}$ would take signifincatly less time at decompression.

**Example 5** *Suppose that*

$$bI=01011010011010010110000101100010,$$

*as before. bI has 18 0’s and 14 1’s. Assumming sl = 8, APPROXv5 works as follows.*

*First, we compute the pairs:*(*a*_{1}= 1,*b*_{1}= 0), (*a*_{2}= 1,*b*_{2}= 0), (*a*_{3}= 1,*b*_{3}= 2), (*a*_{4}= 3,*b*_{4}= 4).*The first pair*(*a*,_{i}*b*)_{i}*that satisfies*$${a}_{i}\le (NofZs-NofOs)/2\le {b}_{i}$$

*is*(*a*_{3},*b*_{3}).*Thus, we search for the position q only in the substring bI*_{17:24},*and we find q*= 22.*So, we are in the second case, when*$q\ge \text{\hspace{0.17em}}\left(\frac{8L}{2}+1\right)$,*that is, q is the second half of bI*.*Thus, we consider*$$rI=01000110100001101001011001011010,$$

*the reverse of bI.**The pairs corresponding to rI would be*$({a}_{1}^{\text{'}}=1,\text{\hspace{0.17em}}{b}_{1}^{\text{'}}=2)$, $({a}_{2}^{\text{'}}=1,\text{\hspace{0.17em}}{b}_{2}^{\text{'}}=\text{\hspace{0.17em}}4)$, $({a}_{3}^{\text{'}}=\text{\hspace{0.17em}}3,\text{\hspace{0.17em}}{b}_{3}^{\text{'}}=\text{\hspace{0.17em}}4)$, $({a}_{4}^{\text{'}}=\text{\hspace{0.17em}}5,\text{\hspace{0.17em}}{b}_{4}^{\text{'}}=\text{\hspace{0.17em}}4)$.*The first pair*$({a}_{i}^{\text{'}},{b}_{i}^{\text{'}})$*that satisfies*$${a}_{i}^{\text{'}}\le (NofZs-NofOs)/2\le {b}_{i}^{\text{'}}$$

*is*$({a}_{1}^{\text{'}},{b}_{1}^{\text{'}})$.*Thus, we search for the position q*′*only in the substring rI*_{1:8},*and we find q*′ = 4.*The output is*$x\overline{r{I}_{1:4}}r{I}_{5:32}$,*where x is a binary string of length 256 such that the first 120 bits of x store q′ = 4, the next 3 bits are 101, the next 5 bits are 10000, and the remaining 128 bits are used to make x a binary string with an equal number of 0’s and 1’s.*

Similarly as in the case of APPROXv3, for inputs *bI* of length a few millions bits, both APPROXv4 and APPROXv5 take in our implementations around one second.

The algorithms presented are indeed very fast in practice, which should provide for a real-time and fast working of protein-based memories. This conclusion comes from the fact that, for example, APPROXv5 takes around one second for an input of a few millions bits, so for a protein-based memory page of one thousand bits in size, APPROXv5 would normally take around one millisecond, which is quite fast. We anticipate that the processor latency caused by the execution of one of these algorithms for the correct compression (when writing a page in the protein-based memory) or the correct decompression (when reading a page from the protein-based memory) would be negligible, especially with the current technology, which means that in practice the protein-based memories should be quite fast.

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**D. Trinca** received his BSc degree in 2003, from Facultatea de Informatica, Universitatea Alexandru Ioan Cuza din Iasi, Romania, and his PhD degree in 2008, from the Department of Computer Science and Engineering, University of Connecticut, USA. Currently he is with Sc Piretus Prod Srl in Osoi, jud. iasi, Romania. His research interests are in the area of computational problems in nanotechnology.

**S. Rajasekaran** received his BSc degree in Physics in 1977, from Madurai Kamaraj University, India, and his PhD degree in Computer Science in 1988, from Harvard University, USA. Currently he is with University of Connecticut, USA. His research interests are in the area of applied algorithms.