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Protein folding in HP model on hexagonal lattices with diagonals
BMC Bioinformatics volume 15, Article number: S7 (2014)
Abstract
Three dimensional structure prediction of a protein from its amino acid sequence, known as protein folding, is one of the most studied computational problem in bioinformatics and computational biology. Since, this is a hard problem, a number of simplified models have been proposed in literature to capture the essential properties of this problem. In this paper we introduce the hexagonal lattices with diagonals to handle the protein folding problem considering the well researched HP model. We give two approximation algorithms for protein folding on this lattice. Our first algorithm is a \frac{5}{3}approximation algorithm, which is based on the strategy of partitioning the entire protein sequence into two pieces. Our next algorithm is also based on partitioning approaches and improves upon the first algorithm.
Introduction
Protein folding is one of the most studied computational problems in bioinformatics. Many approximation solutions for this problem are given in the literature by using simplified, abstract models. There exist a variety of models attempting to simplify the problem by abstracting only the "essential physical properties" of real proteins. A lattice model for folding amino acids is represented by connected beads in two dimensional lattices or three dimensional cubic lattices and considers a simplified energy function. We can categorize the lattice models into two different classes: Simplified Lattice Models (e.g. [1]) and Realistic Lattice Models [2]. One of the widely used simplified lattice model is the HP model which was first introduced by Dill [1]. In HP model, there are only two types of beads: H represents a hydrophobic or nonpolar bead and P represents a polar or hydrophilic one. The main force in the folding process is the hydrophobichydrophobic force, i.e., HH contacts. For optimal embedding, our main goal is to maximize the HH contacts.
The protein folding problem in HP model is NPhard [3]. So a number of approximation algorithms have been developed for the HP model on different lattice structures. Hart and Istrail gave the first 4approximation algorithm for the problem on the 2D square lattice [4]. Later on, Newman [5] improved the approximation ratio to 3 considering the conformation as a folded loop. A \frac{8}{3}approximation algorithm for the problem on the 3D square lattice was given by Hart and Istrail [4]. In [6], the authors introduced square lattice with diagonals and presented algorithms that give an approximation ratio of \frac{26}{15} for the twodimensional and \frac{8}{5} for the threedimensional lattice. Later, Newman and Ruhl improved this based on different geometric ideas; they achieved an improved approximation ratio of 0.37501 [7]. To remove the parity problem of the square and cubic lattices Agarwala et al. first proposed the triangular lattice [8]. There, they gave a \frac{11}{6} approximation algorithm. For a more generalized version, namely, the 3D FCC lattice, Agarwala et al. [8] gave an approximation algorithm having an approximation ratio of \frac{5}{3}. To alleviate the problem of sharp turns, Jiang and Zhu introduced the hexagonal lattice model and gave an approximation algorithm having approximation ratio of 6 [9]. A linear time approximation algorithm for protein folding in the HP side chain model on extended cubic lattice having an approximation ratio of 0.84 was presented by Heun [10].
A number of heuristic and metaheuristic techniques have also been applied to tackle the protein folding problem in the literature. A genetic algorithm for the protein folding problem in HP model in 2D square lattice was proposed in [11]. In [12, 13], a hybrid genetic algorithm was presented for HP model in 2D triangular lattice and 3D FCC lattice. The authors in [14] first proposed the pull move set for rectangular lattices, which is used in the HP model under a variety of local search methods. They also showed the completeness and reversibility of the pull move set for rectangular grid lattices. In [15], the authors extended the idea of the pull move set in the local search approach for finding an optimal embedding in 2D triangular grid and the FCC lattice in 3D.
In this paper, we introduce the hexagonal lattices with diagonals for protein folding. The motivation for introducing hexagonal lattice comes from the secondary structure of a protein as follows. The secondary structure of a protein suggests that, in real protein folding, sharp turn does not occur frequently. Hexagonal model alleviates this sharp turn problem [9]. On the other hand, in the square lattice HP model there is a serious shortcoming, namely, the parity problem as follows. Due to a grid structure in a square lattice, contact can be established between two hydrophobic atoms only if they both are either on even positions or on odd positions of the sequence. To address this parity problem, we came up with the idea of this new lattice model, i.e., hexagonal lattice model with diagonals. In this model contacts may exist through diagonals (see Figure 1). Notably, these issues have also been partially alleviated in square lattice with diagonals and triangular lattice. To this end, our new model opens a new avenue for further research for this long standing problem. We present two novel approximation algorithms for protein folding on this lattice. Our first algorithm is a \frac{5}{3}approximation algorithm for k > 10 where k is the number of sequences of H's in the HP string. This algorithm is based on a strategy of partitioning the entire protein sequence into two pieces. Our second algorithm partitions the HP string into four pieces and employs the idea of the first algorithm on the two halves. This gives a better approximation ratio of \frac{5}{4} for k > 22. The latter result is applicable to HP strings where all the sequences of H's are of even length greater than two. The expected approximation ratio of this algorithm would be \frac{5}{4} for n > 260 when both odd and even length sequences of H's having length greater than two are allowed. Here, n is the number of total H's in the HP string. We also present the idea of folding HP strings with sequences of H having length less than two. Notably, in the literature the best approximation ratio for the hexagonal lattice is 6, which is due to [9], and that for the square lattice with diagonals is \frac{25}{16}[6]. Clearly, the approximation ratio of our algorithm is better than the above results.
The rest of the paper is organized as follows. In Section 'Preliminaries', we introduce the hexagonal lattice with diagonals and define some related notions. Section 'Our Approaches' describes our algorithms and relevant results. We briefly conclude in Section 'Conclusion.
Preliminaries
In this section, we present the required notions and notations to describe the hexagonal lattice model with diagonals.
Definition The twodimensional hexagonal lattice with diagonals is an infinite graph G = (V, E) in Euclidian Space with vertex set V = ℝ^{2} and edge set E = {(x, x')x, x' ∈ ℝ^{2}, x  x'  ≤ 2}, where . denotes the Euclidean norm. An edge e ≡ (x, x') ∈ E is a non diagonal edge iff x  x'  = 1; otherwise it is a diagonal edge.
We use the well known notion of neighbourhood or adjacency of graph theory: two vertices are adjacent/neighbour to each other if they are connected through an edge. In this connection, the difference between the usual hexagonal model and our propose model lies in the fact that a vertex in the former has three neighbours, whereas in the latter it has additional 9 neighbours, i.e., a total of twelve neighbours (see Figure 1).
Although the lattice is defined as an infinite graph, we will be concerned with only a finite subgraph of it for each conformation of a protein. The input to the protein folding problem is a finite string p over the alphabet {P, H} where p = {P}*b_{1}{P}^{+}b_{2}{P}^{+}...{P}^{+}b_{ k }{P}*. Here b_{ i } ∈ {H}^{+} for 1 ≤ i ≤ k and let n={\sum}_{i=1}^{k}\left{b}_{i}\right. Here, H denotes nonpolar and P denotes polar amino acids respectively. Often, in what follows, the input string in our problem will be refer to as an HP string. An Hrun in an HP string denotes the consecutive H's and a Prun denotes consecutive P's. So, the total number of Hruns is k and total number of H is n. An Hrun of even (odd) length is said to be an even Hrun (odd Hrun). We will now define the valid embeddings and conformation of a protein into this lattice. An embedding is a selfavoiding walk inside the grid.
Definition Let p = p_{1} ... p_{ t } be an HP string of length t and let G = (V, E) be a lattice. An embedding of p into G is a mapping function f: {1, ..., t} → V from the positions of the string to the vertices of the lattice. It assigns adjacent positions in p to adjacent vertices in G, (f(i), f(i + 1)) ∈ E for all 1 ≤ i ≤ t  1. The edges (f(i), f (i + 1)) ∈ E for 1 ≤ i ≤ t  1 are called binding edges. An embedding of p into G is called a conformation, if no two binding edges cross each other (see Figure 2).
In an conformation, a vertex occupied by an H (P) will often be referred to as an Hvertex (a Pvertex). Figure 3 shows an example of a conformation. Throughout the paper, Hvertices are indicated by filled circle and Pvertices are indicated by blank circles.
Definition Given a conformation ϕ, an edge (x, x') of G is called a contact edge, if it is not a binding edge, but there exist i, j ∈ {1, ..., t} such that f(i) = x, f(j) = x', and p_{ i } = p_{ j } = H. The vertices of the lattice which are not occupied by an H or a P are called unused vertices. A binding edge connecting an H with a P is called an alternating edge. Loss edge is a nonbinding edge incident to an H that is not a contact edge (see Figure 4).
Now, we define the neighbourhood of an edge in the lattice.
Definition Let e = (x, y) be any edge in G. We define the neighbourhood N(e) of e as the intersection of the neighbour of its endpoints x and y.
Neighbourhood of an edge e = (x, y) is shown in Figure 5 for nondiagonal edges, and in Figure 6 for diagonal edges. As can be seen from the figure for a nondiagonal edge, the number of possible neighbours is 8 whereas for a diagonal one, it is 4.
Our approaches
In this section, we present two approximation algorithms for protein folding in a hexagonal lattice with diagonal. We start with deducing two upper bounds on the number of possible contacts for any H in the HP string.
An upper bound
We will deduce a bound based on a simple counting argument: we will count the number of neighbours of a vertex in the lattice. We start with the following useful lemmas.
Lemma 0.1 Let p be an HP string and G = (V, E) is a hexagonal lattice with diagonals. If p has a conformation in G, then any H in p can have at most ten contact edges.
Proof: Every vertex in the lattice G has exactly twelve neighbours comprising 3 nondiagonal neighbours and 9 diagonal neighbours (see Figure 1). In this conformation, every Hvertex has exactly two binding edges. Hence 10 edges remain, which could potentially be contact edges. And hence the result follows. □
Lemma 0.2 Let p be an input string for the problem and ϕ be a conformation of p. Let e = (x, y) be a loss edge with respect to ϕ. Then there are at most four alternating edges in N (e).
Proof: If e is a nondiagonal edge, then the neighbourhood of e contains eight vertices (see Figure 5). If e is a diagonal edge, then the neighbourhood of e contains only four vertices (see Figure 6). Now, each of x and y can be incident to at most two binding edges. So, there are at most four binding edges in N (e). It follows immediately that there can be at most four alternating edges adjacent to e. □
Now we are ready to present the upper bound.
Lemma 0.3 For a given HP string p, the the total number of contacts in a conformation ϕ is at most \mathsf{\text{1}}0n\frac{\mathsf{\text{1}}}{2}k, where k is the total number of Hruns and n is the total number of H.
Proof : From Lemma 0.1, we know that the number of contacts is at most 10n. In a confirmation one loss edge incident to H means that it would lose one contact edge. In what follows we will show that there will be at least \frac{\mathsf{\text{1}}}{2}k loss edges in ϕ. Since every Hrun is preceded and followed by a total of two alternating edges, it is sufficient to prove that, for each alternating edge in ϕ for p, we have \frac{\mathsf{\text{1}}}{4} loss edge on average. From Lemma 0.2 we know that, for every loss edge there will be at most four alternating edges in its neighbourhood. Alternatively, we can say that, for every four alternating edges there will be at least one loss edge, assuming that the alternating edges are in the neighbourhood of that loss edge. Clearly, if the alternating edges are not within the neighbourhood then the number of loss edges will increase. So, for every alternating edge there will be at least \frac{\mathsf{\text{1}}}{4} loss edge. There are a total of 2k alternating edges. So, the total number of loss edges will be, \frac{\mathsf{\text{1}}}{4}\times \mathsf{\text{2}}\times k=\frac{\mathsf{\text{1}}}{2}k. Hence, the result follows. □
Algorithms and lower bounds
In this section, we present two novel approximation algorithms for the problem. The idea of the first algorithm is to arrange all H's occurring in the input string along the two chains. We arrange the H's in the prefix of the string up to the \u230a\frac{n}{2}\u230bth H on the left chain and arrange the rest of those on the right one (see Figure 7). Then we arrange the P's between H's outside these two chains. The arrangements of the Pruns along the sidearms of the two chains are shown in Figure 7. The arrangement in the left (right) chain can be further divided into four regions, namely, the left region, the right region, the up region and the down region (see Figure 8 and Figure 9). Now we formally present our algorithm in the form of Algorithm ChainArrangement.
Algorithm ChainArrangement
Input: An HP string p.

1.
Set f=\u230a\frac{n}{2}\u230b.

2.
Suppose F denotes the position in p after the fth H. Denote by pref F(p) the prefix of p up to position F and by suff F(p) the suffix, that starts right after it. Now,
Approximation ratio for Algorithm ChainArrangement
Now we focus on deducing an approximation ratio for Algorithm ChainArrangement. Suppose that {m}_{1}=\u230a\frac{n}{2}\u230b. So, according to Algorithm ChainArrangement, the left (right) chain will contain m_{1} (m_{1} or m_{1} + 1) H's. We need to consider two cases, namely, where m_{1} = 2x + 1 and m_{1} = 2x, with an integer x > 0. In what follows, we will use vwleft chain (vwright chain) to denote a particular region of the left (right) chain. So, vw could be one of the 4 options, namely, lR (left region), rR (right region), uR (up region) and dR (down region). We also use ϕ_{ CA } to refer to the conformation given by Algorithm ChainArrangement.
case 1: m_{1} = 2x + 1
The analysis for this case will be easier to understand with the help of Figure 8. Suppose n is even. In ϕ_{ CA }, every vertex in the lRleft chain has at least 5 contacts. There are a total of x  2 such vertices (see Figure 8 and Table 1). Every vertex in the rRleft chain has at least 7 contacts. There are a total of x  1 such vertices. The two vertices in the uRleft chain each has at least 4 contacts. One vertex in the dRleft chain has at least 4 contacts while the other has at least 3 contacts.
So, the total number of contacts \left(\mathcal{C}\right) of all the vertices in the left chain, can be computed as follows:
Since the right chain is symmetric to the left one, both chains will have the same number of vertices if n = 2m_{1}, i.e., all the vertices of the right chain will also have at least \phantom{\rule{0.5em}{0ex}}\mathcal{C} contacts. So the total number of contacts will be at least \mathsf{\text{2}}\mathcal{C} or 6n  16.
If n = 2m_{1} + 1 then let n_{1} = n  1. This n_{1} vertices will have at least 6n_{1}  16 contacts. The remaining vertex will have at least 2 contacts. So the total number of contacts will be at least 6(n  1)  16 + 2 or 6n  20.
case 2: m_{1} = 2x
The analysis of this case will be easy to understand with the help of Figure 9. Let n is even. In ϕ_{ CA }, every vertex in the lRleft chain has at least 5 contacts. There are a total of x  2 such vertices (see Figure 9). Every vertex in the rRleft chain has at least 7 contacts. There are a total of x  2 such vertices. The two vertices in the uRleft chain each has at least 4 contacts. One vertex in the dRleft chain has at least 5 contacts while the other has at least 2 contacts.
So, the total number of contacts \left(\mathcal{C}\right) of all the vertices of the left chain can be computed as follows:
Since the right chain is symmetric to the left one, both chains will have the same number of vertices if n = 2m_{1}. So all the vertices of the right chain will also have at least \phantom{\rule{0.5em}{0ex}}\mathcal{C} contacts. So the total number of contacts will be at least \mathsf{\text{2}}\mathcal{C} or 6n  18.
If n = 2m_{1} + 1 and m_{1} = 2x then let n_{1} = n  1. This n_{1} vertices will have at least 6n_{1}  18 contacts. The remaining vertex will have at least 2 contacts. So the total number of contacts will be at least 6(n  1)  18 + 2 or 6n  22.
So, combining the two cases, we get that the total number of contacts is at least 6n  22. Now we need to take the alternating edges into our consideration. For every alternating edge we get two extra contacts for the two vertices (each having one). So, for n H's and k alternating edges we get a total of at least 6n  22 + 2k contacts. Hence we get the following approximation ratio A_{1}:
From Equation 1 it can be seen that for large n, A_{1} tends to reach \frac{10}{6}. So we compute the value of k so that our approximation ratio is at most \frac{10}{6} as shown below.
So, if the total number of Hruns is greater than 9, then Algorithm ChainArrangement will achieve an approximation ratio of \frac{10}{6} or \frac{5}{3}.
Note that, the value of k is dependent on n and the HP string. We now deduce the expected value of k for a given HP string. This problem can be mapped into the problem of Integer Partitioning as defined below. Notably, similar mapping has recently been utilized in [16] for deriving an expected approximation ratio of another algorithm.
Problem 0.4 Given an integer Y , the problem of Integer Partitioning aims to provide all possible ways of writing Y , as a sum of positive integers.
Note that the ways that differ only in the order of their summands are considered to be the same partition. A summand in a partition is called a part. Now, if we consider n as the input of Problem 0.4 (i.e., Y) then each length of Hruns can be viewed as parts of the partition. So if we can find the expected number of partitions we could in turn get the expected value of k. Kessler and Livingston [17] showed that to get an integer partition of an integer Y , expected number of required parts is:
where γ is the famous Euler's constant.
For our problem Y = n. If we denote E[P ] as the expected number of Hruns then,
Now, as \left(\frac{1}{2}\mathsf{\text{log}}\phantom{\rule{2.77695pt}{0ex}}n+\gamma \mathsf{\text{log}}\sqrt{\frac{\pi}{6}}\right)\le \left(\sqrt{\frac{2\pi}{3}}\times \frac{\mathsf{\text{1}}}{2}\mathsf{\text{log}}\phantom{\rule{0.3em}{0ex}}n\right) for n ≥ 5, we can say that
So the expected value of k is less than or equal to \sqrt{n}\times \mathsf{\text{log}}\phantom{\rule{0.3em}{0ex}}n which implies that \sqrt{n}\times \text{log}n\ge \frac{220}{3} or n ≥ 16. The above findings are summarized in the form of the following theorems.
Theorem 0.5 For any given HP string, Algorithm ChainArrangement gives a \frac{5}{3}approximation ratio for k > 10, where k is the total number of Hruns. □
Theorem 0.6 For any given HP string, Algorithm ChainArrangement is expected to achieve an approximation ratio of \frac{5}{3}for n ≥ 16, where n is the total number of H. □
An improved algorithm
From Equation 1 we can see that a higher value of k will give us a better approximation ratio. So, if there are many short Hruns then we will get better results. This interesting insight provides us with an idea of an improved algorithm. However, as will be discussed later, the better approximation ratio will be applicable only if every Hrun is of length greater than 2. For this improved algorithm we introduce the notion of innerleft chains, outerleft chains, innerright chains and outerright chains as shown in Figure 10. Recall that, unlike the current algorithm, there were only two chains in our previous algorithm. The arrangement in the outerleft (outerright) chain can be further divided into four regions, namely, the left region, the right region, the up region and the down region. The arrangement in the innerleft (innerright) chain can be further divided into three regions, namely, the middle region, the up region and the down region (see Figure 11). We apply the following procedures. We first put an H of an Hrun in the outerleft chain; the next two H's of the Hrun is placed in the innerleft chain. Rest of the H's of the Hrun are placed alternatively on the innerleft chain (innerright chain) and on the outerleft chain (outerright chain) (see Figure 10). At this point, a brief discussion on the difference between the arrangements done by the two algorithm is in order. In Algorithm ChainArrangement, we can place all P's of an HP string in the side arms. However in the current algorithm we may have to arrange some P's of an HP string in the outerleft chain or outerright chain also (see Figure 10). The algorithm is finally presented below.
Algorithm ImprovedChainArrangement
Input: An HP string p such that every Hrun is of length greater than two.

1.
Set f=\u230a\frac{n+{k}_{1}}{4}\u230b, where k_{1} denotes total number of odd Hruns.

2.
Suppose F denotes the position in p after the fth H. Denote by pref F(p) the prefix of p up to position F and by suff F(p) the suffix, that starts right after it. Now, place the Hruns of pref F(p) in the outerleft chain and the innerleft chain as follows.

(a)
First put an H of an Hrun in the outerleft chain; then put the next two H's of it in the innerleft chain.

(b)
Arrange the rest of the H's alternatively, between the outerleft chain and the innerleft chain.

(c)
If the current Hrun ends at the outerleft chain, the Prun following it is placed in the sidearms of the outerleft chain; otherwise, the Hrun ends at the innerleft chain (i.e., odd Hruns), and hence the first P of the Prun following it is placed at the outerleft chain. Finally the rest of the P's of the Prun are arranged in the sidearms of the outerleft chain (see Figure 10).

(a)
And place the Hruns of suff F(p) in the outerright chain and the innerright chain as follows.

(a)
First put an H of an Hrun in the outerright chain; then put the next two H's of it in the innerright chain.

(b)
Arrange the rest of the H's alternatively, between the outerright chain and the innerright chain.

(c)
If the current Hrun ends at the outerright chain, the Prun following it is placed in the sidearms of the outerright chain; otherwise the Hrun ends at the innerright chain (i.e. odd Hruns), and hence the first P of the Prun following it is placed at outerright chain. Finally the rest of the P's of the Prun are arranged in the sidearms of outerright chain (see Figure 10).
Approximation ratio for Algorithm ImprovedChainArrangement
In this section, we deduce the approximation ratio for Algorithm ImprovedChainArrangement. We present our analysis in two separate cases. In Case 1, we only have even Hruns in HP strings. In Case 2 we may also have odd Hruns in HP strings. In what follows, we will use vwouterleft chain (vwouterright chain) to denote a particular region of the outerleft (outerright) chain. So, vw could be one of the 4 options, namely, lR (left region), rR (right region), uR (up region) and dR (down region). We also use vwinnerleft chain (vwinnerright chain) to denote a particular region of the innerleft (innerright) chain. So, vw could be one of the 3 options, namely, mR (middle region), uR (up region) and dR (down region). Furthermore we use ϕ_{ ICA } to refer to the conformation given by Algorithm ImprovedChainArrangement.
HP string contains only even Hruns
Suppose that all the Hruns are of even length. Let n = 4m_{2} and m_{2} = 2x + 1, where x > 0 is an integer. In ϕ_{ ICA } every vertex in the mRinnerleft chain has at least 10 contacts (see Figure 11 and Table 2). There are a total of 2x  3 such vertices. One vertex in the uRinnerleft chain and one vertex in the dRinnerleft chain has at least 4 contacts each. The other vertices in the uRinnerleft chain and dRinnerleft chain has at least 7 contacts each.
So, the total number of contacts of all the vertices of the innerleft chain, {\mathcal{C}}_{1} can be computed as follows:
Since the innerleft chain and the innerright chain are symmetric to each other, all the vertices of the innerright chain will also have at least {\mathcal{C}}_{1} contacts. So the total number of contacts in the innerleft chain and the innerright chain will be at least \mathsf{\text{2}}{\mathcal{C}}_{1} or 20m_{2}  36.
Now, let us consider the outerleft chain and outerright chain. Every vertex in the lRouterleft chain has at least 5 contacts (see Figure 11 and Table 2). There are a total of x  1 such vertices. Every vertex in the rRouterleft chain has at least 7 contacts. There are a total of x  2 such vertices. One vertex in the uRouterleft chain and one vertex in the dRouterleft chain has at least 4 contacts each. Each of the other vertices in the uRouterleft chain and dRouterleft chain has at least 5 contacts.
So, the total number of contacts of all the vertices of outerleft chain, {\mathcal{C}}_{2} can be computed as follows:
Since the outerleft chain and the outerright chain are symmetric to each other, all the vertices of the outerright chain will also have at least {\mathcal{C}}_{2} contacts. So total number of contacts in the outerleft chain and the outerright chain will be at least \mathsf{\text{2}}{\mathcal{C}}_{2} or 12m_{2}  14. So, the number of total contacts will be at least 20m_{2}  36 + 12m_{2}  14 = 32m_{2}  50 = 8n  50.
So far we have assumed n = 4m_{2} and m_{2} = 2x + 1. Now we consider the case where n = 4m_{2} and m_{2} = 2x such that x > 0 is an integer. For this case, we can do a similar analysis to compute the total number of contacts, which will be the same, i.e., 8n  50. So when all the Hruns have even length, we get that the total number of contacts is at least 8n  50. Now we consider the alternating edges. For every alternating edge we get two extra contacts for the two corresponding vertices (each having one). So, for n number of H's and k alternating edges we get a total of at least 8n  50 + 2k contacts. Hence we get the following approximation ratio A_{2},
From Equation 2 it can be seen that for large n, A_{2} tends to reach \frac{10}{8}. So we are going to find the value of k for which our approximation ratio will be at most \frac{10}{8} or \frac{5}{4}.
Note that, the value of k is dependent on n and the HP string. We now deduce the expected value of k for a given HP string such that each Hrun is even and has length greater than two. Again, this problem can be mapped into the problem of Integer Partitioning, and hence, as before, the expected value of k is less than or equal to \sqrt{n}\times \mathsf{\text{log}}\phantom{\rule{0.3em}{0ex}}n which implies \sqrt{n}\times \mathsf{\text{log}}\phantom{\rule{0.3em}{0ex}}n\ge \frac{125}{6}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{or}}\phantom{\rule{2.77695pt}{0ex}}n\ge 22. The above results can be summarized in the form of following theorems.
Theorem 0.7 For any given HP string such that each Hrun is even and has length greater than two, Algorithm ImprovedChainArrangement achieves an approximation ratio of \frac{5}{4}for k > 20 where k is the total number of Hruns. □
Theorem 0.8 For any given HP string such that each Hrun is even and has length greater than two, it is expected that Algorithm ImprovedChainArrangement would achieve an approximation ratio of \frac{5}{4}for n ≥ 22 where n is the total number of H's. □
HP string contains both odd Hruns and even Hruns
So far we have assumed that the given HP string contains only even Hruns. Now we are going to consider the case where both odd and even Hruns are present. Let k_{1} is the total number of odd Hruns. According to the Steps 3 and 4 of Algorithm ImprovedChainArrangement, we have to put P in the outerleft chain or outerright chain for each odd Hrun. So, the total number of P's in the outerleft chain or the outerright chain is k_{1}. Let, n_{2} = n + k_{1}. We will loose at most 14 (10) contacts for each P in the left (right) region of the outerleft chain and same will happen for the outerright chain. So, on an average, we lose 12k_{1} contacts for such placement of P due to odd Hruns. So, from Equation 2, we get the following expected approximation ratio A_{3},
Assuming that an Hrun can be odd or even with equal probability, we get k = 2k_{1}. Then we can simplify as follows: {A}_{3}=\frac{10n\frac{1}{2}k}{\left(8n50+2k2k\right)}=\frac{10n\frac{1}{2}k}{\left(8n50\right)}.
This gives us an approximation ratio for the case when Hruns could be both odd and/or even under the assumption that Hrun could be odd or even with equal probability. Now we are going to find the value of k so that our expected approximation ratio will be at most \frac{10}{8} or \frac{5}{4}.
Note that, the value of k is dependent on n. To get an idea on the expected behaviour of our algorithm, we now deduce the expected value of k for a given HP string such that Hruns can be even or odd and has length greater than two. Again, this problem can be mapped into the problem of Integer Partitioning, So the expected value of k is less than or equal to \sqrt{n}\times \mathsf{\text{log}}\phantom{\rule{0.3em}{0ex}}n which implies \sqrt{n}\times \mathsf{\text{log}}\phantom{\rule{0.3em}{0ex}}n\ge 125\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{or}}\phantom{\rule{2.77695pt}{0ex}}n\ge 260. The results discussed above can be summarised in the form of following theorem.
Theorem 0.9 For any given HP string such that Hruns can be even or odd and has length greater than two, it is expected that Algorithm ImproveChainArrangement gives a \frac{5}{4}approximation ratio for n ≥ 260.
Hruns of length 1 and 2
Although in our analysis we excluded the HPstring having Hruns with length less than 3, below we discuss, how we can arrange such Hruns to get a folding using our approach. We can arrange HPstrings with Hruns of length 2 in the innerleft chain (innerright chain) as shown in Figure 12. For each Hrun of length two we will lose 24 contacts. If the total number of such Hruns is k_{2}, then we will lose at most 24k_{2} contacts. If we have HP strings having Hruns of length one, we can arrange this at the outside of the outerleft chain (outerright chain) as shown in Figure 12. For each Hrun of length one we will lose 20 contacts. If the total number of such Hruns is k_{3}, then we will lose at most 20k_{3} contacts.
Conclusion
In this paper, we have introduced hexagonal lattice with diagonals for the protein folding problem in the HP model. We have presented two novel approximation algorithms for protein folding in this lattice. Our first algorithm is a \frac{5}{3}approximation algorithm for k > 10 where k is the number of Hruns in the HP string. Our second algorithm gives a better approximation ratio of \frac{5}{4} for k > 22. The latter result is applicable to HP strings where the Hruns are of even length greater than two. The expected approximation ratio of this algorithm would be \frac{5}{4} for n > 260 when both odd and even length Hruns having length greater than two are allowed (n is the number of total H's in the HP string). Notably the best approximation ratio for hexagonal lattice is 6, which is due to [9], and the approximation ratio for square lattice with diagonal is \frac{25}{16}[6]. Clearly the approximation ratio of our algorithm is better than the above result.
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Acknowledgements
The authors would like to thank Md. Mahbubul Hasan and Shuvasish Karmaker for fruitful discussion. The authors gratefully acknowledge the fruitful comments and suggestions of the anonymous reviewers which aided in improving the presentation of the paper. This research work was partially supported by a CodeCraftersInvestortools Research Grant for CSE BUET.
Declaration
The publication costs for this article were partially funded by a small grant from BRAC University, Bangladesh and a CodeCraftersInvestortools Research Grant. A significant part of the costs have been borne by the authors.
This article has been published as part of BMC Bioinformatics Volume 15 Supplement 2, 2014: Selected articles from the Twelfth Asia Pacific Bioinformatics Conference (APBC 2014): Bioinformatics. The full contents of the supplement are available online at http://www.biomedcentral.com/bmcbioinformatics/supplements/15/S2.
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Both Shaw and Islam perceived the study, proposed the models and the algorithms. Both of them along with Rahman conducted and verified the analysis. The total work was supervised by Rahman and Hasan.
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Shaw, D.L., Shohidull Islam, A., Sohel Rahman, M. et al. Protein folding in HP model on hexagonal lattices with diagonals. BMC Bioinformatics 15 (Suppl 2), S7 (2014). https://doi.org/10.1186/1471210515S2S7
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DOI: https://doi.org/10.1186/1471210515S2S7