Background

Genome rearrangements are studied on the basis of genome-wide analysis of gene orders and important in the evolution of species [1–6]. Since a DNA molecule has two strands, a gene in the genome rearrangement studies is usually denoted by a signed integer, with sign indicating the DNA strand to which the gene belongs, and a chromosome by a series of integers corresponding to those genes on the chromosome. In the last two decades, a variety of rearrangement operations have been proposed to evaluate the differences between gene orders in two or more genomes. Basically, these operations can be classified into two categories: (1) ‘intra-chromosomal’ rearrangements, such as reversals, transpositions and block-interchanges (also called ‘generalized transpositions’), and (2) ‘inter-chromosomal’ rearrangements, such as fusions, fissions and translocations. Reversals, often called inversions in the biological literature, rearrange a segment of continuous integers on the chromosome by reversing the order of the integers and changing their signs [3, 7–11]. Transpositions act on two adjacent and non-overlapping segments on the chromosome by exchanging their locations [10, 12–15]. Block-interchanges function as a generalized transposition that exchanges two non-overlapping but not necessarily adjacent segments on the chromosome [11, 15–19]. Translocations affect two chromosomes by exchanging their end segments [2, 11, 20–22]. Fusions merge two chromosomes into one chromosome and fissions split a chromosome into two chromosomes [2, 11, 13, 18].

Recently, great attention has been paid to the study of genome rearrangement using block-interchanges, since block-interchanges contain transpositions as a special case and, currently, the computational models involving block-interchanges are more tractable than those involving transpositions. More recently, Yancopoulos et al. defined a double cut and join (DCJ) operation that can model all the rearrangement operations described previously [23]. The DCJ is an operation that cuts one or two chromosomes in two places and rejoins the four broken ends in a new way. Intriguingly, block-interchanges, as well as transpositions, can be modeled by two consecutive DCJ operations, while others by one DCJ operation. In fact, as mentioned in [24], the two consecutive DCJ operations can be viewed as the following procedure to model transpositions or block-interchanges. (1) Excision: cut a segment from a chromosome that can be linear or circular. (2) Circularization: join the ends of the excised segment into a temporary circle. (3) Linearization: cut the temporary circle in any place as a linear segment. (4) Reincorporation: paste the linearized segment back to the remaining chromosome at a new site. As also pointed out in [24], this process of fragment excision, circularization, linearization and reincorporation indeed occurs in the immune response of higher animals. Here, we make a little modification to the reincorporation step in the above process by allowing the linearized segment to be possibly inverted before its reinsertion and also allowing inverted or non-inverted linearized segment to be pasted back to the remaining chromosome at any site (see Figure 1 for the modified model). This modification enables the above cut-circularize-linearize-and-paste (CCLP for short) operation to model seven different kinds of rearrangements, as will be detailed below. It is interesting to note that in addition to transposition and block-interchange, a CCLP operation can model reversal, inverted transposition (or transversal) [10] and others that are currently not reported in the biological literature. The seven rearrangements modeled by the CCLP operation are described as follows (see Figure 1 for a reference).

Figure 1
figure 1

Illustration of a cut-circularize-linearize-and-paste operation. A modified cut-circularize-linearize-and-paste operation that can model seven different kinds of rearrangement, where the cutting site of the temporary circle with genes 2, 3 and 4 can be either a, b or c, and the inserting place of the linearized segment at the remaining chromosome can be either d, e, f or g.

Full size image

• Case I – reversal:

As illustrated in Figure 1, a segment with genes 2, 3 and 4 is cut from a chromosome (1,2,3,4,5,6) and joined as a temporary circle, which is then cut in the same place as it was created by the join (i.e., the a site in Figure 1), and inverted and pasted back to the chromosome at the cutting site (i.e., the e site in Figure 1). As a result, this CCLP operation performs as a reversal that changes the chromosome (1,2,3,4,5,6) into (1,-4,-3,-2,5,6).

• Case II – transposition:

The temporary circle is cut in a new place (e.g., the b site in Figure 1) and pasted back to the chromosome at the cutting site. This CCLP operation performs as a transposition that changes (1,2,3,4,5,6) into (1,3,4,2,5,6).

• Case III – two consecutive, adjacent reversals:

The temporary circle is cut in a new place (e.g., the b site in Figure 1), and then inverted and pasted back to the chromosome at the cutting site. This CCLP operation changes (1,2,3,4,5,6) into (1,-2,-4,-3,5,6), which is equivalent to that (1,2,3,4,5,6) is first changed into (1,2,-4,-3,5,6) by a reversal, which is further changed into (1,-2,-4,-3,5,6) by another reversal. Note that the chromosomal regions affected by these two consecutive reversals are adjacent.

• Case IV – transposition:

The temporary circle is cut in the same place as it was joined and then pasted back to the chromosome at a new site (e.g., the f site in Figure 1). This CCLP operation performs as a transposition that changes (1,2,3,4,5,6) into (1,5,2,3,4,6).

• Case V – transversal:

The temporary circle is cut in the same place as it was joined, and then inverted and pasted back to the chromosome at a new site (e.g., the f site in Figure 1). This CCLP operation performs as an inverted transposition (i.e., transversal) that changes (1,2,3,4,5,6) into (1,5,-4,-3,-2,6).

• Case VI – block-interchange:

The temporary circle is cut in a new place (e.g., the b site in Figure 1) and then pasted back to the chromosome at a new site (e.g., the f site in Figure 1). This CCLP operation performs as a block-interchange that changes (1,2,3,4,5,6) into (1,5,3,4,2,6).

• Case VII – two consecutive, overlapping reversals:

The temporary circle is cut in a new place (e.g., the b site in Figure 1), and then inverted and pasted back to the chromosome at a new site (e.g., the f site in Figure 1). This CCLP operation changes (1,2,3,4,5,6) into (1,5,-2,-4,-3,6), which is equivalent to that (1,2,3,4,5,6) is first changed into (1,2,-5,-4,-3,6) by a reversal, which is further changed into (1,5,-2,-4,-3,6) by another reversal. Note that the chromosomal regions affected by these two consecutive reversals are overlapping.

All these seven rearrangements described above are simply called CCLP operations. But, to distinguish those CCLP operations from the reversal, we call them as non-reversal CCLP operations in the sequel of this paper. In this article, we are interested in designing efficient algorithms to solve the genome rearrangement problem involving all the seven CCLP operations. If all these CCLP operations are weighted equally, the problem aims to find a minimum number of operations to sort a signed permutation of representing a chromosome. In this case, however, non-reversal CCLP operations are favored in the rearrangement scenario of the optimal solution, as will be clear later, which contradicts with the observation made by biologists that in most organisms, reversals are observed much more frequently when compared with other rearrangements. Therefore, it may require a reversal to be weighted differently from other CCLP operations. In this circumstance, the problem is then called weighted sorting problem by CCLP operations, which is to find a series of CCLP operations whose weight sum is minimum. In this study, we pay our attention on the case in which the weight ratio between reversals and non-reversal CCLP operations is 1:2 and use the permutation group in algebra to design an O(δn) time algorithm for solving the problem, where n is the number of genes in the given chromosome and δ is the number of needed CCLP operations.

Preliminaries

Below, we introduce some definitions about the basics of permutation groups, as well as a couple of lemmas from Huang and Lu [11], that are useful for the study of genome rearrangements. Let E = {1, 2, …, n} be a set of n positive integers. Then a permutation of E is defined as a one-to-one function from E into itself and can simply be denoted by a product of some cycles. For example, a permutation expressed as α = (1, 6, 4) (2, 5, 3) means that α(1) = 6, α(6) = 4, α(4) = 1, α(2) = 5, α(5) = 3 and α(3) = 2. Basically, a cycle is cyclic and hence it does not matter which element in the cycle is written as the first. If the cycles in a permutation are all disjoint (i.e., any two cycles have no common elements), then their product is called the cycle decomposition. If a cycle has k elements, then it is called a k-cycle. The element in a 1-cycle is usually called fixed. It is a convention that the 1-cycles in a permutation are not written explicitly. If all the elements in E are fixed in a permutation, then this permutation is called an identity permutation and simply denoted by 1 = (1)(2)⋯(n).

The composition (or product) of two given permutations α and β of E is a permutation, denoted by αβ, such that αβ(e) = α(β(e)) for all e ∈ E. For example, suppose that α = (1,6,4)(2,5,3) and β = (4,3) are two given permutations of E = {1,2, …,6}. Then αβ = (1,6,4,2,5,3). It is not hard to see that if α and β are disjoint, then αβ = βα. The inverse of α, denoted by α–1, is a permutation such that αα–1 = α–1α = 1. The conjugation of β by α, denoted by α ⋅ β, is the permutation αβα–1.

As demonstrated in [11, 17, 18], the permutation groups can serve as a useful tool for studying genome rearrangement, because a genome can be expressed using a permutation, in which each cycle corresponds to a chromosome in the genome, and a fusion or a fission acting on the genome can be simulated by the product of a 2-cycle and the corresponding, as detailed as follows. Let α = (a1, a2) be a 2-cycle and β be an any permutation of E. If both a1 and a2 belong to the same cycle of β, then the effect of αβ (or βα) is equivalent to a fission acting on β and hence α is called a split operation of β. For instance, suppose that α = (1, 2) and β = (1, 6, 4, 2, 5, 3) . Then αβ = (1, 6, 4)(2, 5, 3) and βα = (5, 3, 1)(6, 4, 2). On the other hand, if a1 and a2 belong to two different cycles of β, then the effect of αβ (or βα) equals to a fusion acting on β and α is called a join operation of β. For instance, if α = (1,2) and β = (1, 6, 4)(2, 5, 3), then αβ = (1, 6, 4, 2, 5, 3) and βα = (6, 4, 1, 5, 3, 2).

In fact, any permutation α of E can be written as a composition of 2-cycles in many ways [11]. The norm of α, denoted by ||α||, is the minimum number k such that α can be expressed by a composition of k 2-cycles. The number of disjoint cycles in the cycle decomposition of α is denoted by n c (α), which needs to count those non-expressed 1-cycles in α. For instance, if α = (1, 3, 2)(5,6) and E = {1, 2, …,6}, then n c (α) = 3, rather than n c (α) = 2, because α = (1, 3, 2)(4)(5, 6). For any permutation α of E, it can be shown that ||α|| = |E| – n c (α) [11, 17]. For any two permutations α and β of E, α divides β, denoted by α|β, if and only if ||βα–1|| = ||β|| – ||α||. Actually, whether α divides β or not can be easily determined using the following lemma from [11].

Lemma 1[11]. Let e1, e2, …, e k ∈ E and β be any permutation of E. Then e1, e2, …,e k appear in the same cycle of β in the order of e1, e2, …, e k if and only if (e1, e2, …, e k )|β.

It is required to further extend the definition of E as E = {±1, ±2, …, ±n} for properly modeling reversals using the permutation groups, as described in Lemma 3 below. Let Γ = (1, –1)(2, –2) ··· (n, –n). It is not difficult to verify that Γ2 = 1 and Γ–1 = Γ. If a cycle contains no e and –e at the same time, where e ∈ E, then it is called admissible and can be used to denote a DNA strand. Let π+ denote a strand of a DNA molecule π. Then π– = Γ · (π+)–1 is the reverse complement of π+, representing another strand of π. Note that π+ and π– are disjoint. For the purpose of modeling reversals using the permutation groups, the DNA molecule π is represented by the composition of its two strands π+ and π– (i.e., π = π+π– = π–π+), as demonstrated in [11].

Lemma 2 [11]. Let π and σ be two different chromosomes. Suppose that α is a cycle in σπ–1. Then (πΓ) · α–1is also a cycle in σπ–1.

Actually, α and (π Γ) · α–1 are mate cycles for each other in σπ–1 according to Lemma 2.

Lemma 3[11]. Let u and v be in the different strands of a chromosome π, that is, (u, v) ł π. Then γ = (πΓ(v), πΓ(u)) (u, v) affects π as a reversal.

Note that in Lemma 3, (u, v) acts on π as a join operation and (π Γ(v),π Γ(u)) acts on (u, v)π as a split operation, indicating that a reversal acting on π can be implemented using the product of two 2-cycles and π. Actually, other non-reversal CCLP operations can be implemented by multiplying four 2-cycles (π Γ(x),π Γ(w))(w, x)(π Γ(v),π Γ(u))(u, v) with the given chromosome π if the following conditions are satisfied: (1) (u, v)|π, (2) (w, x) ł (u, v)π (3) w ≠ Γ(x) or Γ(w) ≠ x and (4) (w, Γ(x)) ł (u, v)π or (Γ(w), x) ł (u, v)π. The first condition is to make sure that (u, v) and (π Γ(v),π Γ(u)) respectively act on the two strands of π as splits, which lead to two temporary circles excised from π. Note that these two temporary circles are complement to each other. The second condition is to make sure that (w, x) and (π Γ(x), π Γ(w)) respectively act on the two temporary circles and the cycles of the remaining π as joins, which paste back the two temporary circles into the remaining π. It is worth mentioning that the joins also fulfill the process of linearization with possible inversion. The inversion is performed when the temporary circles are reinserted into the chromosome strands different from the ones they come from. The third and fourth conditions are to make sure that the resulting π are admissible (i.e., no e and –e from E are in the same chromosome strand). Therefore, we have the following lemma.

Lemma 4. Let π be a chromosome and β = (πΓ(x), πΓ(w))(w, x)(πΓ(v),πΓ(u))(u, v). Suppose that the following four conditions are satisfied: (1) (u, v)|π, (2) (w, x) ł (u, v)π (3) w ≠ Γ(x) or Γ(w) ≠ x and (4) (w, Γ(x)) ł (u, v)π or (Γ(w), x) ł (u, v)π. Then β affects π as a non-reversal CCLP operation.

Algorithmic result

In this section, we design an efficient algorithm on the basis of the permutation groups that sorts a given chromosome π into I = (1, 2, …, n)(–n, …, –2, –1) using the CCLP operations when the weight ratio between reversals and non-reversal CCLP operations is 1:2. The basic idea behind this algorithm is as follows. As mentioned before, any permutation can be written as a product of 2-cycles and the effect of a reversal (respectively, non-reversal CCLP operation) acting on π can be simulated by multiplying two (respectively, four) 2-cycles with π. Moreover, the product of Iπ–1 and π equals to I. All these facts indicate that one can derive a product of 2-cycles from Iπ–1 such that these 2-cycles perform as a sequence of CCLP operations to optimally transform π into I. Below, for simplicity of describing our algorithm, x and y are said to be adjacent in a permutation α if α(x) = y or α(y) = x.

Lemma 5. Let π = π+π–be a chromosome. Suppose that (x, y)|Iπ–1and (x, y)|π, that is, there are two elements x and y in a cycle of Iπ–1such that (x, y) acts on π as a split. Let β = (π Γ(y), π Γ(x))(x, y). Then there are two adjacent elements x′ and y′ in a cycle of I(βπ)–1such that (x′ ,y′) and (βπ Γ(y′),βπ Γ(x′)) act on βπ as joins. Moreover, the cycles in β′βπ are admissible, where β′ = (βπ Γ(y′), βπ Γ(x′))(x′ ,y′).

Proof. For convenience, let π = π+π– = (a1, a2, … a n )(–a n , –a n –1, …, –a1). The assumption (x, y)|π indicates that x and y are in the same cycle of π, say in π+, and hence π Γ(x) and π Γ(y) are in π–. Hence, both (x, y) and (π Γ(y),π Γ(x)) act on π as splits and β = (π Γ(y), π Γ(x))(x, y) divides π into four cycles. Let . For simplicity of our further discussion, we assume that a i <a i +1 <n for 1 ≤ i ≤ k – 2. This indicates that a k –1 is the maximum in and hence a k –1 + 1 is not in . Moreover, I(βπ)–1(a1) = I(a k –1) = a k –1 + 1, meaning that a1 and a k –1 + 1 are adjacent in I(βπ)–1. In other words, there are two adjacent elements a1 and a k– 1 + 1 in I(βπ)–1 such that (a1,a k –1 + 1), as well as (βπ Γ(a k –1 + l), βπ Γ((a1)), acts on βπ as a join. If the two cycles in (βπ Γ(a k –1 + 1),βπ Γ(a1))(a1, a k –1 + 1)βπ are admissible (i.e., they represent a chromosome), then we have completed the proof of this lemma based on Lemma 4. Now, suppose that the two cycles in (βπ Γ(a k –1 + 1),βπ Γ(a1))(a1, a k –1 + l)βπ are not admissible (i.e., for some 1 ≤ i ≤ n, both i and –i are in the same cycle). We then show below that we can still find two other adjacent elements x′ and y′ in a cycle of I(βπ)–1 such that (x′ ,y′) and (βπ Γ(y′),βπ Γ(x′)) can join βπ into two admissible cycles. First of all, a k –1 + 1 must be in (otherwise, (βπ Γ(a k –1 + 1),βπ Γ(a1))(a1,a k –1 + 1)βπ is an admissible chromosome), leading to that the cycle created by joining using (a1, a k –1 + 1) is not admissible. Further suppose that a j is the minimum in . Then Γ(a j ) = –a j , which is the maximum in . Therefore, we have –a j ≥ a k –1 + 1 (since a k –1 + 1 is also in ). In addition, –a j –1 and I(–a j ) are adjacent in I(βπ)–1 because I(βπ)–1(–a j –1) = I(–a j ). In the following, we consider five possibilities.

Case 1. a j ≠ –n and a j ≠ 1. Then I(–a j ) = –a j + 1, which is not in since –a j is the maximum in . If –a j + 1 is in , then a k –1 cannot be the maximum in , since –a j ≥ a k –1 + 1 and hence –a j + 1 >a k –1 which contradicts to our assumption that a k –1 is the maximum in . In other words, I(–a j ) belongs to either or and hence (–a j –1, I(–a j )) acts on βπ as a join and the cycles in (βπ ΓI(–a j ),βπ Γ(–a j –1))(–a j –1,I(–a j ))βπ are admissible.

Case 2. a j = –n and both 1 and –1 are not in . Then I(–a j ) = 1 (instead of I(–a j ) = –a j + 1 = n + 1). Because and are complement to each other from chromosomal point of view, both of them contains no 1 and –1, as a result, I(–a j ) belongs to either or . Therefore, (–a j –1,I(–a j )) acts on βπ as a join and (βπ ΓI(–a j ),βπ Γ(–a j– 1))(–a j –1, I(–a j ))βπ contains only admissible cycles.

Case 3. a j = 1 and both n and –n are not in . Then I(–a j ) = –n (instead of I(–a j ) = –a j + 1 = 0 ). Clearly, I(–a j ) belongs to either or . Therefore, (–a j –1,I(–a j )) acts on βπ as a join and (βπ ΓI(–a j ), βπ Γ(–a j –1))(–a j –1, I(–a j ))βπ have two admissible cycles.

Case 4. a j = –n and 1 or –1 is in . Because and are complement strands, 1 is in if and only if –1 is in . Hence, both and contains no –n, 1 and –1. Then we can exchange the roles of and with and , respectively, and follow the similar discussion as given in Case 1 to show that we can still find two adjacent elements x′ and y′ in a cycle of I(βπ)–1 such that (x′ ,y′) and (βπ Γ(y′), βπ Γ(x′)) can join the four cycles of βπ into two admissible cycles.

Case 5. a j = 1 and n or –n is in . Actually, we need not consider this case, because we have initially assumed that all the elements in are less than n and among them, a j is the smallest.

According to the above discussion, we have completed the proof of this lemma.

Theorem 1. Let Φ denote a minimum weighted sequence of CCLP operations required to transform π into I. Then the weight of Φ is great than or equal to.

Proof. Let Φ contain a reversals and b non-reversal CCLP operations. It is not hard to see that a + 2b is the weight of Φ. Recall that the effect of a reversal can be simulated using two 2-cycles and a non-reversal CCLP operation using four 2-cycles. It indicates that Φ can be written by a composition of 2a + 4b 2-cycles such that Φπ = I, which equals to that Iπ–1 can be expressed as a composition of 2a + 4b 2-cycles. In other words, ||Iπ–1|| ≤ 2a + 4b. As mentioned before, we also have ||Iπ–1|| = |E| – n c (Iπ–1), which bases on the lemma proposed in [11, 17]. Therefore, |E| – n c (Iπ–1) ≤ 2a + 4b and, as a result, the weight of Φ is great than or equal to .

Assume that there are at least two adjacent elements x and y in a cycle of Iπ–1 such that (x, y)|π. Then, according to Lemma 5, we can always find a non-reversal CCLP operation β′β from Iπ–1 to rearrange π into β′βπ, where β = (π Γ(y), π Γ(x))(x, y) and β′ = (βπ Γ(y′),βπ Γ(x′))(x′, y′). Assume that there are no any two adjacent elements x and y in a cycle of Iπ–1 such that (x, y)|π, which implies that (x, y) ł π. Then based on Lemma 3, (π Γ(y), π Γ(x))(x,y) can serve as a reversal to transform π into (π Γ(y), π Γ(x))(x, y)π. Using these properties, we design Algorithm 1 to sort π into I by CCLP operations. It is not hard to see that a non-reversal CCLP operation derived in Algorithm 1 decreases the norm of Iπ–1 by 4 and a reversal by 2. Since non-reversal CCLP operations are weighted 2 and reversals are weighted 1, Algorithm 1 decreases the norm of Iπ–1 by 1 at the weight of and hence its total weight equals to , which is optimal according to Theorem 1.

Theorem 2. Given a chromosome π, the weighted sorting problem by CCLP operations can be solved in O(δn) time when with weight ratio between reversals and non-reversal CCLP operations is 1:2, where δ is the number of CCLP operations needed to transform π into I. Moreover, the weight of the optimal solution isthat can be calculated in O(n) time in advance.

Proof. As discussed before, Algorithm 1 transforms π into I by a minimum weighted sequence of δ CCLP operations, whose total weight is that can be calculated in O(n) time. Below, the time-complexity of Algorithm 1 is analyzed. Basically, the computation in steps 1 and 2 can be done in O(n) time. As for step 3, there are δ iterations to perform. For each such iteration, it takes O(n) time to find (x, y) and (x′, y′) by determining every pair of adjacent elements in all the cycles of Iπ–1 and Iπ–1β, respectively, and a constant time to perform other operations in step 3.1, and also takes O(n) time to perform step 3.2. Therefore, the cost of step 3 is O(δn). Step 4 is executed in constant time. Totally, the time-complexity of Algorithm 1 is O(δn).

It is worth mentioning here that our algorithm is applicable to both circular and linear chromosomes. Actually, using similar discussion as in [17], one can prove that given a gene x on a circular chromosome, a CCLP operation acting on x has an equivalent one without acting on x. Based on this property, one can further prove that the problem of sorting by CCLP operations is equivalent for circular and linear chromosomes.

Conclusion

In this article, we have introduced and studied the sorting problem by CCLP operations, where CCLP is a cut-circularize-linearize-and-paste operation that can model several known and unknown rearrangements. In addition, we have proposed an O(δn) time algorithm for solving the weighted sorting problem by CCLP operations when the weight ratio between reversals and non-reversal CCLP operations is 1:2, where n is the number of genes and δ is the number of needed CLLP operations. As described in this article, this algorithm is very simple so that it can be easily implemented using 1-dimensional arrays and useful in the studies of phylogenetic tree reconstruction and human immune response to tumors. It would be an interesting future work to design efficient algorithms for solving the problem of sorting by CCLP operations when all the CCLP operations are weighted equally.