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Sequence alignment. When a new gene is discovered, a standard approach to understanding its function is to look through a database of known genes and find close matches. The closeness of two genes is measured by the extent to which they are aligned. To formalize this, think of a gene as being a long string over an alphabet ={A,C,G,T}. Consider two genes (strings) x=ATGCCand y=TACGCA. An alignment of x and y is a way of matching up these two strings by writing them in columns, for instance:

A-T-GCCTA-CGC

Here the “_” indicates a “gap.” The characters of each string must appear in order, and each column must contain a character from at least one of the strings. The score of an alignment is specified by a scoring matrixδof size (+1)×(+1), where the extra row and column are to accommodate gaps. For instance the preceding alignment has the following score:

δ(-T)+δ(A,A)+δ(T,-)+δ(G,G)+δ(C,C)+δ(C,A)

Give a dynamic programming algorithm that takes as input two strings X[1K n] and Y {1K m} and a scoring matrix δand returns the highest-scoring alignment. The running time should be O(mn) .

Short Answer

Expert verified

The dynamic algorithm that runs in O(nm) time has been obtained.

Step by step solution

01

Explain the given problem

Consider the two strings x and y with the length n and m respectively. Define a 2-Dimension array or a matrix that stores the score of aligning the string. Each value of the matrix signifies the score of aligning the sequence .

02

Defining the Recursive Relation

Based on he given condition, The matrix should be constructed using the following condition:

  • Align the strings as it is in input but omit the last element of these strings and then align characters xpand yp
  • The recurrence will be:Si,j=maxSi-1,j+δxi,-Si,j-1,j+δ-,yiSi-1,j-1+0δxi,yj
03

Analysis of the Recurrence Relation

Since string x1x2x3Kxphas n characters and string y1y2y3kyqhas characters, The recursion relation will run up to n*mtimes. Hence time complexity is: O (nm).

Therefore, the dynamic algorithm that runs in O (nm) time has been obtained.

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Most popular questions from this chapter

Time and space complexity of dynamic programming. Our dynamic programming algorithm for computing the edit distance between strings of length m and n creates a table of size n×mand therefore needs O (mn) time and space. In practice, it will run out of space long before it runs out of time. How can this space requirement be reduced?

  1. Show that if we just want to compute the value of the edit distance (rather than the optimal sequence of edits), then only O(n) space is needed, because only a small portion of the table needs to be maintained at any given time.
  2. Now suppose that we also want the optimal sequence of edits. As we saw earlier, this problem can be recast in terms of a corresponding grid-shaped dag, in which the goal is to find the optimal path from node (0,0) to node (n,m). It will be convenient to work with this formulation, and while we’re talking about convenience, we might as well also assume that is a power of 2.
    Let’s start with a small addition to the edit distance algorithm that will turn out to be very useful. The optimal path in the dag must pass through an intermediate node (k,m2) for some k; show how the algorithm can be modified to also return this value k.
  3. Now consider a recursive scheme:
    Procedure find-path((0,0)(n,m))
    Compute the value kabove
    find-path ((0,0)k,m2)
    find-path k,m2n,m
    concatenate these two paths, with kin the middle.
    Show that this scheme can be made to run inO (mn) time and O(n) space.

Suppose two teams, A and B, are playing a match to see who is the first to win games (for some particular n). We can suppose that A and B are equally competent, so each has a 50% chance of winning any particular game. Suppose they have already played i+j games, of which A has won i and B has won j. Give an efficient algorithm to compute the probability that A will go on to win the match. For example, if i=n-1 and j=n-3 then the probability that A will win the match is 78, since it must win any of the next three games.

You are given a string of n characters s[1...n], which you believe to be a corrupted text document in which all punctuation has vanished (so that it looks something like “itwasthebestoftimes...”). You wish to reconstruct the document using a dictionary, which is available in the form of a Boolean function dict(.): for any string w,

dict(w)={trueifwisavalidwordfalseotherwise

Give a dynamic programming algorithm that determines whether the string s[.]can be reconstituted as a sequence of valid words. The running time should be at mostO(n2) , assuming calls to dict take unit time.

In the event that the string is valid, make your algorithm output the corresponding sequence of words.

Given two strings x=x1x2···xnand y=y1y2···ym, we wish to find the length of their longest common substring, that is, the largest k for which there are indices i and j with xixi+1···xi+k-1=yjyj+1···yj+k-1. Show how to do this in time0(mn)

Local sequence alignment. Often two DNA sequences are significantly different, but contain regions that are very similar and are highly conserved. Design an algorithm that takes an input two strings x[1Kn]and y[1Km]and a scoring matrix δ(as defined in Exercise 6.26), and outputs substrings x'andy'of x and y respectively, that have the highest-scoring alignment over all pairs of such substrings. Your algorithm should take time O(mn).

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