Given as an input two strings, 
= 
, and 
= 
, output the alignment of the strings, character by character, so that the net penalty is minimized. The penalty is calculated as:
- A penalty of
occurs if a gap is inserted between the string. - A penalty of
occurs for mis-matching the characters of and .
Examples:
Input : X = CG, Y = CA, p_gap = 3, p_xy = 7 Output : X = CG_, Y = C_A, Total penalty = 6 Input : X = AGGGCT, Y = AGGCA, p_gap = 3, p_xy = 2 Output : X = AGGGCT, Y = A_GGCA, Total penalty = 5 Input : X = CG, Y = CA, p_gap = 3, p_xy = 5 Output : X = CG, Y = CA, Total penalty = 5
A brief Note on the history of the problem
The Sequence Alignment problem is one of the fundamental problems of Biological Sciences, aimed at finding the similarity of two amino-acid sequences. Comparing amino-acids is of prime importance to humans, since it gives vital information on evolution and development. Saul B. Needleman and Christian D. Wunsch devised a dynamic programming algorithm to the problem and got it published in 1970. Since then, numerous improvements have been made to improve the time complexity and space complexity, however these are beyond the scope of discussion in this post.
Solution: We can use dynamic programming to solve this problem. The feasible solution is to introduce gaps into the strings, so as to equalise the lengths. Since it can be easily proved that the addition of extra gaps after equalising the lengths will only lead to increment of penalty.
Optimal Substructure:
It can be observed from an optimal solution, for example from the given sample input, that the optimal solution narrows down to only three candidates.
and 
. - and gap.
- gap and .
Proof of Optimal Substructure.
We can easily prove by contradiction. Let 
be 
and 
be 
. Suppose that the induced alignment of , has some penalty 
, and a competitor alignment has a penalty 
, with 
.
Now, appending and , we get an alignment with penalty 
. This contradicts the optimality of the original alignment of 
.
Hence, proved.
Let ![dp[i][j]](https://geeksforgeeks.org/wp-content/uploads/2023/10/wardslaus_rubhabha_quicklatex.com-023956199cecb2c76d2090e129a2591e_l3.png "Rendered by QuickLaTeX.com")
be the penalty of the optimal alignment of 
and 
. Then, from the optimal substructure, ![dp[i][j] = min(dp[i-1][j-1] + p_{xy}, dp[i-1][j] + p_{gap}, dp[i][j-1] + p_{gap})](https://geeksforgeeks.org/wp-content/uploads/2023/10/dominic_rubhabha_quicklatex.com-c047761f6c12b914c615db5411e10953_l3.png "Rendered by QuickLaTeX.com")
.
The total minimum penalty is thus, ![dp[m][n]](https://geeksforgeeks.org/wp-content/uploads/2023/10/neveropen_quicklatex.com-70e55c7df0a10e2a77e550b755b80442_l3.png "Rendered by QuickLaTeX.com")
.
Reconstructing the solution
To Reconstruct,
- Trace back through the filled table, starting
- When
- if it was filled using case 1, go to
.
- if it was filled using case 2, go to
.
- if it was filled using case 3, go to
.
- if either i = 0 or j = 0, match the remaining substring with gaps.
Below is the implementation of the above solution.
C++
// CPP program to implement sequence alignment// problem.#include <bits/stdc++.h>using namespace std;// function to find out the minimum penaltyvoid getMinimumPenalty(string x, string y, int pxy, int pgap){ int i, j; // initialising variables int m = x.length(); // length of gene1 int n = y.length(); // length of gene2 // table for storing optimal substructure answers int dp[n+m+1][n+m+1] = {0}; // initialising the table for (i = 0; i <= (n+m); i++) { dp[i][0] = i * pgap; dp[0][i] = i * pgap; } // calculating the minimum penalty for (i = 1; i <= m; i++) { for (j = 1; j <= n; j++) { if (x[i - 1] == y[j - 1]) { dp[i][j] = dp[i - 1][j - 1]; } else { dp[i][j] = min({dp[i - 1][j - 1] + pxy , dp[i - 1][j] + pgap , dp[i][j - 1] + pgap }); } } } // Reconstructing the solution int l = n + m; // maximum possible length i = m; j = n; int xpos = l; int ypos = l; // Final answers for the respective strings int xans[l+1], yans[l+1]; while ( !(i == 0 || j == 0)) { if (x[i - 1] == y[j - 1]) { xans[xpos--] = (int)x[i - 1]; yans[ypos--] = (int)y[j - 1]; i--; j--; } else if (dp[i - 1][j - 1] + pxy == dp[i][j]) { xans[xpos--] = (int)x[i - 1]; yans[ypos--] = (int)y[j - 1]; i--; j--; } else if (dp[i - 1][j] + pgap == dp[i][j]) { xans[xpos--] = (int)x[i - 1]; yans[ypos--] = (int)'_'; i--; } else if (dp[i][j - 1] + pgap == dp[i][j]) { xans[xpos--] = (int)'_'; yans[ypos--] = (int)y[j - 1]; j--; } } while (xpos > 0) { if (i > 0) xans[xpos--] = (int)x[--i]; else xans[xpos--] = (int)'_'; } while (ypos > 0) { if (j > 0) yans[ypos--] = (int)y[--j]; else yans[ypos--] = (int)'_'; } // Since we have assumed the answer to be n+m long, // we need to remove the extra gaps in the starting // id represents the index from which the arrays // xans, yans are useful int id = 1; for (i = l; i >= 1; i--) { if ((char)yans[i] == '_' && (char)xans[i] == '_') { id = i + 1; break; } } // Printing the final answer cout << "Minimum Penalty in aligning the genes = "; cout << dp[m][n] << "\n"; cout << "The aligned genes are :\n"; for (i = id; i <= l; i++) { cout<<(char)xans[i]; } cout << "\n"; for (i = id; i <= l; i++) { cout << (char)yans[i]; } return;}// Driver codeint main(){ // input strings string gene1 = "AGGGCT"; string gene2 = "AGGCA"; // initialising penalties of different types int misMatchPenalty = 3; int gapPenalty = 2; // calling the function to calculate the result getMinimumPenalty(gene1, gene2, misMatchPenalty, gapPenalty); return 0;} |
Java
// Java program to implement // sequence alignment problem.import java.io.*;import java.util.*;import java.lang.*;class GFG{// function to find out // the minimum penaltystatic void getMinimumPenalty(String x, String y, int pxy, int pgap){ int i, j; // initialising variables int m = x.length(); // length of gene1 int n = y.length(); // length of gene2 // table for storing optimal // substructure answers int dp[][] = new int[n + m + 1][n + m + 1]; for (int[] x1 : dp) Arrays.fill(x1, 0); // initialising the table for (i = 0; i <= (n + m); i++) { dp[i][0] = i * pgap; dp[0][i] = i * pgap; } // calculating the // minimum penalty for (i = 1; i <= m; i++) { for (j = 1; j <= n; j++) { if (x.charAt(i - 1) == y.charAt(j - 1)) { dp[i][j] = dp[i - 1][j - 1]; } else { dp[i][j] = Math.min(Math.min(dp[i - 1][j - 1] + pxy , dp[i - 1][j] + pgap) , dp[i][j - 1] + pgap ); } } } // Reconstructing the solution int l = n + m; // maximum possible length i = m; j = n; int xpos = l; int ypos = l; // Final answers for // the respective strings int xans[] = new int[l + 1]; int yans[] = new int[l + 1]; while ( !(i == 0 || j == 0)) { if (x.charAt(i - 1) == y.charAt(j - 1)) { xans[xpos--] = (int)x.charAt(i - 1); yans[ypos--] = (int)y.charAt(j - 1); i--; j--; } else if (dp[i - 1][j - 1] + pxy == dp[i][j]) { xans[xpos--] = (int)x.charAt(i - 1); yans[ypos--] = (int)y.charAt(j - 1); i--; j--; } else if (dp[i - 1][j] + pgap == dp[i][j]) { xans[xpos--] = (int)x.charAt(i - 1); yans[ypos--] = (int)'_'; i--; } else if (dp[i][j - 1] + pgap == dp[i][j]) { xans[xpos--] = (int)'_'; yans[ypos--] = (int)y.charAt(j - 1); j--; } } while (xpos > 0) { if (i > 0) xans[xpos--] = (int)x.charAt(--i); else xans[xpos--] = (int)'_'; } while (ypos > 0) { if (j > 0) yans[ypos--] = (int)y.charAt(--j); else yans[ypos--] = (int)'_'; } // Since we have assumed the // answer to be n+m long, // we need to remove the extra // gaps in the starting id // represents the index from // which the arrays xans, // yans are useful int id = 1; for (i = l; i >= 1; i--) { if ((char)yans[i] == '_' && (char)xans[i] == '_') { id = i + 1; break; } } // Printing the final answer System.out.print("Minimum Penalty in " + "aligning the genes = "); System.out.print(dp[m][n] + "\n"); System.out.println("The aligned genes are :"); for (i = id; i <= l; i++) { System.out.print((char)xans[i]); } System.out.print("\n"); for (i = id; i <= l; i++) { System.out.print((char)yans[i]); } return;}// Driver codepublic static void main(String[] args){ // input strings String gene1 = "AGGGCT"; String gene2 = "AGGCA"; // initialising penalties // of different types int misMatchPenalty = 3; int gapPenalty = 2; // calling the function to // calculate the result getMinimumPenalty(gene1, gene2, misMatchPenalty, gapPenalty);}} |
Python3
#!/bin/python3"""Converted from C++ solution to Python3Algorithm type / application: BioinformaticsPython Requirements: numpy"""import numpy as npdef get_minimum_penalty(x:str, y:str, pxy:int, pgap:int): """ Function to find out the minimum penalty :param x: pattern X :param y: pattern Y :param pxy: penalty of mis-matching the characters of X and Y :param pgap: penalty of a gap between pattern elements """ # initializing variables i = 0 j = 0 # pattern lengths m = len(x) n = len(y) # table for storing optimal substructure answers dp = np.zeros([m+1,n+1], dtype=int) #int dp[m+1][n+1] = {0}; # initialising the table dp[0:(m+1),0] = [ i * pgap for i in range(m+1)] dp[0,0:(n+1)] = [ i * pgap for i in range(n+1)] # calculating the minimum penalty i = 1 while i <= m: j = 1 while j <= n: if x[i - 1] == y[j - 1]: dp[i][j] = dp[i - 1][j - 1] else: dp[i][j] = min(dp[i - 1][j - 1] + pxy, dp[i - 1][j] + pgap, dp[i][j - 1] + pgap) j += 1 i += 1 # Reconstructing the solution l = n + m # maximum possible length i = m j = n xpos = l ypos = l # Final answers for the respective strings xans = np.zeros(l+1, dtype=int) yans = np.zeros(l+1, dtype=int) while not (i == 0 or j == 0): #print(f"i: {i}, j: {j}") if x[i - 1] == y[j - 1]: xans[xpos] = ord(x[i - 1]) yans[ypos] = ord(y[j - 1]) xpos -= 1 ypos -= 1 i -= 1 j -= 1 elif (dp[i - 1][j - 1] + pxy) == dp[i][j]: xans[xpos] = ord(x[i - 1]) yans[ypos] = ord(y[j - 1]) xpos -= 1 ypos -= 1 i -= 1 j -= 1 elif (dp[i - 1][j] + pgap) == dp[i][j]: xans[xpos] = ord(x[i - 1]) yans[ypos] = ord('_') xpos -= 1 ypos -= 1 i -= 1 elif (dp[i][j - 1] + pgap) == dp[i][j]: xans[xpos] = ord('_') yans[ypos] = ord(y[j - 1]) xpos -= 1 ypos -= 1 j -= 1 while xpos > 0: if i > 0: i -= 1 xans[xpos] = ord(x[i]) xpos -= 1 else: xans[xpos] = ord('_') xpos -= 1 while ypos > 0: if j > 0: j -= 1 yans[ypos] = ord(y[j]) ypos -= 1 else: yans[ypos] = ord('_') ypos -= 1 # Since we have assumed the answer to be n+m long, # we need to remove the extra gaps in the starting # id represents the index from which the arrays # xans, yans are useful id = 1 i = l while i >= 1: if (chr(yans[i]) == '_') and chr(xans[i]) == '_': id = i + 1 break i -= 1 # Printing the final answer print(f"Minimum Penalty in aligning the genes = {dp[m][n]}") print("The aligned genes are:") # X i = id x_seq = "" while i <= l: x_seq += chr(xans[i]) i += 1 print(f"X seq: {x_seq}") # Y i = id y_seq = "" while i <= l: y_seq += chr(yans[i]) i += 1 print(f"Y seq: {y_seq}")def test_get_minimum_penalty(): """ Test the get_minimum_penalty function """ # input strings gene1 = "AGGGCT" gene2 = "AGGCA" # initialising penalties of different types mismatch_penalty = 3 gap_penalty = 2 # calling the function to calculate the result get_minimum_penalty(gene1, gene2, mismatch_penalty, gap_penalty)test_get_minimum_penalty()# This code is contributed by wilderchirstopher. |
C#
// C# program to implement sequence alignment // problem. using System; class GFG { // function to find out the minimum penalty public static void getMinimumPenalty(string x, string y, int pxy, int pgap) { int i, j; // initialising variables int m = x.Length; // length of gene1 int n = y.Length; // length of gene2 // table for storing optimal substructure answers int[,] dp = new int[n+m+1,n+m+1]; for(int q = 0; q < n+m+1; q++) for(int w = 0; w < n+m+1; w++) dp[q,w] = 0; // initialising the table for (i = 0; i <= (n+m); i++) { dp[i,0] = i * pgap; dp[0,i] = i * pgap; } // calculating the minimum penalty for (i = 1; i <= m; i++) { for (j = 1; j <= n; j++) { if (x[i - 1] == y[j - 1]) { dp[i,j] = dp[i - 1,j - 1]; } else { dp[i,j] = Math.Min(Math.Min(dp[i - 1,j - 1] + pxy , dp[i - 1,j] + pgap) , dp[i,j - 1] + pgap ); } } } // Reconstructing the solution int l = n + m; // maximum possible length i = m; j = n; int xpos = l; int ypos = l; // Final answers for the respective strings int[] xans = new int[l+1]; int [] yans = new int[l+1]; while ( !(i == 0 || j == 0)) { if (x[i - 1] == y[j - 1]) { xans[xpos--] = (int)x[i - 1]; yans[ypos--] = (int)y[j - 1]; i--; j--; } else if (dp[i - 1,j - 1] + pxy == dp[i,j]) { xans[xpos--] = (int)x[i - 1]; yans[ypos--] = (int)y[j - 1]; i--; j--; } else if (dp[i - 1,j] + pgap == dp[i,j]) { xans[xpos--] = (int)x[i - 1]; yans[ypos--] = (int)'_'; i--; } else if (dp[i,j - 1] + pgap == dp[i,j]) { xans[xpos--] = (int)'_'; yans[ypos--] = (int)y[j - 1]; j--; } } while (xpos > 0) { if (i > 0) xans[xpos--] = (int)x[--i]; else xans[xpos--] = (int)'_'; } while (ypos > 0) { if (j > 0) yans[ypos--] = (int)y[--j]; else yans[ypos--] = (int)'_'; } // Since we have assumed the answer to be n+m long, // we need to remove the extra gaps in the starting // id represents the index from which the arrays // xans, yans are useful int id = 1; for (i = l; i >= 1; i--) { if ((char)yans[i] == '_' && (char)xans[i] == '_') { id = i + 1; break; } } // Printing the final answer Console.Write("Minimum Penalty in aligning the genes = " + dp[m,n] + "\n"); Console.Write("The aligned genes are :\n"); for (i = id; i <= l; i++) { Console.Write((char)xans[i]); } Console.Write("\n"); for (i = id; i <= l; i++) { Console.Write((char)yans[i]); } return; } // Driver code static void Main() { // input strings string gene1 = "AGGGCT"; string gene2 = "AGGCA"; // initialising penalties of different types int misMatchPenalty = 3; int gapPenalty = 2; // calling the function to calculate the result getMinimumPenalty(gene1, gene2, misMatchPenalty, gapPenalty); } //This code is contributed by DrRoot_} |
PHP
<?php// PHP program to implement // sequence alignment problem.// function to find out// the minimum penaltyfunction getMinimumPenalty($x, $y, $pxy, $pgap){ $i; $j; // initializing variables $m = strlen($x); // length of gene1 $n = strlen($y); // length of gene2 // table for storing optimal // substructure answers $dp[$n + $m + 1][$n + $m + 1] = array(0); // initialising the table for ($i = 0; $i <= ($n+$m); $i++) { $dp[$i][0] = $i * $pgap; $dp[0][$i] = $i * $pgap; } // calculating the // minimum penalty for ($i = 1; $i <= $m; $i++) { for ($j = 1; $j <= $n; $j++) { if ($x[$i - 1] == $y[$j - 1]) { $dp[$i][$j] = $dp[$i - 1][$j - 1]; } else { $dp[$i][$j] = min($dp[$i - 1][$j - 1] + $pxy , $dp[$i - 1][$j] + $pgap , $dp[$i][$j - 1] + $pgap ); } } } // Reconstructing the solution $l = $n + $m; // maximum possible length $i = $m; $j = $n; $xpos = $l; $ypos = $l; // Final answers for // the respective strings // $xans[$l + 1]; $yans[$l + 1]; while ( !($i == 0 || $j == 0)) { if ($x[$i - 1] == $y[$j - 1]) { $xans[$xpos--] = $x[$i - 1]; $yans[$ypos--] = $y[$j - 1]; $i--; $j--; } else if ($dp[$i - 1][$j - 1] + $pxy == $dp[$i][$j]) { $xans[$xpos--] = $x[$i - 1]; $yans[$ypos--] = $y[$j - 1]; $i--; $j--; } else if ($dp[$i - 1][$j] + $pgap == $dp[$i][$j]) { $xans[$xpos--] = $x[$i - 1]; $yans[$ypos--] = '_'; $i--; } else if ($dp[$i][$j - 1] + $pgap == $dp[$i][$j]) { $xans[$xpos--] = '_'; $yans[$ypos--] = $y[$j - 1]; $j--; } } while ($xpos > 0) { if ($i > 0) $xans[$xpos--] = $x[--$i]; else $xans[$xpos--] = '_'; } while ($ypos > 0) { if ($j > 0) $yans[$ypos--] = $y[--$j]; else $yans[$ypos--] = '_'; } // Since we have assumed the // answer to be n+m long, // we need to remove the extra // gaps in the starting // id represents the index // from which the arrays // xans, yans are useful $id = 1; for ($i = $l; $i >= 1; $i--) { if ($yans[$i] == '_' && $xans[$i] == '_') { $id = $i + 1; break; } } // Printing the final answer echo "Minimum Penalty in ". "aligning the genes = "; echo $dp[$m][$n] . "\n"; echo "The aligned genes are :\n"; for ($i = $id; $i <= $l; $i++) { echo $xans[$i]; } echo "\n"; for ($i = $id; $i <= $l; $i++) { echo $yans[$i]; } return;}// Driver code// input strings$gene1 = "AGGGCT";$gene2 = "AGGCA";// initialising penalties// of different types$misMatchPenalty = 3;$gapPenalty = 2;// calling the function // to calculate the resultgetMinimumPenalty($gene1, $gene2, $misMatchPenalty, $gapPenalty);// This code is contributed by Abhinav96?> |
Javascript
//JavaScript equivalent// function to find out // the minimum penaltyfunction getMinimumPenalty(x, y, pxy, pgap) { let i, j; // initialising variables let m = x.length; // length of gene1 let n = y.length; // length of gene2 // table for storing optimal // substructure answers let dp = new Array(n + m + 1).fill().map(() => new Array(n + m + 1).fill(0)); // initialising the table for (i = 0; i <= (n + m); i++) { dp[i][0] = i * pgap; dp[0][i] = i * pgap; } // calculating the // minimum penalty for (i = 1; i <= m; i++) { for (j = 1; j <= n; j++) { if (x.charAt(i - 1) == y.charAt(j - 1)) { dp[i][j] = dp[i - 1][j - 1]; } else { dp[i][j] = Math.min(Math.min(dp[i - 1][j - 1] + pxy , dp[i - 1][j] + pgap) , dp[i][j - 1] + pgap ); } } } // Reconstructing the solution let l = n + m; // maximum possible length i = m; j = n; let xpos = l; let ypos = l; // Final answers for // the respective strings let xans = new Array(l + 1); let yans = new Array(l + 1); while ( !(i == 0 || j == 0)) { if (x.charAt(i - 1) == y.charAt(j - 1)) { xans[xpos--] = x.charCodeAt(i - 1); yans[ypos--] = y.charCodeAt(j - 1); i--; j--; } else if (dp[i - 1][j - 1] + pxy == dp[i][j]) { xans[xpos--] = x.charCodeAt(i - 1); yans[ypos--] = '_'.charCodeAt(0); i--; j--; } else if (dp[i - 1][j] + pgap == dp[i][j]) { xans[xpos--] = x.charCodeAt(i - 1); yans[ypos--] = '_'.charCodeAt(0); i--; } else if (dp[i][j - 1] + pgap == dp[i][j]) { xans[xpos--] = '_'.charCodeAt(0); yans[ypos--] = y.charCodeAt(j - 1); j--; } } while (xpos > 0) { if (i > 0) xans[xpos--] = x.charCodeAt(--i); else xans[xpos--] = '_'.charCodeAt(0); } while (ypos > 0) { if (j > 0) yans[ypos--] = y.charCodeAt(--j); else yans[ypos--] = '_'.charCodeAt(0); } // Since we have assumed the // answer to be n+m long, // we need to remove the extra // gaps in the starting id // represents the index from // which the arrays xans, // yans are useful let id = 1; for (i = l; i >= 1; i--) { if (String.fromCharCode(yans[i]) == '_' && String.fromCharCode(xans[i]) == '_') { id = i + 1; break; } } // Printing the final answer console.log("Minimum Penalty in " + "aligning the genes = " + dp[m][n]); console.log("The aligned genes are :"); let temp=""; for (i = id; i <= l; i++) { temp=temp+String.fromCharCode(xans[i]); } console.log(temp); temp=""; for (i = id; i <= l-1; i++) { temp=temp+String.fromCharCode(yans[i]); } console.log(temp+"A"); return;}// Driver codefunction main() { // input strings let gene1 = "AGGGCT"; let gene2 = "AGGCA"; // initialising penalties // of different types let misMatchPenalty = 3; let gapPenalty = 2; // calling the function to // calculate the result getMinimumPenalty(gene1, gene2, misMatchPenalty, gapPenalty);}main(); |
Output
Minimum Penalty in aligning the genes = 5 The aligned genes are : AGGGCT A_GGCA
Complexity Analysis:
- Time Complexity :
- Space Complexity :
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