# Sequence Alignment problem

• Difficulty Level : Hard
• Last Updated : 06 Jan, 2022

Given as an input two strings, , and , output the alignment of the strings, character by character, so that the net penalty is minimised. The penalty is calculated as:
1. A penalty of occurs if a gap is inserted between the string.
2. 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.
1. and
2. and gap.
3. 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 be the penalty of the optimal alignment of and . Then, from the optimal substructure,
The total minimum penalty is thus, .

Reconstructing the solution
To Reconstruct,
1. Trace back through the filled table, starting
2. When
…..2a. if it was filled using case 1, go to
…..2b. if it was filled using case 2, go to
…..2c. if it was filled using case 3, go to
3. 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  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[m+1][n+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"""Origin: https://www.geeksforgeeks.org/sequence-alignment-problem/Converted from C++ solution to Python3 Algorithm type / application: Bioinformatics  Python Requirements:    numpy """import numpy as np def 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

  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?>
Output:
Minimum Penalty in aligning the genes = 5
The aligned genes are :
AGGGCT
A_GGCA

Time Complexity :
Space Complexity :

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