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 <bits/stdc++.h> using namespace std; // function to find out the minimum penalty void getMinimumPenalty(string x, string y, int pxy, int pgap) { int i, j; // intialising 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}; // intialising the table for (i = 0; i <= (n+m); i++) { dp[i][0] = i * pgap; dp[0][i] = i * pgap; } // calcuting 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 code int main(){ // input strings string gene1 = "AGGGCT" ; string gene2 = "AGGCA" ; // intialsing 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 penalty static void getMinimumPenalty(String x, String y, int pxy, int pgap) { int i, j; // intialising 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 ); // intialising the table for (i = 0 ; i <= (n + m); i++) { dp[i][ 0 ] = i * pgap; dp[ 0 ][i] = i * pgap; } // calcuting 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 code public static void main(String[] args) { // input strings String gene1 = "AGGGCT" ; String gene2 = "AGGCA" ; // intialsing penalties // of different types int misMatchPenalty = 3 ; int gapPenalty = 2 ; // calling the function to // calculate the result getMinimumPenalty(gene1, gene2, misMatchPenalty, gapPenalty); } } |
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; // intialising 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; // intialising the table for (i = 0; i <= (n+m); i++) { dp[i,0] = i * pgap; dp[0,i] = i * pgap; } // calcuting 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" ; // intialsing 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 penalty function getMinimumPenalty( $x , $y , $pxy , $pgap ) { $i ; $j ; // intializing 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); // intialising the table for ( $i = 0; $i <= ( $n + $m ); $i ++) { $dp [ $i ][0] = $i * $pgap ; $dp [0][ $i ] = $i * $pgap ; } // calcuting 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" ; // intialsing penalties // of different types $misMatchPenalty = 3; $gapPenalty = 2; // calling the function // to calculate the result getMinimumPenalty( $gene1 , $gene2 , $misMatchPenalty , $gapPenalty ); // This code is contributed by Abhinav96 ?> |
Minimum Penalty in aligning the genes = 5 The aligned genes are : AGGGCT A_GGCA
Time Complexity :
Space Complexity :