Given as an input two strings,
- 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
Now, appending
Hence, proved.
Let
The total minimum penalty is thus,
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.
// 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; // 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 code int 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 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; // 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 code public 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);
} } |
#!/bin/python3 """ 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# 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 program to implement // sequence alignment problem. // function to find out // the minimum penalty function 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 result getMinimumPenalty( $gene1 , $gene2 ,
$misMatchPenalty , $gapPenalty );
// This code is contributed by Abhinav96 ?> |
//JavaScript equivalent // function to find out // the minimum penalty function 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 code function 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 :