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I have matrix each cell contain number I need to calculate the cell and its neighbor cells and come out with an average of that cells.
Look at the code that I wrote.
public Matrix imageFilterAverage() {
for (int i=0; i < _array.length;i++)
for (int j=0; i < _array[i].length;j++){
_array[i][j] = (_array[i][j] + _array[i][j+1] + _array[i+1][j] + _array[i+1][j+1]) / 4;
}
return this;
}
My code return error on index issue when j+1 reach 3 it is out of bound because the cells are 0 1 2.
So for example, if I have a matrix like this
10 5 7 3
50 3 2 1
60 2 5 2
The results matrix should be like this.
17 12 3 3
21 16 2 3
28 20 2 2
I have posted the images one matrix source and matrix results
Thank you very much for your time and help.
/**
* Takes the given array and transforms each slot in the array as an average of the slots around it.
* #return an array with each where each slot in the array is "blurred" by the slots around it.
*/
public Matrix imageFilterAverage() {
int avgArray[][] = new int[_twoDiPicture.length][];
int numOfCellsAround = 0;
int cellsSum = 0;
for (int y = 0; y < _twoDiPicture.length; y++) {
avgArray[y] = new int[_twoDiPicture[y].length];
for (int x = 0; x < _twoDiPicture[y].length; x++) {
numOfCellsAround = 0;
cellsSum = 0;
numOfCellsAround += cellsAround(y, x);
cellsSum += cellsSum(y, x);
avgArray[y][x] = cellsSum / numOfCellsAround;
}
}
return new Matrix(avgArray);
}
/* a private method that deals with index out of bound exceptions. */
private boolean isInBounds(int y, int x) {
return y < _twoDiPicture.length && y >= 0 && x < _twoDiPicture[y].length && x >= 0;
}
/* A private methods that uses "isInBounds" to find how many cells are surrounding the target array. */
private int cellsAround(int y, int x) {
int cells = 1;
if (isInBounds(y + 1, x)) {
cells++;
}
if (isInBounds(y - 1, x)) {
cells++;
}
if (isInBounds(y, x + 1)) {
cells++;
}
if (isInBounds(y, x - 1)) {
cells++;
}
if (isInBounds(y - 1, x + 1)) {
cells++;
}
if (isInBounds(y - 1, x - 1)) {
cells++;
}
if (isInBounds(y + 1, x - 1)) {
cells++;
}
if (isInBounds(y + 1, x + 1)) {
cells++;
}
return cells;
}
/*A private method that returns the sum of all the adjacent cells around target cell. */
private int cellsSum(int y, int x) {
int sum = _twoDiPicture[y][x];
if (isInBounds(y + 1, x)) {
sum += _twoDiPicture[y + 1][x];
}
if (isInBounds(y - 1, x)) {
sum += _twoDiPicture[y - 1][x];
}
if (isInBounds(y, x + 1)) {
sum += _twoDiPicture[y][x + 1];
}
if (isInBounds(y, x - 1)) {
sum += _twoDiPicture[y][x - 1];
}
if (isInBounds(y - 1, x + 1)) {
sum += _twoDiPicture[y - 1][x + 1];
}
if (isInBounds(y - 1, x - 1)) {
sum += _twoDiPicture[y - 1][x - 1];
}
if (isInBounds(y + 1, x - 1)) {
sum += _twoDiPicture[y + 1][x - 1];
}
if (isInBounds(y + 1, x + 1)) {
sum += _twoDiPicture[y + 1][x + 1];
}
return sum;
}
I have an ugly solution which can be improved:
public static void main(String[] args) {
int[][] matrix = {{10, 5, 7, 3},
{50, 3, 2, 1},
{60, 2, 5, 2}};
int[][] average = new int[matrix.length][matrix[0].length];
for(int i = 0; i< matrix.length; i++){
for(int j = 0; j< matrix[0].length; j++){
int sum = 0;
int div = ((i==0 && j ==0) ||
(i==0 && j == matrix[0].length-1) ||
(i== matrix.length-1 && j ==0)||
(i== matrix.length-1 && j == matrix[0].length-1)) ? 4 :
((i==0 && j > 0) ||
(i>0 && j == 0) ||
(i== matrix.length-1 && j >0)||
(i> 0 && j == matrix[0].length-1))? 6 : 9;
for(int k = Math.max(i-1, 0); k <= Math.min(i+1, matrix.length-1); k++){
for(int t = Math.max(j-1, 0); t <= Math.min(j+1, matrix[0].length-1); t++){
sum += matrix[k][t];
}
}
average[i][j] = sum / div;
}
}
for(int[] r:average){
System.out.println(Arrays.toString(r));
}
}
I made an implementation of Wagner Fischer algorithm in java with input cost, but I want to display all steps.
I search but can't find any idea.After a long time I tried to keep each transformation in matrix alongside cost and to go through back to first solution then reverse it... is this a good idea, if it is, how should I set condition?
kitten -> sitting
1.replace k with s
2.keep i
3.keep t
4.keep t
5.replace t
6.add g
I tried to make function for display steps and can't figure out how to solve it.
import java.io.File;
import java.io.FileNotFoundException;
import java.util.Scanner;
public class Principal {
static int c1, c2, c3;
static String word1, word2;
public static void main(String[] args) throws FileNotFoundException {
Scanner data_in = new Scanner(new File("data.in"));
word1 = data_in.next();
word2 = data_in.next();
c1 = data_in.nextInt();
c2 = data_in.nextInt();
c3 = data_in.nextInt();
System.out.printf("\nInsert: %s, Delete: %s, Replace: %s\n", c1, c2, c3);
System.out.printf("\nLevenstheinDistance: %s", LevenshteinDistance(word1, word2, c1, c2, c3));
}
private static int LevenshteinDistance(String str1, String str2, int InsCost, int DelCost, int ReplCost)
{
if(word1.length() == 0)
return InsCost*str1.length();
if(word2.length() == 0)
return DelCost*str2.length();
int substitutionCost = ReplCost;
if(ReplCost > InsCost + DelCost)
ReplCost = InsCost + DelCost;
Solution[][] ManageSol = new Solution[str1.length()+1][str2.length()+1];
for(int i = 0; i <= str1.length(); i++)
{
for(int j = 0; j <= str2.length(); j++){
ManageSol[i][j] = new Solution();
}
}
System.out.printf("\nLungime str1: %s", str1.length());
System.out.printf("\nLungime str2: %s", str2.length());
for(int i = 0; i <= str1.length(); i++)
{
for (int j = 0; j <= str2.length(); j++)
{
if (i == 0)
ManageSol[i][j].solution = j;
else if (j == 0)
ManageSol[i][j].solution = i;
else if (str1.charAt(i - 1) == str2.charAt(j - 1))
{
substitutionCost = 0;
ManageSol[i][j].ecqualTo(minimum(
ManageSol[i][j - 1].solution + InsCost,
ManageSol[i - 1][j].solution + DelCost,
ManageSol[i - 1][j - 1].solution + substitutionCost));
System.out.printf("\nManagerSol[%s, %s]: ch1: %s, ch2: %s", i, j, str1.charAt(i - 1), str2.charAt(j - 1));
}
else
{
substitutionCost = 1;
ManageSol[i][j].ecqualTo(minimum(
ManageSol[i][j - 1].solution + InsCost,
ManageSol[i - 1][j].solution + DelCost,
ManageSol[i - 1][j - 1].solution + substitutionCost));
System.out.printf("\nManagerSol[%s, %s]: ch1: %s, ch2: %s", i, j, str1.charAt(i - 1), str2.charAt(j - 1));
}
}
}
System.out.printf("\n");
path(ManageSol, str1.length(), str2.length(), str1, str2);
System.out.printf("\n");
for(int i = 0; i <= str1.length(); i++)
{
for (int j = 0; j <= str2.length(); j++)
{
System.out.printf("[%s,%s]:(%s,%s) ", i, j, ManageSol[i][j].solution, ManageSol[i][j].operation);
}
System.out.printf("\n");
}
return ManageSol[str1.length()][str2.length()].solution;
}
static int minimum(int x, int y)
{
if(x >= y)
return x;
return y;
}
static Solution minimum(int Ins, int Del, int Replace)
{
Solution solution = null;
if(Ins <= Del && Ins <= Replace)
{
solution = new Solution();
solution.operation = 1;
solution.solution = Ins;
return solution;
}
else if(Del <= Ins && Del <= Replace)
{
solution = new Solution();
solution.operation = 2;
solution.solution = Del;
return solution;
}
else
{
solution = new Solution();
solution.solution = Replace;
solution.operation = 0;
return solution;
}
}
//my failure, function that should display steps
static void path(Solution[][] ManagerSol, int i, int j, String str1, String str2)
{
if(ManagerSol[i][j].operation == 0)
{
System.out.printf("\nReplace %s -> %s", str1.charAt(i-1), str2.charAt(j-1));
if(i > 1 && j > 1)
path(ManagerSol, i-1,j-1, str1, str2);
}
if(ManagerSol[i][j].operation == 1)
{
System.out.printf("\nAdd %s on position %s", str2.charAt(j-1), i-1);
if(j > 1)
path(ManagerSol, i,j-1, str1, str2);
}
if(ManagerSol[i][j].operation == 2)
{
System.out.printf("\nDelete %s", str1.charAt(i-1));
if(j > 1)
path(ManagerSol, i-1,j, str1, str2);
}
}
}
Output for kitten to sitting:
[0,0]:(0,3) [0,1]:(1,3) [0,2]:(2,3) [0,3]:(3,3) [0,4]:(4,3) [0,5]:(5,3) [0,6]:(6,3) [0,7]:(7,3)
[1,0]:(1,3) [1,1]:(1,0) [1,2]:(2,1) [1,3]:(3,1) [1,4]:(4,1) [1,5]:(5,1) [1,6]:(6,1) [1,7]:(7,1)
[2,0]:(2,3) [2,1]:(2,2) [2,2]:(1,0) [2,3]:(2,1) [2,4]:(3,1) [2,5]:(4,1) [2,6]:(5,1) [2,7]:(6,1)
[3,0]:(3,3) [3,1]:(3,2) [3,2]:(2,2) [3,3]:(1,0) [3,4]:(2,1) [3,5]:(3,1) [3,6]:(4,1) [3,7]:(5,1)
[4,0]:(4,3) [4,1]:(4,2) [4,2]:(3,2) [4,3]:(2,2) [4,4]:(1,0) [4,5]:(2,1) [4,6]:(3,1) [4,7]:(4,1)
[5,0]:(5,3) [5,1]:(5,2) [5,2]:(4,2) [5,3]:(3,2) [5,4]:(2,2) [5,5]:(2,0) [5,6]:(3,1) [5,7]:(4,1)
[6,0]:(6,3) [6,1]:(6,2) [6,2]:(5,2) [6,3]:(4,2) [6,4]:(3,2) [6,5]:(3,2) [6,6]:(2,0) [6,7]:(3,1)
In general your idea is correct. It's even simpler than that. You don't need to store any additional information.
You can go backwards (starting from the end of the given strings) and use your dynamic programming values in the following way:
If one of the indices is 0, there is only one way to go.
Otherwise, you can look at 3 possible transitions "backwards" (from (i, j) to (i - 1, j - 1), (i - 1, j) and (i, j - 1)) and choose the one which yields the actual value for (i, j). If there are several possible options, you can choose any of them.
Once you know where to go from the given pair of positions, the operation is uniquely determined.
I'm not versed in Java but here is an illustration in JavaScript:
var a = 'kitten',
b = 'sitting';
var m = new Array(a.length + 1);
for (var i=0; i<m.length; i++){
m[i] = new Array(b.length + 1);
for (var j=0; j<m[i].length; j++){
if (i === 0) m[i][j] = j;
if (j === 0) m[i][j] = i;
}
}
for (var i=1; i<=a.length; i++){
for (var j=1; j<=b.length; j++){
// no change needed
if (a[i - 1] === b[j - 1]){
m[i][j] = m[i - 1][j - 1];
// choose deletion or insertion
} else {
m[i][j] = Math.min(m[i - 1][j], m[i][j - 1], m[i - 1][j - 1]) + 1;
}
}
}
console.log('a: ' + JSON.stringify(a));
console.log('b: ' + JSON.stringify(b));
var i = a.length,
j = b.length,
steps = '';
while (i !== 0 && j !== 0){
if (a[i - 1] === b[j - 1]){
steps = 'no change; ' + steps;
i--;
j--;
} else if (m[i - 1][j] < m[i][j - 1]){
steps = 'delete \'' + a[i - 1] + '\'; ' + steps;
i--;
} else if (m[i - 1][j] === m[i][j - 1]){
steps = 'replace \'' + a[i - 1] + '\' with \'' + b[j - 1] + '\'; ' + steps;
i--;
j--;
} else {
steps = 'insert \'' + b[j - 1] + '\'; ' + steps;
j--;
}
}
if (i === 0 && j > 0){
steps = 'insert first ' + j + ' elements from b; ' + steps;
} else if (j === 0 && i > 0){
steps = 'delete first ' + i + ' elements from a; ' + steps;
}
console.log('\n' + steps[0].toUpperCase() + steps.substr(1));
console.log('\nMatrix:\n');
for (var i in m){
console.log(JSON.stringify(m[i]));
}
i'm try find most similar string in a array, and i found a code in c sharp that is this one
public static int LevenshteinDistance(string s, string t)
{
int n = s.Length;
int m = t.Length;
int[,] d = new int[n + 1, m + 1];
if (n == 0)
{
return m;
}
if (m == 0)
{
return n;
}
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= m; j++)
{
int cost = (t[j - 1] == s[i - 1]) ? 0 : 1;
Console.WriteLine(cost);
d[i, j] = Math.Min(
Math.Min(d[i - 1, j] + 1, d[i, j - 1] + 1),
d[i - 1, j - 1] + cost);
}
}
return d[n, m];
}
and i'm trying to convert it into java but i get 1 error this is my code in java
public static int LevenshteinDistance(String s, String t)
{
int n = s.length();
int m = t.length();
int[][] d = new int[n + 1][ m + 1];
if (n == 0)
{
return m;
}
if (m == 0)
{
return n;
}
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= m; j++)
{
int cost = (t[j - 1] == s[i - 1])? 0 : 1;
d[i][ j] = Math.min(
Math.min(d[i - 1][ j] + 1, d[i][ j - 1] + 1),
d[i - 1][ j - 1]+cost );
}
}
return d[n] [m];
}
i get the error in this line of code
int cost = (t[j - 1] == s[i - 1]) ? 0 : 1;
the error that i have is "Array is required,but string found" this is what i have in my main
String []ray ={"food","drinks","stuffs"};
String fa="drink";
for (int i = 0; i < ray.length; i++)
{
System.out.print(LevenshteinDistance(fa, ray[i]));
}
i would appreciate any help
Use t.charAt(j-1) == s.charAt(i-1) as to access characters (letters) in string You cannot access them directly via index (brackets []).
int cost = (t.charAt(j - 1) == s.charAt(i - 1))? 0 : 1;
You are accessing the strings as arrays here with the [] array operator:
t[j - 1] == s[i - 1]
to get the nth char of a string, instead use .charAt(n)
so in this case change it to:
t.charAt(j - 1) == s.charAt(i - 1)
the same applies to the rest of the code.
I have the following code translated as best I could from Java to C#:
public double maxProfit(double[] prices, int K)
{
if (K == 0 || prices.Length == 0)
{
return 0;
}
var dp = new double[K + 1, prices.Length];
for (int i = 1; i < K + 1; i++)
{
double maxDiff = -prices[0];
for (int j = 1; j < prices.Length; j++)
{
dp[i, j] = Math.Max(dp[i, j - 1], prices[j] + maxDiff);
maxDiff = Math.Max(maxDiff, dp[i - 1, j] - prices[j]);
}
}
printTrans(dp, prices, K);
return dp[K, prices.Length - 1];
}
public void printTrans(double[,] dp, double[] prices, int K)
{
int i = K - 1;
int j = prices.Length;
var priceList = new List<double>();
while (true)
{
if (i == 0 || j == 0)
{
break;
}
if (dp[i, j] == dp[i, j - 1])
{
j = j - 1;
}
else
{
priceList.Add(j);
double maxDiff = dp[i, j] - prices[j];
for (int z = j - 1; z >= 0; z--)
{
if (dp[i - 1, z] - prices[z] == maxDiff)
{
i = i - 1;
j = z;
priceList.Add(j);
break;
}
}
}
}
while (priceList.Count > 0)
{
Console.WriteLine("Buy # " + prices[priceList.IndexOf(0)]);
Console.WriteLine("Sell # " + prices[priceList.IndexOf(0)]);
}
}
Error occurs in the second method on lines:
if (dp[i, j] == dp[i, j - 1])
and
for (int z = j - 1; z >= 0; z--)
{
if (dp[i - 1, z] - prices[z] == maxDiff)
I am getting an Index was outside the bounds of the array. error. I understand what this error means but I have no clue on my to fix it. It took me quite a bit to understand the first part of this code but for the second part, I am at a loss.
Also what is the C# equivalent of the Java pollFirst() method?
Probably this line is the cause
public void printTrans(double[,] dp, double[] prices, int K)
{
int i = K - 1;
int j = prices.Length; // <=== this line is the cause
its causing the j to refer an index outside the bounds of the 2D array.
If you have ported from java recheck your java code.
Either make that line
int j = prices.Length - 1;
Or you need to make changes to how you create your array
var dp = new double[K + 1, prices.Length]; // <-- prices.Length would have to change here
I have a program that I am writing in Java and have to do 2 things, find the longest common sub-sequence and align the common characters. The LCS works just fine but the align part just loops away or do nothing.
I try to do this algorithm which I found on Wikipedia
function printDiff(C[0..m,0..n], X[1..m], Y[1..n], i, j)
if i > 0 and j > 0 and X[i] = Y[j]
printDiff(C, X, Y, i-1, j-1)
print " " + X[i]
else if j > 0 and (i = 0 or C[i,j-1] ≥ C[i-1,j])
printDiff(C, X, Y, i, j-1)
print "+ " + Y[j]
else if i > 0 and (j = 0 or C[i,j-1] < C[i-1,j])
printDiff(C, X, Y, i-1, j)
print "- " + X[i]
else
print ""
Here is the code I wrote (I removed the LCS part)
static char[] input1 = "ABCDE".toCharArray();
static char[] input2 = "ACDC".toCharArray();
static int M = input1.length;
static int N = input2.length;
static int[][] opt = new int[M + 1][N + 1];
public static void printDiff(int opt[][], char input1[], char input2[]) {
int i = 0, j = 0;
while (i < input1.length && j < input2.length) {
if (i > 0 && j > 0 && input1[i] == input2[j]) {
System.out.print(" " + input1[i]);
i++;
j++;
} else if (j > 0 && (i == 0 || opt[i][j - 1] >= opt[i - 1][j])) {
System.out.print("+ " + input2[j]);
j++;
} else if (i > 0 && (j == 0 || opt[i][j - 1] < opt[i - 1][j])) {
System.out.print("- " + input1[i]);
i++;
} else {
System.out.print("");
}
}
}
I rewrote your code to use the Wikipedia algorithm. In other words, I used recursion rather than a where clause. I had to change one of the if conditions because Java is zero index based and the Wikipedia algorithm is one index based.
I had to add the LCS function back in so that I could calculate the int[][]opt.
I added parenthesis to the if statements to make sure that the operations were done in the order I wanted them done.
I also fixed the output. The Wikipedia algorithm had "+ " and "- " as output. That appears to be a typo. The output should be " +" and " -", respectively.
Here's my version of the code.
public class PrintDiff {
char[] input1 = "ABCDE".toCharArray();
char[] input2 = "ACDC".toCharArray();
int M = input1.length;
int N = input2.length;
public void run() {
int[][] opt = lcsLength(input1, input2);
printDiff(opt, input1, input2, M - 1, N - 1);
}
public int[][] lcsLength(char[] input1, char[] input2) {
int[][] opt = new int[M][N];
for (int i = 1; i < input1.length; i++) {
for (int j = 1; j < input2.length; j++) {
if (input1[i] == input2[j]) {
opt[i][j] = opt[i - 1][j - 1] + 1;
} else {
opt[i][j] = Math.max(opt[i][j - 1], opt[i - 1][j]);
}
}
}
return opt;
}
public void printDiff(int opt[][], char input1[], char input2[], int i,
int j) {
if ((i >= 0) && (j >= 0) && (input1[i] == input2[j])) {
printDiff(opt, input1, input2, i - 1, j - 1);
System.out.print(" " + input1[i]);
} else if ((j > 0) && ((i == 0) || (opt[i][j - 1] >= opt[i - 1][j]))) {
printDiff(opt, input1, input2, i, j - 1);
System.out.print(" +" + input2[j]);
} else if ((i > 0) && ((j == 0) || (opt[i][j - 1] < opt[i - 1][j]))) {
printDiff(opt, input1, input2, i - 1, j);
System.out.print(" -" + input1[i]);
} else {
System.out.print("");
}
}
public static void main(String[] args) {
new PrintDiff().run();
}
}
And here's my output.
A -B C D -E +C
Here is a version which returns the diffs of all the longest common subsequences (basically backtracks using the cached table - similar to the approach taken to get to all longest common subsequences in All Longest Common Subsequences) (or, you may refer to my blog#: http://codingworkout.blogspot.com/2014/07/longest-common-subsequence.html)
for ex, for GAC and AGCAT, it returns => { { "[G][A]C", "[G]A[C]", "G[A][C]" }, {"A[G]C[A]T", "A[G][C]AT", "[A]G[C]AT"} where GA, GC and AC are longest common subsequences...
string[][] GetDiffs(string A, string B, int aIndex, int bIndex,
int[][] DP_LCS_AllPrefixes_Cache)
{
if((aIndex == 0) && (bIndex ==0))
{
return null;
}
if (DP_LCS_AllPrefixes_Cache[aIndex][bIndex] == 0)
{
var r = new string[2][];
r[0] = new string[] { A.Substring(0, aIndex) };
r[1] = new string[] { B.Substring(0, bIndex) };
return r;
}
if (A[aIndex - 1] == B[bIndex - 1])
{
var r = this.GetDiffs(A, B, aIndex - 1, bIndex - 1,
DP_LCS_AllPrefixes_Cache);
string ch = string.Format("[{0}]", A[aIndex - 1]);
if (r == null)
{
r = new string[2][];
r[0] = new string[] { ch };
r[1] = new string[] { ch };
}
else
{
r[0] = r[0].Select(s => s + ch).ToArray();
r[1] = r[1].Select(s => s + ch).ToArray();
}
return r;
}
int lcs_up_direction = DP_LCS_AllPrefixes_Cache[aIndex - 1][bIndex];
int lcs_left_direction = DP_LCS_AllPrefixes_Cache[aIndex][bIndex - 1];
string[][] lcs_up = null, lcs_left = null;
if (lcs_up_direction == lcs_left_direction)
{
lcs_up = this.GetDiffs(A, B, aIndex - 1, bIndex,
DP_LCS_AllPrefixes_Cache);
lcs_left = this.GetDiffs(A, B, aIndex, bIndex - 1,
DP_LCS_AllPrefixes_Cache);
}
else if (lcs_up_direction > lcs_left_direction)
{
lcs_up = this.GetDiffs(A, B, aIndex - 1, bIndex,
DP_LCS_AllPrefixes_Cache);
}
else
{
lcs_left = this.GetDiffs(A, B, aIndex, bIndex - 1, DP_LCS_AllPrefixes_Cache);
}
char a = A[aIndex - 1], b = B[bIndex - 1];
string[][] rl = new string[2][];
rl[0] = new string[0];
rl[1] = new string[0];
if(lcs_up != null)
{
//we moved upward, that is we accepted that they differ with 'A' at aIndex-1 (a)
rl[0] = lcs_up[0].Select(s => s + a.ToString()).ToArray();
rl[1] = lcs_up[1];
}
if (lcs_left != null)
{
//we moved left, that is we accepted that they differ with 'B' at bIndex-1 (b)
rl[0] = rl[0].Union(lcs_left[0]).ToArray(); ;
rl[1] = rl[1].Union(lcs_left[1].Select(s => s + b.ToString())).ToArray();
}
return rl.ToArray();
}
where the caller is
string[][] GetDiffs(string A, string B, int[][] DP_LCS_AllPrefixes_Cache)
{
var r = this.GetDiffs(A, B, A.Length, B.Length,
DP_LCS_AllPrefixes_Cache);
return r;
}
And the DP method which captures LCS lengths to backtrack
public int[][] LCS_OfAllPrefixes_Length(string A, string B)
{
A.ThrowIfNullOrWhiteSpace("a");
B.ThrowIfNullOrWhiteSpace("b");
int[][] DP_LCS_AllPrefixes_Cache = new int[A.Length+1][];
for(int i = 0;i<DP_LCS_AllPrefixes_Cache.Length; i++)
{
DP_LCS_AllPrefixes_Cache[i] = new int[B.Length + 1];
}
for (int rowIndexOfCache = 1; rowIndexOfCache <= A.Length; rowIndexOfCache++)
{
for (int columnIndexOfCache = 1; columnIndexOfCache <= B.Length; columnIndexOfCache++)
{
//LCS(Ai, Bj) = 0 if i <=0, or j <= 0
// LCS(Ai, Bj) + 1 if Ai == Bj
// Max(LCS(Ai-1, Bj), LCS(Ai, Bj-1))
if(A[rowIndexOfCache-1] == B[columnIndexOfCache-1])
{
DP_LCS_AllPrefixes_Cache[rowIndexOfCache][columnIndexOfCache] = DP_LCS_AllPrefixes_Cache[rowIndexOfCache - 1][columnIndexOfCache - 1] + 1;
}
else
{
DP_LCS_AllPrefixes_Cache[rowIndexOfCache][columnIndexOfCache] = Utilities.Max(DP_LCS_AllPrefixes_Cache[rowIndexOfCache - 1][columnIndexOfCache],
DP_LCS_AllPrefixes_Cache[rowIndexOfCache][columnIndexOfCache - 1]);
}
}
}
return DP_LCS_AllPrefixes_Cache;
}
TestMethod
[TestMethod]
public void LCS_Tests()
{
string A = "GAC", B = "AGCAT";
var DP_LCS_AllPrefixes_Cache = this.LCS_OfAllPrefixes_Length(A, B);
Assert.IsTrue(DP_LCS_AllPrefixes_Cache[A.Length][B.Length] == 2);
var lcs_sequences = this.GetLongestCommonSubsequences(A, B, DP_LCS_AllPrefixes_Cache);
Assert.IsNotNull(lcs_sequences);
var diffs = this.GetDiffs(A, B, DP_LCS_AllPrefixes_Cache);
Assert.IsNotNull(diffs);
Assert.IsTrue(diffs.Length == 2);
Assert.IsTrue(diffs[0].Length == lcs_sequences.Length);
Assert.IsTrue(diffs[1].Length == lcs_sequences.Length);
Assert.IsTrue(lcs_sequences.Any(s => "AC".Equals(s)));
Assert.IsTrue(lcs_sequences.Any(s => "GC".Equals(s)));
Assert.IsTrue(lcs_sequences.Any(s => "GA".Equals(s)));
var DP_LCS_AllPrefixes_Subsequences_Cache = this.LCS_OfAllPrefixes_Subsequences(A, B);
Assert.IsTrue(DP_LCS_AllPrefixes_Subsequences_Cache[A.Length][B.Length].Length == 2);
Assert.IsTrue(DP_LCS_AllPrefixes_Subsequences_Cache[A.Length][B.Length].Subsequences
.Any(s => "AC".Equals(s)));
Assert.IsTrue(DP_LCS_AllPrefixes_Subsequences_Cache[A.Length][B.Length].Subsequences
.Any(s => "GC".Equals(s)));
Assert.IsTrue(DP_LCS_AllPrefixes_Subsequences_Cache[A.Length][B.Length].Subsequences
.Any(s => "GA".Equals(s)));
A = "ABCDGH"; B = "AEDFHR";
DP_LCS_AllPrefixes_Cache = this.LCS_OfAllPrefixes_Length(A, B);
Assert.IsTrue(DP_LCS_AllPrefixes_Cache[A.Length][B.Length] == 3);
lcs_sequences = this.GetLongestCommonSubsequences(A, B, DP_LCS_AllPrefixes_Cache);
Assert.IsNotNull(lcs_sequences);
diffs = this.GetDiffs(A, B, DP_LCS_AllPrefixes_Cache);
Assert.IsNotNull(diffs);
Assert.IsTrue(diffs.Length == 2);
Assert.IsTrue(diffs[0].Length == lcs_sequences.Length);
Assert.IsTrue(diffs[1].Length == lcs_sequences.Length);
Assert.IsTrue(lcs_sequences.Any(s => "ADH".Equals(s)));
DP_LCS_AllPrefixes_Subsequences_Cache = this.LCS_OfAllPrefixes_Subsequences(A, B);
Assert.IsTrue(DP_LCS_AllPrefixes_Subsequences_Cache[A.Length][B.Length].Length == 3);
Assert.IsTrue(DP_LCS_AllPrefixes_Subsequences_Cache[A.Length][B.Length].Subsequences
.Any(s => "ADH".Equals(s)));
A = "AGGTAB"; B = "GXTXAYB";
DP_LCS_AllPrefixes_Cache = this.LCS_OfAllPrefixes_Length(A, B);
Assert.IsTrue(DP_LCS_AllPrefixes_Cache[A.Length][B.Length] == 4);
lcs_sequences = this.GetLongestCommonSubsequences(A, B, DP_LCS_AllPrefixes_Cache);
Assert.IsNotNull(lcs_sequences);
diffs = this.GetDiffs(A, B, DP_LCS_AllPrefixes_Cache);
Assert.IsNotNull(diffs);
Assert.IsTrue(diffs.Length == 2);
Assert.IsTrue(diffs[0].Length == 2);
Assert.IsTrue(diffs[1].Length == lcs_sequences.Length);
Assert.IsTrue(lcs_sequences.Any(s => "GTAB".Equals(s)));
DP_LCS_AllPrefixes_Subsequences_Cache = this.LCS_OfAllPrefixes_Subsequences(A, B);
Assert.IsTrue(DP_LCS_AllPrefixes_Subsequences_Cache[A.Length][B.Length].Length == 4);
Assert.IsTrue(DP_LCS_AllPrefixes_Subsequences_Cache[A.Length][B.Length].Subsequences
.Any(s => "GTAB".Equals(s)));
A = "ABCDEF"; B = "UVWXYZ";
DP_LCS_AllPrefixes_Cache = this.LCS_OfAllPrefixes_Length(A, B);
Assert.IsTrue(DP_LCS_AllPrefixes_Cache[A.Length][B.Length] == 0);
lcs_sequences = this.GetLongestCommonSubsequences(A, B, DP_LCS_AllPrefixes_Cache);
diffs = this.GetDiffs(A, B, DP_LCS_AllPrefixes_Cache);
Assert.IsNotNull(diffs);
Assert.IsTrue(diffs.Length == 2);
Assert.IsTrue(diffs[0].Length == 1);
Assert.IsTrue(diffs[1].Length == 1);
}