OJ Board • 526 - String Distance and Transform Process - Page 2

Page 2 of 3

Why wrong?

Posted: Mon May 16, 2005 5:35 am

by wanderley2k

I do this algorithm and I don

Posted: Sun Jun 05, 2005 4:07 pm

by wanderley2k

I got AC. My problem is with input. Have one space more in input.

Code: Select all

while(cin>>str1>>str2) // wrong

while(!cin.eof()) {
 cin>>str1;
 if (cin.eof()) break;
 cin>>str2;
} // It is correct

Edit String 526 trace back operations

Posted: Tue Aug 02, 2005 7:55 am

by temper_3243

Hi,
I have implemented the number of changes required for edit distance. I cannot trace back the number of changes like positions and insertions. I tried a lot but in vain.
The positions are also required. I am getting confused when there are insertions and deletions. Please do help.

False tries

Code: Select all

void
mytrace (int i, int j)
{
  /* Need this function so that it works for 526 */
  if (i == 0 || j == 0)
    return;
  if (final[i - 1][j] + 1 == final[i][j])
    {
    mytrace (i - 1, j):
      printf ("insert \n");
    }
  else if (final[i][j - 1] + 1 == final[i][j])
    {
    mytrace (i, j - 1):
      printf ("delete\n");
    }
  else
    {
      mytrace (i - 1, j - 1);
      if (s[i - 1] != y[j - 1])
	printf ("Replace\n");
    }
  return;
}

-------code for only number of changes -------------------

Code: Select all

#include<stdio.h>
int final[100][100];
char s[100], y[100];
void
mytrace (int i, int j)
{
  /* Need this function so that it works for 526 */
  return;
}

int
main ()
{
  int len1, len2, i, j, val, p, q, r;
  while (gets (s))
    {
      gets (y);
      len1 = strlen (s);
      len2 = strlen (y);
      printf ("%s:%s\n", s, y);

      for (i = 0; i <= len1 + 2; i++)
	for (j = 0; j <= len2 + 2; j++)
	  {
	    final[i][j] = 0;
	  }

      for (i = 0; i <= len1; i++)
	final[i][0] = i;

      for (j = 0; j <= len2; j++)
	final[0][j] = j;

      for (i = 1; i <= len1; i++)
	for (j = 1; j <= len2; j++)
	  {
	    val = ((s[i - 1] == y[j - 1]) ? 0 : 1);

	    p = final[i - 1][j - 1] + val;
	    q = final[i][j - 1] + 1;
	    r = final[i - 1][j] + 1;

	    if (p <= q && p <= r)
	      {
		final[i][j] = p;
	      }
	    else if (q <= p && q <= r)
	      {
		final[i][j] = q;
	      }
	    else if (r <= p && r <= q)
	      {
		final[i][j] = r;
	      }
	  }


      printf ("%d\n", final[len1][len2]);

      printf ("***************\n");

      for (i = 1; i <= len1; i++)
	{
	  for (j = 1; j <= len2; j++)
	    printf ("%d ", final[i][j]);

	  printf ("\n");
	}


      mytrace (len1, len2);
      printf ("***************\n");

    }
  return 0;
}

Code: Select all

526 edit string trace back operations

Posted: Tue Aug 02, 2005 7:56 am

by temper_3243

Hi,
I have implemented the number of changes required for edit distance. I cannot trace back the number of changes like positions and insertions. I tried a lot but in vain.
The positions are also required. I am getting confused when there are insertions and deletions. Please do help.

False tries

Code: Select all

void
mytrace (int i, int j)
{
  /* Need this function so that it works for 526 */
  if (i == 0 || j == 0)
    return;
  if (final[i - 1][j] + 1 == final[i][j])
    {
    mytrace (i - 1, j):
      printf ("insert \n");
    }
  else if (final[i][j - 1] + 1 == final[i][j])
    {
    mytrace (i, j - 1):
      printf ("delete\n");
    }
  else
    {
      mytrace (i - 1, j - 1);
      if (s[i - 1] != y[j - 1])
	printf ("Replace\n");
    }
  return;
}

-------code for only number of changes -------------------

Code: Select all

#include<stdio.h>
int final[100][100];
char s[100], y[100];
void
mytrace (int i, int j)
{
  /* Need this function so that it works for 526 */
  return;
}

int
main ()
{
  int len1, len2, i, j, val, p, q, r;
  while (gets (s))
    {
      gets (y);
      len1 = strlen (s);
      len2 = strlen (y);
      printf ("%s:%s\n", s, y);

      for (i = 0; i <= len1 + 2; i++)
	for (j = 0; j <= len2 + 2; j++)
	  {
	    final[i][j] = 0;
	  }

      for (i = 0; i <= len1; i++)
	final[i][0] = i;

      for (j = 0; j <= len2; j++)
	final[0][j] = j;

      for (i = 1; i <= len1; i++)
	for (j = 1; j <= len2; j++)
	  {
	    val = ((s[i - 1] == y[j - 1]) ? 0 : 1);

	    p = final[i - 1][j - 1] + val;
	    q = final[i][j - 1] + 1;
	    r = final[i - 1][j] + 1;

	    if (p <= q && p <= r)
	      {
		final[i][j] = p;
	      }
	    else if (q <= p && q <= r)
	      {
		final[i][j] = q;
	      }
	    else if (r <= p && r <= q)
	      {
		final[i][j] = r;
	      }
	  }


      printf ("%d\n", final[len1][len2]);

      printf ("***************\n");

      for (i = 1; i <= len1; i++)
	{
	  for (j = 1; j <= len2; j++)
	    printf ("%d ", final[i][j]);

	  printf ("\n");
	}


      mytrace (len1, len2);
      printf ("***************\n");

    }
  return 0;
}

Code: Select all

edit string 526 traceback operations

Posted: Wed Aug 03, 2005 5:07 am

by temper_3243

Hi,
I should have posted this in C and C++

I have implemented the number of changes required for edit distance. I cannot trace back the number of changes like positions and insertions. I tried a lot but in vain.
The positions are also required. I am getting confused when there are insertions and deletions. Please do help.

False tries

Code: Select all

void
mytrace (int i, int j)
{
  /* Need this function so that it works for 526 */
  if (i == 0 || j == 0)
    return;
  if (final[i - 1][j] + 1 == final[i][j])
    {
    mytrace (i - 1, j):
      printf ("insert \n");
    }
  else if (final[i][j - 1] + 1 == final[i][j])
    {
    mytrace (i, j - 1):
      printf ("delete\n");
    }
  else
    {
      mytrace (i - 1, j - 1);
      if (s[i - 1] != y[j - 1])
	printf ("Replace\n");
    }
  return;
}

-------code for only number of changes -------------------

Code: Select all

#include<stdio.h>
int final[100][100];
char s[100], y[100];
void
mytrace (int i, int j)
{
  /* Need this function so that it works for 526 */
  return;
}

int
main ()
{
  int len1, len2, i, j, val, p, q, r;
  while (gets (s))
    {
      gets (y);
      len1 = strlen (s);
      len2 = strlen (y);
      printf ("%s:%s\n", s, y);

      for (i = 0; i <= len1 + 2; i++)
	for (j = 0; j <= len2 + 2; j++)
	  {
	    final[i][j] = 0;
	  }

      for (i = 0; i <= len1; i++)
	final[i][0] = i;

      for (j = 0; j <= len2; j++)
	final[0][j] = j;

      for (i = 1; i <= len1; i++)
	for (j = 1; j <= len2; j++)
	  {
	    val = ((s[i - 1] == y[j - 1]) ? 0 : 1);

	    p = final[i - 1][j - 1] + val;
	    q = final[i][j - 1] + 1;
	    r = final[i - 1][j] + 1;

	    if (p <= q && p <= r)
	      {
		final[i][j] = p;
	      }
	    else if (q <= p && q <= r)
	      {
		final[i][j] = q;
	      }
	    else if (r <= p && r <= q)
	      {
		final[i][j] = r;
	      }
	  }


      printf ("%d\n", final[len1][len2]);

      printf ("***************\n");

      for (i = 1; i <= len1; i++)
	{
	  for (j = 1; j <= len2; j++)
	    printf ("%d ", final[i][j]);

	  printf ("\n");
	}


      mytrace (len1, len2);
      printf ("***************\n");

    }
  return 0;
}

Code: Select all

edit string 526 traceback operations

Posted: Wed Aug 03, 2005 5:08 am

by temper_3243

Hi,
I should have posted this in C and C++

I have implemented the number of changes required for edit distance. I cannot trace back the number of changes like positions and insertions. I tried a lot but in vain.
The positions are also required. I am getting confused when there are insertions and deletions. Please do help.

False tries

Code: Select all

void
mytrace (int i, int j)
{
  /* Need this function so that it works for 526 */
  if (i == 0 || j == 0)
    return;
  if (final[i - 1][j] + 1 == final[i][j])
    {
    mytrace (i - 1, j):
      printf ("insert \n");
    }
  else if (final[i][j - 1] + 1 == final[i][j])
    {
    mytrace (i, j - 1):
      printf ("delete\n");
    }
  else
    {
      mytrace (i - 1, j - 1);
      if (s[i - 1] != y[j - 1])
	printf ("Replace\n");
    }
  return;
}

-------code for only number of changes -------------------

Code: Select all

#include<stdio.h>
int final[100][100];
char s[100], y[100];
void
mytrace (int i, int j)
{
  /* Need this function so that it works for 526 */
  return;
}

int
main ()
{
  int len1, len2, i, j, val, p, q, r;
  while (gets (s))
    {
      gets (y);
      len1 = strlen (s);
      len2 = strlen (y);
      printf ("%s:%s\n", s, y);

      for (i = 0; i <= len1 + 2; i++)
	for (j = 0; j <= len2 + 2; j++)
	  {
	    final[i][j] = 0;
	  }

      for (i = 0; i <= len1; i++)
	final[i][0] = i;

      for (j = 0; j <= len2; j++)
	final[0][j] = j;

      for (i = 1; i <= len1; i++)
	for (j = 1; j <= len2; j++)
	  {
	    val = ((s[i - 1] == y[j - 1]) ? 0 : 1);

	    p = final[i - 1][j - 1] + val;
	    q = final[i][j - 1] + 1;
	    r = final[i - 1][j] + 1;

	    if (p <= q && p <= r)
	      {
		final[i][j] = p;
	      }
	    else if (q <= p && q <= r)
	      {
		final[i][j] = q;
	      }
	    else if (r <= p && r <= q)
	      {
		final[i][j] = r;
	      }
	  }


      printf ("%d\n", final[len1][len2]);

      printf ("***************\n");

      for (i = 1; i <= len1; i++)
	{
	  for (j = 1; j <= len2; j++)
	    printf ("%d ", final[i][j]);

	  printf ("\n");
	}


      mytrace (len1, len2);
      printf ("***************\n");

    }
  return 0;
}

Code: Select all

Posted: Mon Feb 13, 2006 9:13 am

by sclo

This question is a very standard dp that should be in every textbooks. If you get WA on this one, probably you didn't read the input correctly. You have to make sure to read the whole line, including the spaces.

Posted: Wed Apr 05, 2006 11:28 pm

by Moha

It is obvious, because you move in backward from your dynamic/memoization, so you can fill up your dynamic tables forward and go backward for showing the changes and make the wanted string.

Code: Select all

void print(int a,int b)
{
	switch(trace[a][b])
	{
	case 'g':
		print(a-1,b-1);
		return;
	case 'c':
		print(a-1,b-1);
		str[a+temp-1]=s2[b-1];
		cout<<++cc<<" Replace "<<a+temp<<','<<s2[b-1]<<endl;
		return;

I think this portion of my code helps you.

Posted: Mon Jun 05, 2006 4:54 am

by andre_abrantes

Even finding this kind of problem in some text books and using their code, I keep getting WA in this one.
I think I've checked the input correctly and the code seems ok with the one found in chapter 11 of Programming Challenges (it looks the same actually).
I've compared my output with the one posted by Caesum and it seems only the lines with Insert are different.
Could anyone, please, give some light to my thoughts!?

Thank you!

Code: Select all

#include <stdio.h>
#include <string.h>

#define MATCH 0
#define INSERT 1
#define DELETE 2

#define MAXLEN 128

typedef struct {
    int cost;
    int parent;
} cell;

cell m[MAXLEN][MAXLEN];
int cmd;

void row_init (int i) {
    m[0][i].cost = i;
    if (i>0)
        m[0][i].parent=INSERT;
    else
        m[0][i].parent=-1;
}

void column_init (int i) {
    m[i][0].cost = i;
    if (i>0)
        m[i][0].parent=DELETE;
    else
        m[i][0].parent=-1;
}

int match (char c, char d) {
    if (c==d) return 0;
    return 1;
}

int string_compare (char *s, char *t) {
    int sizes, sizet;
    int i, j, k;
    int opt[3];
    
    for (i=0; i<MAXLEN; i++) {
        row_init(i);
        column_init(i);
    }

    sizes=strlen(s);
    sizet=strlen(t);
    for (i=1; i<sizes; i++)
        for (j=1; j<sizet; j++) {
            opt[MATCH] = m[i-1][j-1].cost + match(s[i], t[j]);
            opt[INSERT] = m[i][j-1].cost + 1;
            opt[DELETE] = m[i-1][j].cost + 1;

            m[i][j].cost = opt[MATCH];
            m[i][j].parent = MATCH;
            for (k=INSERT; k<=DELETE; k++)
                if (opt[k] < m[i][j].cost) {
                    m[i][j].cost = opt[k];
                    m[i][j].parent = k;
                }
        }

    return m[sizes-1][sizet-1].cost;
}

void insert_out (char *t, int j) {
    printf("%d Insert %d,%c\n", cmd++, j, t[j]);
}
void delete_out (char *s, int i) {
    printf("%d Delete %d\n", cmd++, i);
}
void match_out (char *s, char *t, int i, int j) {
    if (s[i]!=t[j])
        printf("%d Replace %d,%c\n", cmd++, i, t[j]);
}

void reconstruct_path (char *s, char *t, int i, int j) {

    if (m[i][j].parent == -1) return;

    if (m[i][j].parent == MATCH) {
        reconstruct_path(s, t, i-1, j-1);
        match_out(s, t, i, j);
        return;
    }
    if (m[i][j].parent == INSERT) {
        reconstruct_path (s, t, i, j-1);
        insert_out(t, j);
        return;
    }
    if (m[i][j].parent == DELETE) {
        reconstruct_path(s, t, i-1, j);
        delete_out(s, i);
        return;
    }
}

int main (void) {
    char s[MAXLEN], t[MAXLEN];
    
    s[0]=t[0]=' ';
    while (gets(&s[1])) {
        gets(&t[1]);
        printf("%d\n", string_compare (s, t));
        cmd=1;
        reconstruct_path(s, t, strlen(s)-1, strlen(t)-1);
        printf("\n");
    }
    return 0;
}

Posted: Wed Jun 28, 2006 12:36 pm

by karu

I was getting WA on this problem until I realised the that it didn't say anywhere in the input that the strings weren't empty. I got AC after I made my program handle empty lines (and spaces). Here is some input and the output generated by my AC program.

Code: Select all

Input:
testing

a
b

Output:
7
1 Delete 7
2 Delete 6
3 Delete 5
4 Delete 4
5 Delete 3
6 Delete 2
7 Delete 1

1
1 Replace 1,b

Posted: Mon Aug 06, 2007 1:22 pm

by andysoft

Hello everyone.
I am trying to solve this problem. I use String Distance Dynamic Programming algorithm. I fill DP array ok, but I have no idea on how to restore path. Can anyone help me with that??
Thanx.

Posted: Mon Aug 06, 2007 4:12 pm

by emotional blind

how to restore path? depends on how your code is implemented. But i don't know how is your code.

normally path is showed recursively,,

for example in a graph

of 6 node
if there is a path 0->2-> 4-> 3-> 5
just store the previous node in the array..
and restore path recursively
_

Code: Select all

-------------------------------
0      1     2      3    4      5
-------------------------------
-1    -1    0       4    2      3
-------------------------------

Code: Select all

show(int n){
    if(prev[n]!=-1)
     show(prev[n]);
     printf(" %d",n);
}

It works like this:

Code: Select all

show(5)
 |
show(3)    5    // called show prev[5]=3 and prints 5
 |
show(4)    3    5        // show (3) calles show(prev[3]=4) and prints 4
 |
show(2)   4    3   5    //.......
 |
show(0)   2  4  3  5  //   
|
0  2 4 3 5

hope it helps

Posted: Tue Aug 07, 2007 10:43 am

by andysoft

I think our solutions are implemented differently.
I understand, that it would be nice to restore the path recursively from some path-array. But I have problems on what to write to this path-array. Here is my commented code, I think you will understand what I meant there.

Code: Select all

#include <cstdlib>
#include <iostream>
#include <vector>
#include <string>
#include <sstream>
#include <algorithm>
using namespace std;
#define F(i, e) for(int i = 0; i < e; i++)
#define FB(i, s) for (int i = s-1; i>=0; i--)
#define G(i, s, e) for(int i = s; i <= e; i++)
#define OUT(a) for (int ttx=0; ttx<a.size(); ttx++) cout << a[ttx] << " "; cout << endl;

int main(int argc, char* argv[]) {
	int ci=0,i,j,k,l,m,n;
	int a[81][81],b[81][81];
	string s1,s2;
	while (!cin.eof()) {
		cin >> s1;
		if (cin.eof())
			break;
		cin >> s2;
//INITIALIZING ARRAYS...
		ci++;
		F(i,81)
			F(j,81)
				a[i][j] = 0;

		a[0][0] = 0; //ARRAY a IS FOR THE EDIT COST (distance) (a[i,j] -  cost (s1[1..i]->s2[1..j])

//HERE IS THE DYNAMIC PROGRAMMING PART.
		G(i,1,s1.length())
			a[i][0] = i;

		G(i,1,s2.length())
			a[0][i] = i;
		int val,cm;
		G(i,0,s1.length()-1)
			G(j,0,s2.length()-1) {
				val = (s1[i]==s2[j]) ? 0 : 1;
				cm = min (a[i][j]+val,min(a[i][j+1]+1,a[i+1][j]+1)); //cm - store what we choose: REPLACE, DELETE or INSERT
				a[i+1][j+1] = cm;
				if (j>i)   //!!
					b[i+1][j+1]=1; //INSERT (if length of initial string s1 is less than length of s2
				else
				if (cm==a[i][j]+val)
					b[i+1][j+1] = (s1[i]==s2[j]) ? 0 : 3; //REPLACE if s1[i]!=s2[j] or DO NOTHING if they are equal
				else if (cm==a[i][j+1]+1)
					b[i+1][j+1] = 2; //DELETE
				else
					b[i+1][j+1] = 1; //INSERT
			}

		cout << a[s1.length()][s2.length()] << endl;
                //UPPER STRING OUTPUTS DISTANCE (answer) correctly

                //BELOW IS THE WRONG PART OF THE CODE: RESTORING THE PATH
		i = s1.length();
		j = s2.length();
		F(k,a[s1.length()][s2.length()]) {
			l = b[i][j];
			if (l==1) {
				cout << "Insert " << i << "," << s2[j-1] << endl;
				i--;j--;
			}else if (l==2) {
				cout << "Delete " << i << endl;
				j--;i--;
			}else if (l==3) {
				cout << "Replace " << i << "," << s2[j-1] << endl;
				i--;j--;
			}else{
				i--;j--;
			}
		}

	}

	return 0;
}

Hope you might help me

Re: 526 String Editing

Posted: Sun Aug 07, 2011 2:58 am

by willmetalufg

Code: Select all

 Code Removed - Got AC

I tried with many examples I found in the forum and it seems correct.
But I'm still getting WA

I used a DP with backtracing and I don't know where'e my code is wrong.
Could someone help me please?

PS: It was a silly mistake, I forgot a +1 in the code

Re: 526 String Editing

Posted: Tue Jul 24, 2012 3:43 pm

by mostafiz93

In the output specification it is said that

Actually many command lists can satisfy the request, but only one of them is required.

so, are there multiple outputs?

is this output valid for input-2:

Code: Select all

4
1 Insert 3,b
2 Insert 5,a
3 Insert 6,a
4 Insert 7,a