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whitehatmiracle@gmail.com 09-16-2006 05:20 PM

Richard Heathfield's TSP permutation algorithm
 
Dear Sir I couldnt quite figure out wat your permute function does
exactly... could you please throw some light on it?
void Permute(char *Perm,
size_t n,
size_t unchanged)
{
size_t outer = 0;
size_t inner = 0;
int temp = 0;


if(unchanged < n)
{
for(outer = unchanged; outer < n; outer++)
{
temp = Perm[outer];
for(inner = outer; inner > unchanged; inner--)
{
Perm[inner] = Perm[inner - 1];
}
Perm[unchanged] = temp;
Permute(Perm,
n,
unchanged + 1);


for(inner = unchanged; inner < outer; inner++)
{
Perm[inner] = Perm[inner + 1];
}
Perm[outer] = temp;
}
}
else
{
printf("%s\n", Perm);
}



}


int main(int argc, char **argv)
{
char Input[256] = {0};
size_t len = 0;

if(argc > 1)
{
len = strlen(argv[1]);
if(len > sizeof Input - 1)
{
fprintf(stderr, "word too long for demo - truncating\n");
argv[1][sizeof Input - 1] = '\0';
}
strcpy(Input, argv[1]);
len = strlen(Input);
}
else
{
len = 3;
strcpy(Input, "ABC");
}


Permute(Input, len, 0);


return 0;

}

Thanking you
The whitehat


Richard Heathfield 09-16-2006 05:29 PM

Re: Richard Heathfield's TSP permutation algorithm
 
whitehatmiracle@gmail.com said:

> Dear Sir I couldnt quite figure out wat your permute function does
> exactly... could you please throw some light on it?


It just permutes an array. If you want to turn it into a brute-forcer, I
suggest you hack it to accept a function that will be called whenever a
permutation is found.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at above domain (but drop the www, obviously)

whitehatmiracle@gmail.com 09-17-2006 10:38 AM

Re: Richard Heathfield's TSP permutation algorithm
 
What does inner outer and unchanged do?


Richard Heathfield 09-17-2006 12:43 PM

Re: Richard Heathfield's TSP permutation algorithm
 
whitehatmiracle@gmail.com said:

> What does inner outer and unchanged do?


Partition the array into two parts, an empty part, and the part you want to
permute.

| ABCDE

Move each element on the right-hand side to the left-hand side, in turn.

A | BCDE
B | CDEA
C | DEAB
D | EABC
E | ABCD

All you're really doing is rotating the array, yes?

For each of those partitioned array arrangements, we now have to solve the
problem of permuting the right-hand side, and incorporating the left-hand
side, ***unchanged***, into the final result.

Take A | BCDE, for example. We have 1 unchanged element, which will prefix
each solution, and an array, BCDE. We can solve this problem by recursing.
How? Well, like this...

Move each element on the right-hand side to the left-hand side, in turn.

AB | CDE
AC | DEB
AD | EBC
AE | BCD

All you're really doing is rotating the BCDE part of the array, yes?

For each of those partitioned array arrangements, we now have to solve the
problem of permuting the right-hand side, and incorporating the left-hand
side, ***unchanged***, into the final result.

Take AB | CDE, for example. We have 2 unchanged elements, which will prefix
each solution, and an array, CDE. We can solve this problem by recursing.
How? Well, like this...

Move each element on the right-hand side to the left-hand side, in turn.

ABC | DE
ABD | EC
ABE | CD

All you're really doing is rotating the CDE part of the array, yes?

For each of those partitioned array arrangements, we now have to solve the
problem of permuting the right-hand side, and incorporating the left-hand
side, ***unchanged***, into the final result.

Take ABC | DE, for example. We have 3 unchanged elements, which will prefix
each solution, and an array, DE. We can solve this problem by recursing.
How? Well, like this...

Move each element on the right-hand side to the left-hand side, in turn.

ABCD | E
ABCE | D

All you're really doing is rotating the DE part of the array, yes?

For each of those partitioned array arrangements, we now have to solve the
problem of permuting the right-hand side, and incorporating the left-hand
side, ***unchanged***, into the final result. But since each right-hand
side only has 1 element therein, there is only one permutation.

So for each of them we just display the result. ABCDE, and ABCED.

Then we drop out of this recursion, and continue with the next, and so on
until everything is done.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at above domain (but drop the www, obviously)

CBFalconer 09-17-2006 02:07 PM

Re: Richard Heathfield's TSP permutation algorithm
 
Richard Heathfield wrote:
> whitehatmiracle@gmail.com said:
>
>> What does inner outer and unchanged do?

>
> Partition the array into two parts, an empty part, and the part
> you want to permute.
>
> | ABCDE
>
> Move each element on the right-hand side to the left-hand side,
> in turn.
>
> A | BCDE
> B | CDEA
> C | DEAB
> D | EABC
> E | ABCD
>
> All you're really doing is rotating the array, yes?
>
> For each of those partitioned array arrangements, we now have to
> solve the problem of permuting the right-hand side, and incorporating
> the left-hand side, ***unchanged***, into the final result.
>
> Take A | BCDE, for example. We have 1 unchanged element, which will
> prefix each solution, and an array, BCDE. We can solve this problem
> by recursing. How? Well, like this...
>
> Move each element on the right-hand side to the left-hand side,
> in turn.
>
> AB | CDE
> AC | DEB
> AD | EBC
> AE | BCD
>
> All you're really doing is rotating the BCDE part of the array, yes?
>
> For each of those partitioned array arrangements, we now have to solve
> the problem of permuting the right-hand side, and incorporating the
> left-hand side, ***unchanged***, into the final result.
>
> Take AB | CDE, for example. We have 2 unchanged elements, which will
> prefix each solution, and an array, CDE. We can solve this problem
> by recursing. How? Well, like this...
>
> Move each element on the right-hand side to the left-hand side, in
> turn.
>
> ABC | DE
> ABD | EC
> ABE | CD
>
> All you're really doing is rotating the CDE part of the array, yes?
>
> For each of those partitioned array arrangements, we now have to
> solve the problem of permuting the right-hand side, and incorporating
> the left-hand side, ***unchanged***, into the final result.
>
> Take ABC | DE, for example. We have 3 unchanged elements, which will
> prefix each solution, and an array, DE. We can solve this problem by
> recursing. How? Well, like this...
>
> Move each element on the right-hand side to the left-hand side, in
> turn.
>
> ABCD | E
> ABCE | D
>
> All you're really doing is rotating the DE part of the array, yes?
>
> For each of those partitioned array arrangements, we now have to
> solve the problem of permuting the right-hand side, and
> incorporating the left-hand side, ***unchanged***, into the final
> result. But since each right-hand side only has 1 element therein,
> there is only one permutation.
>
> So for each of them we just display the result. ABCDE, and ABCED.
>
> Then we drop out of this recursion, and continue with the next,
> and so on until everything is done.


And that very nice description applies equally to my 'jumble'
routine, which I published here earlier as a complete program.
Jumble also allows you to suppress any output past a preselected
length value.

Richards description, and my lack of one, illustrates why I do not
make a good teacher. I am repeating my actual heart code for
illustration below:

/* exchange 0th and ith char in wd */
void trade(char *wd, unsigned int i)
{
char c;

c = *wd;
*wd = wd[i];
wd[i] = c;
} /* trade */

/* ------------------ */

/* Form all n char permutations of the characters in the
string wd of length lgh into outstring at index ix.
Output the results to stdout. */
void jumble(char *wd, unsigned int lgh,
unsigned int ix, /* output place to fill */
unsigned int n, /* max out places to fill */
char *outstring)
{
unsigned int i;

if (0 == n) {
outstring[ix] = '\0';
puts(outstring);
}
else
for (i = 0; i < lgh; i++) {
trade(wd, i); /* nop when (0 == i) */
outstring[ix] = *wd;
jumble(wd+1, lgh-1, ix+1, n-1, outstring);
trade(wd, i); /* restore the wd string */
}
} /* jumble */

--
"The most amazing achievement of the computer software industry
is its continuing cancellation of the steady and staggering
gains made by the computer hardware industry..." - Petroski



--
Posted via a free Usenet account from http://www.teranews.com


whitehatmiracle@gmail.com 09-18-2006 01:23 PM

Re: Richard Heathfield's TSP permutation algorithm
 
WOW
Thanx a millions to Mr Richard Heathfield and to Mr CBFalconer
Now both your algorithms are clear, and even the ideas are clearer.

Thanx once again......


whitehatmiracle@gmail.com 09-20-2006 01:56 PM

Re: Richard Heathfield's TSP permutation algorithm
 
Just one last clarification:

for
A | BCDE
B | CDEA
inner is the right hand side of the bar |?
and outer is it the left hand side of the | bar?

The external most for loop does it permute only the first letter of the
combination??


if(unchanged < n)
{
for(outer = unchanged; outer < n; outer++)
{
temp = Perm[outer];
for(inner = outer; inner > unchanged; inner--)
{
Perm[inner] = Perm[inner - 1];
}
Perm[unchanged] = temp;
Permute(Perm,
n,
unchanged + 1);


for(inner = unchanged; inner < outer; inner++)
{
Perm[inner] = Perm[inner + 1];
}
Perm[outer] = temp;
}
}
else
{
printf("%s\n", Perm);
}


Richard Heathfield 09-20-2006 02:32 PM

Re: Richard Heathfield's TSP permutation algorithm
 
whitehatmiracle@gmail.com said:

> Just one last clarification:
>
> for
> A | BCDE
> B | CDEA
> inner is the right hand side of the bar |?
> and outer is it the left hand side of the | bar?


No. You leave everything to the left of the bar alone.

> The external most for loop does it permute only the first letter of the
> combination??


Nothing special about it at all. It just rotates the array and then recurses
to permute each rotation in turn.

> if(unchanged < n)


Have we finished yet? No? Okay, let's rotate the array...

> {
> for(outer = unchanged; outer < n; outer++)
> {
> temp = Perm[outer];
> for(inner = outer; inner > unchanged; inner--)
> {
> Perm[inner] = Perm[inner - 1];
> }
> Perm[unchanged] = temp;


This bit changes, say, BCDE into EBCD

> Permute(Perm,
> n,
> unchanged + 1);


This bit submits AEBCD to recursion.

> for(inner = unchanged; inner < outer; inner++)
> {
> Perm[inner] = Perm[inner + 1];
> }
> Perm[outer] = temp;


And this bit jiggles the array back into the right order so that the next
recursion level up from here works properly.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at above domain (but drop the www, obviously)

whitehatmiracle@gmail.com 09-21-2006 01:00 PM

Re: Richard Heathfield's TSP permutation algorithm
 
So actually what would be the appropriate name for outer and inner?


Richard Heathfield 09-21-2006 02:58 PM

Re: Richard Heathfield's TSP permutation algorithm
 
whitehatmiracle@gmail.com said:

> So actually what would be the appropriate name for outer and inner?


In the absence of contextual information to the contrary, the most
appropriate name for outer is outer.

In the absence of contextual information to the contrary, the most
appropriate name for inner is inner.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at above domain (but drop the www, obviously)


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