I'm trying to write a D2 app and need it to list every combination of a
series of letters (or numbers).

A formula that could do strings as well as integers would be nice, but not
necessary.  If it could take a number/string of any length would also be
nice, but i could do that myself once i figure out the formula.

example:
  if you started with ABCD, then i want:
  ABCD
  ABDC
  ACBD
  ACDB
  etc.

i don't care what order the letters are in as long as i get all of them

I know there must be a formula for this, but i can't seem to figure it out.

BTW there should be 120 different combonations of 4 letters (4! = 120).
--

Eli{*word*106}inson
Lightning Computer Techologies, Inc.
el...@nyc.pipeline.com
http://www.nethosting.com/~eli/

This will probably help in the debuging process ;-)

Best regards,
Frank
--
// Frank Mikalsen, System Developer, Finale a.s      
// Homepage: http://home.sol.no/frankm
// Author of ShareWare: Silent Partner Backup Screensaver v2.60
// Download: http://www.winsite.com/pc/win3/desktop/spbck260.zip

In article <52a2p4$...@news1.t1.usa.pipeline.com>,
   el...@nyc.pipeline.com(Eli{*word*106}inson) wrote:
]-I'm trying to write a D2 app and need it to list every combination of a
]-series of letters (or numbers).
]-
]-A formula that could do strings as well as integers would be nice, but not
]-necessary.  If it could take a number/string of any length would also be
]-nice, but i could do that myself once i figure out the formula.
]-
]-example:
]-  if you started with ABCD, then i want:
]-  ABCD
]-  ABDC
]-  ACBD
]-  ACDB
]-  etc.
]-
]-i don't care what order the letters are in as long as i get all of them
]-
]-I know there must be a formula for this, but i can't seem to figure it out.
]-
]-BTW there should be 120 different combonations of 4 letters (4! = 120).

check with SimTel or Garbo, and download the latest update
to SWAG.  in the math section there are several code snippets
for obtaining all the permutations of a set of [integers,
characters, whatever...]

if you want a thoroughly documented solution, check Chapter 6,
Section 5 of "Scientific Pascal" by Harley Flanders
( http://www.birkhauser.com/cgi-win/ISBN/0-8176-3760-5 )

Mark Vaughan

      4 * 1 =  4
      4 * 2 =  8
      4 * 3 = 12
      4 * 4 = 16
              --
              120

Lawrence C. Thurman Jr.

        Seems like some sort of recursive gizmo is called for - the exact
syntax would depend on what sort of format you want for output. (And then
it might or might not be worthwhile to unwrap the recursion into a loop.)

--
David Ullrich

?his ?s ?avid ?llrich's ?ig ?ile
(Someone undeleted it for me...)

CFC <cfis...@mail.idt.net> wrote in article <3249722A.4...@mail.idt.net>...

x! = x * (x-1) * (x-2) * ... * 1

ergo,

4! = 4 * 3 * 2 * 1 = 24

Verify with the Calculator that comes with Windows if you don't believe me!

Where did *you* learn about factorials?  ;)

-------------
Steve Neumann
sneum...@bridge.com

On Sep 25, 1996 15:11:18 in article <Re: could someone help me? (mostly
math problem)>, '"Steven C. Neumann" <sneum...@bridge.com>' wrote:

Eli{*word*106}inson
Lightning Computer Techologies, Inc.
el...@nyc.pipeline.com
http://www.nethosting.com/~eli/

Sorry, but 4! = 24.

I didn't give this much thought. There is probably a
nice recursive algorithm but anyway...

Here is one possible way:
If you generate all possibilities with the first letter
ABCD
ABDC
ACBD
ACDB
ADBC
ADCB
Then all you have to do is rotate each of these 3 (n-1) times.
ABCD->BCDA->CDAB->DABC
ABDC->BDCA->DCAB->CABD
ACBD->CBDA->BDAC->DACB
...

Good Luck,
Al

.

A factorial is the product of numbers from 1 to n. So 4! = 24 and 5! =
120.

--
The reason people get lost in thought...
is because its unfamiliar territory.

Here is a little Pascal function I made up that will return the Nth
combination for an original string of any length.  Calling it with an
increasing N value until you get your original string back will yield
all possible combinations.

Function Shuffle(S:String; N:Longint):String
Var
   p,r,j:integer;
   C:char;
begin
   for p:=lenght(S) downto 2 do
     begin
        r:= N mod p;
        N:= N div p;
        for j:= p-r to p-1 do
           begin
             C:=S[j];
             S[j]:=S[j+1];
             S[j+1]:=C;
           end;
      end;
     Shuffle:=S;
end;

To use the function call it several times with your original string and
with an increasing N value.

Here is an example calling loop:

S0:='ABCDE';  {S0 and S1 are strings.  SO is the original}
N:=0;
repeat
  begin
    N:=N+1;
    S1:=shuffle(S0,N);
    ListBox1.items.add(S1);
  end
until S1=S0;
ListBox1.Sorted:=true;  {This makes it easyer to check}

Hope this helps...

BTW feel free to submit more little problems like this.
That one was fun.

Alain Toutant.

Hello,

Here is a little Pascal function I made up that will return the Nth
combination for an original string of any length.  Calling it with an
increasing N value until you get your original string back will yield
all possible combinations.

Function Shuffle(S:String; N:Longint):String
Var
   p,r,j:integer;
   C:char;
begin
   for p:=lenght(S) downto 2 do
     begin
        r:= N mod p;
        N:= N div p;
        for j:= p-r to p-1 do
           begin
             C:=S[j];
             S[j]:=S[j+1];
             S[j+1]:=C;
           end;
      end;
     Shuffle:=S;
end;

To use the function call it several times with your original string and
with an increasing N value.

Here is an example calling loop:

S0:='ABCDE';  {S0 and S1 are strings.  SO is the original}
N:=0;
repeat
  begin
    N:=N+1;
    S1:=shuffle(S0,N);
    ListBox1.items.add(S1);
  end
until S1=S0;
ListBox1.Sorted:=true;  {This makes it easyer to check}

Hope this helps...

BTW feel free to submit more little problems like this.
That one was fun.

Alain Toutant.

Here is a little Pascal function I made up that will return the Nth
combination for an original string of any length.  Calling it with an
increasing N value until you get your original string back will yield
all possible combinations.

Function Shuffle(S:String; N:Longint):String
Var
   p,r,j:integer;
   C:char;
begin
   for p:=lenght(S) downto 2 do
     begin
        r:= N mod p;
        N:= N div p;
        for j:= p-r to p-1 do
           begin
             C:=S[j];
             S[j]:=S[j+1];
             S[j+1]:=C;
           end;
      end;
     Shuffle:=S;
end;

To use the function call it several times with your original string and
with an increasing N value.

Here is an example calling loop:

S0:='ABCDE';  {S0 and S1 are strings.  SO is the original}
N:=0;
repeat
  begin
    N:=N+1;
    S1:=shuffle(S0,N);
    ListBox1.items.add(S1);
  end
until S1=S0;
ListBox1.Sorted:=true;  {This makes it easyer to check}

Hope this helps...

BTW feel free to submit more little problems like this.
That one was fun.

Alain Toutant.

Here is a little Pascal function I made up that will return the Nth
combination for an original string of any length.  Calling it with an
increasing N value until you get your original string back will yield
all possible combinations.

Function Shuffle(S:String; N:Longint):String
Var
   p,r,j:integer;
   C:char;
begin
   for p:=lenght(S) downto 2 do
     begin
        r:= N mod p;
        N:= N div p;
        for j:= p-r to p-1 do
           begin
             C:=S[j];
             S[j]:=S[j+1];
             S[j+1]:=C;
           end;
      end;
     Shuffle:=S;
end;

To use the function call it several times with your original string and
with an increasing N value.

Here is an example calling loop:

S0:='ABCDE';  {S0 and S1 are strings.  SO is the original}
N:=0;
repeat
  begin
    N:=N+1;
    S1:=shuffle(S0,N);
    ListBox1.items.add(S1);
  end
until S1=S0;
ListBox1.Sorted:=true;  {This makes it easyer to check}

Hope this helps...

BTW feel free to submit more little problems like this.
That one was fun.

Alain Toutant.

Here's something I found in PC Magazine's Tutbo Pascal 6.0 Techniques and
Utilities:
(Modified by me)

type
  PermString = string[13];

function Factorial( B : byte ) : extended;
  var n: byte;  
begin
   Result := 1;
   for n := 1 to B
   do Result := Result * N;
end;

procedure XChg( var A, B : char )
  var T : char;
begin
   T := A;
   A := B;
   B := T;
end;

function PermOf( Org : PermString; L : longint ) : PermString;
  var P, Q : byte;
      Nump : longint;
begin
  Result := '';
  if L < 1
  then exit;

  Nump := round( Factorial( length( Org )));
  if L > Nump
  then exit;

  dec( L );
  P := 0;
  for Q := length( Org ) downto 2
  do begin
       inc( P );
       Nump := Nump div Q;
       XChg( Org[P], Org[P + (L DIV Nump)]);
       L := L mod Nump;
     end;
  Result := Org;
end;

To use:

var N : longint;
begin
  for N := 1 to round( Factorial( 5 ))
  do write( PermOf( 'abcde', N) : 8);
end;

Eli{*word*106}inson <el...@nyc.pipeline.com> wrote in article
<52a2p4$...@news1.t1.usa.pipeline.com>...

Go ahead and find 120 different permutations of ABCD, and
drop me a note when you have them all ;-)

Best regards,
Frank
--
// Frank Mikalsen, System Developer, Finale a.s      
// Homepage: http://home.sol.no/frankm
// Author of ShareWare: Silent Partner Backup Screensaver v2.60
// Download: http://www.winsite.com/pc/win3/desktop/spbck260.zip