Generate character combinations based on arbitrary string and length - similar to permutations - algorithm

Question

This question was asked before in other languages but not delphi after searching SO.
see this question:How to Generate Permutations With Repeated Characters and this question: Generate all combinations of arbitrary alphabet up to arbitrary length and this one: How to generate combination of fix length strings using a set of characters?
so the question is not new but I am having a hard time translating any of this to delphi.
What I'm trying to do is generate combinations that does include repeats of characters such as this:
if we have a string of characters (specified by user): ABC and we want to generate length of three characters (also length specified by user) I would get:
AAA AAB AAC ABA ABB ABC ACA ACB ACC BAA BAB BAC etc...
This code seems to do this but in C++:
int N_LETTERS = 4;
char alphabet[] = {'a', 'b', 'c', 'd'};
std::vector<std::string> get_all_words(int length)
{
std::vector<int> index(length, 0);
std::vector<std::string> words;
while(true)
{
std::string word(length);
for (int i = 0; i < length; ++i)
word[i] = alphabet[index[i]];
words.push_back(word);
for (int i = length-1; ; --i)
{
if (i < 0) return words;
index[i]++;
if (index[i] == N_LETTERS)
index[i] = 0;
else
break;
}
}
}
This also seems to do this:
#include <iostream>
#include <string>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
using namespace std;
void displayPermutation(string permutation[], int length){
int i;
for (i=0;i<length;i++){
cout<<permutation[i];
}
cout << endl;
}
void getPermutations(string operatorBank[], int operatorCount,
string permutation[],int permutationLength, int curIndex){
int i;
//stop recursion condition
if(curIndex == permutationLength){
displayPermutation(permutation,permutationLength);
}
else{
for(i = 0; i < operatorCount; i++){
permutation[curIndex] = operatorBank[i];
getPermutations(operatorBank,operatorCount,permutation,
permutationLength,curIndex+1);
}
}
}
int main ()
{
int operatorCount = 4;
int permutationLength = 3;
string operatorBank[] = {"+","-","*","/"};
string permutation[] = {"","","",""}; //empty string
int curIndex = 0;
getPermutations(operatorBank,operatorCount,permutation,
permutationLength,curIndex);
return 0;
}
closest to what I want in delphi is found here but does not allow AAA for example:
http://www.swissdelphicenter.ch/torry/showcode.php?id=1032
And no this is not homework in case you are guessing. No other motive but just learning.
EDIT3:
Removed all irrelevant code from question to make it easier for other people to read it and get to the answers below. Look under answers for 2 different methods to accomplish this: one using recursion and the other by using a counter function.

Answer 1

The examples you show make this considerably more complex than necessary, at least IMO.
What you're really looking at is a 3 digit, base 3 number. You can just count from 0 to 33 = 27, then convert each number to base 3 (using 'A', 'B' and 'C' as your digits instead of '0', '1' and '2').
In C++, the conversion could look something like this:
std::string cvt(int in) {
static const int base = 3;
static const int digits = 3;
std::string ret;
for (int i = 0; i<digits; i++) {
ret.push_back('A' + in % base);
in /= base;
}
return std::string(ret.rbegin(), ret.rend());
}
With the conversion in place, producing all the combinations becomes utterly trivial:
for (int i = 0; i < 27; i++)
std::cout << cvt(i) << "\t";
I believe converting that to Delphi should be barely short of purely mechanical -- assignments change from = to :=, % becomes mod, the integer division changes to div, the for loop changes to something like for i = 0 to 27 do, and so on. The most tedious (but ultimately quite simple) part will probably be dealing with the fact that in C++, char is simply a small integer type, on which you can do normal integer math. At least if memory serves, in Pascal (or a derivative like Delphi) you'll need ord to convert from a character to an ordinal, and chr to convert back from ordinal to character. So the 'A' + in % base; will end up something more like chr(ord('A') + in mod base);
Like I said though, it seems like nearly the entire translation could/should end up almost completely mechanical, with no requirement for real changes in how the basic algorithms work, or anything like that.

Answer 2

Not exactly following your sequence of output, but following a sequence similar to the way binary numbers add up...
0001
0010
0011
0100
...
The idea is simple: loop index values in an array indicating which character to use at the respective position to compose the output combination string. No recursion required.
NextCombination updates the index array so the next combination is defined, it returns true as long as not all combinations are formed. False when back to all 0's.
DefineCombinations accepts a string with chars to use (for example 'ABC') and a size of the combined string (eg: 3): this adds the following sequence to a memo:
AAA, AAB, AAC, ABA, ABB, ABC, ACA, ACB, ACC, BAA, BAB, BAC, BBA, BBB, BBC, BCA, BCB, BCC, CAA, CAB, CAC, CBA, CBB, CBC, CCA, CCB, CCC
Adapt as you wish.
function TForm1.NextCombination(var aIndices: array of Integer; const MaxValue: Integer): Boolean;
var Index : Integer;
begin
Result:=False;
Index:=High(aIndices);
while not(Result) and (Index >= Low(aIndices)) do
begin
if (aIndices[Index] < MaxValue) then
begin
{ inc current index }
aIndices[Index]:=aIndices[Index] + 1;
Result:=True;
end
else
begin
{ reset current index, process next }
aIndices[Index]:=0;
Dec(Index);
end;
end;
end;
procedure TForm1.DefineCombinations(const Chars: String; const Size: Integer);
var aIndices : array of Integer;
Index : Integer;
sData : String;
begin
try
SetLength(sData, Size);
SetLength(aIndices, Size);
repeat
for Index:=Low(aIndices) to High(aIndices) do
sData[Index + 1]:=Chars[aIndices[Index] + 1];
memo1.Lines.Add(sData);
until not(NextCombination(aIndices, Length(Chars) - 1));
finally
SetLength(aIndices, 0);
SetLength(sData, 0);
end;
end;
Let me know if I missed something from the original question.

Answer 3

Here it is done with Recursion (Credit to this post's accepted answer:How to Generate Permutations With Repeated Characters
program Combinations;
{$APPTYPE CONSOLE}
{$R *.res}
uses
System.SysUtils;
procedure displayPermutation(const permutation : array of char; ilength: integer);
var
i: integer;
begin
for i := 0 to ilength - 1 do
begin
if i mod iLength = 0 then
writeln('')
else write(permutation[i]);
end;
end;
procedure getPermutations(const operatorBank: array of char; operatorCount: integer;
permutation: array of char; permutationLength, curIndex: integer);
var
i : integer;
begin
//stop recursion condition
if(curIndex = permutationLength)then
displayPermutation(permutation, permutationLength)
else
for i := 0 to operatorCount - 1 do
begin
permutation[curIndex] := operatorBank[i];
getPermutations(operatorBank,operatorCount,permutation,
permutationLength,curIndex+1);
end;
end;
var
operatorBank,permutation : array of char;
i, permutationLength, curIndex, operatorCount: integer;
Q, S : String;
begin
try
Q := ' ';
S := ' ';
while (Q <> '') and (S <> '') do
begin
Writeln('');
Write('P(N,R) N=? : ');
ReadLn(Q);
operatorCount := Length(Q);
setLength(operatorBank,operatorCount);
for i := 0 to operatorCount - 1 do
operatorBank[i] := Q[i+1];
Write('P(N,R) R=? : ');
ReadLn(S);
if S <> '' then permutationLength := StrToInt(S) + 1;
SetLength(permutation,operatorCount);
curIndex := 0;
Writeln('');
getPermutations(operatorBank, operatorCount, permutation,
permutationLength, curIndex );
end;
except
on E: Exception do
Writeln(E.ClassName, ': ', E.Message);
end;
end.