Generating all n-letter permutations - c++

Question

I am attempting to calculate all the possible 3 letter permutations, using the 26 letters (Which amounts to only 26*25*24=15,600). The order of the letters matters, and I don't want repeating letters. (I wanted the permutations to be generated in lexicographical order, but that isn't necessary)
So far I attempted to nest for loops, but I ended up iterating through every combination possible. So there are repeating letters, which I do not want, and the for loops can become difficult to manage if I want more than 3 letters.
I can flip through the letters until I get a letter that has not been used, but it isn't in lexicographical order and it is much slower than using next_permutation (I cannot use this std method because I'm left calculating all of the subsets of the 26 letters).
Is there a more efficient way to do this?
To put in perspective of the inefficiency, next_permutation iterates through the first 6 digits instantaneously. However, it takes several seconds to get all the three letter permutations using this method, and next_permutation still quickly becomes inefficient with the 2^n subsets I must calculate.
Here is what I have for the nested for loops:
char key[] = {'a','b','c','d','e','f','g','h','i','j','k',
'l','m','n','o','p','r','s','t','u','v','w','x','y','z'};
bool used[25];
ZeroMemory( used, sizeof(bool)*25 );
for( int i = 0; i < 25; i++ )
{
while( used[i] == true )
i++;
if( i >= 25 )
break;
used[i] = true;
for( int j = 0; j < 25; j++ )
{
while( used[j] == true )
j++;
if( j >= 25 )
break;
used[j] = true;
for( int k = 0; k < 25; k++ )
{
while( used[k] == true )
k++;
if( k >= 25 )
break;
used[k] = true;
cout << key[i] << key[j] << key[k] << endl;
used[k] = false;
}
used[j] = false;
}
used[i] = false;
}

Answer 1

Make a root which represents the start of a combination, so it has no value.
calculate all the possible children (26 letter, 26 children...)
for each root child calculate possible children (so: remaining letters)
use a recursive limited-depth search to find your combinations.

Answer 2

This is a solution I would try if i just want a "simple" solution. I'm not sure how recource intensive this is so I suggest you start trying with a small set of letters.
a = {a...z}
b = {a...z}
c = {a...z}
for each(a)
{
for each(b)
{
for each(c)
{
echo a + b + c;
}
}
}

Answer 3

For a specific and small, n, manual loops like you have is the easiest way. However, your code can be highly simplified:
for(char a='a'; a<='z'; ++a) {
for(char b='a'; b<='z'; ++b) {
if (b==a) continue;
for(char c='a'; c<='z'; ++c) {
if (c==a) continue;
if (c==b) continue;
std::cout << a << b << c << '\n';
}
}
}
For a variable N, obviously we need a different strategy. And, it turns out, it needs an incredibly different strategy. This is based on DaMachk's answer, of using recursion to generate subsequent letters
template<class func_type>
void generate(std::string& word, int length, const func_type& func) {
for(char i='a'; i<='z'; ++i) {
bool used = false;
for(char c : word) {
if (c==i) {
used = true;
break;
}
}
if (used) continue;
word.push_back(i);
if (length==1) func(word);
else generate(word, length-1, func);
word.pop_back();
}
}
template<class func_type>
void generate(int length, const func_type& func) {
std::string word;
generate(word, length, func);
}
You can see it here
I also made an unrolled version, which turned out to be incredibly complicated, but is significantly faster. I have two helper functions: I have a function to "find the next letter" (called next_unused) which increases the letter at an index to the next unused letter, or returns false if it cannot. The third function, reset_range "resets" a range of letters from a given index to the end of the string to the first unused letter it can. First we use reset_range to find the first string. To find subsequent strings, we call next_unused on the last letter, and if that fails, the second to last letter, and if that fails the third to last letter, etc. When we find a letter we can properly increase, we then "reset" all the letters to the right of that to the smallest unused values. If we get all the way to the first letter and it cannot be increased, then we've reached the end, and we stop. The code is frightening, but it's the best I could figure out.
bool next_unused(char& dest, char begin, bool* used) {
used[dest] = false;
dest = 0;
if (begin > 'Z') return false;
while(used[begin]) {
if (++begin > 'Z')
return false;
}
dest = begin;
used[begin] = true;
return true;
}
void reset_range(std::string& word, int begin, bool* used) {
int count = word.size()-begin;
for(int i=0; i<count; ++i)
assert(next_unused(word[i+begin], 'A'+i, used));
}
template<class func_type>
void doit(int n, func_type func) {
bool used['Z'+1] = {};
std::string word(n, '\0');
reset_range(word, 0, used);
for(;;) {
func(word);
//find next word
int index = word.size()-1;
while(next_unused(word[index], word[index]+1, used) == false) {
if (--index < 0)
return; //no more permutations
}
reset_range(word, index+1, used);
}
}
Here it is at work.
And here it is running in a quarter of the time as the simple one

Answer 4

I was doing a similar thing in powershell. Generating all the possible combinations of 9 symbols. After a bit of trial and error this is what I came up with.
$S1=New-Object System.Collections.ArrayList
$S1.Add("a")
$S1.Add("b")
$S1.Add("c")
$S1.Add("d")
$S1.Add("e")
$S1.Add("f")
$S1.Add("g")
$S1.Add("h")
$S1.Add("i")
$S1 | % {$a = $_
$S2 = $S1.Clone()
$S2.Remove($_)
$S2 | % {$b = $_
$S3 = $S2.Clone()
$S3.Remove($_)
$S3 | % {$c = $_
$S4 = $S2.Clone()
$S4.Remove($_)
$S4 | % {$d = $_
$S5 = $S4.Clone()
$S5.Remove($_)
$S5 | % {$e = $_
$S6 = $S5.Clone()
$S6.Remove($_)
$S6 | % {$f = $_
$S7 = $S6.Clone()
$S7.Remove($_)
$S7 | % {$g = $_
$S8 = $S7.Clone()
$S8.Remove($_)
$S8 | % {$h = $_
$S9 = $S8.Clone()
$S9.Remove($_)
$S9 | % {$i = $_
($a+$b+$c+$d+$e+$f+$g+$h+$i)
}
}
}
}
}
}
}
}
}