Words of length k generation from a DFA - c++

Question

Good day,
I'm having a serious problem into generating a list with all the words of length k (the generate function is the function intended to generate all the words of length k, the other function is used to find out if a word is accepted or not) from a DFA just by using the DFS algorithm, here's my attempt:
#include<vector>
#include<iostream>
#include<fstream>
#include<string.h>
using namespace std;
vector < pair <int,char> > a[100];
int viz[100], v[100];
char s1[100];
void accepted (int q, char c, char s[100], int x) {
c = s[x];
viz[q] = 1;
cout << q;
for (int i = 0; i < a[q].size(); i++)
if (a[q][i].second == c) {
x++;
accepted (a[q][i].first, a[q][i+1].second, s, x);
}
}
void generate (int q, int k) {
int x = 0;
v[q] = 1;
while (x < k) {
cout << a[q][0].second;
for (int i = 0; i < a[q].size(); i++)
if (v[a[q][i].first] == 0)
{
cout << a[q][i].second;
generate(a[q][i].first, k);
}
x++;
}
}
int main() {
ifstream f ("input.txt");
int n, m, x, y, i, j, k;
char c;
char s[100];
f >> n >> m;
for (i = 0; i < m; i++) {
f >> x >> y;
f >> c;
a[x].push_back (make_pair(y,c));
}
for (i = 0; i < n; i++) {
cout << i << ": ";
for (j = 0; j < a[i].size(); j++)
cout << a[i][j].first << a[i][j].second << " ";
cout << endl;
}
cout << endl << "Fiite states: 2, 3" << endl << endl;
cin.get (s,100);
accepted(0, s[0], s, 0);
if (viz[2] == 1) cout << endl << "Accepted";
cout << endl;
cin >> k;
generate (0, k);
cout << endl;
return 0;
}
Also here's how my input looks like:
4 6
0 0 a
0 1 b
1 2 c
2 2 a
2 3 c
3 3 c
Here's how the DFA and the output would look like:
The serious problem I'm facing is that I can't find a way to ouput properly all the obtained words to the screen by calling the generate function.

Answer 1

I changed your generate function as below. Following that is an explanation as to what I thought and how I changed it.
void generate (int q, int k, string &s) {
if (k > 0) {
for (int i = 0; i < a[q].size(); i++)
{
s += a[q][i].second;
generate(a[q][i].first, k-1, s);
s.pop_back();
}
}
else {
cout << s << endl;
}
}
First and foremost, you were attempting a mixture of recursive and repetitive version of DFS, but you had no structure for keeping the stack if you were, and I doubt, going for a repetitive version using an explicit stack. Basically, your outer while loop was wrong, as the depth should increase as you recursively traverse the graph and not repetitively at a single level of recursion using a while loop as you did. You could also, as I mentioned, have a repetitive approach and use an explicitly defined stack other than the one implicitly used by memory when you implement DFS recursively. But it was easier and more intuitive to get a grasp of DFS with its recursive implementation, so I left out the outer loop.
Secondly, keeping a list of visited nodes is not a good idea as you want to list all k-length strings and your DFA is not a simple graph. (i.e. there may exist edges from node u to node u) So I removed the if statement inside the for loop as you could visit the sane node multiple times. Your approach is exponential based on the branching factor of the DFA, but if your k is small enough, it should work regardless of that. And the approach being exponential is not the problem to the solution of which you are looking with this question.
Thirdly, probably due to your usage of while loop, there was a mixup with printing a single character at each level, which is incorrect. Remember, at every node of depth k, you have to print all the characters you encountered starting from the root of the tree. That is why I added a string as the third parameter to your function. Don't worry, though, it's passed by reference and it will only cause an addition of O(k) space complexity to your algorithm, which should be negligible.
If in your main function you start traversing using the call below, you will find that it works properly.
string S;
generate(0, k, S);