efficient permuation algorithm for multiple lists - c++

Question

I have a variable number of lists. Each contains different number of elements.
For instance with four lists,
array1 = {1, 2, 3, 4};
array2 = {a, b, c};
array3 = {X};
array4 = {2.10, 3.5, 1.2, 6.2, 0.3};
I need to find all possible tuples whose ith element is from ith list, e.g. {1,a,X,2.10}, {1,a,X,3.5}, ...
Currently I am using a recursive implementation which has performance issue. Therefore, I want to find a noniterative way that can perform faster.
Any advice? Is there any efficient algorithms (or some pseudo code). Thanks!
Some pseudo code of what I implemented so far:
Recusive version:
vector<size_t> indices; // store current indices of each list except for the last one)
permuation (index, numOfLists) { // always called with permutation(0, numOfLists)
if (index == numOfLists - 1) {
for (i = first_elem_of_last_list; i <= last_elem_of_last_list; ++i) {
foreach(indices.begin(), indices.end(), printElemAtIndex());
printElemAtIndex(last_list, i);
}
}
else {
for (i = first_elem_of_ith_list; i <= last_elem_of_ith_list; ++i) {
update_indices(index, i);
permutation(index + 1, numOfLists); // recursive call
}
}
}
non-recursive version:
vector<size_t> indices; // store current indices of each list except for the last one)
permutation-iterative(index, numOfLists) {
bool forward = true;
int curr = 0;
while (curr >= 0) {
if (curr < numOfLists - 1){
if (forward)
curr++;
else {
if (permutation_of_last_list_is_done) {
curr--;
}
else {
curr++;
forward = true;
}
if (curr > 0)
update_indices();
}
}
else {
// last list
for (i = first_elem_of_last_list; i <= last_elem_of_last_list; ++i) {
foreach(indices.begin(), indices.end(), printElemAtIndex());
printElemAtIndex(last_list, i);
}
curr--;
forward = false;
}
}
}

Answer 1

There are O(l^n)1 different such tuples, where l is the size of a list and n is the number of lists.
Thus, generating all of them cannot be done efficiently polynomially.
There might be some local optimizations that can be made - but I doubt swtiching between iterative and (efficient) recursive will do a lot of difference if any, especially if the iterative version is trying to mimic a recursive solution using a stack + loop, which is likely less optimized for this purpose then the hardware stack.
A possible recursive approach is:
printAll(list<list<E>> listOfLists, list<E> sol):
if (listOfLists.isEmpty()):
print sol
return
list<E> currentList <- listOfLists.removeAndGetFirst()
for each element e in currentList:
sol.append(e)
printAll(listOfLists, sol) //recursively invoking with a "smaller" problem
sol.removeLast()
listOfLists.addFirst(currentList)
(1) To be exact, there are l1 * l2 * ... * ln tuples, where li is the size of the ith list. for lists of equal length it decays to l^n.