# Understanding Backtracking in C++

I have a good basic understanding of the fundamentals of C++, I also have an understanding of how recursion works too. I came across certain problems like the classic eight queens problem and solving a Sudoku with Backtracking.

I realize that I'm quite lost when it comes to this, I can't seem to be able to get my mind around the concept of going back in the recursion stack and starting again in order to solve the problem. It seems easy with a pen and paper but when it comes to writing code for this, I'm confused on how to begin attacking these problems.

It would be helpful if there were a tutorial aimed at beginners to backtracking or if there were a good book where this was covered. If somebody can shed light on this topic or give me some links to decent references, I'd be really grateful.

And yes I do know that it would be easier in functional languages but I'd like to understand the implementation in imperative languages too.

• I think this is a good question, but I think it'd be better to emphasize the request for someone to explain backtracking over asking for tutorials or other resources. An in-depth explanation kind of answer beats a list of references any day. Jun 28, 2011 at 21:42
• It would be perfect if somebody could give a detailed explanation, but i wouldn't mind reading up references either. Its just that I don't know where to start from. Jun 28, 2011 at 21:55

... I can't seem to be able to get my mind around the concept of going back in the recursion stack and starting again in order to solve the problem.

In backtracking, you are not starting again. Instead, you iterate through all options at the current situation.

Think about finding solution for a maze. At one point where you have two different paths, you try the left one first. If the left one does not lead you to the exit, you return to the point and try the other path. That's how backtracking works. In 8 Q and other problems where backtracking can be used, the confusing part is in the problem domain - how to iterate through your options in a given situation in a deterministic way.

EDIT: the following is a pseudo code helping understanding backtracking.

``````# depending on the problem, backtracking is not necessarily calling the
# method itself directly. for now, let's just stick with the simple case.

def backtracking(state)
option_list = state.get_all_options
option_list.each {|option|
state.apply option
return resolved if state.is_resolved
return resolved if backtracking(state) == resolved
state.undo option
}
return not_resolved
end
``````

For 8Q question:

• state.get_all_options would return a list of the possible positions for the next queen
• state.is_resolved would test if all queens are on the board and if they are good with each other.
• state.apply and state.undo will modify the board to apply or undo a positioning.
• The first recursive code I wrote (in 1984 using Pascal) for an assignment was a maze solving algorithm. Jun 28, 2011 at 23:49
• Know of some simple assignment where I can actually write code to get the actual feel of this stuff. Jun 29, 2011 at 8:09
• @nikhil: are you asking if there are some simple problem? It is better to write some pseudo code to demonstrate the generic routing of backtracking. I will try one later in a reply. Jun 29, 2011 at 15:23
• Yes exactly, that'll be most helpful. Jun 29, 2011 at 15:34
• Thank you so much, I've been reading some stuff lately. Slowly but steadily my understanding is improving. Jul 1, 2011 at 20:02

You have seen a program to walk a binary tree, right? It looks like this:

``````void walk(node* p){
if (p == NULL) return;  // this is backtracking
else if (WeWin(p)){
// print We Win !!
// do a Throw, or otherwise quit
}
else {
walk(p->left);   // first try moving to the left
walk(p->right);  // if we didn't win, try moving to the right
// if we still didn't win, just return (i.e. backtrack)
}
}
``````

• can't you return a bool/int to check if the solution is found in the subtree? `else{return walk(p->left)||walk(p->right));}` no need for throw for the expected result Jun 28, 2011 at 22:30