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Solving Combinatorial Problems with STL and Backtracking

February 2000/Solving Combinatorial Problems with STL and Backtracking/Listing 1

Listing 1: An STL-style backtracking functor

#ifndef BackTrack_h
#define BackTrack_h

template <class T, class I, class V>
class BackTrack {

   // precondition: first <= last
   BackTrack(const T&  first, const T&  last);

   // Finds the next solution to the problem. Repeated calls 
   // will find all solutions to a problem if multiple solutions 
   // exist.
   // Returns true if a solution was found.
   // Set first to true to get the first solution.
   bool operator() (const I& begin, I end, bool& first);

   // Finds the next valid sibling of the leaf (end-1). 
   // Returns true is if a valid sibling was found.
   bool FindValidSibling (const I& begin, const I& end);

   // Backtracks through the decision tree until it finds a node
   // that hasn't been visited. Returns true if an unvisited 
   // node was found.
   bool VisitNewNode (const I& begin, I& end);

   void CreateLeftLeaf (I& end);

   T left_child;
   T right_child;

   V IsValid;

template <class T, class I, class V>
BackTrack<T,I,V>::BackTrack(const T& first, const T& last) 
   : left_child (first), right_child (last)

template <class T, class I, class V>
bool BackTrack<T,I,V>::VisitNewNode(const I& begin, I& end)
   // If the current node is the rightmost child we must 
   // backtrack one level because there are no more children at 
   // this level. So we back up until we find a non-rightmost 
   // child, then generate the child to the right. If we back 
   // up to the top without finding an unvisted child, then all 
   // nodes have been generated.

   // Back up as long as the node is the rightmost child of 
   // its parent.
   while (end-begin > 0 && *(end-1) == right_child)

   I back = end-1;
   if (end-begin > 0 && *back != right_child) {
      return true;
   } else
      return false;

template <class T, class I, class V> 
bool BackTrack<T,I,V>::FindValidSibling
(const I& begin, const I& end)
   // Note that on entry to this routine the leaf has not yet 
   // been used or tested for validity, so the leaf is 
   // considered a "sibling" of itself.

   I back = end-1;
   while (true) {
      if (IsValid (begin, end))
         return true;
      else if (*back != right_child)
         return false;

template <class T, class I, class V>
inline void BackTrack<T,I,V>::CreateLeftLeaf (I& end)
   *(end++) = left_child;

template <class T, class I, class V> 
bool BackTrack<T,I,V>::operator () 
(const I& begin, I end, bool& first)
   const int size = end - begin;

   // If first time, need to create a root.
   // Otherwise need to visit new node since vector
   // contains the last solution.
   if (first) {
      first = false;
      end = begin;
      CreateLeftLeaf (end);
   } else if (!VisitNewNode (begin, end))
      return false;

   while (true) {
      if (FindValidSibling (begin, end)) {
         if (end - begin < size)
            CreateLeftLeaf (end);
            return true;

      } else if (!VisitNewNode (begin, end))
         return false; // the tree has been exhausted, 
                       // so no solution exists.

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