如何计算回溯算法的时间复杂度？

Question

How to calculate time complexity for these backtracking algorithms and do they have same time complexity? If different how? Kindly explain in detail and thanks for the help.

1. Hamiltonian cycle:

        bool hamCycleUtil(bool graph[V][V], int path[], int pos) {
            /* base case: If all vertices are included in Hamiltonian Cycle */
            if (pos == V) {
                // And if there is an edge from the last included vertex to the
                // first vertex
                if ( graph[ path[pos-1] ][ path[0] ] == 1 )
                    return true;
                else
                    return false;
            }

            // Try different vertices as a next candidate in Hamiltonian Cycle.
            // We don't try for 0 as we included 0 as starting point in in hamCycle()
            for (int v = 1; v < V; v++) {
                /* Check if this vertex can be added to Hamiltonian Cycle */
                if (isSafe(v, graph, path, pos)) {
                    path[pos] = v;

                    /* recur to construct rest of the path */
                    if (hamCycleUtil (graph, path, pos+1) == true)
                        return true;

                    /* If adding vertex v doesn't lead to a solution, then remove it */
                    path[pos] = -1;
                }
            }

            /* If no vertex can be added to Hamiltonian Cycle constructed so far, then return false */
            return false;
        }

2. Word break:

       a. bool wordBreak(string str) {
            int size = str.size();

            // Base case
            if (size == 0)
                return true;

            // Try all prefixes of lengths from 1 to size
            for (int i=1; i<=size; i++) {
                // The parameter for dictionaryContains is str.substr(0, i)
                // str.substr(0, i) which is prefix (of input string) of
                // length 'i'. We first check whether current prefix is in
                // dictionary. Then we recursively check for remaining string
                // str.substr(i, size-i) which is suffix of length size-i
                if (dictionaryContains( str.substr(0, i) ) && wordBreak( str.substr(i, size-i) ))
                    return true;
            }

            // If we have tried all prefixes and none of them worked
            return false;
        }
    b. String SegmentString(String input, Set<String> dict) {
           if (dict.contains(input)) return input;
           int len = input.length();
           for (int i = 1; i < len; i++) {
               String prefix = input.substring(0, i);
               if (dict.contains(prefix)) {
                   String suffix = input.substring(i, len);
                   String segSuffix = SegmentString(suffix, dict);
                   if (segSuffix != null) {
                       return prefix + " " + segSuffix;
                   }
               }
           }
           return null;
      }


3. N Queens:

        bool solveNQUtil(int board[N][N], int col) {
            /* base case: If all queens are placed then return true */
            if (col >= N)
                return true;

            /* Consider this column and try placing this queen in all rows one by one */
            for (int i = 0; i < N; i++) {
                /* Check if queen can be placed on board[i][col] */
                if ( isSafe(board, i, col) ) {
                    /* Place this queen in board[i][col] */
                    board[i][col] = 1;

                    /* recur to place rest of the queens */
                    if ( solveNQUtil(board, col + 1) == true )
                        return true;

                    /* If placing queen in board[i][col] doesn't lead to a solution then remove queen from board[i][col] */
                    board[i][col] = 0; // BACKTRACK
                }
            }
        }

I am actually confused a bit, as for Word Break(b) the complexity is O(2 ⁿ ) but for Hamiltonian cycle its different and so does for printing different permutations of the same string and then again for solving n queens problem.

Answer 1

In short:

1. Hamiltonian cycle : O(N!) in the worst case
2. WordBreak and StringSegment : O(2^N)
3. NQueens : O(N!)

Note: For WordBreak there is an O(N^2) dynamic programming solution.

---

More details:

1. In Hamiltonian cycle, in each recursive call one of the remaining vertices is selected in the worst case. In each recursive call the branch factor decreases by 1. Recursion in this case can be thought of as n nested loops where in each loop the number of iterations decreases by one. Hence the time complexity is given by:

   T(N) = N*(T(N-1) + O(1))
T(N) = N*(N-1)*(N-2).. = O(N!)

2. Similarly in NQueens, each time the branching factor decreases by 1 or more, but not much, hence the upper bound of O(N!)

3. For WordBreak it is more complicated but I can give you an approximate idea. In WordBreak each character of the string has two choices in the worst case, either to be the last letter in the previous word, or to be the first letter of a new word, hence the branching factor is 2. Therefore for both WordBreak & SegmentString T(N) = O(2^N)

Answer 2

Backtracking algo:

n-queen problem: O(n!)

graph coloring problem: O(nm^n) //where n=no. of vertex,m=no. of color used

hamilton cycle: O(N!)

WordBreak and StringSegment: O(2^N)

subset sum problem: O(nW)