Question

The ISBN-10 ID consists of ten

Answer 1

Backtracking

DFS { stack<locations> S; mark start visited; S.push(start); While (S is not empty) { t = S.top(); if (t == goal) { Success(S); return; } if (t has unvisited neighbors) { choose a next unvisited neighbor n; mark n visited; S.push(n); } else BackTrack(S); } Failure(S); } BackTrack(S) { while (!S.empty() && S.top() has no unvisited neighbors) S.pop(); } Success(S) { cout << "Path from " << goal << " to " << start << ": "; while (!S.empty()) { output (S.top()); S.pop(); } } Failure(S) { cout << "No path from " << start << " to " << goal; while (!S.empty()) S.pop(); }

Outcome. If there is a path in G from start to goal, DFS finds one such path.

vector<int> permutation(N); vector<int> used(N,0); void try(int which, int what) { // try taking the number "what" as the "which"-th element permutation[which] = what; used[what] = 1; if (which == N-1) outputPermutation(); else // try all possibilities for the next element for (int next=0; next<N; next++) if (!used[next]) try(which+1, next); used[what] = 0; } int main() { // try all possibilities for the first element for (int first=0; first<N; first++) try(0,first); }

In this case a trivial lower bound on the time complexity is the number of possible solutions. Backtracking algorithms are usually used to solve hard problems