CS533 Midterm Exam Study Guide

The Midterm will cover everything we have discussed in class up through October 27. This includes the following:

I am worried that I didn't do a very good job of explaining A* search in class. Here is another attempt. Suppose that we are trying to find the shortest feasible schedule in a job-shop scheduling problem space. At each state x in the problem space we have two pieces of information. First, we have g(x) which gives the length of the current schedule x. Second, we have a function h(x) that attempts to estimate how much longer the schedule will become if we apply the best possible operators to x from this state onward. This function h(x) is called the heuristic function, and it always under-estimates how much longer the schedule will become.

Under these assumptions, f(x) = g(x) + h(x) is our estimate of best schedule length that can be achieved by starting at x and applying the best possible operators until a feasible solution is found. With these assumptions, A* can be written as

class state
  list<state> * expand();  // method for expanding a state
  float g();             // returns value of g()
  float h();             // returns value of h()
  float f() {return g() + h();};

  // heap uses < to find the smallest element, so states will be
  // expanded in this order.
  int operator < (state & other) {
     return f() < other.f();

// main search routine:
heap<state> frontier();
while (! frontier.isEmpty()) {
   state current = frontier.deleteMin();   
   if (state.feasible()) {
      cout << "The solution is " << state << endl;
   // expand state and add all children to the frontier
   list<state> * l = state.expand();
   listIterator<state> itr(*l);
   for (itr.init(); !itr; ++itr) frontier.add(itr());
Consider the following search tree. Next to each node are values for g and h. A* will expand the nodes in the order A, C, F, B, E, K, and find the optimal solution. Best-first search would only use g to order the nodes. It would expand the nodes in the order A, C, F, B, D, E, K, so it would waste one expansion step.