Question

I’ve been practicing for an upcoming programming competition and I have stumbled across a question that I am just completely bewildered at. However, I feel as though it’s a concept I should learn now rather than cross my fingers that it never comes up.

Basically, it deals with a knight piece on a chess board. You are given two inputs: starting location and ending location. The goal is to then calculate and print the shortest path that the knight can take to get to the target location.

I’ve never dealt with shortest-path-esque things, and I don’t even know where to start. What logic do I employ to go about tackling this?

P.S. If it’s of any relevance, they want you to supplement the knight’s normal moves by also allowing it to move to the four corners of the square formed by the (potentially) eight moves a knight can make, given that the center of the square is the knight’s location.

Answer 1

You have a graph here, where all available moves are connected (value=1), and unavailable moves are disconnected (value=0), the sparse matrix would be like:

(a1,b3)=1,
(a1,c2)=1,
  .....

And the shortest path of two points in a graph can be found using http://en.wikipedia.org/wiki/Dijkstra's_algorithm

Pseudo-code from wikipedia-page:

function Dijkstra(Graph, source):
   for each vertex v in Graph:           // Initializations
       dist[v] := infinity               // Unknown distance function from source to v
       previous[v] := undefined          // Previous node in optimal path from source
   dist[source] := 0                     // Distance from source to source
   Q := the set of all nodes in Graph
   // All nodes in the graph are unoptimized - thus are in Q
   while Q is not empty:                 // The main loop
       u := vertex in Q with smallest dist[]
       if dist[u] = infinity:
          break                         // all remaining vertices are inaccessible from source
       remove u from Q
       for each neighbor v of u:         // where v has not yet been removed from Q.
           alt := dist[u] + dist_between(u, v) 
           if alt < dist[v]:             // Relax (u,v,a)
               dist[v] := alt
               previous[v] := u
   return dist[]

EDIT:

1. as moron, said using the
   http://en.wikipedia.org/wiki/A*_algorithm
   can be faster.
2. the fastest way, is
   to pre-calculate all the distances
   and save it to a 8×8 full matrix.
   well, I would call that cheating,
   and works only because the problem
   is small. But sometimes competitions
   will check how fast your program
   runs.
3. The main point is that if you are preparing
   for a programming competition, you must know
   common algorithms including Dijkstra’s.
   A good starting point is reading
   Introduction to Algorithms ISBN 0-262-03384-4.
   Or you could try wikipedia, http://en.wikipedia.org/wiki/List_of_algorithms