Solver for TSP-like Puzzle, perhaps in Javascript

I have created a puzzle which is a derivative of the travelling salesman problem, which I call Trace Perfect.

It is essentially an undirected graph with weighted edges. The goal is to traverse every edge at least once in any direction using mi开发者_开发知识库nimal weight (unlike classical TSP where the goal is to visit every vertex using minimal weight).

As a final twist, an edge is assigned two weights, one for each direction of traversal.

I create a new puzzle instance everyday and publish it through a JSON interface.

Now I know TSP is NP-hard. But my puzzles typically have only a good handful of edges and vertices. After all they need to be humanly solvable. So a brute force with basic optimization might be good enough.

I would like to develop some (Javascript?) code that retrieves the puzzle from the server, and solves with an algorithm in a reasonable amount of time. Additionally, it may even post the solution to the server to be registered in the leader board.

I have written a basic brute force solver for it in Java using my back-end Java model on the server, but the code is too fat and runs out of heap-space quick, as expected.

Is a Javascript solver possible and feasible?

The JSON API is simple. You can find it at: http://service.traceperfect.com/api/stov?pdate=20110218 where pdate is the date for the puzzle in yyyyMMdd format.

Basically a puzzle has many lines. Each line has two vertices (A and B). Each line has two weights (timeA for when traversing A -> B, and timeB for when traversing B -> A). And this should be all you need to construct a graph data structure. All other properties in the JSON objects are for visual purposes.

If you want to become familiar with the puzzle, you can play it through a flash client at http://www.TracePerfect.com/

If anyone is interested in implementing a solver for themselves, then I will post detail about the API for submitting the solution to the server, which is also very simple.

Thank you for reading this longish post. I look forward to hear your thoughts about this one.

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If you are running out of heap space in Java, then you are solving it wrong.

The standard way to solve something like this is to do a breadth-first search, and filter out duplicates. For that you need three data structures. The first is your graph. The next is a queue named todo of "states" for work you have left to do. And the last is a hash that maps the possible "state" you are in to the pair (cost, last state).

In this case a "state" is the pair (current node, set of edges already traversed).

Assuming that you have those data structures, here is pseudocode for a full algorithm that should solve this problem fairly efficiently.

foreach possible starting_point:
  new_state = state(starting_point, {no edges visited})
  todo.add(new_state)
  seen[new_state] = (0, null)

while todo.workleft():
  this_state = todo.get()
  (cost, edges) = seen[this_state]
  foreach directed_edge in graph.directededges(this_state.current_node()):
    new_cost = cost + directed_edge.cost()
    new_visited = directed_edge.to()
    new_edges = edges + directed_edge.edge()
    new_state = state(new_visited, new_edges)
    if not exists seen[new_state] or new_cost < seen[new_state][0]:
      seen[new_state] = (new_cost, this_state)
      queue.add(new_state)

best_cost = infinity
full_edges = {all possible edges}
best_state
foreach possible location:
  end_state = (location, full_edges)
  (cost, last_move) = seen[end_state]
  if cost < best_cost:
    best_state = end_state
    best_cost = cost

# Now trace back the final answer.
path_in_reverse = []
current_state = best_state
while current_state[1] is not empty:
    previous_state = seen[current_state][1]
    path_in_reverse.push(edge from previous_state[0] to current_state[0])
    current_state = previous_state

And now reverse(path_in_reverse) gives you your optimal path.

Note that the hash seen is critical. It is what prevents you from getting into endless loops.

Looking at today's puzzle, this algorithm will have a maximum of a million or so states that you need to figure out. (There are 2**16 possible sets of edges, and 14 possible nodes you could be at.) That is likely to fit into RAM. But most of your nodes only have 2 edges connected. I would strongly advise collapsing those. This will reduce you to 4 nodes and 6 edges, for an upper limit of 256 states. (Not all are possible, and note that multiple edges now connect two nodes.) This should be able to run very quickly with little use of memory.

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For most parts of graph you can apply http://en.wikipedia.org/wiki/Seven_Bridges_of_K%C3%B6nigsberg.

This way you can obtain number of lines that you should repeat in order to solve.

At beginning you should not start at nodes which has short vertices over which you should travel two times. If I summarize:

• start at node whit odd number of edges.
• do not travel over lines which sit on even node more than once.
• use shortest path to travel from one odd node to another.

Simple recursive brute force solver whit this heuristic might be good way to start.

Or another way.
Try to find shortest vertices that if you remove them from graph remining graph will have only two odd numbered nodes and will be considered solvable as Koningsberg bridge. Solution is solving graph without picking up pencil on this reduced graph and once you hit node whit "removed" edge you just go back and forward.

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On your java backend you might be able to use this TSP code (work in progress) which uses Drools Planner (open source, java).