Covering all edges in graph

I'm experiencing a serious problem. I would appreciate
your help.
The task is to cover all edges in oriented graph, starting
from selected node A and ending in selected node B.
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May I ask if you already have a data structure which represents the graph? Or can I choose the data structure? I mean, I would then have procedures like DefineNode and ConnectNodes or similar. The decision of the representation depends on the usage. I would choose a different implementation if enumeration was the most important thing.
pavlovskyAuthor Commented:
You can choose any data structure. Actually I need the algorithm.
The usual representation is an adjaceny list. You keep a list or an array of nodes. Each node has a list whose elements are chained records. The only important field, other then the next pointer, is a pointer to the node (or node number if you use an array).
   My next question starts is as follows. There are basically two graph search algorithms, breath-first and depth-first. These visit each node only once. The depth-first is the easiest, you keep a visitor marker with each node and don't search further if you have visited the node. Else you process the adjaceny list for the node recursively. If you remember in a global array the place you are at on each recursion you then have a path to the node. Clearly when you arrive at your destination node, path gives you the edge list and you can do what you want. So :-
    dfs(v) :
     is-end-node(v) -> do what you want; Exit
     is-visited(v)     -> Exit;
         mark v as visited
         add v to path
         FOR EACH x IN adjaceny list of v DO dfs(x);
         remove v from path;
OK? or do you want something different?
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Pavlovsky means not visiting all nodes but all !edges! in oriented graph
(BTW there may be cycles in graph so you CAN visit some nodes and
edges more than once).
I actually added a futher comment in here this afternoon. Now ITS GONE!!!!

My comment was: In your graph have you got multiple edges eg: A->B A->B and/or have you got node cycles eg: A->A?
Here is some pseudocode for an algorithm to find a path from A to B covering all nodes

var visited
var visits

function GetPath1 (graph, A, B, Solution)  
    Var curpath
    GetPath graph, a, b, curpath, solution
end function

function GetPath1 (graph, A, B, curPath, Solution)
   GetPath1 = false
   visited(A) = true
   curpath[visits] = A   {Administration}

   if (A = B) and (visits = nrofnodes) then {test for solution}
      solution = curpath {copy solution}
      GetPath1 = true
     for each node in Graph
        if (not visited (node)) then
           found = GetPath (graph, A, B, Path, solution)
           if found then break; {ready}    
        end if
   end if

   visited(A) = false  {Administration}
end function

pavlovskyAuthor Commented:
You should cover not all nodes but all EDGES!
And there can be cycles, multiple edges( A->B,A->B) and cycles(A->A) in graph.
Since the adjaceny lists contain one record which represents a path from node A to B (even to itself) it is quite simple to modify the3 depth first algorithm to remember the entry (its just a pointer). The "visted" code in my example comes out and we now look at the "edge list" to see if an entry has been made :-
   FOR EACH e IN adjacenylist(v) DO   /* look at the entry */
       IF NOT is-in-set(e,EdgeSet) THEN
          dfs(n OF e);
This will ensure that we look at each edge only once in any path from A to B.
   However may I ask why you need this (unusual) variation to the algorithm?
What you want is to see if there is a hamilton path from A to B.

See also
for source code examples.

pavlovskyAuthor Commented:
Thanks guys for helping me, but I've already solved that problem.
Let me see, how does that make my answer wrong????
I've sent you LISP-code. Of course, I hate LISP, but I found only LISP example, so see for some compiler

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pavlovskyAuthor Commented:
That's OK.
Hmmm, this smells like fraud.
Does the P in jack_p stand for Pavlovsky?

Yes IT DOES!!!!

Well two accounts less at EE.
An extremely nasty weasle! Linda should have it destroyed and NOT humanely
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