# Dijkstra's Algorithm

Started by cryo75, Oct 22 2003 10:35 PM

32 replies to this topic

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#1
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Posted 22 October 2003 - 10:35 PM

Hi all,
I have implemented the algorithm and it works fine when returning the shortest path from A to B. Now I need something more...
How is it possible to return the 2nd, 3rd, 4th, etc... shortest paths using the same algorithm??
Thanks,
Ivan

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#2
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Posted 23 October 2003 - 12:05 AM

This is a much harder (as in NP-complete) problem to solve. Let's consider this for a while. That would mean you must at some point choose a node that's not optimal, but how do you know when to choose that node in order to get the 2nd shortest path? You don't, so the only thing left would be to test all (wow). That is, you have to run the full algo for all nodes choosing one non-optimal node to finally find from which node you should choose the non-optimal node.

[edited by - amag on October 23, 2003 7:06:29 AM]

[edited by - amag on October 23, 2003 7:06:29 AM]

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#3
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Posted 23 October 2003 - 12:38 AM

Yeah that''s my problem and I don''t know how to solve it.

I have node A (eg. Paris) and node B (eg. Frankfurt). The 1st shortest path would be A to B over C (Cologne). So the algorithm gives me that one.

Now I want also the second shortest path (which will be a little longer than the 1st), example A to B over C and D (Bonn)...

Any ideas??

I have node A (eg. Paris) and node B (eg. Frankfurt). The 1st shortest path would be A to B over C (Cologne). So the algorithm gives me that one.

Now I want also the second shortest path (which will be a little longer than the 1st), example A to B over C and D (Bonn)...

Any ideas??

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#4
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Posted 23 October 2003 - 01:00 AM

Dijkstra''s algorithm requires you to remember the "best" path to a node, and then use this best path to reach another node, like this.

A -(m)-> B

A is reached in minimum n

B will be reached in minimum n+m from A

A variant of this would be to store a linked list of the n best paths, instead of just the best one.

A -(m)-> B

A is reached in {n1, n2, n3}

B will be reached in minimum {n1+m,n2+m,n3+m}

Now, how would this apply to the following case ?

A -(3)-> B

A -(5)-> C

C -(2)-> D

B -(1)-> C

B -(6)-> D

Start with A = {0}

Study node A : B = {3} C = {5}

Study node B : C = {4, 5} D = {9}

Study node C : D = {6, 7, 9}

Final state : A = {0} B = {3} C = {4, 5} D = {6, 7, 9}

So you''d have three ways of reaching D, of respecitve lengths 6, 7, 9. (Note : this only computes paths that go only once through any node).

From there, to find out where each came from: say we want to backtrace the second best path (length 7).

Test D -> B : 7-6 = 1, B = {3} so this didn''t come from B.

Test D -> C : 7-2 = 5, C = {4, 5} so this came from C.

Test C -> B : 5-1 = 4, B = {3} so this didn''t come from B.

Test C -> A : 5-5 = 0, A = {0} so this came from A.

Backtrace results in A -> C -> D which is of length 5+2 = 7.

A few hints :

- keep the list of possible paths sorted.

- do not use duplicate values.

Problems : this method will only find the "second best" path if it is LONGER than the best path. Generally, if more than one path is of a certain length, this method will only find one. But you could easily get past this by remembering from where each n1,n2,n3,... came.

ToohrVyk

A -(m)-> B

A is reached in minimum n

B will be reached in minimum n+m from A

A variant of this would be to store a linked list of the n best paths, instead of just the best one.

A -(m)-> B

A is reached in {n1, n2, n3}

B will be reached in minimum {n1+m,n2+m,n3+m}

Now, how would this apply to the following case ?

A -(3)-> B

A -(5)-> C

C -(2)-> D

B -(1)-> C

B -(6)-> D

Start with A = {0}

Study node A : B = {3} C = {5}

Study node B : C = {4, 5} D = {9}

Study node C : D = {6, 7, 9}

Final state : A = {0} B = {3} C = {4, 5} D = {6, 7, 9}

So you''d have three ways of reaching D, of respecitve lengths 6, 7, 9. (Note : this only computes paths that go only once through any node).

From there, to find out where each came from: say we want to backtrace the second best path (length 7).

Test D -> B : 7-6 = 1, B = {3} so this didn''t come from B.

Test D -> C : 7-2 = 5, C = {4, 5} so this came from C.

Test C -> B : 5-1 = 4, B = {3} so this didn''t come from B.

Test C -> A : 5-5 = 0, A = {0} so this came from A.

Backtrace results in A -> C -> D which is of length 5+2 = 7.

A few hints :

- keep the list of possible paths sorted.

- do not use duplicate values.

Problems : this method will only find the "second best" path if it is LONGER than the best path. Generally, if more than one path is of a certain length, this method will only find one. But you could easily get past this by remembering from where each n1,n2,n3,... came.

ToohrVyk

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#5
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Posted 23 October 2003 - 02:27 AM

Hello cryo75,

One way to do what you whant is to for you to Mark the shost path with a very high cost for each node or remove the node form your check.

if as you say.

"I have node A (eg. Paris) and node B (eg. Frankfurt). The 1st shortest path would be A to B over C (Cologne). So the algorithm gives me that one."

then remove C (Cologne) form your vaild nodes to search and then the next path might include D (luxemburg).

Lord Bart

One way to do what you whant is to for you to Mark the shost path with a very high cost for each node or remove the node form your check.

if as you say.

"I have node A (eg. Paris) and node B (eg. Frankfurt). The 1st shortest path would be A to B over C (Cologne). So the algorithm gives me that one."

then remove C (Cologne) form your vaild nodes to search and then the next path might include D (luxemburg).

Lord Bart

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#8
Members - Reputation: **313**

Posted 23 October 2003 - 03:29 AM

It will mess up...

There was a similar problem where someone wanted Dijkstra's to randomly pick a path that's pretty good, turned out a good solution was to multiply all edge costs by

George D. Filiotis

[b]I am a signature virus. Please add me to your signature so that I may multiply.

[edited by - symphonic on October 23, 2003 10:29:47 AM]

There was a similar problem where someone wanted Dijkstra's to randomly pick a path that's pretty good, turned out a good solution was to multiply all edge costs by

**1+Rand(1)**(in floating point) if you're just looking for similarly good paths this might do it for you, if you want absolute order, it is NP-Complete and will not be usefull in a real-time app unless the number of nodes in your graph is VERY small (less than 10).George D. Filiotis

[b]I am a signature virus. Please add me to your signature so that I may multiply.

[edited by - symphonic on October 23, 2003 10:29:47 AM]

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#10
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Posted 23 October 2003 - 04:50 AM

How about looking for the shortest route, and then finding the node which is ''closest'' to that route, i.e. I can go ABC, which is shortest, but I can also go ADC which is not shortest. If you found all those single-node alternate paths, you''d be able to pick the shortest one of them (producing the best path after the main one). I think it builds up to finding all the paths, but I don''t know.

The labelling nature of Dijikstra''s has to come into this somewhere. Hmm.....

Richard "Superpig" Fine

- saving pigs from untimely fates, and when he''s not doing that, runs The Binary Refinery.

Enginuity1 | Enginuity2 | Enginuity3 | Enginuity4

ry. .ibu cy. .y''ybu. .abu ry. dy. "sy. .ubu py. .ebu ry. py. .ibu gy." fy. .ibu ny. .ebu

OpenGL is a language

The labelling nature of Dijikstra''s has to come into this somewhere. Hmm.....

Richard "Superpig" Fine

- saving pigs from untimely fates, and when he''s not doing that, runs The Binary Refinery.

Enginuity1 | Enginuity2 | Enginuity3 | Enginuity4

ry. .ibu cy. .y''ybu. .abu ry. dy. "sy. .ubu py. .ebu ry. py. .ibu gy." fy. .ibu ny. .ebu

OpenGL is a language

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#11
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Posted 23 October 2003 - 06:18 AM

If it helps at all, there are algorithms that find the k shortest paths (somewhat) efficiently. This isn''t exactly what you want since it will return multiple paths and ties, but perhaps you can still utilize it for what you want.

One paper on this:

http://citeseer.nj.nec.com/eppstein97finding.htm

It runs in O(m + n + log(n + k)), where m is the number of edges, and n is the number of edges.

I didn''t check it thoroughly, but I don''t belive it uses Dijkstra''a algorithm, so you would have to implement a completely new algorithm if you decide this could be useful.

One paper on this:

http://citeseer.nj.nec.com/eppstein97finding.htm

It runs in O(m + n + log(n + k)), where m is the number of edges, and n is the number of edges.

I didn''t check it thoroughly, but I don''t belive it uses Dijkstra''a algorithm, so you would have to implement a completely new algorithm if you decide this could be useful.

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#14
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Posted 25 October 2003 - 04:22 AM

Actually using a priority-queue in Dijkstra is only an optimization to find the lowest-cost node faster. It wouldn''t help solve this problem ASFAIK.

You also cannot remove a path or change the weights of the edges, since that would change the whole graph and you will most likely get different results.

I also think what Symphonic suggests might be a good idea, or at least you might want to look for a non-optimal solution on this one.

If you are certain you don''t have any cycles in your graph (not likely for a bus&train route) you can use a modified version of Dijkstra, where you don''t mark nodes as visited. That way you''ll get all paths and costs.

You also cannot remove a path or change the weights of the edges, since that would change the whole graph and you will most likely get different results.

I also think what Symphonic suggests might be a good idea, or at least you might want to look for a non-optimal solution on this one.

If you are certain you don''t have any cycles in your graph (not likely for a bus&train route) you can use a modified version of Dijkstra, where you don''t mark nodes as visited. That way you''ll get all paths and costs.

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#16
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Posted 25 October 2003 - 05:24 AM

quote:Why would anyone want a non-optimal bus/train route?Original post by cryo75

I need this algorithm for a bus & train route for quite a big area in Germany so there will be quite a lot of nodes in there.

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#17
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Posted 25 October 2003 - 05:59 AM

quote:

Why would anyone want a non-optimal bus/train route?

Beats me, maybe he just wants to give the user a choice?

quote:

He mentioned having quite a bit of nodes, thats what I was talking about.

Oh, sorry I thought you were refering to the main problem.

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#18
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Posted 25 October 2003 - 03:13 PM

Does the bus/train schedule change often? If not, you could use a really really slow algorithm(like ''try every possible path up to the same length as the optimal path +3'') and just store that and then recompute it overnight or something whenever the schedules do change.

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#19
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Posted 25 October 2003 - 07:29 PM

quote:Original post by cryo75

I need this algorithm for a bus & train route for quite a big area in Germany so there will be quite a lot of nodes in there.

Would there be a different algorithm instead of Dijkstra which does what I need?

Ivan

Yes, this problem is iso-morphic to the Salesman Problem, and as mentioned it is NP-complete _and_ the listing algorithm is already O(n

^{2}) - this is horribly time-consuming - O(n

^{2}!). If that is not bad enough, now you want to find the 3rd, 4th etc... this adds another O(n

^{2}) worst-case factor; you''re looking at a O(2 n

^{2}!) problem. If you can solve this problem in RT, you will be a rich man

I see two options; if the routes in question are fixed and do not change, or change very infrequently then pre-calculate all the common routes. Or, you cannot find _THE_ second best route, just another good one.

The base solution to your problem is simple to code though; whenver you test a node with Dijkstra algorithm, you check to see if the path matches the best path already found. If it does, you reject it and continue searching. For 3rd, 4th, you just check and reject against the 1st & 2nd, etc...

With some clever coding you can reduce the amoritized complexity of this check from n

^{2}to n, but it hardly seems worth the effort.

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#20
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Posted 25 October 2003 - 09:17 PM

Another thing you could do is just cost the nodes in the first path to be (original value)+(minimum value of all nodes / K) or something like that, so that they are less likely to get chosen (and if it still returns the original path, keep increasing the value until its a different path). If you want the first N paths, keep the original average cost around and continually increase the value of the costs until you get all the different paths you needed. It definitely isn''t optimal, but if you keep the value of the cost increase low, it should return reasonable results.

You might want to try variations as well - if the list is long, maybe only increase the cost of the last node or two on the path so you can get very simmilar but still different paths. If you want to do it in realtime, you might just return the best path then allow the person to choose which places they don''t want to go and give those nodes very high weights and recompute the path.

You might want to try variations as well - if the list is long, maybe only increase the cost of the last node or two on the path so you can get very simmilar but still different paths. If you want to do it in realtime, you might just return the best path then allow the person to choose which places they don''t want to go and give those nodes very high weights and recompute the path.