Timeline for House of Santa Claus
Current License: CC BY-SA 3.0
24 events
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| Feb 16, 2017 at 15:16 | history | edited | Mr.Wizard | CC BY-SA 3.0 |
it seems that this is presently not a correct solution and I wish to lift my vote
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| Oct 18, 2016 at 20:08 | comment | added | nadlr | @J.M. I believe FindPostmanTour gives tours that traverse each edge at least once. It seems to me the OP wants tours that traverse each edge exactly once. Does it suffice to select from the tours you found those that have no duplicates? | |
| Oct 18, 2016 at 12:04 | comment | added | yode | @Dr.WolfgangHintze I have given a vote. :) | |
| Oct 17, 2016 at 9:51 | comment | added | Dr. Wolfgang Hintze | @J.M. To be honest, the valuations of the contributions to this problem look a bit strange to me. What do they measure? | |
| Oct 16, 2016 at 10:58 | comment | added | yode | Yes,I can't.As the of the state of You must not repeat a line. Maybe this answer have some problem or maybe I miss something.I think this post want find a eulerian path from 1 to 2.I mean this is half eulerian graph | |
| Oct 16, 2016 at 10:39 | comment | added | J. M.'s missing motivation♦ |
@yode, were you able to make FindEulerianCycle[] work in this case?
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| Oct 16, 2016 at 10:10 | comment | added | yode |
As this article,I think we should use FindEulerianCycle but not FindPostmanTour?
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| Oct 16, 2016 at 5:34 | comment | added | J. M.'s missing motivation♦ | @Dr. Hintze, apparently so; let me think about it. | |
| Oct 15, 2016 at 21:13 | comment | added | Dr. Wolfgang Hintze | @mrz IMHO the solution of J.M. is not correct as it draws some edges more than once. Please see my solution. | |
| Oct 15, 2016 at 18:20 | comment | added | Dr. Wolfgang Hintze | @J.M. I'm not sure anymore that this solution is correct. The first tour is {{1 -> 2, 2 -> 4, 4 -> 5, 5 -> 3, 3 -> 4, 4 -> 1, 1 -> 3, 3 -> 2, 2 -> 4, 4 -> 1},. It containes the edges 2->4 and 4->1 twice. Generally, we should have only 8 items in a tour, rather than 10. | |
| Oct 15, 2016 at 17:44 | comment | added | J. M.'s missing motivation♦ |
@mrz, as I said, you can try relabeling the vertices so that 1 is some other vertex. Have you tried it already?
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| Oct 15, 2016 at 16:58 | comment | added | mrz | @J. M. see my update ... | |
| Oct 15, 2016 at 16:00 | comment | added | Dr. Wolfgang Hintze | @J.M. No problem. Unfortunately this type of changes from version to version are not uncommon in Mathematica. But, again, very nice code the results of which I showed to my litle granddaughter who had the ambition of solving the puzzle by herself. | |
| Oct 15, 2016 at 15:45 | comment | added | J. M.'s missing motivation♦ | @Dr. Hintze, unfortunately I did not have version 8 to check. The code as presented is working in 11, tho. | |
| Oct 15, 2016 at 15:40 | comment | added | Dr. Wolfgang Hintze | @J.M. Beautiful. But in version 8 the code does not run. Here's the correction. The line after house = ... should read Show[house], and as FindPostmanTour requires a graph object as the first parameter, the tours are calculated by FindPostmanTour[house,All]. | |
| Oct 15, 2016 at 9:27 | comment | added | J. M.'s missing motivation♦ |
@mrz, 1 -> 4 means a directed path going from node 1 to node 4, which seems to be the reverse of what's in your diagram.
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| Oct 15, 2016 at 9:26 | vote | accept | mrz | ||
| Oct 16, 2016 at 22:01 | |||||
| Oct 15, 2016 at 9:25 | comment | added | mrz |
So it is wrong when I would define edges = {1 -> 2, 1 -> 3, 1 -> 4, 2 -> 4, 2 -> 3, 2 -> 1, 3 -> 2, 3 -> 4, 3 -> 1, 3 -> 5, 4 -> 1, 4 -> 2, 4 -> 3, 4 -> 5, 5 -> 3, 5 -> 4}; and then use FindPostmanTour?
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| Oct 15, 2016 at 9:01 | comment | added | J. M.'s missing motivation♦ |
@mrz, you can rename the vertices so that 1 corresponds to your desired starting point. That, I'll leave for you as an exercise.
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| Oct 15, 2016 at 8:34 | comment | added | mrz | Thank you ... this is very elegant. You show the possible solutions over 1->2->4->... . How do you find all solutions from all 5 starting points? | |
| Oct 15, 2016 at 2:09 | history | edited | J. M.'s missing motivation♦ | CC BY-SA 3.0 |
added 1 character in body
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| Oct 15, 2016 at 2:08 | comment | added | J. M.'s missing motivation♦ | @user6014, thanks. Remembering the name of the problem surprisingly took more effort than coding it up. :D (Aside: please consider changing your username.) | |
| Oct 15, 2016 at 2:03 | comment | added | ktm |
Neat! I spent a lot of time looking for the FindPostmanTour function and couldn't seem to in my naive documentation search. Great reply.
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| Oct 15, 2016 at 2:01 | history | answered | J. M.'s missing motivation♦ | CC BY-SA 3.0 |