Talk:Puzzles/Geometric Puzzles/Connecting Utilities/Solution

^ Connecting Utilities ^

This problem can be analyzed using graph theory.

The problem is essentially showing that the bipartite graph K3,3 is planar. However, Kuratowski ' s theorem tells us that this graph must not be planar.

Essentially, there is no solution and the required construction cannot be done! Sorry! :)


 * That's right. However, if you absolutely want to find a solution you have to let the buildings be on a more complicated topology (with a plane, or even the surface of a ball it won't work). The problem is solvable if the buildings are located on a donut-shaped "planet". Thomas

/¯¯¯¯¯¯¯¯¯\       //¯¯¯¯¯¯¯¯¯\\       ///¯¯¯¯¯¯¯¯¯\\\       |||         |||      /---\'      /---\'      /---\'      | A |       | B |       | C | -      -       -       |||         |||         |||       |||         \\\_________///       |||          \\_________//       |||           \_________/       \\\        \\¯¯¯¯¯¯¯¯¯¯\___________        | ¯¯¯¯¯¯¯¯¯¯\           \       [G]         [W]         [E]

hmmm, I'm not sure if this picture is really a solution since B and C should be connected directly to G,W, and E. The solution with the donut (or more mathematically: torus) is quite hard to draw on a 2D screen I think :) Thomas

Here is the solution lying on a torus. The V signifes a looping around the torus, with the "border" marked in for clarity. Dysprosia __                               __   \    |   \     |      /  |     /  ==V===V====V====V=====V===V====V= : |  |     \   |    /    |   / :  : |  /-\'   |  /-\'  |   /-\' | :  : |  |A|    |  |B|   |   |C|  | : : | -  | -  |  - | :  : \__| |     \_| |___/   | |__/ :  :      \______     ______/      :  :             \   /             :  :              | |              :  :    [G]       [W]       [E]    : :  / | \       |       / | \   :  ===V==V==V======V======V==V==V=== __/  |   \     |     /   |   \__


 * Does G connect to C, and E to A, without crossing, in that? Cyp 23:58, 21 Oct 2003 (UTC)

Yes, it loops around the torus from the lower left to the upper right, same for the lower right to the upper left (Dysprosia, not logged in)


 * I made it a bit clearer that the G-C and E-A connections have to go around the whole torus (I assume this perspective is a frontal view on the torus). Also the connections G-C and E-A have to "cross without actually crossing" which can be done if one path does a turn around the torus tube. Thomas

Here is a version that is a bit simpler, I think. Thomas E_                               __G \       \            /        /  ==V========V==========V========V= : |        \        /        / :  : |  /-\'   |  /-\'  |   /-\' | :  : |  |A|    |  |B|   |   |C|  | : : | -  | -  |  - | :  : \__|||     \_|||___/   |||__/ :  :     |\______  |  ______/|     :  :     |       \ | /       |     :  :     |        |||        |     :  :    [G]       [W]       [E]    : :  /   \               /   \   :  ===V=====V=============V=====V=== C__/      \           /       \__A

Yep, that's fine too! I think they're rather equivalent in simplicity, though Dysprosia 05:21, 22 Oct 2003 (UTC)


 * Doesn't A-E and C-G cross? Here's the above, just repeated twice. Cyp 08:27, 22 Oct 2003 (UTC)

E_                               __G \       \            /        /E                                __G |       |          |        |  \        \            /        /  ==V========V==========V========V==V========V==========V========V= : |        \        /        / : |         \        /        / :  : |  /-\'   |  /-\'  |   /-\' | : |  /-\'   |  /-\'  |   /-\' | :  : |  |A|    |  |B|   |   |C|  | : |  |A|    |  |B|   |   |C|  | : : | -  | -  |  - | : | -   | -  |  - | :  : \__|||     \_|||___/   |||__/ : \__|||     \_|||___/   |||__/ :  :     |\______  |  ______/|     :     |\______  |  ______/|     :  :     |       \ | /       |     :     |       \ | /       |     :  :     |        |||        |     :     |        |||        |     :  :    [G]       [W]       [E]    :    [G]       [W]       [E]    : :  /   \               /   \   :   /   \               /   \   :  ===V=====V=============V=====V=====V=====V=============V=====V=== C__/      \           /       \__A |     |             |     | C__/      \           /       \__A

Hmmm, I think you're right Cyp, I'm going to redo my diagram: \   | |  |    |      /  |       ==V===V=V==V====V=====V===V====== : |  |  \  \   |    /    |     :  : |  /-\' | |  /-\'  |   /-\'   :  : |  |A|  | |  |B|   |   |C|    : : | - | | - |  -   :  : |__/ |  |  \_| |___/   | |    :  :      |   \____________/  |    :  :      \______     ________/    :  :             \   /             :  :              | |              :  :    [G]---\   [W]       [E]    : :    | \  |    |       / | \   :  ======V=V==V=====V=====V==V==V=== | | |    |     /   |   \

I think this solves the doubleedge problem! Dysprosia 08:42, 22 Oct 2003 (UTC)

That looks right. Ascii art is wonderful. Cyp 15:11, 22 Oct 2003 (UTC)

__G \           /        /  ===========V==========V========V= :          \        /        / :  :    /-\'   |  /-\'  |   /-\' | :  :    |A|    |  |B|   |   |C|  | : :  -   | -  |  - | :  :    |||     \_|||___/   |||__/ : _V____/|\______  |  ______/|  ___V_ :    |       \ | /       | /   :  :     |        |||        ||    :  :    [G]       [W]       [E]    : :  /   \               /       :  ===V=====V=============V========= C__/      \           /


 * A bit cleaner solution I think24.187.18.164 07:02, 6 Sep 2004 (UTC)