matchstick question
 

on pg. 230 of number freak 1 to 200 it talks of the smallest 4 regular matchstick graph outside of one technicality I think i can get it down from 104 to 32 (possibly less actually) if one thing is allowed. the attached image is my possible solution the catch I think the x's shown would need either .5 matches or the allowing of matchsticks to cross each other. anyone for a look ?

[QUOTE=science_man_88;221012]on pg. 230 of number freak 1 to 200 it talks of the smallest 4 regular matchstick graph outside of one technicality I think i can get it down from 104 to 32 (possibly less actually) if one thing is allowed. the attached image is my possible solution the catch I think the x's shown would need either .5 matches or the allowing of matchsticks to cross each other. anyone for a look ?[/QUOTE]

The overall shape should be hexagonal, I'm afraid. Otherwise you would need matchsticks of different lengths.
Edit: And presumably the graph should be planar, i.e. no crossings.

Note that four matchsticks with common end point, together defining three adjacent equilateral triangles, by construction enclose three adjacent 60-degree angles, i. e. 180 degrees. This means: If the matchsticks must have same length, your ring construction actually is straight. You cannot bend it into a finite polygon. You need to reduce the connectivity of the boundary in order to allow for "gaps" with sub-60-degree angles. These are needed in order to bend the boundary to some finite perimeter.

[QUOTE=ccorn;221036]The overall shape should be hexagonal, I'm afraid. Otherwise you would need matchsticks of different lengths.
Edit: And presumably the graph should be planar, i.e. no crossings.[/QUOTE]

I think you are wrong about the hexagon shape as it shows the record holder on that page of the book and it's a irregular octagon if my eyes do not deceive me.

[QUOTE=science_man_88;221059]I think you are wrong about the hexagon shape as it shows the record holder on that page of the book and it's a irregular octagon if my eyes do not deceive me.[/QUOTE]
OK, scrap my post #2. I have refined what I mean in post #3. In order to clarify things, it could be helpful if you presented the problem with its original specification.

[QUOTE=ccorn;221068]OK, scrap my post #2. I have refined what I mean in post #3. In order to clarify things, it could be helpful if you presented the problem with its original specification.[/QUOTE]

I tried to look it up online I couldn't find it. I found a review of the book but nothing for e books.

[url]http://www.ebooks.com/ebooks/book_display.asp?IID=449939#[/url], found it but the page isn't available without buying.

got my webcam stuff working. that's my best picture.

[QUOTE=science_man_88;221091]got my webcam stuff working. that's my best picture.[/QUOTE]

Some of those sure look shorter than others.

[QUOTE=lfm;221126]Some of those sure look shorter than others.[/QUOTE]

I actually measured the image in the book they are about 4/16th of a inch long each.

[QUOTE=science_man_88;221091]got my webcam stuff working. that's my best picture.[/QUOTE]
Wow. Look at those boundary points that are used as hinges (with less-than-60-degree internal angles). You need some of these, otherwise you cannot get a curved boundary. That's what I meant in post #3.

If 3D were allowed, you could simply build an octaeder. 12 matches only :smile:

the book shows the solution for 3 at each and tells what the solution for 2 is the solution for 2 is a equilateral triangle(though those hinges work as well).

the solution for 3 is 12 matches 2 diamonds split by a match each and then connected into a hexagon(does this work for 4 with the three match solution?).

can you draw the octahedron in 2-d ? if so it's a possible lowest solution.

There are some [url=http://www2.stetson.edu/~efriedma/papers/matchstick.ppt]slides by Erich Friedman[/url] on the topic. The slides also show the [url=http://mathworld.wolfram.com/HarborthGraph.html]Harborth graph[/url] depicted in your copy of Number Freak. Using the string shown at the bottom of page 22 of Friedman's slides, you could elegantly build a 4-regular matchstick graph with 3*21 = 63 vertices and 126 edges. (page 28 contains it as a subgraph.) Yet Harborth's solution is the smallest known 4-regular matchstick graph with 52 vertices and 104 edges.