matchstick question

science_man_88 · 2010-07-10, 21:50

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 ?

ccorn · 2010-07-11, 06:13

Quote:

Originally Posted by science_man_88

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 ?

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.

ccorn · 2010-07-11, 15:04

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.

science_man_88 · 2010-07-11, 16:01

Quote:

Originally Posted by ccorn

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.

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.

ccorn · 2010-07-11, 17:31

Quote:

Originally Posted by science_man_88

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.

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.

science_man_88 · 2010-07-11, 19:50

Quote:

Originally Posted by ccorn

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.

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

http://www.ebooks.com/ebooks/book_di...sp?IID=449939#, found it but the page isn't available without buying.

science_man_88 · 2010-07-11, 22:12

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

lfm · 2010-07-12, 05:41

Quote:

Originally Posted by science_man_88

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

Some of those sure look shorter than others.

science_man_88 · 2010-07-12, 18:10

Quote:

Originally Posted by lfm

Some of those sure look shorter than others.

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

ccorn · 2010-07-12, 20:50

Quote:

Originally Posted by science_man_88

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

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

science_man_88 · 2010-07-12, 21:37

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.

2010-07-10, 21:50	#1
science_man_88 "Forget I exist" Jul 2009 Dumbassville 3×2,797 Posts	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 ? Attached Thumbnails

2010-07-11, 22:12	#7
science_man_88 "Forget I exist" Jul 2009 Dumbassville 8391₁₀ Posts	got my webcam stuff working. that's my best picture. Attached Thumbnails

2010-07-11, 15:04	#3
ccorn Apr 2010 3²·17 Posts	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.

2010-07-12, 21:37	#11
science_man_88 "Forget I exist" Jul 2009 Dumbassville 10000011000111₂ Posts	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.