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 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, 06:13   #2
ccorn

Apr 2010

32·17 Posts

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.

Last fiddled with by ccorn on 2010-07-11 at 06:15

 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.
2010-07-11, 16:01   #4
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

839110 Posts

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.

2010-07-11, 17:31   #5
ccorn

Apr 2010

32×17 Posts

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.

2010-07-11, 19:50   #6
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

3×2,797 Posts

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.

Last fiddled with by science_man_88 on 2010-07-11 at 19:56

 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
2010-07-12, 05:41   #8
lfm

Jul 2006
Calgary

52·17 Posts

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.

2010-07-12, 18:10   #9
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

100000110001112 Posts

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.

2010-07-12, 20:50   #10
ccorn

Apr 2010

32×17 Posts

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

Last fiddled with by ccorn on 2010-07-12 at 21:30

 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.

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