# Message #2407

From: Melinda Green <melinda@superliminal.com>

Subject: Which is the most difficult for it’s size?

Date: Thu, 04 Oct 2012 22:13:45 -0700

I don’t see Andrey’s {8,3} 10-color FT entry in the wiki HOF. How do you

guys feel about it’s difficulty compared to the. {8,4} 9-color ET? Which

is the baddest boy of the bunch?

On 10/4/2012 9:38 PM, Melinda Green wrote:

>

>

> Definitely interesting. 2 questions come to mind.

>

> 1. Can you construct a puzzle in which all the octagonal edges

> contain digons, and

> 2. Can you flip some or all edges in order to create non-orientable

> versions?

>

> -Melinda

>

> On 10/4/2012 8:15 PM, Roice Nelson wrote:

>> Cool stuff! Taking a look, the underlying abstract shape has 10

>> faces, 24 edges, and 16 vertices. So its Euler Characteristic is 2,

>> and it has the topology of the sphere. This means the graph of it

>> can be drawn on the plane:

>>

>> http://www.gravitation3d.com/magictile/pics/83/83-10_graph.png

>>

>> Here is the unrolled version for reference:

>>

>> http://www.gravitation3d.com/magictile/pics/83/83-10_unrolled.png

>>

>> The first pic nicely shows how by starting your solution with the

>> digons (the "order 2" faces), it will be similar to solving a 3^3

>> starting with the middle layer.

>>

>> The "irregular" octagonal faces are interesting. I initially thought

>> these faces were hexagons in the abstract, until I realized they

>> shared multiple disjoint edges with the same neighbor. I hadn’t seen

>> anything like that before.

>>

>> Cheers,

>> Roice

>>

>>

>> On Fri, Sep 28, 2012 at 4:02 AM, Andrey <andreyastrelin@yahoo.com

>> <mailto:andreyastrelin@yahoo.com>> wrote:

>>

>> {8,3} 10 colors puzzle is an another strange beast. It has four

>> faces of order 2 (i.e. each of them has only 2 neighbors), two

>> faces of order 4 and 4 "irregular" faces. And if you start to

>> solve it from order 2 faces (that is good idea because puzzle is

>> the most dense there), you find yourself in situation where you

>> have two disjoint unsolved "layers" - around order 4 faces - and

>> have to sort pieces and exchange parity/orientation between them

>> (like when you solve 3^3 starting with the middle layer).

>> And there is a chance to meet parity problem: some 2C pieces

>> are identical and if odd number of pairs are swapped, you’ll need

>> to solve it (by swapping some pair once more). And repeat sorting

>> of order 4 layers again :)

>> Nice thing :)

>>

>> Andrey

>>

>

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