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March 18

36 cube puzzle and Graeco-Latin squares

Wouldn't a solution to the 36 cube be an order 6 Graeco-Latin square, which doesn't exist. Yet the article says " ThinkFun's 36 cube puzzle does have a number of solutions". Bubba73 ^{You talkin' to me?} 00:20, 18 March 2013 (UTC)[reply]

I don't understand exactly how this puzzle works, but it appears from the picture that the colored pieces are of different heights, and the base has various elevations, so part of the puzzle is to place a piece of the correct height in each position on the base so that the tops of the pieces are level. If that's what is going on, then the base must have two positions in the same row or column that are the same elevation, so the solution includes two pieces in the same row or column that have the same height. —Bkell (talk) 03:01, 18 March 2013 (UTC)[reply]

Yes, the bases do have different elevations, so the colored pieces are 1 through 6 units in height, for each of the six colors. Actually it is restricted from the general Graeco-Latin square, since the bases already determine a Latin square on the heights. Bubba73 ^{You talkin' to me?} 03:07, 18 March 2013 (UTC)[reply]

But what I'm saying is that the base might not be a Latin square. In fact it can't be, because there's no solution to the 36-officers problem. Right? —Bkell (talk) 03:14, 18 March 2013 (UTC)[reply]

No, the base heights do form a Latin square, as opposed to Graeco-Latin. Bubba73 ^{You talkin' to me?} 04:00, 18 March 2013 (UTC)[reply]

Well, then it would seem the solution is impossible. How do you know the base heights form a Latin square? It doesn't say that in the Wikipedia article. Do you have this puzzle? —Bkell (talk) 04:08, 18 March 2013 (UTC)[reply]

Thanks for the photo. It's difficult to see what's going on there, though—can you describe the base heights as a 6-by-6 matrix of numbers from 1 through 6? —Bkell (talk) 04:14, 18 March 2013 (UTC)[reply]

Yes, in each row and each column, each of the six possible base heights are represented once and only once, a Latin square. Bubba73 ^{You talkin' to me?} 04:16, 18 March 2013 (UTC)[reply]

Front to back, as it is in the photo, # of cubes in the base:

Bubba73 ^{You talkin' to me?} 04:21, 18 March 2013 (UTC)[reply]

(edit conflict) Also, is there a reason that the narrower "spires" of the gray towers are of different heights? Is it true that all gray towers of the same height have spires of the same height? —Bkell (talk) 04:17, 18 March 2013 (UTC)[reply]

The colored pieces fit over the thin top part of the spires, so that the number of "blocks" in the spire plus the number of "blocks" in the colored piece totals six, making them all the same height. And yes to the second question. Bubba73 ^{You talkin' to me?} 04:25, 18 March 2013 (UTC)[reply]

My daughter has been working on this, and today I realized that it was a Graeco-Latin square. I remembered my Martin Gardner book, and I thought that the order 6 was the one that was hard to get, but it was the order 10 and the order 6 is impossible. Bubba73 ^{You talkin' to me?} 04:30, 18 March 2013 (UTC)[reply]

Okay. So let's look at this chain of reasoning:

The base heights form a Latin square, using the numbers 0 through 5.
There are 36 colored pieces; for each of six colors, there is one colored piece of each height from 1 through 6.
When a colored piece is placed on a base tower, the height of the base tower plus the height of the colored piece always add together to the height of the combination of the two.
In a solution to the puzzle, all of these "combination heights" are equal.
Therefore, in a solution to the puzzle, the heights of the colored pieces form a Latin square.
In a solution to the puzzle, there is one piece of each color in every row and every column.
Therefore, in a solution to the puzzle, the colors of the pieces form a Latin square.
Thus, a solution to the puzzle yields a Graeco-Latin square of order 6, which is a solution to the thirty-six officers problem.
A solution to the puzzle exists.
A solution to the thirty-six officers problem does not exist.

Obviously there is a contradiction here, so one of these claims (or one of the steps of reasoning) must be wrong. I have confidence in claim 10, since that has been proved. The step that seems most suspicious to me at the moment is claim 3 (and claim 9, but supposedly we are assured that claim 9 is true). —Bkell (talk) 04:48, 18 March 2013 (UTC)[reply]

I wondered if there was a trick, such as a set of colors having two the same height and a row of spires having two the same height, but no. Bubba73 ^{You talkin' to me?} 04:58, 18 March 2013 (UTC)[reply]

Well, I googled and found this: "Hint #2 (show): The assumption that all towers of the same size only differ in color is wrong." So there is a trick to it. Later, because it is past my bedtime. Bubba73 ^{You talkin' to me?} 05:32, 18 March 2013 (UTC)[reply]

Yeah, I suspected something like that. I think claim 3 above is bogus. For example, what happens if you put a colored block of height 6 on a tower of height 5? I bet you don't get a thing of height 11. —Bkell (talk) 05:48, 18 March 2013 (UTC)[reply]

No, it doesn't fit right. But I've examined the pieces after seeing that hint, and I don't see the trick. Notice that the article is carefully worded to avoid revealing much. Bubba73 ^{You talkin' to me?} 14:29, 18 March 2013 (UTC)[reply]

But as far as I can tell, a piece only fits correctly on the base of the right height, since they all go to the same height, as in the second photo. Bubba73 ^{You talkin' to me?} 14:39, 18 March 2013 (UTC)[reply]

And most puzzles come in a solved state - this one does not, see the photo in the article about it. Bubba73 ^{You talkin' to me?} 15:00, 18 March 2013 (UTC)[reply]

What do you mean when you say "it doesn't fit right"? I'm guessing it fits over some portion of the "large part" of the gray tower, right? Does it do so in such a way that you can get another combination of the same height in some other way, maybe by fudging again? —Bkell (talk) 18:46, 18 March 2013 (UTC)[reply]

I don't have this puzzle, but the corresponding article on the German Wikipedia says "Tatsächlich gibt es einen Führungsstift für Farbstäbe der Höhe 6, auf den auch ein bestimmter Farbstab der Höhe 5 passt. Im Gegenzug passt ein bestimmter Farbstab der Höhe 6 auf einen Führungsstift für die Höhe 5". Translation: "Indeed there is a spire for pieces of height 6, on which also a certain piece of height 5 fits. Correspondingly there is certain piece of height 6 that fits on a spire for pieces of height 5". Without having access to the puzzle, I can only assume that this is the key to a solution. 86.136.42.134 (talk) 22:12, 18 March 2013 (UTC)[reply]

Yes, #3 is where the trick is. There are little tabs inside most of the pieces. On one of height 5 the tabs are a little bigger. One piece of height 6 doesn't have the tabs. And there is one spire of height 0 where the piece of height 5 fits and one spire of height 1 where the piece of height 6 fits. Sneaky!! It really throws off someone who knows that there shouldn't be a solution. Bubba73 ^{You talkin' to me?} 00:11, 19 March 2013 (UTC)[reply]

Resolved

Isosceles trapezoid definition

Please go to Talk:Isosceles trapezoid and study the bottom section very carefully; it's talking about what it would mean if we define an isosceles trapezoid as having congruent legs. Georgia guy (talk) 18:56, 18 March 2013 (UTC)[reply]

I think David Eppstein has solved the problem by restoring the alternative definition in the article (from February 2012), but other opinions are welcomed. The definitions should match the taxonomy at the end of Quadrilateral. Dbfirs 09:24, 19 March 2013 (UTC)[reply]