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Alexandre Muñiz
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Alexandre Muñiz

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My exchange gift for the next Gathering for Gardner is a "magic" die. All numbers between 1 and 24 occur once. Faces sum to 50, bands around the cube sum to 100, and diagonal bands sum to 75.

After I had the dice in hand I thought of a way that I might have been able to make them more awesome. The upper left number on each face is a number between 1 and 6, so that the die can be used as a standard d6. We could think each of the positions on each face as 'modes' for rolling the dice. It would be really cool if the other three modes formed a set of non-transitive 'dice'. (Non-transitive dice are sets of three[1] dice where A usually beats B, B usually beats C, and C usually beats A.) It happens that the other modes in these dice do not form a non-transitive triple. I intend to search for a numbering of the dice that does make a non-transitive triple. This might require relaxing the constraints I put in for symmetry (of the positions used for the d6 mode) or for the diagonal bands. A "good" non-transitive triple would have margins of victory in the three contests that are balanced and relatively pronounced.

[1] Or more, but for our purposes three.
#g4g12
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Alexandre Muñiz

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Here's a new type of puzzle I'm working on. Given five transparent 15-omino pieces, make a complete set of pentominoes by overlapping them. The solution for this particular set of pieces is shown below.  There are a zillion ways to arrange the pentominoes into five and seven piece clusters that could produce a puzzle of this form.  The real problem is, can we find a particularly good set of 15-ominoes to use?

What makes a particularly good set? As much as possible, the 15-ominoes should be partitionable into three different pentominoes in multiple ways. (These partitions must have a middle pentomino that leaves two other pentominoes when it is cut out.) Also, as many as possible of the pentominoes should be present in dissections of more than one 15-omino. It'd also be nice to have at least one false partial solution, like a pair of 15-ominoes that can make 5 different pentominoes by overlapping that isn't present in the correct solution.

Then there are some practical matters related to making this puzzle as a physical lasercut set. It would be bad if the centers of gravity of the pieces in the pair were not within the pentomino of overlap, because the top piece could fall over. I'd also want the 15-omino pieces to have an efficient packing into a square, in order to minimize wasted material and cut length.
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Alexandre Muñiz

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This is a fragment of a double tessellation. A double tessellation is a set of polygons that covers the plane such that every (non-boundary) point in the plane is in two polygons. Of course, you can create a double tessellation simply by overlaying two single tessellations, but this isn't interesting, so we look for proper double tessellations, which can't be decomposed into single tessellations. (It's also nice if none of the segments overlap.) This tessellation contains two different polygons: 150°–90°–150°–90°–150°–90° hexagons and regular hexagons. There are also two types of vertices: ones where six irregular hexagons meet, and ones where one regular and two irregular hexagons meet. The former are "double vertices" since one goes twice around the vertex before getting to the polygon you started with, and the latter are single vertices.

Puzzles:

Can you find a double tessellation with only double vertices and no single vertices?

Can you find such a double tessellation that is also "vertex transitive"? (i.e. any vertex can be moved to any other vertex by a symmetry of the tessellation?)

Can you find a double tessellation with only one type of polygon? (i.e. all polygons are congruent?)
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Alexandre Muñiz

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A magic figure. All lines sum to 50.
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Oh. Huh.
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Sicherman dice are pairs of dice numbered [1, 2, 2, 3, 3, 4] and [1, 3, 4, 5, 6, 8] respectively. They have the nice feature that the distribution of results when you roll them and add them together is the same as for a normal pair of six sided dice. But some games care about rolling doubles, so they don't work the same as regular dice for that purpose.

This suggests a puzzle: color the faces of a pair of Sicherman dice so that rolling a matching color pair works the same as rolling doubles on normal dice. (That is, there is exactly one matching color pair for each even sum: 2, 4, 6, 8, 10, and 12.) How many distinct ways are there to do this? (Up to isomorphism, in other words, the choice of particular colors isn't what's important.)

Some faces might not match any face on the other die. It's convenient to designate one color to be the color that never matches, so we don't have spurious extra solutions where we choose various ways of splitting the never-matching faces into different colors. So let's use black for that. All faces that match no others are required to be black, and black faces are considered to not match each other, in case there happen to be non-matching faces on both dice.

Given this formulation of the problem, I have found four solutions. I think I have them all, but I'm not entirely certain. I hope to get physical copies of some of these solutions made; last year I backed a Kickstarter to produce custom engraved dice, and some of these will probably make it into my order.
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Ah. Just goes to show you; I lack the focus to be a diligent puzzle solver. I did find the one you consider best. It has a nice mnemonic: 1 on the low die, odd on the high die; 4 on the low die, even on the high die. You could easily use it without even coloring the dice.

The other one I like is:
[1A, 2B, 2_, 3_, 3C, 4D], [1A, 3A, 4B, 5C, 6D, 8D]
(Your 4th solution, but with letters swapped.)
In addition to only having non-matching faces on one die, this has a couple nice properties:
· Each of the 36 rolls is distinct: No roll has a pair of faces with the same numbers and colors as another roll. As far as I can tell, only your first four solutions have this property, of which this and the first are the ones with non-matching faces on one die only.
· The colors (excluding black) can be given a strict ordering that follows the numerical order of the faces on both dice.

I'll probably make a set of these and a set of the one you thought was best.
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Here's the Mate panel applet I wrote to display and change the name of the current workspace. It's useful if you're someone like me who uses a dozen workspaces and has a different project on each one. This works with Mate 1.8.x. It's likely to require you to install some packages you don't have installed in order to work. If you want to run this and have trouble, let me know, and I can try to help sort it out. (Theoretically, there would be checks in the configure script that catch that, but I'm not there yet.) There's a bug that breaks dragging the applet around on the panel. The workaround is to drag another applet while holding down shift to push it around as necessary.
mate-workspace-name-applet - Mate panel applet for viewing and setting the name of the current workspace.
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What happens when we forget about units.
 
Numbers are hard.
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Alexandre Muñiz

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There are 12 ways to make an integer length segment of length 7 or less with a mark at a positive integer distance from the endpoints. A reasonable puzzle is to try to form a figure where the pieces are connected as in the lower left. I noodled around a bit, and my best attempt was the one on the lower right, where the remaining piece was the right length, but the mark was in the wrong place. I have no idea if this puzzle is actually solvable.
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Oh, wow. I deleted the comment about the other figure shortly after I posted it because I realized that I hadn't thought it through at all, and there might be some fairly simple proof that it couldn't work. But there it is. Nice!
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I made a laptop. Materials: Samsung Galaxy Note 8 tablet, Keycool 84 keyboard, USB On-The-Go adapter cable, two custom lasercut white delrin pieces, Southco adjustable friction hinges, #8-32 3/8" machine screws and nuts, velcro squares with sticky backs (so that the tablet can be used separately) and high strength molding tape (connecting the lower delrin piece to the keyboard.)
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I was thinking about the famous 15-puzzle, and wondering what it  would be if it were a 12-puzzle, and this doodle is the result. (The object of the puzzle, by analogy with the impossible version of the 15-puzzle, is to restore the 11 and 12 to numerical order by sliding pieces in sequence.) I do not feel that this is a particularly good or interesting puzzle, and I'm not sure what one would have to do to make it so. The pieces' ability to rotate is problematic, although it could be avoided (or controlled) if this were a touchscreen game, but that's something that would have a bit of a learning curve to accomplish, and I have a bunch of other projects at the moment.
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As I mentioned before,  I am going to be getting some custom laser-engraved dice made, and I have been looking for nice mathy design ideas. One promising idea was the magic cube found by Mirko Dobnik, and shown at http://www.magic-squares.net/c-t-htm/c_unusual.htm#order%202

This cube has faces divided into 2×2 grids, where each face sums to 50, and each ring around the cube sums to 100. In order to turn Dobnik's cube into a usable d6, I would need to find a solution that placed the numbers 1 to 6 on different faces of the cube.

I decided this was a good time to learn how to use a constraint solver. I picked Numberjack because it uses Python, which is the language I am most comfortable with, and there was a magic square example that I could tweak. With face sum and ring sum constraints, and constraints to put the numbers 1 to 6 on the proper faces, (plus symmetry breaking constraints) I was getting at least hundreds of thousands of solutions. So I added constraints for the four diagonals that traverse all six faces to sum to 75, and I fixed the numbers 1 to 6 in the positions I liked best. That gave just eight solutions. Of those eight, the one shown had odds and evens in checkered positions on all six faces. (Note that I did not use the 'zig-zag' line constraints that Dobnik's solution satisfies.)
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Here's a problem +Brendan Barnwell came up with and told me about. Given an N × N square grid, where the edges wrap around in a toroidal fashion, color N squares in each of N different colors such that for any pair (a, b) of colors, (a and b can be the same color) there are exactly 4 ways to go from a cell of color a to an adjacent cell of color b. Because there are four ways to go from a cell to another of the same color, the area of a given color must contain either a tromino, (either L or straight) or two dominoes. All other cells of that color must be disconnected single squares.

Some solutions for N=3, 4, and 5 are shown below. (You can think of the unshaded area as being a single image of the grid, surrounded by additional shaded images of the grid where it wraps around.)

Some questions: for small N an exhaustive enumeration of solutions should be possible: how many are there? In all of the cases shown, the squares of any color can be translated onto the squares of  any other color. Are there solutions with no symmetry? Is there  a quick algorithm for generating a solution for a given size?
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For N odd, at least, here's a quick algorithm: let the colour of square (x, y) be (x-y)^2 + x mod N.  Many other quadratic functions should work as well.
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