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Alexandre Muñiz
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Alexandre Muñiz

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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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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 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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A puzzle: make six of the 12 pentominoes out of the collection of pieces below. (Each pentomino will be formed from two pieces.) Then make the remaining six pentominoes out of a second copy of these pieces. Each pentomino is individually trivial to make; the trickiness of the puzzle comes in finding a way to partition the pentominoes into two sets of six that will work.
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Note: flipping the pieces over is allowed.
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A "perfect polyomino" discovered by +Justin Lanier. Excepting itself, only the shown polyominoes tile the whole. A set of one of each of these also tiles the whole. This is analogous to the idea of perfect numbers, (numbers whose proper divisors sum to themselves.)
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I'm working on a blog post about World Cup group results. It turns out you can model them with oriented graphs: directed edges correspond here to a victory for the team that the edge points away from, and absent edges stand for ties. There are 42 distinct oriented graphs with four vertices, but only 40 different group scores; in two cases the same score can be achieved with two different graphs. (Wins score three points, ties one, losses zero.) The graphs are not equally likely to appear; those with more symmetry can be made in a smaller number of different ways.

A puzzle: In the 2010 men's soccer World Cup, the eight groups all had different scores. Is this what you would expect, or would you expect there to be at least one repeat? (For the sake of a simple model, you can assume that for each game, the three possible results have equal probability. This is obviously not actually true, and yet it's close enough to the truth to make a reasonable rough model!)
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+Joseph DeVincentis  So, I was intending to get around to writing such a program, (I did write one to get the numbers of ways to get each score) but I haven't yet gotten around to it. I did write a program to simulate the results, and it shows 8 different group scores about 400 times in 1000 trials. So one of us has something wrong with our program!
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Alexandre Muñiz

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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 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.
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I wrote up the blog post I was threatening.
Here's an article from the BBC about the so-called “Birthday Paradox” applied to the teams of the 2014 FIFA World Cup. The Birthday Paradox is the term for a common failure of intuition. It's rather unlikely for two people to share a birthday, and people usually expect it to remain unlikely for ...
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Have him in circles
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