Posts Tagged ‘cube’

This is how I roll: Magic dice, part 2

April 11th, 2015

One formulation of a magic cube simply numbers the vertices of a cube from one to eight, and requires that, for each face, the set of vertices that frame it have the same sum. There are three distinct magic cubes of this type:

magic-vertex-cubes-all
Now, first of all, in applying this to a design of dice, it seems that the ideal application would be octahedral dice, since the octahedron is the dual of the cube, and thus the numbers could simply be printed on different faces. The sets of faces with magic sums would then be the ones surrounding a vertex of the octahedron. But I don’t have an order for custom d8’s, I have one for custom d6’s. Given that, what is the best way to represent one of these figures on a cube? As before, we can use a bold font to highlight the number that is used as the result of a roll. This time, since we can’t print directly on the vertices, the numbers will be repeated on each face that borders a vertex.

I have two answers. The first places the numbers on their traditional faces (with opposite faces summing to seven.) Only the middle numbering above can be used in this manner.

magic-vertex-cube-1

The second is suggested by the figure on the right. Since the numbers we aren’t using (that is, 7 and 8) appear at opposite vertices, we can highlight a diagonal ring around the cube and catch all of the numbers from 1 to 6. This is nicely symmetrical, but it does not put the numbers in their traditional positions.

magic-vertex-cube-2

Wanderings on a Six-Sided Die

August 27th, 2010

Here’s a little doodle on a grid based on a standard six-sided die:

I started by deciding that the pip positions should all connect North to south and East to West. It followed logically that I could have a puzzle where the solver could choose one of two possibilities for each empty cell: connecting North to West and South to East, or North to East and South to West. Because there are 33 non-pip squares, there would therefore be 233=8,589,934,592 ways to fill the grid. The lines on the outside of the grid show how the squares would connect when folded into a cube.

As an exercise, I found a way to make a single circuit, which is shown above. While that turned out to be about the difficulty of puzzle I can handle in something I am solving by hand, I’m sure there are more interesting specimens to be found.

Unfortunately, because the number of pips is odd, it’s impossible to have two circuits where each go through all of the pips. The circuits would have to cross each other an even number of times. But we could have one of the circuits cross itself once, and then have both circuits go through the remaining 20 pips. (Let’s call that problem number, oh what are we on, #12. By the way, the problem numbers are so that I can keep track of solvers of numbered problems and give them the fame they deserve. Nobody has solved any yet. You can be the first!)

Another possibility would be to use three circuits, each crossing itself once, and each visiting 12 pips in addition to the self-crossing. That’s #13.

I like big, wide ranging circuits here, so a constraint I like is to have circuits that visit all 6 sides. So bonus points on the preceding problem for having all three circuits do that. And that suggests #14: Maximize the number of circuits in a solution where all circuits present visit all 6 sides.

I thought early on that I could take this in the direction of a knot theoretical puzzle, but then one would have to keep track of which thread went under which in the crossings, which seemed like an unnecessary complication. I also think it would be interesting to make a multi-state maze (See Robert Abbott’s site for some good examples) using this template, but I haven’t yet had any good ideas for how that would work. If you have a good idea for a variation on this puzzle, I’d love to hear it. (This is of course true for all of my puzzles.)

For your solving convenience, I have an empty grid image here, and the Inkscape SVG file I used to produce it and the image above is here.