The math behind it
Four cubes, four colors, and a stack that has to look "all different" on every side. It sounds easy. Here's why it isn't, and how a little graph theory solves it in a few minutes with pencil and paper.
1. One cube has 24 orientations
Pick which of the 6 faces points up. Then you can still spin the cube about the vertical axis into 4 quarter-turn positions. Every position of a cube is exactly one of those choices:
6 faces that could be on top × 4 quarter-turns = 24 orientationsSpinning, tilting and rolling never separate two opposite faces, so each cube always keeps the same three opposite pairs. That fact is the key to the whole puzzle.
2. Counting stacks
| Cubes in a fixed order, each in any of 24 orientations | 244 = 331,776 |
| …and the order of the cubes may also change (4! = 24 orders) | 4! × 244 = 7,962,624 |
| "Essentially different" stacks: divide by the tower's 8 symmetries and the 24 orders | 7,962,624 ÷ (8 × 24) = 41,472 |
Why 8 symmetries? Turn the whole tower about its vertical axis (4 ways), and optionally flip it upside down (×2). You're looking at the same stack, so those shouldn't count as different. And the cube order doesn't matter for winning: each side just needs four different colors, wherever they sit. So 41,472 = 331,776 ÷ 8 = 3 × 243.
3. How many winners?
| Up to symmetry | exactly 1 |
| With the cube order fixed | 8 of 331,776 |
| If order counts too | 192 of 7,962,624 |
The chance a random stack wins is 1 in 41,472, about 0.0024%.
Imagine building a fresh random stack once per second. If you might repeat stacks, you'd expect about 41,472 tries, roughly 11.5 hours. If you're organized and never repeat a stack, on average you'd hit the winner halfway through the list: about 20,736 tries, or 5.75 hours. Either way, guessing is a bad plan.
How far is any stack from a win?
Count one quarter-turn of one cube as a move. A single cube can reach any of its 24 orientations in at most 3 moves (1 orientation needs 0 moves, 6 need 1, 11 need 2, 6 need 3). Checking all 331,776 stacks by computer: every stack can be solved in 8 moves or fewer, and the average is about 5.4. Only 2,688 stacks actually need all 8.
4. The graph theory solution
Make a graph with 4 vertices, one per color. For each cube, draw 3 edges, one for each pair of opposite faces, joining the two colors on that pair. Label the edge with the cube's number. If both faces of a pair are the same color you get a loop.
What we're looking for
Two edge-disjoint subgraphs, each one
- uses exactly one edge from every cube, and
- gives every color degree 2.
The first tells you what goes on the front and back of each cube; the second, the left and right. Degree 2 means that color shows up exactly once on the front and once on the back. Edge-disjoint means no opposite pair is asked to face two ways at once.
Front/back: R –1– G –4– W –2– B –3– R
Left/right: R –4– G –3– W –1– B –2– R
To build the stack, walk around each cycle in one direction. An edge going R –1– G means "cube 1 has R in front and G behind." For left/right, the walk gives cube 4 R on the left and G on the right, and so on. Each cube's remaining pair goes on top and bottom. Done!
Fun check
Count each vertex's degree (a loop counts 2). You get R 7, G 6, B 5, W 6, exactly the number of faces of each color across all four cubes. That always works: every face is the end of exactly one edge.
5. A little history
- 1900: a version called the Katzenjammer puzzle was patented.
- Later versions were sold as The Great Tantalizer.
- 1967: Franz Armbruster's version became Parker Brothers' Instant Insanity, which sold more than 12 million copies.
6. The bigger picture
With n cubes and n colors, deciding whether a set of cubes has a solution is NP-complete (Robertson & Munro, 1978). That puts it with the famously hard problems that no one knows how to solve quickly in general. The graph trick works great for 4 cubes, but it doesn't scale to big ones.
Across all possible 4-cube puzzles, the most solutions any set can have is 72. The classic set is designed to have just one.
Try it yourself
1. A cube sits with red on top. How many ways can you turn it so red stays on top?
4: the four quarter-turns about the vertical axis. Do that for each of the 6 faces and you get 6 × 4 = 24.
2. Why does every color need degree exactly 2 in a solution subgraph?
The front/back subgraph has one edge per cube, so it describes 4 front faces and 4 back faces. Each color must appear once on the front and once on the back: 2 edge-ends per color.
3. Why can't a loop be in a solution subgraph here?
A loop already gives its color degree 2, so the other three cubes would have to form a triangle on the other three colors. Cube 1's R–R loop would need edges G–B, B–W and W–G, but no cube has a G–B pair. Cube 2's G–G loop would need R–B, B–W and R–W from cubes 1, 3 and 4. Cube 1 can only give B–W, and then cube 4 can only give B–W too. Both are dead ends.
4. Why must the two subgraphs be edge-disjoint?
Each edge is one opposite pair of one cube. That pair can face front/back or left/right, but not both at once.
5. Quick calculation: what is 245? Why does it equal 4! × 244?
245 = 7,962,624. It matches because 4! = 24, so 4! × 244 = 24 × 244.