How to find out the number of ways to solve Instant Insanity

Question

Problem : We are given 4 cubes. The 6 faces of every cube are variously colored - Blue, Green, Red or White. Stack the cubes on top of another in such a way that no color appears twice on any of the four sides of this column.

The book says that by a trial-and-error method, a person can try 41,472 ways.
It is given as : 41,472 = 3 * 24 * 24 * 24

How did they do it? I know that by using proper Permutations and Combinations, one can deduce the number of ways. Can anyone please guide me through this?

Answer 1

I am puzzled by the absence of information on how the cubes are actually coloured.

• what colour is used for the remaining faces?

• are the colours applied the same for all dices (like standard dice have the dots painted in the same relative arrangement)?

• how are the colours distributed? Are there all blue cubes?

To compensate for this lack of information, I came up with this story:

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Assuming that we use a simple bot without camera to perform the manual labour of stacking the dice.

There are 6 ways to choose the front face, times 4 orientations = 24 configurations.

This applies to four cubes and gives $$ 24 \times 24 \times 24 \times 24 $$

trial configurations.

The control mechanism has colour detection, and it inspects only the four sides.

However

• The validity of the pattern along the sides is the same (either matching the criterion or not) if we rotate the column as a whole in one of four ways, reduction by factor 4 is possible.

• The validity of the pattern is the same, if we turn the whole column upside down or not, reduction by factor 2 is possible.

So we would try to provide a software update to the bot, which just tries $$ 3 \times 24 \times 24 \times 24 $$ times and is guaranteed to stack up a valid configuration, if it exists.

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Update: I found a page about this puzzle Instant Insanity, the cubes are coloured indeed by $4$ colours applied to the 6 faces. It turns out that the $4^6 = 4096$ ways to colour $6$ faces with $4$ colours reduce to $240$ possibilites if one considers rotations. (Link) The producers of this puzzle pick a specific set of $4$ out of the $240$ possible cubes which has exact one solution.

Answer 2

I was recently was wondering about the same problem and came across your thread. Many thanks for posting this. It's a really interesting problem because it seems as though it should be simple, but isn't at all obvious how to perform this calculation: likewise, the apparent simplicity but fiendish difficulty of this puzzle explains why it's been so popular for over 50 years! I'm sorry I don't have an answer, but I can contribute the following information which might help.

For brevity, I'm using C1 to C4 to indicate each of the four cubes, with lower case {f,b,t,u,l,r} to indicate the faces {Front, Back, Top, Underside (= bottom), Left, Right} ; and {R,Y,G,B,V,W} to indicate 6 different possible colours {Red, Blue, Green, Yellow, Violet, White}. There's probably a better standard notation out there: if so please let me know.

1) Consider C1. Colour the front of it Red. Choose a second colour for the opposite face, e.g. Blue. i.e. C1f = R, C1b = B. The remaining 4 colours must now be permuted around the remaining faces (t,u,l,r), You'd expect 4! = 24 permutations, but in fact it is 4!/4 = 6 permutations if we divide by 4 to account for symmetrical patterns (? 4-fold rotation of each square face) which produce the same end result. If we keep C1f = R, but choose each of the 5 colours (B,G,Y,V,W) in turn for C1b, there are again going to be 6 distinct ways in which the remaining 4 faces can be coloured using the remaining 4 colours in each case. i.e. there will be 5 * 6 = 30 distinct ways in which one cube can be painted using 6 different colours, one colour per face. (Bonus points if you can work out how many distinct ways one cube can be painted using a choice of 7 colours, one colour per face!). (Adapted from: Askew M, Ebbutt S. The Bedside Book of Geometry. New Burlington Books, London: 80-81).

So you might naively expect the number of combinations of the 4 cubes used in Instant Insanity to be 30^4 = 810,000; however this figure must be reduced to take into account the additional constraints that: a) the cubes do not have to be placed in order C1 to C4, so there is some symmetry in the sequence of cubes; b) we are not interested in the sides of the cubes: the sides of adjacent cubes are touching and not visible, and the ends are not visible in the completed puzzle; c) Instant Insanity only uses 4 colours, not 6 colours, to colour all 6 faces of each of the 4 cubes. Consequently every cube has either 2 instances of two colours, or 3 instances of one colour.

2) The actual face colours and arrangements of the 4 cubes (make a set yourself!) are: In order {b,t,f,u,l,r}: C1: {G,B,Y,G,R,R}, C2: {R,B,Y,G,Y,Y}, C3: {G,B,Y,R,Y,B}, C4: {R,B,Y,R,B,G}.

Ivars Peterson gives a really neat solution to the puzzle using graph theory - you can solve it in 5 minutes!: http://www.cs.brandeis.edu/~storer/JimPuzzles/MATCH/InstantInsanity/NOTES/InstantInsanitySolutionPeterson.pdf

How to combine the above information with the rest of the thread to come up with the answer 3 * 24 * 24 * 24 = 41,472 is beyond my high school level maths skills; but if anyone can explain step by step where this calculation comes from, I'd be really keen to learn. Best wishes, Dave.