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calendar cubes

I believe the aha!!!! factor is as follows...

for single digit days hide one cube behind the other, so you can only see the front of one cube

Because we have days like 11 and 22 we must have 1 and 2 on both cubes.  (Also note that 1 and 2 precede all digits at some point and those digits must be split evenly)

Because we have days like 10, 20, 30, we must have 0 opposite the cube with 1,2,3...

cubes:
1  1
2  2
3  4
5  6
7  8
9  0

Cheers.

Will Smith
Friday, January 07, 2005

Close, but what about the 5th or 8th? You don't have any way to do 05 or 08. Or can we hide a cube if needed?

Let's try puting 0, 1, and 2 on both:

0  0
1  1
2  2
3  6
4  7
5  8

I'll put 3, 4, 5 on one, and 6, 7, 8 on the other.

But wait, where's 9??

Aha! Just turn the 6 upside down!

BradC
Wednesday, January 12, 2005

Actually, the 0 on the 3 cube could also be a blank. (The other one could not, you need it for 10, 20 and 30.) This would make things slightly inconsistent (_7 vs 05), so we won't do that.

You can actually spread 3, 4, 5, 6/9, 7, 8 around as you wish on the two cubes.

Just for fun, lets figure out how many possible ways we could make these cubes:

The three has to go on one of the two blocks, so if we randomly select 2 more from the remaining 5 choices, we end up with 5 * 4 = 20 possible combinations, or only 10 if we ignore order.

The 10 unique block sets:
0 1 2 3 4 5 - 0 1 2 6 7 8 <-- this is the easiest to use
0 1 2 3 4 6 - 0 1 2 5 7 8
0 1 2 3 4 7 - 0 1 2 5 6 8
0 1 2 3 4 8 - 0 1 2 5 6 7
0 1 2 3 5 6 - 0 1 2 4 7 8
0 1 2 3 5 7 - 0 1 2 4 6 8
0 1 2 3 5 8 - 0 1 2 4 6 7
0 1 2 3 6 7 - 0 1 2 4 5 8
0 1 2 3 6 8 - 0 1 2 4 5 7
0 1 2 3 7 8 - 0 1 2 4 5 6

Of course, for each of these, there are different ways to arrange the numbers on each cube.

6! = 720, but we have to account for cube rotations and reflections that are really equivalent. A cube has six faces, and four rotations of each, so 720 / 24 = 30 possible arrangements.

For two cubes, 30*30 = 900, x 10 block sets = 9000 unique ways to construct these blocks.

Oh, and each number can be written 4 possible directions on each face :)

(Except 0, which looks the same upside down. Some fonts might also make the 1 or 8 look the same upside down.)

So 4^12 = 16,777,216, divided by 4 for the (two) zeros = 4,194,304 * 9,000 = 37,748,736,000 or only 4,718,592,000 if a font is used where 1 and 8 both look the same upside down.

BradC
Wednesday, January 12, 2005

Granted, the 6 - 9 approach works perfectly, but what is wrong with my solution.  The question doesn't state that on cube cannot be aligned behind the other in such a what that it's front face doesn't show.

So any single digit number could be

Back
?    or  ?
5          8
Front

Any two digit number can be

Back
1 8  or  2 2
Front

Cheers.

Will Smith
Wednesday, January 12, 2005

"a man has two cubes on his desk. every day he arranges both cubes so that the front faces show the current day of the month. what numbers are on the faces of the cubes to allow this?"

I think the precise wording of the question precludes hiding a cube behind the other or sticking it in your top desk drawer for a day, but if you want to allow that, then that's up to you. It is a variation of the puzzle, though, and therefore will have a diff different solution set.

BradC
Thursday, January 13, 2005

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