Class is back in session
It’s the start of another school year, so let’s get some homework assigned.
One of the presents I got for Father’s Day this year was a bear-themed perpetual calendar. If you’ve never seen such a thing before, its a means of displaying the current month and date using a number of blocks. Specifically, there are four skinny rectangular blocks (on whose sides are printed the months of the year) and two cubes (on whose sides are printed Arabic digits). By arranging the blocks appropriately, any date fron January 1st through December 25th — even the infrequent leap year date of February 29! — can be shown. For example, to display July 7th, the blocks would be arranged as [July][0][7]: this is shown on my calendar below:
This type of perpetual calendar was patented in the UK by John Stephenson Singleton on March 17, 1958 as an “improvement in and relating to perpetual calendar devices.” One criterion for a British patent is that the invention be non-obvious, and that’s the challenge:
How do you distribute the digits 0, 1, 2, etc. on to the faces of two cubes so that every possible date — from [0][1], [0][2], and [0][3] all the way through [3][0] and [3][1] — can be displayed?
At first blush, this doesn’t seem like a challenge at all. You’ve got ten digits to work with — 0 through 9 — and twelve faces on which to put them. Of course, a 1 and a 2 need to be on both cubes in order to get the dates [1][1] and [2][2], but this still leaves a total of eight faces available on which to place the eight remaining digits 0, 3, 4, etc. Nothin’ to it.
…Well, not quite.
The problem is that a 0 also needs to be on both cubes. Granted, the pair [0][0] never appears as a date, but the other nine pairings of [0][1] through [0][9] do. If 0 was on only one cube, then the [0] block could be paired with at most six other digits, which means three dates at the start of each month would be unattainable. But now with 0, 1, and 2 on both cubes, there are only six faces available, but seven digits left unaccounted for. Therein lies the challenge.
Your homework is to solve the perpetual calendar problem.
Update!
The comments contain spoilers!
I believe you posited this too me shortly be for a graduated and fled South Dakota. For those of you reading who want to figure it out I’ll follow the observed online protocol and tell you that what follows is a SPOILER. So SPOILER ALERT!
The trick here is with the font the numbers are printed on the cubes. If it is an agreeable font in which 9 is simply a 6 upside down, you have it made in the shade. Now we only have 6 digits to put on the 6 remaining sides. Solution below.
Cube 1: 0 1 2 3 4 5
Cube 2: 0 1 2 6or9 7 8
Ha ha. I win.
Comment by Ben — 10.19.2007 @
No mention that the Queen B figured this one out in less than 10 seconds…about one hour, 29 minutes and 50 seconds before her sometime mathematically challenged husband? Oh, sorry for the reminder. Smooches!
Comment by Queen B — 10.21.2007 @
When I first thought about this problem (which I read on Defective Yeti sometime back), I plunkered down in my office and figured it out in probably a half-hour. When I found the solution that Ben mentions above, I thought that the problem was especially clever — what appears to be solely a pigeon-hole-principle argumnet involving numbers in fact also involves utilizing fortuitous geometric symmetries — that I posed it to several other professors in the Math Department, and the usual solution time was in the range of 15-45 minutes.
So when I come home that night, I decided to let the Queen B take a crack at it. I gave her the run-down on the problem, and she looks up for a moment in thought, and then says:
“Well, you don’t need a 6 and a 9…”
Comment by Travis — 10.22.2007 @