# 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.

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