==> geometry/calendar.s <==
First note that there are *nineteen* different letters in the
month abbreviations (abcdef gjlmno prstuv y) so to get them all on the
eighteen faces of 3 cubes, you know right away you're going to have to
resort to trickery.
So I wrote them all down and looked at which ones could be
reversed to make another letter in the set. The only pair that jumped
out at me was the d/p pair. Now I knew that it was at least feasible,
as long as it wasn't necessary to duplicate any letters.
Then I scanned the abbreviations to find ones that had a lot of
common letters. The jan-jun-jul series looked like a good place to
start:
j a n
u l
was a good beginning but I realized
right away that I had no room for duplicate letters and the second cube
had both a and u so aug was going to be impossible. In fact I almost
posted that answer. Then I realized that if Martin Gardner wrote about
it, it must have a solution. :-) So I went back to the letter list.
I don't put tails on my u's so it didn't strike me the first
time through that n and u could be combined.
Cube 1 Cube 2 Cube 3
j a n/u
n/u l
would let me get away with putting the g
on the first cube to get aug, so I did.
j a n/u
g n/u l (1)
Now came the fun part. The a was placed so I had to work around
it for the other months that had an a in them (mar, apr, may).
m a r
d/p y (2)
Now the d/p was placed so I had to work around that for sep and dec.
This one was easy since they shared an e as well.
d/p e s
c (3)
Now the e was placed so feb had to be worked in.
f e b (4)
The two months left (oct, nov) were far more complex. Not only
did they have two "set" letters (c, n/u), there were two possible n/u's
to be set with. That's why I left them for last.
o t c
n/u v (5)
So now I had five pieces to fit together, so that no set would
have more than six letters in it. Trial and error provided:
j a n/u a b e
g n/u l or, c d/p g
r s m alphabetically: f l j
y c d/p n/u m o
e v t s n/u r
o f b v t y
Without some gimmick the days cannot be done. Because of the dates 11 and
22, there must be a 1 and a 2 on each cube. Thus there are 8 remaining spaces
for the 8 remaining numbers, and because of 30, we put 3 and 0 on different
cubes. I don't think the way you allocate the others matter. Now 6 numbers on
each cube can produce at most 36 distinct pairs, and we need 31 distinct pairs
to represent all possible dates. But since 3 each of {4,5,6,7,8,9} are on each
cube, there are at least 9 representable numbers which can't be dates.
Therefore there are at most 27 distinct numbers which are dates on the two
cubes, and it can't be done. In particular, not all of {04,05,06,07,08,09} can
be represented.
The gimmick solution would be to represent the numbers in a stylised format
(like say, on a digital clock or on a computer screen) such that the 6 can be
turned upside down to be a 9. Then you can have 012 on both cubes, and three
each of {3,4,5,6,7,8} on the other faces. Done.
Example: 012468 012357