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On July 3, 1957, John Stephenson Singleton filed for a patent with the UK Patent Office. His invention was called "Improvements in and relating to perpetual calendar devices," and described a way by which two cubes could be used to display all the days in a month.

If you're thirty or older, you may remember these calendars from the bank. There was typically a barrier at the back of the check writing station, with three wells on the top of it and three windows on the side facing the patron. The first of the three wells was rectangular and the remaining wells were square. The bank employee could drop a wooden block into first slot and two wooden cubes into the second and third. The block bore the name of the month; each side of the cubes showed a digit; between the three of them, they could display the current date, e.g., [April][2][4].

Mr. Singleton received his patent on March 17, 1958. But I want you to consider something.

One of the criteria for a patent is that the invention be "non-obvious." On the face of it, Mr.Singleton "improvements in and relating to perpetual calendar devices" seems like a no-brainer: you have three blocks (each with the names of four months on their rectangular-sides, and their square-sides blank) and two cubes with the digits distributed amongst them in such a way that every possible day from 01 to 31 can be shown -- what's so innovative about that. In truth, that final bit -- the part about distributing the digits amongst a pair of cubes such that every possible day can be displayed using only the two of them -- is considerably more "non-obvious" than it seems. Can you figure out how to do it?

The patent can be seen here -- but viewing it (or the comments to this post) will ruin the fun of trying to solve the puzzle. Wait until you're stumped or, better yet, confident that you have sussed out the answer -- you'll be glad you did.

Posted on February 08, 2006 to Storytelling


This was the puzzler on Car Talk about four or five months ago. Pretty neat -- took me about 15 minutes to figure it out, IIRC. I won't ruin it for anyone (yet), but I will say it's pretty funny that someone actually patented that.

Here's a very obscure hint: think of a digital alarm clock.

Posted by: Chris Lawson on February 8, 2006 10:23 PM

There are six sides to a cube and twelve mmonths in a year, so if the name of each month is stamped baseline-outward on one half of one side, any given month can be made to peek through a 2:1 viewport.

Meanwhile two cubes have twelve sides altogether. Since the only numbers which would need to be doubled for the day of the month are one and two (11th, 22nd) one cube would be stamped with the numberals 0-2 and 7-9 while the other would be stamped with the numberals 1-6.

I hope you're moderating these comments.

Posted by: ben on February 8, 2006 10:23 PM

one cube would be stamped with the numberals 0-2 and 7-9 while the other would be stamped with the numberals 1-6.

And the days 07-09 would be displayed how?

Posted by: Matthew on February 8, 2006 10:28 PM

I have a plastic version of his calendar! Instead of a block for the months though there are 6 plastic "cards" with the months on them and they go in a slot on the base underneath the date cubes. I inherited it from my grandmother because I used to love switching the days over as a child. Wow, now I feel special.

Posted by: Deena on February 8, 2006 10:33 PM

Never posted here before but this was far too entertaining. Another combination that works:
Not much of a change, but ... think I could patent it?

Posted by: MatthewButNotThatMatthew on February 8, 2006 10:45 PM

If 6 turned out to be nine, I wouldn't mind.

Posted by: eric on February 8, 2006 10:59 PM


MatthewButNotThatMatthew appears to have it right, though he doesn't explain much. Of course, there are multiple combinations that could work, as long as the base rules are followed. Without further ado...

Each block needs a 1 and a 2, because you need to represent 11 and 22. You don't need two 0's to appear at the same time, and 3, though it does occur as a first digit, will only appear with 0, 1 or 2. 0 does, however, need to appear with every other digit. This would appear to be a problem: 0 must appear on both blocks, unless you can put every other number on one block, which, well, you can't. Now with 0, 1 and 2 on each block, you have 6 faces to play with. But -- you have 7 digits: 3,4,5,6,7,8,9. You're stuck.

But wait! The creators of the Hindu-Arabic numeral set, in their infinite wisdom, saw fit to give us a pair of numerals that are rotated versions of each other: 6 and 9. Truly a brilliant coup by the same people who brought us the Kama Sutra. So now, you can fill up your remaining 6 faces with 3, 4, 5, 6/9, 7 and 8. Ta-da.

You may be starting to think, at this point, gee, wouldn't our number system be a lot more efficient if we had more rotationally equivalent mumeral pairs. But ask yourself, just how is this useful? I mean, really? We've got our calendar dates compressed into two blocks. What more could you want out of life?

Posted by: Bob S on February 9, 2006 12:31 AM

I had this as an interview question once. It took me about 30 seconds to catch the 6/9 relevance (I tried 2/5 first, but they are mirrored, not just upside down).

Posted by: Ryan Waddell on February 9, 2006 12:58 AM

I'm a newish lurker here (I think this might be my first comment?)

I have to say I love the random stuff you post. And I've probably directed about 500 people to your 'I'm a start some drama' post. Gold.

So it is with no great joy that I say: it's 'criteria' in that case. But I applaud you for being one of the few people in the world who are actually aware of the singular version of the word!

Oh god, I sound like a total prat. Sorry, I'll go back to lurking now.

Posted by: Lucy on February 9, 2006 4:44 AM

We have one so I didn't really have to wonder.

Posted by: Rachel on February 9, 2006 6:08 AM

I came up with a different combination that works (by doing it my head, writing it down, figuring out I was wrong, and switching two numbers).

1,2,3,7,8,9 and 1,2,4,5,6,0 gives you 1 through 31 well enough.

Posted by: Conner on February 9, 2006 6:25 AM

You don't get 05 that way.

Posted by: Isaac Kelley on February 9, 2006 6:30 AM

Nice. Almost clicked through when I thought I was stumped, but decided to keep trying. As MbntM points out, there is no way to do it without inverting 6 to make 9.

Now, dammit, tell us the answer to the puzzle about the chemotherapy pills A and B!

Posted by: ajay on February 9, 2006 6:37 AM

I went with the 123456, 120789 combo, thinking that for 01-09, you could just leave the first block out and have an empty space. Obviously, that would confuse the general population, which is why I don't have a patent.

Posted by: Crash on February 9, 2006 7:42 AM

I got this as an interview question too (a few years ago). I got it eventually but I think more by luck/intuition/whatever than logical thinking.

My wife got one of these so-called "calendars" as a gift last week. I don't understand the point of it. You have to know the date to set the calendar; and if you know the date, what's the point of a stupid calendar? Isn't that what computers are for, to do these trivial tasks like changing what date it is?

Posted by: Steve Dupree on February 9, 2006 8:29 AM

I can't believe that in the above 15 comments, TWO people saw fit to post their incorrect answers. Who does that?

Posted by: rob Cockerham on February 9, 2006 9:26 AM

Sorry, I couldn't get past this:
"One of the criterion for an patent..." !!!

One of the criteria for a patent...

Posted by: Larry on February 9, 2006 9:35 AM

Maybe I've got the wrong answer, but my solution came to me instantly...basically, you just need both cubes to have the numbers 1 and 2, and the rest (0, 3, 4, 5, 6, 7, 8, 9) can go anywhere. Well, as long as 3 and 0 are on different cubes.

Posted by: Anne on February 9, 2006 11:12 AM

Crap. I forgot about 0 needing to be against all numbers. Back to the drawing board. I knew mine was too easy!

Posted by: Anne on February 9, 2006 11:13 AM

Frankly, I find the insistence on a leading zero to be rather pedantic (though in being so quick to post, I might've considered that the second block would need to be in there to keep the first from sliding around all the time, or somesuch). ...In that case, yes, MatthewButNotMatthew's right.

P.S. to rob Crockerham: IRL and in public do you break into conversation to point out that somebody is wrong (without contributing anything useful to the conversation), or do you just save that sort of thing for your online conversations?

Posted by: ben on February 9, 2006 2:36 PM

What is the font like? Can 2 be used upside down as a 5 (ignoring the more obvious one that could be done regardless of font)?

Posted by: Matthew on February 9, 2006 3:09 PM

To accommodate the repeat-digit days 11 and 22 you must have a 1 and a 2 on each cube. And to accommodate 01 through 09 you must have a 0 on each cube as well.
That leaves six remaining sides with seven remaining digits. Thus you must use one side to represent two digits, which can only be 6 and 9.
Since 0 and 1 are on each cube it doesn't matter which has the 3 for 30 and 31.
You can then distribute the 3,4,5,6/9,7,8 in any way between the two cubes. If my math is correct there are 20 unique solutions.

Posted by: Rabid Child on February 9, 2006 7:35 PM

Very good puzzle. I love the "non-obvious" part. Think "outside of the box". ;)

Now, back to my Su Doku puzzle already in progress.

Posted by: dr.andros on February 9, 2006 9:26 PM

While we're quibbling, criterion may now have been corrected, but "an patent" ?

Posted by: George on February 10, 2006 5:22 AM

Sorry Ben. I didn't mean to attack you personally. As a peace offering, I won't point out the error on your latest post.

I must also admit that it was the type of puzzle that could fool a person into thinking he had solved it. A fun puzzle!

Posted by: rob cockerham on February 10, 2006 7:40 AM

Seeing as that challenge couldn't withstand the combined brainpower of this group, here's an extension;

What is the minimum number of cubes needed to display all digits from 01 to 99?

Bonus: What is you also include 00?

Posted by: Duncan D'Nuts on February 10, 2006 8:19 AM

Three cubes will let you display any 2-digit number (including 00). One solution is 0/1/2/3/4/5, 0/1/2/6/7/8, 3/4/5/6/7/8 - every digit appears on two cubes.

To forestall the obvious next question, 5 cubes are needed for all 3-digit numbers. Unless there is some trick I am not thinking of (always a possibility), N*3/2 cubes (rounded up) are needed for N-digit numbers (since cubes have 6 faces, there are effectively 9 digits treating the "6" and "9" as the same, and we need to be able to show N copies of each digit simultaneously).

Posted by: Dan Blum on February 10, 2006 9:27 AM

Ha - Dan Blum - your formula fails to take into account the obvious need for commas beyond 3 digit numbers. You lose. You get nothing. Good day sir.

Posted by: pick my nits on February 10, 2006 12:37 PM

this post reminded me of the time cube website (www.timecube.com), which seems to have been updated since last I viewed it. Now Wikipedia is evil too.

I can't stop reading this page, because I feel like if I read it enough times, eventually it'll make sense. maybe. It also creeps me out.

Sorry, I know this has nothing to do with anything.

Posted by: pips on February 10, 2006 2:15 PM

I'd like to offficially patent my binary calendar and my Roman Numeral calendar, also using cubes. Now I just sit back and wait for royalties, right?

Posted by: Crash on February 10, 2006 2:39 PM

This is all fine and dandy but why does it still take an hour to get a check cashed?

Posted by: cwilsongo on February 10, 2006 4:24 PM

You numerical brainiacs have all hurt my head. Anyone for Scrabble?

Posted by: Belinda on February 11, 2006 5:37 AM

Very interesting. I am enlightened! ;o)

Posted by: Saur on February 11, 2006 12:16 PM

Well, I'm disappointed in myself. I immediately thought of the 6/9 inversion (too easy, thought I) but confused myself and believed I had an extra number to allocate somewhere. So I spent a good 20 minutes thinking about it and came up with this:

If the recess for the cubes and the square faceplate are slightly yet deliberately misaligned it is possible to make the '8' appear as a '3' - the edge of the square faceplate can obscure part of the '8' in one orientation and rotating the block 180 degrees around the side's normal will show a complete 8.

This allows you to collapse the 8 and the 3 into a single side and gives you a spare one, perhaps for a smiley? :)

I believe I get extra points for being a smart arse.

Posted by: David on February 12, 2006 4:25 PM

wow! is it like so! looking slightly tricky but interesting huh! I would definitely go for it.

Posted by: sam on February 13, 2006 4:17 AM

cwilsongo: You think an hour is slow? Move to the UK, for some inexplicable reason it takes 3-5 days here.

Posted by: Simon on February 13, 2006 3:22 PM