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 Post subject: Set Theory Puzzler
PostPosted: Sun Aug 08, 2004 7:19 pm 
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Nightstar Graveyard Daemon
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My coworker has a little wooden calendar curio thing that tells the date with two six-sided dice. You pick up the two dice, pick the two digits for the date, and set them back down.

There are only two dice. That means that you only have 12 sides and you could need as much as 31 different sides. Obviously, there's some reuse of numbers, and you can swap the dice, so the 12th and the 21st could use the same sides, but swap the dice.

So here's your puzzler. What are the numbers painted on each of the two dice? There are many combinations that work, so to reduce those I'll give you a starting hint: assume one of the dice has 4, 5, and 6 on it (and the other one doesn't since you don't need a 44th, 55th, or 66th day of the month).

You don't need to write out how you would make the 31 different numbers. Just list the numbers on the two dice.

Bonus question: what's the first number you can't make with those dice?


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 Post subject: Re: Set Theory Puzzler
PostPosted: Sun Aug 08, 2004 11:28 pm 
I'm pretty sure this can't be done in real base 10. The closest I can get is:

0, 1, 2, 3, 4, 5
0, 1, 2, 6, 7, 8, 9

The problem is that the first digit can be 0, 1, or 2 (or 3, but that's a special case). So what we end up with is we must have 3 on the opposite die as 0 and 1. But we also need to be able to reproduce 0*, 1* and 2*, which means each of those digits must be on both dice. Sadly, using base 10 with two 6-sided dice only allows us to duplicate 2 numbers.

Therefore, I'm assuming the first digit may be absent to signify 0*:

0, 1, 2, 4, 5, 6
1, 2, 3, 7, 8, 9

This allows combinations such as "1", "30", and so forth. The first number that can't be created is 33.

There are also several ways to do it outside base 10 (using base 6, 7 or 8) with larger contiguous segments, but I'm guessing by your wording that this isn't what you were looking for. However, I'll give some anyway. For all of these solutions I'm assuming I can't do the missing digit trick, which is why I'm not allowing 9 (it requires the missing digit); if the missing bit were allowed, then we'd just stick with base 10 anyway.

Base 6:
0, 1, 2, 3, 4, 5
0, 1, 2, 3, 4, 5

Base 7:
0, 1, 2, 3, 4, 5
0, 1, 2, 3, 4, 6

Base 8:
0, 1, 2, 3, 4, 5
0, 1, 2, 3, 6, 7

Base six allows us to go up to 35 (55 in base six), base seven up to 40 (55) and base eight up to 36 (44). In the end, this makes base seven our "optimal" solution, but it's not intuitive for most people to work with. On the other hand, base eight is quite common in the realm of computer science and it happens to be the next best solution, so why not? :)

So, do I win anything?


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 Post subject:
PostPosted: Mon Aug 09, 2004 4:03 am 
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Energizer Bunny
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Posts: 1634
0 1 2 3 4 5
0 1 2 6 7 8

Oh if six... turned out to be nine... I don't mind, I don't mind...

:)

Vorn


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 Post subject:
PostPosted: Mon Aug 09, 2004 7:17 am 
Ah, but can you do the same trick in binary? HUH! HUH!


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 Post subject:
PostPosted: Mon Aug 09, 2004 9:20 am 
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Nightstar Graveyard Daemon
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Posts: 1071
Location: Wouldn't you rather observe my Velocity?
Vorn the Unspeakable wrote:
Oh if six... turned out to be nine... I don't mind, I don't mind...


Hehe, that was the catch. Interestingly enough, he has one die that has all the month names printed on it. The trick? The die sits in a bit of a recess. Each face has two months on it, so that on one face, say, January is on the top edge and February is upside-down on the bottom edge. Roll the dice over to get to February. The recess in the base hides the upside-down month name.


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