AlefBet
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Put half a million lightbulbs on one side of the scale and the rest on the other side. Take the side that's heavier and put 250,000 bulbs on each side. If they balance.... Oh, wait. No scales? Hmm. I'll have to think of another way, then.

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Fafnir
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Wouldn't light be refracted slightly upon exiting and re-entering the air? Could you shine a laser through and use a sophisticated instrument of some description to measure the change?

DISCLAIMER: I may be unintentionally talking bullmanure here (looooong week - me no think good sleep without).

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agaricus5
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Fafnir wrote:
Wouldn't light be refracted slightly upon exiting and re-entering the air? Could you shine a laser through and use a sophisticated instrument of some description to measure the change?

DISCLAIMER: I may be unintentionally talking bullmanure here (looooong week - me no think good sleep without).

Actually, that sounds like a viable solution, since different pressures (and so densities) of air will create different amount of refraction when light is passed through them. You'd need to position the laser carefully, though, or it may be spread out by refraction in several directions by the curved glass of the bulb.

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Maurog
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File: assume.JPG (56.9 KB)
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That's good thinking there, but I'm sure the differences in the structure of the glass between lightbulbs, tiny as they may be, will create more refraction than the pressure difference inside.

We need a better way to measure the pressure inside the bulbs without making holes in them.

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[Last edited by Maurog at 02-12-2011 07:18 PM]

Mattcrampy
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I keep thinking a bucket of water should be involved somewhere.

Ooh, what if they could find a material that had just a bit less pressure than the usual amount for bulbs or something, so that the good bulbs would sink and the bad ones would float?

Can you tell I've completely forgotten all my physics?

Matt

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Maurog
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You got it, Mattcrampy. The solution is to use a liquid with exactly the right amount of pressure. Game on

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AlefBet
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Matt, are you going to grace us with the next puzzle?

[Edited by AlefBet at Local Time:12-06-2004 at 01:26 AM]

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Mattcrampy
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Okay, here's one.

Say I have a whole collection of, I don't know, computer games. I want to arrange them in a perfect square, and it just so happens that my collection has exactly the amount of games needed so I can do that.

But now that I'm done, I want to get more ambitious. Not only do I want to arrange my games in a perfect square, I want to arrange them so the games on the left are the oldest, and the games on the right are the newest, and each row has only one genre of game in it. I can do that too, and I note with a smirk that each column only has games from one year of publication in it.

So I've got a perfect square, with each row having a different genre of game, and each column having its own year of game, and I realise, oh my god, I'm sorting my games into rows and columns. The psychologist told me that I had to work hard to not turn out obsessive-compulsive, and so I scramble them up. I wonder, while I'm idly scrambling, if I can get it so that no two games from the same year or genre are in the same row or column. But I can't do it, no matter where I put them.

What is the least amount of games I could have? And for extra credit, which genres do I have?

Matt

[Edited by Mattcrampy at Local Time:12-07-2004 at 10:44 AM: Spelling errors and added cuteness]

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Watcher
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Four games will do. For example, you might have DROD: Architect's Edition (puzzle, 2003), Seiklus (adventure, 2003), Fish Fillets (puzzle, 1998), and Sanitarium (adventure, 1998). At first, you arrange them like this:

Fish Fillets DROD
Sanitarium   Seiklus

But no matter how you arrange them, DROD will be in the same row or column as Fish Fillets (same genre) or Seiklus (same year). So you can't arrange them so that no games with same genre or year are in the same row or column.

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Abbyzzmal
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It seems that you'd only need one. If your only game was DROD, the oldest game would be on the left, the newest on the right, one genre per row and one year per column.



Regardless, that's all my current collection of computer games consists of.

TripleM
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Abbyzzmal wrote:
It seems that you'd only need one. If your only game was DROD, the oldest game would be on the left, the newest on the right, one genre per row and one year per column.



Regardless, that's all my current collection of computer games consists of.

Mattcrampy wrote:
I wonder, while I'm idly scrambling, if I can get it so that no two games from the same year or genre are in the same row or column. But I can't do it, no matter where I put them.

Thats where the 1-game solution doesn't work. With just the 1 game there, no games from the same year or genre are in the same row or column

Abbyzzmal
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Ah, I didn't see that because of the added cuteness.

Mattcrampy
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Aah, didn't notice that solution.

Watcher will have to take it, but I'll give 2 rank points to anyone who can find the next one, which is the answer I was looking for.

Matt

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Maurog
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It still can be solved with only DROD it seems. (It appears most things can be solved with DROD)
Suppose you have n^2 copies of DROD, n>1.
Just arrange them in a perfect square (nxn), and behold:
- The games on the left are the oldest.
- The games on the right are the newest.
- Each row has only one genre in it (puzzle).
- Each column has only one year in it (2003).

Now scramble it up, haha.
No matter how good you scramble it, two games from the same year or genre are in the same row or column.

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KevG
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Nice try, but:

Mattcrampy wrote:
So I've got a perfect square, with each row having a different genre of game,

Maurog
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Ahh, indeed I missed that one.
However, this is a restriction only on genre.
pick n games of n different genres, all of the same year.
place n copies of each game in a row.

- The games on the left are the oldest.
- The games on the right are the newest.
- Each row has only one genre in it.
- Each column has only one year in it.

Ok, now scramble it.
No matter how good you scramble it, two games from the same year are in the same row, column, diagonal, etc.


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Watcher
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Actually, there's also a restriction on years:

So I've got a perfect square, with each row having a different genre of game, and each column having its own year of game

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Doom
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How about this?

8 Games:

A1  A2  A3            A1  C2  ?? 

B1      B3    ----->  C3      A2

C1  C2  C3            ??  A3  C1

This seems to be the only way to arrange all A and C games like that.
The missing B2 makes it impossible to place all games.

Maurog
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File: guess.JPG (63.6 KB)
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Incorrect. Look, no B2!

A2  C1  B3

C3      A1

B1  A3  C2

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[Last edited by Maurog at 02-12-2011 07:06 PM]

Mattcrampy
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And it has to be a perfect square. So 4, 9, 16, etc.

Matt

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Abbyzzmal
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I'm pretty sure that any perfect square made of even numbers will match the criteria of your puzzle. I've been playing around with this a bit, and it seems the only way to place all the elements of one criteria in the array is to arrange them in a sort of sequence.

I.E. in the nine set:

123
312
231

The only way to keep things from being in the same row or column is to keep them in the same diagonal (imagine the array wrapping around) The permutations can be shifted or flipped, but this will still stand. It works when the second set of exclusive criteria are added:

c1 b2 a3
b3 a1 c2
a2 c3 b1

because the other diagonal has numbers that are all exclusive. It is the overlap between two series like the first one.

123
312
231

and

321
213
132

In the even case:

1234
4123
3412
2341

The second criteria will not work with the first because it is impossible to make the second diagonal function in a way that does not repeat. So, the next answer would be 16.

I hope my logic can be comprehended.

Maurog
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Sorry to be the realist again, but you're too obsessed with diagonals.
You don't have to follow diagonals at all! Behold:

1234     adcb
2143     cbad
4321     dabc
3412     bcda

If you don't follow diagonals, you have more freedom...
Now we'll just combine these two and voila!

a1 d2 c3 b4
c2 b1 a4 d3
d4 a3 b2 c1
b3 c4 d1 a2

I'm starting to think that 2x2 is the only perfect square that fits the rules.

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Maurog
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Hmmz, now I can prove that for any square NxN where N is prime and N > 2, there is a proper build, so it cannot be the solution. It is based on the qualities of ZN, for N prime.

Start with the first row of numbers and mark them 123...n
for all the next rows, let each row be a shift by 2 steps right of the previous row. There will be no conflicts. Now the letters, start with the first row again and mark it abc...x
now again, lets shift each row, but this time so that there will be a2 in the second row, a3 in the thirs row, etc (move the row until a matches tha place of the number of the row in the first matrix). There won't be any conflicts again. Now you can combine the two matrices and get a result.

Examples (N = 5 and N = 7):

12345   abcde     a1 b2 c3 d4 e5
45123   cdeab     c4 d5 e1 a2 b3
23451 + eabcd  =  e2 a3 b4 c5 d1
51234   bcdea     b5 c1 d2 e3 a4
34512   deabc     d3 e4 a4 b1 c2

1234567   abcdefg     a1 b2 c3 d4 e5 f6 g7
6712345   efgabcd     e6 f7 g1 a2 b3 c4 d5
4567123   bcdefga     b4 c5 d6 e7 f1 g2 a3
2345671 + fgabcde  =  f2 g3 a4 b5 c6 d7 e1
7123456   cdefgab     c7 d1 e2 f3 g4 a5 b6
5671234   gabcdef     g5 a6 b7 c1 d2 e3 f4
3456712   defgabc     d3 e4 f5 g6 a7 b1 c2

Therefore all NxN, N prime are not the solution, except N = 2.

All we need now is an insight on a how to build non-prime squares out of prime squares to maintain consistency and we
will prove that the only solution is N = 2.

[Edited by Maurog at Local Time:12-19-2004 at 01:24 PM]

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wackhead_uk
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Well, 2 is dodgy territory with rime numbers, like 1. It is technically a prime number, but only because it is so low, and that is why 1 isnt, because it is only divisible by 1 number, and a prime is divisible by two numbers - no more, no less. So, It might just be the pure strangeness of the number 2 which makes it possible. And I have only done up to binomial expansion in maths so far, so I don't think I can add much more, or is this was even helpful.

Abbyzzmal
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I admit it, I am obsessed with diagonals. They're so beautiful. Unfortunately, they won't return my calls.

Anyway, since this isn't really my field of expertise, I was wondering what algorithm you would use for even numbers, say the n=8 case. You cannot move it two to the each side (obviously), and three didn't seem to work either, which means five wouldn't work. I'm dropping out of this puzzle, but there is obviously some formula we're not quite hitting upon.

Mattcrampy
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There's some impressive maths going on here, I must say.

It may be easier to start finding ways to arrange the sets of games so that there is only class of game in each row and column, and work on the method of elimination.

Another hint: I am able to play all the games in my collection. You should be able to discern reasonable limits for how many games I could have based on the size of the games industry earlier in its life and how many and how prolific each genre is. (For instance, I couldn't have 20 first-person shooters, as first-person shooters didn't exist in 1984.)

Matt

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TripleM
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I know the answer to this, but I also know that the only way to prove it can't work is to exhaustively list combinations, and that there was a famous mathematician who came up with the wrong conjecture about other values and it took a long time for the actual answer to be found..

So if I were you, I wouldn't try proving it too much All someone needs to do is come up with a list of all the options for the right square size..

Maurog
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Hmm, he limited it at 20, and all the primes are out. Actually I doubt there are more than 10 distinct genres.
I proved that 4 isn't it.

That only leaves 6, 8, 9, and 10.

I'm tempted to write a really small program that just checks all the options.


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Mattcrampy
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I don't need you to prove it.

Matt

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DiMono
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Maybe I'm missing something, but doesn't the use of diagonals render everything above 2 possible, and thus not a valid solution? Send the years diagonally down to the left, and the genres diagonally down to the right, and there will be no overlap.

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