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Math(s) and the Mosaic Frame: Competition!

It’s Monday - and we thought we'd help jump-start your brain with a brainteaser. Back at the beginning of October, we introduced the Mosaic Frame, which made it pretty easy to transform your MiniCards in to an art piece. Based on our Flickr pool, it looks like people have already started making some early Christmas gifts and are experimenting with the teaser, without even knowing it.

Jon Walters used the Frame for his wedding photos:

While Yaelfran has used it for her artwork:

Now, let's get back to the Brain Teaser. It all started when Paul (our product designer) first showed me the design, my brain started to hurt. Why? Well, I kept trying to calculate how many different ways I could put together a set of cards in a frame. I must admit my University Math class was ages ago. So, can you help me out?

The Challenge:
How many different ways can you arrange the cards in the frame given the following:
1. you have 20 different MiniCards
2. each MiniCard looks equally good vertically or horizontally
3. each MiniCard must be included in the frame arrangement only one time

Remember, the frame can also hang vertically or horizontally (perhaps that adds another 2 variations?) Here's a little video we made to show you how it works. It should give you an idea of how the little frames fit in to the bigger frame...

So, if you think you know the answer, leave your answer and calculations in our comments. Based on collective understanding here at MOO we will pick the winner from the selection of answers we receive by end of week. The winner will receive a Mosaic Frame for free. (Your answer doesn't have to be right but needs to be well thought out - remember 'Part Marks' ). We'll announce the winner next Monday.
Good Luck!

PS If you want tips on how to put together the frame Paul has provided his expert advice!


( 11 comments — Leave a comment )
17th Nov, 2008 12:04 (UTC)
oh, that's easy.



Ok, you want to see some working? Here we go.

Ignoring the various ways of actually arranging the frame itself for now, working on just *one* layout, you've got a metric ton of ways of arranging 20 cards:

Take the first 'cell' - you've got 20 options for that one. Then 19 for the second cell, 18 for the third and so on, right down to the very last cell, which you have to put the last card in.

All of which comes to 20 factorial possible arrangements, or 20*19*18*17*....*1 = 2.43290201 × 1018

or 128,047,474,114,560,000 possible arrangements of 20 cards in just *one* frame layout.

At one arrangement per second (boy, you can lay out cards *fast*!), it would take you 4,060,358,762.2 *years* to arrange those 20 cards in all their permutations.

Or about the age of the earth, give or take the odd half billion years.

For *one* frame layout. At which point I'm guessing you don't *really* need to know about the other frame layouts.

So, I'm going with *lots*

17th Nov, 2008 15:21 (UTC)
I like 'lots' as a technical term :)
(And damn! You people are quick!)

Edited at 2008-11-17 15:21 (UTC)
17th Nov, 2008 15:26 (UTC)
faced with the choice of working this out, or doing some actual 'work', it was no contest, really.

17th Nov, 2008 16:27 (UTC)
But does this solution account for the fact that you can put the second, third... nth card in different positions? I'd thought of it this way, but then realized that if I did it like that, I believe it would only account for the position if it were upright. I could be wrong, though.

Love to hear your thoughts :)
17th Nov, 2008 16:32 (UTC)
Hmm. It counts for putting the cards in a certain order, into a single set layout for the cells.

So, the first card can go in 20 different locations in the grid, the 2nd card has 19 free locations and so on.

If you start tinkering with the orientation of the cards (ie. putting cards 'up' or 'down') then the number gets even bigger - the first card has 20 different locations and each of those has 2 different orientations. Second card has 19 positions each with two orientations.

Can I therefore refine my 'lots' to 'really, really lots'?

17th Nov, 2008 13:46 (UTC)
I tried, but it's way beyond me.
Let's see, you have 20 (1 unit X 2 unit) unique rectangles to fit into a 5 unit by 8 unit 'box'.

Assuming the box to be configured 8 across and 5 down:
Each row across has 7 possible positions for the first rectangle (if you could only pack in one direction there would be only 4 possibilities, but gaps may be filled by a rectangle going the other direction, so the possible positions overlap.)

So 7 X 5= 35 across possibilities for the first rectangle.

Each column down has 4 possibilities for the first rectangle.
So 4 X 8= 32 down possibilities for the first rectangle.

So the first rectangle has 67 possible positions.
You'd think the second rectangle would have 66 possible positions (every one except the one being used by the first.) and so you could solve the problem by giving each succeeding rectangle one less placement option. But no, because each corner position can only be in one of two rectangles, each non-corner edge position can be only in one of three rectangles, and each center position can only be in one of 4 rectangles, so depending on where you place the first one, the number of options remaining varies. (And then there is the fact that you need to eliminate *impossible* layouts where rectangles isolate a space too small to place a rectangle.)

This problem is probably solvable by a mathematician and a very good computer programmer. I am neither. So I'll see what I can do empirically.

Assume the first rectangle goes horizontally across the first possible position in the top row. That leaves the second rectangle only 5 possibilities (instead of 6) across in the first row, and only 3 possibilities down for the first column, leaving it with 64 total possibilities.

At this point, my brain went *tilt*. I think the answer to how many Moo cards can dance on the Moosaic is As Many As Your Imagination Can Conceive.

Edited at 2008-11-17 13:53 (UTC)
17th Nov, 2008 16:05 (UTC)
As Many As Your Imagination Can Conceive.
17th Nov, 2008 22:09 (UTC)
I have calculated 2.83189794 × 10^21 so far but I am not done calculating yet. If I didn't have that Biology exam I would finish it now it I will have to figuring it out later and then write a nice, hopefully easy to read report, with pictures :)

Edited at 2008-11-17 22:16 (UTC)
17th Nov, 2008 22:43 (UTC)
new calculated least amount: [ (30 x 30 x 3)+(24 x 24) ] x 20!
Which google says is 7.97018698 × 10^21 (which is LOTS, but not "a lot" see http://sodiumeyes.com/2008/03/24/a-lot-to-consider/ for math logical goodness?)

Numbers I got will be explained later when I have time to make visuals to expalin my logic.

But there is still more. And I am working on it. Makes me wish I had some mad math skillz.

Edited at 2008-11-17 22:44 (UTC)
19th Nov, 2008 11:04 (UTC)
Ok - just make sure you're doing all the stuff you *should* be doing. You're making us worry about your biology exam now! Hope it went well.
24th Nov, 2008 19:18 (UTC)
just curious - will you be picking a winnner from here *and* the moo blog on moo.com, or just one between the two?

( 11 comments — Leave a comment )

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