For you math freaks out there, help!

[Johnmclane]

Ok, finished building my modular board and seem to forgotten some basic math.
Each square is 30x60 cm (1'x2') and different.
That means each square have four different positions (up,down,left,right). The whole board consist of twelve squares.
How many combinations will that possibly make?

Thanx in advance

[punisher_1983]

Lots .....

[Bunnahabhain]

That is a really hard calculation, as there are huge number of possible lay outs .

for instance, you can lay them out... ( Ascii diagrams ahoy...)

=l=l
l=l=
or
===
===
or
llllll
llllll
or
===
llllll
or
---
llllll
---
etc, etc
let alone the many, many ones I can't do in ASCII.

then we have to do the permutations of tiles within the layouts, which is 12!^4, I think, or 5.26 x 10 ^34, *

So we have several hundred tile lay outs ( estimate), x 5.26 x10^34, so, we're on 10^37 to 10^38 possible layouts, at an educated guess.

* for the non maths people. 10^10 means 1 followed by 10 zeros. so 10^34 is 1 followed by 34 zeros.

[Johnmclane]

Oh.... Forget the thing i said about basic math. Been wrecking my brain with this and ... Yep, better let this question die unless someone really want to solve it.
Thanks for the work you put in. (guess you could do something where you just use halv a tile so the don't really allign but overlap to make it even more complicated... And can't draw that i ascii either .

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[Ace Rimmer]

Well, assuming that rotating each tile gives a unique alteration each time, so each tile has 4 unique layouts based on facing, I'd say around 5.264387585885x10^34? Assuming they aren't double sided :P

[Isambard]

Quick reply: they are not squares.

[sycopat]

I believe the answer should be given by this: (12 options in 12 slots, each with 4 possible facings)
(12x4)x(11x4)x(10x4)x(9x4)x(8x4)x(7x4)x(6x4)x(5x4) x(4x4)x(3x4)x(2x4)x(1x4)

which is a very large number.

Shall we just leave it at 'enough'?

Ah it's actually not this because they are not squares of course, Isambard is right on that

[BooTMGSG]

Lets go with Troll counting, Lots!
really beyond the first 20 arangement, who cares?

[The Marshel]

the answer is quite simple really: Far more then you realistically need.

honestly, if you can get more then 10 different layouts, any more wont have any effect on if and/or when you get board of the board
Personally I'd be fine with 5 alt layouts myself

[Isambard]

OK, there are three combined problems here.

Firstly, there are multiple ways to lay out the 12 board sections, as Bunns above mentioned. Secondly, you then have the problem of what order the tiles are then placed in. Finally, each order can be changed by rotating through 180 degrees one or more tiles.
.
There are lots of different ways the board pieces can be laid out in, as Bunns stated. Of course, if the sections are not symmetrical, you could simply rotate one piece round to make a new board (eg !! turns into i! or !i); this would give 12 more designs just for that layout of pieces. Or you could rotate 2 pieces, giving 66 permutations. Or you could rotate any 3 bits, giving 220 combinations. In total, from rotating 0 pieces all 12 pieces you can make 4096 variations for each of the different board layouts.

If all 12 pieces are different the the order in which they are put down matters; you have 12 options for your first piece, 11 for your second, 10 for your third and so on, giving 12*11*10*9*8*7*6*5*4*3*2*1 = 12! = 479 001 600 different orders of board pieces. That means for each different layout option there are 1 961 990 553 600; nearly 2 trillion possible options PER board set up.

I will have to put some more thought into this when it is not 1 am though...

[Sir_Turalyon]

then we have to do the permutations of tiles within the layouts, which is 12!^4, I think, or 5.26 x 10 ^34, *

<mad mathemathican voice>

12!^2, as each tile can fill its "slot" in only two ways (other two sides are wrong length).

As of number of tilings, assuming we want a 120x180 cm tables... we have a 6x4 chessbboard like one below, and want to count its coverings with domino pieces (or perfect matchings of chessboards x-o bipartite graph).
oxoxox
xoxoxo
oxoxox
xoxoxo

Let's ask google, it shoudn't be too difficult, probably somebody did it...
http://people.math.gatech.edu/~ciucu/symmgraphs.pdf
... ok, so it is difficult in general case. This one states it's #P-complete in general, but gives an algorithm for planar graphs, which should suffice....
http://www.cs.bris.ac.uk/~montanar/p.../matchings.pdf

So one may run it on computer to check... Wait, page 22 of the last one states an 4x4 board (using 8 tiles) has 32 tilings, 8x8 would have 12.988.916 and 6x4 is not listed but gives an approximation... about 1.339^24. That's 1,792921^12, 3,214565712241^6, 10,33343271831549^3... about 1100 tilings.

This one has the algorithm and a formula on page 28 (4.16)... not going to apply it now, too convoluted, someone with Maple licence should do it... and it says this is an old problem, only in physics instead of wargaming and with arranging molecules rather than table tiles :

http://www.science.uva.nl/onderwijs/...f887198315.pdf

Nice area of discret math you pointed me at. I wonder if there is anything to be done there, most papers are from 1990s...

But without going for precision, it's about 1100 * 12!^2 combinations. Or 1100*479.001.600*479.001.600, windows calculator refuses to go further. Slightly less than 10^3 * 10^8 *10^8 *25, 25 * 10^19, or 250 000 000 000 000 000 000 combinations.

Fun fact: if we believe wikipedia (we don't), age of universe is less than 5 * 10^17 seconds:

http://en.wikipedia.org/wiki/Age_of_the_universe

So if you started at Big Bang and tried new combination each second, you'd already have tried out 1/500 th of them all by now, probably less.
</mad mathemathican voice>

Congratulations on your table. <checks the clock> You did it to me on purpose, didn't you?

[de Selby]

I'm not going to try to work this out, but if the table is finished I want to see it!

In one or more of its various permutations...

[Johnmclane]

Haha I love you guys !

Good thing for me I actually made some 6 extra tiles as well to keep the board fresh and interesting .

(edit): IŽll take pictures as soon as I get it to my gaming place (which is at my work, benefits of owning your own business).

These are the different ones btw:




high tile: Church //// low tile: graveyard
high tile: Defence laser ///// low tile: bombed out fortifications
high tile: Manefactorum //// low tile: chemical waste
high tile: Administratorum //// low tile: plaza of heroes
high tile: House of commerce //// low tile: plaza of death
high tile: House of knowledge //// low tile: Hill
high tile: Defence missile battery //// low tile: 30x2 flat ground
high tile: PDF Comm centre //// low tile: 30x2 flat ground

[Bunnahabhain]

<mad mathemathican voice>
12!^2, as each tile can fill its "slot" in only two ways (other two sides are wrong length).

Doh! If I was awake last night to forget that, I wonder what else I did....

[Aliarzathanil]

Without seeing the tiles this is not solvable. If a tile is completely flat, rotating it doesn't actually produce a unique layout; same goes for a tile that has at least two identical sides. Either way, if you get into to double digits it's more than adequate.

[Beppo1234]

why am I thinking 24! ? Each piece has 2 options of alignment, so their are 24 initial placements for a first tile down... then all options stem from there.

[Baaltor]

Ok, finished building my modular board and seem to forgotten some basic math.
Each square is 30x60 cm (1'x2') and different.
That means each square have four different positions (up,down,left,right). The whole board consist of twelve squares.
How many combinations will that possibly make?

Thanx in advance

This just gave me an absurd idea: you know those puzzles where one piece is missing and you move the pieces to form a picture? What if you played a game where each turn the player got to move a piece of terrain in a similar way? Or maybe they got like say 4 moves or something.

[Bunnahabhain]

This just gave me an absurd idea: you know those puzzles where one piece is missing and you move the pieces to form a picture? What if you played a game where each turn the player got to move a piece of terrain in a similar way? Or maybe they got like say 4 moves or something.

Only works so long as there is a gap, therefore the board needs a gap--- a vortex of doom! That sounds a great idea for a fun game.

[RandomThoughts]

First of all, we'll definitely end up with twelve pieces fitting into twelve slots on the table, independent of the layout of said slots. That alone amounts to 12*11*10*9*8*7*6*5*4*3*2*1 which is 479001600 or half a billion.

Next, no matter how you arrange the slots, each piece will always fit into its slot forward or backward, so we add another 2^12, which is 4096.

Combining those two, we get 479001600*4096 different combination of tiles for each single layout/arrangement of spots on the table, which is 1961990553600 or ~2000 billions.

We'd now have to multiply that number with the number of different tile arrangements possible (which I haven't figured out yet) and then divide the result by two (since we'll have each possible combination twice, once facing towards the door and once facing away from the door).

I think I got the number of possible tile arrangements now, which is 280. The process by which I got that number was a bit complicated. I started by dividing the table into 6 square areas of 2'x2'. I then looked at several cases separately: How many different tile arrangements are there, if there is no overlap between these six areas; how many different tile arrangements are there, if individual tiles overlap between two of these squares; how many tile arrangements are there if individual tiles overlap between three of these tiles, etc. In the end I added all of those together, which resulted in the number 280. I'm feeling petty sure about those results, although it would be a hell lot of work to put them into a computer and post them for independent confirmation.

Anyway, our total number of potential combinations is now 280 tile arrangements x 479001600 piece-to-slot assignment x 4096 ways the individual tiles can face in their assigned spots x 1/2 because we got each possible combination twice now, once facing towards the window and once facing away from the window.

That makes a total possible combinations of: 1125083863056380000 or 1.1 quintillion (US & UK; in Europe's counting system it's 1.1 trillion).

Assuming you play two games on two different tables each day, you'll run out of new combinations in 1.5 Quadrillion years.

Of course, if all those tiles are simple dessert area with a single rock in the center, each battlefield will still feel exactly like the same...

[RandomThoughts]

Oh, I just found out about the six extra tiles. I think that will increase the number significantly, without making it that much harder to calculate. Basically, all we have to do is replace the part of my formula which says 12*11*10*9*8*7*6*5*4*3*2*1 with 18*17*16*15*14*13*12*11*10*9*8*7、which leads to a total of:

5099134969184260000 or 5 quintillion different combinations.

But like I said, if they are all open dessert, it won't really matter...