EventHorizon

Exactly. Economics saw discrepancies, so they added more to game theory.

Not exactly. Game theory has a lot to do with "what you predict other, potentially irrational, people will do." Game theory is closely related to economics, psychology, and AI. In fact, most of what is now game theory was developed by economists. You can be risk-averse, risk-neutral, or risk-seeking. You can look at worst case payoffs and maximize that over your actions (a maximin strategy). You can play equilibria. etc.

This game is a multi-agent form of the matrix game "stag hunt." Essentially, all players can work together and get the stag, or just hunt rabbits by themselves (and causing the ones hunting the deer to fail to get it). There are two Nash equilibria. They are when all people take the $1,000, and when all people take the $50,000. Unfortunately, the 50000 is not a stable equilibria (if we know one person will choose the 1000, then the optimal strategy for all players is to do the same). So.....how would you respond if the payoffs were such that if you choose 50000 and someone chose 1000, that you needed to pay the 1000 (or at least the people who chose 50000 needed to split the cost of giving 1000 to all that chose it)? What if the people that chose 50000 and didn't get it needed to pay 50000 instead? What if the people that chose 50000 and didn't get it is executed? I think that even seeing the other players face to face may change outcomes. If everyone was in their own rooms and didn't see each other, I think 1000's would be taken more often then if all were in the same room. If you want to see more games like this, look up "list_of_games_in_game_theory" on wikipedia (some possibilities are chicken, the prisoner's dilemma, battle of the sexes, ultimatum game, volunteer's dilemma, matching pennies). At the bottom of the page are also links to strategies.

I just happened to check wikipedia and was able to verify some numbers (2d, 3d, and 3d_flip) on it. (search for polycube or polyomino) They also say that the problem of finding a function for the number of n-ominos is unsolved. So all that means is that when we solve it we'll be rich and famous!

Alright. I've made my code let you choose to find duplicates by only rotation or both rotation and mirroring. I've included a number of text files. The ones named dX_nY_flip.txt are the ones that removed duplicates found by rotations and mirroring. xominos_both.zip Here's a table of total arrangements (by dimensions and number of squares/cubes/etc)... num 2d 3d 4d 5d 2d_flip 3d_flip 4d_flip 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 3 2 2 2 2 2 2 2 4 7 8 7 7 5 7 7 5 18 29 27 26 12 23 26 6 60 166 164 154 35 112 147 7 196 1023 1316 1172 108 607 1019 8 704 6922 369 3811 9 2500 1285 10 9189 4655[/codebox] One part in the table troubles me. Look at going from 3 to 4 dimensions with 4,5,or 6 cubes. The number actually goes down. Out of curiosity, I just added 5d, and the numbers decrease going from 4d to 5d with 5,6,and 7 hypercubes. I can see how this can happen to specific objects (extra unused dimension allows a rotation that can essentially flip it), but totals would intuitively go up. But my intuition has steered me wrong before... (It's beginning to make sense....since the number of cubes is so close to the number of dimensions...) Ideas? Intuition? An all encompassing formula?

Depends on whether you count mirrored objects that cannot be achieved by rotation as unique or not. My old code didn't. And it turns out that it is only one extra "if" statement to make it only do rotations. I'll post some of the text files now, and will post code / exe tomorrow. Here's some data, though I will give a bigger list tomorrow (better computer in the lab...and it's late). 2d -> 1,1,2,7,18,60,196,704,2500 3d -> 1,1,2,8,29,166,1023,6922 Oh, and kswack you show 2 identical ones in your 5 block 2d. The two you are missing are numbers 7 and 15 in my d2_n5.txt file. xxx x x and xx xx x Good job with the 6 block 2d ones.

My code assumes mirrored (flipped over an axis) objects aren't counted for each mirroring. This is the way I worked with 2d ones back in 6th grade, and that's what I assumed the original poster meant. If this is not the case, then I'll need to figure out how to find all possible orthogonal rotations of n-dimensional objects.....yay.... :-/

are you sure there is more than 23? That would mean there is a bug in my code...so let me know if you are certain. I'm almost positive that the 2d numbers are right up to 7 squares. Here's what I got for totals number 2d 3d 4d 2 1 1 1 3 2 2 2 4 5 7 7 5 12 23 26 6 35 112 147 7 108 607 1019 8 369 3811 9 1285 10 4655[/codebox] Here's the code, executable, and text files it printed out showing all the unique configurations (so you don't have to run it unless you want to). If you have any questions about what the code is doing, let me know. The text files give a unique number to each (1 to the first, etc), show the dimensions of the object (length,width,depth,etc), then show 2d slices (third dimension is to the right separated by |, and 4+ d's are down separated by blank spaces).

I know it's lame, but....

Working with the 2d analog of the problem, I couldn't find a simple solution. You'd think that the equation for one would be similar to the one for the other (if only because each of the 2d arrangements are included in the 3d ones). Since this is a problem that I had worked with ages ago and never solved, I may write some code to generate all of them (both 2d and 3d (and 4+ d?)). I think I could write it without too much difficulty, and having the correct numbers would definitely help in determining possible equations. (I did a similar thing with the towers of hanoi problem with more than 3 pegs.....yes it is easier than with 3, but the question is....how much easier? The answer turned out to be fairly interesting....search for "hanoi" if you're interested)

the 2-d analog to this problem is interesting as well. I remember working on it way back in 6th grade. Tetrominos (like tetris pieces), pentominos (5 squares), hexominos, etc. I don't think I came up with a formula for it either. Though I cut them out of paper (up to heptominos (septominos?)) Perhaps working on the 2d version would be better first. Or is it a solved problem?

do all pirates vote simultaneously?....or do they vote by seniority?

Very nice...though it seems you didn't come up with it yourself. Nice googleing though

That shows that you cannot get 2,3 or 5 squares by using hmmm...'s algorithm and method of breaking up the square, but you need to show it is not possible regardless of the algorithm or how you try and break up the square. Keep trying! I'm sure someone will stumble on it. I'd rather not post my solution if someone can else can solve it.

http://brainden.com/forum/index.php?showto...8&hl=pirate repeated puzzle...

Good job. Even simpler than my solution. Any ideas on the other part?

Yeah....it would definitely take an uploaded picture to show it. A sequence of moves would simply be too long and hard to tell if it is correct.

Show that a knight can move on a chessboard to every space exactly once. (you do not need to end on the space you begin on....but bonus points if you can!)

Again, this is taken from another website...but can't post the link (because it would be removed). Prove that the only numbers of non-overlapping squares that you can not break a unit square into are 2, 3, and 5. (Obviously, there can be no left over space...the areas of the smaller squares need to add up to 1 as well.)

what if you want to use your ladder afterwards?

Of course....he could stay 15 nights with 15.3 cm of gold... Yay for binary! This reminds me of another puzzle....basically the same kinda thing, though I cannot remember the situation/wording exactly...so I'll copy yours somewhat. You have a gold chain with 7 links. You don't have enough rent, but your landlord agrees to allow you to give him a gold link every day until you have the rent to get them back (he must be holding 2 links on day 2, 5 on day 5 etc). You don't want to make too many cuts, since you will be getting them back. How many cuts do you need to make if you are going get the money for rent in a week? Follow-up questions: 1. If you have a chain of length 23 and will only make 2 cuts, how long could you delay rent? 2. If you have a chain of length 365 and want to delay rent a whole year....how many cuts are needed? 3. If you have a chain of length 4000 and want to delay rent for a decade....how many cuts are needed? 4. If you have a chain of length F(x) and want to delay rent for F(x) days.....only x cuts are needed, but x+1 cuts are needed if you had one link more and one day longer to delay. What is F(x)?

no communication...it would be way too easy otherwise