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2

Game Development / Artificial Intelligence / Re: Hidden Markov Models, how to train using BaumWelch?

on: 20151019 19:45:36

The entire problem is taking in a stream of emissions/observations and predicting the next emission. The underlying states are irrelevant. You mean it's irrelevant to your post because they've given you a structure? Out of interest, how many states are there and how many transitions with nonzero probability? If it's up to you to make the structure then this is a pretty critical step. BaumWelch seems to suck. If I repeatedly train my HMM with the entire sequence of emissions/observations, it eventually explodes into a HMM filled with 0s. The scarce information I've found says that this is due to overfitting. I don't think that should be possible. The transition probabilities out of each state have to sum to 1, though that is including the "stay" transition back in to the state itself. Overfitting can generate some states with only a single nonzero probability transition in and out but you'll either need to have tons of states or be training it on the same sequence again and again. What is the optimal way of training an HMM given a stream of data? BaumWelch is pretty good. I'd stick with that until you get it to work. Given the sequence of emissions [a, b, c], to calculate the probability of each possible next emission I compute the probability of the emission [a, b, c] occurring given the current HMM state and then the probabilities of [a, b, c, <X>] occurring, where X is each possible emission. The conditional probability of each emission is then p([a, b, c, <X>]) / p([a, b, c]) Is this the correct way of calculating the probability of the next emission? After seeing [a,b,c] you could be in any of your states. Call the set of states <N>. You first need to calculate the probability of being in each n in <N>: P([a,b,c,n]). This is the "forward" step that you'll find documented (Wikipedia is fine for this). In each of the states there will also be a probability of emitting each x in <X>: P(xn). That's straight from the current model parameters. For an emission x, P([a,b,c,x]) is the sum of all the ways x could be emitted, i.e. sum of P([a,b,c,n])*P(xn). If you follow through with the method above to just predict the next state I think you'll be doing Forward/Backward. If all you want is the single most likely state then it's much simpler to do Viterbi, which is just plain old dynamic programming applied to the HMM (and for me at least, is easier to understand).



4

Discussions / Miscellaneous Topics / Re: advanced 'Knapsack problem'

on: 20150422 12:46:33

This also sounds like something you could potentially solve with a theorem prover like Z3.
Yeah, this is an SMT problem http://en.wikipedia.org/wiki/Satisfiability_modulo_theories. What you are doing is called constraint programming. If you really want to roll your own, I'd do it this way:  first make a solver that finds a solution to the hard (boolean) constraints  generate a single valid solution  traverse the space of valid solutions, testing the overall value of the soft constraints as you go  after some time or after exhausting the search space, return the best result found so far From your description of the problem it shouldn't be too hard to define a function that takes a valid solution and generates similar, valid solutions for the traversal, so this kind of approach should be ok. If you can make the first step nondeterministic then keep repeating with greedy searches from new valid start points  this is much less prone to getting stuck in local optima than evolutionary algorithms and will be quicker too.



5

Game Development / Shared Code / Re: Curve fitting

on: 20150401 16:08:27

I find the easiest way to understand the DFT is as least squares: http://en.wikipedia.org/wiki/Linear_least_squares_%28mathematics%29. You're finding the best coefficients for a set of sinusoids so their sum is closest to your data. By using sufficient sinusoids of different frequencies these form a basis for all 1D signals, so you can get a perfect answer. The X in that page are all the sinusoids as a matrix (one per column). beta are the coefficients  they're complex so you get magnitude and phase. y is the data (your 1D signal).



8

Discussions / Miscellaneous Topics / Re: What I did today

on: 20150216 22:26:51

I finally have a solid initial implementation of a Goal planning system. It allows for entities to have goals and subgoals and to replan as circumstances change. A lot of GOAP material available is fiarly complicated from an implementation side, and it took a while for me to get to a simpler system that meets my needs.
I worked on a planning system too, but mine's overly complicated. Today I fixed some bugs exposed by the fallinlove and decidesexuality actions.



10

Game Development / Game Play & Game Design / Re: The impossible 15Puzzle

on: 20150212 21:52:02

My apologies, there was a bug after all. Using "0" to represent the gap makes the indexing a bit confusing and the even/odd state check was wrong. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
 public static boolean isSolvable(int[] state){ int[] positions = new int[16]; for(int i = 0; i < state.length; i++){ positions[state[i]] = (i+1)%16; } int row = (positions[0]1)/4; int col = (positions[0]1)row*4; boolean isEvenState = positions[0] == 0  row % 2 == col %2; int evenCount = 0; boolean[] visited = new boolean[16]; for(int i = 0; i < positions.length; i++){ if(visited[i]) continue; int cycleLength = 0; int nextTile = i; while(!visited[nextTile]){ cycleLength++; visited[nextTile] = true; nextTile = positions[nextTile]; } if(cycleLength % 2 == 0) evenCount++; } return isEvenState == (evenCount % 2 == 0); } 
If you have code that makes a whole lot of random swaps from the goal that would be a good way to test the correctness.



11

Discussions / Miscellaneous Topics / Re: What I did today

on: 20150212 14:02:18

For someone normal living its food and warm house and time for doing what hi love to do)
Aged 17 to 20? At that age, you might be at uni and all you need is enough money for a roof over your head, some food and beer. You get a little older and you want to travel. You just need enough money for a roof over your head, food, beer, a plane ticket per year and some hostel fees. You want to go out a little more with your girlfriend or boyfriend. Roof, food, beer, travel, and a bit of entertainment each week. You get a little older and want to have a house  no point paying rent forever. So you need to add a few thousand savings per year for a deposit. You get a little older and have a kid. Drop the beer and travel. Roof, food, savings, family entertainment and a good education for your child. If you want to be an indie game developer, chances are you'll have to go back through that last list and choose at least two things to throw out. That's "half a living".



12

Game Development / Game Play & Game Design / Re: The impossible 15Puzzle

on: 20150212 08:52:34

One thing I should have mentioned: I was assuming the blank space would be at the bottom right. Everything above is still correct, but the rule "even number of even cycles > solvable" holds when the blank space is at an even position (e.g. bottom right is position 16, in your other example the blank space is at position 2). When the blank space is in an odd position, the rule is "odd number of even cycles > solvable". If I find time today I'll put it in code.. Edit: here you go: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37
 public static boolean isSolvable(int[] state){ int[] positions = new int[16]; for(int i = 0; i < state.length; i++){ positions[state[i]] = i; } boolean isEvenState = positions[0] % 2 == 0; int evenCount = 0; boolean[] visited = new boolean[16]; for(int i = 0; i < positions.length; i++){ if(visited[i]) continue; int cycleLength = 0; int nextTile = i; while(!visited[nextTile]){ cycleLength++; visited[nextTile] = true; nextTile = positions[nextTile]; } if(cycleLength % 2 == 0) evenCount++; } return isEvenState == (evenCount % 2 == 0); } 
Examples: Is the goal state solvable? 1
 System.out.println(isSolvable(new int[]{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,0})); 
"true" Is one move from the goal state solvable (having moved tile 15)? 1
 System.out.println(isSolvable(new int[]{1,2,3,4,5,6,7,8,9,10,11,12,13,14,0,15})); 
"true" If you swap tiles 14 and 15 is it solvable? 1
 System.out.println(isSolvable(new int[]{1,2,3,4,5,6,7,8,9,10,11,12,13,15,14,0})); 
"false" Is your other example solvable? 1
 System.out.println(isSolvable(new int[]{1,0,3,4,6,2,11,10,5,8,7,9,14,12,15,13};)); 
"false" Those are the only tests I've done, but the code seems to be bug free.



13

Game Development / Game Play & Game Design / Re: The impossible 15Puzzle

on: 20150211 14:23:08

Here's how the counting goes:  The cycle starting at 1 is (1) because it is already in the correct location. Even cycle count is still: 0  The cycle starting at 2 is (2 6 5 9 12 14 13 *). Even cycle count is now: 1  The cycle starting at 3 is (3). Even cycle count is still: 1  The cycle starting at 4 is (4). Even cycle count is still: 1  The cycle starting at 7 is (7 11). Even cycle count is now: 2  The cycle starting at 8 is (8 10). Even cycle count is now: 3  The cycle starting at 15 is (15). Even cycle count is still: 3
There are no more tiles that aren't in a cycle so we are done. The number of even cycles is 3, which is odd. So the arrangement is not solvable.
To get the long cycle, you look at the first tile which is 2. It's position in the arrangement is the one that 6 should be in in the solution. So 6 is next in the cycle. Then we look at tile 6. It's position in the arrangement is the one that 5 should be in in the solution. etc. When we get to the blank spot *, it is in the position that 2 should be in in the solution, but we've already processed 2 so the cycle has finished.
Hope that helps.
Edit: Your last example has cycles (1)(2)(3)(4)(5)(6)(7)(8 )(9)(10)(11)(12)(13)(14 15)(*). There is one even cycle. That's an odd number, so not solvable either.



14

Game Development / Game Play & Game Design / Re: The impossible 15Puzzle

on: 20150210 20:40:50

The first link you posted with the permutations is probably the easiest test. In their first 15puzzle example they describe the board position as: 1
 (1 11 10 9 2 3 4 7 12 13 8 * 6 15)(5 14) 
The first set of parentheses says that tile 1 is in the position that would have 11 in the solution. 11 is in the position that 10 would have. 10 is in position 9 etc. The second set of parentheses says that tile 5 is in position 14 and tile 14 is in position 5. You can count the size of these cycles by picking one tile to start with and just following the trail of permutations. Then picking an unvisited tile and repeating until no more tiles are left unvisited. Count the number of cycles that have an even number of tiles in them. If this count is even, your arrangement is solvable.



16

Game Development / Newbie & Debugging Questions / Re: Moving player tiletotile (A* Pathfinding finally done)

on: 20140908 21:10:36

(Near)zerocost is not required, the effect is already observable if we have a parallel dirt road and paved road. Let's say the edges of the dirt road have cost 1.1 and the paved road has cost 0.9, all other edges (grass) have a cost of 2.0.
Hmm.. actually I think we're in disagreement on this bit. The heuristic distance I'd use is Manhattan*0.9, because the most optimistic estimate of the distance is that there is some direct path by paved road. A* using this will expand more nodes than if it used unweighted Manhattan distance but it will be optimal again. Regardless, at least the OP decided on a solution in the meantime



18

Game Development / Newbie & Debugging Questions / Re: Moving player tiletotile (A* Pathfinding finally done)

on: 20140908 14:41:00

Oh, and A* is not guaranteed to yield the fastest path, as it is heavily biased to a path that lies on the line src→dst, it might not even visit the 'instant portal' only a few cells away (in the opposite direction). Dijkstra would handle this odd case just fine, at the cost of worse performance in the general case.
If there are portals then instead of just using Euclidean distance from A to B for the heuristic, use (Euclidean distance from A to nearest portal) + (Euclidean distance from B to nearest portal) if that is shorter. Then A* won't be incorrectly biased and will still guarantee the shortest path.



21

Game Development / Newbie & Debugging Questions / Re: Stuck in recursion hell!

on: 20140828 07:36:36

I'd suggest you move to using a queue and perform a breadthfirst fill.
Tail recursion is just depthfirst traversal where you use the program stack instead of explicitly using a stack object. This makes the code more succinct and maps nicely to recurrence relations, but doesn't buy you much else.
For the breadthfirst fill, start by putting all sources in the queue then just dequeue and process one by one as you are now.
Edit: you can do fancier stuff with nontail recursion, but that doesn't really apply here



22

Game Development / Shared Code / Re: Triebased dictionary

on: 20140709 11:14:34

The idea was just to sort both dictionary and query so that you're working with sets of letters instead of series. Alphabetic order was just what first popped in to my head, but going in least or most popular order sounds like a good idea.
If keeping the words' strings in memory as well as trie nodes is too much then I guess this wouldn't work. I'm pretty sure it's guaranteed to be no slower than the unsorted version (with the worst case being when there are no anagrams sharing leaves) and could be a lot quicker when there are anagrams sharing nodes.



24

Game Development / Shared Code / Re: Triebased dictionary

on: 20140708 12:44:28

Did you consider sorting the characters of each word in the dictionary before building the trie? Then each node would need to additionally contain a list/set of the valid, unsorted, words. You sort the query characters and you would be searching for all subsets rather than all permutations. That's down to 2^9 queries. I might have misunderstood what you're doing. If so, ignore this.



25

Discussions / Community & Volunteer Projects / Re: game idea: "space wars" on Klein bottle surface?

on: 20140518 12:11:21

Actually it's pretty trivial to do this by just tweaking the way screen wrapping works. Right at the the top of the wikipedia page it shows how to write the mapping from leftside to rightside and top to bottom: http://en.wikipedia.org/wiki/Klein_bottle. It's more fun when you make top map to left and bottom to right though. I've used that before as a quick way to model a planet's surface on a diamond. It has some nice properties where the longitudinal lines at the equator are longer than those near the poles which is a bit more realistic than the cylinder maps.



27

Discussions / Java Gaming Wiki / Re: Smoothing Algorithm Question

on: 20130531 06:55:53

One last shot on the basic sliding rectangle. Instead of smoothing the whole volume just apply the function when a cell's value is requested. Because the rays will be continuous you can still use some of the simple optimisations and slide the rectangle down the ray.
This would require three copies of the data though: the original, the sparse smoothed version, and the result.
The efficiency would then depend on how many cells are typically hit by rays between smoothing operations.



28

Discussions / Java Gaming Wiki / Re: Smoothing Algorithm Question

on: 20130530 10:59:00

I think we both mean rectangular function though in 3 dimensions. As a filter in the frequency domain that's ... sinc (thanks wikipedia!). This is also the Haar scaling function, so the roughly opposite effect of the Haar wavelet that turns discontinuities into high value spikes.
I'm still only thinking smoothing, but since the filter is compact you can process the data in overlapping pieces and ignore the edge values. I'd guess this would be in "run overnight" territory even for very large data sets, much less for less than a billion cells.



29

Discussions / Java Gaming Wiki / Re: Smoothing Algorithm Question

on: 20130530 07:08:34

If you use a rectangular filter (i.e. all cells are equally weighted) then just calculating the differences as the block shifts should be the easiest to program and reasonably efficient. If you're doing a nonrectangular filter then philfrei's suggestion of performing it as a convolution is probably the right way. This requires a 3D Fourier transform, but a quick Google gives http://www.mathworks.com/matlabcentral/fileexchange/24504Which should run in Octave if you don't have Matlab.



30

Discussions / Java Gaming Wiki / Re: Smoothing Algorithm Question

on: 20130528 13:19:09

As the 100x50x50 block slides around you don't need to recalculate all the values. A simple optimisation is to only calculate the next 50x50 slice that it is shifting to then update the previous mean.
If you can afford to have a duplicate of the data for writing to then you can do better again. Just calculate the mean for whole block once for the very first position. Then replace the remaining cells with their contribution to change in mean (the difference between their value and one 50 or 100 places distant, depending on their axis). Then you should be able to walk the block around using only 50 (or 100) calculations per step.





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