I built conway's game of life and its working fine but my game after many generations is either ending with no lives or reaching a stable pattern which it can't escape.
For example I have followed these rules
Any live cell with fewer than two live neighbours dies, as if by underpopulation.
(live_cell = True and neighourhood < 2)
Any live cell with two or three live neighbours lives on to the next generation.
(live_cell = True and neighourhood == 2 or neighourhood == 3)
Any live cell with more than three live neighbours dies, as if by overpopulation.
(live_cell = True and neighourhood > 3)
Any dead cell with exactly three live neighbours becomes a live cell, as if by reproduction.
(live_cell = False and neighourhood == 3)
This is my game of life matrix where 1 is life and 0 not
000000
001000
010100
001000
000000
000000
and this is its corresponding neighbourhood maps created by my programe
011100
122210
124210
122210
011100
000000
After reaching this pattern even after thousands of generation its still stuck in this pattern itself. I dont know how to escape this pattern ?
If the space is finite, then the number of possible configurations is finite and then the GoL will end either in a stable pattern or in a loop. If the space is very small (as it looks like) then you will observe only stupid behavior. You need at least to use a much bigger space (500x500), fill it with 1's at many places and look; that is the basic play with GoL. The next step is to build interesting configurations, and there exists a lot that have been discovered over time, for examples see GoL Pattern Library. Basic well-known patterns are gliders, glider-guns, oscillators... You will discover that GoL is in fact a very interesting way of programming: the initial configuration is a program code executed by the machine that you can see evolving on your screen. But that programming is not so easy, especially if you want to obtain a specific behavior!
I've been given as an assignment to write using prolog a solver for
the battleships solitaire puzzle. To those unfamiliar, the puzzle deals
with a 6 by 6 grid on which a series of ships are placed according to the provided
constraints on each row and column, i.e. the first row must contain 3 squares with ships, the second row must contain 1 square with a ship, the third row must contain 0 squares etc for the other rows and columns.
Each puzzle comes with it's own set of constraints and revealed squares, typically two. An example can be seen here:
battleships
So, here's what I've done:
step([ShipCount,Rows,Cols,Tiles],[ShipCount2,Rows2,Cols2,Tiles2]):-
ShipCount2 is ShipCount+1,
nth1(X,Cols,X1),
X1\==0,
nth1(Y,Rows,Y1),
Y1\==0,
not(member([X,Y,_],Tiles)),
pairs(Tiles,TilesXY),
notdiaglist(X,Y,TilesXY),
member(T,[1,2,3,4,5,6]),
append([X,Y],[T],Tile),
append([Tile],Tiles,Tiles2),
dec_elem1(X,Cols,Cols2),dec_elem1(Y,Rows,Rows2).
dec_elem1(1,[A|Tail],[B|Tail]):- B is A-1.
dec_elem1(Count,[A|Tail],[A|Tail2]):- Count1 is Count-1,dec_elem1(Count1,Tail,Tail2).
neib(X1,Y1,X2,Y2) :- X2 is X1,(Y2 is Y1 -1;Y2 is Y1+1; Y2 is Y1).
neib(X1,Y1,X2,Y2) :- X2 is X1-1,(Y2 is Y1 -1;Y2 is Y1+1; Y2 is Y1).
neib(X1,Y1,X2,Y2) :- X2 is X1+1,(Y2 is Y1 -1;Y2 is Y1+1; Y2 is Y1).
notdiag(X1,Y1,X2,Y2) :- not(neib(X1,Y1,X2,Y2)).
notdiag(X1,Y1,X2,Y2) :- neib(X1,Y1,X2,Y2),((X1 == X2,t(Y1,Y2));(Y1 == Y2,t(X1,X2))).
notdiaglist(X1,Y1,[]).
notdiaglist(X1,Y1,[[X2,Y2]|Tail]):-notdiag(X1,Y1,X2,Y2),notdiaglist(X1,Y1,Tail).
t(X1,X2):- X is abs(X1-X2), X==1.
pairs([],[]).
pairs([[X,Y,Z]|Tail],[[X,Y]|Tail2]):-pairs(Tail,Tail2).
I represent a state with a list: [Count,Rows,Columns,Tiles]. The last state must be
[10,[0,0,0,0,0,0],[0,0,0,0,0,0], somelist]. A puzzle starts from an initial state, for example
initial([1, [1,3,1,1,1,2] , [0,2,2,0,0,5] , [[4,4,1],[2,1,0]]]).
I try to find a solution in the following manner:
run:-initial(S),step(S,S1),step(S1,S2),....,step(S8,F).
Now, here's the difficulty: if i restrict myself to one type of ship parts by using member(T,[1])
instead of
member(T,[1,2,3,4,5,6])
it works fine. However, when I use the full range of possible values for T which are needed
later, the query never ends since it runs for too long. this puzzles me, since :
(a) it works for 6 types of ships but only for 8 steps instead of 9
(b) going from a single type of ship to 6 types increases the number
of options for just the last step by a factor of 6, which
shouldn't have such a dramatic effect.
So, what's going on?
To answer your question directly, what's going on is that Prolog is trying to sift through an enormous space of possibilities.
You're correct that altering that line increases the search space of the last call by a factor of six, note that the size of the search space of, say, nine calls, isn't proportional to 9 times the size of one call. Prolog will backtrack on failure, so it's proportional (bounded above, actually) to the size of the possible results of one call raised to the ninth power.
That means we can expect the size of the space Prolog needs to search to grow by at most a factor of 6^9 = 10077696 when we allow T to take on 6 times as many values.
Of course, it doesn't help that (as far as I was able to tell) a solution doesn't exist if we call step 9 times starting with initial anyways. Since that last call is going to fail, Prolog will keep trying until it's exhausted all possibilities (of which there are a great many) before it finally gives up.
As far as a solution goes, I'm not sure I know enough about the problem. If the value if T is the kind of ship that fits in the grid (e.g. single square, half of a 2-square-ship, part of a 3-square-ship) you should note that that gives you a lot more information than the numbers on the rows/columns.
Right now, in pseudocode, your step looks like this:
Find a (X,Y) pair that has non-zero markings on its row/column
Check that there isn't already a ship there
Check that it isn't diagonal to a ship
Pick a kind of ship-part for it to be.
I'd suggest you approach like this:
Finish any already placed ship-bits to form complete ships (if we can't: fail)
Until we're finished:
Find acceptable places to place ship
Check that the markings on the row/column aren't zero
Try to place an entire ship here. (instead of a single part)
By using the most specific information that we have first (in this case, the previously placed parts), we can reduce the amount of work Prolog has to do and make things return reasonably fast.
I tried coming up with a compression algorithm. I do little bit about compression theories and so am aware that this scheme that I have come up with could very well never achieve compression at all.
Currently it works only for a string with no consecutive repeating letters/digits/symbols. Once properly established I hope to extrapolate it to binary data etc. But first the algorithm:
Assuming there are only 4 letters: a,b,c,d; we create a matrix/array corresponding to the letters. Whenever a letter is encountered, the corresponding index is incremented so that the index of the last letter encountered is always largest. We incremement an index by 2 if it was originally zero. If it was not originally zero then we increment it by 2+(the second largest element in the matrix). An example to clarify:
Array = [a,b,c,d]
Initial state = [0,0,0,0]
Letter = a
New state = [2,0,0,0]
Letter = b
New state = [2,4,0,0]
.
.c
.d
.
New state = [2,4,6,8]
Letter = a
New state = [12,4,6,8]
//Explanation for the above state: 12 because Largest - Second Largest - 2 = Old value
Letter = d
New state = [12,4,6,22]
and so on...
Decompression is just this logic in reverse.
A rudimentary implementation of compression (in python):
(This function is very rudimentary so not the best kind of code...I know. I can optimize it once I get the core algorithm correct.)
def compress(text):
matrix = [0]*95 #we are concerned with 95 printable chars for now
for i in text:
temp = copy.deepcopy(matrix)
temp.sort()
largest = temp[-1]
if matrix[ord(i)-32] == 0:
matrix[ord(i)-32] = largest+2
else:
matrix[ord(i)-32] = largest+matrix[ord(i)-32]+2
return matrix
The returned matrix is then used for decompression. Now comes the tricky part:
I can't really call this compression at all because each number in the matrix generated from the function are of the order of 10**200 for a string of length 50000. So storing the matrix actually takes more space than storing the original string. I know...totally useless. But I had hoped prior to doing all this that I can use the mathematical properties of a matrix to effectively represent it in some kind of mathematical shorthand. I have tried many possibilities and failed. Some things that I tried:
Rank of the matrix. Failed because not unique.
Denote using the mod function. Failed because either the quotient or the remainder
Store each integer as a generator using pickle.
Store the matrix as a bitmap file but then the integers are too large to be able to store as color codes.
Let me iterate again that the algorithm could be optimized. e.g. instead of adding 2 we could add 1 and proceed. But don't really result in any compression. Same for the code. Minor optimizations later...first I want to improve the main algorithm.
Furthermore, it is very likely that this product of a mediocre and idle mind like myself could never be able to achieve compression after all. In which case, I would then like your help and ideas on what this could probably be useful in.
TL;DR: Check coded parts which depict a compression algorithm. The compressed result is longer than the original string. Can this be fixed? If yes, how?
PS: I have the entire code on my PC. Will create a repo on github and upload in some time.
Compression is essentially a predictive process. Look for patterns in the input and use them to encode the more likely next character(s) more efficiently than the less likely. I can't see anything in your algorithm that tries to build a predictive model.
Recently I wrote a Ruby program to determine solutions to a "Scramble Squares" tile puzzle:
I used TDD to implement most of it, leading to tests that looked like this:
it "has top, bottom, left, right" do
c = Cards.new
card = c.cards[0]
card.top.should == :CT
card.bottom.should == :WB
card.left.should == :MT
card.right.should == :BT
end
This worked well for the lower-level "helper" methods: identifying the "sides" of a tile, determining if a tile can be validly placed in the grid, etc.
But I ran into a problem when coding the actual algorithm to solve the puzzle. Since I didn't know valid possible solutions to the problem, I didn't know how to write a test first.
I ended up writing a pretty ugly, untested, algorithm to solve it:
def play_game
working_states = []
after_1 = step_1
i = 0
after_1.each do |state_1|
step_2(state_1).each do |state_2|
step_3(state_2).each do |state_3|
step_4(state_3).each do |state_4|
step_5(state_4).each do |state_5|
step_6(state_5).each do |state_6|
step_7(state_6).each do |state_7|
step_8(state_7).each do |state_8|
step_9(state_8).each do |state_9|
working_states << state_9[0]
end
end
end
end
end
end
end
end
end
So my question is: how do you use TDD to write a method when you don't already know the valid outputs?
If you're interested, the code's on GitHub:
Tests: https://github.com/mattdsteele/scramblesquares-solver/blob/master/golf-creator-spec.rb
Production code: https://github.com/mattdsteele/scramblesquares-solver/blob/master/game.rb
This isn't a direct answer, but this reminds me of the comparison between the Sudoku solvers written by Peter Norvig and Ron Jeffries. Ron Jeffries' approach used classic TDD, but he never really got a good solution. Norvig, on the other hand, was able to solve it very elegantly without TDD.
The fundamental question is: can an algorithm emerge using TDD?
From the puzzle website:
The object of the Scramble Squares®
puzzle game is to arrange the nine
colorfully illustrated square pieces
into a 12" x 12" square so that the
realistic graphics on the pieces'
edges match perfectly to form a
completed design in every direction.
So one of the first things I would look for is a test of whether two tiles, in a particular arrangement, match one another. This is with regard to your question of validity. Without that method working correctly, you can't evaluate whether the puzzle has been solved. That seems like a nice starting point, a nice bite-sized piece toward the full solution. It's not an algorithm yet, of course.
Once match() is working, where do we go from here? Well, an obvious solution is brute force: from the set of all possible arrangements of the tiles within the grid, reject those where any two adjacent tiles don't match. That's an algorithm, of sorts, and it's pretty certain to work (although in many puzzles the heat death of the universe occurs before a solution).
How about collecting the set of all pairs of tiles that match along a given edge (LTRB)? Could you get from there to a solution, quicker? Certainly you can test it (and test-drive it) easily enough.
The tests are unlikely to give you an algorithm, but they can help you to think about algorithms, and of course they can make validating your approach easier.
dunno if this "answers" the question either
analysis of the "puzzle"
9 tiles
each has 4 sides
each tile has half a pattern / picture
BRUTE FORCE APPROACH
to solve this problem
you need to generate 9! combinations ( 9 tiles X 8 tiles X 7 tiles... )
limited by the number of matching sides to the current tile(s) already in place
CONSIDERED APPROACH
Q How many sides are different?
IE how many matches are there?
therefore 9 X 4 = 36 sides / 2 ( since each side "must" match at least 1 other side )
otherwise its an uncompleteable puzzle
NOTE: at least 12 must match "correctly" for a 3 X 3 puzzle
label each matching side of a tile using a unique letter
then build a table holding each tile
you will need 4 entries into the table for each tile
4 sides ( corners ) hence 4 combinations
if you sort the table by side and INDEX into the table
side,tile_number
ABcd tile_1
BCda tile_1
CDab tile_1
DAbc tile_1
using the table should speed things up
since you should only need to match 1 or 2 sides at most
this limits the amount of NON PRODUCTIVE tile placing it has to do
depending on the design of the pattern / picture
there are 3 combinations ( orientations ) since each tile can be placed using 3 orientations
- the same ( multiple copies of the same tile )
- reflection
- rotation
God help us if they decide to make life very difficult
by putting similar patterns / pictures on the other side that also need to match
OR even making the tiles into cubes and matching 6 sides!!!
Using TDD,
you would write tests and then code to solve each small part of the problem,
as outlined above and write more tests and code to solve the whole problem
NO its not easy, you need to sit and write tests and code to practice
NOTE: this is a variation of the map colouring problem
http://en.wikipedia.org/wiki/Four_color_theorem
I'm working on a game where on each update of the game loop, the AI is run. During this update, I have the chance to turn the AI-controlled entity and/or make it accelerate in the direction it is facing. I want it to reach a final location (within reasonable range) and at that location have a specific velocity and direction (again it doesn't need to be exact) That is, given a current:
P0(x, y) = Current position vector
V0(x, y) = Current velocity vector (units/second)
θ0 = Current direction (radians)
τmax = Max turn speed (radians/second)
αmax = Max acceleration (units/second^2)
|V|max = Absolute max speed (units/second)
Pf(x, y) = Target position vector
Vf(x, y) = Target velocity vector (units/second)
θf = Target rotation (radians)
Select an immediate:
τ = A turn speed within [-τmax, τmax]
α = An acceleration scalar within [0, αmax] (must accelerate in direction it's currently facing)
Such that these are minimized:
t = Total time to move to the destination
|Pt-Pf| = Distance from target position at end
|Vt-Vf| = Deviation from target velocity at end
|θt-θf| = Deviation from target rotation at end (wrapped to (-π,π))
The parameters can be re-computed during each iteration of the game loop. A picture says 1000 words so for example given the current state as the blue dude, reach approximately the state of the red dude within as short a time as possible (arrows are velocity):
Pic http://public.blu.livefilestore.com/y1p6zWlGWeATDQCM80G6gaDaX43BUik0DbFukbwE9I4rMk8axYpKwVS5-43rbwG9aZQmttJXd68NDAtYpYL6ugQXg/words.gif
Assuming a constant α and τ for Δt (Δt → 0 for an ideal solution) and splitting position/velocity into components, this gives (I think, my math is probably off):
Equations http://public.blu.livefilestore.com/y1p6zWlGWeATDTF9DZsTdHiio4dAKGrvSzg904W9cOeaeLpAE3MJzGZFokcZ-ZY21d0RGQ7VTxHIS88uC8-iDAV7g/equations.gif
(EDIT: that last one should be θ = θ0 + τΔt)
So, how do I select an immediate α and τ (remember these will be recomputed every iteration of the game loop, usually > 100 fps)? The simplest, most naieve way I can think of is:
Select a Δt equal to the average of the last few Δts between updates of the game loop (i.e. very small)
Compute the above 5 equations of the next step for all combinations of (α, τ) = {0, αmax} x {-τmax, 0, τmax} (only 6 combonations and 5 equations for each, so shouldn't take too long, and since they are run so often, the rather restrictive ranges will be amortized in the end)
Assign weights to position, velocity and rotation. Perhaps these weights could be dynamic (i.e. the further from position the entity is, the more position is weighted).
Greedily choose the one that minimizes these for the location Δt from now.
Its potentially fast & simple, however, there are a few glaring problems with this:
Arbitrary selection of weights
It's a greedy algorithm that (by its very nature) can't backtrack
It doesn't really take into account the problem space
If it frequently changes acceleration or turns, the animation could look "jerky".
Note that while the algorithm can (and probably should) save state between iterations, but Pf, Vf and θf can change every iteration (i.e. if the entity is trying to follow/position itself near another), so the algorithm needs to be able to adapt to changing conditions.
Any ideas? Is there a simple solution for this I'm missing?
Thanks,
Robert
sounds like you want a PD controller. Basically draw a line from the current position to the target. Then take the line direction in radians, that's your target radians. The current error in radians is current heading - line heading. Call it Eh. (heading error) then you say the current turn rate is going to be KpEh+d/dt EhKd. do this every step with a new line.
thats for heading
acceleration is "Accelerate until I've reached max speed or I wont be able to stop in time". you threw up a bunch of integrals so I'm sure you'll be fine with that calculation.
I case you're wondering, yes I've solved this problem before, PD controller works. don't bother with PID, don't need it in this case. Prototype in matlab. There is one thing I've left out, you need to have a trigger, like "i'm getting really close now" so I should start turning to get into the target. I just read your clarification about "only accelerating in the direction we're heading". that changes things a bit but not too much. that means to need to approach the target "from behind" meaning that the line target will have to be behind the real target, when you get near the behind target, follow a new line that will guide you to the real target. You'll also want to follow the lines, rather than just pick a heading and try to stick with it. So don't update the line each frame, just say the error is equal to the SIGNED DISTANCE FROM THE CURRENT TARGET LINE. The PD will give you a turn rate, acceleration is trivial, so you're set. you'll need to tweak Kd and Kp by head, that's why i said matlab first. (Octave is good too)
good luck, hope this points you in the right direction ;)
pun intended.
EDIT: I just read that...lots of stuff, wrote real quick. this is a line following solution to your problem, just google line following to accompany this answer if you want to take this solution as a basis to solving the problem.
I would like to suggest that yout consider http://en.wikipedia.org/wiki/Bang%E2%80%93bang_control (Bang-bang control) as well as a PID or PD. The things you are trying to minimise don't seem to produce any penalty for pushing the accelerator down as far as it will go, until it comes time to push the brake down as far as it will go, except for your point about how jerky this will look. At the very least, this provides some sort of justification for your initial guess.