When I was a child, I had a square board tetris puzzle, and the instruction said there are >15 thousand solutions. I've always wondered how they counted the solutions, and I'd like to replicate that.
So I would like to search for divisions of NxN boards into 4-cell and 5-cell fragments. Rotation is allowed, flipping is not.
Problem A: Determine if a set of blocks can be assembled into a NxN grid.
Problem B: Find divisions without rectangular subdivisions, except for single fragments (4 cells, that is, 2x2 and 1x4).
I'm thinking of constraint programming. But if I encode the presence of each wall as boolean, how can I efficiently count the blocks? Is it better to work in terms of fragments and check for rectangles later?
What other technique could help?
note: this is just for fun, and not homework or anything.
I once challenged solving pentomino variant in Prolog.
Program:
http://www2.koyahatataku.com/programming/pent.txt
Result:
http://www2.koyahatataku.com/programming/pent_out.txt
I tried with naive brute force Depth-first search algorithm.
Related
I have a problem I am working on involving breadth first search and iterative deepening search. I understand the search mechanisms for trees but I don't understand how to apply it to grid and geometric shapes. If I wanted to perform a BFS(breadth first search) how would I apply it to this problem where I have to move the pieces so that they would fit perfectly in the square on the right hand side. My attempt is to first take two pieces and place them in the square and then branch out from each side. The problem is, there are many ways I can place the pieces in level 1 of the BFS tree. By looking at the image I know the solution but do not know how I would go about it in terms of the searches
I am going to assume from your post history that we are enrolled in the same class and have the same assignment due on Monday.
The way I thought to approach this problem is:
Case 0 is the empty board
Case 1 is the multitude of different position that a shape such as the 3x1 could fit in the rectangle
Case 2 is the multitude of different positions that another shape such as the U shaped one could fit in while taking the 3x1 into account.
As you go on, some shapes aren't going to fit anymore so those branches aren't prolonged anymore.
I haven't figure it out fully, if you want to ponder upon it further or if you have figured out another way to do this, I guess we could 'team up' and try to figure it out.
So I'm working on simulating a large number of n-dimensional particles, and I need to know the distance between every pair of points. Allowing for some error, and given the distance isn't relevant at all if exceeds some threshold, are there any good ways to accomplish this? I'm pretty sure if I want dist(A,C) and already know dist(A,B) and dist(B,C) I can bound it by [dist(A,B)-dist(B,C) , dist(A,B)+dist(B,C)], and then store the results in a sorted array, but I'd like to not reinvent the wheel if there's something better.
I don't think the number of dimensions should greatly affect the logic, but maybe for some solutions it will. Thanks in advance.
If the problem was simply about calculating the distances between all pairs, then it would be a O(n^2) problem without any chance for a better solution. However, you are saying that if the distance is greater than some threshold D, then you are not interested in it. This opens the opportunities for a better algorithm.
For example, in 2D case you can use the sweep-line technique. Sort your points lexicographically, first by y then by x. Then sweep the plane with a stripe of width D, bottom to top. As that stripe moves across the plane new points will enter the stripe through its top edge and exit it through its bottom edge. Active points (i.e. points currently inside the stripe) should be kept in some incrementally modifiable linear data structure sorted by their x coordinate.
Now, every time a new point enters the stripe, you have to check the currently active points to the left and to the right no farther than D (measured along the x axis). That's all.
The purpose of this algorithm (as it is typically the case with sweep-line approach) is to push the practical complexity away from O(n^2) and towards O(m), where m is the number of interactions we are actually interested in. Of course, the worst case performance will be O(n^2).
The above applies to 2-dimensional case. For n-dimensional case I'd say you'll be better off with a different technique. Some sort of space partitioning should work well here, i.e. to exploit the fact that if the distance between partitions is known to be greater than D, then there's no reason to consider the specific points in these partitions against each other.
If the distance beyond a certain threshold is not relevant, and this threshold is not too large, there are common techniques to make this more efficient: limit the search for neighbouring points using space-partitioning data structures. Possible options are:
Binning.
Trees: quadtrees(2d), kd-trees.
Binning with spatial hashing.
Also, since the distance from point A to point B is the same as distance from point B to point A, this distance should only be computed once. Thus, you should use the following loop:
for point i from 0 to n-1:
for point j from i+1 to n:
distance(point i, point j)
Combining these two techniques is very common for n-body simulation for example, where you have particles affect each other if they are close enough. Here are some fun examples of that in 2d: http://forum.openframeworks.cc/index.php?topic=2860.0
Here's a explanation of binning (and hashing): http://www.cs.cornell.edu/~bindel/class/cs5220-f11/notes/spatial.pdf
You have 12 shapes:
which you can make each out of five identical squares.
You need to combine the 12 pieces to one rectangle.
You can form four different rectangles:
2339 solutions (6x10), 2 solutions (3x20), 368 solutions (4x15), 1010 solutions (5x12).
I need to build the 3X20 rectangle:
My question what is the maximum number of states (i.e., the branching factor) that is possible?
My half way calculation:
The way I see it, there are 4 operations on each shape: turn 90/180/270 degrees and mirroring (turning it upside down).
Then, you have to put the shape on the board, somewhere on the 3X20 board.
Illegal states will be one that the shape doesn't fit in the board, but they are still states.
For the first move, you can chose each shape in 4 ways which is 4X12 ways, and then you need to multiply in the number of positions the shape can be in, and that is the number of states you have. But how can I calculate the number of positions?
Please help me with this calculation it is very important, it is not some kind of homework which I'm trying to avoid.
I think there is no easy & 'intelligent' way to list solutions (or states) to pentomino puzzles. You have to try all possibilities. Recursive programming or backtracking is the way to do it. You should check this solution that also has java source code available. Hopefully that points you to the right direction.
There is also a python solution that is perhaps more readable.
A naive approach is to find, for each edge in the polygon, the point on that edge closest to the given point, and then take the one that's closest. Is there a faster algorithm? My goal is to implement a 2D Super Mario Galaxy-style platformer.
Apparently this can be done with Voronoi regions, as in this video: http://www.youtube.com/watch?v=Ldh2YKobuWo
However, I can't find any Voronoi algorithms that deal with edges as well as points. Ideas?
Calculate the point-line distance for each of the edges, then pick the shortest one. There is no shortcut. This site has a good explanation and even implementations in various languages.
However, finding "the point on that edge closest to the given point" is a computationally unnecessary intermediate result.
If the polygon is convex, then the overhead of the voronoi calculation far exceeds that of the naive approach.
If this is run many times, and each time the point changes slightly, you only need to check 3 segments (think about it: as you move around, assuming many checks, then the closest edge will only change to an adjacent edge)
This question is a little involved. I wrote an algorithm for breaking up a simple polygon into convex subpolygons, but now I'm having trouble proving that it's not optimal (i.e. minimal number of convex polygons using Steiner points (added vertices)). My prof is adamant that it can't be done with a greedy algorithm such as this one, but I can't think of a counterexample.
So, if anyone can prove my algorithm is suboptimal (or optimal), I would appreciate it.
The easiest way to explain my algorithm with pictures (these are from an older suboptimal version)
What my algorithm does, is extends the line segments around the point i across until it hits a point on the opposite edge.
If there is no vertex within this range, it creates a new one (the red point) and connects to that:
If there is one or more vertices in the range, it connects to the closest one. This usually produces a decomposition with the fewest number of convex polygons:
However, in some cases it can fail -- in the following figure, if it happens to connect the middle green line first, this will create an extra unneeded polygon. To this I propose double checking all the edges (diagonals) we've added, and check that they are all still necessary. If not, remove it:
In some cases, however, this is not enough. See this figure:
Replacing a-b and c-d with a-c would yield a better solution. In this scenario though, there's no edges to remove so this poses a problem. In this case I suggest an order of preference: when deciding which vertex to connect a reflex vertex to, it should choose the vertex with the highest priority:
lowest) closest vertex
med) closest reflex vertex
highest) closest reflex that is also in range when working backwards (hard to explain) --
In this figure, we can see that the reflex vertex 9 chose to connect to 12 (because it was closest), when it would have been better to connect to 5. Both vertices 5 and 12 are in the range as defined by the extended line segments 10-9 and 8-9, but vertex 5 should be given preference because 9 is within the range given by 4-5 and 6-5, but NOT in the range given by 13-12 and 11-12. i.e., the edge 9-12 elimates the reflex vertex at 9, but does NOT eliminate the reflex vertex at 12, but it CAN eliminate the reflex vertex at 5, so 5 should be given preference.
It is possible that the edge 5-12 will still exist with this modified version, but it can be removed during post-processing.
Are there any cases I've missed?
Pseudo-code (requested by John Feminella) -- this is missing the bits under Figures 3 and 5
assume vertices in `poly` are given in CCW order
let 'good reflex' (better term??) mean that if poly[i] is being compared with poly[j], then poly[i] is in the range given by the rays poly[j-1], poly[j] and poly[j+1], poly[j]
for each vertex poly[i]
if poly[i] is reflex
find the closest point of intersection given by the ray starting at poly[i-1] and extending in the direction of poly[i] (call this lower bound)
repeat for the ray given by poly[i+1], poly[i] (call this upper bound)
if there are no vertices along boundary of the polygon in the range given by the upper and lower bounds
create a new vertex exactly half way between the lower and upper bound points (lower and upper will lie on the same edge)
connect poly[i] to this new point
else
iterate along the vertices in the range given by the lower and upper bounds, for each vertex poly[j]
if poly[j] is a 'good reflex'
if no other good reflexes have been found
save it (overwrite any other vertex found)
else
if it is closer then the other good reflexes vertices, save it
else
if no good reflexes have been found and it is closer than the other vertices found, save it
connect poly[i] to the best candidate
repeat entire algorithm for both halves of the polygon that was just split
// no reflex vertices found, then `poly` is convex
save poly
Turns out there is one more case I didn't anticipate: [Figure 5]
My algorithm will attempt to connect vertex 1 to 4, unless I add another check to make sure it can. So I propose stuffing everything "in the range" onto a priority queue using the priority scheme I mentioned above, then take the highest priority one, check if it can connect, if not, pop it off and use the next. I think this makes my algorithm O(r n log n) if I optimize it right.
I've put together a website that loosely describes my findings. I tend to move stuff around, so get it while it's hot.
I believe the regular five pointed star (e.g. with alternating points having collinear segments) is the counterexample you seek.
Edit in response to comments
In light of my revised understanding, a revised answer: try an acute five pointed star (e.g. one with arms sufficiently narrow that only the three points comprising the arm opposite the reflex point you are working on are within the range considered "good reflex points"). At least working through it on paper it appears to give more than the optimal. However, a final reading of your code has me wondering: what do you mean by "closest" (i.e. closest to what)?
Note
Even though my answer was accepted, it isn't the counter example we initially thought. As #Mark points out in the comments, it goes from four to five at exactly the same time as the optimal does.
Flip-flop, flip flop
On further reflection, I think I was right after all. The optimal bound of four can be retained in a acute star by simply assuring that one pair of arms have collinear edges. But the algorithm finds five, even with the patch up.
I get this:
removing dead ImageShack link
When the optimal is this:
removing dead ImageShack link
I think your algorithm cannot be optimal because it makes no use of any measure of optimality. You use other metrics like 'closest' vertices, and checking for 'necessary' diagonals.
To drive a wedge between yours and an optimal algorithm, we need to exploit that gap by looking for shapes with close vertices which would decompose badly. For example (ignore the lines, I found this on the intertubenet):
concave polygon which forms a G or U shape http://avocado-cad.wiki.sourceforge.net/space/showimage/2007-03-19_-_convexize.png
You have no protection against the centre-most point being connected across the concave 'gap', which is external to the polygon.
Your algorithm is also quite complex, and may be overdoing it - just like complex code, you may find bugs in it because complex code makes complex assumptions.
Consider a more extensive initial stage to break the shape into more, simpler shapes - like triangles - and then an iterative or genetic algorithm to recombine them. You will need a stage like this to combine any unnecessary divisions between your convex polys anyway, and by then you may have limited your possible decompositions to only sub-optimal solutions.
At a guess something like:
decompose into triangles
non-deterministically generate a number of recombinations
calculate a quality metric (number of polys)
select the best x% of the recombinations
partially decompose each using triangles, and generate a new set of recombinations
repeat from 4 until some measure of convergence is reached
but vertex 5 should be given preference because 9 is within the range given by 4-5 and 6-5
What would you do if 4-5 and 6-5 were even more convex so that 9 didn't lie within their range? Then by your rules the proper thing to do would be to connect 9 to 12 because 12 is the closest reflex vertex, which would be suboptimal.
Found it :( They're actually quite obvious.
*dead imageshack img*
A four leaf clover will not be optimal if Steiner points are allowed... the red vertices could have been connected.
*dead imageshack img*
It won't even be optimal without Steiner points... 5 could be connected to 14, removing the need for 3-14, 3-12 AND 5-12. This could have been two polygons better! Ouch!