I have a circular data structure that has a reading direction and a starting point. (Internally it could be a directed graph, linked list or just an array that gets rotated using modulo.) I want to have a way to normalize the starting point of the data structure in a way that the same data input yields the same starting point.
A = [1,2,3]
B = [2,3,1]
C = [3,1,2]
assert(Normalized(A) == Normalized(B))
assert(Normalized(B) == Normalized(C))
I don't care so much about a specific rotation. (Just, of corse, the circular order and direction needs to be persisted :P )
Here is a cool answer about comparing circular structures:
Interview question: Check if one string is a rotation of other string
[EDIT: Removed non-working approach]
I feel, that smart people thought about this before. What is the terminology I am missing or what are known algorithms for normalizing the rotation of circular data structures?
what are known algorithms for normalizing the rotation of circular data structures?
The goal is a rotation, so that the same circle input ends up on the same rotation. One way is to simply look for the group that describes the biggest number, if you look at the whole group. (The rotation with the greatest number(s) first)
[1,2,3] -> [3,1,2]
[2,3,1] -> [3,1,2]
(Needless to say, this works with the smallest group, as well.)
What is the terminology I am missing
https://en.wikipedia.org/wiki/Circular_buffer
https://en.wikipedia.org/wiki/Modular_arithmetic
Related
This is a follow-up to another stack overflow question, here:
3D Correspondences from fundamental matrix
Just like in that question, I am trying to get a camera matrix from a fundamental matrix, the ultimate goal being 3d reconstruction from 2d points. The answer given there is good, and correct. I just don't understand it. It says, quote, "If you have access to Hartley and Zisserman's textbook, you can check section 9.5.3 where you will find what you need." He also provides a link to source code.
Now, here's what section 9.5.3 of the book, among other things, says:
Result 9.12. A non-zero matrix F is the fundamental matrix
corresponding to a pair of camera matrices P and P if and only if PTFP
is skew-symmetric.
That, to me, is gibberish. (I looked up skew-symmetric - it means the inverse is its negative. I have no idea how that is relevant to anything.)
Now, here is the source code given (source):
[U,S,V] = svd(F);
e = U(:,3);
P = [-vgg_contreps(e)*F e];
This is also a mystery.
So what I want to know is, how does one explain the other? Getting that code from that statement seems like black magic. How would I, or anyone, figure out that "A non-zero matrix F is the fundamental matrix corresponding to a pair of camera matrices P and P if and only if PTFP is skew-symmetric." means what the code is telling you to do, which is basically
'Take the singular value decomposition. Take the first matrix. Take the third column of that. Perform some weird re-arrangment of its values. That's your answer.' How would I have come up with this code on my own?
Can someone explain to me the section 9.5.3 and this code in plain English?
Aha, that "PTFP" is actually something I have also wondered about and could not find the answer in literature. However, this is what I figured out:
The 4x4 skew-symmetric matrix you are mentioning is not just any matrix. It is actually the dual Plücker Matrix of the baseline (see also https://en.wikipedia.org/wiki/Pl%C3%BCcker_matrix). In other words, it only gives you the line on which the camera centers are located, which is not useful for reconstruction tasks as such.
The condition you mention is identical to the more popularized fact that the fundamental matrix for the view 1 & 0 is the negative transpose of the fundamental matrix for the views 0 & 1 (using MATLAB/Octave syntax here)
Consider first that the fundamental matrix maps a point x0 in one image to line l1 in the other
l1=F*x0
Next, consider that the transpose of the projection matrix back-projects a lines l1 in the image to a plane E in space
E=P1'*l1
(I find this beautifully simple and understated in most geometry / computer vision classes)
Now, I will use a geometric argument: Two lines are corresponding epipolar lines iff they correspond to the same epipolar plane i.e. the back-projection of either line gives the same epipolar plane. Algebraically:
E=P0'*l0
E=P1'*l1
thus (the important equation)
P0'*l0=P1'*l1
Now we are almost there. Let's assume we have a 3D point X and its two projections
x0=P0*X
x1=P1*X
and the epipolar lines
l1=F*x0
l0=-F'*x1
We can just put that into the important equation and we have for all X
P0'*-F'*P1*X=P1'*F*P0*X
and finally
P0'*-F'*P1=P1'*F*P0
As you can see, the left-hand-side is the negative transpose of the right-hand-side. So this matrix is a skew symmetric 4x4 matrix.
I also published these thoughts in Section II B (towards the end of the paragraph) in the following paper. It should also explain why this matrix is a representation of the baseline.
Aichert, André, et al. "Epipolar consistency in transmission imaging."
IEEE transactions on medical imaging 34.11 (2015): 2205-2219.
https://www.ncbi.nlm.nih.gov/pubmed/25915956
Final note to #john ktejik : skew-symmetry means that a matrix is identical to its negative transpose (NOT inverse transpose)
I have some data in the form of Line objects (eg Line1(start, end), where start and end are coordinates in the form of point objects). Is there a quick way to go through all the lines to see if any of them form a triangle? By quick I mean anything better than going through all nC3 possibilities.
Edit: Just realised I may not understand all the replies (I'm no Adrian Lamo). Please try and explain wrt Python.
1) geometric step: enter all line segments in a dictionary, with the first endpoint as the key and the second endpoint as the value. There will be duplicate keys, so you will keep a list of values rather than single values. In principle there will be no duplicates in the lists (unless you enter the same edge twice).
2) topological step: for all entries P in the dictionary, consider all the pairs of elements from its list, let (Q, R). Lookup Q and check if R belongs to the list of Q. If yes, you have found the triangle (P, Q, R).
By symmetry, all six permutations of every triangle will be reported. You can avoid that by enforcing that P<Q<R in the lexicographical sense.
Is there a quick way to go through all the lines to see if any of them form a triangle?
Yes. Assuming that your Points are integers (or can be easily converted to such, because they have fixed significant digits or similar):
Being creative for you here:
Make two fastly searchable storage structures (e.g. std::multimap<int>), one for each of the x and y coordinates of the endpoints, associating the coordinates with a pointer to the respective line
In the first structure, search for elements with the same x coordinate. Those are the only "candidates" for being a corner of a triangle. Searching for duplicate entries is fast, because of you using an appropriate data structure.
For each of these, verify whether they are actually an edge. If they are not, discard.
For each of the remaining corners, verify that both "opposite" line ends are part of the same edge. Discard the others. Done.
I'm trying to optimize a sort for an isometric renderer. What is proving tricky is that the comparitor can return "A > B", "A < B", or "the relationship between A and B is ambiguous." All of the standard sorting algorithms expect that you can compare the values of any two objects, but I cannot do that here. Are there known algorithms that are made for this situation?
For example, there can be a situation where the relationship between A and C is ambiguous, but we know A > B, and B > C. The final list should be A,B,C.
Note also that there may be multiple solutions. If A > B, but C is ambiguous to both A and B, then the answer may be C,A,B or A,B,C.
Edit: I did say it was for sorting in an isometric renderer, but several people asked for more info so here goes. I've got an isometric game and I'm trying to sort the objects. Each object has a rectangular, axis-aligned footprint in the world, which means that they appears as a sort of diamond shapes from the perspective of the camera. Height of the objects is not known, so an object in front of another object is assumed to be capable of occluding the one in the back, and therefore must be drawn after (sorted later in the list than) the one in the back.
I also left off an important consideration, which is that a small number of objects (the avatars) move around.
Edit #2: I finally have enough rep to post pictures! A and B...well, they aren't strictly ambiguous because in each case they have a definite relationship compared to each other. However, that relationship cannot be known by looking at the variables of A and B themselves. Only when we look at C as well can we know what order they go in.
I definitely think topographical sort is the way to go, so I'm considering the question answered, but I'm happy to make clarifications to the question for posterity.
You may want to look at sorts for partial orders.
https://math.stackexchange.com/questions/55891/algorithm-to-sort-based-on-a-partial-order links to the right papers on this topic (specifically topological sorting).
Edit:
Given the definition of the nodes:
You need to make sure objects never can cause mutual occlusion. Take a look at the following grid, where the camera is in the bottom left corner.
______
____X_
_YYYX_
_YXXX_
_Y____
As you can see, parts of Y are hidden by X and parts of X are hidden by Y. Any drawing order will cause weird rendering. This can be solved in many ways, the simplest being to only allow convex, hole-free shapes as renderable primitives. Anything concave needs to be broken into chunks.
If you do this, you can then turn your partial order into a total order. Here's an example:
def compare(a,b):
if a.max_x < b.min_x:
return -1
elif a.max_y < b.min_y:
return -1
elif b.max_x < a.min_x:
return 1
elif b.max_y < a.min_y:
return 1
else:
# The bounding boxes intersect
# If the objects are both rectangular,
# this should be impossible
# If you allow non-rectangular convex shapes,
# like circles, you may need to do something fancier here.
raise NotImplementedException("I only handle non-intersecting bounding boxes")
And use any old sorting algortithm to give you your drawing order.
You should first build a directed graph, using that graph you will be able to find the relationships, by DFS-ing from each node.
Once you have relationships, some pairs might still be ambiguous. In that case look for partial sorting.
The problem
I have multiple groups which specify the relationships of symbols.. for example:
[A B C]
[A D E]
[X Y Z]
What these groups mean is that (for the first group) the symbols, A, B, and C are related to each other. (The second group) The symbols A, D, E are related to each other.. and so forth.
Given all these data, I would need to put all the unique symbols into a 1-dimension array wherein the symbols which are somehow related to each other would be placed closer to each other. Given the example above, the result should be something like:
[B C A D E X Y Z]
or
[X Y Z D E A B C]
In this resulting array, since the symbol A has multiple relationships (namely with B and C in one group and with D and E in another) it's now located between those symbols, somewhat preserving the relationship.
Note that the order is not important. In the result, X Y Z can be placed first or last since those symbols are not related to any other symbols. However, the closeness of the related symbols is what's important.
What I need help in
I need help in determining an algorithm that takes groups of symbol relationships, then outputs the 1-dimension array using the logic above. I'm pulling my hair out on how to do this since with real data, the number of symbols in a relationship group can vary, there is also no limit to the number of relationship groups and a symbol can have relationships with any other symbol.
Further example
To further illustrate the trickiness of my dilemma, IF you add another relationship group to the example above. Let's say:
[C Z]
The result now should be something like:
[X Y Z C B A D E]
Notice that the symbols Z and C are now closer together since their relationship was reinforced by the additional data. All previous relationships are still retained in the result also.
The first thing you need to do is to precisely define the result you want.
You do this by defining how good a result is, so that you know which is the best one. Mathematically you do this by a cost function. In this case one would typically choose the sum of the distances between related elements, the sum of the squares of these distances, or the maximal distance. Then a list with a small value of the cost function is the desired result.
It is not clear whether in this case it is feasible to compute the best solution by some special method (maybe if you choose the maximal distance or the sum of the distances as the cost function).
In any case it should be easy to find a good approximation by standard methods.
A simple greedy approach would be to insert each element in the position where the resulting cost function for the whole list is minimal.
Once you have a good starting point you can try to improve it further by modifying the list towards better solutions, for example by swapping elements or rotating parts of the list (local search, hill climbing, simulated annealing, other).
I think, because with large amounts of data and lack of additional criteria, it's going to be very very difficult to make something that finds the best option. Have you considered doing a greedy algorithm (construct your solution incrementally in a way that gives you something close to the ideal solution)? Here's my idea:
Sort your sets of related symbols by size, and start with the largest one. Keep those all together, because without any other criteria, we might as well say their proximity is the most important since it's the biggest set. Consider every symbol in that first set an "endpoint", an endpoint being a symbol you can rearrange and put at either end of your array without damaging your proximity rule (everything in the first set is an endpoint initially because they can be rearranged in any way). Then go through your list and as soon as one set has one or more symbols in common with the first set, connect them appropriately. The symbols that you connected to each other are no longer considered endpoints, but everything else still is. Even if a bigger set only has one symbol in common, I'm going to guess that's better than smaller sets with more symbols in common, because this way, at least the bigger set stays together as opposed to possibly being split up if it was put in the array later than smaller sets.
I would go on like this, updating the list of endpoints that existed so that you could continue making matches as you went through your set. I would keep track of if I stopped making matches, and in that case, I'd just go to the top of the list and just tack on the next biggest, unmatched set (doesn't matter if there are no more matches to be made, so go with the most valuable/biggest association). Ditch the old endpoints, since they have no matches, and then all the symbols of the set you just tacked on are the new endpoints.
This may not have a good enough runtime, I'm not sure. But hopefully it gives you some ideas.
Edit: Obviously, as part of the algorithm, ditch duplicates (trivial).
The problem as described is essentially the problem of drawing a graph in one dimension.
Using the relationships, construct a graph. Treat the unique symbols as the vertices of the graph. Place an edge between any two vertices that co-occur in a relationship; more sophisticated would be to construct a weight based on the number of relationships in which the pair of symbols co-occur.
Algorithms for drawing graphs place well-connected vertices closer to one another, which is equivalent to placing related symbols near one another. Since only an ordering is needed, the symbols can just be ranked based on their positions in the drawing.
There are a lot of algorithms for drawing graphs. In this case, I'd go with Fiedler ordering, which orders the vertices using a particular eigenvector (the Fiedler vector) of the graph Laplacian. Fiedler ordering is straightforward, effective, and optimal in a well-defined mathematical sense.
It sounds like you want to do topological sorting: http://en.wikipedia.org/wiki/Topological_sorting
Regarding the initial ordering, it seems like you are trying to enforce some kind of stability condition, but it is not really clear to me what this should be from your question. Could you try to be a bit more precise in your description?
Say I have two maps, each represented as a 2D array. Each map contains several distinct features (rocks, grass, plants, trees, etc.). I know the two maps are of the same general region but I would like to find out: 1.) if they overlap and 2.) if so, where does this overlap occur. Does anyone know of any algorithms which would help me do this?
[EDIT]
Each feature is contained entirely inside an array index. Although it is possibly to discern (for example) a rock from a patch of grass, it is not possible to discern one rock from another (or one patch of grass from another).
When doing this in 1D, I would for each index in the first collection (a string, really), try find the largest match in the second collection. If the match goes to the end, I have an overlap (like in action and ionbeam).
match( A on B ):
for each i in length(A):
see if A[i..] matches B[0..]
if no match found: do the same for B on A.
For 2D, you do the same thing, basically: find an 'edge' of A that overlaps with the opposite edge of B. Only the edges aren't 1D, but 2D:
for each point xa,ya in A:
find a row yb in B that has a match( A[ya] on B[yb] )
see if A[ya..] matches B[yb..]
You need to do this for the 2 diagonals, in each sense.
For one map, go through each feature and find the nearest other feature to it. Record these in a list, storing the type of each of the two features and the dx dy between them. Store in a hash table or sorted list. These are now location invariant, since they record only relative distances.
Now for your second map, start doing the same: pick any feature, find its closest neighbor, find the delta. Look for the same correspondence in the original map list. If the features are shared between the maps, you'll find it in the list and you now know one correspondence between the maps. Repeat for many features if necessary. The results will give you a decent answer of if the maps overlap, and if so, at what offset.
Sounds like image registration (wikipedia), finding a transformation (translation only, in your case) which can align two images. There's a pile of software that does this sort of thing linked off the wikipedia page.