I have a directed graph (more specifically, a control flow graph), and each of the vertices has a set of properties.
I'd like to find (or write) an algorithm that, given a vertex V with property P, finds the longest path to some vertex E such that all vertices along all possible paths from V to E contain the property P.
Example 1
Say I had the following graph. (Please excuse the bad ascii drawing.)
+----+
+--------+P +--------+
| +----+ |
| V1 |
| |
| |
+--v--+ |
+----+P ++ |
| +-----++ +--v--+
| | +----+P |
| | | +-----+
+--v--+ +--v--+ |
|P +-+ +-+P | |
+-----+ | | +-----+ |
| | |
| | |
+v-v--+ |
V6 |P +---------+ |
+-----+ | |
| |
| |
| |
| |
+-v--v-+
V7 |P |
+---+--+
|
|
+---v--+
V8 |!P |
+------+
Starting at V1, the longest path where P always holds on all possible paths is V1 -> V7. Note that the other paths, say V1 -> V6, are "valid" in that P always holds, but V1 -> V7 is the longest.
Example 2
This example is the same as above, except now the P doesn't hold in V3:
+----+
+--------+P +--------+
| +----+ |
| V1 |
| |
| |
+--v--+ |
+----+P ++ |
| +-----++ +--v--+
| | +----+!P | V3
| | | +-----+
+--v--+ +--v--+ |
|P +-+ +-+P | |
+-----+ | | +-----+ |
| | |
| | |
+v-v--+ |
V6 |P +---------+ |
+-----+ | |
| |
| |
| |
| |
+-v--v-+
V7 |P |
+---+--+
|
|
+---v--+
V8 |!P |
+------+
In this case, starting at V1, the longest path where P always holds in all possible paths is V1 -> V6. The path V1 -> V7 is not valid, because there is a path between V1 and V7 in which P does not hold.
Further notes about my situation
The graph could be cyclic.
The graph will be of a "small to medium" size, with maybe 1000 vertices or less.
The graph does not necessarily always have one root and one leaf, like my examples above.
Question
Is there a standard algorithm for computing such paths?
The problem has no known efficient solution, as it is easily reduceable from Hamiltonian Path Problem, which says - given a graph - is there a path that goes through all vertices exactly once?
The reduction is simple - Given Hamiltonian Path problem, label all nodes with p, and find longest path. Since Hamiltonian path is NP-Complete, so is this problem, and there is no known polynomial solution to it.
An alternative is using a brute-force search (simplest form is generate all permutations and chose the best valid one) - but that will become impossible for large graphs. You might also need to consider using a heuristic approach (that finds a "good" solution, but not the optimal), like Genetic Algorithms.
Another possible solution is to reduce the problem to a Traveling Salesman Problem, and use some existing TSP solver. Note that while this problem is also NP-hard, since it is well-studied, there are some pretty efficient solutions for medium size graphs.
Also, if your graph happens to be somehow 'special' (a DAG for example), you might utilize some smart techniques to achieve significant speed up to polynomial time, like Dynamic Programming.
Related
Let a 3D grid, just like a checkerboard, with an extra dimension. Now let's say that I have a certain amount of cubes into that grid, each cube occupying 1x1x1 cells. Let's say that each of these cubes is an item.
What I would like to do is replace/combine these cubes into larger boxes occupying any number of cells on the X, Y and Z axes, so that the resulting number of boxes is as small as possible while preserving the overall "appearance".
It's probably unclear so I'll give a 2D example. Say I have a 2D grid containing several squares occupying 1x1 cells. A letter represents the cells occupied by a given item, each item having a different letter from the other ones. In the first example we have 10 different items, each of them occupying 1x1x1 cells.
+---+---+---+---+---+---+
| | | | | | |
+---+---+---+---+---+---+
| | A | B | C | D | |
+---+---+---+---+---+---+
| | E | F | G | H | |
+---+---+---+---+---+---+
| | | K | L | | |
+---+---+---+---+---+---+
| | | | | | |
+---+---+---+---+---+---+
That's my input data. I could now optimize it, i.e reduce the number of items while still occupying the same cells, by multiple possible ways, one of which could be :
+---+---+---+---+---+---+
| | | | | | |
+---+---+---+---+---+---+
| | A | B | B | C | |
+---+---+---+---+---+---+
| | A | B | B | C | |
+---+---+---+---+---+---+
| | | B | B | | |
+---+---+---+---+---+---+
| | | | | | |
+---+---+---+---+---+---+
Here, instead of 10 items, I only have 3 (i.e A, B and C). However it can be optimized even more :
+---+---+---+---+---+---+
| | | | | | |
+---+---+---+---+---+---+
| | A | A | A | A | |
+---+---+---+---+---+---+
| | A | A | A | A | |
+---+---+---+---+---+---+
| | | B | B | | |
+---+---+---+---+---+---+
| | | | | | |
+---+---+---+---+---+---+
Here I only have two items, A and B. This is as optimized as this can be.
What I am looking for is an algorithm capable of finding the best item sizes and arrangement, or at least a reasonably good one, so that I have as few items as possible while occupying the same cells, and in 3D !
Is there such an algorithm ? I'm sure there are some domains where that kind of algorithm would be useful, and I need it for a video game. Thanks !!
Perhaps a simpler algorithm is possible, but a set partition should suffice.
Min x1 + x2 + x3 + ... //where x1 is 1 if the 1th partition is chosen, 0 otherwise
such that x1 + + x3 = 1// if 1st and 3rd partition contain 1st item
x2 + x3 = 1//if 2nd and 3rd partition contain 2nd item and so on.
x1, x2, x3,... are binary
You have 1 constraint for each item. Each constraint stipulates that each item can be part of exactly one box. The objective minimizes the total number of boxes.
This is an NP Hard integer programming however.
The number of variables in this problem could be exponential. You need to have an efficient way of enumerating them -- that is figuring out when a contiguous box can be found that is capable of including all points in it. It is here that you have to take into account information such as whether the grid is 2d or 3d, how you define a contiguous "box", etc.
Such problems are usually solved by column-generation, where these columns of the integer program are dynamically generated on the fly.
If I understand David Eppstein's1 explanation (see section 3) then a solution can be found in a maximal independent set in the bipartite intersection graph of axis-aligned diagonals connecting one concave vertex to another. (This would be 2d. I'm not sure about 3d, although perhaps it involves evaluating hyperplanes instead of lines?)
In your example, there is only one such diagonal:
________
| |
|_x....x_|
|____|
The two xs represent connected concave vertices. The maximal independent set of edges here contains only one edge, splitting the polygon in two.
Here's another with only one axis-parallel edge connecting two concave vertices, x and x. This polygon, though, also has two concave vertices, a and b, that do not have an opposite, axis-parallel match. In that case, it seems to me, each concave vertex without a partner would split the polygon it's on in two (either vertically or horizontally):
____________
| |
| |x
| . |
| . |a
|___ . |
b| . |
| .___|
|________|x
results in 4 rectangles:
____________
| |
| |x
| . |
| ..|a
|___.......... |
b| . |
| .___|
|________|x
Here's one with two intersecting axis-parallel diagonals, each connecting two concave vertices, (x,x) and (y,y):
____________
| |
| |x_
| . |
| . |
|___ . . . .z. .|y
y| . |
| .____|
|________|x
In this case, as I understand, the intersection graph again contains only one independent set:
(y,z) (z,y) (x,z) (z,x)
yielding 4 rectangles as a solution.
Since I'm not completely sure how the "intersection graph" in the paper is defined, I would welcome any clarifying comments.
1. Graph-Theoretic Solutions to Computational Geometry Problems, David Eppstein (Submitted on 26 Aug 2009)
Lucas Kanade Feature Tracker Refer Page 6 I am implementing the Lucas Kanade Feature Tracker in C++.
One thing is unclear in implementing the equation 23 which is mentioned in attached paper. I think Matrix G calculation should happened inside K loop, not outside K loop. In case when Patch B is present at the border in frame j, That time it is not useful to use full G Spatial Gradient Matrix which is calculated before K loop (as mentioned in paper). For Frame j, Matrix G should calculate for the showed patch B portion only.
Patch A Patch B
| |
| |
-----|--- -|-------
| |---| | | | |
| | | | |--| |
| |---| | | | |
| | |--| |
--------- ---------
Frame i Frame j
I need to write a program to draw all possible paths in a given matrix that can be had by moving in only left, right and up direction.
One should not cross the same location more than once. Note also that on a particular path, we may or may not use motion in all possible directions.
Path will start in the bottom-left corner in the matrix and will reach the top-right corner.
Following symbols are used to denote the direction of the motion in the current position:
+---+
| > | right
+---+
+---+
| ^ | up
+---+
+---+
| < | left
+---+
The symbol * is used in the final location to indicate end of path.
Example:
For a 5x8 matrix, using left, right and up directions, 2 different paths are shown below.
Path 1:
+---+---+---+---+---+---+---+---+
| | | | | | | | * |
+---+---+---+---+---+---+---+---+
| | | > | > | > | > | > | ^ |
+---+---+---+---+---+---+---+---+
| | | ^ | < | < | | | |
+---+---+---+---+---+---+---+---+
| | > | > | > | ^ | | | |
+---+---+---+---+---+---+---+---+
| > | ^ | | | | | | |
+---+---+---+---+---+---+---+---+
Path 2
+---+---+---+---+---+---+---+---+
| | | | > | > | > | > | * |
+---+---+---+---+---+---+---+---+
| | | | ^ | < | < | | |
+---+---+---+---+---+---+---+---+
| | | | | | ^ | | |
+---+---+---+---+---+---+---+---+
| | | > | > | > | ^ | | |
+---+---+---+---+---+---+---+---+
| > | > | ^ | | | | | |
+---+---+---+---+---+---+---+---+
Can anyone help me with this?
I tried to solve using lists. It i soon realized that i am making a disaster. Here is the code i tried with.
solution x y = travel (1,1) (x,y)
travelRight (x,y) = zip [1..x] [1,1..] ++ [(x,y)]
travelUp (x,y) = zip [1,1..] [1..y] ++ [(x,y)]
minPaths = [[(1,1),(2,1),(2,2)],[(1,1),(1,2),(2,2)]]
travel startpos (x,y) = rt (x,y) ++ up (x,y)
rt (x,y) | odd y = map (++[(x,y)]) (furtherRight (3,2) (x,2) minPaths)
| otherwise = furtherRight (3,2) (x,2) minPaths
up (x,y) | odd x = map (++[(x,y)]) (furtherUp (2,3) (2,y) minPaths)
| otherwise = furtherUp (2,3) (2,y) minPaths
furtherRight currpos endpos paths | currpos == endpos = (travelRight currpos) : map (++[currpos]) paths
| otherwise = furtherRight (nextRight currpos) endpos ((travelRight currpos) : (map (++[currpos]) paths))
nextRight (x,y) = (x+1,y)
furtherUp currpos endpos paths | currpos == endpos = (travelUp currpos) : map (++[currpos]) paths
| otherwise = furtherUp (nextUp currpos) endpos ((travelUp currpos) : (map(++[currpos]) paths))
nextUp (x,y) = (x,y+1)
identify lst = map (map iden) lst
iden (x,y) = (x,y,1)
arrows lst = map mydir lst
mydir (ele:[]) = "*"
mydir ((x1,y1):(x2,y2):lst) | x1==x2 = '>' : mydir ((x2,y2):lst)
| otherwise = '^' : mydir ((x2,y2):lst)
surroundBox lst = map (map createBox) lst
bar = "+ -+"
mid x = "| "++ [x] ++" |"
createBox chr = bar ++ "\n" ++ mid chr ++ "\n" ++ bar ++ "\n"
This ASCII grids are much more confusing than enlightening. Let me describe a better way to represent each possible path.
Each non-top row will have exactly one cell with UP. I claim that once each of the UP cells has been chosen that the LEFT and RIGHT and EMPTY cells can be determined. I claim that all possible cells in each of the non-top rows can be UP in all combination.
Each path is thus isomorphic to a (rows-1) length list of numbers in the range (1..columns) that determine the UP cells. The number of allowed paths is thus columns^(rows-1) and enumerating the possible paths in this format should be easy.
Then you could make a printer that converts this format to the ASCII art. This may be annoying, depending on skill level.
Looks like a homework so I will try to give enough hints
Try first filling number of paths from a cell to your goal.
So
+---+---+---+---+---+---+---+---+
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | * |
+---+---+---+---+---+---+---+---+
| | | | | | | | |
+---+---+---+---+---+---+---+---+
| | | | | | | | |
+---+---+---+---+---+---+---+---+
| | | | | | | | |
+---+---+---+---+---+---+---+---+
| | | | | | | | |
+---+---+---+---+---+---+---+---+
The thing to note here is from the cell in the top level there will always be one path to the *.
Number of possible path from cells in the same row will be same. You can realize this as all the paths will ultimately have to move up as there is no down action so in any path the cell above the current row can be reached by any cell in the current row.
You can feel the all possible paths from the current cell has its relation with the possible paths from the cell left,right and above. But as we know we can find all possible paths from only one cell in a row and rest of cells' possible paths will be some movements in the same row followed by a suffix of possible paths from that cell.
Maybe I will give you a example
+---+---+---+
| 1 | 1 | * |
+---+---+---+
| | | |
+---+---+---+
| | | |
+---+---+---+
You know all possible paths from cells in the first row. You need to find the same in the second row. So a good strategy would be to do it for the right most cell
+---+---+---+
| > | > | * |
+---+---+---+
| ^ | < | < |
+---+---+---+
| | | |
+---+---+---+
+---+---+---+
| | > | * |
+---+---+---+
| | ^ | < |
+---+---+---+
| | | |
+---+---+---+
+---+---+---+
| | | * |
+---+---+---+
| | | ^ |
+---+---+---+
| | | |
+---+---+---+
Now finding this for rest of the cells in the same row is trivial using these as I have told before.
In the end if you have m X n matrix the number of paths from bottom-left corner to top-right corner will be n^(m-1).
Another way
This way is not very optimal but easy to implement. Consider m X n grid
Find the path of longest length. You dont need the exact path just the number of <,>,^.
You can find the direct formula in terms of m and n.
Like
^ = m - 1
< = (n-1) * floor((m-1)/2)
> = (n-1) * (floor((m-1)/2) + 1)
Any valid path will be a prefix of the permutations of this which you can search exhaustively. Use permutations from Data.List to get all possible permutations. Then make a function which given a path strips a valid path from this. map this over the list of permutations and remove duplicates. The thing to note is path will be a prefix of what you get from permutation, so there can be several permutations for the same path.
Can you create that matrix and define the "fields"? Even if you can't (a specific matrix is given), you can map an [(Int, Int)] matrix (which sounds reasonable for this kind of task) to your own representation.
Since you didn't specify what your skill level was, I hope you don't mind that I suggest that you first try to create some kind of a grid in order to have something to work on:
data Status = Free | Left | Right | Up
deriving (Read, Show, Eq)
type Position = (Int, Int)
type Field = (Position, Status)
type Grid = [Field]
grid :: Grid
grid = [((x, y), stat) | x <- [1..10], y <- [1..10], let stat = Free]
Of course there are other ways to achieve this. Afterwards you can define some movement, map Position to Grid index and Statuses to printable characters... Try to fiddle with it and you might get some ideas.
There is a N x M board we should paint. We can paint either an entire row or an entire column at once. Given an N x M matrix of colours of all board cells find the minimal number of painting operations to paint the board.
For example: we should paint a 3 x 3 board as follows (R - red, B - blue, G - green):
B, B, B
B, R, R
B, G, G
The minimal number of painting operations is 4:
Paint row 0 with Blue
Paint row 1 with Red
Paint row 2 with Green
Paint column 0 with Blue
How would you solve it ?
This looks like a fun problem. Let me take a shot at it with some pseudocode.
Function MinPaints(Matrix) Returns Integer
If the matrix is empty return 0
Find all rows and columns which have a single color
If there are none, return infinity, since there is no solution
Set the current minimum to infinity
For each row or column with single color:
Remove the row/column from the matrix
Call MinPaints with the new matrix
If the result is less than the current minimum, set the current minimum to the result
End loop
Return the current minimum + 1
End Function
I think that will solve your problem, but I didn't try any optimization or anything. This may not be fast enough though, I don't know. I doubt this problem is solvable in sub-exponential time.
Here is how this algorithm would solve the example:
BBB
BRR
BGG
|
+---BRR
| BGG
| |
| +---RR
| | GG
| | |
| | +---GG
| | | |
| | | +---[]
| | | | |
| | | | Solvable in 0
| | | |
| | | Solvable in 1
| | |
| | +---RR
| | | |
| | | +---[]
| | | | |
| | | | Solvable in 0
| | | |
| | | Solvable in 1
| | |
| | Solvable in 2
| |
| Solvable in 3
| BB
+---Another branch with RR ...
| GG
Solvable in 4
For starters, you can try an informed exhaustive search.
Let your states graph be: G=(V,E) where V = {all possible boards} and E = {(u,v) | you can move from board u to v within a single operation}.
Note that you do not need to generate the graph in advance - you can generate it on the fly, using a successors(board) function, that return all the successors of the given board.
You will also need h:V->R - an admissible heuristic function that evaluates the board1.
Now, you can run A*, or bi-directional BFS search [or combination of both], your source will be a white board, and your target is the requested board. Because we use admissible heuristic function - A* is both complete (always finds a solution if one exists) and optimal (finds the shortest solution), it will find the best solution. [same goes for bi-directional BFS].
drawbacks:
Though the algorithm is informed, it will have exponential behavior.
But if it is an interview question, I believe a non-efficient
solution is better then no solution.
Though complete and optimal - if there is no solution - the algorithm may be stuck in an infinite loop, or a very long loop at the very least until it realizes it has exuahsted all possibilities.
(1) example for admissible heuristic is h(board) = #(miscolored_squares)/max{m,n}
A number of students want to get into sections for a class, some are already signed up for one section but want to change section, so they all get on the wait lists. A student can get into a new section only if someone drops from that section. No students are willing to drop a section they are already in unless that can be sure to get into a section they are waiting for. The wait list for each section is first come first serve.
Get as many students into their desired sections as you can.
The stated problem can quickly devolve to a gridlock scenario. My question is; are there known solutions to this problem?
One trivial solution would be to take each section in turn and force the first student from the waiting list into the section and then check if someone end up dropping out when things are resolved (O(n) or more on the number of section). This would work for some cases but I think that there might be better options involving forcing more than one student into a section (O(n) or more on the student count) and/or operating on more than one section at a time (O(bad) :-)
Well, this just comes down to finding cycles in the directed graph of classes right? each link is a student that wants to go from one node to another, and any time you find a cycle, you delete it, because those students can resolve their needs with each other. You're finished when you're out of cycles.
Ok, lets try. We have 8 students (1..8) and 4 sections. Each student is in a section and each section has room for 2 students. Most students want to switch but not all.
In the table below, we see the students their current section, their required section and the position on the queue (if any).
+------+-----+-----+-----+
| stud | now | req | que |
+------+-----+-----+-----+
| 1 | A | D | 2 |
| 2 | A | D | 1 |
| 3 | B | B | - |
| 4 | B | A | 2 |
| 5 | C | A | 1 |
| 6 | C | C | - |
| 7 | D | C | 1 |
| 8 | D | B | 1 |
+------+-----+-----+-----+
We can present this information in a graph:
+-----+ +-----+ +-----+
| C |---[5]--->1| A |2<---[4]---| B |
+-----+ +-----+ +-----+
1 | | 1
^ | | ^
| [1] [2] |
| | | |
[7] | | [8]
| V V |
| 2 1 |
| +-----+ |
\--------------| D |--------------/
+-----+
We try to find a section with a vacancy, but we find none. So because all sections are full, we need a dirty trick. So lets take a random section with a non empty queue. In this case section A and assume, it has an extra position. This means student 5 can enter section A, leaving a vacancy at section C which is taken by student 7. This leaves a vacancy in section D which is taken by student 2. We now have a vacancy at section A. But we assumed that section A has an extra position, so we can remove this assumption and have gained a simpler graph.
If the path never returned to section A, undo the moves and mark A as an invalid startingpoint. Retry with another section.
If there are no valid sections left we are finished.
Right now we have the following situation:
+-----+ +-----+ +-----+
| C | | A |1<---[4]---| B |
+-----+ +-----+ +-----+
| 1
| ^
[1] |
| |
| [8]
V |
1 |
+-----+ |
| D |--------------/
+-----+
We repeat the trick with another random section, and this solves the graph.
If you start with several students currently not assigned, you add an extra dummy section as their startingpoint. Of course, this means that there must be vacancies in any sections or the problem is not solvable.
Note that due to the order in the queue, it can be possible that there is no solution.
This is actually a Graph problem. You can think of each of these waiting list dependencies as edges on a directed graph. If this graph has a cycle, then you have one of the situations you described. Once you have identified a cycle, you can chose any point to "break" the cycle by "over filling" one of the classes, and you will know that things will settle correctly because there was a cycle in the graph.