Related
Here's the problem:
X is a positive integer (include 0) set which has n different elements I know in advance. All of them is less equal than m. And I want to have an occ-free hash function as simple as possible to map them to 0-n-1.
For example:
X = [31,223,121,100,123,71], so n = 6, m = 223.
I want to find a hash function to map them to [0, 1, 2, 3, 4, 5].
If mapping to 0-n-1 is too difficult, then how to mapping X to a small range is also a problem.
Finding such a function is not too difficult, but to be simple and easy to be generated is hard.
It's better to preserve the order of the X.
Any clues?
My favorite perfect hash is pretty easy.
The hash function you generate has the form:
hash = table1[h1(key)%N] + table2[h2(key)%N]
h1 and h2 are randomly generated hash functions. In your case, you can generate random constants and then have h1(key)=key*C1/m and h2(key)=key*C2/m or something similarly simple
To generated the perfect hash:
Generate random constants C1 and C2
Imagine the bipartite graph, with table1 slots and table2 slots as vertices and an edge for each key between table1[h1(key)%N] and table2[h2(key)%N]. Run a DFS to see if the graph is acyclic. If not, go back to step 1.
Now that you have an acyclic graph, start at any key/edge in each connected component, and set its slots in table1 and table2 however you like to give it whatever hash you like.
Traverse the tree starting at the vertices adjacent to the edge you just set. For every edge you traverse, one of its slots will already be set. Set the other one to make the hash value come out however you like.
That's it. All of steps (2), (3) and (4) can be combined into a single DFS traversal pretty easily.
The complete description and analysis is in this paper.
I am trying to implement the Parallel Algorithm for Longest Common Subsequence Problem described in http://www.iaeng.org/publication/WCE2010/WCE2010_pp499-504.pdf
But i am having a problem with the variable C in Equation 6 on page 4
The paper refered to C on at the end of page 3 as
C as Let C[1 : l] bethe finite alphabet
I am not sure what is ment by this, as i guess it would it with the 2 strings ABCDEF and ABQXYEF be ABCDEFQXY. But what if my 2 stings is a list of objects (Where my match test for an example is obj1.Name = obj2.Name), what would my C be here? just a union on the 2 arrays?
Having read and studied the paper, I can say that C is supposed to be an array holding the alphabet of your strings, where the alphabet size (and, thus, the size of C) is l.
By the looks of your question, however, I feel the need to go deeper on this, because it looks like you didn't get the whole picture yet. What is P[i,j], and why do you need it? The answer is that you don't really need it, but it's an elegant optimization. In page 3, a little bit before Theorem 1, it is said that:
[...] This process ends when j-k = 0 at the k-th step, or a(i) =
b(j-k) at the k-th step. Assume that the process stops at the k-th
step, and k must be the minimum number that makes a(i) = b(j-k) or j-k
= 0. [...]
The recurrence relation in (3) is equivalent to (2), but the fundamental difference is that (2) expands recursively, whereas with (3) you never have recursive calls, provided that you know k. In other words, the magic behind (3) not expanding recursively is that you somehow know the spot where the recursion on (2) would stop, so you look at that cell immediately, rather than recursively approaching it.
Ok then, but how do you find out the value for k? Since k is the spot where (2) reaches a base case, it can be seen that k is the amount of columns that you have to "go back" on B until you are either off the limits (i.e., the first column that is filled with 0's) OR you find a match between a character in B and a character in A (which corresponds to the base case conditions in (2)). Remember that you will be matching the character a(i-1), where i is the current row.
So, what you really want is to find the last position in B before j where the character a(i-1) appears. If no such character ever appears in B before j, then that would be equivalent to reaching the case i = 0 or j-1 = 0 in (2); otherwise, it's the same as reaching a(i) = b(j-1) in (2).
Let's look at an example:
Consider that the algorithm is working on computing the values for i = 2 and j = 3 (the row and column are highlighted in gray). Imagine that the algorithm is working on the cell highlighted in black and is applying (2) to determine the value of S[2,2] (the position to the left of the black one). By applying (2), it would then start by looking at a(2) and b(2). a(2) is C, b(2) is G, to there's no match (this is the same procedure as the original, well-known algorithm). The algorithm now wants to find the value of S[2,2], because it is needed to compute S[2,3] (where we are). S[2,2] is not known yet, but the paper shows that it is possible to determine that value without refering to the row with i = 2. In (2), the 3rd case is chosen: S[2,2] = max(S[1, 2], S[2, 1]). Notice, if you will, that all this formula is doing is looking at the positions that would have been used to calculate S[2,2]. So, to rephrase that: we're computing S[2,3], we need S[2,2] for that, we don't know it yet, so we're going back on the table to see what's the value of S[2,2] in pretty much the same way we did in the original, non-parallel algorithm.
When will this stop? In this example, it will stop when we find the letter C (this is our a(i)) in TGTTCGACA before the second T (the letter on the current column) OR when we reach column 0. Because there is no C before T, we reach column 0. Another example:
Here, (2) would stop with j-1 = 5, because that is the last position in TGTTCGACA where C shows up. Thus, the recursion reaches the base case a(i) = b(j-1) when j-1 = 5.
With this in mind, we can see a shortcut here: if you could somehow know the amount k such that j-1-k is a base case in (2), then you wouldn't have to go through the score table to find the base case.
That's the whole idea behind P[i,j]. P is a table where you lay down the whole alphabet vertically (on the left side); the string B is, once again, placed horizontally in the upper side. This table is computed as part of a preprocessing step, and it will tell you exactly what you will need to know ahead of time: for each position j in B, it says, for each character C[i] in C (the alphabet), what is the last position in B before j where C[i] is found (note that i is used to index C, the alphabet, and not the string A. Maybe the authors should have used another index variable to avoid confusion).
So, you can think of the semantics for an entry P[i,j] as something along the lines of: The last position in B where I saw C[i] before position j. For example, if you alphabet is sigma = {A, E, I, O, U}, and B = "AOOIUEI", thenP` is:
Take the time to understand this table. Note the row for O. Remember: this row lists, for every position in B, where is the last known "O". Only when j = 3 will we have a value that is not zero (it's 2), because that's the position after the first O in AOOIUEI. This entry says that the last position in B where O was seen before is position 2 (and, indeed, B[2] is an O, the one that follows A). Notice, in that same row, that for j = 4, we have the value 3, because now the last position for O is the one that correspnds to the second O in B (and since no more O's exist, the rest of the row will be 3).
Recall that building P is a preprocessing step necessary if you want to easily find the value of k that makes the recursion from equation (2) stop. It should make sense by now that P[i,j] is the k you're looking for in (3). With P, you can determine that value in O(1) time.
Thus, the C[i] in (6) is a letter of the alphabet - the letter that we are currently considering. In the example above, C = [A,E,I,O,U], and C[1] = A, C[2] = E, etc. In equaton (7), c is the position in C where a(i) (the current letter of string A being considered) lives. It makes sense: after all, when building the score table position S[i,j], we want to use P to find the value of k - we want to know where was the last time we saw an a(i) in B before j. We do that by reading P[index_of(a(i)), j].
Ok, now that you understand the use of P, let's see what's happening with your implementation.
About your specific case
In the paper, P is shown as a table that lists the whole alphabet. It is a good idea to iterate through the alphabet because the typical uses of this algorithm are in bioinformatics, where the alphabet is much, much smaller than the string A, making the iteration through the alphabet cheaper.
Because your strings are sequences of objects, your C would be the set of all possible objects, so you'd have to build a table P with the set of all possible object instance (nonsense, of course). This is definitely a case where the alphabet size is huge when compared to your string size. However, note that you will only be indexing P in those rows that correspond to letters from A: any row in P for a letter C[i] that is not in A is useless and will never be used. This makes your life easier, because it means you can build P with the string A instead of using the alphabet of every possible object.
Again, an example: if your alphabet is AEIOU, A is EEI and B is AOOIUEI, you will only be indexing P in the rows for E and I, so that's all you need in P:
This works and suffices, because in (7), P[c,j] is the entry in P for the character c, and c is the index of a(i). In other words: C[c] always belongs to A, so it makes perfect sense to build P for the characters of A instead of using the whole alphabet for the cases where the size of A is considerably smaller than the size of C.
All you have to do now is to apply the same principle to whatever your objects are.
I really don't know how to explain it any better. This may be a little dense at first. Make sure to re-read it until you really get it - and I mean every little detail. You have to master this before thinking about implementing it.
NOTE: You said you were looking for a credible and / or official source. I'm just another CS student, so I'm not an official source, but I think I can be considered "credible". I've studied this before and I know the subject. Happy coding!
This question already has answers here:
Dynamic programming - Largest square block
(7 answers)
Closed 1 year ago.
I would like to know the different algorithms to find the biggest square in a two dimensions map dotted with obstacles.
An example, where o would be obstacles:
...........................
....o......................
............o..............
...........................
....o......................
...............o...........
...........................
......o..............o.....
..o.......o................
The biggest square would be (if we choose the first one):
.....xxxxxxx...............
....oxxxxxxx...............
.....xxxxxxxo..............
.....xxxxxxx...............
....oxxxxxxx...............
.....xxxxxxx...o...........
.....xxxxxxx...............
......o..............o.....
..o.......o................
What would be the fastest algorithm to find it? The one with the smallest complexity?
EDIT: I know that people are interested on the algorithm explained in the accepted answer, so I made a document that explains it a bit more, you can find it here:
https://docs.google.com/document/d/19pHCD433tYsvAor0WObxa2qusAjKdx96kaf3z5I8XT8/edit?usp=sharing
Here is how to do this in the optimal amount of time, O(nm). This is built on top of #dukeling's insight that you never need to check a solution of size less than your current known best solution.
The key is to be able to build a data structure that can answer this query in O(1) time.
Is there an obstacle in the square whose top left corner is at r, c and has size k?
To solve that problem, we'll support answering a slightly harder question, also in O(1).
What is the count of items in the rectangle from r1, c1 to r2, c2?
It's easy to answer the square existence question with an answer from the rectangle count question.
To answer the rectangle count question, note that if you had pre-computed the answer for every rectangle that starts in the top left, then you could answer the general question for from r1, c1 to r2, c2 by a kind of clever/inclusion exclusion tactic using only rectangles that start in the top left
c1 c2
-----------------------
| | | |
| A | B | |
|_____________|____| | r1
| | | |
| C | D | |
|_____________|____| | r2
|_____________________|
We want the count of stuff inside D. In terms of our pre-computed counts from the top left.
Count(D) = Count(A ∪ B ∪ C ∪ D) - Count(A ∪ C) - Count(A ∪ B) + Count(A)
You can pre-compute all the top left rectangles in O(nm) by doing some clever row/column partial sums, but I'll leave that to you.
Then to answer the to the problem you want just involves checking possible solutions, starting with solutions that are at least as good as your known best. Your known best will only get better up to min(n, m) times total, so the best_possible increment will happen very rarely and almost all squares will be rejected in O(1) time.
best_possible = 0
for r in range(n):
for c in range(m):
while True:
# this looks O(min(n, m)), but it's amortized O(1) since best_possible
# rarely increased.
if possible(r, c, best_possible+1):
best_possible += 1
else:
break
One idea, making use of binary search.
The basic idea:
Start off in the top-left corner. See if a 1x1 square would work.
If it will work, increase the sides lengths of the square by 1 and repeat.
If it won't work, move right and repeat. If you've reached the right-most position, move to the next line.
The native approach:
We can simply check every possible cell of every square at each step, but this is fairly inefficient.
The optimized approach:
When increasing the square size, we can just do a binary search over the next row and column to see if that row / column contains an obstacle at any of those positions.
When moving to the right, we can do a binary search for each next column to determine if that column contains an obstacle at any of those positions.
When moving down, we can do a similar binary on each of the columns in the target position.
Implementation note:
To start off, we'd need to go through all the rows and columns and set up arrays containing the positions of the obstacles for each of them, which we can use for the binary searches.
Running time:
We do 2 binary searches to increase the square size, and the square size is maximum the size of the grid, so that is fairly small (O(min(m,n) log max(m,n))) and gets dominated by the below.
Beyond that, for each position, we do a single binary search on a column.
So, for a grid with m columns and n rows, the overall complexity is O(mn log m).
But note how little we're actually searching below when the grid is sparse.
Example:
For your example:
012345678901234567890123456
0...........................
1....o......................
2............o..............
3...........................
4....o......................
5...............o...........
6...........................
7......o..............o.....
8..o.......o................
We'd first try a 1x1 square in the top-left corner, which works.
Then a 2x2 square. For this, we do a binary search for the range [0,1] on the row 1, which can be represented simply by {4} - an array of a single position corresponding to where the obstacle is. And we also do a binary search for the range [0,1] on the column 1, which contains no obstacles, thus an empty array - {}.
Then a 3x3 square. For this, we do a binary search for [0,2] on the row 2, which contains 1 obstacles at position 12, thus {12}. And we also do a binary search for [0,2] on the column 2, which contains an obstacle at position 8, thus {8}.
Then a 4x4 square. For this, we do a binary search for [0,3] on the row 3 - {}. And for [0,3] on column 3 - {}.
Then a 5x5 square. For this, we do a binary search for [0,4] on the row 4 - {4}. And for [0,4] column 4 - {1,4}.
Here is the first one we actually find. In the range [0,4], we find 4 in both the row and the column (we only really need to find one of them). So this indicates a fail.
From here we do a binary search on column 4 (again - not really necessary) for [0,4]. Then binary search columns 5-8 for [0,4], none of them found, so a square starting at position 5,0 is the next possible candidate.
So from here we try to increase the square size to 5x5, which works, then 6x6 and 7x7, which works.
Then we try 8x8, which doesn't work.
And so on.
I know binary search, but how does yours work?
So we're basically doing a range search within a set of values. This is fairly easy to do. First search for the starting value of the range, then the end value. If we get to the same point, there are no values in the range.
We don't really care what values exist in the range, just whether or not there are any.
So here's one rough approach.
Store the x-y positions of all the obstacles.
For each obstacle O
find obstacle C that is nearest to it column-wise.
find obstacle R-top that is nearest to it row-wise from the top.
find obstacle R-bottom that is nearest to it row-wise from the bottom.
if (|R-top.y - R-bottom.y| != |O.x - C.x|) continue
Size of the square = Abs((R-top.y - R-bottom.y) * (O.x - C.x))
Keep track of the sizes and positions to find the largest square
Complexity is roughly O(k^2) where k is the number of obstacles. You could reduce it to O(k * log k) if you use binary search.
The following SO articles are identical/similar to the problem you're trying to solve. You may want to look over those answers as well as the responses to your question.
Dynamic programming - Largest square block
dynamic programming: finding largest non-overlapping squares
Dynamic programming: Find largest diamond (rhombus)
Here's the baseline case I'd use, written in simplified Python/pseudocode.
# obstacleMap is a list of list of MapElements, stored in row-major order
max([find_largest_rect(obstacleMap, element) for row in obstacleMap for element in row])
def find_largest_rect(obstacleMap, upper_left_elem):
size = 0
while not has_obstacles(obstacleMap, upper_left_elem, size+1):
size += 1
return size
def has_obstacles(obstacleMap, upper_left_elem, size):
#determines if there are obstacles on the on outside square layer
#for example, if U is the upper left element and size=3, then has_obstacles checks the elements marked p.
# .....
# ..U.p
# ....p
# ..ppp
periphery_row = obstacleMap[upper_left_elem.row][upper_left_elem.col:upper_left_elem.col+size]
periphery_col = [row[upper_left_elem.col+size] for row in obstacleMap[upper_left_elem.row:upper_left_elem.row+size]
return any(is_obstacle(elem) for elem in periphery_row + periphery_col)
def is_obstacle(elem):
return elem.value == 'o'
class MapElement(object):
def __init__(self, row, col, value):
self.row = row
self.col = col
self.value = value
here is an approach using recurrence relation :-
isSquare(R,C1,C2) = noObstacle(R,C1,R,C2) && noObstacle(R,C2,R-(C2-C1),C2) && isSquare(R-1,C1,C2-1)
isSquare(R,C1,C2) = square that has bottom side (R,C1) to (R,C2)
noObstacle(R1,C1,R2,C2) = checks whether there is no obstacle in line segment (R1,C1) to (R2,C2)
Find Max (C2-C1+1) which where isSquare(R,C1,C2) = true
You can use dynamic programming to solve this problem in polynomial time. Use suitable data structure for searching obstacle.
Let CFG G be:
S −→ AB|BA|AC|BD|EE
A −→ a
B −→ b
C −→ EB
D −→ EA
E −→ AB|BA|AC|BD|EE
How do I use the CYK algorithm to determine if the string aabbab is part of the language?
This is the pseudo code I have in my notes:
for i in 1 .. n
V[i,1] = { A | A -> x[i] }
for j in 2..n
for i in 1 .. n-j+1
{
V[i,j] = phi
for k in 1 .. j-1
V[i,j] = V[i,j] union { A | A -> BC where B in V[i,k]
and C in V[i+k,j-k]}
}
But I am not understanding how the answer got to be in an upside down triangular shape.
For example,
V[i,j] i
1(b) 2(a) 3(a) 4(b) 5(a)
1 B A,C A,C B A,C
2 S,A B S,C S,A
j
3 phi B B
4 phi S,A,C
5 S,A,C
^
|_ accept
The pseudocode[*] describes how to apply the algorithm to create the chart.
The [i, j] pair refers to a substring of the input that starts at the ith symbol and extends for j symbols. So [2, 3] refers to a 3-symbol substring, starting at symbol 2. If your input is baaba, then [2, 3] refers to the aab in the middle. (The indexes are 1-based, not 0-based.)
The chart forms a triangle because you can't have a substring that's longer than the input. If the input is 5 symbols long, then you can have a value in [1, 5], but you can't have [2, 5] because that wouldn't refer to a substring anymore. So each row is one box shorter than the row before it, forming the triangle.
V[i, j] refers to a box in the chart. Each box is the set of non-terminals that may have produced the substring described by [i, j].
The algorithm relies on the grammar being in Chomsky Normal Form. In CNF, the right side of each production is either one terminal symbol or two non-terminal symbols. (There's another algorithm that can transform a context-free grammar into CNF.)
Basically, you start with all the 1-symbol substrings of the input. The first loop in your pseudocode fills out the top row (j == 1) of your chart. It looks at all the productions in the grammar, and, if the right side of a production corresponds to that symbol, then the non-terminal on the left side of that production is added to the set V[i, 1]. (Your example seems to have some bogus entries in the first row. The {A, C} sets should be just {A}.)
The algorithm then proceeds through the rest of the rows, looking for all the possible productions that can produce the corresponding substring. For each possible way to split the current substring into two, it looks for a corresponding production. This involves combining pairs of non-terminals from certain boxes on previous rows and checking if there are any productions that produce that pair, thus building a set of non-terminals for that box.
If the box in the last row ends up with a set that contains the start symbol, then the input is valid according to the grammar. Intuitively, it says that the start symbol is a valid production for making the substring that starts at the first symbol and proceeds for the entire length.
[*] It looks like the pseudocode shown in the question contains some transcription errors. You'll want to consult an authoritative source to get the details right.
I have a symmetric matrix like shown in the image attached below.
I've made up the notation A.B which represents the value at grid point (A, B). Furthermore, writing A.B.C gives me the minimum grid point value like so: MIN((A,B), (A,C), (B,C)).
As another example A.B.D gives me MIN((A,B), (A,D), (B,D)).
My goal is to find the minimum values for ALL combinations of letters (not repeating) for one row at a time e.g for this example I need to find min values with respect to row A which are given by the calculations:
A.B = 6
A.C = 8
A.D = 4
A.B.C = MIN(6,8,6) = 6
A.B.D = MIN(6, 4, 4) = 4
A.C.D = MIN(8, 4, 2) = 2
A.B.C.D = MIN(6, 8, 4, 6, 4, 2) = 2
I realize that certain calculations can be reused which becomes increasingly important as the matrix size increases, but the problem is finding the most efficient way to implement this reuse.
Can point me in the right direction to finding an efficient algorithm/data structure I can use for this problem?
You'll want to think about the lattice of subsets of the letters, ordered by inclusion. Essentially, you have a value f(S) given for every subset S of size 2 (that is, every off-diagonal element of the matrix - the diagonal elements don't seem to occur in your problem), and the problem is to find, for each subset T of size greater than two, the minimum f(S) over all S of size 2 contained in T. (And then you're interested only in sets T that contain a certain element "A" - but we'll disregard that for the moment.)
First of all, note that if you have n letters, that this amounts to asking Omega(2^n) questions, roughly one for each subset. (Excluding the zero- and one-element subsets and those that don't include "A" saves you n + 1 sets and a factor of two, respectively, which is allowed for big Omega.) So if you want to store all these answers for even moderately large n, you'll need a lot of memory. If n is large in your applications, it might be best to store some collection of pre-computed data and do some computation whenever you need a particular data point; I haven't thought about what would work best, but for example computing data only for a binary tree contained in the lattice would not necessarily help you anything beyond precomputing nothing at all.
With these things out of the way, let's assume you actually want all the answers computed and stored in memory. You'll want to compute these "layer by layer", that is, starting with the three-element subsets (since the two-element subsets are already given by your matrix), then four-element, then five-element, etc. This way, for a given subset S, when we're computing f(S) we will already have computed all f(T) for T strictly contained in S. There are several ways that you can make use of this, but I think the easiest might be to use two such subset S: let t1 and t2 be two different elements of T that you may select however you like; let S be the subset of T that you get when you remove t1 and t2. Write S1 for S plus t1 and write S2 for S plus t2. Now every pair of letters contained in T is either fully contained in S1, or it is fully contained in S2, or it is {t1, t2}. Look up f(S1) and f(S2) in your previously computed values, then look up f({t1, t2}) directly in the matrix, and store f(T) = the minimum of these 3 numbers.
If you never select "A" for t1 or t2, then indeed you can compute everything you're interested in while not computing f for any sets T that don't contain "A". (This is possible because the steps outlined above are only interesting whenever T contains at least three elements.) Good! This leaves just one question - how to store the computed values f(T). What I would do is use a 2^(n-1)-sized array; represent each subset-of-your-alphabet-that-includes-"A" by the (n-1) bit number where the ith bit is 1 whenever the (i+1)th letter is in that set (so 0010110, which has bits 2, 4, and 5 set, represents the subset {"A", "C", "D", "F"} out of the alphabet "A" .. "H" - note I'm counting bits starting at 0 from the right, and letters starting at "A" = 0). This way, you can actually iterate through the sets in numerical order and don't need to think about how to iterate through all k-element subsets of an n-element set. (You do need to include a special case for when the set under consideration has 0 or 1 element, in which case you'll want to do nothing, or 2 elements, in which case you just copy the value from the matrix.)
Well, it looks simple to me, but perhaps I misunderstand the problem. I would do it like this:
let P be a pattern string in your notation X1.X2. ... .Xn, where Xi is a column in your matrix
first compute the array CS = [ (X1, X2), (X1, X3), ... (X1, Xn) ], which contains all combinations of X1 with every other element in the pattern; CS has n-1 elements, and you can easily build it in O(n)
now you must compute min (CS), i.e. finding the minimum value of the matrix elements corresponding to the combinations in CS; again you can easily find the minimum value in O(n)
done.
Note: since your matrix is symmetric, given P you just need to compute CS by combining the first element of P with all other elements: (X1, Xi) is equal to (Xi, X1)
If your matrix is very large, and you want to do some optimization, you may consider prefixes of P: let me explain with an example
when you have solved the problem for P = X1.X2.X3, store the result in an associative map, where X1.X2.X3 is the key
later on, when you solve a problem P' = X1.X2.X3.X7.X9.X10.X11 you search for the longest prefix of P' in your map: you can do this by starting with P' and removing one component (Xi) at a time from the end until you find a match in your map or you end up with an empty string
if you find a prefix of P' in you map then you already know the solution for that problem, so you just have to find the solution for the problem resulting from combining the first element of the prefix with the suffix, and then compare the two results: in our example the prefix is X1.X2.X3, and so you just have to solve the problem for
X1.X7.X9.X10.X11, and then compare the two values and choose the min (don't forget to update your map with the new pattern P')
if you don't find any prefix, then you must solve the entire problem for P' (and again don't forget to update the map with the result, so that you can reuse it in the future)
This technique is essentially a form of memoization.