Given a matrix of 1's and 0's, I want to find a combination of rows and columns with least or none 0's, maximizing the n_of_rows * n_of_columns picked.
For example, rows (0,1,2) and columns (0,1,3) have only one zero in col #0 row #1, and the rest 8 values are 1's.
1 1 0 1 0
0 1 1 1 0
1 1 0 1 1
0 0 1 0 0
Pracical task is to search over 1000's to 1000000's of rows and columns, finding the maximal biclique in a bipartite graph – rows and cols can be viewed as verticles, and values as connections.
The problem in NP-complete, as far as I learned.
Please advice an approach / algorithm that would speed up the task and reduce requirements to CPU and memory.
Not sure you could minimise thism
However, easy way to work this out would be...
Multiple your matrix by a 1 column and n rows full of 1's. This will give you number of ones in each row. Next do a 1 row by n columns multiplcation (at frot of) your matrix full of 1's. This will give you totals of 1's for each column, From there it's a pretty easy compairson........
ie original matrix...
1 0 1
0 1 1
0 0 0
do
1 0 1 x 1 = 2 (row totals)
o 1 1 1 2
0 0 0 1 0
do
1 1 1 x 1 0 1 = 1 (Column totals)
0 1 1 2
0 0 0 0
nb max sum is 2 (which you would keep track of as you work it out.
Actually given the following assumptions:
1. You don't care how many 0's are in each row or column
2. You don't need to keep track of their order....
Then you only really need to store values to count the total in each row/column as you read the values in and don't actually store the matrix itself.
If you are given the number of rows and columns prior to reading in the matrix you can do the following heuristics to reduce computational time...
Keep track of the current max. If the current row cannot reach this potential max stop counting for the row (but continue in the columns). Vice versa is true for the columns
But you still have a worst case scenario in which all rows and columns have sme number of 1's and 0's.... :)
Related
An algorithm that need process a matrix n x m that is scalable.
E.g. I have a time series of 3 seconds containing the values: 2,1,4.
I need to decompose it to take a 3 x 4 matrix, where 3 is the number of elements of time series and 4 the maximum value. The resulting matrix that would look like this:
1 1 1
1 0 1
0 0 1
0 0 1
Is this a bad solution or is it only considered a data entry problem?
The question is,
do I need to distribute information from each row of the matrix for various elements without losing the values?
I've been searching for an algorithm for the solution of all possible matrices of dimension 'n' that can be obtained with two arrays, one of the sum of the rows, and another, of the sum of the columns of a matrix. For example, if I have the following matrix of dimension 7:
matriz= [ 1 0 0 1 1 1 0
1 0 1 0 1 0 0
0 0 1 0 1 0 0
1 0 0 1 1 0 1
0 1 1 0 1 0 1
1 1 1 0 0 0 1
0 0 1 0 1 0 1 ]
The sum of the columns are:
col= [4 2 5 2 6 1 4]
The sum of the rows are:
row = [4 3 2 4 4 4 3]
Now, I want to obtain all possible matrices of "ones and zeros" where the sum of the columns and the rows fulfil the condition of "col" and "row" respectively.
I would appreciate ideas that can help solve this problem.
One obvious way is to brute-force a solution: for the first row, generate all the possibilities that have the right sum, then for each of these, generate all the possibilities for the 2nd row, and so on. Once you have generated all the rows, you check if the sum of the columns is right. But this will take a lot of time. My math might be rusty at this time of the day, but I believe the number of distinct possibilities for a row of length n of which k bits are 1 is given by the binomial coefficient or nchoosek(n,k) in Matlab. To determine the total number of possibilities, you have to multiply this number for every row:
>> n = 7;
>> row= [4 3 2 4 4 4 3];
>> prod(arrayfun(#(k) nchoosek(n, k), row))
ans =
3.8604e+10
This is a lot of possibilities to check! Doing the same for the columns gives
>> col= [4 2 5 2 6 1 4];
>> prod(arrayfun(#(k) nchoosek(n, k), col))
ans =
555891525
Still a large number, but 'only' a factor 70 smaller.
It might be possible to improve this brute-force method a little bit by seeing if the later rows are already constrained by the previous rows. If in your example, for a particular combination of the first two rows, both rows have a 1 in the second column, the rest of this column should all be 0, since the sum must be 2. This reduces the number of possibilities for the remaining rows a bit. Implementing such checks might complicate things a bit, but they might make the difference between a calculation that takes 2 days or one that takes just 1 hour.
An optimized version of this might alternatively generate rows and columns, and start with those for which the number of possibilities is the lowest. I don't know if there is a more elegant solution than this brute-force method, I would be interested to hear one.
I need to assign random papers to students of a class, but I have the constraints that:
Each student should have two papers assigned.
Each paper should be assigned to (approximately) the same number of students.
Is there an elegant way to generate a matrix that has this property? i.e. it is shuffled but the row and column sums are constant? As an illustration:
Student A 1 0 0 1 1 0 | 3
Student B 1 0 1 0 0 1 | 3
Student C 0 1 1 0 1 0 | 3
Student D 0 1 0 1 0 1 | 3
----------------
2 2 2 2 2 2
I thought of first building an "initial matrix" with the right row/column sum, then randomly permuting first the rows, then the colums, but how do I generate this initial matrix? The problem here is that I'd be choosing between (e.g.) the following alternatives, and the fact that there are two students with the same pair of papers assigned (in the left setup) won't change through row/column shuffling:
INITIAL (MA): OR (MB):
A 1 1 1 0 0 0 || 1 1 1 0 0 0
B 1 1 1 0 0 0 || 0 1 1 1 0 0
C 0 0 0 1 1 1 || 0 0 0 1 1 1
D 0 0 0 1 1 1 || 1 0 0 0 1 1
I know I could come up with something quick/dirty and just tweak where necessary but it seemed like a fun exercise.
If you want to make permutations, what about:
Chose randomly a student, say student 1
For this student, chose a random paper he has, say paper A
Chose randomly another student
For this student, chose a random paper he has, say paper B (different from A)
Give paper B to student 1 and paper A to student 2.
That way, you preserve both the number of different papers and the number of papers per student. Indeed, both students give one paper and receive one back. Moreover, no paper is created nor deleted.
In term of table, it means finding two pairs of indices(i1,i2) and (j1,j2) such that A(i1,j1) = 1, A(i2,j2)=1, A(i1,j2)=0 and A(i2,j1)=0 and changing the 0s for 1s and the 1s for 0s => The sums of the rows and columns do not change.
Remark 1: If you do not want to proceed by permutations, you can simply put in a vector all the paper (put 2 times paper A, 2 times paper B,...). Then, random shuffle the vector and attribute the k first to the first student, the k next ones to student 2, ... However, you can end with a student having several times the same paper. In this case, make some permutations starting with the surnumerary papers.
You can generate the initial matrix as follows (pseudo-Python syntax):
column_sum = [0] * n_students
for i in range(n_students):
if column_sum[i] < max_allowed:
for j in range(i + 1, n_students):
if column_sum[j] < max_allowed:
generate_row_with_ones_at(i, j)
column_sum[i] += 1
column_sum[j] += 1
if n_rows == n_wanted:
return
This is a straightforward iteration over all n choose 2 distinct rows, but with the constraint on column sums enforced as early as possible.
This was one of my interview questions.
We have a matrix containing integers (no range provided). The matrix is randomly populated with integers. We need to devise an algorithm which finds those rows which match exactly with a column(s). We need to return the row number and the column number for the match. The order of of the matching elements is the same. For example, If, i'th row matches with j'th column, and i'th row contains the elements - [1,4,5,6,3]. Then jth column would also contain the elements - [1,4,5,6,3]. Size is n x n.
My solution:
RCEQUAL(A,i1..12,j1..j2)// A is n*n matrix
if(i2-i1==2 && j2-j1==2 && b[n*i1+1..n*i2] has [j1..j2])
use brute force to check if the rows and columns are same.
if (any rows and columns are same)
store the row and column numbers in b[1..n^2].//b[1],b[n+2],b[2n+3].. store row no,
// b[2..n+1] stores columns that
//match with row 1, b[n+3..2n+2]
//those that match with row 2,etc..
else
RCEQUAL(A,1..n/2,1..n/2);
RCEQUAL(A,n/2..n,1..n/2);
RCEQUAL(A,1..n/2,n/2..n);
RCEQUAL(A,n/2..n,n/2..n);
Takes O(n^2). Is this correct? If correct, is there a faster algorithm?
you could build a trie from the data in the rows. then you can compare the columns with the trie.
this would allow to exit as soon as the beginning of a column do not match any row. also this would let you check a column against all rows in one pass.
of course the trie is most interesting when n is big (setting up a trie for a small n is not worth it) and when there are many rows and columns which are quite the same. but even in the worst case where all integers in the matrix are different, the structure allows for a clear algorithm...
You could speed up the average case by calculating the sum of each row/column and narrowing your brute-force comparison (which you have to do eventually) only on rows that match the sums of columns.
This doesn't increase the worst case (all having the same sum) but if your input is truly random that "won't happen" :-)
This might only work on non-singular matrices (not sure), but...
Let A be a square (and possibly non-singular) NxN matrix. Let A' be the transpose of A. If we create matrix B such that it is a horizontal concatenation of A and A' (in other words [A A']) and put it into RREF form, we will get a diagonal on all ones in the left half and some square matrix in the right half.
Example:
A = 1 2
3 4
A'= 1 3
2 4
B = 1 2 1 3
3 4 2 4
rref(B) = 1 0 0 -2
0 1 0.5 2.5
On the other hand, if a column of A were equal to a row of A then column of A would be equal to a column of A'. Then we would get another single 1 in of of the columns of the right half of rref(B).
Example
A=
1 2 3 4 5
2 6 -3 4 6
3 8 -7 6 9
4 1 7 -5 3
5 2 4 -1 -1
A'=
1 2 3 4 5
2 6 8 1 2
3 -3 -7 7 4
4 4 6 -5 -1
5 6 9 3 -1
B =
1 2 3 4 5 1 2 3 4 5
2 6 -3 4 6 2 6 8 1 2
3 8 -7 6 9 3 -3 -7 7 4
4 1 7 -5 3 4 4 6 -5 -1
5 2 4 -1 -1 5 6 9 3 -1
rref(B)=
1 0 0 0 0 1.000 -3.689 -5.921 3.080 0.495
0 1 0 0 0 0 6.054 9.394 -3.097 -1.024
0 0 1 0 0 0 2.378 3.842 -0.961 0.009
0 0 0 1 0 0 -0.565 -0.842 1.823 0.802
0 0 0 0 1 0 -2.258 -3.605 0.540 0.662
1.000 in the top row of the right half means that the first column of A matches on of its rows. The fact that the 1.000 is in the left-most column of the right half means that it is the first row.
Without looking at your algorithm or any of the approaches in the previous answers, but since the matrix has n^2 elements to begin with, I do not think there is a method which does better than that :)
IFF the matrix is truely random...
You could create a list of pointers to the columns sorted by the first element. Then create a similar list of the rows sorted by their first element. This takes O(n*logn).
Next create an index into each sorted list initialized to 0. If the first elements match, you must compare the whole row. If they do not match, increment the index of the one with the lowest starting element (either move to the next row or to the next column). Since each index cycles from 0 to n-1 only once, you have at most 2*n comparisons unless all the rows and columns start with the same number, but we said a matrix of random numbers.
The time for a row/column comparison is n in the worst case, but is expected to be O(1) on average with random data.
So 2 sorts of O(nlogn), and a scan of 2*n*1 gives you an expected run time of O(nlogn). This is of course assuming random data. Worst case is still going to be n**3 for a large matrix with most elements the same value.
I'm looking for some pointers here as I don't quite know where to start researching this one.
I have a 2D matrix with 0 or 1 in each cell, such as:
1 2 3 4
A 0 1 1 0
B 1 1 1 0
C 0 1 0 0
D 1 1 0 0
And I'd like to sort it so it is as "upper triangular" as possible, like so:
4 3 1 2
B 0 1 1 1
A 0 1 0 1
D 0 0 1 1
C 0 0 0 1
The rows and columns must remain intact, i.e. elements can't be moved individually and can only be swapped "whole".
I understand that there'll probably be pathological cases where a matrix has multiple possible sorted results (i.e. same shape, but differ in the identity of the "original" rows/columns.)
So, can anyone suggest where I might find some starting points for this? An existing library/algorithm would be great, but I'll settle for knowing the name of the problem I'm trying to solve!
I doubt it's a linear algebra problem as such, and maybe there's some kind of image processing technique that's applicable.
Any other ideas aside, my initial guess is just to write a simple insertion sort on the rows, then the columns and iterate that until it stabilises (and hope that detecting the pathological cases isn't too hard.)
More details: Some more information on what I'm trying to do may help clarify. Each row represents a competitor, each column represents a challenge. Each 1 or 0 represents "success" for the competitor on a particular challenge.
By sorting the matrix so all 1s are in the top-right, I hope to then provide a ranking of the intrinsic difficulty of each challenge and a ranking of the competitors (which will take into account the difficulty of the challenges they succeeded at, not just the number of successes.)
Note on accepted answer: I've accepted Simulated Annealing as "the answer" with the caveat that this question doesn't have a right answer. It seems like a good approach, though I haven't actually managed to come up with a scoring function that works for my problem.
An Algorithm based upon simulated annealing can handle this sort of thing without too much trouble. Not great if you have small matrices which most likely hae a fixed solution, but great if your matrices get to be larger and the problem becomes more difficult.
(However, it also fails your desire that insertions can be done incrementally.)
Preliminaries
Devise a performance function that "scores" a matrix - matrices that are closer to your triangleness should get a better score than those that are less triangle-y.
Devise a set of operations that are allowed on the matrix. Your description was a little ambiguous, but if you can swap rows then one op would be SwapRows(a, b). Another could be SwapCols(a, b).
The Annealing loop
I won't give a full exposition here, but the idea is simple. You perform random transformations on the matrix using your operations. You measure how much "better" the matrix is after the operation (using the performance function before and after the operation). Then you decide whether to commit that transformation. You repeat this process a lot.
Deciding whether to commit the transform is the fun part: you need to decide whether to perform that operation or not. Toward the end of the annealing process, you only accept transformations that improved the score of the matrix. But earlier on, in a more chaotic time, you allow transformations that don't improve the score. In the beginning, the algorithm is "hot" and anything goes. Eventually, the algorithm cools and only good transforms are allowed. If you linearly cool the algorithm, then the choice of whether to accept a transformation is:
public bool ShouldAccept(double cost, double temperature, Random random) {
return Math.Exp(-cost / temperature) > random.NextDouble();
}
You should read the excellent information contained in Numerical Recipes for more information on this algorithm.
Long story short, you should learn some of these general purpose algorithms. Doing so will allow you to solve large classes of problems that are hard to solve analytically.
Scoring algorithm
This is probably the trickiest part. You will want to devise a scorer that guides the annealing process toward your goal. The scorer should be a continuous function that results in larger numbers as the matrix approaches the ideal solution.
How do you measure the "ideal solution" - triangleness? Here is a naive and easy scorer: For every point, you know whether it should be 1 or 0. Add +1 to the score if the matrix is right, -1 if it's wrong. Here's some code so I can be explicit (not tested! please review!)
int Score(Matrix m) {
var score = 0;
for (var r = 0; r < m.NumRows; r++) {
for (var c = 0; c < m.NumCols; c++) {
var val = m.At(r, c);
var shouldBe = (c >= r) ? 1 : 0;
if (val == shouldBe) {
score++;
}
else {
score--;
}
}
}
return score;
}
With this scoring algorithm, a random field of 1s and 0s will give a score of 0. An "opposite" triangle will give the most negative score, and the correct solution will give the most positive score. Diffing two scores will give you the cost.
If this scorer doesn't work for you, then you will need to "tune" it until it produces the matrices you want.
This algorithm is based on the premise that tuning this scorer is much simpler than devising the optimal algorithm for sorting the matrix.
I came up with the below algorithm, and it seems to work correctly.
Phase 1: move rows with most 1s up and columns with most 1s right.
First the rows. Sort the rows by counting their 1s. We don't care
if 2 rows have the same number of 1s.
Now the columns. Sort the cols by
counting their 1s. We don't care
if 2 cols have the same number of
1s.
Phase 2: repeat phase 1 but with extra criterions, so that we satisfy the triangular matrix morph.
Criterion for rows: if 2 rows have the same number of 1s, we move up the row that begin with fewer 0s.
Criterion for cols: if 2 cols have the same number of 1s, we move right the col that has fewer 0s at the bottom.
Example:
Phase 1
1 2 3 4 1 2 3 4 4 1 3 2
A 0 1 1 0 B 1 1 1 0 B 0 1 1 1
B 1 1 1 0 - sort rows-> A 0 1 1 0 - sort cols-> A 0 0 1 1
C 0 1 0 0 D 1 1 0 0 D 0 1 0 1
D 1 1 0 0 C 0 1 0 0 C 0 0 0 1
Phase 2
4 1 3 2 4 1 3 2
B 0 1 1 1 B 0 1 1 1
A 0 0 1 1 - sort rows-> D 0 1 0 1 - sort cols-> "completed"
D 0 1 0 1 A 0 0 1 1
C 0 0 0 1 C 0 0 0 1
Edit: it turns out that my algorithm doesn't give proper triangular matrices always.
For example:
Phase 1
1 2 3 4 1 2 3 4
A 1 0 0 0 B 0 1 1 1
B 0 1 1 1 - sort rows-> C 0 0 1 1 - sort cols-> "completed"
C 0 0 1 1 A 1 0 0 0
D 0 0 0 1 D 0 0 0 1
Phase 2
1 2 3 4 1 2 3 4 2 1 3 4
B 0 1 1 1 B 0 1 1 1 B 1 0 1 1
C 0 0 1 1 - sort rows-> C 0 0 1 1 - sort cols-> C 0 0 1 1
A 1 0 0 0 A 1 0 0 0 A 0 1 0 0
D 0 0 0 1 D 0 0 0 1 D 0 0 0 1
(no change)
(*) Perhaps a phase 3 will increase the good results. In that phase we place the rows that start with fewer 0s in the top.
Look for a 1987 paper by Anna Lubiw on "Doubly Lexical Orderings of Matrices".
There is a citation below. The ordering is not identical to what you are looking for, but is pretty close. If nothing else, you should be able to get a pretty good idea from there.
http://dl.acm.org/citation.cfm?id=33385
Here's a starting point:
Convert each row from binary bits into a number
Sort the numbers in descending order.
Then convert each row back to binary.
Basic algorithm:
Determine the row sums and store
values. Determine the column sums
and store values.
Sort the row sums in ascending order. Sort the column
sums in ascending order.
Hopefully, you should have a matrix with as close to an upper-right triangular region as possible.
Treat rows as binary numbers, with the leftmost column as the most significant bit, and sort them in descending order, top to bottom
Treat the columns as binary numbers with the bottommost row as the most significant bit and sort them in ascending order, left to right.
Repeat until you reach a fixed point. Proof that the algorithm terminates left as an excercise for the reader.