Imagine you have N distinct people and that you have a record of where these people are, exactly M of these records to be exact.
For example
1,50,299
1,2,3,4,5,50,287
1,50,299
So you can see that 'person 1' is at the same place with 'person 50' three times. Here M = 3 obviously since there's only 3 lines. My question is given M of these lines, and a threshold value (i.e person A and B have been at the same place more than threshold times), what do you suggest the most efficient way of returning these co-occurrences?
So far I've built an N by N table, and looped through each row, incrementing table(N,M) every time N co occurs with M in a row. Obviously this is an awful approach and takes 0(n^2) to O(n^3) depending on how you implent. Any tips would be appreciated!
There is no need to create the table. Just create a hash/dictionary/whatever your language calls it. Then in pseudocode:
answer = []
for S in sets:
for (i, j) in pairs from S:
count[(i,j)]++
if threshold == count[(i,j)]:
answer.append((i,j))
If you have M sets of size of size K the running time will be O(M*K^2).
If you want you can actually keep the list of intersecting sets in a data structure parallel to count without changing the big-O.
Furthermore the same algorithm can be readily implemented in a distributed way using a map-reduce. For the count you just have to emit a key of (i, j) and a value of 1. In the reduce you count them. Actually generating the list of sets is similar.
The known concept for your case is Market Basket analysis. In this context, there are different algorithms. For example Apriori algorithm can be using for your case in a specific case for sets of size 2.
Moreover, in these cases to finding association rules with specific supports and conditions (which for your case is the threshold value) using from LSH and min-hash too.
you could use probability to speed it up, e.g. only check each pair with 1/50 probability. That will give you a 50x speed up. Then double check any pairs that make it close enough to 1/50th of M.
To double check any pairs, you can either go through the whole list again, or you could double check more efficiently if you do some clever kind of reverse indexing as you go. e.g. encode each persons row indices into 64 bit integers, you could use binary search / merge sort type techniques to see which 64 bit integers to compare, and use bit operations to compare 64 bit integers for matches. Other things to look up could be reverse indexing, binary indexed range trees / fenwick trees.
In a recent campus Facebook interview i have asked to divide an array into 3 equal parts such that the sum in each array is roughly equal to sum/3.My Approach1. Sort The Array2. Fill the array[k] (k=0) uptil (array[k]<=sum/3)3. After that increment k and repeat the above step for array[k]Is there any better algorithm for this or it is NP Hard Problem
This is a variant of the partition problem (see http://en.wikipedia.org/wiki/Partition_problem for details). In fact a solution to this can solve that one (take an array, pad with 0s, and then solve this problem) so this problem is NP hard.
There is a dynamic programming approach that is pseudo-polynomial. For each i from 0 to the size of the array, you keep track of all possible combinations of current sizes for the sub arrays, and their current sums. As long as there are a limited number possible sums of subsets of the array, this runs acceptably fast.
The solution that I would have suggested is to just go for "good enough" closeness. First let's consider the simpler problem with all values positive. Then sort by value descending. Take that array in threes. Build up the three subsets by always adding the largest of the triple to the one with the smallest sum, the smallest to the one with the largest, and the middle to the middle. You will end up dividing the array evenly, and the difference will be no more than the value of the third smallest element.
For the general case you can divide into positive and negative, use the above approach on each, and then brute force all combinations of a group of positives, a group of negatives, and the few leftover values in the middle that did not divide evenly.
Here are details on a dynamic programming solution if you are interested. The running time and memory usage is O(n*(sum)^2) where n is the size of your array and sum is the sum of absolute values of your array values. For each array index j from 1 to n, store all the possible values you can get for your 3 subset sums when you split the array from index 1 to j into 3 subsets. Also for each possibility, store one possible way to split the array to get the 3 sums. Then to extend this information for 1 to (j+1) given the information from 1 to j, simply take each possible combination of 3 sums for splitting 1 to j and form the 3 combinations of 3 sums you get when you choose to add the (j+1)th array element to any one of the 3 subsets. Finally, when you are done and reach j = n, go through the set of all combinations of 3 subset sums you can get when you split array positions 1 to n into 3 sets, and choose the one whose maximum deviation from sum/3 is minimized. At first this may seem like O(n*(sum)^3) complexity, but for each j and each combination of the first 2 subset sums, the 3rd subset sum is uniquely determined. (because you are not allowed to omit any elements of the array). Thus the complexity really is O(n*(sum)^2).
I need to randomly generate an NxN matrix of integers in the range 1 to K inclusive such that all rows and columns individually have the property that their elements are pairwise distinct.
For example for N=2 and K=3
This is ok:
1 2
2 1
This is not:
1 3
1 2
(Notice that if K < N this is impossible)
When K is sufficiently larger than N an efficient enough algorithm is just to generate a random matrix of 1..K integers, check that each row and each column is pairwise distinct, and if it isn't try again.
But what about the case where K is not much larger than N?
This is not a full answer, but a warning about an intuitive solution that does not work.
I am assuming that by "randomly generate" you mean with uniform probability on all existing such matrices.
For N=2 and K=3, here are the possible matrices, up to permutations of the set [1..K]:
1 2 1 2 1 2
2 1 2 3 3 1
(since we are ignoring permutations of the set [1..K], we can assume wlog that the first line is 1 2).
Now, an intuitive (but incorrect) strategy would be to draw the matrix entries one by one, ensuring for each entry that it is distinct from the other entries on the same line or column.
To see why it's incorrect, consider that we have drawn this:
1 2
x .
and we are now drawing x. x can be 2 or 3, but if we gave each possibility the probability 1/2, then the matrix
1 2
3 1
would get probability 1/2 of being drawn at the end, while it should have only probability 1/3.
Here is a (textual) solution. I don't think it provides good randomness, but nevertherless it could be ok for your application.
Let's generate a matrix in the range [0;K-1] (you will do +1 for all elements if you want to) with the following algorithm:
Generate the first line with any random method you want.
Each number will be the first element of a random sequence calculated in such a manner that you are guarranteed to have no duplicate in subsequent rows, that is for any distinct column x and y, you will have x[i]!=y[i] for all i in [0;N-1].
Compute each row for the previous one.
All the algorithm is based on the random generator with the property I mentioned. With a quick search, I found that the Inversive congruential generator meets this requirement. It seems to be easy to implement. It works if K is prime; if K is not prime, see on the same page 'Compound Inversive Generators'. Maybe it will be a little tricky to handle with perfect squares or cubic numbers (your problem sound like sudoku :-) ), but I think it is possible by creating compound generators with prime factors of K and different parametrization. For all generators, the first element of each column is the seed.
Whatever the value of K, the complexity is only depending on N and is O(N^2).
Deterministically generate a matrix having the desired property for rows and columns. Provided K > N, this can easily be done by starting the ith row with i, and filling in the rest of the row with i+1, i+2, etc., wrapping back to 1 after K. Other algorithms are possible.
Randomly permute columns, then randomly permute rows.
Let's show that permuting rows (i.e. picking up entire rows and assembling a new matrix from them in some order, with each row possibly in a different vertical position) leaves the desired properties intact for both rows and columns, assuming they were true before. The same reasoning then holds for column permutations, and for any sequence of permutations of either kind.
Trivially, permuting rows cannot change the property that, within each row, no element appears more than once.
The effect of permuting rows on a particular column is to reorder the elements within that column. This holds for any column, and since reordering elements cannot produce duplicate elements where there were none before, permuting rows cannot change the property that, within each column, no element appears more than once.
I'm not certain whether this algorithm is capable of generating all possible satisfying matrices, or if it does, whether it will generate all possible satisfying matrices with equal probability. Another interesting question that I don't have an answer for is: How many rounds of row-permutation-then-column-permutation are needed? More precisely, is any finite sequence of row-perm-then-column-perm rounds equivalent to a bounded number of (or in particular, one) row-perm-then-column-perm round? If so then nothing is gained by further permutations after the first row and column permutations. Perhaps someone with a stronger mathematics background can comment. But it may be good enough in any case.
the problem requires us to find out the number of ways of placing R coins on a N*M grid such that each row and column has at least one coin. Constraints given are N , M < 200 , R < N*M. I initially thought of backtracking, but i was made to realise that it would never finish in time . Can someone guide me to another solution? (DP , closed form formula.) any pointers would be nice. Thanks.
Answer
According to OEIS sequence A055602 one possible solution to this is:
Let a(m, n, r) = Sum_{i=0..m} (-1)^i*binomial(m, i)*binomial((m-i)*n, r)
Answer = Sum_{i=0..N} (-1)^i*binomial(N, i)*a(M, N-i, R)
You will need to evaluate N+1 different values for a.
Assuming you have precomputed binomial coefficients, each evaluation of a is O(M) so the total complexity is O(NM).
Interpretation
This formula can be derived using the inclusion-exclusion principle twice.
a(m,n,r) is the number of ways of putting r coins on a grid of size m*n such that every one of the m columns is occupied, but not all the rows are necessarily occupied.
Inclusion-Exclusion turns this into the correct answer. (The idea is that we get our first estimate from a(M,N,R). This overestimates the correct answer because not all rows are occupied so we subtract cases a(M,N-1,R) where we only occupy N-1 rows. This then underestimates so we need to correct again...)
Similarly we can compute a(m,n,r) by considering b(m,n,r) which is the number of ways of placing r coins on a grid where we don't care about rows or columns being occupied. This can be derived simply from the number of ways of choosing r places in a grid size m*n , i.e. binomial(m*n,r). We use IE to turn this into the function a(m,n,r) where we know that all columns are occupied.
If you want to allow different conditions on the number of coins on each square, then you can just change b(m,n,r) to the appropriate counting function.
This is tough, but if you begin by working out how many ways you can have at least one coin on each row and column (call them reserve coins). The answer will be the product of #1 (n! / r! (n - r)!) *, where #2 n = N*M - NUMBER_OF_RESERVE_COINS and #3 r = (R - NUMBER_OF_RESERVE_COINS) for #4 each arrangement of reserving one coin on each row/column.
#4 is where the trickier stuff takes place. For N*M where N!=M, abs(N-M) tells you how many reserve coins will be on a single rows/columns. I'm having trouble on identifying the correct way of proceeding to the next step, mainly due to lack of time (though I can return to this on the weekend), but I hope I have provided you with useful information, and if what I have said is correct that you will be able to complete the process.
Given an n×n matrix of real numbers. You are allowed to erase any number (from 0 to n) of rows and any number (from 0 to n) of columns, and after that the sum of the remaining entries is computed. Come up with an algorithm which finds out which rows and columns to erase in order to maximize that sum.
The problem is NP-hard. (So you should not expect a polynomial-time algorithm for solving this problem. There could still be (non-polynomial time) algorithms that are slightly better than brute-force, though.) The idea behind the proof of NP-hardness is that if we could solve this problem, then we could solve the the clique problem in a general graph. (The maximum-clique problem is to find the largest set of pairwise connected vertices in a graph.)
Specifically, given any graph with n vertices, let's form the matrix A with entries a[i][j] as follows:
a[i][j] = 1 for i == j (the diagonal entries)
a[i][j] = 0 if the edge (i,j) is present in the graph (and i≠j)
a[i][j] = -n-1 if the edge (i,j) is not present in the graph.
Now suppose we solve the problem of removing some rows and columns (or equivalently, keeping some rows and columns) so that the sum of the entries in the matrix is maximized. Then the answer gives the maximum clique in the graph:
Claim: In any optimal solution, there is no row i and column j kept for which the edge (i,j) is not present in the graph. Proof: Since a[i][j] = -n-1 and the sum of all the positive entries is at most n, picking (i,j) would lead to a negative sum. (Note that deleting all rows and columns would give a better sum, of 0.)
Claim: In (some) optimal solution, the set of rows and columns kept is the same. This is because starting with any optimal solution, we can simply remove all rows i for which column i has not been kept, and vice-versa. Note that since the only positive entries are the diagonal ones, we do not decrease the sum (and by the previous claim, we do not increase it either).
All of which means that if the graph has a maximum clique of size k, then our matrix problem has a solution with sum k, and vice-versa. Therefore, if we could solve our initial problem in polynomial time, then the clique problem would also be solved in polynomial time. This proves that the initial problem is NP-hard. (Actually, it is easy to see that the decision version of the initial problem — is there a way of removing some rows and columns so that the sum is at least k — is in NP, so the (decision version of the) initial problem is actually NP-complete.)
Well the brute force method goes something like this:
For n rows there are 2n subsets.
For n columns there are 2n subsets.
For an n x n matrix there are 22n subsets.
0 elements is a valid subset but obviously if you have 0 rows or 0 columns the total is 0 so there are really 22n-2+1 subsets but that's no different.
So you can work out each combination by brute force as an O(an) algorithm. Fast. :)
It would be quicker to work out what the maximum possible value is and you do that by adding up all the positive numbers in the grid. If those numbers happen to form a valid sub-matrix (meaning you can create that set by removing rows and/or columns) then there's your answer.
Implicit in this is that if none of the numbers are negative then the complete matrix is, by definition, the answer.
Also, knowing what the highest possible maximum is possibly allows you to shortcut the brute force evaluation since if you get any combination equal to that maximum then that is your answer and you can stop checking.
Also if all the numbers are non-positive, the answer is the maximum value as you can reduce the matrix to a 1 x 1 matrix with that 1 value in it, by definition.
Here's an idea: construct 2n-1 n x m matrices where 1 <= m <= n. Process them one after the other. For each n x m matrix you can calculate:
The highest possible maximum sum (as per above); and
Whether no numbers are positive allowing you to shortcut the answer.
if (1) is below the currently calculate highest maximum sum then you can discard this n x m matrix. If (2) is true then you just need a simple comparison to the current highest maximum sum.
This is generally referred to as a pruning technique.
What's more you can start by saying that the highest number in the n x n matrix is the starting highest maximum sum since obviously it can be a 1 x 1 matrix.
I'm sure you could tweak this into a (slightly more) efficient recursive tree-based search algorithm with the above tests effectively allowing you to eliminate (hopefully many) unnecessary searches.
We can improve on Cletus's generalized brute-force solution by modelling this as a directed graph. The initial matrix is the start node of the graph; its leaves are all the matrices missing one row or column, and so forth. It's a graph rather than a tree, because the node for the matrix without both the first column and row will have two parents - the nodes with just the first column or row missing.
We can optimize our solution by turning the graph into a tree: There's never any point exploring a submatrix with a column or row deleted that comes before the one we deleted to get to the current node, as that submatrix will be arrived at anyway.
This is still a brute-force search, of course - but we've eliminated the duplicate cases where we remove the same rows in different orders.
Here's an example implementation in Python:
def maximize_sum(m):
frontier = [(m, 0, False)]
best = None
best_score = 0
while frontier:
current, startidx, cols_done = frontier.pop()
score = matrix_sum(current)
if score > best_score or not best:
best = current
best_score = score
w, h = matrix_size(current)
if not cols_done:
for x in range(startidx, w):
frontier.append((delete_column(current, x), x, False))
startidx = 0
for y in range(startidx, h):
frontier.append((delete_row(current, y), y, True))
return best_score, best
And here's the output on 280Z28's example matrix:
>>> m = ((1, 1, 3), (1, -89, 101), (1, 102, -99))
>>> maximize_sum(m)
(106, [(1, 3), (1, 101)])
Since nobody asked for an efficient algorithm, use brute force: generate every possible matrix that can be created by removing rows and/or columns from the original matrix, choose the best one. A slightly more efficent version, which most likely can be proved to still be correct, is to generate only those variants where the removed rows and columns contain at least one negative value.
To try it in a simple way:
We need the valid subset of the set of entries {A00, A01, A02, ..., A0n, A10, ...,Ann} which max. sum.
First compute all subsets (the power set).
A valid subset is a member of the power set that for each two contained entries Aij and A(i+x)(j+y), contains also the elements A(i+x)j and Ai(j+y) (which are the remaining corners of the rectangle spanned by Aij and A(i+x)(j+y)).
Aij ...
. .
. .
... A(i+x)(j+y)
By that you can eliminate the invalid ones from the power set and find the one with the biggest sum in the remaining.
I'm sure it can be improved by improving an algorithm for power set generation in order to generate only valid subsets and by that avoiding step 2 (adjusting the power set).
I think there are some angles of attack that might improve upon brute force.
memoization, since there are many distinct sequences of edits that will arrive at the same submatrix.
dynamic programming. Because the search space of matrices is highly redundant, my intuition is that there would be a DP formulation that can save a lot of repeated work
I think there's a heuristic approach, but I can't quite nail it down:
if there's one negative number, you can either take the matrix as it is, remove the column of the negative number, or remove its row; I don't think any other "moves" result in a higher sum. For two negative numbers, your options are: remove neither, remove one, remove the other, or remove both (where the act of removal is either by axing the row or the column).
Now suppose the matrix has only one positive number and the rest are all <=0. You clearly want to remove everything but the positive entry. For a matrix with only 2 positive entries and the rest <= 0, the options are: do nothing, whittle down to one, whittle down to the other, or whittle down to both (resulting in a 1x2, 2x1, or 2x2 matrix).
In general this last option falls apart (imagine a matrix with 50 positives & 50 negatives), but depending on your data (few negatives or few positives) it could provide a shortcut.
Create an n-by-1 vector RowSums, and an n-by-1 vector ColumnSums. Initialize them to the row and column sums of the original matrix. O(n²)
If any row or column has a negative sum, remove edit: the one with the minimum such and update the sums in the other direction to reflect their new values. O(n)
Stop when no row or column has a sum less than zero.
This is an iterative variation improving on another answer. It operates in O(n²) time, but fails for some cases mentioned in other answers, which is the complexity limit for this problem (there are n² entries in the matrix, and to even find the minimum you have to examine each cell once).
Edit: The following matrix has no negative rows or columns, but is also not maximized, and my algorithm doesn't catch it.
1 1 3 goal 1 3
1 -89 101 ===> 1 101
1 102 -99
The following matrix does have negative rows and columns, but my algorithm selects the wrong ones for removal.
-5 1 -5 goal 1
1 1 1 ===> 1
-10 2 -10 2
mine
===> 1 1 1
Compute the sum of each row and column. This can be done in O(m) (where m = n^2)
While there are rows or columns that sum to negative remove the row or column that has the lowest sum that is less than zero. Then recompute the sum of each row/column.
The general idea is that as long as there is a row or a column that sums to nevative, removing it will result in a greater overall value. You need to remove them one at a time and recompute because in removing that one row/column you are affecting the sums of the other rows/columns and they may or may not have negative sums any more.
This will produce an optimally maximum result. Runtime is O(mn) or O(n^3)
I cannot really produce an algorithm on top of my head, but to me it 'smells' like dynamic programming, if it serves as a start point.
Big Edit: I honestly don't think there's a way to assess a matrix and determine it is maximized, unless it is completely positive.
Maybe it needs to branch, and fathom all elimination paths. You never no when a costly elimination will enable a number of better eliminations later. We can short circuit if it's found the theoretical maximum, but other than any algorithm would have to be able to step forward and back. I've adapted my original solution to achieve this behaviour with recursion.
Double Secret Edit: It would also make great strides to reduce to complexity if each iteration didn't need to find all negative elements. Considering that they don't change much between calls, it makes more sense to just pass their positions to the next iteration.
Takes a matrix, the list of current negative elements in the matrix, and the theoretical maximum of the initial matrix. Returns the matrix's maximum sum and the list of moves required to get there. In my mind move list contains a list of moves denoting the row/column removed from the result of the previous operation.
Ie: r1,r1
Would translate
-1 1 0 1 1 1
-4 1 -4 5 7 1
1 2 4 ===>
5 7 1
Return if sum of matrix is the theoretical maximum
Find the positions of all negative elements unless an empty set was passed in.
Compute sum of matrix and store it along side an empty move list.
For negative each element:
Calculate the sum of that element's row and column.
clone the matrix and eliminate which ever collection has the minimum sum (row/column) from that clone, note that action as a move list.
clone the list of negative elements and remove any that are effected by the action taken in the previous step.
Recursively call this algorithm providing the cloned matrix, the updated negative element list and the theoretical maximum. Append the moves list returned to the move list for the action that produced the matrix passed to the recursive call.
If the returned value of the recursive call is greater than the stored sum, replace it and store the returned move list.
Return the stored sum and move list.
I'm not sure if it's better or worse than the brute force method, but it handles all the test cases now. Even those where the maximum contains negative values.
This is an optimization problem and can be solved approximately by an iterative algorithm based on simulated annealing:
Notation: C is number of columns.
For J iterations:
Look at each column and compute the absolute benefit of toggling it (turn it off if it's currently on or turn it on if it's currently off). That gives you C values, e.g. -3, 1, 4. A greedy deterministic solution would just pick the last action (toggle the last column to get a benefit of 4) because it locally improves the objective. But that might lock us into a local optimum. Instead, we probabilistically pick one of the three actions, with probabilities proportional to the benefits. To do this, transform them into a probability distribution by putting them through a Sigmoid function and normalizing. (Or use exp() instead of sigmoid()?) So for -3, 1, 4 you get 0.05, 0.73, 0.98 from the sigmoid and 0.03, 0.42, 0.56 after normalizing. Now pick the action according to the probability distribution, e.g. toggle the last column with probability 0.56, toggle the second column with probability 0.42, or toggle the first column with the tiny probability 0.03.
Do the same procedure for the rows, resulting in toggling one of the rows.
Iterate for J iterations until convergence.
We may also, in early iterations, make each of these probability distributions more uniform, so that we don't get locked into bad decisions early on. So we'd raise the unnormalized probabilities to a power 1/T, where T is high in early iterations and is slowly decreased until it approaches 0. For example, 0.05, 0.73, 0.98 from above, raised to 1/10 results in 0.74, 0.97, 1.0, which after normalization is 0.27, 0.36, 0.37 (so it's much more uniform than the original 0.05, 0.73, 0.98).
It's clearly NP-Complete (as outlined above). Given this, if I had to propose the best algorithm I could for the problem:
Try some iterations of quadratic integer programming, formulating the problem as: SUM_ij a_ij x_i y_j, with the x_i and y_j variables constrained to be either 0 or 1. For some matrices I think this will find a solution quickly, for the hardest cases it would be no better than brute force (and not much would be).
In parallel (and using most of the CPU), use a approximate search algorithm to generate increasingly better solutions. Simulating Annealing was suggested in another answer, but having done research on similar combinatorial optimisation problems, my experience is that tabu search would find good solutions faster. This is probably close to optimal in terms of wandering between distinct "potentially better" solutions in the shortest time, if you use the trick of incrementally updating the costs of single changes (see my paper "Graph domination, tabu search and the football pool problem").
Use the best solution so far from the second above to steer the first by avoiding searching possibilities that have lower bounds worse than it.
Obviously this isn't guaranteed to find the maximal solution. But, it generally would when this is feasible, and it would provide a very good locally maximal solution otherwise. If someone had a practical situation requiring such optimisation, this is the solution that I'd think would work best.
Stopping at identifying that a problem is likely to be NP-Complete will not look good in a job interview! (Unless the job is in complexity theory, but even then I wouldn't.) You need to suggest good approaches - that is the point of a question like this. To see what you can come up with under pressure, because the real world often requires tackling such things.
yes, it's NP-complete problem.
It's hard to easily find the best sub-matrix,but we can easily to find some better sub-matrix.
Assume that we give m random points in the matrix as "feeds". then let them to automatically extend by the rules like :
if add one new row or column to the feed-matrix, ensure that the sum will be incrementive.
,then we can compare m sub-matrix to find the best one.
Let's say n = 10.
Brute force (all possible sets of rows x all possible sets of columns) takes
2^10 * 2^10 =~ 1,000,000 nodes.
My first approach was to consider this a tree search, and use
the sum of positive entries is an upper bound for every node in the subtree
as a pruning method. Combined with a greedy algorithm to cheaply generate good initial bounds, this yielded answers in about 80,000 nodes on average.
but there is a better way ! i later realised that
Fix some choice of rows X.
Working out the optimal columns for this set of rows is now trivial (keep a column if its sum of its entries in the rows X is positive, otherwise discard it).
So we can just brute force over all possible choices of rows; this takes 2^10 = 1024 nodes.
Adding the pruning method brought this down to 600 nodes on average.
Keeping 'column-sums' and incrementally updating them when traversing the tree of row-sets should allow the calculations (sum of matrix etc) at each node to be O(n) instead of O(n^2). Giving a total complexity of O(n * 2^n)
For slightly less than optimal solution, I think this is a PTIME, PSPACE complexity issue.
The GREEDY algorithm could run as follows:
Load the matrix into memory and compute row totals. After that run the main loop,
1) Delete the smallest row,
2) Subtract the newly omitted values from the old row totals
--> Break when there are no more negative rows.
Point two is a subtle detail: subtracted two rows/columns has time complexity n.
While re-summing all but two columns has n^2 time complexity!
Take each row and each column and compute the sum. For a 2x2 matrix this will be:
2 1
3 -10
Row(0) = 3
Row(1) = -7
Col(0) = 5
Col(1) = -9
Compose a new matrix
Cost to take row Cost to take column
3 5
-7 -9
Take out whatever you need to, then start again.
You just look for negative values on the new matrix. Those are values that actually substract from the overall matrix value. It terminates when there're no more negative "SUMS" values to take out (therefore all columns and rows SUM something to the final result)
In an nxn matrix that would be O(n^2)Log(n) I think
function pruneMatrix(matrix) {
max = -inf;
bestRowBitField = null;
bestColBitField = null;
for(rowBitField=0; rowBitField<2^matrix.height; rowBitField++) {
for (colBitField=0; colBitField<2^matrix.width; colBitField++) {
sum = calcSum(matrix, rowBitField, colBitField);
if (sum > max) {
max = sum;
bestRowBitField = rowBitField;
bestColBitField = colBitField;
}
}
}
return removeFieldsFromMatrix(bestRowBitField, bestColBitField);
}
function calcSumForCombination(matrix, rowBitField, colBitField) {
sum = 0;
for(i=0; i<matrix.height; i++) {
for(j=0; j<matrix.width; j++) {
if (rowBitField & 1<<i && colBitField & 1<<j) {
sum += matrix[i][j];
}
}
}
return sum;
}