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 asked whether this problem was NP-complete on the Computer Science forum, but asking for programming heuristics seems better suited for this site. So here it goes.
You are given an NxN grid of unit squares and 2N binary strings of length N. The goal is to fill the grid with 0's and 1's so that each string appears once and only once in the grid, either horizontally (left to right) or vertically (top down). Or determine that no such solution exists. If N is not fixed I suspect this is an NP-complete problem. However are there any heuristics that can hopefully speed up the search to faster than brute force trying all ways to fill in the grid with N vertical strings?
I remember programming this for my friend that had the 5x5 physical version of this game, but I used brute force back then. I can only think of this heuristic:
Consider a 4x4 map with these 8 strings (read each from left to right):
1 1 0 1
1 0 0 1
1 0 1 1
1 0 1 0
1 1 1 1
1 0 0 0
0 0 1 1
1 1 1 0
(Note that this is already solved, since the second 4 is the first 4 transposed)
First attempt:
We will choose columns from left to right. Since 7 of 8 strings start with 1, we will try to put the one with most 1s to the first column (so that we can lay rows more easily when columns are done).
In the second column, most string have 0, so you can also try putting a string with most zeros to the second row, and so on.
This i would call a wide-1 prediction, since it only looks at one column at a time
(Possible) Improvement:
You can look at 2 columns at a time (a wide-2 prediction, if i may call it like that). In this case, from the 8 strings, the most common combination of first two bits is 10 (5/8), so you would like to choose first two columns so the the combination 10 occurring as much as possible (in this case, 1111 followed by 1000 has 3 of 4 10 at start).
(Of course you don't have to stop at 2)
Weaknesses:
I don't know if this would work. I just made it up and thought it might work.
If you choose to he wide-X prediction, the number of possibilities is exponential with X
This can absolutely fail if the distribution of combinations if even.
What you can do:
As i said, this game has physical 5x5 adaptation, only there you can also lay the string from right-to-left and bottom-to-top, if you found that name, you could google further. I unfortunately don't remember it.
Sounds like you want the crossword grid filling algorithm:
First, build 2N subsets of your 2N strings -- each subset has all the strings with a particular bit at a particular postion. So subset(0,3) is all the strings that have a 0 in the 3rd position and subset(1,5) is all the strings that have a 1 in the 5th position.
The algorithm is a basic brute-force depth fist search trying all possible mappings of strings to slots in the grid, with severe pruning of impossible branches
Your search state is a set of assignments of strings to slots and a set of sets of possible assignments to the remaining slots. The initial state has 0 assignments and 2N sets, all of which contain all 2N strings.
At each step of the search, pick the most constrained set (the set with the fewest elements) from the set of possible sets. Try each element of the set in turn in that slot (adding it to the assigments and removing it from the set of sets), and constrain all the remaining sets of sets by removing the chosen string and intersecting the crossing sets with subset(X,N) (computed in step 1) where X is the bit from the chosen string and N is the row/column number of the chosen string
If you find an empty set when picking above, there is no solution with the choices so far, so backtrack up the tree to a different choice
This is still EXPTIME, but it is about as fast as you can get it. Since the main time consuming step is the set intersections, using 2N bit binary strings for your set representation is very fast -- for N=32, the sets fit in a 64-bit word and can be intersected with a single AND instruction. It also helps to have a POPCOUNT instruction, since you also need set sizes.
This can be solved as a 0/1 integer linear program with O(N^2) variables and constraints. First there are variables Xij which are 1 if string i is assigned to line j (where j=1 to N are rows and j = (N+1) to 2N are columns). Then there is a variable for each square in the grid, which indicates if the entry is 0 or 1. If the position of the square is (i,j) with variable Yij then the sum of all X variables for line j that correspond to strings that have a 1 in position i is equal to Yij, and the sum of all X variables for line j that correspond to strings that have a 0 in position i is equal to (1 - Yij). And similarly for line i and position j. Finally, the sum of all X variables Xij for each string i (summed over all lines j) is equal to 1.
There has been a lot of research in speeding up solvers for 0/1 integer programming so this may be able to often handle fairly large N (like N=100) for many examples. Also, in some cases, solving the relaxed non-integer linear program and rounding the solution off to 0/1 may produce a valid solution, in polynomial time.
We could choose the first lg 2N rows out of the 2N strings, and then since 2^(lg 2N) = 2N, in a lot of cases there shouldn't be very many ways to assign the N columns so that the prefixes of length lg 2N are respected. Then all the rows are filled in so they can be checked to see if a solution has been found. We can also try assigning more rows in the beginning, and fill in different combinations of rows besides the initial rows. (e.g. we can try filling in contiguous rows starting anywhere in the grid).
Running time for assigning lg 2N rows out of 2N strings is O((2N)^(lg 2N)) = O(2^((lg 2N)^2)), which grows slower than 2^N. Assigning columns to match the prefixes is the part that's the hardest to predict run time. If a prefix occurs K times among the assigned rows, and there are M remaining strings that have the prefix, then the number of assignments for this prefix is M*(M-1)...(M-K+1). The total number of possible column assignments is the product of these terms over all prefixes that occur among the rows. If this gets to be too large, the number of rows initially assigned can be increased. But it's hard to predict the worst-case run time unless an assumption is made like the NxN grid is filled in randomly.
I'm looking for a variation on the Hungarian algorithm (I think) that will pair N people to themselves, excluding self-pairs and reverse-pairs, where N is even.
E.g. given N0 - N6 and a matrix C of costs for each pair, how can I obtain the set of 3 lowest-cost pairs?
C = [ [ - 5 6 1 9 4 ]
[ 5 - 4 8 6 2 ]
[ 6 4 - 3 7 6 ]
[ 1 8 3 - 8 9 ]
[ 9 6 7 8 - 5 ]
[ 4 2 6 9 5 - ] ]
In this example, the resulting pairs would be:
N0, N3
N1, N4
N2, N5
Having typed this out I'm now wondering if I can just increase the cost values in the "bottom half" of the matrix... or even better, remove them.
Is there a variation of Hungarian that works on a non-square matrix?
Or, is there another algorithm that solves this variation of the problem?
Increasing the values of the bottom half can result in an incorrect solution. You can see this as the corner coordinates (in your example coordinates 0,1 and 5,6) of the upper half will always be considered to be in the minimum X pairs, where X is the size of the matrix.
My Solution for finding the minimum X pairs
Take the standard Hungarian algorithm
You can set the diagonal to a value greater than the sum of the elements in the unaltered matrix — this step may allow you to speed up your implementation, depending on how your implementation handles nulls.
1) The first step of the standard algorithm is to go through each row, and then each column, reducing each row and column individually such that the minimum of each row and column is zero. This is unchanged.
The general principle of this solution, is to mirror every subsequent step of the original algorithm around the diagonal.
2) The next step of the algorithm is to select rows and columns so that every zero is included within the selection, using the minimum number of rows and columns.
My alteration to the algorithm means that when selecting a row/column, also select the column/row mirrored around that diagonal, but count it as one row or column selection for all purposes, including counting the diagonal (which will be the intersection of these mirrored row/column selection pairs) as only being selected once.
3) The next step is to check if you have the right solution — which in the standard algorithm means checking if the number of rows and columns selected is equal to the size of the matrix — in your example if six rows and columns have been selected.
For this variation however, when calculating when to end the algorithm treat each row/column mirrored pair of selections as a single row or column selection. If you have the right solution then end the algorithm here.
4) If the number of rows and columns is less than the size of the matrix, then find the smallest unselected element, and call it k. Subtract k from all uncovered elements, and add it to all elements that are covered twice (again, counting the mirrored row/column selection as a single selection).
My alteration of the algorithm means that when altering values, you will alter their mirrored values identically (this should happen naturally as a result of the mirrored selection process).
Then go back to step 2 and repeat steps 2-4 until step 3 indicates the algorithm is finished.
This will result in pairs of mirrored answers (which are the coordinates — to get the value of these coordinates refer back to the original matrix) — you can safely delete half of each pair arbitrarily.
To alter this algorithm to find the minimum R pairs, where R is less than the size of the matrix, reduce the stopping point in step 3 to R. This alteration is essential to answering your question.
As #Niklas B, stated you are solving Weighted perfect matching problem
take a look at this
here is part of document describing Primal-dual algorithm for weighted perfect matching
Please read all and let me know if is useful to you
I have an array of N elements (representing the N letters of a given alphabet), and each cell of the array holds an integer value, that integer value meaning the number of occurrences in a given text of that letter. Now I want to randomly choose a letter from all of the letters in the alphabet, based on his number of appearances with the given constraints:
If the letter has a positive (nonzero) value, then it can be always chosen by the algorithm (with a bigger or smaller probability, of course).
If a letter A has a higher value than a letter B, then it has to be more likely to be chosen by the algorithm.
Now, taking that into account, I've come up with a simple algorithm that might do the job, but I was just wondering if there was a better thing to do. This seems to be quite fundamental, and I think there might be more clever things to do in order to accomplish this more efficiently. This is the algorithm i thought:
Add up all the frequencies in the array. Store it in SUM
Choosing up a random value from 0 to SUM. Store it in RAN
[While] RAN > 0, Starting from the first, visit each cell in the array (in order), and subtract the value of that cell from RAN
The last visited cell is the chosen one
So, is there a better thing to do than this? Am I missing something?
I'm aware most modern computers can compute this so fast I won't even notice if my algorithm is inefficient, so this is more of a theoretical question rather than a practical one.
I prefer an explained algorithm rather than just code for an answer, but If you're more comfortable providing your answer in code, I have no problem with that.
The idea:
Iterate through all the elements and set the value of each element as the cumulative frequency thus far.
Generate a random number between 1 and the sum of all frequencies
Do a binary search on the values for this number (finding the first value greater than or equal to the number).
Example:
Element A B C D
Frequency 1 4 3 2
Cumulative 1 5 8 10
Generate a random number in the range 1-10 (1+4+3+2 = 10, the same as the last value in the cumulative list), do a binary search, which will return values as follows:
Number Element returned
1 A
2 B
3 B
4 B
5 B
6 C
7 C
8 C
9 D
10 D
The Alias Method has amortized O(1) time per value generated, but requires two uniforms per lookup. Basically, you create a table where each column contains one of the values to be generated, a second value called an alias, and a conditional probability of choosing between the value and its alias. Use your first uniform to pick any of the columns with equal likelihood. Then choose between the primary value and the alias based on your second uniform. It takes a O(n log n) work to initially set up a valid table for n values, but after the table's built generating values is constant time. You can download this Ruby gem to see an actual implementation.
Two other very fast methods by Marsaglia et al. are described here. They have provided C implementations.
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;
}