I'm wondering how to solve "One-dimensional binpacking problem with relative costs". We pack N volumes (with given sizes) into M bins (with given capacities) and have the matrix (NxM) of costs of each volume per each bin. So, the total cost should be minimized.
Can you advice any algorithm for solving this? Or, probably, any open-source lib for doing this?
Thanks!
If the problem under consideration is the generalized assignment problem, it is NP-hard but admits an approximation algorithm. From a brief look, the approximation ratio depends on the approximation ratio of an approximation algorithm for the knapsack problem, which in turn admits a fully polynomial time approximation scheme. In total. the generalized assignment problem also admits a fully polynomial time approximation scheme.
algorithm,coordinates,coordinate-systems,coordinate
So here's my question: does this algorithm exists already? Has it a name? This mapping is called the Z-order curve or Morton code: In mathematical analysis and computer science, Z-order, Morton order, or Morton code is a function which maps multidimensional data to one dimension while preserving locality of...
Let's say the domain is as following String domain[] = { a, b, .., z, A, B, .. Z, 0, 1, 2, .. 9 }; Let's say the password size is 8 ArrayList allCombinations = getAllPossibleStrings2(domain,8); This is going to generate SIZE(domain) * LENGTH number of combinations, which is in...
If you look at your inserstion sort As you already put count =1 because as for exits on exit condition of for loop. for same reason then it also make sense that when while loop cancels the count++ inside will not get executed but there was a comparison made. but...
MillenialMedia - Pretty decent engineering. Pretty good revenue Inmobi - Good revenue. Sometimes really dodgy ads. I've experiemented with half a dozen others, but they provided insignificant revenue, or provided many bugs. But your mileage may vary....
3. is indeed correct, as you will need to go through the algorithm and terminate at the "worst" stop clause, where the list is empty, needed log(n) iterations. 1. is not correct. The best case is NOT when the first element is the target, it is when the middle element...
c#,xml,csv,optimization,type-conversion
IEnumerable<string> values = new List<string>(); values = … Probably not going to be a big deal, but why create a new List<string>() just to throw it away. Replace this with either: IEnumerable<string> values; values = … If you need values defined in a previous scope, or else just: Enumerable<string> values...
c++,algorithm,parallel-processing,c++14
Parallel prefix sum is a classical distributed programming algorithm, which elegantly uses a reduction followed by a distribution (as illustrated in the article). The key observation is that you can compute parts of the partial sums before you know the leading terms.
c,performance,opencv,optimization,sse
As harold said, delta is used to make unsigned comparsion. Let's describe this implementation by steps: __m128i x0 = _mm_sub_epi8(_mm_loadu_si128((const __m128i*)(ptr + pixel[0])), delta); __m128i x1 = _mm_sub_epi8(_mm_loadu_si128((const __m128i*)(ptr + pixel[4])), delta); __m128i x2 = _mm_sub_epi8(_mm_loadu_si128((const __m128i*)(ptr + pixel[8])), delta); __m128i x3 = _mm_sub_epi8(_mm_loadu_si128((const __m128i*)(ptr + pixel[12])), delta); m0 =...
c++,algorithm,inheritance,time-complexity
The first is supposedly in O(M*logN) time, where M is the size of the list, and N = number of concrete derived classes of Base It's not though. unordered_map is a hashtable, lookup and insertion have constant complexity on average. So the first is still O(M). Just with more...
undefined is not a boolean value so when you use ! operator, your value will be converted to boolean at first. but == operator just checking your values. so if you want to get true from undefined == false you should do it like Boolean(undefined) == false ? 't' :...
I think this should work for square matrices: void printHemisphere(int matrix[N][M], int n, int m) { int mid = n / 2; for(int i = 1; i < mid; i++) { for (int j = n - i; j < m; ++j) { std::cout << matrix[i][j] << " "; }...
java,optimization,jvm,jvm-hotspot
The fact that what you see is the result of some JIT optimization should be clear by now looking at all the comments you received. But what is really happening and why that code is optimized always nearly after the same amount of iterations of the outer for? I'll try...
You can try that out yourself using php's microtime function: $saleprice = 1; $start = microtime(true); if(!empty($saleprice)) echo $saleprice . '<br/>'; echo 'empty: ' . number_format(( microtime(true) - $start), 30) . '<br/>'; $start = microtime(true); if(!is_null($saleprice)) echo $saleprice . '<br/>'; echo 'is_null: ' . number_format(( microtime(true) - $start), 30) ....
IIUC, this is the classic Minimum Vertex Cover problem, which is, unfortunately, NP Complete. Fortunately, the most intuitive and greedy possible algorithm is as good as it gets in this case....
python,algorithm,time-complexity,longest-substring
It's O(N) 'why to use DP of O(N2)' : You don't need to for this problem. Note, though, that you take advantage of the fact that your sequence tokens (letters) are finite - so you can set up a list to hold all the possible starting values (26) and...
If and only if, at some point during kahn's algorithm has no source to choose (and the remaining graph is still none empty), there is a cycle Proof: Direction1 <--: If there is a cycle, let it be v1->v2->v3->vk->v1. At each step of the algorithm, none of v1,v2,...,vk is a...
Here are my results so far: L=1-7: "aabbaba" -> "aabbabaa" (or the mirror, confirming your result) L=8: "aaabbaba" -> "aaabbabaa" (or the mirror) L=9: "aaaabbbaba" -> "aaaabbbabaa" (or the mirror) All futher L can be solved just by prefixing an additional a to the starting string....
Basically, you are finding all permutations of the array using a recursive permutation algorithm. There are 4 things you need to change: First, start your loop from pos, not 0 Second, swap elements back after recursing (backtracking) Third, only test once you have generated each complete permutation (when pos =...
Dijkstra should pass, I just make a submission using JAVA, and it took less than a second to complete each task. As I have mentioned, each value in the matrix can go up to 10^9, your solution can encounter a number overflow problem, which can effect the running time. My...
java,c++,algorithm,recursion,dfs
In the line res.add(temp); temp is a reference. You are adding a reference to the same list (itemList) every time you add it. Try changing it to something list res.add(new ArrayList(temp)); so that it copies the list instead....
algorithm,discrete-mathematics
The loops in fact only go up to the cube root of N. (i^3 < n, etc.) The 3 nested loops of this length, give O(cube root of N, cubed). This O(N) Of note, if you were correct and they each went to one third of N, then cubing this...
algorithm,math,statistics,variance,standard-deviation
Given the forward formulas Mk = Mk-1 + (xk – Mk-1) / k Sk = Sk-1 + (xk – Mk-1) * (xk – Mk), it's possible to solve for Mk-1 as a function of Mk and xk and k: Mk-1 = Mk - (xk - Mk) / (k - 1)....
javascript,dom,optimization,asynchronous,requestanimationframe
Both types of DOM operations (read/write) have their own job queues. Each queue is flushed (e.g. all jobs in it are ran/executed) every requestAnimationFrame If you add 100 read operations all within 5ms, for example, (during a loop for instance), all of those read operations will (most likely) occur the...
algorithm,dynamic-programming,knapsack-problem
One possibility would be to provide a suitable number of multiplicities of the items. For item i, there can be at most m_i := K / w_i choices of that item, where K denotes the knapsack capacity and w_i denotes the weight of the i-th item. Furthermore, for each weight...
algorithm,optimization,dynamic-programming,frequency
There is no need for Dynamic Programming for this problem. It is a simple sorting problem, with a slight twist. Find frequency / length for each file. This gets you, how frequent is each unit length access for the file. Sort these in descending order since each file is "penalized"...
See I know you will feel bad when I will tell you what is wrong in your answer. But the only thing wrong is you using gets(). Its 2015 no one uses it. I changed that to a scanf statement i.e delete this part of your code and get(s); get(s);...
