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.
The merge function is wrong, you did not consider when some part of a or b has leftover elements. void merge(int * a, int ac, int * b, int bc, int * out) { int i = 0, j = 0; while (i < ac && j < bc) {...
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...
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 =...
To clarify @Dan Getz and add @collapsar answer I will add the following: Dan's Formula is correct: (score1 * weight1 + ... + scoreN * weightN) / (weight1 + ... + weightN) The beauty of the weighted average is you get to choose the weights! So we choose days since...
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...
Here is the graph I propose: Two kinds of vertices: departure vertex: airport+departure time arrival vertex: airport + arrival time. Two kind of edges: flight edge: from a departure vertex to an arrival vertex wait edge: from an arrival vertex to a departure vertex of later time in the same...
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....
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...
What's important to realize is that it's easy to take big steps: 1 digit numbers: 123456789 - 9 * 1 digit 2 digit numbers: 101112...9899 - 90 * 2 digits 3 digit numbers: 100101102...998999 - 900 * 3 digits 4 digit numbers: ... Now you can do a recursive solution...
python,regex,algorithm,python-2.7,datetime
What about fuzzyparsers: Sample inputs: jan 12, 2003 jan 5 2004-3-5 +34 -- 34 days in the future (relative to todays date) -4 -- 4 days in the past (relative to todays date) Example usage: >>> from fuzzyparsers import parse_date >>> parse_date('jun 17 2010') # my youngest son's birthday datetime.date(2010,...
algorithm,graph,tree,runtime,big-o
Don't know if your algorithm is correct, but it doesn't seem O(|V|) at least, because getting the "smallest edge not in T" of a vertex cannot be done in O(1). The most usual way to add an edge e=(u, v) into a MST T is: Run a BFS in T...
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...
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);...
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...
javascript,algorithm,recursion
I'm going to extend @georg's answer and provide a full implementation var additivePersistance = (function () { function sumOfDigits (n) { var ret = 0; n.toString().split('').forEach(function (i) { ret += parseInt(i, 10); }); return ret; } return function additivePersistance (n) { if (n < 10) { return 0; } return...
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' :...
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...
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...
l = [1,'a',12,'b','poppy'] def p(l,t): return [l[i-1] for i in t] print(p(l,(3,4,5,2,1))) [12, 'b', 'poppy', 'a', 1] indexing is 0 based so if you actually wanted to use the indexes for a 5 element list it would be (2,3,4,1,0) and [l[i] for i in t]...
I would go with performing Arithmetic Right Shift(till the length of the binary number) two at a time. This >> used in my logic is for arithmetic shift. (Note: In C language, right shifts may or may not be arithmetic!) Like, int count=0; bit=extractLastbit(binary_representation_of_the_number); while(count!=binaryLength){ // binaryLength is the length...
c,algorithm,security,math,encryption
This is not a power operator. It is the XOR operator. The thing that you notice for the XOR operator is that x ^ k ^ k == x. That means that your encryption function is already the decryption function when called with the same key and the ciphertext instead...
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"...
c++,arrays,algorithm,recursion
Do a regular binary search but with the (array[i] == i) condition instead of searching for a particular value. If (array[i] > i) move left else move right Of course this requires the values to be sorted, but your example indicates that is the case....
Assume task 0 finishes at day 0. Add a new column to your table, "completion day". Go down the list of tasks, add the duration of the current task onto the completion day of the task it is dependent on. Store that as the current task's completion day. Find the...
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...
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...
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) ....
Big O,Theta or Omega notation all refer to how a solution scales asymptotically as the size of the problem tends to infinity, however, they should really be prefaced with what you are measuring. Usually when one talks about big O(n) one usually means that the worst case complexity is O(n),...
ruby,algorithm,search,recursion
The first time you enter the adjacencies[to_append].each, you immediately return from the method, therefore the loop will never be executed more than once. You need to return a list of phone numbers instead of just a single phone number build that list somehow in your recursive call ...
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] << " "; }...
r,optimization,circular,maximization
I would compute all the pairs of rows in df: (pairs <- cbind(1:nrow(df), c(2:nrow(df), 1))) # [,1] [,2] # [1,] 1 2 # [2,] 2 3 # [3,] 3 4 # [4,] 4 5 # [5,] 5 6 # [6,] 6 1 You can find the best pairing with which.max:...
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)....
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...
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...
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....
Your logic does not work if(i%(ans%i)==0)ans*=(i/(ans%i)); else ans*=i; For example, if ans = 10 and i = 14, so, the lcm should be 70, but in your code, it is 140. The reason is, between ans and i , there are common divisors, but your code cannot detect that. For...