I am looking for an algorithm that finds minimal subset of vertices such that by removing this subset (and edges connecting these vertices) from graph all other vertices become unconnected (i.e. the graph won't have any edges).
I have a basic knowledge of graph theory so please excuse any incorrectness.
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.
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] << " "; }...
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....
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...
Replacing the following code: if(movei == (N*N)+1){ return true; } ...with a hardcoded value... if(movei == 62){ return true; } ...gave me a good result after 0.1 seconds. (A field with only three "zeroes" remaining.) So your overall algorithm works. Hint for better looks of the output: #include <iomanip> cout...
python,numpy,matplotlib,graph,plot
You can use the condition z=='some tag' to index the x and y array Here's an example (based on the code in your previous question) that should do it. Use a set to automate the creation of tags: import csv import datetime as dt import numpy as np import matplotlib.pyplot...
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),...
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...
This exact scenario is in the documentation: https://github.com/Readify/Neo4jClient/wiki/cypher-examples#create-a-user
python,graph,dataframes,networkx
You could use B.add_weighted_edges_from( [(row['domain'], row['subdomain'], 1) for idx, row in df.iterrows()], weight='weight') to add weighted edges, or you could use B.add_edges_from( [(row['domain'], row['subdomain']) for idx, row in df.iterrows()]) to add edges without weights. You may not need weights since the node degree is the number of edges adjacent to...
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...
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...
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...
python,csv,matplotlib,graph,plot
you need to turn x and y into type np.array before you calculate above_threshold and below_threshold, and then it works. In your version, you don't get an array of bools, but just False and True. I added comma delimiters to your input csv file to make it work (I assume...
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...
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...
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,...
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 ...
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) {...
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...
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)....
algorithm,numerical-methods,numerical,numerical-integration
numerical integration always return just a number if you do not want the number but function instead then you can not use numerical integration for this task directly Polynomial approach you can use any approximation/interpolation technique to obtain a polynomial representing f(x) then integrate as standard polynomial (just change...
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....
Both of your links are to the 'development' part of networkx (check the url), so I don't think it's in the standard release yet. Probably your best option is to copy the relevant code into a file in your working directory and import it separately. Otherwise, look to download the...
This would be a hard real-time system - if a deadline is missed then a patient might die. As such, Algorithm B is the preferred algorithm because you can design the hardware around the n^2 worst case, whereas with Algorithm A the hardware must account for the n^4 worst case...
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....
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...
Or you could place a rectangle on the region of interest: rect(xleft=1994,xright = 1998,ybottom=range(CVD$cvd)[1],ytop=range(CVD$cvd)[2], density=10, col = "blue") ...
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 =...
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"...
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...
Basically you can model it using each city/airport as a node and the flights as the connections in between. The weight of the connections/flights is then the time. If you assume that all stopovers are 30 minutes (in a simplified model) then you can add an additional cost per visited...
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...
python,sql,r,graph,connected-components
In R, you could use the package igraph: library(igraph) gg <- graph.edgelist(as.matrix(d), directed=F) split(V(gg)$name, clusters(gg)$membership) #$`1` #[1] "a" "b" "c" "u" "e" # #$`2` #[1] "f" "g" "h" "j" # #$`3` #[1] "z" "y" And you can look at the graph using: plot(gg) This is based on an excellent answer...
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...
c,string,algorithm,data-structures
Yes, this is in O(n) in the average and worst case, where n is the length of the shorter of both given strings. You could also express that as O(min(m,n)) with m and n being the lengths of both strings, respectively. But no, O(n) doesn't mean that it needs exactly...
algorithm,recursion,permutation
A common approach is as follows. If you have, say, K bins, then add K-1 special values to your initial array. I will use the -1 value assuming that it never occurs in the initial array; you can choose a different value. So for your example the initial array becomes...
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...
javascript,svg,d3.js,graph,charts
This isn't possible with D3. The axis component will generate an axis that corresponds to the associated range, i.e. to make it any particular size, you have to modify the output range of the associated scale. You can do this quite easily in a responsive manner though by computing the...
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' :...
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...
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...
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...
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]...
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);...
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...
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++,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.