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  • Algorithm: how to check intersections of recurring events definitions?

    - by glaz666
    The question comes from MS Outlook calendar behavior. Imagine I have two recurring events (starting from today): "each second Monday" and "every odd date". Is there any way to check intersections and/or find the first intersecting date algorithmically without brute-forcing over each date? Definitions can be made in CRON's notations or ICal notation. I think it doesn't matter. Are there any solutions for this in Gregorian calendar?

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  • Algorithm: How to tell if an array is a permutation in O(n)?

    - by Iulian Serbanoiu
    Hello, Input: A read-only array of N elements containing integer values from 1 to N. And a memory zone of a fixed size (10, 100, 1000 etc - not depending on N). How to tell in O(n) if the array represents a permutation? --What I achieved so far:-- I use the limited memory area to store the sum and the product of the array. I compare the sum with N*(N+1)/2 and the product with N! I know that if condition (2) is true I might have a permutation. I'm wondering if there's a way to prove that condition (2) is sufficient to tell if I have a permutation. So far I haven't figured this out ... Thanks, Iulian

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  • How to find minimum weight with maximum cost in 0-1 Knapsack algorithm?

    - by Nitin9791
    I am trying to solve a spoj problem Party Schedule the problem statement is- You just received another bill which you cannot pay because you lack the money. Unfortunately, this is not the first time to happen, and now you decide to investigate the cause of your constant monetary shortness. The reason is quite obvious: the lion's share of your money routinely disappears at the entrance of party localities. You make up your mind to solve the problem where it arises, namely at the parties themselves. You introduce a limit for your party budget and try to have the most possible fun with regard to this limit. You inquire beforehand about the entrance fee to each party and estimate how much fun you might have there. The list is readily compiled, but how do you actually pick the parties that give you the most fun and do not exceed your budget? Write a program which finds this optimal set of parties that offer the most fun. Keep in mind that your budget need not necessarily be reached exactly. Achieve the highest possible fun level, and do not spend more money than is absolutely necessary. Input The first line of the input specifies your party budget and the number n of parties. The following n lines contain two numbers each. The first number indicates the entrance fee of each party. Parties cost between 5 and 25 francs. The second number indicates the amount of fun of each party, given as an integer number ranging from 0 to 10. The budget will not exceed 500 and there will be at most 100 parties. All numbers are separated by a single space. There are many test cases. Input ends with 0 0. Output For each test case your program must output the sum of the entrance fees and the sum of all fun values of an optimal solution. Both numbers must be separated by a single space. Example Sample input: 50 10 12 3 15 8 16 9 16 6 10 2 21 9 18 4 12 4 17 8 18 9 50 10 13 8 19 10 16 8 12 9 10 2 12 8 13 5 15 5 11 7 16 2 0 0 Sample output: 49 26 48 32 now I know that it is an advance version of 0/1 knapsack problem where along with maximum cost we also have to find minimum weight that is less than a a given weight and have maximum cost. so I have used dp to solve this problem but still get a wrong awnser on submission while it is perfectly fine with given test cases. My code is typedef vector<int> vi; #define pb push_back #define FOR(i,n) for(int i=0;i<n;i++) int main() { //freopen("input.txt","r",stdin); while(1) { int W,n; cin>>W>>n; if(W==0 && n==0) break; int K[n+1][W+1]; vi val,wt; FOR(i,n) { int x,y; cin>>x>>y; wt.pb(x); val.pb(y); } FOR(i,n+1) { FOR(w,W+1) { if(i==0 || w==0) { K[i][w]=0; } else if (wt[i-1] <= w) { if(val[i-1] + K[i-1][w-wt[i-1]]>=K[i-1][w]) { K[i][w]=val[i-1] + K[i-1][w-wt[i-1]]; } else { K[i][w]=K[i-1][w]; } } else { K[i][w] = K[i-1][w]; } } } int a1=K[n][W],a2; for(int j=0;j<W;j++) { if(K[n][j]==a1) { a2=j; break; } } cout<<a2<<" "<<a1<<"\n"; } return 0; } Could anyone suggest what am I missing??

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  • Can SHA-1 algorithm be computed on a stream? With low memory footprint?

    - by raoulsson
    I am looking for a way to compute SHA-1 checksums of very large files without having to fully load them into memory at once. I don't know the details of the SHA-1 implementation and therefore would like to know if it is even possible to do that. If you know the SAX XML parser, then what I look for would be something similar: Computing the SHA-1 checksum by only always loading a small part into memory at a time. All the examples I found, at least in Java, always depend on fully loading the file/byte array/string into memory. If you even know implementations (any language), then please let me know!

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  • What is the best data structure and algorithm for comparing a list of strings?

    - by Chiraag E Sehar
    I want to find the longest possible sequence of words that match the following rules: Each word can be used at most once All words are Strings Two strings sa and sb can be concatenated if the LAST two characters of sa matches the first two characters of sb. In the case of concatenation, it is performed by overlapping those characters. For example: sa = "torino" sb = "novara" sa concat sb = "torinovara" For example, I have the following input file, "input.txt": novara torino vercelli ravenna napoli liverno messania noviligure roma And, the output of the above file according to the above rules should be: torino novara ravenna napoli livorno noviligure since the longest possible concatenation is: torinovaravennapolivornovilligure Can anyone please help me out with this? What would be the best data structure for this?

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  • Is there any algorithm for determining 3d position in such case? (images below)

    - by Ole Jak
    So first of all I have such image (and ofcourse I have all points coordinates in 2d so I can regenerate lines and check where they cross each other) But hey, I have another Image of same lines (I know thay are same) and new coords of my points like on this image So... now Having points (coords) on first image, How can I determin plane rotation and Z depth on second image (asuming first one's center was in point (0,0,0) with no rotation)?

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  • Algorithm for assigning a unique series of bits for each user?

    - by Mark
    The problem seems simple at first: just assign an id and represent that in binary. The issue arises because the user is capable of changing as many 0 bits to a 1 bit. To clarify, the hash could go from 0011 to 0111 or 1111 but never 1010. Each bit has an equal chance of being changed and is independent of other changes. What would you have to store in order to go from hash - user assuming a low percentage of bit tampering by the user? I also assume failure in some cases so the correct solution should have an acceptable error rate. I would an estimate the maximum number of bits tampered with would be about 30% of the total set. I guess the acceptable error rate would depend on the number of hashes needed and the number of bits being set per hash. I'm worried with enough manipulation the id can not be reconstructed from the hash. The question I am asking I guess is what safe guards or unique positioning systems can I use to ensure this happens.

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  • Is there a name for this type of algorithm?

    - by rehanift
    I have a 2 dimensional array forming a table: [color][number][shape ] ------------------------- [black][10 ][square ] [black][10 ][circle ] [red ][05 ][triangle] [red ][04 ][triangle] [green][11 ][oval ] and what I want to do is group largest common denominators, such that we get: 3 groups group #1: color=black, number=10, shapes = [square, circle] group #2: color=red, shape=triange, numbers = [05,04] group #3: color=green, number=11, shape = oval I wrote code that will handle a 2 "column" scenario, then I needed to adjusted it for 3 and I was figuring I might as well do it for n. I wanted to check first if there is some literature around this but I can't think of what to start looking for!

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  • Is this a bad version of the Merge Sort algorithm?

    - by SebKom
    merge1(int low, int high, int S[], U[]) { int k = (high - low + 1)/2 for q (from low to high) U[q] = S[q] int j = low int p = low int i = low + k while (j <= low + k - 1) and (i <= high) do { if ( U[j] <= U[i] ) { S[p] := U[j] j := j+1 } else { S[p] := U[i] i := i+1 } p := p+1 } if (j <= low + k - 1) { for q from p to high do { S[q] := U[j] j := j+1 } } } merge_sort1(int low, int high, int S[], U[]) { if low < high { int k := (high - low + 1)/2 merge_sort1(low, low+k-1, S, U) merge_sort1(low+k, high, S, U) merge1(low, high, S, U) } } I am really sorry for the terrible formating, as you can tell I am not a regular visitor here. So, basically, this is on my lecture notes. I find it quite confusing in general but I understand the biggest part of it. What I don't understand is the need of the "if (j <= low + k - 1)" part. It looks like it checks if there are any elements "left" in the left part. Is that even possible when mergesorting?

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  • Algorithm design: can you provide a solution to the multiple knapsack problem?

    - by MalcomTucker
    I am looking for a pseudo-code solution to what is effectively the Multiple Knapsack Problem (optimisation statement is halfway down the page). I think this problem is NP Complete so the solution doesn't need to be optimal, rather if it is fairly efficient and easily implemented that would be good. The problem is this: I have many work items, with each taking a different (but fixed and known) amount of time to complete. I need to divide these work items into groups so as to have the smallest number of groups (ideally), with each group of work items taking no longer than a given total threshold - say 1 hour. I am flexible about the threshold - it doesnt need to be rigidly applied, though should be close. My idea was to allocate work items into bins where each bin represents 90% of the threshold, 80%, 70% and so on. I could then match items that take 90% to those that take 10%, and so on. Any better ideas?

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  • algorithm to generate maximum number of character 'A' using keystrokes 'A', CTRL + 'A', CTRL + 'C' and CTRL + 'V'

    - by munda
    This is an interview question from google. I am not able to solve it by myself. Can somebody throw some light? The question goes like this. Write a program to print the sequence of keystrokes such that it generates the maximum number of character 'A's. You are allowed to use only 4 keys: 'A', CTRL + 'A', CTRL + 'C' and CTRL + 'V'. Only N keystrokes are allowed. All CTRL+ characters are considered as one keystroke, so CTRL+A is one keystroke. e.g.: A, ctrl+A, ctrl+C, ctrl+V generates two As in 4 keystrokes. Edit: CTRL + A is Select All CTRL + C is copy CTRL + V is paste I did some mathematics. For any N, using x numbers of A's , one CTRL+A, one CTRL+C and y CTRL+V, we can generate max ((N-1)/2)^2 numbers of A's. But for some N M, it is better to use as many ^A, ^C and ^V as it doubles the number of As. Edit 2: ^A, ^V and ^C will not overwrite on the existing selection but it will append the copied selection to selected one.

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  • Fastest algorithm to check if a number is pandigital?

    - by medopal
    Pandigital number is a number that contains the digits 1..number length. For example 123, 4312 and 967412385. I have solved many Project Euler problems, but the Pandigital problems always exceed the one minute rule. This is my pandigital function: private boolean isPandigital(int n){ Set<Character> set= new TreeSet<Character>(); String string = n+""; for (char c:string.toCharArray()){ if (c=='0') return false; set.add(c); } return set.size()==string.length(); } Create your own function and test it with this method int pans=0; for (int i=123456789;i<=123987654;i++){ if (isPandigital(i)){ pans++; } } Using this loop, you should get 720 pandigital numbers. My average time was 500 millisecond. I'm using Java, but the question is open to any language.

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  • how to get started with TopCoder to update/develop algorithm skills ?

    - by KaluSingh Gabbar
    at workplace, the work I do is hardly near to challenging and doing that I think I might be loosing the skills to look at a completely new problem and think about different ideas to solve it. A friend suggested TopCoder.com to me, but looking at the overwhelming number of problems I can not decide how to get started? what I want is to sharpen my techniques ( not particular language or framework ).

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  • How do I implement an higher lower game algorithm?

    - by lazorde
    The computer will guess a player’s number between 1 and 100. After each guess the human player should respond “higher”, “lower” or “correct”. Your program should be able to guess the player’s number in no more than 7 tries. Begin by explaining the game to the player, telling him/her to think of a number between 1 and 100. Make the computer do what you would normally do to guess a number in a certain range. Allow the user to respond with “higher”, “lower”, or “correct” after each computer guess. Output the number of tries it took the computer to guess the number. Make the game as user friendly as you can.

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  • Worse is better. Is there an example?

    - by J.F. Sebastian
    Is there a widely-used algorithm that has time complexity worse than that of another known algorithm but it is a better choice in all practical situations (worse complexity but better otherwise)? An acceptable answer might be in a form: There are algorithms A and B that have O(N**2) and O(N) time complexity correspondingly, but B has such a big constant that it has no advantages over A for inputs less then a number of atoms in the Universe. Examples highlights from the answers: Simplex algorithm -- worst-case is exponential time -- vs. known polynomial-time algorithms for convex optimization problems. A naive median of medians algorithm -- worst-case O(N**2) vs. known O(N) algorithm. Backtracking regex engines -- worst-case exponential vs. O(N) Thompson NFA -based engines. All these examples exploit worst-case vs. average scenarios. Are there examples that do not rely on the difference between the worst case vs. average case scenario? Related: The Rise of ``Worse is Better''. (For the purpose of this question the "Worse is Better" phrase is used in a narrower (namely -- algorithmic time-complexity) sense than in the article) Python's Design Philosophy: The ABC group strived for perfection. For example, they used tree-based data structure algorithms that were proven to be optimal for asymptotically large collections (but were not so great for small collections). This example would be the answer if there were no computers capable of storing these large collections (in other words large is not large enough in this case). Coppersmith–Winograd algorithm for square matrix multiplication is a good example (it is the fastest (2008) but it is inferior to worse algorithms). Any others? From the wikipedia article: "It is not used in practice because it only provides an advantage for matrices so large that they cannot be processed by modern hardware (Robinson 2005)."

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