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  • algorithm to combinatorics

    - by peiska
    I am trying to solve a combinatorics problem, it seems easy, but i am having some trouble with it. If i have at most X tables, and N persons to sit on the tables, Each table can have 1 to N seating places, and I can only sit persons in one side of a rectangular table( so the order how people sit matters). I want to make a code that can calculate all the distributions of seating places from 1 up to K tables. For example, if I have 12 persons and 1 table i have 479001600 ways of seating persons( thats easy to calculate I've used Factorial of 12). But if I have 12 persons and 3 tables i have 4390848000 ways of seating persons. I've tried different solutions but i was not able to find the correct one. I've tried to divided the 12 in 3, then o use factorial of the result (it didnt work), i've tried to use 12! * 3( it didn't work too). Can some one give me a tip in a algorithm that i can use?

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  • Combinatorics grouping problem

    - by Harry Pap
    I'm looking for an algorithm in c# that solves a combinatorics problem: Assume i have the objects 1,2,3,4 I want to get all possible ways to group these object in multiple groups, that each time contain all objects. Order is not important. Example: <1,2,3,4 <1,2 / 3,4 <1,3 / 2,4 <1,4 / 3,2 <1,2,3 / 4 <1,2,4 / 3 <1,3,4 / 2 <2,3,4 / 1 <1 / 2 / 3 / 4 In the first case there is one group that contain all 4 objects. Next are cases with 2 groups that contain all objects in many different ways. The last case is 4 groups, that each one contains only one object.

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  • Combinatorics, probability, dice

    - by TarGz
    A friend of mine asked: if I have two dice and I throw both of them, what is the most frequent sum (of the two dice' numbers)? I wrote a small script: from random import randrange d = dict((i, 0) for i in range(2, 13)) for i in xrange(100000): d[randrange(1, 7) + randrange(1, 7)] += 1 print d Which prints: 2: 2770, 3: 5547, 4: 8379, 5: 10972, 6: 13911, 7: 16610, 8: 14010, 9: 11138, 10: 8372, 11: 5545, 12: 2746 The question I have, why is 11 more frequent than 12? In both cases there is only one way (or two, if you count reverse too) how to get such sum (5 + 6, 6 + 6), so I expected the same probability..?

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  • Help me finish this Python 3.x self-challenge.

    - by Hamish Grubijan
    This is not a homework. I saw this article praising Linq library and how great it is for doing combinatorics stuff, and I thought to myself: Python can do it in a more readable fashion. After half hour of dabbing with Python I failed. Please finish where I left off. Also, do it in the most Pythonic and efficient way possible please. from itertools import permutations from operator import mul from functools import reduce glob_lst = [] def divisible(n): return (sum(j*10^i for i,j in enumerate(reversed(glob_lst))) % n == 0) oneToNine = list(range(1, 10)) twoToNine = oneToNine[1:] for perm in permutations(oneToNine, 9): for n in twoToNine: glob_lst = perm[1:n] #print(glob_lst) if not divisible(n): continue else: # Is invoked if the loop succeeds # So, we found the number print(perm) Thanks!

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  • Help me finish this Python self-challenge.

    - by Hamish Grubijan
    This is not a homework. I saw this article praising Linq library and how great it is for doing combinatorics stuff, and I thought to myself: Python can do it in a more readable fashion. After half hour of dabbing with Python I failed. Please finish where I left off. Also, do it in the most Pythonic and efficient way possible please. from itertools import permutations from operator import mul from functools import reduce glob_lst = [] def divisible(n): return (sum(j*10^i for i,j in enumerate(reversed(glob_lst))) % n == 0) oneToNine = list(range(1, 10)) twoToNine = oneToNine[1:] for perm in permutations(oneToNine, 9): for n in twoToNine: glob_lst = perm[1:n] #print(glob_lst) if not divisible(n): continue else: # Is invoked if the loop succeeds # So, we found the number print(perm) Thanks!

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  • Minimal-change algorithm which maximises 'swapping'

    - by Kim Bastin
    This is a question on combinatorics from a non-mathematician, so please try to bear with me! Given an array of n distinct characters, I want to generate subsets of k characters in a minimal-change order, i.e. an order in which generation n+1 contains exactly one character that was not in generation n. That's not too hard in itself. However, I also want to maximise the number of cases in which the character that is swapped out in generation n+1 is the same character that was swapped in in generation n. To illustrate, for n=7, k=3: abc abd abe* abf* abg* afg aeg* adg* acg* acd ace* acf* aef adf* ade bde bdf bef bcf* bce bcd* bcg* bdg beg* bfg* cfg ceg* cdg* cde cdf* cef def deg dfg efg The asterisked strings indicate the case I want to maximise; e.g. the e that is new in generation 3, abe, replaces a d that was new in generation 2, abd. It doesn't seem possible to have this happen in every generation, but I want it to happen as often as possible. Typical array sizes that I use are 20-30 and subset sizes around 5-8. I'm using an odd language, Icon (or actually its derivative Unicon), so I don't expect anyone to post code that I can used directly. But I will be grateful for answers or hints in pseudo-code, and will do my best to translate C etc. Also, I have noticed that problems of this kind are often discussed in terms of arrays of integers, and I can certainly apply solutions posted in such terms to my own problem. Thanks Kim Bastin

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  • Calculate the number of ways to roll a certain number

    - by helloworld
    I'm a high school Computer Science student, and today I was given a problem to: Program Description: There is a belief among dice players that in throwing three dice a ten is easier to get than a nine. Can you write a program that proves or disproves this belief? Have the computer compute all the possible ways three dice can be thrown: 1 + 1 + 1, 1 + 1 + 2, 1 + 1 + 3, etc. Add up each of these possibilities and see how many give nine as the result and how many give ten. If more give ten, then the belief is proven. I quickly worked out a brute force solution, as such int sum,tens,nines; tens=nines=0; for(int i=1;i<=6;i++){ for(int j=1;j<=6;j++){ for(int k=1;k<=6;k++){ sum=i+j+k; //Ternary operators are fun! tens+=((sum==10)?1:0); nines+=((sum==9)?1:0); } } } System.out.println("There are "+tens+" ways to roll a 10"); System.out.println("There are "+nines+" ways to roll a 9"); Which works just fine, and a brute force solution is what the teacher wanted us to do. However, it doesn't scale, and I am trying to find a way to make an algorithm that can calculate the number of ways to roll n dice to get a specific number. Therefore, I started generating the number of ways to get each sum with n dice. With 1 die, there is obviously 1 solution for each. I then calculated, through brute force, the combinations with 2 and 3 dice. These are for two: There are 1 ways to roll a 2 There are 2 ways to roll a 3 There are 3 ways to roll a 4 There are 4 ways to roll a 5 There are 5 ways to roll a 6 There are 6 ways to roll a 7 There are 5 ways to roll a 8 There are 4 ways to roll a 9 There are 3 ways to roll a 10 There are 2 ways to roll a 11 There are 1 ways to roll a 12 Which looks straightforward enough; it can be calculated with a simple linear absolute value function. But then things start getting trickier. With 3: There are 1 ways to roll a 3 There are 3 ways to roll a 4 There are 6 ways to roll a 5 There are 10 ways to roll a 6 There are 15 ways to roll a 7 There are 21 ways to roll a 8 There are 25 ways to roll a 9 There are 27 ways to roll a 10 There are 27 ways to roll a 11 There are 25 ways to roll a 12 There are 21 ways to roll a 13 There are 15 ways to roll a 14 There are 10 ways to roll a 15 There are 6 ways to roll a 16 There are 3 ways to roll a 17 There are 1 ways to roll a 18 So I look at that, and I think: Cool, Triangular numbers! However, then I notice those pesky 25s and 27s. So it's obviously not triangular numbers, but still some polynomial expansion, since it's symmetric. So I take to Google, and I come across this page that goes into some detail about how to do this with math. It is fairly easy(albeit long) to find this using repeated derivatives or expansion, but it would be much harder to program that for me. I didn't quite understand the second and third answers, since I have never encountered that notation or those concepts in my math studies before. Could someone please explain how I could write a program to do this, or explain the solutions given on that page, for my own understanding of combinatorics? EDIT: I'm looking for a mathematical way to solve this, that gives an exact theoretical number, not by simulating dice

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  • Algorithm for dynamic combinations

    - by sOltan
    My code has a list called INPUTS, that contains a dynamic number of lists, let's call them A, B, C, .. N. These lists contain a dynamic number of Events I would like to call a function with each combination of Events. To illustrate with an example: INPUTS: A(0,1,2), B(0,1), C(0,1,2,3) I need to call my function this many times for each combination (the input count is dynamic, in this example it is three parameter, but it can be more or less) function(A[0],B[0],C[0]) function(A[0],B[1],C[0]) function(A[0],B[0],C[1]) function(A[0],B[1],C[1]) function(A[0],B[0],C[2]) function(A[0],B[1],C[2]) function(A[0],B[0],C[3]) function(A[0],B[1],C[3]) function(A[1],B[0],C[0]) function(A[1],B[1],C[0]) function(A[1],B[0],C[1]) function(A[1],B[1],C[1]) function(A[1],B[0],C[2]) function(A[1],B[1],C[2]) function(A[1],B[0],C[3]) function(A[1],B[1],C[3]) function(A[2],B[0],C[0]) function(A[2],B[1],C[0]) function(A[2],B[0],C[1]) function(A[2],B[1],C[1]) function(A[2],B[0],C[2]) function(A[2],B[1],C[2]) function(A[2],B[0],C[3]) function(A[2],B[1],C[3]) This is what I have thought of so far: My approach so far is to build a list of combinations. The element combination is itself a list of "index" to the input arrays A, B and C. For our example: my list iCOMBINATIONS contains the following iCOMBO lists (0,0,0) (0,1,0) (0,0,1) (0,1,1) (0,0,2) (0,1,2) (0,0,3) (0,1,3) (1,0,0) (1,1,0) (1,0,1) (1,1,1) (1,0,2) (1,1,2) (1,0,3) (1,1,3) (2,0,0) (2,1,0) (2,0,1) (2,1,1) (2,0,2) (2,1,2) (2,0,3) (2,1,3) Then I would do this: foreach( iCOMBO in iCOMBINATIONS) { foreach ( P in INPUTS ) { COMBO.Clear() foreach ( i in iCOMBO ) { COMBO.Add( P[ iCOMBO[i] ] ) } function( COMBO ) --- (instead of passing the events separately) } } But I need to find a way to build the list iCOMBINATIONS for any given number of INPUTS and their events. Any ideas? Is there actually a better algorithm than this? any pseudo code to help me with will be great. C# (or VB) Thank You

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  • a non recursive approach to the problem of generating combinations at fault

    - by mark
    Hi, I wanted a non recursive approach to the problem of generating combination of certain set of characters or numbers. So, given a subset k of numbers n, generate all the possible combination n!/k!(n-k)! The recursive method would give a combination, given the previous one combination. A non recursive method would generate a combination of a given value of loop index i. I approached the problem with this code: Tested with n = 4 and k = 3, and it works, but if I change k to a number 3 it does not work. Is it due to the fact that (n-k)! in case of n = 4 and k = 3 is 1. and if k 3 it will be more than 1? Thanks. int facto(int x); int len,fact,rem=0,pos=0; int str[7]; int avail[7]; str[0] = 1; str[1] = 2; str[2] = 3; str[3] = 4; str[4] = 5; str[5] = 6; str[6] = 7; int tot=facto(n) / facto(n-k) / facto(k); for (int i=0;i<tot;i++) { avail[0]=1; avail[1]=2; avail[2]=3; avail[3]=4; avail[4]=5; avail[5]=6; avail[6]=7; rem = facto(i+1)-1; cout<<rem+1<<". "; for(int j=len;j>0;j--) { int div = facto(j); pos = rem / div; rem = rem % div; cout<<avail[pos]<<" "; avail[pos]=avail[j]; } cout<<endl; } int facto(int x) { int fact=1; while(x0) fact*=x--; return fact; }

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  • Generating all unique combinations for "drive ya nuts" puzzle

    - by Yuval A
    A while back I wrote a simple python program to brute-force the single solution for the drive ya nuts puzzle. The puzzle consists of 7 hexagons with the numbers 1-6 on them, and all pieces must be aligned so that each number is adjacent to the same number on the next piece. The puzzle has ~1.4G non-unique possibilities: you have 7! options to sort the pieces by order (for example, center=0, top=1, continuing in clockwise order...). After you sorted the pieces, you can rotate each piece in 6 ways (each piece is a hexagon), so you get 6**7 possible rotations for a given permutation of the 7 pieces. Totalling: 7!*(6**7)=~1.4G possibilities. The following python code generates these possible solutions: def rotations(p): for i in range(len(p)): yield p[i:] + p[:i] def permutations(l): if len(l)<=1: yield l else: for perm in permutations(l[1:]): for i in range(len(perm)+1): yield perm[:i] + l[0:1] + perm[i:] def constructs(l): for p in permutations(l): for c in product(*(rotations(x) for x in p)): yield c However, note that the puzzle has only ~0.2G unique possible solutions, as you must divide the total number of possibilities by 6 since each possible solution is equivalent to 5 other solutions (simply rotate the entire puzzle by 1/6 a turn). Is there a better way to generate only the unique possibilities for this puzzle?

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  • List all possible combinations of k integers between 1...n (n choose k)

    - by Asaf R
    Hi, Out of no particular reason I decided to look for an algorithm that produces all possible choices of k integers between 1...n, where the order amongst the k integer doesn't matter (the n choose k thingy). From the exact same reason, which is no reason at all, I also implemented it in C#. My question is: Do you see any mistake in my algorithm or code? And, more importantly, can you suggest a better algorithm? Please pay more attention to the algorithm than the code itself. It's not the prettiest code I've ever written, although do tell if you see an error. EDIT: Alogirthm explained - We hold k indices. This creates k nested for loops, where loop i's index is indices[i]. It simulates k for loops where indices[i+1] belongs to a loop nested within the loop of indices[i]. indices[i] runs from indices[i - 1] + 1 to n - k + i + 1. CODE: public class AllPossibleCombination { int n, k; int[] indices; List<int[]> combinations = null; public AllPossibleCombination(int n_, int k_) { if (n_ <= 0) { throw new ArgumentException("n_ must be in N+"); } if (k_ <= 0) { throw new ArgumentException("k_ must be in N+"); } if (k_ > n_) { throw new ArgumentException("k_ can be at most n_"); } n = n_; k = k_; indices = new int[k]; indices[0] = 1; } /// <summary> /// Returns all possible k combination of 0..n-1 /// </summary> /// <returns></returns> public List<int[]> GetCombinations() { if (combinations == null) { combinations = new List<int[]>(); Iterate(0); } return combinations; } private void Iterate(int ii) { // // Initialize // if (ii > 0) { indices[ii] = indices[ii - 1] + 1; } for (; indices[ii] <= (n - k + ii + 1); indices[ii]++) { if (ii < k - 1) { Iterate(ii + 1); } else { int[] combination = new int[k]; indices.CopyTo(combination, 0); combinations.Add(combination); } } } } I apologize for the long question, it might be fit for a blog post, but I do want the community's opinion here. Thanks, Asaf

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  • Fast permutation -> number -> permutation mapping algorithms

    - by ijw
    I have n elements. For the sake of an example, let's say, 7 elements, 1234567. I know there are 7! = 5040 permutations possible of these 7 elements. I want a fast algorithm comprising two functions: f(number) maps a number between 0 and 5039 to a unique permutation, and f'(permutation) maps the permutation back to the number that it was generated from. I don't care about the correspondence between number and permutation, providing each permutation has its own unique number. So, for instance, I might have functions where f(0) = '1234567' f'('1234567') = 0 The fastest algorithm that comes to mind is to enumerate all permutations and create a lookup table in both directions, so that, once the tables are created, f(0) would be O(1) and f('1234567') would be a lookup on a string. However, this is memory hungry, particularly when n becomes large. Can anyone propose another algorithm that would work quickly and without the memory disadvantage?

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  • Open space sitting optimization algorithm

    - by Georgy Bolyuba
    As a result of changes in the company, we have to rearrange our sitting plan: one room with 10 desks in it. Some desks are more popular than others for number of reasons. One solution would be to draw a desk number from a hat. We think there is a better way to do it. We have 10 desks and 10 people. Lets give every person in this contest 50 hypothetical tokens to bid on the desks. There is no limit of how much you bid on one desk, you can put all 50, which would be saying "I want to sit only here, period". You can also say "I do not care" by giving every desk 5 tokens. Important note: nobody knows what other people are doing. Everyone has to decide based only on his/her best interest (sounds familiar?) Now lets say we obtained these hypothetical results: # | Desk# >| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 1 | Alise | 30 | 2 | 2 | 1 | 0 | 0 | 0 | 15 | 0 | 0 | = 50 2 | Bob | 20 | 15 | 0 | 10 | 1 | 1 | 1 | 1 | 1 | 0 | = 50 ... 10 | Zed | 5 | 5 | 5 | 5 | 5 | 5 | 5 | 5 | 5 | 5 | = 50 Now, what we need to find is that one (or more) configuration(s) that gives us maximum satisfaction (i.e. people get desks they wanted taking into account all the bids and maximizing on the total of the group. Naturally the assumption is the more one bade on the desk the more he/she wants it). Since there are only 10 people, I think we can brute force it looking into all possible configurations, but I was wondering it there is a better algorithm for solving this kind of problems?

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  • How to generate all variations with repetitions of a string?

    - by Svenstaro
    I want to generate all variations with repetitions of a string in C++ and I'd highly prefer a non-recursive algorithm. I've come up with a recursive algorithm in the past but due to the complexity (r^n) I'd like to see an iterative approach. I'm quite surprised that I wasn't able to find a solution to this problem anywhere on the web or on StackOverflow. I've come up with a Python script that does what I want as well: import itertools variations = itertools.product('ab', repeat=4) for variations in variations: variation_string = "" for letter in variations: variation_string += letter print variation_string Output: aaaa aaab aaba aabb abaa abab abba abbb baaa baab baba babb bbaa bbab bbba bbbb Ideally I'd like a C++ program that can produce the exact output, taking the exact same parameters. This is for learning purposes, it isn't homework. I wish my homework was like that.

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  • average case running time of linear search algorithm

    - by Brahadeesh
    Hi all. I am trying to derive the average case running time for deterministic linear search algorithm. The algorithm searches an element x in an unsorted array A in the order A[1], A[2], A[3]...A[n]. It stops when it finds the element x or proceeds until it reaches the end of the array. I searched on wikipedia and the answer given was (n+1)/(k+1) where k is the number of times x is present in the array. I approached in another way and am getting a different answer. Can anyone please give me the correct proof and also let me know whats wrong with my method? E(T)= 1*P(1) + 2*P(2) + 3*P(3) ....+ n*P(n) where P(i) is the probability that the algorithm runs for 'i' time (i.e. compares 'i' elements). P(i)= (n-i)C(k-1) * (n-k)! / n! Here, (n-i)C(k-1) is (n-i) Choose (k-1). As the algorithm has reached the ith step, the rest of k-1 x's must be in the last n-i elements. Hence (n-i)C(k-i). (n-k)! is the total number of ways of arranging the rest non x numbers, and n! is the total number of ways of arranging the n elements in the array. I am not getting (n+1)/(k+1) on simplifying.

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  • Program contains 2 nested loops which contain 4 if conditionals. How many paths?

    - by Student4Life
    In Roger Pressman's book, there is an example described of a program with 2 nested loops, the inner loop enclosing four if statements. The two loops can execute up to 20 times. He states that this makes about 10^14 paths. To get a number this large, it seems the paths inside the loops are multipllied by 2^40, i.e. 2^20 times 2^20 to account for all the possibilities of going through the two loops. I can't see why this factor is not just 400, i.e. 20 times 20. Can someone shed some light? It will help if you have the ppt slides and can see the program graph. Thanks.

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  • Algorithm to determine coin combinations

    - by A.J.
    I was recently faced with a prompt for a programming algorithm that I had no idea what to do for. I've never really written an algorithm before, so I'm kind of a newb at this. The problem said to write a program to determine all of the possible coin combinations for a cashier to give back as change based on coin values and number of coins. For example, there could be a currency with 4 coins: a 2 cent, 6 cent, 10 cent and 15 cent coins. How many combinations of this that equal 50 cents are there? The language I'm using is C++, although that doesn't really matter too much. edit: This is a more specific programming question, but how would I analyze a string in C++ to get the coin values? They were given in a text document like 4 2 6 10 15 50 (where the numbers in this case correspond to the example I gave)

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  • C++: compute a number's complement and its number of possible mismatches

    - by Eagle
    I got a bit stuck with my algorithm and I need some help to solve my problem. I think an example would explain better my problem. Assuming: d = 4 (maximum number of allowed bits in a number, 2^4-1=15). m_max = 1 (maximum number of allowed bits mismatches). kappa = (maximum number of elements to find for a given d and m, where m in m_max) The main idea is for a given number, x, to compute its complement number (in binary base) and all the possible combinations for up to m_max mismatches from x complement's number. Now the program start to scan from i = 0 till 15. for i = 0 and m = 0, kappa = \binom{d}{0} = 1 (this called a perfect match) possible combinations in bits, is only 1111 (for 0: 0000). for i = 0 and m = 1, kappa = \binom{d}{1} = 4 (one mismatch) possible combinations in bits are: 1000, 0100, 0010 and 0001 My problem was to generalize it to general d and m. I wrote the following code: #include <stdlib.h> #include <iomanip> #include <boost/math/special_functions/binomial.hpp> #include <iostream> #include <stdint.h> #include <vector> namespace vec { typedef std::vector<unsigned int> uint_1d_vec_t; } int main( int argc, char* argv[] ) { int counter, d, m; unsigned num_combination, bits_mask, bit_mask, max_num_mismatch; uint_1d_vec_t kappa; d = 4; m = 2; bits_mask = 2^num_bits - 1; for ( unsigned i = 0 ; i < num_elemets ; i++ ) { counter = 0; for ( unsigned m = 0 ; m < max_num_mismatch ; m++ ) { // maximum number of allowed combinations num_combination = boost::math::binomial_coefficient<double>( static_cast<unsigned>( d ), static_cast<unsigned>(m) ); kappa.push_back( num_combination ); for ( unsigned j = 0 ; j < kappa.at(m) ; j++ ) { if ( m == 0 ) v[i][counter++] = i^bits_mask; // M_0 else { bit_mask = 1 << ( num_bits - j ); v[i][counter++] = v[i][0] ^ bits_mask } } } } return 0; } I got stuck in the line v[i][counter++] = v[i][0] ^ bits_mask since I was unable to generalize my algorithm to m_max1, since I needed for m_max mismatches m_max loops and in my original problem, m is unknown until runtime.

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  • Combinations and Permutations in F#

    - by Noldorin
    I've recently written the following combinations and permutations functions for an F# project, but I'm quite aware they're far from optimised. /// Rotates a list by one place forward. let rotate lst = List.tail lst @ [List.head lst] /// Gets all rotations of a list. let getRotations lst = let rec getAll lst i = if i = 0 then [] else lst :: (getAll (rotate lst) (i - 1)) getAll lst (List.length lst) /// Gets all permutations (without repetition) of specified length from a list. let rec getPerms n lst = match n, lst with | 0, _ -> seq [[]] | _, [] -> seq [] | k, _ -> lst |> getRotations |> Seq.collect (fun r -> Seq.map ((@) [List.head r]) (getPerms (k - 1) (List.tail r))) /// Gets all permutations (with repetition) of specified length from a list. let rec getPermsWithRep n lst = match n, lst with | 0, _ -> seq [[]] | _, [] -> seq [] | k, _ -> lst |> Seq.collect (fun x -> Seq.map ((@) [x]) (getPermsWithRep (k - 1) lst)) // equivalent: | k, _ -> lst |> getRotations |> Seq.collect (fun r -> List.map ((@) [List.head r]) (getPermsWithRep (k - 1) r)) /// Gets all combinations (without repetition) of specified length from a list. let rec getCombs n lst = match n, lst with | 0, _ -> seq [[]] | _, [] -> seq [] | k, (x :: xs) -> Seq.append (Seq.map ((@) [x]) (getCombs (k - 1) xs)) (getCombs k xs) /// Gets all combinations (with repetition) of specified length from a list. let rec getCombsWithRep n lst = match n, lst with | 0, _ -> seq [[]] | _, [] -> seq [] | k, (x :: xs) -> Seq.append (Seq.map ((@) [x]) (getCombsWithRep (k - 1) lst)) (getCombsWithRep k xs) Does anyone have any suggestions for how these functions (algorithms) can be sped up? I'm particularly interested in how the permutation (with and without repetition) ones can be improved. The business involving rotations of lists doesn't look too efficient to me in retrospect. Update Here's my new implementation for the getPerms function, inspired by Tomas's answer. Unfortunately, it's not really any fast than the existing one. Suggestions? let getPerms n lst = let rec getPermsImpl acc n lst = seq { match n, lst with | k, x :: xs -> if k > 0 then for r in getRotations lst do yield! getPermsImpl (List.head r :: acc) (k - 1) (List.tail r) if k >= 0 then yield! getPermsImpl acc k [] | 0, [] -> yield acc | _, [] -> () } getPermsImpl List.empty n lst

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  • Algorithm for finding symmetries of a tree

    - by Paxinum
    I have n sectors, enumerated 0 to n-1 counterclockwise. The boundaries between these sectors are infinite branches (n of them). The sectors live in the complex plane, and for n even, sector 0 and n/2 are bisected by the real axis, and the sectors are evenly spaced. These branches meet at certain points, called junctions. Each junction is adjacent to a subset of the sectors (at least 3 of them). Specifying the junctions, (in pre-fix order, lets say, starting from junction adjacent to sector 0 and 1), and the distance between the junctions, uniquely describes the tree. Now, given such a representation, how can I see if it is symmetric wrt the real axis? For example, n=6, the tree (0,1,5)(1,2,4,5)(2,3,4) have three junctions on the real line, so it is symmetric wrt the real axis. If the distances between (015) and (1245) is equal to distance from (1245) to (234), this is also symmetric wrt the imaginary axis. The tree (0,1,5)(1,2,5)(2,4,5)(2,3,4) have 4 junctions, and this is never symmetric wrt either imaginary or real axis, but it has 180 degrees rotation symmetry if the distance between the first two and the last two junctions in the representation are equal. Edit: This is actually for my research. I have posted the question at mathoverflow as well, but my days in competition programming tells me that this is more like an IOI task. Code in mathematica would be excellent, but java, python, or any other language readable by a human suffices. Here are some examples (pretend the double edges are single and we have a tree) http://www2.math.su.se/~per/files.php?file=contr_ex_1.pdf http://www2.math.su.se/~per/files.php?file=contr_ex_2.pdf http://www2.math.su.se/~per/files.php?file=contr_ex_5.pdf Example 1 is described as (0,1,4)(1,2,4)(2,3,4)(0,4,5) with distances (2,1,3). Example 2 is described as (0,1,4)(1,2,4)(2,3,4)(0,4,5) with distances (2,1,1). Example 5 is described as (0,1,4,5)(1,2,3,4) with distances (2). So, given the description/representation, I want to find some algorithm to decide if it is symmetric wrt real, imaginary, and rotation 180 degrees. The last example have 180 degree symmetry. (These symmetries corresponds to special kinds of potential in the Schroedinger equation, which has nice properties in quantum mechanics.)

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  • Combinatorial optimisation of a distance metric

    - by Jose
    I have a set of trajectories, made up of points along the trajectory, and with the coordinates associated with each point. I store these in a 3d array ( trajectory, point, param). I want to find the set of r trajectories that have the maximum accumulated distance between the possible pairwise combinations of these trajectories. My first attempt, which I think is working looks like this: max_dist = 0 for h in itertools.combinations ( xrange(num_traj), r): for (m,l) in itertools.combinations (h, 2): accum = 0. for ( i, j ) in itertools.izip ( range(k), range(k) ): A = [ (my_mat[m, i, z] - my_mat[l, j, z])**2 \ for z in xrange(k) ] A = numpy.array( numpy.sqrt (A) ).sum() accum += A if max_dist < accum: selected_trajectories = h This takes forever, as num_traj can be around 500-1000, and r can be around 5-20. k is arbitrary, but can typically be up to 50. Trying to be super-clever, I have put everything into two nested list comprehensions, making heavy use of itertools: chunk = [[ numpy.sqrt((my_mat[m, i, :] - my_mat[l, j, :])**2).sum() \ for ((m,l),i,j) in \ itertools.product ( itertools.combinations(h,2), range(k), range(k)) ]\ for h in itertools.combinations(range(num_traj), r) ] Apart from being quite unreadable (!!!), it is also taking a long time. Can anyone suggest any ways to improve on this?

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  • Python script to calculate aded combinations from a dictionary

    - by dayde
    I am trying to write a script that will take a dictionary of items, each containing properties of values from 0 - 10, and add the various elements to select which combination of items achieve the desired totals. I also need the script to do this, using only items that have the same "slot" in common. For example: item_list = { 'item_1': {'slot': 'top', 'prop_a': 2, 'prop_b': 0, 'prop_c': 2, 'prop_d': 1 }, 'item_2': {'slot': 'top', 'prop_a': 5, 'prop_b': 0, 'prop_c': 1, 'prop_d':-1 }, 'item_3': {'slot': 'top', 'prop_a': 2, 'prop_b': 5, 'prop_c': 2, 'prop_d':-2 }, 'item_4': {'slot': 'mid', 'prop_a': 5, 'prop_b': 5, 'prop_c':-5, 'prop_d': 0 }, 'item_5': {'slot': 'mid', 'prop_a':10, 'prop_b': 0, 'prop_c':-5, 'prop_d': 0 }, 'item_6': {'slot': 'mid', 'prop_a':-5, 'prop_b': 2, 'prop_c': 3, 'prop_d': 5 }, 'item_7': {'slot': 'bot', 'prop_a': 1, 'prop_b': 3, 'prop_c':-4, 'prop_d': 4 }, 'item_8': {'slot': 'bot', 'prop_a': 2, 'prop_b': 2, 'prop_c': 0, 'prop_d': 0 }, 'item_9': {'slot': 'bot', 'prop_a': 3, 'prop_b': 1, 'prop_c': 4, 'prop_d':-4 }, } The script would then need to select which combinations from the "item_list" dict that using 1 item per "slot" that would achieve a desired result when added. For example, if the desired result was: 'prop_a': 3, 'prop_b': 3, 'prop_c': 8, 'prop_d': 0, the script would select 'item_2', 'item_6', and 'item_9', along with any other combination that worked. 'item_2': {'slot': 'top', 'prop_a': 5, 'prop_b': 0, 'prop_c': 1, 'prop_d':-1 } 'item_6': {'slot': 'mid', 'prop_a':-5, 'prop_b': 2, 'prop_c': 3, 'prop_d': 5 } 'item_9': {'slot': 'bot', 'prop_a': 3, 'prop_b': 1, 'prop_c': 4, 'prop_d':-4 } 'total': 'prop_a': 3, 'prop_b': 3, 'prop_c': 8, 'prop_d': 0 Any ideas how to accomplish this? It does not need to be in python, or even a thorough script, but just an explanation on how to do this in theory would be enough for me. I have tried working out looping through every combination, but that seems to very quickly get our of hand and unmanageable. The actual script will need to do this for about 1,000 items using 20 different "slots", each with 8 properties. Thanks for the help!

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  • Creating combinations that have no more one intersecting element

    - by khuss
    I am looking to create a special type of combination in which no two sets have more than one intersecting element. Let me explain with an example: Let us say we have 9 letter set that contains A, B, C, D, E, F, G, H, and I If you create the standard non-repeating combinations of three letters you will have 9C3 sets. These will contain sets like ABC, ABD, BCD, etc. I am looking to create sets that have at the most only 1 common letter. So in this example, we will get following sets: ABC, ADG, AEI, AFH, BEH, BFG, BDI, CFI, CDH, CEG, DEF, and GHI - note that if you take any two sets there are no more than 1 repeating letter. What would be a good way to generate such sets? It should be scalable solution so that I can do it for a set of 1000 letters, with sub set size of 4. Any help is highly appreciated. Thanks

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  • What category of combinatorial problems appear on the logic games section of the LSAT?

    - by Merjit
    There's a category of logic problem on the LSAT that goes like this: Seven consecutive time slots for a broadcast, numbered in chronological order I through 7, will be filled by six song tapes-G, H, L, O, P, S-and exactly one news tape. Each tape is to be assigned to a different time slot, and no tape is longer than any other tape. The broadcast is subject to the following restrictions: L must be played immediately before O. The news tape must be played at some time after L. There must be exactly two time slots between G and P, regardless of whether G comes before P or whether G comes after P. I'm interested in generating a list of permutations that satisfy the conditions as a way of studying for the test and as a programming challenge. However, I'm not sure what class of permutation problem this is. I've generalized the type problem as follows: Given an n-length array A: How many ways can a set of n unique items be arranged within A? Eg. How many ways are there to rearrange ABCDEFG? If the length of the set of unique items is less than the length of A, how many ways can the set be arranged within A if items in the set may occur more than once? Eg. ABCDEF = AABCDEF; ABBCDEF, etc. How many ways can a set of unique items be arranged within A if the items of the set are subject to "blocking conditions"? My thought is to encode the restrictions and then use something like Python's itertools to generate the permutations. Thoughts and suggestions are welcome.

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