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• binary quicksort

  - by davit-datuashvili

  hi i want implement Binary quicksort algorithm from robert sedgewick book it looks like this public class quickb{ public static final int bitsword=32; public static void quicksortB(int a[],int l,int r,int d){ int i=l; int j=r-1; if (r<=l || d>bitsword) return ; while (j!=i) { while (digit(a[i],d)==0 && (i<j)) i++; while (digit(a[j],d)==1 && (j>i)) j++; int t=a[i]; a[i]=a[j]; a[j]=t; } if (digit(a[r-1],d)== 0) j++; quicksortB(a,l,j-1,d+1); quicksortB(a,j,r,d+1); } public static void main(String[]args){ int a[]=new int[]{4,7,3,9,8,2}; quicksortB(a,0,a.length-1,0); for (int i=0;i<a.length;i++){ System.out.println(a[i]); } } public static int digit(int m,int d){ return (m>>d)&1; } } but it show me error: java.lang.ArrayIndexOutOfBoundsException: 6 at quickb.quicksortB(quickb.java:13) at quickb.main(quickb.java:32) what is wrong?

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• Iterative Reduction to Null Matrix

  - by user1459032

  Here's the problem: I'm given a matrix like Input: 1 1 1 1 1 1 1 1 1 At each step, I need to find a "second" matrix of 1's and 0's with no two 1's on the same row or column. Then, I'll subtract the second matrix from the original matrix. I will repeat the process until I get a matrix with all 0's. Furthermore, I need to take the least possible number of steps. I need to print all the "second" matrices in O(n) time. In the above example I can get to the null matrix in 3 steps by subtracting these three matrices in order: Expected output: 1 0 0 0 1 0 0 0 1 0 0 1 1 0 0 0 1 0 0 1 0 0 0 1 1 0 0 I have coded an attempt, in which I am finding the first maximum value and creating the second matrices based on the index of that value. But for the above input I am getting 4 output matrices, which is wrong: My output: 1 0 0 0 1 0 0 0 1 0 1 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 1 0 My solution works for most of the test cases but fails for the one given above. Can someone give me some pointers on how to proceed, or find an algorithm that guarantees optimality? Test case that works: Input: 0 2 1 0 0 0 3 0 0 Output 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0

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• Fastest container or algorithm for unique reusable ids in C++

  - by gman

  I have a need for unique reusable ids. The user can choose his own ids or he can ask for a free one. The API is basically class IdManager { public: int AllocateId(); // Allocates an id void FreeId(int id); // Frees an id so it can be used again bool MarkAsUsed(int id); // Let's the user register an id. // returns false if the id was already used. }; Assume ids happen to start at 1 and progress, 2, 3, etc. This is not a requirement, just to help illustrate. IdManager mgr; mgr.MarkAsUsed(3); printf ("%d\n", mgr.AllocateId()); printf ("%d\n", mgr.AllocateId()); printf ("%d\n", mgr.AllocateId()); Would print 1 2 4 Because id 3 has already been declared used. What's the best container / algorithm to both remember which ids are used AND find a free id? If you want to know the a specific use case, OpenGL's glGenTextures, glBindTexture and glDeleteTextures are equivalent to AllocateId, MarkAsUsed and FreeId

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• Learning Java and logic using debugger. Did I cheat?

  - by centr0

  After a break from coding in general, my way of thinking logically faded (as if it was there to begin with...). I'm no master programmer. Intermediate at best. I decided to see if i can write an algorithm to print out the fibonacci sequence in Java. I got really frustrated because it was something so simple, and used the debugger to see what was going on with my variables. solved it in less than a minute with the help of the debugger. Is this cheating? When I read code either from a book or someone else's, I now find that it takes me a little more time to understand. If the alghorithm is complex (to me) i end up writing notes as to whats going on in the loop. A primitive debugger if you will. When you other programmers read code, do you also need to write things down as to whats the code doing? Or are you a genius and and just retain it?

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• Android - How to approach fall detection algorithm

  - by bobby123

  I want to be able to feature a fairly simple fall detection algorithm in my application. At the moment in onSensorChanged(), I am getting the absolute value of the current x,x,z values and subtracting SensorManager.GRAVITY_EARTH (9.8 m/s) from this. The resulting value has to be bigger than a threshold value 10 times in a row to set a flag saying a fall has been detected by the accelerometer, the threshold value is about 8m/s. Also I'm comparing the orientation of the phone as soon as the threshold has been passed and the orienation of it when the threshold is no longer being passed, this sets another flag saying the orientation sensor has detected a fall. When both flags are set, an event occurs to check is user ok, etc etc. My problem is with the threshold, when the phone is held straight up the absolute value of accelerometer is about 9.8 m/s, but when i hold it still at an angle it can be over 15m/s. This is causing other events to trigger the fall detection, and if i increase the threshold to avoid that, it won't detect falls. Can anyone give me some advice here with what possible values i should use or how to even improve my method? Many thanks.

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• Given a vector of maximum 10 000 natural and distinct numbers, find 4 numbers(a, b, c, d) such that

  - by king_kong

  Hi, I solved this problem by following a straightforward but not optimal algorithm. I sorted the vector in descending order and after that substracted numbers from max to min to see if I get a + b + c = d. Notice that I haven't used anywhere the fact that elements are natural, distinct and 10 000 at most. I suppose these details are the key. Does anyone here have a hint over an optimal way of solving this? Thank you in advance! Later Edit: My idea goes like this: '<<quicksort in descending order>>' for i:=0 to count { // after sorting, loop through the array int d := v[i]; for j:=i+1 to count { int dif1 := d - v[j]; int a := v[j]; for k:=j+1 to count { if (v[k] > dif1) continue; int dif2 := dif1 - v[k]; b := v[k]; for l:=k+1 to count { if (dif2 = v[l]) { c := dif2; return {a, b, c, d} } } } } } What do you think?(sorry for the bad indentation)

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• Finding the closest match

  - by doublescorpio

  I Have an object with a set of parameters like: var obj = new {Param1 = 100; Param2 = 212; Param3 = 311; param4 = 11; Param5 = 290;} On the other side i have a list of object: var obj1 = new {Param1 = 1221; Param2 = 212; Param3 = 311; param4 = 11; Param5 = 290;} var obj3 = new {Param1 = 35; Param2 = 11; Param3 = 319; param4 = 211; Param5 = 790;} var obj4 = new {Param1 = 126; Param2 = 218; Param3 = 2; param4 = 6; Param5 = 190;} var obj5 = new {Param1 = 213; Param2 = 121; Param3 = 61; param4 = 11; Param5 = 29;} var obj7 = new {Param1 = 161; Param2 = 21; Param3 = 71; param4 = 51; Param5 = 232;} var obj9 = new {Param1 = 891; Param2 = 58; Param3 = 311; param4 = 21; Param5 = 590;} var obj11 = new {Param1 = 61; Param2 = 212; Param3 = 843; param4 = 89; Param5 = 210;} What is the best (easiest) algorithm to find the closest match for the first obj in the listed objects?

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• The perverse hangman problem

  - by Shalmanese

  Perverse Hangman is a game played much like regular Hangman with one important difference: The winning word is determined dynamically by the house depending on what letters have been guessed. For example, say you have the board _ A I L and 12 remaining guesses. Because there are 13 different words ending in AIL (bail, fail, hail, jail, kail, mail, nail, pail, rail, sail, tail, vail, wail) the house is guaranteed to win because no matter what 12 letters you guess, the house will claim the chosen word was the one you didn't guess. However, if the board was _ I L M, you have cornered the house as FILM is the only word that ends in ILM. The challenge is: Given a dictionary, a word length & the number of allowed guesses, come up with an algorithm that either: a) proves that the player always wins by outputting a decision tree for the player that corners the house no matter what b) proves the house always wins by outputting a decision tree for the house that allows the house to escape no matter what. As a toy example, consider the dictionary: bat bar car If you are allowed 3 wrong guesses, the player wins with the following tree: Guess B NO -> Guess C, Guess A, Guess R, WIN YES-> Guess T NO -> Guess A, Guess R, WIN YES-> Guess A, WIN

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• Writing a recursive sorting algorithm of an array of integers

  - by 12345

  I am trying to write a recursive sorting algorithm for an array of integers. The following codes prints to the console: 3, 5, 2, 1, 1, 2, 6, 7, 8, 10, 20 The output should be sorted but somehow "it doesn't work". public static void main(String[] args) { int[] unsortedList = {20, 3, 1, 2, 1, 2, 6, 8, 10, 5, 7}; duplexSelectionSort(unsortedList, 0, unsortedList.length-1); for (int i = 0; i < unsortedList.length; i++) { System.out.println(unsortedList[i]); } } public static void duplexSelectionSort( int[] unsortedNumbers, int startIndex, int stopIndex) { int minimumIndex = 0; int maximumIndex = 0; if (startIndex < stopIndex) { int index = 0; while (index <= stopIndex) { if (unsortedNumbers[index] < unsortedNumbers[minimumIndex]) { minimumIndex = index; } if (unsortedNumbers[index] > unsortedNumbers[maximumIndex]) { maximumIndex = index; } index++; } swapEdges(unsortedNumbers, startIndex, stopIndex, minimumIndex, maximumIndex); duplexSelectionSort(unsortedNumbers, startIndex + 1, stopIndex - 1); } } public static void swapEdges( int[] listOfIntegers, int startIndex, int stopIndex, int minimumIndex, int maximumIndex) { if ((minimumIndex == stopIndex) && (maximumIndex == startIndex)) { swap(listOfIntegers, startIndex, stopIndex); } else { if (maximumIndex == startIndex) { swap(listOfIntegers, maximumIndex, stopIndex); swap(listOfIntegers, minimumIndex, startIndex); } else { swap(listOfIntegers, minimumIndex, startIndex); swap(listOfIntegers, maximumIndex, stopIndex); } } } public static void swap(int[] listOfIntegers, int index1, int index2) { int savedElementAtIndex1 = listOfIntegers[index1]; listOfIntegers[index1] = listOfIntegers[index2]; listOfIntegers[index2] = savedElementAtIndex1; }

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• Finding k elements of length-n list that sum to less than t in O(nlogk) time

  - by tresbot

  This is from Programming Pearls ed. 2, Column 2, Problem 8: Given a set of n real numbers, a real number t, and an integer k, how quickly can you determine whether there exists a k-element subset of the set that sums to at most t? One easy solution is to sort and sum the first k elements, which is our best hope to find such a sum. However, in the solutions section Bentley alludes to a solution that takes nlog(k) time, though he gives no hints for how to find it. I've been struggling with this; one thought I had was to go through the list and add all the elements less than t/k (in O(n) time); say there are m1 < k such elements, and they sum to s1 < t. Then we are left needing k - m1 elements, so we can scan through the list again in O(n) time looking for all elements less than (t - s1)/(k - m1). Add in again, to get s2 and m2, then again if m2 < k, look for all elements less than (t - s2)/(k - m2). So: def kSubsetSumUnderT(inList, k, t): outList = [] s = 0 m = 0 while len(outList) < k: toJoin = [i for i in inList where i < (t - s)/(k - m)] if len(toJoin): if len(toJoin) >= k - m: toJoin.sort() if(s0 + sum(toJoin[0:(k - m - 1)]) < t: return True return False outList = outList + toJoin s += sum(toJoin) m += len(toJoin) else: return False My intuition is that this might be the O(nlog(k)) algorithm, but I am having a hard time proving it to myself. Thoughts?

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• Change value of adjacent vertices and remove self loop

  - by StereoMatching

  Try to write a Karger’s algorithm with boost::graph example (first column is vertice, other are adjacent vertices): 1 2 3 2 1 3 4 3 1 2 4 4 2 3 assume I merge 2 to 1, I get the result 1 2 3 2 1 1 3 4 2 1 3 4 3 1 2 4 4 2 3 first question : How could I change the adjacent vertices("2" to "1") of vertice 1? my naive solution template<typename Vertex, typename Graph> void change_adjacent_vertices_value(Vertex input, Vertex value, Graph &g) { for (auto it = boost::adjacent_vertices(input, g); it.first != it.second; ++it.first){ if(*it.first == value){ *(it.first) = input; //error C2106: '=' : left operand must be l-value } } } Apparently, I can't set the value of the adjacent vertices to "1" by this way The result I want after "change_adjacent_vertices_value" 1 1 3 1 1 1 3 4 2 1 3 4 3 1 2 4 4 2 3 second question : How could I pop out the adjacent vertices? Assume I want to pop out the consecutive 1 from the vertice 1 The result I expected 1 1 3 1 3 4 2 1 3 4 3 1 2 4 4 2 3 any function like "pop_adjacent_vertex" could use?

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• Statistical approach to chess?

  - by Chinmay Kanchi

  Reading about how Google solves the translation problem got me thinking. Would it be possible to build a strong chess engine by analysing several million games and determining the best possible move based largely (completely?) on statistics? There are several such chess databases (this is one that has 4.5 million games), and one could potentially weight moves in identical (or mirrored or reflected) positions using factors such as the ratings of the players involved, how old the game is (to factor in improvements in chess theory) etc. Any reasons why this wouldn't be a feasible approach to building a chess engine?

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• How to make concept representation with the help of bag of words

  - by agazerboy

  Hi All, Thanks for stoping to read my question :) this is very sweet place full of GREAT peoples ! I have a question about "creating sentences with words". NO NO it is not about english grammar :) Let me explain, If I have bag of words like "person apple apple person person a eat person will apple eat hungry apple hungry" and it can generate some kind of following sentence "hungry person eat apple" I don't in which field this topic will relate. Where should I try to find an answer. I tried to search google but I only found english grammar stuff :) Any body there who can tell me which algo can work in this problem? or any program Thanks P.S: It is not an assignment :) if it would be i would ask for source code ! I don't even know in which field I should look for :)

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• Mysql results sorted by list which is unique for each user

  - by ADAM

  Ive got a table of thousands of products and 50 or so authenticated users. These users all show the products on their own web sites and they all require the ability to have them ordered differently. Im guesing i need some kind of seperate table for the orders which contains the product_id, user_id and order column? How do i do this the most efficiently in mysql so as to be very fast, and not slow down if i get millions of products in the database. Is it even wise to do it in mysql or should i be using some kind of other index like solr/lucene? My Product table is called "products" My User table is called "users" A good example of the functionality i need is google search where you can order/supress the results if you are logged in. edit: the product results will be paginated and the users have the authority to edit the products, so its not just ready only

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• Is there a name for the technique of using base-2 numbers to encode a list of unique options?

  - by Lunatik

  Apologies for the rather vague nature of this question, I've never been taught programming and Google is rather useless to a self-help guy like me in this case as the key words are pretty ambiguous. I am writing a couple of functions that encode and decode a list of options into a Long so they can easily be passed around the application, you know this kind of thing: 1 - Apple 2 - Orange 4 - Banana 8 - Plum etc. In this case the number 11 would represent Apple, Orange & Plum. I've got it working but I see this used all the time so assume there is a common name for the technique, and no doubt all sorts of best practice and clever algorithms that are at the moment just out of my reach.

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• Where can I find soft-multiply and divide algorithms?

  - by srking

  I'm working on a micro-controller without hardware multiply and divide. I need to cook up software algorithms for these basic operations that are a nice balance of compact size and efficiency. My C compiler port will employ these algos, not the the C developers themselves. My google-fu is so far turning up mostly noise on this topic. Can anyone point me to something informative? I can use add/sub and shift instructions. Table lookup based algos might also work for me, but I'm a bit worried about cramming so much into the compiler's back-end...um, so to speak. Thanks!

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• Dynamic Programming resources in C?

  - by EsotericMe

  Hi everyone, I'll be writing the online Google test tomorrow as a fresher. Apparently, they definitely ask one problem on Dynamic Programming? Does anyone know of a good resource for collection of DP problems in C along with solutions? I know what DP is & have used it on an occasion or twice. However I feel to crack a DP problem in test, prior practice of typical problems will make it easier to approach. Any good resources or problem sets with solutions in C will be highly appreciated. Thanks.

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• Generate a valid array key from an URL string in PHP

  - by John Riche

  I have a PHP array with some predefined values: $aArray = array( 0 => 'value0', 1 => 'value1' ); I need to create a function where the string input will always return the same, valid, array key so that when I call: GiveMeAKey('http://www.google.com'); // May return 0 or 1 I receive always the same key (I don't care which one) from the array. Obvisously I can't store the relationship in a database and the string passed to the GiveMeAKey method can be any URL. I wonder if there is a way of doing that ?

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• Enumerate all k-partitions of 1d array with N elements?

  - by user301217

  This seems like a simple request, but google is not my friend because "partition" scores a bunch of hits in database and filesystem space. I need to enumerate all partitions of an array of N values (N is constant) into k sub-arrays. The sub-arrays are just that - a starting index and ending index. The overall order of the original array will be preserved. For example, with N=4 and k=2: [ | a b c d ] (0, 4) [ a | b c d ] (1, 3) [ a b | c d ] (2, 2) [ a b c | d ] (3, 1) [ a b c d | ] (4, 0) I'm pretty sure this isn't an original problem (and no, it's not homework), but I'd like to do it for every k <= N, and it'd be great if the later passes (as k grows) took advantage of earlier results. If you've got a link, please share.

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• [Java] RSA BadPaddingException : data must start with zero

  - by Robin Monjo

  Hello everyone. I try to implement an RSA algorithm in a Java program. I am facing the "BadPaddingException : data must start with zero". Here are the methods used to encrypt and decrypt my data : public byte[] encrypt(byte[] input) throws Exception { Cipher cipher = Cipher.getInstance("RSA/ECB/PKCS1Padding");// cipher.init(Cipher.ENCRYPT_MODE, this.publicKey); return cipher.doFinal(input); } public byte[] decrypt(byte[] input) throws Exception { Cipher cipher = Cipher.getInstance("RSA/ECB/PKCS1Padding");/// cipher.init(Cipher.DECRYPT_MODE, this.privateKey); return cipher.doFinal(input); } privateKey and publicKey attributes are read from files this way : public PrivateKey readPrivKeyFromFile(String keyFileName) throws IOException { PrivateKey key = null; try { FileInputStream fin = new FileInputStream(keyFileName); ObjectInputStream ois = new ObjectInputStream(fin); BigInteger m = (BigInteger) ois.readObject(); BigInteger e = (BigInteger) ois.readObject(); RSAPrivateKeySpec keySpec = new RSAPrivateKeySpec(m, e); KeyFactory fact = KeyFactory.getInstance("RSA"); key = fact.generatePrivate(keySpec); ois.close(); } catch (Exception e) { e.printStackTrace(); } return key; } Private key and Public key are created this way : public void Initialize() throws Exception { KeyPairGenerator keygen = KeyPairGenerator.getInstance("RSA"); keygen.initialize(2048); keyPair = keygen.generateKeyPair(); KeyFactory fact = KeyFactory.getInstance("RSA"); RSAPublicKeySpec pub = fact.getKeySpec(keyPair.getPublic(), RSAPublicKeySpec.class); RSAPrivateKeySpec priv = fact.getKeySpec(keyPair.getPrivate(), RSAPrivateKeySpec.class); saveToFile("public.key", pub.getModulus(), pub.getPublicExponent()); saveToFile("private.key", priv.getModulus(), priv.getPrivateExponent()); } and then saved in files : public void saveToFile(String fileName, BigInteger mod, BigInteger exp) throws IOException { FileOutputStream f = new FileOutputStream(fileName); ObjectOutputStream oos = new ObjectOutputStream(f); oos.writeObject(mod); oos.writeObject(exp); oos.close(); } I can't figured out how the problem come from. Any help would be appreciate ! Thanks in advance.

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• The Collatz Sequence problem

  - by Gandalf StormCrow

  I'm trying to solve this problem, its not a homework question, its just code I'm submitting to uva.onlinejudge.org so I can learn better java trough examples. Here is the problem sample input : 3 100 34 100 75 250 27 2147483647 101 304 101 303 -1 -1 Here is simple output : Case 1: A = 3, limit = 100, number of terms = 8 Case 2: A = 34, limit = 100, number of terms = 14 Case 3: A = 75, limit = 250, number of terms = 3 Case 4: A = 27, limit = 2147483647, number of terms = 112 Case 5: A = 101, limit = 304, number of terms = 26 Case 6: A = 101, limit = 303, number of terms = 1 The thing is this has to execute within 3sec time interval otherwise your question won't be accepted as solution, here is with what I've come up so far, its working as it should just the execution time is not within 3 seconds, here is code : import java.util.Scanner; class Main { public static void main(String[] args) { Scanner stdin = new Scanner(System.in); int start; int limit; int terms; int a = 0; while (stdin.hasNext()) { start = stdin.nextInt(); limit = stdin.nextInt(); if (start > 0) { terms = getLength(start, limit); a++; } else { break; } System.out.println("Case "+a+": A = "+start+", limit = "+limit+", number of terms = "+terms); } } public static int getLength(int x, int y) { int length = 1; while (x != 1) { if (x <= y) { if ( x % 2 == 0) { x = x / 2; length++; }else{ x = x * 3 + 1; length++; } } else { length--; break; } } return length; } } And yes here is how its meant to be solved : An algorithm given by Lothar Collatz produces sequences of integers, and is described as follows: Step 1: Choose an arbitrary positive integer A as the first item in the sequence. Step 2: If A = 1 then stop. Step 3: If A is even, then replace A by A / 2 and go to step 2. Step 4: If A is odd, then replace A by 3 * A + 1 and go to step 2. And yes my question is how can I make it work inside 3 seconds time interval?

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• Are there any working implementations of the rolling hash function used in the Rabin-Karp string sea

  - by c14ppy

  I'm looking to use a rolling hash function so I can take hashes of n-grams of a very large string. For example: "stackoverflow", broken up into 5 grams would be: "stack", "tacko", "ackov", "ckove", "kover", "overf", "verfl", "erflo", "rflow" This is ideal for a rolling hash function because after I calculate the first n-gram hash, the following ones are relatively cheap to calculate because I simply have to drop the first letter of the first hash and add the new last letter of the second hash. I know that in general this hash function is generated as: H = c1ak - 1 + c2ak - 2 + c3ak - 3 + ... + cka0 where a is a constant and c1,...,ck are the input characters. If you follow this link on the Rabin-Karp string search algorithm , it states that "a" is usually some large prime. I want my hashes to be stored in 32 bit integers, so how large of a prime should "a" be, such that I don't overflow my integer? Does there exist an existing implementation of this hash function somewhere that I could already use? Here is an implementation I created: public class hash2 { public int prime = 101; public int hash(String text) { int hash = 0; for(int i = 0; i < text.length(); i++) { char c = text.charAt(i); hash += c * (int) (Math.pow(prime, text.length() - 1 - i)); } return hash; } public int rollHash(int previousHash, String previousText, String currentText) { char firstChar = previousText.charAt(0); char lastChar = currentText.charAt(currentText.length() - 1); int firstCharHash = firstChar * (int) (Math.pow(prime, previousText.length() - 1)); int hash = (previousHash - firstCharHash) * prime + lastChar; return hash; } public static void main(String[] args) { hash2 hashify = new hash2(); int firstHash = hashify.hash("mydog"); System.out.println(firstHash); System.out.println(hashify.hash("ydogr")); System.out.println(hashify.rollHash(firstHash, "mydog", "ydogr")); } } I'm using 101 as my prime. Does it matter if my hashes will overflow? I think this is desirable but I'm not sure. Does this seem like the right way to go about this?

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• J: Self-reference in bubble sort tacit implementation

  - by Yasir Arsanukaev

  Hello people! Since I'm beginner in J I've decided to solve a simple task using this language, in particular implementing the bubblesort algorithm. I know it's not idiomatically to solve such kind of problem in functional languages, because it's naturally solved using array element transposition in imperative languages like C, rather than constructing modified list in declarative languages. However this is the code I've written: (((<./@(2&{.)), $:@((>./@(2&{.)),2&}.)) ^: (1<#)) ^: # Let's apply it to an array: (((<./@(2&{.)), $:@((>./@(2&{.)),2&}.)) ^: (1<#)) ^: # 5 3 8 7 2 2 3 5 7 8 The thing that confuses me is $: referring to the statement within the outermost parentheses. Help says that: $: denotes the longest verb that contains it. The other book (~ 300 KiB) says: 3+4 7 5*20 100 Symbols like + and * for plus and times in the above phrases are called verbs and represent functions. You may have more than one verb in a J phrase, in which case it is constructed like a sentence in simple English by reading from left to right, that is 4+6%2 means 4 added to whatever follows, namely 6 divided by 2. Let's rewrite my code snippet omitting outermost ()s: ((<./@(2&{.)), $:@((>./@(2&{.)),2&}.)) ^: (1<#) ^: # 5 3 8 7 2 2 3 5 7 8 Reuslts are the same. I couldn't explain myself why this works, why only ((<./@(2&{.)), $:@((>./@(2&{.)),2&}.)) ^: (1<#) is treated as the longest verb for $: but not the whole expression ((<./@(2&{.)), $:@((>./@(2&{.)),2&}.)) ^: (1<#) ^: # and not just (<./@(2&{.)), $:@((>./@(2&{.)),2&}.), because if ((<./@(2&{.)), $:@((>./@(2&{.)),2&}.)) ^: (1<#) is a verb, it should also form another verb after conjunction with #, i. e. one might treat the whole sentence (first snippet) as a verb. Probably there's some limit for the verb length limited by one conjunction. Look at the following code (from here): factorial =: (* factorial@<:) ^: (1&<) factorial 4 24 factorial within expression refers to the whole function, i. e. (* factorial@<:) ^: (1&<). Following this example I've used a function name instead of $:: bubblesort =: (((<./@(2&{.)), bubblesort@((>./@(2&{.)),2&}.)) ^: (1<#)) ^: # bubblesort 5 3 8 7 2 2 3 5 7 8 I expected bubblesort to refer to the whole function, but it doesn't seem true for me since the result is correct. Also I'd like to see other implementations if you have ones, even slightly refactored. Thanks.

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• Drawing Bresenham’s Line- Algorithm in all quadrants

  - by Yoyo2965259

  I am newbie for OpenGL. I am practicing the exercises from my textbook but I could not get the outputs which is should be in Bresenham's Line Algorithm in all quadrants. Here's the coding: #include <Windows.h> #include <GL/glut.h> void init(void) { glClearColor(0.0, 0.0, 0.0, 0.0); glShadeModel(GL_FLAT); } void BresnCir(void) { int delta, deltadash; glClear(GL_COLOR_BUFFER_BIT); glPointSize(3.0); int r = 150; int x = 0; int y = r; int D = 2 * (1 - r); glBegin(GL_POINTS); do { glVertex2i(x, y); if (D < 0) { delta = 2 * D + 2 * y - 1; if (delta <= 0) { x++; Right(x); } else { x++; y--; Diagonal(x, y); } glVertex2i(x, y); } else { deltadash = 2 * D - 2 * x - 1; if (deltadash <= 0) { x++; y--; Diagonal(x, y); } else { y--; Down(y); } glVertex2i(x, y); } if (D == 0) { x++; y--; Diagonal(x, y); glVertex2i(x, y); } } while (y > 0); glEnd(); glFlush(); } int main(int argc, char** argv) { glutInit(&argc, argv); glutInitDisplayMode(GLUT_SINGLE | GLUT_RGB); glutInitWindowSize(400, 150); glutInitWindowPosition(100, 100); glutCreateWindow(argv[0]); init(); glutDisplayFunc(BresnCir); glutMainLoop(); return 0; } But, it keep comes out with errors C3861.

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• Miller-rabin exception number?

  - by nightcracker

  Hey everyone. This question is about the number 169716931325235658326303. According to http://www.alpertron.com.ar/ECM.HTM it is prime. According to my miller-rabin implementation in python with 7 repetitions is is composite. With 50 repetitions it is still composite. With 5000 repetitions it is STILL composite. I thought, this might be a problem of my implementation. So I tried GNU MP bignum library, which has a miller-rabin primality test built-in. I tested with 1000000 repetitions. Still composite. This is my implementation of the miller-rabin primality test: def isprime(n, precision=7): if n == 1 or n % 2 == 0: return False elif n < 1: raise ValueError("Out of bounds, first argument must be > 0") d = n - 1 s = 0 while d % 2 == 0: d //= 2 s += 1 for repeat in range(precision): a = random.randrange(2, n - 2) x = pow(a, d, n) if x == 1 or x == n - 1: continue for r in range(s - 1): x = pow(x, 2, n) if x == 1: return False if x == n - 1: break else: return False return True And the code for the GMP test: #include <gmp.h> #include <stdio.h> int main(int argc, char* argv[]) { mpz_t test; mpz_init_set_str(test, "169716931325235658326303", 10); printf("%d\n", mpz_probab_prime_p(test, 1000000)); mpz_clear(test); return 0; } As far as I know there are no "exceptions" (which return false positives for any amount of repetitions) to the miller-rabin primality test. Have I stumpled upon one? Is my computer broken? Is the Elliptic Curve Method wrong? What is happening here? EDIT I found the issue, which is http://www.alpertron.com.ar/ECM.HTM. I trusted this applet, I'll contact the author his applet's implementation of the ECM is faulty for this number. Thanks. EDIT2 Hah, the shame! In the end it was something that went wrong with copy/pasting on my side. NOR the applet NOR the miller-rabin algorithm NOR my implementation NOR gmp's implementation of it is wrong, the only thing that's wrong is me. I'm sorry.

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