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• Page replacement algorithm in Java

  - by SamSol

  Is there utility program for Page replacement algorithm in Java?

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• Please explain how Trial Division works for Primality Test

  - by mister_dani

  I came across this algorithm for testing primality through trial division I fully understand this algorithm static boolean isPrime(int N) { if (N < 2) return false; for (int i = 2; i <= Math.sqrt(N); i++) if (N % i == 0) return false; return true; } It works just fine. But then I came across this other one which works just as good but I do not fully understand the logic behind it. static boolean isPrime(int N) { if (N < 2) return false; for (int i = 2; i * i<N; i++) if (N % i == 0) return false; return true; } It seems like i *i < N behaves like i <= Math.sqrt(N). If so, why?

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• question on revrse array

  - by davit-datuashvili

  we know algorithm how reverse array of n integers for (int i=0;i<n/2;i++){ swap(a[i],a[n-1-i]): } is this method better according the speed of algorithm or not because swap using xor is more fast then in other method here is code public class swap{ public static void main(String[]args){ int a[]=new int[]{2,4,5,7,8,11,13,12,14,24}; System.out.println(" array at the begining:"); for (int i=0;i<a.length;i++){ System.out.println(a[i]); } for (int j=0;j<a.length/2;j++){ a[j]^=a[a.length-1-j]; a[a.length-1-j]^=a[j]; a[j]^=a[a.length-1-j]; } System.out.println("reversed array:"); for (int j=0;j<a.length;j++){ System.out.println(a[j]); } } } //result array at the begining: 2 4 5 7 8 11 13 12 14 24 reversed array: 24 14 12 13 11 8 7 5 4 2

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• Compare two integer arrays with same length

  - by meta

  [Description] Given two integer arrays with the same length. Design an algorithm which can judge whether they're the same, the definition of "same" is that, if these two arrays are in sorted order, the elements in corresponding position should be the same. [Example] <1 2 3 4> = <3 1 2 4> <1 2 3 4> != <3 4 1 1> [Limitation] The algorithm should require constant extra space, and O(n) running time.

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• Small questions on data structure

  - by John Graveston

  Hi, I'm trying to search the parent of a node with Kruskal's algorithm. My program works just fine, but I think I have heard of a method to improve the speed of the algorithm by reconstructing the tree while searching for the parent node and connecting it to the parent node. I'm pretty sure that I've heard of this somewhere, maybe in a lecture. Can anyone refresh my memory? And also, given a number of arrays, when searching for the minimum and the maximum value from a certain section of an array, what is the name of the tree that can calculate the minimum/maximum value from the array by making a binary tree that has the minimum/maximum value of each array in O(log N)?

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• file names based on file content

  - by Mark

  So iow, some algorithm to generate a unique, reasonable length filename based on binary file content. Two files that have the same binary content should have the same name. Obviously there would be limits to this, as presumably you couldn't have unique reasonable length filenames for each of a large set of large files only differing at a handful of bit positions. But presumably there is some heuristic, best approximation to this that for example exploits known attributes of typical image files. If I had the name of some algorithm that does this I can google it and find other approaches as well.

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• question on reverse array

  - by davit-datuashvili

  we know algorithm how reverse array of n integers for (int i=0;i<n/2;i++){ swap(a[i],a[n-1-i]): } is this method better according the speed of algorithm or not because swap using xor is more fast then in other method here is code public class swap { public static void main(String[]args){ int a[]=new int[]{2,4,5,7,8,11,13,12,14,24}; System.out.println(" array at the begining:"); for (int i=0;i<a.length;i++){ System.out.println(a[i]); } for (int j=0;j<a.length/2;j++){ a[j]^=a[a.length-1-j]; a[a.length-1-j]^=a[j]; a[j]^=a[a.length-1-j]; } System.out.println("reversed array:"); for (int j=0;j<a.length;j++){ System.out.println(a[j]); } } } Result: array at the begining: 2 4 5 7 8 11 13 12 14 24 reversed array: 24 14 12 13 11 8 7 5 4 2

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• C# Improved algorithm

  - by generixs

  I have been asked at interview (C# 3.0) to provide a logic to remove a list of items from a list. I responded int[] items={1,2,3,4}; List<int> newList = new List<int>() { 1, 2, 3, 4, 5, 56, 788, 9 }; newList.RemoveAll((int i) => { return items.Contains(i); }); 1) The interviewer replied that the algorithm i had employed will gradually take time if the items grow and asked me to give even better and faster one.What would be the efficient algorithm ? 2) How can i achieve the same using LINQ? 3) He asked me to provide an example for Two-Way-Closure? (General I am aware of closure, what is Two-Way-Closure?, I replied there is no such term exists,but he did not satisfy).

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• How can I estimate the entropy of a password?

  - by Wug

  Having read various resources about password strength I'm trying to create an algorithm that will provide a rough estimation of how much entropy a password has. I'm trying to create an algorithm that's as comprehensive as possible. At this point I only have pseudocode, but the algorithm covers the following: password length repeated characters patterns (logical) different character spaces (LC, UC, Numeric, Special, Extended) dictionary attacks It does NOT cover the following, and SHOULD cover it WELL (though not perfectly): ordering (passwords can be strictly ordered by output of this algorithm) patterns (spatial) Can anyone provide some insight on what this algorithm might be weak to? Specifically, can anyone think of situations where feeding a password to the algorithm would OVERESTIMATE its strength? Underestimations are less of an issue. The algorithm: // the password to test password = ? length = length(password) // unique character counts from password (duplicates discarded) uqlca = number of unique lowercase alphabetic characters in password uquca = number of uppercase alphabetic characters uqd = number of unique digits uqsp = number of unique special characters (anything with a key on the keyboard) uqxc = number of unique special special characters (alt codes, extended-ascii stuff) // algorithm parameters, total sizes of alphabet spaces Nlca = total possible number of lowercase letters (26) Nuca = total uppercase letters (26) Nd = total digits (10) Nsp = total special characters (32 or something) Nxc = total extended ascii characters that dont fit into other categorys (idk, 50?) // algorithm parameters, pw strength growth rates as percentages (per character) flca = entropy growth factor for lowercase letters (.25 is probably a good value) fuca = EGF for uppercase letters (.4 is probably good) fd = EGF for digits (.4 is probably good) fsp = EGF for special chars (.5 is probably good) fxc = EGF for extended ascii chars (.75 is probably good) // repetition factors. few unique letters == low factor, many unique == high rflca = (1 - (1 - flca) ^ uqlca) rfuca = (1 - (1 - fuca) ^ uquca) rfd = (1 - (1 - fd ) ^ uqd ) rfsp = (1 - (1 - fsp ) ^ uqsp ) rfxc = (1 - (1 - fxc ) ^ uqxc ) // digit strengths strength = ( rflca * Nlca + rfuca * Nuca + rfd * Nd + rfsp * Nsp + rfxc * Nxc ) ^ length entropybits = log_base_2(strength) A few inputs and their desired and actual entropy_bits outputs: INPUT DESIRED ACTUAL aaa very pathetic 8.1 aaaaaaaaa pathetic 24.7 abcdefghi weak 31.2 H0ley$Mol3y_ strong 72.2 s^fU¬5ü;y34G< wtf 88.9 [a^36]* pathetic 97.2 [a^20]A[a^15]* strong 146.8 xkcd1** medium 79.3 xkcd2** wtf 160.5 * these 2 passwords use shortened notation, where [a^N] expands to N a's. ** xkcd1 = "Tr0ub4dor&3", xkcd2 = "correct horse battery staple" The algorithm does realize (correctly) that increasing the alphabet size (even by one digit) vastly strengthens long passwords, as shown by the difference in entropy_bits for the 6th and 7th passwords, which both consist of 36 a's, but the second's 21st a is capitalized. However, they do not account for the fact that having a password of 36 a's is not a good idea, it's easily broken with a weak password cracker (and anyone who watches you type it will see it) and the algorithm doesn't reflect that. It does, however, reflect the fact that xkcd1 is a weak password compared to xkcd2, despite having greater complexity density (is this even a thing?). How can I improve this algorithm? Addendum 1 Dictionary attacks and pattern based attacks seem to be the big thing, so I'll take a stab at addressing those. I could perform a comprehensive search through the password for words from a word list and replace words with tokens unique to the words they represent. Word-tokens would then be treated as characters and have their own weight system, and would add their own weights to the password. I'd need a few new algorithm parameters (I'll call them lw, Nw ~= 2^11, fw ~= .5, and rfw) and I'd factor the weight into the password as I would any of the other weights. This word search could be specially modified to match both lowercase and uppercase letters as well as common character substitutions, like that of E with 3. If I didn't add extra weight to such matched words, the algorithm would underestimate their strength by a bit or two per word, which is OK. Otherwise, a general rule would be, for each non-perfect character match, give the word a bonus bit. I could then perform simple pattern checks, such as searches for runs of repeated characters and derivative tests (take the difference between each character), which would identify patterns such as 'aaaaa' and '12345', and replace each detected pattern with a pattern token, unique to the pattern and length. The algorithmic parameters (specifically, entropy per pattern) could be generated on the fly based on the pattern. At this point, I'd take the length of the password. Each word token and pattern token would count as one character; each token would replace the characters they symbolically represented. I made up some sort of pattern notation, but it includes the pattern length l, the pattern order o, and the base element b. This information could be used to compute some arbitrary weight for each pattern. I'd do something better in actual code. Modified Example: Password: 1234kitty$$$$$herpderp Tokenized: 1 2 3 4 k i t t y $ $ $ $ $ h e r p d e r p Words Filtered: 1 2 3 4 @W5783 $ $ $ $ $ @W9001 @W9002 Patterns Filtered: @P[l=4,o=1,b='1'] @W5783 @P[l=5,o=0,b='$'] @W9001 @W9002 Breakdown: 3 small, unique words and 2 patterns Entropy: about 45 bits, as per modified algorithm Password: correcthorsebatterystaple Tokenized: c o r r e c t h o r s e b a t t e r y s t a p l e Words Filtered: @W6783 @W7923 @W1535 @W2285 Breakdown: 4 small, unique words and no patterns Entropy: 43 bits, as per modified algorithm The exact semantics of how entropy is calculated from patterns is up for discussion. I was thinking something like: entropy(b) * l * (o + 1) // o will be either zero or one The modified algorithm would find flaws with and reduce the strength of each password in the original table, with the exception of s^fU¬5ü;y34G<, which contains no words or patterns.

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• Edges on polygon outlines not always correct

  - by user146780

  I'm using the algorithm below to generate quads which are then rendered to make an outline like this http://img810.imageshack.us/img810/8530/uhohz.png The problem as seen on the image, is that sometimes the lines are too thin when they should always be the same width. My algorithm finds the 4 verticies for the first one then the top 2 verticies of the next ones are the bottom 2 of the previous. This creates connected lines, but it seems to not always work. How could I fix this? This is my algorithm: void OGLENGINEFUNCTIONS::GenerateLinePoly(const std::vector<std::vector<GLdouble>> &input, std::vector<GLfloat> &output, int width) { output.clear(); if(input.size() < 2) { return; } int temp; float dirlen; float perplen; POINTFLOAT start; POINTFLOAT end; POINTFLOAT dir; POINTFLOAT ndir; POINTFLOAT perp; POINTFLOAT nperp; POINTFLOAT perpoffset; POINTFLOAT diroffset; POINTFLOAT p0, p1, p2, p3; for(unsigned int i = 0; i < input.size() - 1; ++i) { start.x = static_cast<float>(input[i][0]); start.y = static_cast<float>(input[i][1]); end.x = static_cast<float>(input[i + 1][0]); end.y = static_cast<float>(input[i + 1][1]); dir.x = end.x - start.x; dir.y = end.y - start.y; dirlen = sqrt((dir.x * dir.x) + (dir.y * dir.y)); ndir.x = static_cast<float>(dir.x * 1.0 / dirlen); ndir.y = static_cast<float>(dir.y * 1.0 / dirlen); perp.x = dir.y; perp.y = -dir.x; perplen = sqrt((perp.x * perp.x) + (perp.y * perp.y)); nperp.x = static_cast<float>(perp.x * 1.0 / perplen); nperp.y = static_cast<float>(perp.y * 1.0 / perplen); perpoffset.x = static_cast<float>(nperp.x * width * 0.5); perpoffset.y = static_cast<float>(nperp.y * width * 0.5); diroffset.x = static_cast<float>(ndir.x * 0 * 0.5); diroffset.y = static_cast<float>(ndir.y * 0 * 0.5); // p0 = start + perpoffset - diroffset //p1 = start - perpoffset - diroffset //p2 = end + perpoffset + diroffset // p3 = end - perpoffset + diroffset p0.x = start.x + perpoffset.x - diroffset.x; p0.y = start.y + perpoffset.y - diroffset.y; p1.x = start.x - perpoffset.x - diroffset.x; p1.y = start.y - perpoffset.y - diroffset.y; if(i > 0) { temp = (8 * (i - 1)); p2.x = output[temp + 2]; p2.y = output[temp + 3]; p3.x = output[temp + 4]; p3.y = output[temp + 5]; } else { p2.x = end.x + perpoffset.x + diroffset.x; p2.y = end.y + perpoffset.y + diroffset.y; p3.x = end.x - perpoffset.x + diroffset.x; p3.y = end.y - perpoffset.y + diroffset.y; } output.push_back(p2.x); output.push_back(p2.y); output.push_back(p0.x); output.push_back(p0.y); output.push_back(p1.x); output.push_back(p1.y); output.push_back(p3.x); output.push_back(p3.y); } } Thanks

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• How can I improve my isometric tile-picking algorithm?

  - by Cypher

  I've spent the last few days researching isometric tile-picking algorithms (converting screen-coordinates to tile-coordinates), and have obviously found a lot of the math beyond my grasp. I have come fairly close and what I have is workable, but I would like to improve on this algorithm as it's a little off and seems to pick down and to the right of the mouse pointer. I've uploaded a video to help visualize the current implementation: http://youtu.be/EqwWcq1zuaM My isometric rendering algorithm is based on what is found at this stackoverflow question's answer, with the exception that my x and y axis' are inverted (x increased down-right, while y increased up-right). Here is where I am converting from screen to tiles: // these next few lines convert the mouse pointer position from screen // coordinates to tile-grid coordinates. cameraOffset captures the current // mouse location and takes into consideration the camera's position on screen. System.Drawing.Point cameraOffset = new System.Drawing.Point( 0, 0 ); cameraOffset.X = mouseLocation.X + (int)camera.Left; cameraOffset.Y = ( mouseLocation.Y + (int)camera.Top ); // the camera-aware mouse coordinates are then further converted in an attempt // to select only the "tile" portion of the grid tiles, instead of the entire // rectangle. this algorithm gets close, but could use improvement. mouseTileLocation.X = ( cameraOffset.X + 2 * cameraOffset.Y ) / Global.TileWidth; mouseTileLocation.Y = -( ( 2 * cameraOffset.Y - cameraOffset.X ) / Global.TileWidth ); Things to make note of: mouseLocation is a System.Drawing.Point that represents the screen coordinates of the mouse pointer. cameraOffset is the screen position of the mouse pointer that includes the position of the game camera. mouseTileLocation is a System.Drawing.Point that is supposed to represent the tile coordinates of the mouse pointer. If you check out the above link to youtube, you'll notice that the picking algorithm is off a bit. How can I improve on this?

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• Determining the maximum stack depth

  - by Joa Ebert

  Imagine I have a stack-based toy language that comes with the operations Push, Pop, Jump and If. I have a program and its input is the toy language. For instance I get the sequence Push 1 Push 1 Pop Pop In that case the maximum stack would be 2. A more complicated example would use branches. Push 1 Push true If .success Pop Jump .continue .success: Push 1 Push 1 Pop Pop Pop .continue: In this case the maximum stack would be 3. However it is not possible to get the maximum stack by walking top to bottom as shown in this case since it would result in a stack-underflow error actually. CFGs to the rescue you can build a graph and walk every possible path of the basic blocks you have. However since the number of paths can grow quickly for n vertices you get (n-1)! possible paths. My current approach is to simplify the graph as much as possible and to have less possible paths. This works but I would consider it ugly. Is there a better (read: faster) way to attack this problem? I am fine if the algorithm produces a stack depth that is not optimal. If the correct stack size is m then my only constraint is that the result n is n = m. Is there maybe a greedy algorithm available that would produce a good result here?

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• Sorting tasks to assign

  - by Diego

  I've got a problem that I don't know where to start. I'd realy appreciate some help. The problem: I have several T task that must be done in D days by just 1 employee (let's forget using several resources right now). Each task can be done in some times (not all tasks can be done all time). e.g.: If my employee starts working at 8 o'clock and one task is "call a client". Maybe the client office opens at 9 o'clock. Also each task has a duration (really estimated). It is supposed that the D days are enough to do all task. I've to sort the tasks to the employee. e.g.: at monday 8:00 do task 7, then at 9:30 starts with task 2. In the example task 7 duration would be 1 and a half hour. Thanks for the help! Diego PD: If someone has a way to make this and it is not an algorithm never minds, please answer and I'll manage to think the algorithm. I just don't know how to face the problem. Edit Would Project be usefull? Edit 2 Tasks / Jobs dependency is NOT required

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• Programmatically creating vector arrows in KML

  - by mettadore

  Does anyone have any practical examples of programmatically drawing icons as vectors in KML? Specifically, I have data with a magnitude and an azimuth at given coordinates, and I would like to have icons (or another graphical element) generated based on these values. Some thoughts on how I might approach it: Image directory (a brute force way): Make an image director of 360 different image files (probably batch rotate a single image) each pointing in a cooresponding azimuth. I've seen things like "Excel to KML," but am looking for code that I can use within a program, rather than a web utility. Issue: Arrow does not contain magnitude context, so that would have to be a label. I'd rather dynamically lengthen the arrow. Line creation in KML: Perhaps create a formula that creates a line with the origin at the coordinate points, with the length of the line proportional to the magnitute, and angled according to azimuth. There would then be two more lines, perhaps 30 degrees or so extending from the end of the previous line to make the arrow head. Issues: Not a separate image icon, so not sure how it would work in KML. Also not sure how easy it would be to generate this algorithm. Separate image generation: Perhaps create a PHP file that uses imagemagick or something similar to dynamically generate a .png file in a similar method to the above, and then link to the icon using the URI "domain.tld/imagegen.php?magnitude=magvalue&azimuth=azmvalue". Issue: Still have the problem of actually writing the algorithm for image generation. So, the question: has anyone else come up with solutions for programmatic vector (rather than merely arrow) generation?

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• method for specialized pathfinding?

  - by rlbond

  I am working on a roguelike in my (very little) free time. Each level will basically be a few rectangular rooms connected together by paths. I want the paths between rooms to be natural-looking and windy, however. For example, I would not consider the following natural-looking: B X X X XX XX XX AXX I really want something more like this: B X XXXX X X X X AXXXXXXXX These paths must satisfy a few properties: I must be able to specify an area inside which they are bounded, I must be able to parameterize how windy and lengthy they are, The lines should not look like they started at one path and ended at the other. For example, the first example above looks as if it started at A and ended at B, because it basically changed directions repeatedly until it lined up with B and then just went straight there. I was hoping to use A*, but honestly I have no idea what my heuristic would be. I have also considered using a genetic algorithm, but I don't know how practical that method might end up. My question is, what is a good way to get the results I want? Please do not just specify a method like "A*" or "Dijkstra's algorithm," because I also need help with a good heuristic.

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• Compression algorithm for IEEE-754 data

  - by David Taylor

  Anyone have a recommendation on a good compression algorithm that works well with double precision floating point values? We have found that the binary representation of floating point values results in very poor compression rates with common compression programs (e.g. Zip, RAR, 7-Zip etc). The data we need to compress is a one dimensional array of 8-byte values sorted in monotonically increasing order. The values represent temperatures in Kelvin with a span typically under of 100 degrees. The number of values ranges from a few hundred to at most 64K. Clarifications All values in the array are distinct, though repetition does exist at the byte level due to the way floating point values are represented. A lossless algorithm is desired since this is scientific data. Conversion to a fixed point representation with sufficient precision (~5 decimals) might be acceptable provided there is a significant improvement in storage efficiency. Update Found an interesting article on this subject. Not sure how applicable the approach is to my requirements. http://users.ices.utexas.edu/~burtscher/papers/dcc06.pdf

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• 3x3 Sobel operator and gradient features

  - by pithyless

  Reading a paper, I'm having difficulty understanding the algorithm described: Given a black and white digital image of a handwriting sample, cut out a single character to analyze. Since this can be any size, the algorithm needs to take this into account (if it will be easier, we can assume the size is 2^n x 2^m). Now, the description states given this image we will convert it to a 512-bit feature (a 512-bit hash) as follows: (192 bits) computes the gradient of the image by convolving it with a 3x3 Sobel operator. The direction of the gradient at every edge is quantized to 12 directions. (192 bits) The structural feature generator takes the gradient map and looks in a neighborhood for certain combinations of gradient values. (used to compute 8 distinct features that represent lines and corners in the image) (128 bits) Concavity generator uses an 8-point star operator to find coarse concavities in 4 directions, holes, and lagrge-scale strokes. The image feature maps are normalized with a 4x4 grid. I'm for now struggling with how to take an arbitrary image, split into 16 sections, and using a 3x3 Sobel operator to come up with 12 bits for each section. (But if you have some insight into the other parts, feel free to comment :)

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• Bitwise Interval Arithmetic

  - by KennyTM

  I've recently read an interesting thread on the D newsgroup, which basically asks, Given two signed integers a ∈ [amin, amax], b ∈ [bmin, bmax], what is the tightest interval of a | b? I'm think if interval arithmetics can be applied on general bitwise operators (assuming infinite bits). The bitwise-NOT and shifts are trivial since they just corresponds to -1 − x and 2n x. But bitwise-AND/OR are a lot trickier, due to the mix of bitwise and arithmetic properties. Is there a polynomial-time algorithm to compute the intervals of bitwise-AND/OR? Note: Assume all bitwise operations run in linear time (of number of bits), and test/set a bit is constant time. The brute-force algorithm runs in exponential time. Because ~(a | b) = ~a & ~b and a ^ b = (a | b) & ~(a & b), solving the bitwise-AND and -NOT problem implies bitwise-OR and -XOR are done. Although the content of that thread suggests min{a | b} = max(amin, bmin), it is not the tightest bound. Just consider [2, 3] | [8, 9] = [10, 11].)

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• C++ design question on traversing binary trees

  - by user231536

  I have a binary tree T which I would like to copy to another tree. Suppose I have a visit method that gets evaluated at every node: struct visit { virtual void operator() (node* n)=0; }; and I have a visitor algorithm void visitor(node* t, visit& v) { //do a preorder traversal using stack or recursion if (!t) return; v(t); visitor(t->left, v); visitor(t->right, v); } I have 2 questions: I settled on using the functor based approach because I see that boost graph does this (vertex visitors). Also I tend to repeat the same code to traverse the tree and do different things at each node. Is this a good design to get rid of duplicated code? What other alternative designs are there? How do I use this to create a new binary tree from an existing one? I can keep a stack on the visit functor if I want, but it gets tied to the algorithm in visitor. How would I incorporate postorder traversals here ? Another functor class?

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• Bubble Breaker Game Solver better than greedy?

  - by Gregory

  For a mental exercise I decided to try and solve the bubble breaker game found on many cell phones as well as an example here:Bubble Break Game The random (N,M,C) board consists N rows x M columns with C colors The goal is to get the highest score by picking the sequence of bubble groups that ultimately leads to the highest score A bubble group is 2 or more bubbles of the same color that are adjacent to each other in either x or y direction. Diagonals do not count When a group is picked, the bubbles disappear, any holes are filled with bubbles from above first, ie shift down, then any holes are filled by shifting right A bubble group score = n * (n - 1) where n is the number of bubbles in the bubble group The first algorithm is a simple exhaustive recursive algorithm which explores going through the board row by row and column by column picking bubble groups. Once the bubble group is picked, we create a new board and try to solve that board, recursively descending down Some of the ideas I am using include normalized memoization. Once a board is solved we store the board and the best score in a memoization table. I create a prototype in python which shows a (2,15,5) board takes 8859 boards to solve in about 3 seconds. A (3,15,5) board takes 12,384,726 boards in 50 minutes on a server. The solver rate is ~3k-4k boards/sec and gradually decreases as the memoization search takes longer. Memoization table grows to 5,692,482 boards, and hits 6,713,566 times. What other approaches could yield high scores besides the exhaustive search? I don't seen any obvious way to divide and conquer. But trending towards larger and larger bubbles groups seems to be one approach Thanks to David Locke for posting the paper link which talks above a window solver which uses a constant-depth lookahead heuristic.

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• intelligent path truncation/ellipsis for display

  - by peterchen

  I am looking for an existign path truncation algorithm (similar to what the Win32 static control does with SS_PATHELLIPSIS) for a set of paths that should focus on the distinct elements. For example, if my paths are like this: Unit with X/Test 3V/ Unit with X/Test 4V/ Unit with X/Test 5V/ Unit without X/Test 3V/ Unit without X/Test 6V/ Unit without X/2nd Test 6V/ When not enough display space is available, they should be truncated to something like this: ...with X/...3V/ ...with X/...4V/ ...with X/...5V/ ...without X/...3V/ ...without X/...6V/ ...without X/2nd ...6V/ (Assuming that an ellipsis generally is shorter than three letters). This is just an example of a rather simple, ideal case (e.g. they'd all end up at different lengths now, and I wouldn't know how to create a good suggestion when a path "Thingie/Long Test/" is added to the pool). There is no given structure of the path elements, they are assigned by the user, but often items will have similar segments. It should work for proportional fonts, so the algorithm should take a measure function (and not call it to heavily) or generate a suggestion list. Data-wise, a typical use case would contain 2..4 path segments anf 20 elements per segment. I am looking for previous attempts into that direction, and if that's solvable wiht sensible amount of code or dependencies.

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• reconstructing a tree from its preorder and postorder lists.

  - by NomeN

  Consider the situation where you have two lists of nodes of which all you know is that one is a representation of a preorder traversal of some tree and the other a representation of a postorder traversal of the same tree. I believe it is possible to reconstruct the tree exactly from these two lists, and I think I have an algorithm to do it, but have not proven it. As this will be a part of a masters project I need to be absolutely certain that it is possible and correct (Mathematically proven). However it will not be the focus of the project, so I was wondering if there is a source out there (i.e. paper or book) I could quote for the proof. (Maybe in TAOCP? anybody know the section possibly?) In short, I need a proven algorithm in a quotable resource that reconstructs a tree from its pre and post order traversals. Note: The tree in question will probably not be binary, or balanced, or anything that would make it too easy. Note2: Using only the preorder or the postorder list would be even better, but I do not think it is possible. Note3: A node can have any amount of children. Note4: I only care about the order of siblings. Left or right does not matter when there is only one child.

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• looking for a license key algorithm.

  - by giulio

  There are alot of questions relating to license keys asked on stackoverflow. But they don't answer this question. Can anyone provide a simple license key algorithm that is technology independent and doesn't required a diploma in mathematics to understand ? The license key algorithm is similar to public key encryption. I just need something simple that can be implemented in any platform .Net/Java and uses simple data like characters. Written as Pseudo code is perfect. So if a person presents a string, a complementary string can be generated that is the authorisation code. Below is a common scenario that it would be used for. Customer downloads s/w which generates a unique key upon initial startup/installation. S/w runs during trial period. At end of trial period an authorisation key is required. Customer goes to designated web-site, enters their code and get authorisation code to enable s/w, after paying :) Don't be afraid to describe your answer as though you're talking to a 5 yr old as I am not a mathemtician.

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• Small-o(n^2) implementation of Polynomial Multiplication

  - by AlanTuring

  I'm having a little trouble with this problem that is listed at the back of my book, i'm currently in the middle of test prep but i can't seem to locate anything regarding this in the book. Anyone got an idea? A real polynomial of degree n is a function of the form f(x)=a(n)x^n+?+a1x+a0, where an,…,a1,a0 are real numbers. In computational situations, such a polynomial is represented by a sequence of its coefficients (a0,a1,…,an). Assuming that any two real numbers can be added/multiplied in O(1) time, design an o(n^2)-time algorithm to compute, given two real polynomials f(x) and g(x) both of degree n, the product h(x)=f(x)g(x). Your algorithm should **not** be based on the Fast Fourier Transform (FFT) technique. Please note it needs to be small-o(n^2), which means it complexity must be sub-quadratic. The obvious solution that i have been finding is indeed the FFT, but of course i can't use that. There is another method that i have found called convolution, where if you take polynomial A to be a signal and polynomial B to be a filter. A passed through B yields a shifted signal that has been "smoothed" by A and the resultant is A*B. This is supposed to work in O(n log n) time. Of course i am completely unsure of implementation. If anyone has any ideas of how to achieve a small-o(n^2) implementation please do share, thanks.

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• Intelligent web features, algorithms (people you may follow, similar to you ...)

  - by hilal

  I have 3 main questions about the algorithms in intelligent web (web 2.0) Here the book I'm reading http://www.amazon.com/Algorithms-Intelligent-Web-Haralambos-Marmanis/dp/1933988665 and I want to learn the algorithms in deeper 1. People You may follow (Twitter) How can one determine the nearest result to my requests ? Data mining? which algorithms? 2. How you’re connected feature (Linkedin) Simply algorithm works like that. It draws the path between two nodes let say between Me and the other person is C. Me - A, B - A connections - C . It is not any brute force algorithms or any other like graph algorithms :) 3. Similar to you (Twitter, Facebook) This algorithms is similar to 1. Does it simply work the max(count) friend in common (facebook) or the max(count) follower in Twitter? or any other algorithms they implement? I think the second part is true because running the loop dict{count, person} for person in contacts: dict.add(count(common(person))) return dict(max) is a silly act in every refreshing page. 4. Did you mean (Google) I know that they may implement it with phonetic algorithm http://en.wikipedia.org/wiki/Phonetic_algorithm simply soundex http://en.wikipedia.org/wiki/Soundex and here is the Google VP of Engineering and CIO Douglas Merrill speak http://www.youtube.com/watch?v=syKY8CrHkck#t=22m03s What about first 3 questions? Any ideas are welcome ! Thanks

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