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• DSA signature verification input

  - by calccrypto

  What is the data inputted into DSA when PGP signs a message? From RFC4880, i found A Signature packet describes a binding between some public key and some data. The most common signatures are a signature of a file or a block of text, and a signature that is a certification of a User ID. im not sure if it is the entire public key, just the public key packet, or some other derivative of a pgp key packet. whatever it is, i cannot get the DSA signature to verify here is a sample im testing my program on: -----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 abcd -----BEGIN PGP SIGNATURE----- Version: BCPG v1.39 iFkEARECABkFAk0z65ESHGFiYyAodGVzdCBrZXkpIDw+AAoJEC3Jkh8+bnkusO0A oKG+HPF2Qrsth2zS9pK+eSCBSypOAKDBgC2Z0vf2EgLiiNMk8Bxpq68NkQ== =gq0e -----END PGP SIGNATURE----- Dumped from pgpdump.net Old: Signature Packet(tag 2)(89 bytes) Ver 4 - new Sig type - Signature of a canonical text document(0x01). Pub alg - DSA Digital Signature Algorithm(pub 17) Hash alg - SHA1(hash 2) Hashed Sub: signature creation time(sub 2)(4 bytes) Time - Mon Jan 17 07:11:13 UTC 2011 Hashed Sub: signer's User ID(sub 28)(17 bytes) User ID - abc (test key) <> Sub: issuer key ID(sub 16)(8 bytes) Key ID - 0x2DC9921F3E6E792E Hash left 2 bytes - b0 ed DSA r(160 bits) - a1 be 1c f1 76 42 bb 2d 87 6c d2 f6 92 be 79 20 81 4b 2a 4e DSA s(160 bits) - c1 80 2d 99 d2 f7 f6 12 02 e2 88 d3 24 f0 1c 69 ab af 0d 91 -> hash(DSA q bits) and the public key for it is: -----BEGIN PGP PUBLIC KEY BLOCK----- Version: BCPG v1.39 mOIETTPqeBECALx+i9PIc4MB2DYXeqsWUav2cUtMU1N0inmFHSF/2x0d9IWEpVzE kRc30PvmEHI1faQit7NepnHkkphrXLAoZukAoNP3PB8NRQ6lRF6/6e8siUgJtmPL Af9IZOv4PI51gg6ICLKzNO9i3bcUx4yeG2vjMOUAvsLkhSTWob0RxWppo6Pn6MOg dMQHIM5sDH0xGN0dOezzt/imAf9St2B0HQXVfAAbveXBeRoO7jj/qcGx6hWmsKUr BVzdQhBk7Sku6C2KlMtkbtzd1fj8DtnrT8XOPKGp7/Y7ASzRtBFhYmMgKHRlc3Qg a2V5KSA8PohGBBMRAgAGBQJNM+p5AAoJEC3Jkh8+bnkuNEoAnj2QnqGtdlTgUXCQ Fyvwk5wiLGPfAJ4jTGTL62nWzsgrCDIMIfEG2shm8bjMBE0z6ngQAgCUlP7AlfO4 XuKGVCs4NvyBpd0KA0m0wjndOHRNSIz44x24vLfTO0GrueWjPMqRRLHO8zLJS/BX O/BHo6ypjN87Af0VPV1hcq20MEW2iujh3hBwthNwBWhtKdPXOndJGZaB7lshLJuW v9z6WyDNXj/SBEiV1gnPm0ELeg8Syhy5pCjMAgCFEc+NkCzcUOJkVpgLpk+VLwrJ /Wi9q+yCihaJ4EEFt/7vzqmrooXWz2vMugD1C+llN6HkCHTnuMH07/E/2dzciEYE GBECAAYFAk0z6nkACgkQLcmSHz5ueS7NTwCdED1P9NhgR2LqwyS+AEyqlQ0d5joA oK9xPUzjg4FlB+1QTHoOhuokxxyN =CTgL -----END PGP PUBLIC KEY BLOCK----- the public key packet of the key is mOIETTPqeBECALx+i9PIc4MB2DYXeqsWUav2cUtMU1N0inmFHSF/2x0d9IWEpVzEkRc30PvmEHI1faQi t7NepnHkkphrXLAoZukAoNP3PB8NRQ6lRF6/6e8siUgJtmPLAf9IZOv4PI51gg6ICLKzNO9i3bcUx4ye G2vjMOUAvsLkhSTWob0RxWppo6Pn6MOgdMQHIM5sDH0xGN0dOezzt/imAf9St2B0HQXVfAAbveXBeRoO 7jj/qcGx6hWmsKUrBVzdQhBk7Sku6C2KlMtkbtzd1fj8DtnrT8XOPKGp7/Y7ASzR in radix 64 i have tried many different combinations of sha1(< some data + 'abcd'),but the calculated value v never equals r, of the signature i know that the pgp implementation i used to create the key and signature is correct. i also know that my DSA implementation and PGP key data extraction program are correct. thus, the only thing left is the data to hash. what is the correct data to be hashed?

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• Is there a Java library with 3D spline functions?

  - by Liam

  In particular, I need a way to represent a curve/spline that passes through a set of known 3D points, and a way of finding other points on the curve/spline, by subdivision/interpolation. For example, if I have a set of points P0 to PN, I want to find 100 points between P0 and P1 that are on a spline that passes through P0 and P1. I see that Java3D's KBRotPosScaleSplinePathInterpolator performs such a calculation, but it is tied to that API's scenegraph model and I do not see how to return the values I need.

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• Reverse factorial

  - by dada

  Well, we all know that if N is given it's easy to calculate N!. But what about reversing? N! is given and you are about to find N - Is that possible ? I'm curious.

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• Scrolling a canvas as a shape you're moving approaches its edges

  - by Steven Sproat

  Hi, I develop a Python-based drawing program, Whyteboard. I have tools that the user can create new shapes on the canvas, such as text/images/rectangles/circles/polygons. I also have a Select tool that allows the users to modify these shapes - for example, moving a shape's position, resizing, or editing polygon's points' positions. I'm adding in a new feature where moving or resizing a point near the canvas edge will automatically scroll the canvas. I think it's a good idea in terms of program usability, and annoys me when other program's don't have this feature. I've made some good progress on coding this; below is some Python code to demonstrate what I'm doing. These functions demonstrate how some shapes calculate their "edges": def find_edges(self): """A line.""" self.edges = {EDGE_TOP: min(self.y, self.y2), EDGE_RIGHT: max(self.x, self.x2), EDGE_BOTTOM: max(self.y, self.y2), EDGE_LEFT: min(self. x, self.x2)} def find_edges(self): """An image""" self.edges = {EDGE_TOP: self.y, EDGE_RIGHT: self.x + self.image.GetWidth(), EDGE_BOTTOM: self.y + self.image.GetWidth(), EDGE_LEFT: self.x} def find_edges(self): """Get the bounding rectangle for the polygon""" xmin = min(x for x, y in self.points) ymin = min(y for x, y in self.points) xmax = max(x for x, y in self.points) ymax = max(y for x, y in self.points) self.edges = {EDGE_TOP: ymin, EDGE_RIGHT: xmax, EDGE_BOTTOM: ymax, EDGE_LEFT: xmin} And here's the code I have so far to implement the scrolling when a shape nears the edge: def check_canvas_scroll(self, x, y, moving=False): """ We check that the x/y coords are within 50px from the edge of the canvas and scroll the canvas accordingly. If the shape is being moved, we need to check specific edges of the shape (e.g. left/right side of rectangle) """ size = self.board.GetClientSizeTuple() # visible area of the canvas if not self.board.area > size: # canvas is too small to need to scroll return start = self.board.GetViewStart() # user's starting "viewport" scroll = (-1, -1) # -1 means no change if moving: if self.shape.edges[EDGE_RIGHT] > start[0] + size[0] - 50: scroll = (start[0] + 5, -1) if self.shape.edges[EDGE_BOTTOM] > start[1] + size[1] - 50: scroll = (-1, start[1] + 5) # snip others else: if x > start[0] + size[0] - 50: scroll = (start[0] + 5, -1) if y > start[1] + size[1] - 50: scroll = (-1, start[1] + 5) # snip others self.board.Scroll(*scroll) This code actually works pretty well. If we're moving a shape, then we need to know its edges to calculate when they're coming close to the canvas edge. If we're resizing just a single point, then we just use the x/y coords of that point to see if it's close to the edge. The problem I'm having is a little tricky to describe - basically, if you move a shape to the left, and stop moving it, if you positioned the shape within the 50px from the canvas, then the next time you go to move the shape, the code that says "ok, is this shape close to the end?" gets triggered, and the canvas scrolls to the left, even if you're moving the shape to the right. Can anyone think on how to stop this? I created a youtube video to demonstrate the issue. At about 0:54, I move a polygon to the left of the canvas and position it there. The next time I move it, the canvas scrolls to the left even though I'm moving it right Another thing I'd like to add, but I'm stuck on is the scroll gaining momentum the longer a shape is scrolling? So, with a large canvas, you're not moving a shape for ages, moving 5px at a time, when you need to cover a 2000px distance. Any suggestions there? Thanks all - sorry for the super long question!

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• Examples of monoids/semigroups in programming

  - by jkff

  It is well-known that monoids are stunningly ubiquitous in programing. They are so ubiquitous and so useful that I, as a 'hobby project', am working on a system that is completely based on their properties (distributed data aggregation). To make the system useful I need useful monoids :) I already know of these: Numeric or matrix sum Numeric or matrix product Minimum or maximum under a total order with a top or bottom element (more generally, join or meet in a bounded lattice, or even more generally, product or coproduct in a category) Set union Map union where conflicting values are joined using a monoid Intersection of subsets of a finite set (or just set intersection if we speak about semigroups) Intersection of maps with a bounded key domain (same here) Merge of sorted sequences, perhaps with joining key-equal values in a different monoid/semigroup Bounded merge of sorted lists (same as above, but we take the top N of the result) Cartesian product of two monoids or semigroups List concatenation Endomorphism composition. Now, let us define a quasi-property of an operation as a property that holds up to an equivalence relation. For example, list concatenation is quasi-commutative if we consider lists of equal length or with identical contents up to permutation to be equivalent. Here are some quasi-monoids and quasi-commutative monoids and semigroups: Any (a+b = a or b, if we consider all elements of the carrier set to be equivalent) Any satisfying predicate (a+b = the one of a and b that is non-null and satisfies some predicate P, if none does then null; if we consider all elements satisfying P equivalent) Bounded mixture of random samples (xs+ys = a random sample of size N from the concatenation of xs and ys; if we consider any two samples with the same distribution as the whole dataset to be equivalent) Bounded mixture of weighted random samples Which others do exist?

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• make a 3D spotlight cone matrix

  - by Soubok

  How to create the transformation matrix (4x4) that transforms a cylinder (of height 1 and diameter 1) into a cone that represents my spotlight (position, direction and cutoff angle) ? --edit-- In other words: how to draw the cone that represents my spotlight by drawing a cylinder through a suitable transformation matrix.

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• What statistics can be maintained for a set of numerical data without iterating?

  - by Dan Tao

  Update Just for future reference, I'm going to list all of the statistics that I'm aware of that can be maintained in a rolling collection, recalculated as an O(1) operation on every addition/removal (this is really how I should've worded the question from the beginning): Obvious Count Sum Mean Max* Min* Median** Less Obvious Variance Standard Deviation Skewness Kurtosis Mode*** Weighted Average Weighted Moving Average**** OK, so to put it more accurately: these are not "all" of the statistics I'm aware of. They're just the ones that I can remember off the top of my head right now. *Can be recalculated in O(1) for additions only, or for additions and removals if the collection is sorted (but in this case, insertion is not O(1)). Removals potentially incur an O(n) recalculation for non-sorted collections. **Recalculated in O(1) for a sorted, indexed collection only. ***Requires a fairly complex data structure to recalculate in O(1). ****This can certainly be achieved in O(1) for additions and removals when the weights are assigned in a linearly descending fashion. In other scenarios, I'm not sure. Original Question Say I maintain a collection of numerical data -- let's say, just a bunch of numbers. For this data, there are loads of calculated values that might be of interest; one example would be the sum. To get the sum of all this data, I could... Option 1: Iterate through the collection, adding all the values: double sum = 0.0; for (int i = 0; i < values.Count; i++) sum += values[i]; Option 2: Maintain the sum, eliminating the need to ever iterate over the collection just to find the sum: void Add(double value) { values.Add(value); sum += value; } void Remove(double value) { values.Remove(value); sum -= value; } EDIT: To put this question in more relatable terms, let's compare the two options above to a (sort of) real-world situation: Suppose I start listing numbers out loud and ask you to keep them in your head. I start by saying, "11, 16, 13, 12." If you've just been remembering the numbers themselves and nothing more, and then I say, "What's the sum?", you'd have to think to yourself, "OK, what's 11 + 16 + 13 + 12?" before responding, "52." If, on the other hand, you had been keeping track of the sum yourself while I was listing the numbers (i.e., when I said, "11" you thought "11", when I said "16", you thought, "27," and so on), you could answer "52" right away. Then if I say, "OK, now forget the number 16," if you've been keeping track of the sum inside your head you can simply take 16 away from 52 and know that the new sum is 36, rather than taking 16 off the list and them summing up 11 + 13 + 12. So my question is, what other calculations, other than the obvious ones like sum and average, are like this? SECOND EDIT: As an arbitrary example of a statistic that (I'm almost certain) does require iteration -- and therefore cannot be maintained as simply as a sum or average -- consider if I asked you, "how many numbers in this collection are divisible by the min?" Let's say the numbers are 5, 15, 19, 20, 21, 25, and 30. The min of this set is 5, which divides into 5, 15, 20, 25, and 30 (but not 19 or 21), so the answer is 5. Now if I remove 5 from the collection and ask the same question, the answer is now 2, since only 15 and 30 are divisible by the new min of 15; but, as far as I can tell, you cannot know this without going through the collection again. So I think this gets to the heart of my question: if we can divide kinds of statistics into these categories, those that are maintainable (my own term, maybe there's a more official one somewhere) versus those that require iteration to compute any time a collection is changed, what are all the maintainable ones? What I am asking about is not strictly the same as an online algorithm (though I sincerely thank those of you who introduced me to that concept). An online algorithm can begin its work without having even seen all of the input data; the maintainable statistics I am seeking will certainly have seen all the data, they just don't need to reiterate through it over and over again whenever it changes.

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• Finding coordinates of a point between two points?

  - by Nicros

  Doing some 3D stuff in wpf- want to use a simpler test to see if everything is working (before moving to curves). The basic question is given two points x1,y1,z1 and x2,y2,z2 I have calculated the distance between the points. But how to find the coordinates of another point (x3,y3,z3) that lies on that line at some distance? I.e. if my line is 100 long between -50,0,0 and 50,0,0 what are the coordinates of the point at 100 * 0.1 along the line? I think this is a simple formula but I haven't found it yet....

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• Calculating bounding box a certain distance away from a lat/long coordinate in Java

  - by Bryce Thomas

  Given a coordinate (lat, long), I am trying to calculate a square bounding box that is a given distance (e.g. 50km) away from the coordinate. So as input I have lat, long and distance and as output I would like two coordinates; one being the south-west (bottom-left) corner and one being the north-east (top-right) corner. I have seen a couple of answers on here that try to address this question in Python, but I am looking for a Java implementation in particular. Just to be clear, I intend on using the algorithm on Earth only and so I don't need to accommodate a variable radius. It doesn't have to be hugely accurate (+/-20% is fine) and it'll only be used to calculate bounding boxes over small distances (no more than 150km). So I'm happy to sacrifice some accuracy for an efficient algorithm. Any help is much appreciated. Edit: I should have been clearer, I really am after a square, not a circle. I understand that the distance between the center of a square and various points along the square's perimeter is not a constant value like it is with a circle. I guess what I mean is a square where if you draw a line from the center to any one of the four points on the perimeter that results in a line perpendicular to a side of the perimeter, then those 4 lines have the same length.

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• Algorithm possible amounts (over)paid for a specific price, based on denominations

  - by Wrikken

  In a current project, people can order goods delivered to their door and choose 'pay on delivery' as a payment option. To make sure the delivery guy has enough change customers are asked to input the amount they will pay (e.g. delivery is 48,13, they will pay with 60,- (3*20,-)). Now, if it were up to me I'd make it a free field, but apparantly higher-ups have decided is should be a selection based on available denominations, without giving amounts that would result in a set of denominations which could be smaller. Example: denominations = [1,2,5,10,20,50] price = 78.12 possibilities: 79 (multitude of options), 80 (e.g. 4*20) 90 (e.g. 50+2*20) 100 (2*50) It's international, so the denominations could change, and the algorithm should be based on that list. The closest I have come which seems to work is this: for all denominations in reversed order (large=>small) add ceil(price/denomination) * denomination to possibles baseprice = floor(price/denomination) * denomination; for all smaller denominations as subdenomination in reversed order add baseprice + (ceil((price - baseprice) / subdenomination) * subdenomination) to possibles end for end for remove doubles sort Is seems to work, but this has emerged after wildly trying all kinds of compact algorithms, and I cannot defend why it works, which could lead to some edge-case / new countries getting wrong options, and it does generate some serious amounts of doubles. As this is probably not a new problem, and Google et al. could not provide me with an answer save for loads of pages calculating how to make exact change, I thought I'd ask SO: have you solved this problem before? Which algorithm? Any proof it will always work?

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• sin v/s sinf fucntion in C

  - by user319873

  Hi Guys, I am trying to use sinf function in my C Program and it does give me undefined reference under MSVC 6.0 but sin works fine. This make me curious to find the difference between sin and sinf. What is the logical difference between sin and sinf(). How can I implement my own sinf functionality?

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• Write number N in basis M

  - by VaioIsBorn

  I know how to do it mathematically, but i want it now to do it in c++ using some easy algorithm. Is is possible ? The question is that i need some methods/ideas for writing a number N in basis M, for ex. 14 in basis 3 = 2*(3^0) + 1*(3^1) + 1*(3^2) etc.

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• KD-Trees and missing values (vector comparison)

  - by labratmatt

  I have a system that stores vectors and allows a user to find the n most similar vectors to the user's query vector. That is, a user submits a vector (I call it a query vector) and my system spits out "here are the n most similar vectors." I generate the similar vectors using a KD-Tree and everything works well, but I want to do more. I want to present a list of the n most similar vectors even if the user doesn't submit a complete vector (a vector with missing values). That is, if a user submits a vector with three dimensions, I still want to find the n nearest vectors (stored vectors are of 11 dimensions) I have stored. I have a couple of obvious solutions, but I'm not sure either one seem very good: Create multiple KD-Trees each built using the most popular subset of dimensions a user will search for. That is, if a user submits a query vector of thee dimensions, x, y, z, I match that query to my already built KD-Tree which only contains vectors of three dimensions, x, y, z. Ignore KD-Trees when a user submits a query vector with missing values and compare the query vector to the vectors (stored in a table in a DB) one by one using something like a dot product. This has to be a common problem, any suggestions? Thanks for the help.

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• solving origin of a vectors

  - by Mike

  I have two endpoints (xa,ya) and (xb,yb) of two vectors, respectively a and b, originating from a same point (xo, yo). Also, I know that |a|=|b|+s, where s is a constant. I tried to compute the origin (xo, yo) but seem to fail at some point. How to solve this?

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• Reducing Integer Fractions Algorithm - Solution Explanation?

  - by Andrew Tomazos - Fathomling

  This is a followup to this problem: Reducing Integer Fractions Algorithm Following is a solution to the problem from a grandmaster: #include <cstdio> #include <algorithm> #include <functional> using namespace std; const int MAXN = 100100; const int MAXP = 10001000; int p[MAXP]; void init() { for (int i = 2; i < MAXP; ++i) { if (p[i] == 0) { for (int j = i; j < MAXP; j += i) { p[j] = i; } } } } void f(int n, vector<int>& a, vector<int>& x) { a.resize(n); vector<int>(MAXP, 0).swap(x); for (int i = 0; i < n; ++i) { scanf("%d", &a[i]); for (int j = a[i]; j > 1; j /= p[j]) { ++x[p[j]]; } } } void g(const vector<int>& v, vector<int> w) { for (int i: v) { for (int j = i; j > 1; j /= p[j]) { if (w[p[j]] > 0) { --w[p[j]]; i /= p[j]; } } printf("%d ", i); } puts(""); } int main() { int n, m; vector<int> a, b, x, y, z; init(); scanf("%d%d", &n, &m); f(n, a, x); f(m, b, y); printf("%d %d\n", n, m); transform(x.begin(), x.end(), y.begin(), insert_iterator<vector<int> >(z, z.end()), [](int a, int b) { return min(a, b); }); g(a, z); g(b, z); return 0; } It isn't clear to me how it works. Can anyone explain it? The equivilance is as follows: a is the numerator vector of length n b is the denominator vector of length m

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• How to make an equation span the whole page / line in LaTeX?

  - by Reed Richards

  I have this equation and it's quite big (basically a FDM one) but it aligns with the text and then continues out on the right side to the nothingness. I've tried stuff like \begin{center} and \hspace*{-2.5cm} but to no avail. I want it to use the whole line not just from the left-margin and out to the right. How do I do it and do I need to install some special package for it? I use the \[ instead of the displaymath like this \[ Equation arrays here \] The code \[ \left( \begin{array}{cccccc} -(2\kappa+\frac{hV\rho}{2}) & (\frac{hV\rho}{2}-\kappa) & 0 & \cdots & 0 \\ -\kappa & -(2\kappa+\frac{hV\rho}{2}) & (\frac{hV\rho}{2}-\kappa) & 0 & \cdots \\ 0 & -\kappa & -(2\kappa+\frac{hV\rho}{2}) & (\frac{hV\rho}{2}-\kappa) & 0 & \cdots \\ \vdots & 0 & \ddots & \vdots \\ \vdots & \vdots & \vdots & -\kappa & -(2\kappa+\frac{hV\rho}{2}) & (\frac{hV\rho}{2}-\kappa) \\ 0 & \vdots & \vdots & 0 & \kappa - \frac{2h\kappa_{v}}{\kappa}(\frac{hv\rho}{2} - \kappa) & -2\kappa \\ \end{array} \right) \left( \begin{array}{c} T_{1} \\ T_{2} \\ \vdots \\ T_{n} \\ \end{array} \right) = \left( \begin{array}{c} Q(0) + \kappa T_{0} \\ Q(h) \\ Q(2h) \\ \vdots \\ Q((n-1)h) \\ 2\frac{\kappa_{v}}{\kappa_{v}}T_{out} \\ \end{array} \right) \]

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• Choosing circle radius to fully fill a rectangle

  - by Andy

  Hi, the pixman image library can draw radial color gradients between two circles. I'd like the radial gradient to fill a rectangular area defined by "width" and "height" completely. Now my question, how should I choose the radius of the outer circle? My current parameters are the following: A) inner circle (start of gradient) center pointer of inner circle: (width*0.5|height*0.5) radius of inner circle: 1 color: black B) outer circle (end of gradient) center pointer of outer circle: (width*0.5|height*0.5) radius of outer circle: ??? color: white How should I choose the radius of the outer circle to make sure that the outer circle will entirely fill my bounding rectangle defined by width*height. There shall be no empty areas in the corners, the area shall be completely covered by the circle. In other words, the bounding rectangle width,height must fit entirely into the outer circle. Choosing outer_radius = max(width, height) * 0.5 as the radius for the outer circle is obviously not enough. It must be bigger, but how much bigger? Thanks!

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• draw an arc given 3 points in SWT

  - by Ahmed Kotb

  iam using the swt java library and iam having a problem. the gc draw arc method takes the following arguments GC.drawArc(int x, int y, int width, int height, int startAngle, int endAngle); but i want to be able to draw the arc using 3 arguments : the source ,destination and control points. is there any formula to convert between those parameters ? QuadCurve2D class do exactly what i want but it is not AWT not swt ...and i tried to use java2d under swt but it was very slow .... any solutions ?

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• How to solve such system with given parts of it? (maple)

  - by Kabumbus

  So I had a system #for given koefs k:=3; n:=3; #let us solve system: koefSolution:= solve({ sum(a[i], i = 0 .. k) = 0, sum(a[i], i = 0 .. k)-(sum(b[i], i = 0 .. k)) = 0, sum(i^n*a[i], i = 0 .. k)-(sum(i^(n-1)*b[i], i = 0 .. k)) = 0 }); So I have a vector like koefSolution := { a[0] = 7*a[2]+26*a[3]-b[1]-4*b[2]-9*b[3], a[1] = -8*a[2]-27*a[3]+b[1]+4*b[2]+9*b[3], a[2] = a[2], a[3] = a[3], b[0] = -b[1]-b[2]-b[3], b[1] = b[1], b[2] = b[2], b[3] = b[3]} I have a[0] so I try solve({koefSolution, a[0] = 1}); why it does not solve my system for given a[0]? ( main point here is to fill koefSolution with given a[] and b[] and optimize.)

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• reflection paths between points in2d

  - by Chris H

  Just wondering if there was a nice (already implemented/documented) algorithm to do the following Given any shape (without crossing edges) and two points inside that shape, compute all the paths between the two points such that all reflections are perfect reflections. The path lengths should be limited to a certain length otherwise there are infinite solutions. I'm not interested in just shooting out rays to try to guess how close I can get, I'm interested in algorithms that can do it perfectly. Search based, not guess/improvement based.

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• how do I make a portable isnan/isinf function.

  - by monkeyking

  I've been using isinf,isnan functions on linux platforms which worked perfectly. But this didn't work on osx, so I decided to use std::isinf std::isnan which works on both linux and osx. But the intel compiler doesn't recognize it, and I guess its a bug in the intel compiler according to http://software.intel.com/en-us/forums/showthread.php?t=64188 So now I just want to avoid the hassle and define my own isinf,isnan implementation. Does anyone know how this could be done Thanks edit: I ended up doing this in my sourcecode for making isinf/isnan working #include <iostream> #include <cmath> #ifdef __INTEL_COMPILER #include <mathimf.h> #endif int isnan_local(double x) { #ifdef __INTEL_COMPILER return isnan(x); #else return std::isnan(x); #endif } int isinf_local(double x) { #ifdef __INTEL_COMPILER return isinf(x); #else return std::isinf(x); #endif } int myChk(double a){ std::cerr<<"val is: "<<a <<"\t"; if(isnan_local(a)) std::cerr<<"program says isnan"; if(isinf_local(a)) std::cerr<<"program says isinf"; std::cerr<<"\n"; return 0; } int main(){ double a = 0; myChk(a); myChk(log(a)); myChk(-log(a)); myChk(0/log(a)); myChk(log(a)/log(a)); return 0; }

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• Need some help understanding this problem about maximizing graph connectivity

  - by Legend

  I was wondering if someone could help me understand this problem. I prepared a small diagram because it is much easier to explain it visually. Problem I am trying to solve: 1. Constructing the dependency graph Given the connectivity of the graph and a metric that determines how well a node depends on the other, order the dependencies. For instance, I could put in a few rules saying that node 3 depends on node 4 node 2 depends on node 3 node 3 depends on node 5 But because the final rule is not "valuable" (again based on the same metric), I will not add the rule to my system. 2. Execute the request order Once I built a dependency graph, execute the list in an order that maximizes the final connectivity. I am not sure if this is a really a problem but I somehow have a feeling that there might exist more than one order in which case, it is required to choose the best order. First and foremost, I am wondering if I constructed the problem correctly and if I should be aware of any corner cases. Secondly, is there a closely related algorithm that I can look at? Currently, I am thinking of something like Feedback Arc Set or the Secretary Problem but I am a little confused at the moment. Any suggestions? PS: I am a little confused about the problem myself so please don't flame on me for that. If any clarifications are needed, I will try to update the question.

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• Interview question: f(f(n)) == -n

  - by Hrvoje Prgeša

  A question I got on my last interview: Design a function f, such that: f(f(n)) == -n Where n is a 32 bit signed integer; you can't use complex numbers arithmetic. If you can't design such a function for the whole range of numbers, design it for the largest range possible. Any ideas?

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• Divide not returning the decimal value i expect.

  - by gtas

  if you divide 2 / 3 should return 0,66666666666666667 instead i get in double value type 0.0 in decimal 0. My purpose is to divide even (e.g. 2 / 3) and round to 1 always to the nearest. Any Help?

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• probability of subset sum after rolling dice 4 times

  - by binger

  If we roll 4 dices (fair), what is the probability of "sum of subset" being 5. e.g. 1432,1121, 2344, 2354 have a subset sum of 5. Can you illustrate how to calculate this.

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