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  • dividing double by double gives weird results - Java

    - by Aly
    Hi, I am trying to do the following 33.33333333333333/100.0 to get 0.333333333333333 however when I run System.out.println(33.33333333333333/100.0); I get 0.33333333333333326 as the output, similarly when I run System.out.println(33.33333333333333/1000.0); I get 0.033333333333333326 as the output. Does anyone know why, and how I can get the correct value (without loss of decimal places). Thanks

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  • Graph theory in python

    - by Dan
    I was wondering how people deal with graph theory in python? How is a graph stored? Are there libraries for this? For example how would I input a graph and then find its Chromatic polynomial? Or its girth? Or the number of unique spanning trees? How about problems that involve edge weight like salesman problems? I don't need all of these answered, I'm just looking for a method or tool set that will be able to help me approach solve problems like this. Thanks, Dan

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  • Force-directed graphing

    - by David
    Hello, I'm trying to write a force-directed or force-atlas code base for a graphing application I'm building for myself. Here is an example of what I'm attempting: http://sawamuland.com/flash/graph.html I managed to find some pseudo code to accomplish what I'd like on the Wiki Force-atlas article. I've converted this into ActionScript 3.0 code since it's a Flash application. Here is my source: var timestep:int = 0; var damping:int = 0; var total_kinetic_engery:int = 0; for (var node in list) { var net_force:int = 0; for (var other_node in list) { net_force += coulombRepulsion(node, other_node, nodeList); } for (var spring in list[node].relations) { net_force += hookeAttraction(node, spring, nodeList); } list[node].velocity += (timestep * net_force) * damping; list[node].position += timestep * list[node].velocity; total_kinetic_engery += list[node].mass * (list[node].velocity) ^ 2; } The problem now is finding pseudo code or a function to perform the the coulomb repulsion and hooke attraction code. I'm not exactly sure how to accomplish this. Does anyone know of a good reference I can look at...understand and implement quickly? Best.

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  • Algorithm for computing the inverse of a polynomial

    - by Neville
    I'm looking for an algorithm (or code) to help me compute the inverse a polynomial, I need it for implementing NTRUEncrypt. An algorithm that is easily understandable is what I prefer, there are pseudo-codes for doing this, but they are confusing and difficult to implement, furthermore I can not really understand the procedure from pseudo-code alone. Any algorithms for computing the inverse of a polynomial with respect to a ring of truncated polynomials?

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  • How do I determine a best-fit distribution in java?

    - by Eadwacer
    I have a bunch of sets of data (between 50 to 500 points, each of which can take a positive integral value) and need to determine which distribution best describes them. I have done this manually for several of them, but need to automate this going forward. Some of the sets are completely modal (every datum has the value of 15), some are strongly modal or bimodal, some are bell-curves (often skewed and with differing degrees of kertosis/pointiness), some are roughly flat, and there are any number of other possible distributions (possion, power-law, etc.). I need a way to determine which distribution best describes the data and (ideally) also provides me with a fitness metric so that I know how confident I am in the analysis. Existing open-source libraries would be ideal, followed by well documented algorithms that I can implement myself.

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  • Covering Earth with Hexagonal Map Tiles

    - by carrier
    Many strategy games use hexagonal tiles. One of the main advantages is that the distance between the center of any tile and all its neighboring tiles is the same. I was wondering if anyone has any thoughts on marrying a hexagonal tile system with the traditional geographic system (longitude/latitude). I think it would be interesting to cover a globe with hexagonal tiles and be able to map a geographic coordinate to a tile. Has anyone seen anything remotely close to this before? UPDATE I'm looking for a way to subdivide the surface of a sphere so that each division has the same surface area. Ideally, the centers of adjacent sub-divisions would be equidistant.

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  • Project Euler #15

    - by Aistina
    Hey everyone, Last night I was trying to solve challenge #15 from Project Euler: Starting in the top left corner of a 2×2 grid, there are 6 routes (without backtracking) to the bottom right corner. How many routes are there through a 20×20 grid? I figured this shouldn't be so hard, so I wrote a basic recursive function: const int gridSize = 20; // call with progress(0, 0) static int progress(int x, int y) { int i = 0; if (x < gridSize) i += progress(x + 1, y); if (y < gridSize) i += progress(x, y + 1); if (x == gridSize && y == gridSize) return 1; return i; } I verified that it worked for a smaller grids such as 2×2 or 3×3, and then set it to run for a 20×20 grid. Imagine my surprise when, 5 hours later, the program was still happily crunching the numbers, and only about 80% done (based on examining its current position/route in the grid). Clearly I'm going about this the wrong way. How would you solve this problem? I'm thinking it should be solved using an equation rather than a method like mine, but that's unfortunately not a strong side of mine. Update: I now have a working version. Basically it caches results obtained before when a n×m block still remains to be traversed. Here is the code along with some comments: // the size of our grid static int gridSize = 20; // the amount of paths available for a "NxM" block, e.g. "2x2" => 4 static Dictionary<string, long> pathsByBlock = new Dictionary<string, long>(); // calculate the surface of the block to the finish line static long calcsurface(long x, long y) { return (gridSize - x) * (gridSize - y); } // call using progress (0, 0) static long progress(long x, long y) { // first calculate the surface of the block remaining long surface = calcsurface(x, y); long i = 0; // zero surface means only 1 path remains // (we either go only right, or only down) if (surface == 0) return 1; // create a textual representation of the remaining // block, for use in the dictionary string block = (gridSize - x) + "x" + (gridSize - y); // if a same block has not been processed before if (!pathsByBlock.ContainsKey(block)) { // calculate it in the right direction if (x < gridSize) i += progress(x + 1, y); // and in the down direction if (y < gridSize) i += progress(x, y + 1); // and cache the result! pathsByBlock[block] = i; } // self-explanatory :) return pathsByBlock[block]; } Calling it 20 times, for grids with size 1×1 through 20×20 produces the following output: There are 2 paths in a 1 sized grid 0,0110006 seconds There are 6 paths in a 2 sized grid 0,0030002 seconds There are 20 paths in a 3 sized grid 0 seconds There are 70 paths in a 4 sized grid 0 seconds There are 252 paths in a 5 sized grid 0 seconds There are 924 paths in a 6 sized grid 0 seconds There are 3432 paths in a 7 sized grid 0 seconds There are 12870 paths in a 8 sized grid 0,001 seconds There are 48620 paths in a 9 sized grid 0,0010001 seconds There are 184756 paths in a 10 sized grid 0,001 seconds There are 705432 paths in a 11 sized grid 0 seconds There are 2704156 paths in a 12 sized grid 0 seconds There are 10400600 paths in a 13 sized grid 0,001 seconds There are 40116600 paths in a 14 sized grid 0 seconds There are 155117520 paths in a 15 sized grid 0 seconds There are 601080390 paths in a 16 sized grid 0,0010001 seconds There are 2333606220 paths in a 17 sized grid 0,001 seconds There are 9075135300 paths in a 18 sized grid 0,001 seconds There are 35345263800 paths in a 19 sized grid 0,001 seconds There are 137846528820 paths in a 20 sized grid 0,0010001 seconds 0,0390022 seconds in total I'm accepting danben's answer, because his helped me find this solution the most. But upvotes also to Tim Goodman and Agos :) Bonus update: After reading Eric Lippert's answer, I took another look and rewrote it somewhat. The basic idea is still the same but the caching part has been taken out and put in a separate function, like in Eric's example. The result is some much more elegant looking code. // the size of our grid const int gridSize = 20; // magic. static Func<A1, A2, R> Memoize<A1, A2, R>(this Func<A1, A2, R> f) { // Return a function which is f with caching. var dictionary = new Dictionary<string, R>(); return (A1 a1, A2 a2) => { R r; string key = a1 + "x" + a2; if (!dictionary.TryGetValue(key, out r)) { // not in cache yet r = f(a1, a2); dictionary.Add(key, r); } return r; }; } // calculate the surface of the block to the finish line static long calcsurface(long x, long y) { return (gridSize - x) * (gridSize - y); } // call using progress (0, 0) static Func<long, long, long> progress = ((Func<long, long, long>)((long x, long y) => { // first calculate the surface of the block remaining long surface = calcsurface(x, y); long i = 0; // zero surface means only 1 path remains // (we either go only right, or only down) if (surface == 0) return 1; // calculate it in the right direction if (x < gridSize) i += progress(x + 1, y); // and in the down direction if (y < gridSize) i += progress(x, y + 1); // self-explanatory :) return i; })).Memoize(); By the way, I couldn't think of a better way to use the two arguments as a key for the dictionary. I googled around a bit, and it seems this is a common solution. Oh well.

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  • Efficient 4x4 matrix inverse (affine transform)

    - by Budric
    Hi, I was hoping someone can point out an efficient formula for 4x4 affine matrix transform. Currently my code uses cofactor expansion and it allocates a temporary array for each cofactor. It's easy to read, but it's slower than it should be. Note, this isn't homework and I know how to work it out manually using 4x4 co-factor expansion, it's just a pain and not really an interesting problem for me. Also I've googled and came up with a few sites that give you the formula already (http://www.euclideanspace.com/maths/algebra/matrix/functions/inverse/fourD/index.htm). However this one could probably be optimized further by pre-computing some of the products. I'm sure someone came up with the "best" formula for this at one point or another? Thanks.

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  • nth ugly number

    - by Anil Katti
    Numbers whose only prime factors are 2, 3 or 5 are called ugly numbers. Example: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, ... 1 can be considered as 2^0. I am working on finding nth ugly number. Note that these numbers are extremely sparsely distributed as n gets large. I wrote a trivial program that computes if a given number is ugly or not. For n 500 - it became super slow. I tried using memoization - observation: ugly_number * 2, ugly_number * 3, ugly_number * 5 are all ugly. Even with that it is slow. I tried using some properties of log - since that will reduce this problem from multiplication to addition - but, not much luck yet. Thought of sharing this with you all. Any interesting ideas? Using a concept similar to "Sieve of Eratosthenes" (thanks Anon) for (int i(2), uglyCount(0); ; i++) { if (i % 2 == 0) continue; if (i % 3 == 0) continue; if (i % 5 == 0) continue; uglyCount++; if (uglyCount == n - 1) break; } i is the nth ugly number. Even this is pretty slow. I am trying to find 1500th ugly number.

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  • NTRU Pseudo-code for computing Polynomial Inverses

    - by Neville
    Hello all. I was wondering if anyone could tell me how to implement line 45 of the following pseudo-code. Require: the polynomial to invert a(x), N, and q. 1: k = 0 2: b = 1 3: c = 0 4: f = a 5: g = 0 {Steps 5-7 set g(x) = x^N - 1.} 6: g[0] = -1 7: g[N] = 1 8: loop 9: while f[0] = 0 do 10: for i = 1 to N do 11: f[i - 1] = f[i] {f(x) = f(x)/x} 12: c[N + 1 - i] = c[N - i] {c(x) = c(x) * x} 13: end for 14: f[N] = 0 15: c[0] = 0 16: k = k + 1 17: end while 18: if deg(f) = 0 then 19: goto Step 32 20: end if 21: if deg(f) < deg(g) then 22: temp = f {Exchange f and g} 23: f = g 24: g = temp 25: temp = b {Exchange b and c} 26: b = c 27: c = temp 28: end if 29: f = f XOR g 30: b = b XOR c 31: end loop 32: j = 0 33: k = k mod N 34: for i = N - 1 downto 0 do 35: j = i - k 36: if j < 0 then 37: j = j + N 38: end if 39: Fq[j] = b[i] 40: end for 41: v = 2 42: while v < q do 43: v = v * 2 44: StarMultiply(a; Fq; temp;N; v) 45: temp = 2 - temp mod v 46: StarMultiply(Fq; temp; Fq;N; v) 47: end while 48: for i = N - 1 downto 0 do 49: if Fq[i] < 0 then 50: Fq[i] = Fq[i] + q 51: end if 52: end for 53: {Inverse Poly Fq returns the inverse polynomial, Fq, through the argument list.} The function StarMultiply returns a polynomial (array) stored in the variable temp. Basically temp is a polynomial (I'm representing it as an array) and v is an integer (say 4 or 8), so what exactly does temp = 2-temp mod v equate to in normal language? How should i implement that line in my code. Can someone give me an example. The above algorithm is for computing Inverse polynomials for NTRUEncrypt key generation. The pseudo-code can be found on page 28 of this document. Thanks in advance.

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  • Polynomial operations using operator overloading

    - by Vlad
    I'm trying to use operator overloading to define the basic operations (+,-,*,/) for my polynomial class but when i run the program it crashes and my computer frozes. Update3 Ok i successfully done the first two operations(+,-). Now at multiplication, after multiplying each term of the first polynomial with each of the second i want to sort the poly list descending and then if there are more than one term with the same power to merge them in only one term, but for some reason it doesn't compile because of the sort function which doesn't work. Here's what I got: polinom operator*(const polinom& P) const { polinom Result; constIter i, j, lastItem = Result.poly.end(); Iter it1, it2; int nr_matches; for (i = poly.begin() ; i != poly.end(); i++) { for (j = P.poly.begin(); j != P.poly.end(); j++) Result.insert(i->coef * j->coef, i->pow + j->pow); } sort(Result.poly.begin(), Result.poly.end(), SortDescending()); lastItem--; while (true) { nr_matches = 0; for (it1 = Result.poly.begin(); it < lastItem; it1++) { for (it2 = it1 + 1;; it2 <= lastItem; it2++){ if (it2->pow == it1->pow) { it1->coef += it2->coef; nr_matches++; } } Result.poly.erase(it1 + 1, it1 + (nr_matches + 1)); } return Result; } Also here's SortDescending: struct SortDescending { bool operator()(const term& t1, const term& t2) { return t2.pow < t1.pow; } }; What did i do wrong? Thanks!

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  • VFP Unit Matrix Multiply problem on the iPhone

    - by Ian Copland
    Hi. I'm trying to write a Matrix3x3 multiply using the Vector Floating Point on the iPhone, however i'm encountering some problems. This is my first attempt at writing any ARM assembly, so it could be a faily simple solution that i'm not seeing. I've currently got a small application running using a maths library that i've written. I'm investigating into the benifits using the Vector Floating Point Unit would provide so i've taken my matrix multiply and converted it to asm. Previously the application would run without a problem, however now my objects will all randomly disappear. This seems to be caused by the results from my matrix multiply becoming NAN at some point. Heres the code IMatrix3x3 operator*(IMatrix3x3 & _A, IMatrix3x3 & _B) { IMatrix3x3 C; //C++ code for the simulator #if TARGET_IPHONE_SIMULATOR == true C.A0 = _A.A0 * _B.A0 + _A.A1 * _B.B0 + _A.A2 * _B.C0; C.A1 = _A.A0 * _B.A1 + _A.A1 * _B.B1 + _A.A2 * _B.C1; C.A2 = _A.A0 * _B.A2 + _A.A1 * _B.B2 + _A.A2 * _B.C2; C.B0 = _A.B0 * _B.A0 + _A.B1 * _B.B0 + _A.B2 * _B.C0; C.B1 = _A.B0 * _B.A1 + _A.B1 * _B.B1 + _A.B2 * _B.C1; C.B2 = _A.B0 * _B.A2 + _A.B1 * _B.B2 + _A.B2 * _B.C2; C.C0 = _A.C0 * _B.A0 + _A.C1 * _B.B0 + _A.C2 * _B.C0; C.C1 = _A.C0 * _B.A1 + _A.C1 * _B.B1 + _A.C2 * _B.C1; C.C2 = _A.C0 * _B.A2 + _A.C1 * _B.B2 + _A.C2 * _B.C2; //VPU ARM asm for the device #else //create a pointer to the Matrices IMatrix3x3 * pA = &_A; IMatrix3x3 * pB = &_B; IMatrix3x3 * pC = &C; //asm code asm volatile( //turn on a vector depth of 3 "fmrx r0, fpscr \n\t" "bic r0, r0, #0x00370000 \n\t" "orr r0, r0, #0x00020000 \n\t" "fmxr fpscr, r0 \n\t" //load matrix B into the vector bank "fldmias %1, {s8-s16} \n\t" //load the first row of A into the scalar bank "fldmias %0!, {s0-s2} \n\t" //calulate C.A0, C.A1 and C.A2 "fmuls s17, s8, s0 \n\t" "fmacs s17, s11, s1 \n\t" "fmacs s17, s14, s2 \n\t" //save this into the output "fstmias %2!, {s17-s19} \n\t" //load the second row of A into the scalar bank "fldmias %0!, {s0-s2} \n\t" //calulate C.B0, C.B1 and C.B2 "fmuls s17, s8, s0 \n\t" "fmacs s17, s11, s1 \n\t" "fmacs s17, s14, s2 \n\t" //save this into the output "fstmias %2!, {s17-s19} \n\t" //load the third row of A into the scalar bank "fldmias %0!, {s0-s2} \n\t" //calulate C.C0, C.C1 and C.C2 "fmuls s17, s8, s0 \n\t" "fmacs s17, s11, s1 \n\t" "fmacs s17, s14, s2 \n\t" //save this into the output "fstmias %2!, {s17-s19} \n\t" //set the vector depth back to 1 "fmrx r0, fpscr \n\t" "bic r0, r0, #0x00370000 \n\t" "orr r0, r0, #0x00000000 \n\t" "fmxr fpscr, r0 \n\t" //pass the inputs and set the clobber list : "+r"(pA), "+r"(pB), "+r" (pC) : :"cc", "memory","s0", "s1", "s2", "s8", "s9", "s10", "s11", "s12", "s13", "s14", "s15", "s16", "s17", "s18", "s19" ); #endif return C; } As far as i can see that makes sence. While debugging i've managed to notice that if i were to say _A = C prior to the return and after the ASM, _A will not necessarily be equal to C which has only increased my confusion. I had thought it was possibly due to the pointers I'm giving to the VFPU being incrimented by lines such as "fldmias %0!, {s0-s2} \n\t" however my understanding of asm is not good enough to properly understand the problem, nor to see an alternative approach to that line of code. Anyway, I was hoping someone with a greater understanding than me would be able to see a solution, and any help would be greatly appreciated, thank you :-)

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  • How to embed AsciiMathML in Google Sites?

    - by Joannes Vermorel
    We would need to embed mathematical formulas through AsciiMathML into Google Sites pages (internal wiki for a research team). I am stuck with the limitation of Google Sites. Any idea how to do that? (ps: I have finally found a poorly practical work-around, but better ideas would still be appreciated)

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  • Draw fitted line (OpenCV)

    - by Sunny
    I'm using OpenCV to fit a line from a set of points using cvFitLine() cvFitLine() returns a normalized vector that is co-linear to the line and a point on the line. See details here Using this information how can I get the equation of a line so that I can draw the line?

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  • How to convert latitude or longitude to meters?

    - by Adam Taylor
    Hi, If I have a latitude or longitude reading in standard NMEA format is there an easy way / forumla to convert that reading to meters, which I can then implement in Java (J9)? Edit: Ok seems what I want to do is not possible /easily/, however what I really want to do is: Say I have a lat and long of a way point and a lat and long of a user is there an easy way to compare them to decide when to tell the user they are within a /reasonably/ close distance of the way point? I realise reasonable is subject but is this easily do-able or still overly maths-y? Thanks, Adam

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  • Calculating determinant by hand

    - by ldigas
    Okey, this is only half programming, but let's see how you are on terms with manual calculations. I believe many of you did this on your university's while giving "linear systems" ... the problem is it's been so long I can't remember how to do it any more. I know quite a few algorithms for calculating determinants, and they all work fine ... for large systems, where one would never try to do it manually. Unfortunatelly, I'm soon going on an exam, where I do have to calculate it manually, up to the system of 5. So, I have a K(omega) matrix that looks like this: [2-(omega^2)*c -4 2 0 0] [-2 5-(omega^2)*c -4 1 0] [1 -4 6-(omega^2)*c -4 1] [0 1 -4 5-(omega^2)*c -2] [0 0 2 -4 2-(omega^2)*c] and I need all the omegas which satisfy the det[K(omega)]=0 criteria. What would be a good way to calculate it so it can be repeated in a manual process ?

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  • What is O(n log n) or O(n log(log n))

    - by Mark Tomlin
    What does O, if indeed it is a Oh (As in the letter O) not the number Zero (0) mean? I think the n would be number, but I'm not sure as I'm not a 'real' computer programmer, just a hobbyist. And log would be logarithmic function, but I only know that because of smarter people then I have told me this, while never really explaining what a logarithm is. So please, in plain English, explain what this is, and the differences between the two (such as their applications.

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  • java cosine similarity problem

    - by agazerboy
    Hi again :) I developed some java program to calculate cosine similarity on the basis of TF*IDF. It worked very well. But there is one problem.... :( for example: If I have following two matrix and I want to calculate cosine similarity it does not work as rows are not same in length doc 1 1 2 3 4 5 6 doc 2 1 2 3 4 5 6 7 8 5 2 4 9 if rows and colums are same in length then my program works very well but it does not if rows and columns are not in same length. Any tips ???

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  • Calculating the square of BigInteger

    - by brickner
    Hi, I'm using .NET 4's System.Numerics.BigInteger structure. I need to calculate the square (x^2) of very large numbers. If x is a BigInteger, What is the time complexity of: x*x; or BigInteger.Pow(x,2); ? If it's worse than O(n^2), do you have a better implementation? Maybe something like Schönhage–Strassen algorithm?

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  • Polynomial division overloading operator (solved)

    - by Vlad
    Ok. here's the operations i successfully code so far thank's to your help: Adittion: polinom operator+(const polinom& P) const { polinom Result; constIter i = poly.begin(), j = P.poly.begin(); while (i != poly.end() && j != P.poly.end()) { //logic while both iterators are valid if (i->pow > j->pow) { //if the current term's degree of the first polynomial is bigger Result.insert(i->coef, i->pow); i++; } else if (j->pow > i->pow) { // if the other polynomial's term degree is bigger Result.insert(j->coef, j->pow); j++; } else { // if both are equal Result.insert(i->coef + j->coef, i->pow); i++; j++; } } //handle the remaining items in each list //note: at least one will be equal to end(), but that loop will simply be skipped while (i != poly.end()) { Result.insert(i->coef, i->pow); ++i; } while (j != P.poly.end()) { Result.insert(j->coef, j->pow); ++j; } return Result; } Subtraction: polinom operator-(const polinom& P) const //fixed prototype re. const-correctness { polinom Result; constIter i = poly.begin(), j = P.poly.begin(); while (i != poly.end() && j != P.poly.end()) { //logic while both iterators are valid if (i->pow > j->pow) { //if the current term's degree of the first polynomial is bigger Result.insert(-(i->coef), i->pow); i++; } else if (j->pow > i->pow) { // if the other polynomial's term degree is bigger Result.insert(-(j->coef), j->pow); j++; } else { // if both are equal Result.insert(i->coef - j->coef, i->pow); i++; j++; } } //handle the remaining items in each list //note: at least one will be equal to end(), but that loop will simply be skipped while (i != poly.end()) { Result.insert(i->coef, i->pow); ++i; } while (j != P.poly.end()) { Result.insert(j->coef, j->pow); ++j; } return Result; } Multiplication: polinom operator*(const polinom& P) const { polinom Result; constIter i, j, lastItem = Result.poly.end(); Iter it1, it2, first, last; int nr_matches; for (i = poly.begin() ; i != poly.end(); i++) { for (j = P.poly.begin(); j != P.poly.end(); j++) Result.insert(i->coef * j->coef, i->pow + j->pow); } Result.poly.sort(SortDescending()); lastItem--; while (true) { nr_matches = 0; for (it1 = Result.poly.begin(); it1 != lastItem; it1++) { first = it1; last = it1; first++; for (it2 = first; it2 != Result.poly.end(); it2++) { if (it2->pow == it1->pow) { it1->coef += it2->coef; nr_matches++; } } nr_matches++; do { last++; nr_matches--; } while (nr_matches != 0); Result.poly.erase(first, last); } if (nr_matches == 0) break; } return Result; } Division(Edited): polinom operator/(const polinom& P) const { polinom Result, temp2; polinom temp = *this; Iter i = temp.poly.begin(); constIter j = P.poly.begin(); int resultSize = 0; if (temp.poly.size() < 2) { if (i->pow >= j->pow) { Result.insert(i->coef / j->coef, i->pow - j->pow); temp = temp - Result * P; } else { Result.insert(0, 0); } } else { while (true) { if (i->pow >= j->pow) { Result.insert(i->coef / j->coef, i->pow - j->pow); if (Result.poly.size() < 2) temp2 = Result; else { temp2 = Result; resultSize = Result.poly.size(); for (int k = 1 ; k != resultSize; k++) temp2.poly.pop_front(); } temp = temp - temp2 * P; } else break; } } return Result; } }; The first three are working correctly but division doesn't as it seems the program is in a infinite loop. Final Update After listening to Dave, I finally made it by overloading both / and & to return the quotient and the remainder so thanks a lot everyone for your help and especially you Dave for your great idea! P.S. If anyone wants for me to post these 2 overloaded operator please ask it by commenting on my post (and maybe give a vote up for everyone involved).

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