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  • Polynomial division overloading operator

    - 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) { polinom Result, temp; Iter i = poly.begin(); constIter j = P.poly.begin(); if (poly.size() < 2) { if (i->pow >= j->pow) { Result.insert(i->coef, i->pow - j->pow); *this = *this - Result; } } else { while (true) { if (i->pow >= j->pow) { Result.insert(i->coef, i->pow - j->pow); temp = Result * P; *this = *this - temp; } else break; } } return Result; } The first three are working correctly but division doesn't as it seems the program is in a infinite loop. Update Because no one seems to understand how i thought the algorithm, i'll explain: If the dividend contains only one term, we simply insert the quotient in Result, then we multiply it with the divisor ans subtract it from the first polynomial which stores the remainder. If the polynomial we do this until the second polynomial( P in this case) becomes bigger. I think this algorithm is called long division, isn't it? So based on these, can anyone help me with overloading the / operator correctly for my class? Thanks!

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  • Why isn't my operator overloading working properly?

    - by Mithrax
    I have the following Polynomial class I'm working on: #include <iostream> using namespace std; class Polynomial { //define private member functions private: int coef[100]; // array of coefficients // coef[0] would hold all coefficients of x^0 // coef[1] would hold all x^1 // coef[n] = x^n ... int deg; // degree of polynomial (0 for the zero polynomial) //define public member functions public: Polynomial::Polynomial() //default constructor { for ( int i = 0; i < 100; i++ ) { coef[i] = 0; } } void set ( int a , int b ) //setter function { //coef = new Polynomial[b+1]; coef[b] = a; deg = degree(); } int degree() { int d = 0; for ( int i = 0; i < 100; i++ ) if ( coef[i] != 0 ) d = i; return d; } void print() { for ( int i = 99; i >= 0; i-- ) { if ( coef[i] != 0 ) { cout << coef[i] << "x^" << i << " "; } } } // use Horner's method to compute and return the polynomial evaluated at x int evaluate ( int x ) { int p = 0; for ( int i = deg; i >= 0; i-- ) p = coef[i] + ( x * p ); return p; } // differentiate this polynomial and return it Polynomial differentiate() { if ( deg == 0 ) { Polynomial t; t.set ( 0, 0 ); return t; } Polynomial deriv;// = new Polynomial ( 0, deg - 1 ); deriv.deg = deg - 1; for ( int i = 0; i < deg; i++ ) deriv.coef[i] = ( i + 1 ) * coef[i + 1]; return deriv; } Polynomial Polynomial::operator + ( Polynomial b ) { Polynomial a = *this; //a is the poly on the L.H.S Polynomial c; for ( int i = 0; i <= a.deg; i++ ) c.coef[i] += a.coef[i]; for ( int i = 0; i <= b.deg; i++ ) c.coef[i] += b.coef[i]; c.deg = c.degree(); return c; } Polynomial Polynomial::operator += ( Polynomial b ) { Polynomial a = *this; //a is the poly on the L.H.S Polynomial c; for ( int i = 0; i <= a.deg; i++ ) c.coef[i] += a.coef[i]; for ( int i = 0; i <= b.deg; i++ ) c.coef[i] += b.coef[i]; c.deg = c.degree(); for ( int i = 0; i < 100; i++) a.coef[i] = c.coef[i]; a.deg = a.degree(); return a; } Polynomial Polynomial::operator -= ( Polynomial b ) { Polynomial a = *this; //a is the poly on the L.H.S Polynomial c; for ( int i = 0; i <= a.deg; i++ ) c.coef[i] += a.coef[i]; for ( int i = 0; i <= b.deg; i++ ) c.coef[i] -= b.coef[i]; c.deg = c.degree(); for ( int i = 0; i < 100; i++) a.coef[i] = c.coef[i]; a.deg = a.degree(); return a; } Polynomial Polynomial::operator *= ( Polynomial b ) { Polynomial a = *this; //a is the poly on the L.H.S Polynomial c; for ( int i = 0; i <= a.deg; i++ ) for ( int j = 0; j <= b.deg; j++ ) c.coef[i+j] += ( a.coef[i] * b.coef[j] ); c.deg = c.degree(); for ( int i = 0; i < 100; i++) a.coef[i] = c.coef[i]; a.deg = a.degree(); return a; } Polynomial Polynomial::operator - ( Polynomial b ) { Polynomial a = *this; //a is the poly on the L.H.S Polynomial c; for ( int i = 0; i <= a.deg; i++ ) c.coef[i] += a.coef[i]; for ( int i = 0; i <= b.deg; i++ ) c.coef[i] -= b.coef[i]; c.deg = c.degree(); return c; } Polynomial Polynomial::operator * ( Polynomial b ) { Polynomial a = *this; //a is the poly on the L.H.S Polynomial c; for ( int i = 0; i <= a.deg; i++ ) for ( int j = 0; j <= b.deg; j++ ) c.coef[i+j] += ( a.coef[i] * b.coef[j] ); c.deg = c.degree(); return c; } }; int main() { Polynomial a, b, c, d; a.set ( 7, 4 ); //7x^4 a.set ( 1, 2 ); //x^2 b.set ( 6, 3 ); //6x^3 b.set ( -3, 2 ); //-3x^2 c = a - b; // (7x^4 + x^2) - (6x^3 - 3x^2) a -= b; c.print(); cout << "\n"; a.print(); cout << "\n"; c = a * b; // (7x^4 + x^2) * (6x^3 - 3x^2) c.print(); cout << "\n"; d = c.differentiate().differentiate(); d.print(); cout << "\n"; cout << c.evaluate ( 2 ); //substitue x with 2 cin.get(); } Now, I have the "-" operator overloaded and it works fine: Polynomial Polynomial::operator - ( Polynomial b ) { Polynomial a = *this; //a is the poly on the L.H.S Polynomial c; for ( int i = 0; i <= a.deg; i++ ) c.coef[i] += a.coef[i]; for ( int i = 0; i <= b.deg; i++ ) c.coef[i] -= b.coef[i]; c.deg = c.degree(); return c; } However, I'm having difficulty with my "-=" operator: Polynomial Polynomial::operator -= ( Polynomial b ) { Polynomial a = *this; //a is the poly on the L.H.S Polynomial c; for ( int i = 0; i <= a.deg; i++ ) c.coef[i] += a.coef[i]; for ( int i = 0; i <= b.deg; i++ ) c.coef[i] -= b.coef[i]; c.deg = c.degree(); // overwrite value of 'a' with the newly computed 'c' before returning 'a' for ( int i = 0; i < 100; i++) a.coef[i] = c.coef[i]; a.deg = a.degree(); return a; } I just slightly modified my "-" operator method to overwrite the value in 'a' and return 'a', and just use the 'c' polynomial as a temp. I've put in some debug print statement and I confirm that at the time of computation, both: c = a - b; and a -= b; are computed to the same value. However, when I go to print them, their results are different: Polynomial a, b; a.set ( 7, 4 ); //7x^4 a.set ( 1, 2 ); //x^2 b.set ( 6, 3 ); //6x^3 b.set ( -3, 2 ); //-3x^2 c = a - b; // (7x^4 + x^2) - (6x^3 - 3x^2) a -= b; c.print(); cout << "\n"; a.print(); cout << "\n"; Result: 7x^4 -6x^3 4x^2 7x^4 1x^2 Why is my c = a - b and a -= b giving me different results when I go to print them?

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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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  • 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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  • 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 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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  • 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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  • Good Package for Fitting Polynomial Trend Lines

    - by Rev316
    Given a simple data set, I would like to be able to calculate a trending formula given it's a second order polynomial regression. In fact, it would be great if one could even forecast X periods during calculation (similar to what Excel does). I'm looking for a portable C/C++ package that's relatively easy to use, and allows it to spit out the "best-fit" (highest R^2 value) curve. Any suggestions? Thanks!

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  • Java Polynomial Multiplication with ArrayList

    - by user1506919
    I am having a problem with one of my methods in my program. The method is designed to take 2 arraylists and the perform multiplication between the two like a polynomial. For example, if I was to say list1={3,2,1} and list2={5,6,7}; I am trying to get a return value of 15,28,38,20,7. However, all I can get is an error message that says: Exception in thread "main" java.lang.IndexOutOfBoundsException: Index: 0, Size: 0. I have provided the method below: private static ArrayList<Integer> multiply(ArrayList<Integer> list1,ArrayList<Integer> list2) { ArrayList<Integer> array =new ArrayList<Integer>(list1.size()+list2.size()); for (int i=0;i<array.size();i++) array.add(i, 0); for (int i = 0; i < list1.size(); i++) for (int j = 0; j < list2.size(); j++) array.set(i+j, ((list1.get(i) * list2.get(j))+array.get(i+j))); return array; } Any help with solving this problem is greatly appreciated.

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  • Java and junit: derivative of polynomial method testing issue

    - by Curtis
    Hello all, im trying to finish up my junit testing for finding the derivative of a polynomial method and im having some trouble making it work. here is the method: public Polynomial derivative() { MyDouble a = new MyDouble(0); MyDouble b = this.a.add(this.a); MyDouble c = this.b; Polynomial poly = new Polynomial (a, b, c); return poly; } and here is the junit test: public void testDerivative() { MyDouble a = new MyDouble(2), b = new MyDouble(4), c = new MyDouble(8); MyDouble d = new MyDouble(0), e = new MyDouble(4), f = new MyDouble(4); Polynomial p1 = new Polynomial(a, b, c); Polynomial p2 = new Polynomial(d,e,f); assertTrue(p1.derivative().equals(p2)); } im not too sure why it isnt working...ive gone over it again and again and i know im missing something. thank you all for any help given, appreciate it

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  • polynomial multiplication using fastfourier transform

    - by mawia
    i am going through the above topic from CLRS(CORMEN) (page 834) and I got stuck at this point. Can anybody please explain how the following expression, A(x)=A^{[0]}(x^2) +xA^{[1]}(x^2) follows from, n-1 ` S a_j x^j j=0 Where, A^{[0]} = a_0 + a_2x + a_4a^x ... a_{n-2}x^{\frac{n}{2-1}} A^{[1]} = a_1 + a_3x + a_5a^x ... a_{n-1}x^{\frac{n}{2-1}} WITH REGARDS THANKS

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  • Python fit polynomial, power law and exponential from data

    - by Nadir
    I have some data (x and y coordinates) coming from a study and I have to plot them and to find the best curve that fits data. My curves are: polynomial up to 6th degree; power law; and exponential. I am able to find the best fit for polynomial with while(i < 6): coefs, val = poly.polyfit(x, y, i, full=True) and I take the degree that minimizes val. When I have to fit a power law (the most probable in my study), I do not know how to do it correctly. This is what I have done. I have applied the log function to all x and y and I have tried to fit it with a linear polynomial. If the error (val) is lower than the others polynomial tried before, I have chosen the power law function. Am I correct? Now how can I reconstruct my power law starting from the line y = mx + q in order to draw it with the original points? I need also to display the function found. I have tried with: def power_law(x, m, q): return q * (x**m) using x_new = np.linspace(x[0], x[-1], num=len(x)*10) y1 = power_law(x_new, coefs[0], coefs[1]) popt, pcov = curve_fit(power_law, x_new, y1) but it seems not to work well.

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  • Octave / Matlab: How to plot the roots of a polynomial

    - by Tom
    Hi everyone, Im trying to plot the roots of a polynomial, and i just cant get it. First i create my polynomial p5 = [1 0 0 0 0 -1] %x^5 - 1 r5 = roots(p5) stem (p5) Im using the stem function, but I would like to remove the stems, and just get the circle around the roots. Is this possible, is stem the right command? Thanks in advance, PS: This is not homework, but very close, will tag it if requested.

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  • Orcad / Matlab: How to plot the roots of a polynomial

    - by Tom
    Hi everyone, Im trying to plot the roots of a polynomial, and i just cant get it. First i create my polynomial p5 = [1 0 0 0 0 -1] %x^5 - 1 r5 = roots(p5) stem (p5) Im using the stem function, but I would like to remove the stems, and just get the circle around the roots. Is this possible, is stem the right command? Thanks in advance, PS: This is not homework, but very close, will tag it if requested.

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  • Define polynomial function

    - by user1822707
    How can I define a function - say, def polyToString(poly) - to return a string containing the polynomial poly in standard form? For example: the polynomial represented by [-1, 2, -3, 4, -5] would be returned as: "-5x**4 + 4x**3 -3x**2 + 2x**1 - 1x**0" def polyToString(poly): standard_form='' n=len(poly) - 1 while n >=0: if poly[n]>=0: if n==len(poly)-1: standard_form= standard_form + ' '+ str(poly[n]) + 'x**%d'%n else: standard_form= standard_form + ' + '+str(poly[n]) + 'x**%d'%n else: standard_form= standard_form + ' - ' + str(abs(poly[n])) + 'x**' + str(n) n=n-1 return standard_form

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  • efficiently determining if a polynomial has a root in the interval [0,T]

    - by user168715
    I have polynomials of nontrivial degree (4+) and need to robustly and efficiently determine whether or not they have a root in the interval [0,T]. The precise location or number of roots don't concern me, I just need to know if there is at least one. Right now I'm using interval arithmetic as a quick check to see if I can prove that no roots can exist. If I can't, I'm using Jenkins-Traub to solve for all of the polynomial roots. This is obviously inefficient since it's checking for all real roots and finding their exact positions, information I don't end up needing. Is there a standard algorithm I should be using? If not, are there any other efficient checks I could do before doing a full Jenkins-Traub solve for all roots? For example, one optimization I could do is to check if my polynomial f(t) has the same sign at 0 and T. If not, there is obviously a root in the interval. If so, I can solve for the roots of f'(t) and evaluate f at all roots of f' in the interval [0,T]. f(t) has no root in that interval if and only if all of these evaluations have the same sign as f(0) and f(T). This reduces the degree of the polynomial I have to root-find by one. Not a huge optimization, but perhaps better than nothing.

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  • Traveling Salesman in polynomial time

    - by Andres
    Problem Description: Write a program in any language with the shortest number of characters that solves the following problem: Given a collection of cities and the cost of travel between each pair of them, find the cheapest way of visiting all of the cities and returning to your starting point. All solutions must take at most polynomial time. Input will be in the form of a text file named simply "i". The text file will contain data in the following format: city1# city2# cost$ cityA# cityB# cost$ ... Each element in the file is separated with a space, and there is a newline at the end of every line. Code count does not include whitespace. Solutions that take longer than polynomial time to find will not be accepted

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  • Create a polynomial object from a number using change-class

    - by charlieb
    I have written a polynomial class along the lines described in SICP 2.5.3 (except using defclass). I would like to be able to seamlessly add and multiply polynomials and regular numbers but I can't make change-class accept a number. I tried to simplify the problem by changing class from an integer to a float: (change-class 4 'float) but that gave me the error: There is no applicable method for the generic function #<STANDARD-GENERIC-FUNCTION CHANGE-CLASS (7)> when called with arguments (4 #<BUILT-IN-CLASS FLOAT>). [Condition of type SIMPLE-ERROR] I get an error of the same form from (fyi): (change-class 4 'polynomial) I'm going to go ahead and implement a manual conversion but I would prefer to use the built-in clos facilities.

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  • Modular Reduction of Polynomials in NTRUEncrypt

    - by Neville
    Hello everyone. I'm implementing the NTRUEncrypt algorithm, according to an NTRU tutorial, a polynomial f has an inverse g such that f*g=1 mod x, basically the polynomial multiplied by its inverse reduced modulo x gives 1. I get the concept but in an example they provide, a polynomial f = -1 + X + X^2 - X4 + X6 + X9 - X10 which we will represent as the array [-1,1,1,0,-1,0,1,0,0,1,-1] has an inverse g of [1,2,0,2,2,1,0,2,1,2,0], so that when we multiply them and reduce the result modulo 3 we get 1, however when I use the NTRU algorithm for multiplying and reducing them I get -2. Here is my algorithm for multiplying them written in Java: public static int[] PolMulFun(int a[],int b[],int c[],int N,int M) { for(int k=N-1;k>=0;k--) { c[k]=0; int j=k+1; for(int i=N-1;i>=0;i--) { if(j==N) { j=0; } if(a[i]!=0 && b[j]!=0) { c[k]=(c[k]+(a[i]*b[j]))%M; } j=j+1; } } return c; } It basicall taken in polynomial a and multiplies it b, resturns teh result in c, N specifies the degree of the polynomials+1, in teh example above N=11; and M is the reuction modulo, in teh exampel above 3. Why am I getting -2 and not 1?

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  • Resultant of a polynomial with x^n–1

    - by devin.omalley
    Resultant of a polynomial with x^n–1 (mod p) I am implementing the NTRUSign algorithm as described in http://grouper.ieee.org/groups/1363/lattPK/submissions/EESS1v2.pdf , section 2.2.7.1 which involves computing the resultant of a polynomial. I keep getting a zero vector for the resultant which is obviously incorrect. private static CompResResult compResMod(IntegerPolynomial f, int p) { int N = f.coeffs.length; IntegerPolynomial a = new IntegerPolynomial(N); a.coeffs[0] = -1; a.coeffs[N-1] = 1; IntegerPolynomial b = new IntegerPolynomial(f.coeffs); IntegerPolynomial v1 = new IntegerPolynomial(N); IntegerPolynomial v2 = new IntegerPolynomial(N); v2.coeffs[0] = 1; int da = a.degree(); int db = b.degree(); int ta = da; int c = 0; int r = 1; while (db > 0) { c = invert(b.coeffs[db], p); c = (c * a.coeffs[da]) % p; IntegerPolynomial cb = b.clone(); cb.mult(c); cb.shift(da - db); a.sub(cb, p); IntegerPolynomial v2c = v2.clone(); v2c.mult(c); v2c.shift(da - db); v1.sub(v2c, p); if (a.degree() < db) { r *= (int)Math.pow(b.coeffs[db], ta-a.degree()); r %= p; if (ta%2==1 && db%2==1) r = (-r) % p; IntegerPolynomial temp = a; a = b; b = temp; temp = v1; v1 = v2; v2 = temp; ta = db; } da = a.degree(); db = b.degree(); } r *= (int)Math.pow(b.coeffs[0], da); r %= p; c = invert(b.coeffs[0], p); v2.mult(c); v2.mult(r); v2.mod(p); return new CompResResult(v2, r); } There is pseudocode in http://www.crypto.rub.de/imperia/md/content/texte/theses/da_driessen.pdf which looks very similar. Why is my code not working? Are there any intermediate results I can check? I am not posting the IntegerPolynomial code because it isn't too interesting and I have unit tests for it that pass. CompResResult is just a simple "Java struct".

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