how many times does iterative-fft compute twiddle factors ineach stage? rewrite iterative-fft to compute twiddle factors only 2^ (s-1) times in stage s.
me as iam new to fft and polynomials . iam notsure how to compute twiddle factors for each stage and the solutionthat i see in text book is not detailed enough.

What is the worst-case complexity of the maxrepeats function? assume that the longest string in the names array is at most 25 characters wide (i.e., string comparison can be treated as o( class namecounter { private: int* counts; int nc; string* names; int nn; public: namecounter (int ncounts, int nnames); int maxrepeats() const; }; int namecounter: : maxrepeats () { int maxcount = 0; for (int i = 0; i < nc; ++i) { int count = 1; for (int j = i+1; j < nc; ++j) { if (names[i] == names[j]) ++count; } maxcount = max(count, maxcount); } return maxcount; }

Write a function balancechemical to balance chemical reactions by solving a linear set of equations. the inputs arguments are: reagents: symbols of reagents in string row array products: symbols of products in string row array elements: elements in the reaction in string row array elcmpreag: elemental composition of reactants in two dimensional numeric array elcmpprdcts: elemental composition of prducts in two dimensional numeric array hint: the first part of the problem is setting up the set of linear equations that should be solve. the second part of the problem is to find the integers from the solution. one way to do this is to mulitiply the rational basis for the nullspace by increasing larger integers until both the left-and right-side integers exist. for example, for the reaction that involves reacting with to produce and : reagents=["ch4", "o2"]; products =["co2", "h2o"]; elements =["c","h", "o"] elcmpreag=[1,4,0;

When you interpret the behavior of others according to your experiences and understanding of the world your evaluation is