Search Results

Consequently, the total running time will be extremely efficient, even without trying to split this into three polynomials $a_0(x),a_1(x),a_2(x)$. …

When you have only type-1 equations, this is the graph layout problem. There are a variety of algorithms for finding approximate solutions to that problem; you might try seeing which ones can be adap …

It suffices to describe how to evaluate $P(\omega^u,\omega^v)$ at the roots of unity. Suppose $P(X,Y)=\sum_{i,j} a_{i,j} X^i Y^j$. Let $$F_{b,c}(X,Y) = \sum_{i,j} a_{2i+b,2j+c} X^i Y^j$$ where the s …

Normally, a circuit has one output: it computes a single function. Therefore, your question does not come up, because we cannot have a circuit that outputs both $f(x,y)$ and $g(x,y)$: that would be t …

Your question is answered in the book itself, in the sentence you quote and the next few sentences thereafter. The Verifier doesn't get $s_1$. As the book says: evaluating $s_1(r_1)$ is not an easy …

I don't believe you can compute $\varphi(H_n)(u_1,\dots,u_n)$ from $\varphi(H_n^2)(u_1,\dots,u_n)$ in a black-box manner, because there are two possibilities for the square root, exactly as you wrote. …

The states of a LFSR with characteristic polynomial $p(x)$ correspond to elements of the multiplicative group of $GF(2)[x]/(p(x))$. The period of a LFSR state is the order of the corresponding group …

Is this a practical problem or a theoretical problem? If it a practical problem, it looks to me like standard randomized algorithms for black-box polynomial identity testing should suffice to solve t …

If you put all of this together, I expect it will be possible to construct an algorithm for your problem that is pretty efficient in practice, for polynomials of the size you describe. …

Here is a semi-algorithm. I don't know if it is a good one. If your conjecture is true, it always terminates. In particular, if a solution exists, it will find it; if no solution exists, it will ru …

This can be done in $O(n \log^2 n)$ time, even if the $x_i$ have duplicates, via the following divide-and-conquer method. First compute the coefficients of the polynomial $f_0(x)=(x-x_1) \cdots (x-x_ …

No, $O(n \lg q)$ running time is not achievable. It takes $\Omega(q)$ space even just to write out the answer, so any algorithm will necessarily have running time at least $\Omega(q)$. However, you …

Note that after you've factored the polynomial, there may still be some ambiguity about what the original two polynomials were. … For instance, if you are given the polynomial $x^5+x^3$, the two polynomials might be $x^2+1$ and $x^3$, or they might be $x^3+x$ and $x^2$, etc. You won't be able to tell, given just the product. …

How does the computation scale? Keep reading! The paper describes how this scales for $n$, in the very next line after the excerpt you show. In particular, the next line is Proposition 1, which sta …

The problem can be solved in polynomial time using linear programming. Write $p(x) = c_k x^k + \dots + c_1 x + c_0$, and think of $c_0,\dots,c_k$ as unknowns. Also introduce an unknown $d$, which wi …