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# 数学代写|计算复杂度理论代写Computational Complexity Theory代考|MAST31213 What if P = NP?

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## 数学代写|计算复杂度理论代写Computational Complexity Theory代考|What if P = NP?

If $\mathbf{P}=\mathbf{N P}$ – specifically, if an $\mathbf{N P}$-complete problem like 3SAT had a very efficient algorithm running in say $O\left(n^2\right)$ time – then the world would be mostly a Utopia. Mathematicians could be replaced by efficient theorem-discovering programs (a fact pointed out in Kurt Gödel’s 1956 letter and discovered three decades later). In general for every search problem whose answer can be efficiently verified (or has a short certificate of correctness), we will be able to find the correct answer or the short certificate in polynomial time. AI software would be perfect since we could easily do exhaustive searches in a large tree of possibilities. Inventors and engineers would be greatly aided by software packages that can design the perfect part or gizmo for the job at hand. VLSI designers will be able to whip up optimum circuits, with minimum power requirements. Whenever a scientist has some experimental data, she would be able to automatically obtain the simplest theory (under any reasonable measure of simplicity we choose) that best explains these measurements; by the principle of Occam’s Razor the simplest explanation is likely to be the right one. Of course, in some cases it took scientists centuries to come up with the simplest theories explaining the known data. This approach can be used to solve also non-scientific problems: one could find the simplest theory that explains, say, the list of books from the New York Times’ bestseller list. (NB: All these applications will be a consequence of our study of the Polynomial Hierarchy in Chapter 5.)

Somewhat intriguingly, this Utopia would have no need for randomness. As we will later see, if $\mathbf{P}=\mathbf{N P}$ then randomized algorithms would buy essentially no efficiency gains over deterministic algorithms; see Chapter 7. (Philosophers should ponder this one.)

This Utopia would also come at one price: there would be no privacy in the digital domain. Any encryption scheme would have a trivial decoding algorithm. There would be no digital cash, no SSL, RSA or PGP (see Chapter 10). We would just have to learn to get along better without these, folks.

This utopian world may seem ridiculous, but the fact that we can’t rule it out shows how little we know about computation. Taking the half-full cup point of view, it shows how many wonderful things are still waiting to be discovered.

## 数学代写|计算复杂度理论代写Computational Complexity Theory代考|What if NP = coNP?

If $\mathbf{N P}=\mathbf{c o N P}$, the consequences still seem dramatic. Mostly, they have to do with existence of short certificates for statements that do not seem to have any. To give an example, remember the NP-complete problem of finding whether or not a set of multivariate polynomials has a common root, in other words, deciding whether a system of equations of the following type has a solution:
\begin{aligned} f_1\left(x_1, \ldots, x_n\right) & =0 \ f_2\left(x_1, \ldots, x_n\right) & =0 \ \vdots & \ f_m\left(x_1, \ldots, x_n\right) & =0 \end{aligned}
where each $f_i$ is a quadratic polynomial.

If a solution exists, then that solution serves as a certificate to this effect (of course, we have to also show that the solution can be described using a polynomial number of bits, which we omit). The problem of deciding that the system does not have a solution is of course in coNP. Can we give a certificate to the effect that the system does not have a solution? Hilbert’s Nullstellensatz Theorem seems to do that: it says that the system is infeasible iff there is a sequence of polynomials $g_1, g_2, \ldots, g_m$ such that $\sum_i f_i g_i=1$, where 1 on the right hand side denotes the constant polynomial 1.

What is happening? Does the Nullstellensatz prove coNP $=\mathbf{N P}$ ? No, because the degrees of the $g_i$ ‘s – and hence the number of bits used to represent them- could be exponential in $n, m$. (And it is simple to construct $f_i$ ‘s for which this is necessary.)

## 数学代写|计算复杂度理论代写Computational Complexity Theory代考|What if NP = coNP?

$$f_1\left(x_1, \ldots, x_n\right)=0 f_2\left(x_1, \ldots, x_n\right) \quad=0 \vdots f_m\left(x_1, \ldots, x_n\right) \quad=0$$

## MATLAB代写

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