Fitting a polynomial to a set of points or to a skeleton

Available data

Available to me is a set of points which can be represented as shown in image 1:

Original data

Also available to me is a non-continuous path derived from this data. It is not important how this non-continuous path is obtained. It is however important, that it roughly represents a curve I intent to approximate. This non-continuous path is shown in image 2:

Non-continuous path derived from original data


I want to approximate this data using a polynomial of either second or third degree. Examples of these approximations are shown in images 3, 4:

Fit curve to original data

Fit curve to non-continuous path

Problem / question

Now I am looking for a way to obtain the red curve. Some details confuse me, where my knowledge is likely to be simply lacking. For example, how to fit a polynomial, when technically what I require is not a function, at least not in this coordinate system, because there will be situations where an x value is being assigned two y values.

I thought of possibly using the ends of my path to define a new x-axis, but I consider this approach faulty. Another consideration are Splines.

How should I go about obtaining this red curve from the non-continuous path (preferred) or from original data? What sources should I look into?

Apologies if this is an already answered question which I suspect it might be. However I have been issuing search queries for this without success, hence my question.

Would $\Sigma_i^P \neq \Pi_i^P$ imply that polynomial hierachy cannot collapse to the $i$-th level?

If $ \Sigma_i^P = \Pi_i^P$ , then it follows that the polynomial hierarchy collapses to the $ i$ -th level.

What about the case $ \Sigma_i^P \neq \Pi_i^P$ ? For example, consider the case of $ NP \neq coNP$ . As far as I understand, this would imply the polynomial hierarchy cannot collapse to the first level, since if $ PH =NP$ , then in particular, $ coNP \subseteq NP$ , which means $ NP = coNP$ . Can we expand this idea to proof the general case: $ \Sigma_i^P \neq \Pi_i^P$ implies $ PH$ cannot collapse to $ i$ -th level?

Useful conditions for proving super polynomial time for some kind of recurrences

Given a recurrence of the form $ \forall n,m.\ \ T(n,m)=\begin{cases}1,&,m=1\\sum_i{T(n_i,m_i)}&,\text{else}\end{cases}$

Note: both $ n_i$ and $ m_i$ are dependent on $ n,m$ so they should have been written above as $ n_{i,n,m}$ and $ m_{i,n,m}$ , but they are written above as $ n_i$ and $ m_i$ in a way of abbreviation for more readablity.

I am looking for a useful, necessary and sufficient condition so that $ \exists k. T(n,m)=O\bigg(\Big(\frac{\log(m)}{n}\Big)^k\bigg)$

The importance of such conditions stems from their applicabilty in concluding super-polynomial lower bounds on complex recurrences of the above-mentioned form.

Why does NTM need to derive certificate to prove “If a language is in in NP iff it is decidable by some nondeterministic polynomial time TM”?

enter image description here enter image description here enter image description here (Sipser’s Chapter 7: Time COmplexity, Pgs 294-295)

If we have to prove the forward direction, then we must have the certificate along with the verifier. I don’t get why we are “guessing the certificate”–essentially constructing the verifier– if we already know one exists.

On the other hand, if you don’t have the certificate, you can use a NTM to derive it and check w against c. So you don’t need to pass any input into V. So V is useless.

I don’t really understand this proof.

Polynomial – using Newton’s method, or not?

I have one problem that is expressed as polynomial of 2…n degree. In order to solve it I can use general Newton’s method for all degrees, or only for 5th+, and for 2,3 and 4 I can use algebraic equations (quadratic, cubic and quartic equations).

Since algebraic equations need to use sqrt, acos and similar functions – does it make sense at all to use algebraic equations – or is it better to use Newton’s method for all solutions? I guess it will actually be faster?


This is my polynomial:

$ $ \sum_{i=0}^nf^i\binom{\{b_1,…,b_n\}}{i}(-k+n-i), n>=2$ $

And I need to write code to calculate $ f$ for different values of $ n$ .

Traversing a Graph polynomial time

Given a directed graph $ G = (V, E)$ and a starting vertex $ v_1$ .

The graphs edges is this

$ \{(v_1, v_2), (v_2, v_3), (v_3, v_4)\}$

basically below

$ (v_1) -> (v_2) -> (v_3) -> (v_4)$

Can we traverse this graph and record the path starting at $ v_1$ in $ O(|V|)$ time?

For example Making a list like $ [(v_1, v_2), (v_2, v_3), (v_3, v_4)]$

I’m confusing whether this takes $ O(|V|^2)$ worst case time because the edges are in a set

Coefficient of power $p$ in polynomial expansion : mathematica doesn’t answer me

I would like to access the coefficient in front of $ n^p$ for a polynomial expansion.

I wrote the following code:

f[n_, q1_, q2_] := (n^q1 + (n + 1)^q1)* (n^q2 + (n + 1)^q2)  SeriesCoefficient[f[n, q1, q2], {n, 0, p},   Assumptions -> {Element[q1, Integers] && q1 >= 0,     Element[q2, Integers] && q2 >= 0 }] 

However, mathematica doesn’t compute anything. Why ?

I would expect an expression that would make appear the binomial coefficient (I can do it by hand it is not very complicated thus I wonder why mathematica cannot).

Polynomial size Boolean circuit for counting number of bits

Given a natural number $ n \geq 1$ , I am looking for a Boolean circuit over $ 2n$ variables, $ \varphi(x_1, y_1, \dots, x_n, y_n)$ , such that the output is true if and only if the assignment that makes it true verifies

$ $ \sum_{i = 1}^{i = n} (x_i + y_i) \not\equiv n \bmod 3$ $

I should specify that this I am looking for a Boolean circuit, not necessarily a Boolean formula as it is usually written in Conjunctive Normal Form (CNF). This is because when written in CNF, a formula like the one before has a trivial representation where the number of clauses is approximately $ \frac{4^n}{3}$ , as it contains a clause for every assignment $ (x_1, y_1, \dots, x_n, y_n)$ whose bits sum to a value which is congruent with $ n \bmod 3$ . Constructing such a formula would therefore take exponential time.

I have been told that a Boolean circuit can be found for this formula that accepts a representation of size polynomial in $ n$ . However, so far I have been unable to find it. I would use some help; thanks.