## 8489

BlackHatKings: Proxy Lists
Posted by: Afterbarbag
Post time: June 11, 2019 at 7:10 am

## couples of problems related to Gauss flow and vector analysis

the surface $$S$$ is given by $$r$$ (R,$$theta$$) = (r $$cos theta$$)$$i$$+ (r $$sin theta$$)$$j$$+ $$theta k$$. (0 <= r <= a, 0 <=$$theta$$<=$$frac { pi} {2}$$ ) if the vector feild $$A$$ is given by $$A$$= x$$i$$+ y$$j$$++ z$$k$$,

1. Get the normal unit vector $$n$$ of $$S$$

2.évaluer $$int_ {S} A cdot n$$ and $$int_ {S} left ( bigtriangledown times A right) cdot n dS$$

I am new in vector analysis, any index would be appreciated.

## graphs – Polynomial time algorithm – Couples of corresponding vertices

The idea is to use recursion. Suppose the children of the root $$r$$ are $$v_1, ldots, v_d$$and let $$S_1, ldots, S_d$$ consist of these elements of $$S$$ in the subtree rooted at $$v_1, ldots, v_d$$, respectively.

An easy case, that's when $$r notin S$$ and $$| S_i |$$ is even for everyone $$i$$. In this case, we can simply recurse on the subtrees. There are two complications in the general case:

1. Some $$| S_i |$$ could be strange.
2. The root could belong to $$S$$.

The two complications are related, but let's look at them one by one. Suppose first that $$r notin S$$. Yes $$| S_i |$$ is strange for some $$i$$, so we would like to solve this problem. The only reasonable way is to add $$v_i$$ at $$S_i$$ if $$v_i notin S_i$$and remove $$v_i$$ of $$S_i$$ is $$v_i in S_i$$. We now solve the recursively modified problem and somehow have to find a solution to the initial problem.

Let's introduce some notations: $$O$$ is the set of $$i$$ such as $$| S_i |$$ is strange, and $$S _i$$ is the whole $$S_i$$ after modification (adding or removing $$v_i$$). Consider a solution for the new instance. We will derive a solution for the original instance by associating the indexes in $$O$$. Suppose we choose to twin $$i, j$$and look at solutions for $$S-i, S _j$$. There are three cases to consider:

1. $$v_i in S_i$$ and $$v_j in S_j$$. In this case, we add a path from $$v_i$$ at $$v_j$$ via $$r$$.
2. $$v_i notin S_i$$ and $$v_j notin S_j$$. In this case, we add a path from $$v_i$$ at $$v_j$$ and "erase" $$v_i, v_j$$. That is, in the solution to the new problem, $$v_i$$ is connected to some $$w_i$$ in his subtree, and $$v_j$$ is connected to some $$w_j$$ in his subtree. We connect $$w_i$$ and $$w_j$$ via the path $$w_i-v_i-r-v_j-w_j$$.
3. $$v_i in S_i$$ and $$v_j notin S_j$$. In this case, we add a path from $$v_i$$ at $$v_j$$ and "erase" $$v_j$$. That is, in the solution to the new problem, $$v_j$$ is connected to some $$w_j$$ in the subtree. We connect $$v_i$$ and $$w_j$$ via the path $$v_i-r-v_j-w_j$$.

When $$r in S$$just change the strategy above. We appear arbitrarily $$r$$ with some $$i in O$$, then proceed as above. This time, there are only two cases:

1. $$v_i in S_i$$. In this case, we simply connect $$r$$ and $$v_i$$.
2. $$v_i notin S_i$$. In this case, we add the edge between $$r$$ and $$v_i$$ and "erase" $$v_i$$. That is, in the solution to the new problem, $$v_i$$ is connected to some $$w_i$$ in his subtree. We connect $$r$$ and $$w_i$$ via the path $$r-v_i-w_i$$.

## Technical analysis of all major couples | 24 December 2018 – News and Analysis

Most trade articles aim to minimize weaknesses rather than identify and improve the trader's strengths.

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