Lens – How to approach 180 degrees with a minimum fisheye look on iPhone X or 11?

I am trying to film my rugby practices with my iPhone and I wonder what is the best lens to capture the whole field without so much fish look.

I noticed that the new iPhone 11 (Pro) has an ultra-wide-angle lens of 120 degrees and, it seems, does not look fisheye. Can I buy a new iPhone Pro and add a lens to get a view of 120 to 170/180 degrees with a limited fisheye?

I really have no idea if the lenses are additive; If you add the same wide-angle lens to a 60-degree field of view lens and compare it with the addition to a 120-degree field of view lens, 120 will be incrementally wider than 60 based on the lens attributes ?

Or should I consider an anamorphic type lens, e.g.
https://www.shopmoment.com/shop/anamorphic-lens
Would this give me the desired look, where I can see almost the entire field assuming that the camera is not so far from the side of the field?

Sorry for the ignorance about the lenses, I really don't know much about how they work or if physics will simply make impossible what I am trying to do, but your help is greatly appreciated.

complex analysis – Minimum partial square versus traditional real-world MLR (Biotech)

For 99% of the things I do at work, traditional MLR or logistic regression is suitable for inference tests. What is your opinion on the use of partial least squares regression in sample sets with correlated predictors? What are the disadvantages of PLS ​​versus stratifying and using MLR or simplifying the model to eliminate the correlated variable?

Algorithms: minimum route coverage in a directed acyclic graph

Given a weighted directed acyclic graph G = (V, D, W) and a set of arches RE & # 39; of rewhere the weights of W They are at the vertices. The problem is partitioning Sun on a minimum number of paths separated from vertices that cover all vertices of Sun subject to restrictions that:

  1. the weight of each route is maximum k.
  2. each route must include at least one edge of D & # 39;

What is the complexity of this problem?

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Algorithms: minimum number of rectangles to create a two-dimensional matrix

From this codegolf question.

Consider a $ r $ by $ c $ matrix of non-negative integers, called $ M $. It also has a zero matrix of the same dimensions, called $ N $. A "movement" consists of replacing a rectangle of numbers within $ N $ with any positive integer other than zero.

Some valid "movements":

2 moves:
0 0 0 0      0 0 0 0
0 2 0 0  =>  1 1 1 1
0 2 0 0      0 2 0 0
0 2 0 0      0 2 0 0

2 2 0 0
2 2 0 0
0 0 0 0
0 0 0 0

Invalid Movements:

0 0 0 0
0 2 0 0
0 0 0 0
0 2 0 0

2nd move invalid:
2 2 2 0      2 2 2 0
2 2 2 0  =>  2 0 2 0
2 2 2 0      2 2 2 0

Valid $ M $ The matrices can be:

1 2 3 4 5
5 4 7 0 5

0 0 0 0
0 0 0 0

0

The question is, how do you find the minimum amount of movements needed to recreate $ M $ outside $ N $?

combinatorial – Algorithm to count the number of subsets of size k with the sum of all its minimum possible elements

An array is provided, for example: -1 2 2 2 and we need to count the number of subsets of size k that has the minimum possible sum of elements

Here the subsets of size k = 3 are: –
122
122
122
222

we see that there are 3 subsets that have the minimum sum

I did this in the brute force approach … first I stored the subsets of size k in a vector and then I found the sum of each of the subsets and stored them in another vector and then counted the frequency of the minimum element

How to optimize it?

algorithms: absolute minimum difference

We are given a series of positive numbers. We can perform two types of operations in this sequence of numbers.

  • If the number E is even, then we can replace it with E / 2
  • If the E number is odd, then we can replace it with 2E

We have to find the minimum absolute difference between two elements after performing these operations several times.

Example 1: let the array of numbers be 1.2

Then we can replace 1 with 1 * 2 = 2

then the sequence becomes 2.2

Then the minimum absolute difference between two elements would be 2-2 = 0

Example 2: let the matrix be 5.8

Then we can replace 5 with 10 or (and) 8 with 4.

But for a minimum absolute difference, we will replace 8 with 4.

So the answer would be 5-4 = 1

I have no idea how to write an algorithm for it. Can anyone help me on this?

According to me, the answer should always be 0 or 1.

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dg differential geometry: minimum closed form of a quadratic expression that involves a term of Standard L1

Leave $ X in mathbb {R} ^ {n times d}, w in mathbb {R} ^ d, and in { pm 1 } ^ {n}, alpha in (0, 1), lambda in mathbb {R} $. I have an expression that looks like this

$ frac {1} {2} | Xw -y | 2 ^ 2 + alpha | w | 2 | Xw-y | 1 + frac { lambda} {2} | w | 2 2

where $ lambda> 0 $ It is large enough so that the entire objective is strongly convex. Note that the second term contains a $ L_ {$}-norm term, and that the second term alone can be non-convex, have local minimums in both the space of origin and the solution $ Xw – and $. Also, keep in mind that this goal is limited by the first and last terms,

$ frac {1} {2} | Xw -y | 2 ^ 2 + frac { lambda} {2} | w | 2 2

and thus the geometry of the first objective is limited by a bowl defined by the second objective. The minimum of the second objective is, directly, $ w ^ * = (X ^ { top} X + lambda I) -1 – X ^ { top} and $.

However, this is not, in general, the minimum of the first objective. Are there tools that will help me find the minimum solution of the first objective in closed form?

What is the minimum grade $ d $ required for a tree B with keys of $ 44 * 10 ^ 6 $ so that its height is less than or equal to $ 5 $?

What is the minimum grade? $ d $ required so a B – tree with $ 44 * 10 ^ 6 $ the keys will have a height $ h $such that $ h leq 5 $

My attempt was to build the tallest tree possible with a minimum degree $ d $ Y $ n = 44,000,000 $ keys and then solve for $ d $. That would mean any other tree with a minimum degree $ d & # 39; $ such that $ d & # 39; geq d $ Y $ n $ The keys will be shorter than the one I built:

at depth 0, we have the root and that is $ 1 $ node

in depth 1, we have exactly $ 2 $ nodes

in depth 2, since we go through the highest tree, each node will have a minimum number of keys $ d-1 $ keys each one that means $ d $ children each one so a total of $ 2d $ nodes

in depth 3, following the same reasoning, $ 2d ^ 2 $ nodes

in deep $ h $, exist $ 2d h-1 nodes

The total number of keys is:

$ n = 1+ (d-1) sum_ {k = 0} ^ h-1} {2d ^ k} = 1 + (d-1) frac {2 (d ^ h-1)} {d -1} = 2d ^ h-1 = 44 * 10 ^ 6 $

so:

$ 2d ^ 5-1 = $ 44,000,000

$ d = 29.4 $

$ d geq 30 $

Is that correct?