Hide Safari address bar as part of iOS 13. Workaround for minimal ui?

I know that there have been many contributions to this topic, but I still do not have a solid answer. With the upcoming iOS 13 I would like to know if there is a way to hide the URL address bar. Our product uses the native Safari browser to display PDF files. However, I do not want the URL to be displayed (as our documents are stored in Firebase and the URI of our Firebase address should not be displayed).

I found this link (among many) talking about Apple's minimal UI (used in iOS 8), but it seems to be gone. Any thoughts on how to use something similar in iOS 13? A workaround maybe?

https://stackoverflow.com/questions/24889100/ios-8-removed-minimal-ui-viewport-property-are-theere-other-soft-fullscreen

dnd 3.5e – Reading requires a minimal INT score

Creatures with an Int less than 3 rarely speak a language, but they never do. Speaking a language and lacking the ability to be illiterate (as a Barbarian achieves independently of Int in the first stage) is what is required to read a language, not a certain amount of Int. Even thoughtless beings like constructs can read as long as they speak or understand a language. Therefore, not even & # 39; 1 & # 39; be a requirement, although & # 39; non-zero & # 39; a pretty good limit would be because 0 Int puts you out of action.

Is the minimum focus distance (MFD) really minimal?

Many camera lenses have a mechanical stop that prevents the lens from moving forward beyond a certain distance. This sets the minimum focus distance to about 600 mm. All purpose camera lenses are optimized for near and far subject distances and are slightly compromised when working in close range.

In addition, the f-numbers engraved on the lens barrel become invalid when the task is to work closely together. The f-number is derived by dividing the focal length of the lens by the working aperture diameter. The engraved values ​​are only correct if the camera images objects at infinity (as far as the eye can see ∞). Prompting the camera closer and closer will lengthen the physical distance between the lens and the image plane. With unit (magnification 1 is often specified as 1: 1), the lens is pushed out twice the focal length. A 50 mm lens, which is responsible for working on the unit, thus acts as a 100 mm lens. The engraved panels are therefore two mistaken screens. This is called "bellows factor". The formula to be corrected is (M + 1) ^ 2 (magnification + 1 square). Thus, at unity (magnification 1), the math is (1 + 1) ^ 2 = 4. This value 4 is a multiplier; you correct the underexposure by multiplying the exposure time by 4 or by opening the aperture 2 f-stops.

For this reason, it is common in the industry to stop the forward tracking of the lens when the error reaches 1/3 aperture. All these considerations are controversial if the camera has a transmitted light measurement. Such a design reads the exposure under the influence of the bellows factor error. In this way, the photographer does not have to make any manual compensation. For this reason, many modern cameras do not provide narrow focus restriction.

One more thing: General lenses are designed to work on a curved world (three-dimensional) and project that image onto a flat image sensor or film. If you work very close, the motives are generally flat, and this fact often leads to compromises in sharpness. The countermeasure is to invert the lens since the back elements are optimized for working a flat surface.

A macro lens is optimized for work nearby and compromised when working remotely. Its design is a seamless countermeasure that eliminates the bellows factor.

You can work closely with a general lens by keeping a distance greater than the mechanical limit stops. We use spacers called rings or tubes. In this case, you have to consider the bellows factor manually, unless the camera measures through the lens. You can also use an extra close-up lens to work in close proximity (no bellows factor for this lashing).

Directed Acyclic Graph Partition into minimal subgraphs with one restriction

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I have this problem, I'm not sure if there's a name for it, where a directed acyclic graph has differently colored nodes. The idea is to divide it into a minimum number of subgraphs with the following 2 restrictions:

  1. A subgraph should have nodes of similar color
  2. A subgraph can not depend directly or indirectly on its own output

For example: – In the attached image, the subgraph with the yellow nodes is invalid because the input from the red node violates rule # 2: The fourth node from the top has an input that depends on the output of the second node – over the red one Node (outside the node) subgraph)
Therefore, the algorithm should partition it to # 2 or # 3 so that 2 and 4 are on different nodes

I am sure that this is a fairly common problem in graph theory and must have a name and a standard algorithm for it. Thank you in advance for any hints!

Algorithms – How do I identify a continuous shape and break it down into a minimal number of rectangles?

The input in this case will be coordinates in 2D of the vertices of the shape. There are no curves, but the shape can have holes. The algorithm or program must identify the continuous shape and divide it into rectangles. The number of rectangles created in this way must also be optimized.

I've read about spatial segmentation, but I'm confused how to implement it. Suggestions or help are much appreciated.

Thanks a lot!

Algorithms – Find subset C (the size of k) of a given array with minimal COST

Input: put A from n different numbers and number k <= n Output: (cost of) subset C of A of size k with the minimum COST (C) = max a∈A min c∈C | a – c |

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The solution must define a recursive function and a dynamic programming algorithm depending on the recursive function.

We define that the distance of the element a (∈A) to the subset C min c∈C | a-c | that is, the distance from an element to a subset is equal to the distance between the element and the next element in the subset, e.g. : C = {2,7,9}, a = 5

min c∈C | a-c | = 2

Input: put A from n different numbers and number k <= n Output: (cost of) subset C of A of size k with the minimum COST (C) = max a∈A min c∈C | a – c |

The solution must define a recursive function and a dynamic programming algorithm depending on the recursive function.

We define that the distance of the element a (∈A) to the subset C min c∈C | a-c | that is, the distance from an element to a subset is equal to the distance between the element and the next element in the subset, e.g. : C = {2,7,9}, a = 5

         min c∈C |a-c| = 2  

To illustrate the example A = {3,5,13,8}, K = 2 (input), consider all subsets of A of size k: {3,5}, {3,13}, {3, 8} , {5,13}, {5,8}, {13,8} for each subset we calculate the "distance" of each element to that subset and take the maximum

e.g. for subset {13,8}:

The distance between 3 and the subset is 5

The distance between 5 and the subset is 3

The distance between 8 and the subset is 0

The distance between 13 and the subset is 0

from all this we take the maximum that is 5, we call COST and so on. We do this for all subsets and we want the subset with the minimum COST. (Not the subset itself, it's enough to return the cost of this subgroup)

My solution for the recursive function is:Enter image description here

html – How can I achieve the same result of the animated circle service menu with minimal code / elements (to avoid splitting)?

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Linear Algebra – Minimal polynomial is independent of the choice of base

Suppose I have a linear map $ T: V to V $and two matrix representations, $ M $. $ N $let's say that there is an invertible matrix, $ P $with the same size and $ M = P ^ {- 1} NP $, Say $ m_M $ and $ m_N $ are minimal polynomials for $ M $ and $ N $, How would I show it? $ m_N (M) = 0 $?

May I say that? $ m_N (M) = m_N (P ^ {- 1} NP) = m_N (P ^ {- 1}) m_N (N) m_N (P) = 0 $ and the $ m_M (N) = 0 $ follows in a similar way and therefore is the minimal polynomial independent of the choice of the base?