## script – Cycle Select layers Photoshop – even if they are Hidden

I had this question a while back if its possible to do “Select next layer” and i found photoshop has a shortcut (alt + ( or )).

The PROBLEM is that it cant select hidden layers.

I found a thread before that fixed the issue, but its way to slow when running the script.

(“Select forward layer” keyboard shortcut – include invisible layers)

The Script that I used was created by user327147, https://superuser.com/a/765135/1472583

Is there anyway to speed it up?

What im trying to do is being able to create layer comps per layer (if possible numbered 001, 002, 003…), but I have base layer, which is tha background, and i have a very top layer, which has information that needs to be displayed for every layer comp.

## Can someone explain intuitively why union find works to find a cycle in an undirected graph?

Union find is a data structure (not an algorithm!) that maintains a collection of disjoint sets $$mathcal{S} = { S_1, S_2, dots }$$ under union operations (i.e., replace two sets with their union) and find operations (given an element $$x$$, report the set $$S_i$$ containing $$x$$).

Hopefully it is already clear to you that if, at any point in time, two distinct vertices $$u$$ and $$v$$ lie in the same set $$S_i in mathcal{S}$$ then there must be a path $$pi$$ from $$u$$ to $$v$$ in your input graph $$G$$.

Suppose then that you are examining a new edge $$e=(u,v)$$ and you discover that both $$u$$ and $$v$$ lie in the same set. We can obtain a cycle $$C$$ by adding edge $$e$$ to the path $$pi$$. Notice that $$C$$ lies entirely in $$G$$, as desired.

## samsung – Is it possible to enable the tab key as an indentation instead of “cycle” on android?

I planned to use my phone+external keyboard combo as my primary means of taking notes (replacing my laptop since it is more portable).

Everything works great except for tab key for indentation. Android input it as cycling through options, similar to alt key on desktop.

I heavily use tab indentation in my note-taking since the note-taking-style is a heavily nested markdown file, so the tab key is a very important and frequently used key. For example:

# title
- subtitle 1
- point 1
- subpoint 1
- subpoint 2
- point 2
- subpoint 1
- subsubpoint 1
- ...
- date: 2021-3-3, tuesday
- location: Kyoto, uni cafe
- ... etc. (you got the idea)

Is it possible to make tab key to insert indent / disable shortcut?

## Number of \$n\$ length cycle in \$S_n\$

I don’t fully get why there’s $$(n-1)!$$ of n length cycle in $$Sn$$

## Bellman-Ford algorithm and nodes that not appear on negative cycle

Given directed graph $$G=(V,E,omega,s,t)$$ that $$omega:Eto mathbb{R}$$ is the weight function on $$E$$, and we want to find shortest path from $$s$$ to $$t$$. First we run Bellman-Ford algorithm to finding shortest path from $$s$$ to $$t$$, but after termination of the algorithm, i know that if distance of a node $$v$$ changed, then the graph contain negative cycle that reachable from the source, so, can we conclude that $$v$$ necessarily appear in negative cycle or not?

My attempt:
I think not necessarily $$v$$ appear in negative cycle. Consider below example that $$t$$ not on any negative cycle, but after termination of the algorithm, if we update the distance, its distance changed.
Can we conclude that $$v$$ sometimes appear in negative cycle or sometimes not appear on any cycle?

## cache – RAM access time vs cycle time

About Random Access Memory (SRAM and DRAM), if multiple read or write operation take place, many books calculate the average access time of those operations. Given the definition of access time, I think that it is wrong to talk about access time. If multiple operation take place, I think that would be better to talk about average CYCLE TIME, because the cycle time includes the time that the RAM spent to set-up itself after each operation.
Could someone tell me if this reasoning is correct?
Thank you

## algorithms – reduction for FVS with cycle

problem : for given an undirected graph G and k ∈ N, and the objective is to decide whether there exists a subset X ⊆ V (G) of size at most k such that every connected component in G − X is either a tree or a cycle.

would like to know if I could construct reduction to the graph , that would return new graph G’ such that all his vertex are of degree $$>=3$$ ,

I have algorithm to construct this to FVS :

• Rule 1: if there exist a vertex with self-loop, delete and decrease k by 1
• Rule 2: if there exist a vertex of degree <=1 ,delete v
• Rule 3 : if there exist a vertex v of degree 2 delete v and connect its neighbors

thanks a lot

## reactjs – Cycle in dependencies between targets ‘FirebaseFirestore’ and ‘abseil’;

I have just tried to use Pod Install on React Native Firestore. Building the project produces this error in Release mode. Any ideas? I have already tried to search for this issue but nothing relevant comes up.

Showing All Messages
Cycle in dependencies between targets 'FirebaseFirestore' and 'abseil'; building could produce unreliable results.
Cycle path: FirebaseFirestore → abseil → FirebaseFirestore
Cycle details:
→ Target 'FirebaseFirestore': Libtool /Users/name/Library/Developer/Xcode/DerivedData/Mobile-bsugqrthopspukeoeqiaydzycews/Build/Products/Release-iphoneos/FirebaseFirestore/libFirebaseFirestore.a normal
→ Target 'abseil' has compile command with input '/Users/Name/Project-Mobile/Mobile/ios/Pods/abseil/absl/time/internal/cctz/src/time_zone_lookup.cc'
○ Target 'abseil' has compile command with input '/Users/Name/Project-Mobile/Mobile/ios/Pods/abseil/absl/time/internal/cctz/src/time_zone_libc.cc'

## Software Development Life Cycle

Software development can be a complex process that calls for adequate time, extensive expertise, self-study imagination, and exemplary client-developer communicating. Whether you’re outsourcing the procedure or intending to have it done in-house, it is logical to own a summary of how software development works. To carry out it, you ought to learn about the applications development lifecycle (SDLC).

## Are there any necessary conditions of the existence of a Hamiltonian cycle on directed graphs

I’m trying to prove that one concrete directed graph has no Hamiltonian cycle, but didn’t seem to find any relevant theorems