Viewflow import problem: help me

I need to define the workflow with viewflow, so I wrote the program in python but error when using the command to import the viewflow, can you support me?

proof techniques – Combination of TWO Monte Carlo algorithms to obtain a Las Vegas algorithm that solves the same problem

I have encountered a problem that I have no idea how to solve it.

Consider two Monte Carlo algorithms, called A and B, which both solve the same problem.
A is biased and correct t, while B is biased and correct z.
Show that you can combine A and B to get a Las Vegas algorithm to solve the same problem.

Also, how could I find the best value for R, which is the probability that the Las Vegas algorithm will find the right answer? For this second part, how could I find this fictitious value of R without concrete example or dataset, this question seems completely outside the left field.

Thank you for your time :]

problem gamers – How to balance a move that substitutes a report for a deception

As you pointed out, examples of stunts include many examples of using one skill in place of another in specific circumstances. There are examples of deception and reporting that should help:

  • Mind games (Fate Core, p 104) allows Deceive to replace Provoke for mental attacks "as long as you invent a clever lie as part of the attack".
  • Popular (Fate Core, p 121) allows you to use Report instead of Contacts "if you are in an area where you are popular and appreciated". He points out that this can be established by spending a point of fate to declare a story detail, as well as as a natural consequence of the narrative.
  • Friendly liar (Fate Core, p 300 – Example Zird the Arcane Character Sheet) is pretty close to what your player wants: "Can use Rapport instead of Cheating to create benefits based on a lie."

These are useful examples. Mind games establishes a simple limitation based both on the type of action and on the player and / or character's own actions; the player understands what he can do and how he can justify its use. It also allows a use of the skill which is normally not possible (perform an attack). Popular is broader, but note that it replaces a skill that has similar restrictions (no attacks) but requires a much larger obstacle to satisfy the narrative – something that is either under the control of the GM ( or group collective), or that costs the player a fate point to take control of themselves. And Friendly liar shows that what your player is doing requires a hit, based on the game's default assumptions about the two skills involved – and similarly to Mind games, it has restrictions on the action of the game and on the player's narrative action when it comes into play.

I would recommend offering him a variation on Friendly Liar (perhaps with the narrative restriction being that he is creative with the truth to a friend, rather than lying), and telling him that he can. 39 extend with future stacking or branching cascades, as covered on pages 94 and 95 of Fate Core.

Another option would be to create an Aspect usable by all who see it done: they are not going to trust it after all, and no one will get away with this kind of friendship in good weather forever. If his character truly considers friendships as disposable and as resources, this should have a narrative effect wider than his selection of skills.

Number of gifts received problem

Suppose we have N children who love all dogs. One day, they go to the shelter and there are M dogs of unique breeds. Each child has a preferred dog breed and a second preferred dog breed. They are also aligned in order. The way dogs are selected is as follows. For the first child online, if their first choice breed is available, take it and go. Otherwise if, first unavailable and the second, take it and go. Otherwise, leave to cry without a dog lol. We want to find, for each 0 <= i <= N – 1, to determine how many children would get a dog if the owner if the first i children were removed from the line.

Example. 4 children, 2 breeds of dogs (noted 1 and 2). Let's say that the 4 children have race 1 as the first choice and race 2 as the second choice. The output should be 2,2,2,1. Because, if we remove the first 0 children from the line, only 2 get dogs. If we remove the first child from the line, there are 2 dogs left. If we remove the first 2 children from the line, 2 get dogs. If we remove the first 3 from the line, 1 gets a dog. This is a trivial case.

Obviously, one solution would be to have a list of children's preferences and a set of available dog breeds, then for each i, start from the ith index of the list, refresh the set of dogs and the simulate, but it takes time O (NM). Is there a way to do it smarter and more efficiently?

filtering problem – How to filter mat cards angularly

I'm doing an angular project and I'm wondering how to filter / search the carpet cards containing details (name, description) of different components. I have a search box in home.component.html to filter the cards, but since all the information inside the mat cards comes from different components and from the back-end, I don't know how to do it . any suggestion will be really helpful.

I have a hose to filter and search.

This is how I reuse maps of different components in my home.component.html and ts files.


You currently have 0 public workspaces


export class ExploreComponent implements OnInit, OnDestroy {

 pubWorkspaces: Workspace() = ();
  loading: boolean = false;

  constructor(private workspaceService: WorkspaceService, private router: Router) { }

ngOnInit(): void {

    this.loading = true;
    this.workspaceService.getUserWorkspaces().subscribe((workspaces) =>{

    workspaces.forEach(workspace => {
      if(workspace.type == WorkspaceType.public){
    this.loading = false;
    }, ()=>this.loading = false)

My Pipe Search

import { Pipe, PipeTransform } from '@angular/core';

name: 'search',
pure: true
export class SearchPipe implements PipeTransform {
transform(names: string(), query: string): string() {
  if (!query || !names) {
    return names;

  return names.filter(name => name.toLowerCase().indexOf(query.toLowerCase()) > -1);