The Big Programming Thread - Page 884

On May 25 2017 03:46 shz wrote:
I applied for front end and have my interview on Monday. React <3

On May 26 2017 19:50 SoSexy wrote:
I don't know if this is the right thread, so sorry in advance. I suck hard at science and my knowledge stops to quadratic equations. Would I be able to program an app that displays events in my town, maybe divided into categories such as concerts/dance clubs/etc.? Where do I start and how long would it take?

On May 27 2017 14:36 Blisse wrote:
First addressing the "why would you do this" when this exists in a million ways already - Facebook events, Eventbrite, Bandsintown, Resident Advisor, Songkick, even Ticketmaster, infinitely better than any app you'd be able to make yourself.

On May 30 2017 13:12 TheEmulator wrote:
Finally switched to Arch Linux and surprisingly I got a full basic build together in a little over an hour (thought it would be more difficult). Anyone else use Arch and have any tips that might not be obvious from the wiki?

On May 25 2017 03:46 shz wrote:
I applied for front end and have my interview on Monday. React <3

On May 30 2017 13:12 TheEmulator wrote:
Finally switched to Arch Linux and surprisingly I got a full basic build together in a little over an hour (thought it would be more difficult). Anyone else use Arch and have any tips that might not be obvious from the wiki?

On May 31 2017 04:32 slmw wrote:
I think there might be a really really intuitive solution for problem 3 but I just couldn't think of it on the spot, so I wrote it the long way.

Define f(x) as the solution for having x rooms and x people. Let's think about f(100). If the first guy chooses his own room, we're good. If he chooses the 2nd guy's room, then we have 99 rooms and 99 people. Then the 2nd guy won't be able to choose his own room, but choosing the 1st guy's room is effectively the same: it's a room nobody that's remaining will care about.

If the 1st guy chooses the 3rd guys room, 2nd guy will choose his own, and now we have 98 rooms and 98 people - with the same situation again as the 3rd guy can choose the 1st guy's room and everyone else remaining will have their room available. On every f(x) (down to f(2)) there will be one extra room for the first guy in the remaining line and all the rooms of the remaining people.

The recurrence relation is therefore + Show Spoiler +
A little bit of induction will tell us + Show Spoiler +

def fill_rooms
  available_rooms = *(0..99)
  taken_rooms = []

  100.times do |n|
    chosen_room = random_room(available_rooms)
    
    if n == 0
      take_room(chosen_room, available_rooms, taken_rooms)
      
      next
    end

    redo if taken_rooms.include?(chosen_room)
    
    if available_rooms.include?(n)
      take_room(n, available_rooms, taken_rooms)
      
      next
    end

    take_room(chosen_room, available_rooms, taken_rooms)
  end
end

def take_room(n, from, to)
  to << from.delete(n)
end

def random_room(from)
  from.sample
end

success_count = 0

100_000.times do
  success_count += 1 if fill_rooms.last == 99
end

puts 100 * success_count / 100_000

49
50
49

On May 31 2017 19:20 Hanh wrote:
The last guy can only pick between his room or the first one because if he could pick another room, it means that room is still vacant. That's not possible because it should have been taken by the guy who is assigned to it.

All the possible scenarios are symmetric if you swap the first and last guy rooms. Because the other guys don't care about who is occupying these rooms.

As a consequence, you have exactly as many successes as you have failures and the probability is 1/2.