There's a gap of a whole 25 minutes, why do you want it to be postponed?

last problem was easier than usual , according to me.

The analysis section is open now. You can read the editorial.

Also after the contest you can view anyone's solutions.

Here is my successful code.

It's a similar idea to the editorial, with the following difference: when handling quadrilaterals, I place all points into equivalence classes by the gradient formed when joined to the blue star. In all classes where there are >= 2 stars, and they are on different 'sides' of the blue star (found by checking the signs of the x differences and y differences, I take the pair closest on either side and treat these as a 'candidate pair'. This candidate pair now forms 2 non-adjacent vertices of a potential quadrilateral (since we know that the blue star sits on the intersection of the lines formed by 2 such pairs). I then iterate over all pairs of 'candidate pairs', which gives me all combinations of 4 vertices.

Here's my contest submission code

I used a different idea then the analysis. I viewed the stars as vertices of a graph. I placed an edge from white star x to white star y, if the blue star was to the right of the line through x and y (I did this with a cross product check). The edges have weight equal to the distance between star x and y. If we find a cycle in this graph, it means we went completely around around the blue star, thus we made a polygon. So in this graph we have to find the minimum weight cycle. This can be done with Floyd-Warshall in O(n^3).

Some edge cases that tripped my solution.

  x
oo
oo

 o 
o o  x
 o

Can you explain you approach ( for the first half) for D! I read the editorial and yet not able to understand it.

My solution for C also has time complexity $$$O(N^4)$$$ and I also was very surprised that it passed. I think that it may be hacked by the following testcase:

def testcase(n)
  abort "n must be divisible by 4" unless n % 4 == 0
  puts n
  blue_x, blue_y = 1000, 1000
  (n / 4).times do |i|
    puts "#{blue_x + 1 + i} #{blue_y}"
    puts "#{blue_x - 1 - i} #{blue_y}"
    puts "#{blue_x} #{blue_y + 1 + i}"
    puts "#{blue_x} #{blue_y - 1 - i}"
  end
  puts "#{blue_x} #{blue_y}"
end

t = 100
puts t
10.times { testcase(300) }
90.times { testcase(60) }

Basically place the blue star at x=1000 y=1000. So that it is in the center of a cross spanning 75 points up, 75 points down, 75 points left and 75 points right.

Video explaining the multiset approach for problem 2

kick start is unlike code jam, and each contest is independent. Everyone can register and participate, you don't need to do well in the previous round.