Knight Tournament puzzle suggestion

Feedback & Bugs
#1

In one kingdom a knight tournament takes place every year. 128 knight participate in it. Relative strength of every knight is known from previous tournaments, it is unique so they can be sorted by strength from 1 to 128 (so that we have exactly i - 1 knights weaker than ith knight). In every fight of tournament only two knights meet and stronger one definitely beats weaker one. Stronger knight goes to the next round, weaker one falls off tournament (and goes home). Rounds repeat until only one strongest knight is left, which will be declared the winner of whole tournament.

The smaller difference between knights power is, the longer fight between those knights lasts: ith and jth knights will fight 10 × ( 128 − | i − j | ) seconds.

King decides himself who will fight with who, but he wants that total time spent watching tournament was maximum.

  1. how to make pairs for all tournament to maximise watching time?

  2. how to maximise total watching time spent by all viewers if every kth round is visited by twice more people than round #(k - 1) for k equals 2 … 7?

  3. how to maximise total watching time spent by all viewers if every kth round is visited by 5 times more people than round #(k - 1) for k equals 2 … 7?

  4. how does tournament look like if there are only four participants?

  5. how does king’s time spent watching every round depends on winners of those round?

  6. how to minimise all wiever’s total time spent watching tournament?

#2

Add a example of In and Out, please :stuck_out_tongue:

#4

@SaiksyApo

in:

  1. number of knights

  2. their powers (this can be not just sequent numbers, there can be actual powers, maybe float numbers, they just need to be unique)

  3. coefficient for number of visitors of every next round (one number)

  4. coefficient of fight duration dependency from knight’s strength difference (maybe not)

out:

come on - you are the King! just make pairs for the first round and have fun watching!

#5

the thing is, it looks like horse racing duals, you just have to put every knight with their rival in strength, and you will have the longest tournament possible.

#6

@CvxFous, no :slight_smile: you are the first who falls into unobvious solution
this depends on attendancy coefficient, and can change satisfaction from tournament both ways, so there should be at least two different solutions which should be chosen depending on coefficient’s relative position to some break point of those coefficient values range