Battleship rules question

I am a little confused about how the Battleship tournament is going to be run…

The Readme->Tweaking the Tournament says: “Also, this year the tournament is run by pairing every contestant
against every other one exactly once.”… which indicates to me that it is a “round-robin”… and I would assume that best score wins…

But in the MainWindow.PrepareMatches method it says: " // Prepare the next round of matches in our single-elimination tournament.
// This means simply pairing up the contestants who haven’t yet lost."… The current code seems to use this technique.

How is the actual tournament going to be played?

If it is a “round-robin” there could easily be a tie… will it be broken and if so how?

It depends on how many participants we get. If we have too many people to have enough time for everyone to play everyone, we will have to split it into 2 groups and then the winners of those two groups will play each other. At this time I have no idea how many people will be participating so we are prepared for anything!

OK… great… that will make it even more fun and challenging.

Aside of splitting in two, it’s still unclear if it will a single-elimination tournament or a “round-robin”.

Thanks Norman, I already seen the answer from Dana. And with my limited knowledge of the english language, I understand that every one will play against each other, in one or two groups, depending on how many partecipants.
There is no mention of a single-elimination tournament.
If so, why the provided project works in that way? Is this wrong? Or the instructuons are wrong?

Both are right & we won’t know which method will be used until we get to the conference & see how many participants there will be
If there are 10 then we only need to do 45 matches so everyone plays everyone once
1 vs 10 , 9 ,8, 7 , 6, 5, 4, 3, 2
2 vs 10 , 9 ,8, 7 , 6, 5, 4, 3
3 vs 10 , 9 ,8, 7 , 6, 5, 4
4 vs 10 , 9 ,8, 7 , 6, 5
5 vs 10 , 9 ,8, 7 , 6
6 vs 10 , 9 ,8, 7
7 vs 10 , 9 ,8
8 vs 10 , 9
9 vs 10
But if there are 20 then its 190 matches (n * n-1) / 2
Even at 30 seconds a match 45 rounds takes about 25 minutes
190 matches would take 1.5 hours (give or take a few minutes)
Time becomes the limiting factor

Thanks Norman, now it’s clear.

To do a tournament you need powers of two and in general it requres much less time.
But it’s also less fun.

But 20 partecipants can be split in 4 groups of five plus 1 for the winners of the four groups. Total is 50 matches, about 30 min.

Not always if you do the everyone plays everyone tournament style
If you do double knock out (you have to lose twice before you are eliminated) that can work even with uneven numbers BUT someone will get a “bye” or round where they don’t play - but you can assign that by a draw or something
See Double-elimination tournament - Wikipedia
Basically everyone starts out on the same “side” and in the first round someone wins & some one loses and the losers go to the “B” side and the winners go to the “A” side and in the end the winners of both side meet for the championship because what you have is someone who never lost matched against someone who only lost 1 time

Single knock out is simple but not quite as … gratifying ?

[quote=72251:@Norman Palardy]
If you do double knock out (you have to lose twice before you are eliminated) that can work even with uneven numbers BUT someone will get a “bye” or round where they don’t play[/quote]
I’ve never seen this spelt and I always thought it was “buy” round lol :slight_smile: I guess you learn something new every day, right?