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        riddles >> medium >> Table-tennis
        
(Message started by: ThudanBlunder on Mar 14^th, 2009, 4:16pm)

Title: Table-tennis
Post by ThudanBlunder on Mar 14^th, 2009, 4:16pm

A, B, and C play 'winner stays on' in a series of table-tennis games, with A and B playing the first game and C sitting it out.

i) At the end A has won 10 games and B has won 21 games. How many times did A and B play each other?

ii) At the end A has won 10 games, B has won 15 games, and C has won 17 games. Who lost the last game?

Title: Re: Table-tennis
Post by towr on Mar 16^th, 2009, 10:49am

i) [hide]16[/hide]
ii) [hide]B[/hide]

[hide]10, 21, 2x (or 2x+1):
($A, $B, $C, #AxB, #AxC, #BxC)
5* (2,0,0, 1,1,0)
10*(0,2,0, 1,0,1)
2x*(0,0,2, 0,2,0)
1* (0,1,0, 1,0,0) (or 1* (0,1,1, 1,0,1))

10, 15, 17:
5* (2,0,0, 1,1,0)
7* (0,2,0, 1,0,1)
8* (0,0,2, 0,2,0)
1* (0,1,1, 1,0,1)[/hide]

Title: Re: Table-tennis
Post by ThudanBlunder on Mar 17^th, 2009, 2:53am

Is that your computer print-out?

Title: Re: Table-tennis
Post by towr on Mar 17^th, 2009, 3:31am

No, the tree proved too large to compute. I could only get to about a depth of 10.
[hide]So then I considered that you can concatenate paths that end on the same state they start with (i.e. we start with A vs B and after the last game A plays B again). And if there is a unique answer to these questions, we can probably rearrange these subgames independently.
So ultimately I got 3 cycles we can add however often we want (well, except there's an extra constraint for C's +2 cycle) and a further 4 end games we can conclude the game with.[/hide]

come to think of it [hide](0,0,2, 0,2,0) should obviously be (0,0,2, 0,1,1) if it's to make any sense[/hide]

Title: Re: Table-tennis
Post by Hippo on Mar 17^th, 2009, 8:17am

[hide]Make a finite automaton with states representing waiting persons and transitions winners.
ii: Cycles a²,b²,c²,abc and bca. You have a word of 15b's,17c's and 10a's. If you use odd number of cycles including all three letters, you gain even number of b's and even number of c's and odd number of a's. If you use even number of such cycles, you gain parities reversed.
After applying cycles of length 2 you get either single a or bc. Both paths lead to state where B has to watch following match.

i:there are 31 a's and b's alltogether. Each even returns to the state where C watches A plaing B. So 15 times it returns to that state so 1+15=16.[/hide]