( n) Lower bounds for Selection (cont)

Transcription

1 Lower bounds for Selection bound on the complexity of a problem is a bound on any algorithm that solves the problem Model of omputation : Tournaments Technique: dversary rgument: Evil oracle dynamically decides outcomes of comparisons to thwart any algorithm. Keys (distinct values) omparisons lgorithms Running time Players Matches Schedule of matches Number of matches See handout on dversary arguments. ssumption: Outcomes of matches is consistent Problems omplexity Find the th player out of n = = n n Proof: Play a single elimination tournament. Find the top players out of n Find the st, nd,, th players out of n U W U W 3 ll players except champion lose once n- matches. So n 4

2 = = n n Proof: (continued) To show that suppose you have played fewer than n- matches. Then there are at least two players who have never lost and either could be the top player. So the top player can not be identified with fewer than n- matches. = n n QE 5 W ( ) ( ) = Proof: n W n since if you now the st and nd players than you now the nd player. ( ) ( ) To show n W n assume we now player is nd. Since player is not first it has lost a match, say to player. If any player is better than then could not be second. So must be the top player. W ( ). It is possible for U = n QE 6 W = n + n Proof: Play a single elimination tournament among the n players with as many matches as possible in each round. Uses n- matches. W = n + n Proof: (continued) If the champion is, collect all of the players that lost to into a set S. Play a second single elimination tournament among the players in S. This requires S - matches. S = {,, } n=7 7 8

3 W = n + n Proof: (continued) The champion is the best player since everyone else has lost a match. The players not in S, cannot be second since they have lost to a player who is not the best player. Of the players in S, only the champion of the second tournament has not lost twice. So the champion of the second tournament is the number player. 9 W = n + n Proof: (continued) The number of players in S is the number of matches played by. This is at most the height of the tree of the first tournament. This tree has n leaves and height at most n. The total number of matches played is n n + S = n + n SoW n + 0 W = n + n Proof: (continued) To show we need to devise an oracle which will force this number of matches regardless of how the matches are organized. Notation : n W n + TOP = { currently undefeated players} = # of matches x has played and won against players that were (previously) undefeated W = n + n Proof: (continued) OM(x, 0 ) = {x} OM(x, i) = {y y' s first defeat was OM ( t start: =U i= 0 OM(x, i) x OM( = {x} x = 0 to a player in TOP = { all players} OM(x, i )}

4 W = n + n Proof: (continued) Oracle decides matches as follows: x wins if x, y TOP and y) or if x TOP and y TOP else whatever is consistent laim: fter m matches are played, if x is still in TOP, then OM(. y induction on m. 0 ase: m=0 OM( = { x} = = = 3 W = n + n Proof: (continued) Step: m0. ssume claim holds after m- matches. ase : x does not play in match m. Then OM( and don t change. ase : In match m, x plays y TOP. Then OM( and don t change. ase 3: In match m, x plays y TOP. newom( OM( OM(y) and new +. y) + newom( + = newwin ( 4 W = n + n Proof: (continued) fter m matches if TOP={x}, then n = OM( n ll but one of the losers to x must play a second match. So - players lose at least twice. n # matches n + n st losses nd losses QE 5 Results: = W = = 6 U ( 5) = 5 It s possible to now the top players out 5 without nowing the top player. E These two players are both better than 3 others and so? } are st or nd 6

5 Finding first and last player requires at least matches. See handout. 3n 7