Pennant Race Part I Baseball Pennant Race Pennant Race: Example Another Example Example Example Team Won Left Team Won Left New York 92 2 New York 92 2 Baltimore 91 3 Baltimore 91 3 Toronto 91 3 Toronto 91 3 Boston 89 2 Boston 90 2 Can Boston win the pennant? Can Boston win the pennant? No, because Boston can win at most 91 games. Not clear unless we know what the remaining games are!
Refining the Example Abstracting the Problem Given Example 1. A set of teams S Team Won Left NY Bal Tor Bos 2. For each x ∈ S , the current number of wins w x New York 92 2 − 1 1 0 3. For any x , y ∈ S , the number of remaining games g xy Baltimore 91 3 1 − 1 1 between x and y Toronto 91 3 1 1 − 1 4. A team z Boston 90 2 0 1 1 − Can z win the pennant? Can Boston win the pennant? Suppose Boston does 1. Boston wins both its games to get 92 wins 2. New York must lose both games; now both Baltimore and Toronto have at least 92 3. Winner of Baltimore-Toronto game has 93 wins! Towards a Reduction Flow Network: The basic gadget 1. s : source z can win the pennant if 2. t : sink 1. z wins at least m games 3. x , y : two teams v x 1.1 to maximize z ’s chances we make z win all its remaining games and hence m = w z + � m x ∈ S g xz 4. g xy : number of games − ∞ w x remaining between x 2. no other team wins more than m games s g xy and y . u xy 2.1 for each x , y ∈ S the g xy games between them have to t w y be assigned to either x or y . 5. w x : number of points − ∞ m 2.2 each team x � = z can win at most m − w x − g xz x has. remaining games 6. m : maximum number v y of points x can win before team of interest Is there an assignment of remaining games to teams such that is eliminated. no team x � = z wins more than m − w x games?
Flow Network: An Example Constructing Flow Network Can Boston win? Reduction Construct the flow network G as Team Won Left NY Bal Tor Bos follows Notations New York 90 11 − 1 6 4 1. One vertex v x for each team Baltimore 88 6 1 − 1 4 1. S : set of teams, x ∈ S ′ , one vertex u xy for Toronto 87 11 6 1 − 4 2. w x wins for each each pair of teams x and y in Boston 79 12 4 4 4 − team, and S ′ 3. g xy games left NT N 2. A new source vertex s and between x and y . 6 1 sink t 1. m = 79 + 12 = 91 : 4. m be the maximum 3. Edges ( u xy , v x ) and ( u xy , v y ) 1 4 Boston can get at s NB T t number of wins for of capacity ∞ most 91 points. 1 3 z , 4. Edges ( s , u xy ) of capacity 5. and S ′ = S \ { z } . g xy BT B 5. Edges ( v x , t ) of capacity equal m − w x Correctness of reduction Proof of Correctness Theorem Proof. G ′ has a maximum flow of value g ∗ = � Existence of g ∗ flow ⇒ z can win pennant x , y ∈ S ′ g xy if and only if z can win the most number of games (including possibly tie 1. An integral flow saturating edges out of s , ensures that with other teams). each remaining game between x and y is added to win total of either x or y 2. Capacity on ( v x , t ) edges ensures that no team wins more than m games Conversely, z can win pennant ⇒ flow of value g ∗ 1. Scenario determines flow on edges; if x wins k of the games against y , then flow on ( u xy , v x ) edge is k and on ( u xy , v y ) edge is g xy − k
Theorem Certificate that z cannot win If z cannot win, then maxflow has value less than g ∗ . Theorem By max-flow-min-cut theorem, there is a cut ( S , T ) of Suppose that team z has been eliminated. Then there exists a capacity α < g ∗ . “proof” of this fact of the following form: Let � S be the set of teams x such that v x ∈ � S . The team z can finish with at most m wins. There is a set of teams � S ⊂ S so that � � � � � � �� w x + g xy > m S � . s ∈ � { x , y }⊆ � S S (And hence one of the teams in � S must end with strictly more than m wins.) Helper claim Helper claim proof continued Claim Proof. v x For any two teams x and y for which the vertex u xy exists, we m − w x have that u xy ∈ S if and only if both x and y are in � S . ∞ s g xy u xy t Proof... w y � � − ∞ ∈ � ∈ � m x / S or y / S = ⇒ u xy / ∈ S If x is not in � S then v x is in T . But then, if u xy is in S the v y edge ( u xy → v x ) is in the cut. However, this edge has infinite x ∈ � S and y ∈ � S = ⇒ u xy ∈ S capacity, which implies that ( S , T ) is not a minimum cut. Assume x and y are in � S , then v x and v y are in S . If u xy ∈ T then consider the new cut formed by moving u xy to S . For the new cut ( S ′ , T ′ ) we have � � c ( S ′ , T ′ ) = c ( S , T ) − c ( s → u xy ) .
Proof There are two type of edges in the cut ( S , T ) : (i) ( v x → t ) , for x ∈ � S , and (ii) ( s → u xy ) where at least one of x or y is not in � S . As such, the capacity of the cut ( S , T ) is � � c ( S , T ) = ( m − w x ) + g xy x ∈ � { x , y }�⊂ � S S � � � � � � g ∗ − �� = m S � − w x + g xy . x ∈ � { x , y }⊆ � S S However, c ( S , T ) < g ∗ , and it follows that � � � � � � �� m S � − w x − g xy < = 0 . x ∈ � { x , y }⊆ � S S
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