d ( s i , t i ) = c opt d ( e ) c ( e ) From before: Assign - PowerPoint PPT Presentation

CS270: Lecture 2. Path Routing. Toll problem. Given G = ( V , E ) , ( s 1 , t 1 ) ,..., ( s k , t k ) , find a set of k paths Given G = ( V , E ) , ( s 1 , t 1 ) ,..., ( s k , t k ) , find a set of k paths assign one unit of “toll” to edges to maximize total toll for connecting pairs. connecting s i and t i and minimize max load on any edge. s 2 s 2 Admin: Check Piazza. There is a poll on bspace. s 3 s 3 1 Today: Assign 11 on each of 11 edges. Value: 3 11 + 3 3 11 + 3 11 = 9 ————— s 1 Total toll: ◮ Finish Path Routing. 11 s 1 1 t 1 t 1 Can we do better? 1 2 ◮ Games Value: 2 2 Assign 1 / 2 on these two edges. t 3 t 3 Total toll: 1 2 + 1 2 + 1 2 = 3 2 t 2 t 2 Toll is lower bound on Path Routing. Algorithm. How good is equilibrium? Path is routed along shortest path and d ( e ) ∝ 2 c ( e ) . For e with c ( e ) ≤ c max − 2log m ; 2 c ( e ) ≤ 2 c max − 2log m = 2 cmax m 2 . ≥ ∑ d ( s i , t i ) = ∑ c opt d ( e ) c ( e ) From before: Assign tolls. i e Max bigger than minimum weighted average: How to route? Shortest paths! ∑ e ′ 2 c ( e ′ ) c ( e ) = ∑ e 2 c ( e ) c ( e ) 2 c ( e ) = ∑ max e c ( e ) ≥ ∑ e c ( e ) d ( e ) Let c t = c max − 2log m . Assign routing. ∑ e 2 c ( e ) Total length is total congestion: ∑ e c ( e ) d ( e ) = ∑ i d ( p i ) e How to assign tolls? Higher tolls on congested edges. Each path, p i , in routing has length d ( p i ) ≥ d ( s i , t i ) . ∑ e : c ( e ) > c t 2 c ( e ) c ( e ) Toll: d ( e ) ∝ 2 c ( e ) . ≥ ∑ e : c ( e ) > c t 2 c ( e ) + ∑ e : c ( e ) ≤ c t 2 c ( e ) c ( e ) ≥ ∑ d ( p i ) ≥ ∑ c ( e ) d ( e ) = ∑ Equilibrium: max d ( s i , t i ) . ( c t ) ∑ e : c ( e ) > c t 2 c ( e ) e The shortest path routing has has d ( e ) ∝ 2 c ( e ) . e i i ≥ ( 1 + 1 m ) ∑ e : c ( e ) > c t 2 c ( e ) A toll solution is lower bound on any routing solution. The routing does not change, the tolls do not change. Any routing solution is an upper bound on a toll solution. ( c t ) = c max − 2log m ≥ 1 + 1 ( 1 + 1 m ) m Or c max ≤ ( 1 + 1 m ) c opt + 2log m . (Almost) within 2log m of optimal!

The end: sort of. Getting to equilibrium. An algorithm! Algorithm: reroute paths that are off by a factor of three. (Note: d ( e ) recomputed every rerouting.) Maybe no equilibrium! p : w ( p ) = X Approximate equilibrium: ⇒ w ′ ( p ) = X / 2 − 1 for c ( e ) = Each path is routed along a path with length within a factor of 3 of the shortest path and d ( e ) ∝ 2 c ( e ) . Got to here in class. Feel free to continue reading. + 1 for c ( e ) s i t i p ′ : w ( p ′ ) ≤ X / 3 Lose a factor of three at the beginning. ⇒ w ′ ( p ′ ) ≤ 2 X / 3 = c opt ≥ ∑ i d ( s i , t i ) ≥ 1 3 ∑ e d ( p i ) . Potential function: ∑ e w ( e ) , w ( e ) = 2 c ( e ) We obtain c max = 3 ( 1 + 1 m ) c opt + 2log m . Moving path: This is worse! Divides w ( e ) along long path (with w ( p ) of X ) by two. What do we gain? Multiplies w ( e ) along shorter ( w ( p ) ≤ X / 3) path by two. − X 2 + X 3 = − X 6 . Potential function decreases. = ⇒ termination and existence. Tuning... Wrap up. Strategic Games. N players. Each player has strategy set. { S 1 ,..., S N } . Vector valued payoff function: u ( s 1 ,..., s n ) (e.g., ∈ ℜ N ). Dueling players: Replace d ( e ) = ( 1 + ε ) c ( e ) . Toll player raises tolls on congested edges. Example: Congestion player avoids tolls. Replace factor of 3 by ( 1 + 2 ε ) 2 players Converges to near optimal solution! c max ≤ ( 1 + 2 ε ) c opt + 2log m / ε . . (Roughly) Player 1: { D efect, C ooperate } . A lower bound is “necessary” (natural), Fractional paths? Player 2: { D efect, C ooperate } . and helpful (mysterious?)! Payoff: C D C (3,3) (0,5) D (5,0) (1,1)

Famous because? Digression.. Two Person Zero Sum Games 2 players. What situations? C D Each player has strategy set: C (3,3) (0,5) Prisoner’s dilemma: m strategies for player 1 n strategies for player 2 D (5,0) (.1.1) Two prisoners separated by jailors and asked to betray partner. Payoff function: u ( i , j ) = ( − a , a ) (or just a ). What is the best thing for the players to do? Basis of the free market. “Player 1 pays a to player 2.” Both cooperate. Payoff ( 3 , 3 ) . Companies compete, don’t cooperate. Zero Sum: Payoff for any pair of strategies sums to 0. No Monopoly: If player 1 wants to do better, what does he do? Payoffs by m by n matrix: A . E.G., OPEC, Airlines, . Defects! Payoff ( 5 , 0 ) Row player minimizes, column player maximizes. Should defect. What does player 2 do now? Why don’t they? Roshambo: rock,paper, scissors. Free market economics ...not so much? R P S Defects! Payoff ( . 1 ,. 1 ) . More sophisticated models ,e.g, iterated dominance, coalitions, R 0 1 -1 Stable now! complexity.. P -1 0 1 Nash Equilibrium: neither player has incentive to change Lots of interesting Game Theory! S 1 -1 0 strategy. This class(today): simpler version. Any Nash Equilibrium? ( R , R ) ? no. ( R , P ) ? no. ( R , S ) ? no. Mixed Strategies. Payoffs: Equilibrium. Equilibrium R P S R P S . 33 . 33 . 33 . 33 . 33 . 33 R . 33 0 1 -1 R . 33 0 1 -1 R P S P . 33 -1 0 1 P . 33 -1 0 1 S . 33 1 -1 0 . 33 . 33 . 33 Will Player 1 change strategy? Mixed strategies uncountable! S . 33 1 -1 0 R . 33 0 1 -1 Expected payoffs for pure strategies for player 1. P -1 0 1 . 33 Payoffs? Can’t just look it up in matrix!. S . 33 1 -1 0 Expected payoff of Rock? 1 3 × 0 + 1 3 × 1 + 1 3 × − 1 = 0. Average Payoff. Expected Payoff. How do you play? Expected payoff of Paper? 1 3 × − 1 + 1 3 × 0 + 1 3 × 1 = 0. Sample space: Ω = { ( i , j ) : i , j ∈ [ 1 ,.., 3 ] } Expected payoff of Scissors? 1 3 × 1 + 1 3 × − 1 + 1 3 × 0 = 0. Player 1: play each strategy with equal probability. Random variable X (payoff). Player 2: play each strategy with equal probability. No better pure strategy. = ⇒ No better mixed strategy! Definitions. E [ X ] = ∑ Mixed strat. payoff is weighted av. of payoffs of pure strats. X ( i , j ) Pr [( i , j )] . ( i , j ) Mixed strategies: Each player plays distribution over E [ X ] = ∑ ( i , j ) ( Pr [ i ] × Pr [ j ]) X ( i , j ) = ∑ i Pr [ i ]( ∑ j Pr [ j ] × X ( i , j )) Each player chooses independently: strategies. Mixed strategy can’t be better than the best pure strategy. Pr [( i , j )] = 1 3 × 1 3 = 1 9 . Pure strategies: Each player plays single strategy. Player 1 has no incentive to change! Same for player 2. E [ X ] = 0 . 1 Equilibrium! 1 Remember zero sum games have one payoff.

Another example plus notation. Playing the boss... Equilibrium: play the boss...  0 1 − 1  Rock, Paper, Scissors, prEempt. Row has extra strategy:Cheat. − 1 0 1 PreEmpt ties preEmpt, beats everything else.   A =   1 − 1 0 Ties with rock and scissors, beats paper. (Scissors, or no rock!) Payoffs.   0 0 − 1 Payoff matrix: R P S E Equilibrium: Rock is strategy 1, Paper is 2, Scissors is 3, and Cheat is 4 (for R 0 1 -1 1 Row: ( 0 , 1 3 , 1 6 , 1 2 ) . Column: ( 1 3 , 1 2 , 1 6 ) . row.) P -1 0 1 1 Payoff? Remember: weighted average of pure strategies. S 1 -1 0 1   0 1 − 1 E -1 -1 -1 0 Row Player. − 1 0 1 Equilibrium? (E,E) . Pure strategy equilibrium.   A = Strategy 1: 1 3 × 0 + 1 2 × 1 + 1 6 × − 1 = 1   1 − 1 0 3 Notation: Rock is 1, Paper is 2, Scissors is 3, prEmpt is 4.   Strategy 2: 1 3 ×− 1 + 1 2 × 0 + 1 6 × 1 = − 1 0 0 − 1 6 Payoff Matrix. Strategy 3: 1 3 × 1 + 1 2 ×− 1 + 1 6 × 0 = − 1 6 Note: column knows row cheats. Strategy 4: 1 3 × 0 + 1 2 × 0 + 1 6 ×− 1 = − 1  0 1 − 1 1  6 Why play? Payoff is 0 × 1 3 + 1 3 × ( − 1 6 )+ 1 6 × ( − 1 6 )+ 1 2 × ( − 1 6 ) = − 1 − 1 0 1 1 Row is column’s advisor.   6 A =   1 − 1 0 1 Column player: every column payoff is − 1 ... boss.   6 . − 1 − 1 − 1 0 Both only play optimal strategies! Complementary slackness. Why not play just one? Change payoff for other guy! Two person zero sum games. Equilibrium. m × n payoff matrix A . Row mixed strategy: x = ( x 1 ,..., x m ) . Column mixed strategy: y = ( y 1 ,..., y n ) . Equilibrium pair: ( x ∗ , y ∗ ) ? Payoff for strategy pair ( x , y ) : p ( x , y ) = ( x ∗ ) t Ay ∗ = max y ( x ∗ ) t Ay = min x x t Ay ∗ . Lecture 2 ended here..and Lecture 3 reviewed a few of the p ( x , y ) = x t Ay previous slides and continued into lecture 3 notes. (No better column strategy, no better row strategy.) That is, No row is better: � � � � min i A ( i ) · y = ( x ∗ ) t Ay ∗ . 2 = ∑ ∑ ∑ ∑ x i a i , j y j x i a i , j y j . i j j i No column is better: max j ( A t ) ( j ) · x = ( x ∗ ) t Ay ∗ . Recall row minimizes, column maximizes. Equilibrium pair: ( x ∗ , y ∗ ) ? ( x ∗ ) t Ay ∗ = max y ( x ∗ ) t Ay = min x x t Ay ∗ . (No better column strategy, no better row strategy.) 2 A ( i ) is i th row.