Frugal Path Mechanisms by Aaron Archer and Eva Tardos Presented by Ron Lavi at the seminar: “Topics on the border of CS, Game theory, and Economics” CS.

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Presentation transcript:

Frugal Path Mechanisms by Aaron Archer and Eva Tardos Presented by Ron Lavi at the seminar: “Topics on the border of CS, Game theory, and Economics” CS dept., The Hebrew University, Jerusalem, Israel

The Model b b Each edge is controlled by a selfish agent. The cost of the edge is known only to him. b b The utility of agent i is (his payment minus the cost he incurred) b b Problem: How to find a low-cost s-t path ? st

Possible Solution: VCG b b In VCG: Each edge declares some cost. The shortest path according to the declared costs is chosen The payment to edge e in the chosen path: (the cost of the shortest path in the graph without e) minus (the cost of the chosen path without counting e) b b VCG is truthful (you will be convinced later on). b b Problem: The total payment increases when the chosen path has more edges. When C1, C2 are the 1st, 2nd lowest costs, and the length of the shortest path is k, then:

Main Question: Is there a truthful mechanism that pays less ??

Threshold-based Mechanisms For all truthful mechanisms that pay 0 to “loosing” edges: b b Observation: If edge e “wins” when bidding b e, it must also win with a lower bid b’ e < b e b b Conclusion 1: There is a threshold bid T e : edge e wins with lower bids and looses with higher bids. b b Conclusion 2: The payment to edge e must be exactly T e, its threshold bid. Theorem:Any mechanism for the path selection problem is truthful if and only if it is a Threshold-based Mechanism. b b For example, VCG is Threshold-based (thus it is truthful).

Min-Function Mechanisms Definition: A mechanism is called a Min-Function Mechanism if it defines, for every s-t path P, a (positive real valued) function f P of the vector of bids b P, such that: b b f P is monotonically strictly increasing, and continuous. b b The mechanism always selects the path P with the lowest value f P (b P ). Notice that: b b A Min-Function Mechanism is Threshold-based (and thus truthful). b b VCG (for the path selection problem) is a min function mechanism.

Costly Example for Min-Function Mechanisms b b Take any Min-Function mechanism, and any graph with two node-disjoint s-t paths (P & Q). The following scenario is costly: Define, a vector of bids of the edges in path P : each edge declares L/|P|, except the i’th edge, who declares w.l.o.g : Then, if P bids and Q bids then P wins. In this case, the cost of P is L, and the cost of Q is L(1+d). b b Let us calculate the total payment in this case: Each edge gets paid its threshold bid,. If an edge in P raises its bid by less than, P will still win. So,, and the total payment is ( e.g. for d=1 the payment is (|P|+1)L )

Conclusion: Conclusion: All min-function mechanisms suffer from the same drawback of the VCG mechanism we have seen. What remains to show: What remains to show: “In many cases”, all truthful mechanisms are min-function mechanisms.

Reasonable Mechanism Properties 1. Edge Autonomy: For any edge e, given the bids of the other edges, e has a high enough bid that will ensure that no path using e will not win. (We say that e bids infinity) 2. Path Autonomy: Given the bids of all edges outside P, there is a bid b P such that P will be chosen. 3. Independence: If path P wins, and an edge e not in P raises its bid, then P will still win. Definition: If path P wins, and there is an edge e such that any (small) change in e’s bid causes another path Q to win, then P and Q are tied. 4. Sensitivity: If P wins and Q is tied with P, then increasing the bid of any, or decreasing the bid of any, will cause P to lose.

Comparing Bids Definition: Suppose path P bids b P, path Q bids b Q, and all other edges bid infinity. If P wins, then we write If either then they are comparable. Lemma: Suppose path Q is node-disjoint from paths P and L. Proof: Suppose P bids b P, Q bids b Q, L bids b L,and all other edges bid infinity. Who wins? If Q, then after raising to infinity the bids of L\P, Q still wins (by independence), contradicting (The same for L) Since only P,Q, and L might win, then P wins. P still wins after raising to infinity the bids of Q.

The edge (s,t) is in G Theorem: If G contains the edge (s,t) then any truthful mechanism satisfying the above properties is a min function mechanism. Proof: Let R be the edge (s,t). Define the functions: b b Choose some P and raise all bids besides P and R to infinity. Notice that now, f P (b P ) is exactly R’s threshold bid. b b Therefore, R wins if f R (b R ) is minimal, and looses if not. b b For any P,Q (besides R), if f P (b P ) < f Q (b Q ) then Q will not win: choose some c, f P (b P ) < c < f Q (b Q ), so :

(proof continued) Claim: f P () is strictly increasing. b b Fix some P, e in P, and some bid b P. If R bids f P (b P ) then R and P are tied. Thus any decrease in e’s bid causes P to win (by sensitivity). b b Consider a new bid b’ P, in which e decreases its bid by d. We need to show that f P (b’ P ) < f P (b P ). b b But if f P (b’ P ) = f P (b P ) = b R then R and P are tied again. Thus increasing e’s bid by d/2 will cause P to lose - contradiction. Claim: f P () is continuous. b b Otherwise let (b P-e,b e ) be a discontinuity point for f P (), jumping from x to y. Suppose R bids (x+y)/2. b b Thus R and P are tied. If R wins then small increase in R’s bid makes P the winner - contradiction. (the same if P wins).

Three s-t connected components Theorem: If G contains an s-t connected component and two other s-t paths (disjoint from the rest), then any truthful mechanism satisfying the above properties is a min function mechanism. Proof: Let R,S be the two separate s-t paths. Define: By defining we can show that is strictly increasing and continuous, similar to before. For other bids of R, we define:

(proof continued) Claim: Case 1: Neither P nor Q is R Take c, f P (b P ) < c < f Q (b Q ) and so Case 2a: Q = R and P is not S Take b S such that f P (b P ) < f S (b S ) < f R (b R ) and By case 1, and the claim follows. [ for case 2b, P = R and Q is not S, the proof is similar ] Case 3a: P = S and Q = R Choose some other path L. By autonomy and continuity of f L () there is a bid b L such that f S (b S ) < f L (b L ) < f R (b R ). By case 1 and by case 2, as needed. [ for case 3b, P = R and Q = S, the proof is similar ]

Conclusion: Conclusion: Any truthful mechanism on a graph that contains either an s-t edge or three edge-disjoint s-t paths, and that satisfies the desired properties, can be forced to pay times the cost of the winning path, where k is the length of the winning path. Remark Remark: There is a graph with two disjoint s-t paths and a truthful mechanism for this graph, that is not a min function mechanism. (the next slides, if time permits)

Two s-t disjoint paths The following mechanism is a counter example: b b Given bids b P =(x 1,x 2 ) and b Q =(y 1,y 2 ), draw a line in the x 1 -x 2 plane, with x 1 -intercept: y 1 +(y 2 / 2), and x 2 -intercept: y 2 +(y 1 / 2). b b If b P is strictly above the line, Q wins, otherwise P wins. b b Verify: all four bidders have threshold values. Autonomy, independence and sensitivity hold. x1x1 x2x2 y1y1 y2y2 st P Q

To see that this mechanism is not min-function: b b Take two Q bids: b 1 Q =(2,1 ), b 2 Q =(1,2 ) b b The dashed/solid line is P’s threshold if Q bids b 1 Q / b 2 Q b b From the diagram: and they aren’t tied b b If the mechanism were a min function, we would have: f P (b 1 P ) < f Q (b 1 Q ) < f P (b 2 P ) < f Q (b 2 Q ) < f P (b 1 P ) - a contradiction Two s-t disjoint paths (continue)