1. One-parameter mechanisms, with an application to the SPT problem

2. Given: an undirected graph G=(V,E) such that each edge is owned by a distinct player, and a source node s; we assume that player’s private type is the positive cost (length) of the edge, and her valuation function is equal to her negated type if edge is selected in the solution, and 0 otherwise. Question: design a truthful mechanism in order to find a shortest-path tree rooted at s in G t =(V,E,t). The private-edge Shortest- Paths Tree (SPT) problem

3. More formally… F: set of all spanning trees of G rooted at s For any T  F f(t,T)= arg min  d T (s,v) where ║ e ║ is the set of source-node paths in T containing edge e vVvV On the other hand, v e (t e,T)=-t e if e  E(T), 0 otherwise (this models the multicast protocol)  f(t,T)  argmax  v e (t e,T)  non-utilitarian problem! = arg min  t e ║ e ║ e  E(T) TFTFTFTF TFTF

4. One-parameter MD problems 1. Type owned by each player i is a single parameter t i  2. The valuation function of player i is v i (t i,o)= t i w i (o) where w i (o)  ≥0 is the workload for i w.r.t. output o. Notice that for the sake of simplifying the notation, we are now assuming that the valuation function is positive, and so in the following we will invert – and +, and max with min

5. The SPT problem is one-parameter First of all, the type owned by each player is a single real-value number Second, the valuation function of a player w.r.t. to a tree T is: v e (t e,T)= i.e., v e (t e,T)= t e w e (T), where tete if e  E(T) 0 otherwise 1 if e  E(T) 0 otherwise w e (T)= Multicast protocol  The SPT problem is a one-parameter problem!

6. A necessary condition for designing OP truthful mechanisms Theorem (R.B. Myerson, 1981) A mechanism M= for a minimization one- parameter problem is truthful only if g is monotone, i.e.,  player i, w i (g(r -i,r i )) is non- increasing w.r.t. r i, for every r -i =(r 1,…,r i-1, r i+1,…,r N ).

7. An old result from game theory (R.B. Myerson, 1981) Theorem 1 A mechanism M= for a one-parameter problem is truthful only if g is monotone. Proof: (by contradiction) Assume g is non-monotone, and… …let us show that no any payment scheme can make the mechanism M truthful! If g is non-monotone there exists a player i and a vector r -i s.t. w i (r -i,r i ) is not monotonically non-increasing w.r.t. r i

8. 1. If t i =x and r i =t i  v i (t i,o) = t i w i (o) = x w i (r -i,x) 2. If t i =y and r i =t i  v i (t i,o) = y w i (r -i,y) 3. If t i =x and r i =y  v i (t i,o) = x w i (r -i,y), i.e., i increases her valuation (i.e., her cost) by A 4. If t i =y and r i =x  v i (t i,o) = y w i (r -i,x), i.e., i decreases her cost by A+k Ak xy w i (r -i,y) w i (r -i,x) w i (r -i,r i ) riri Proof (cont’d) cost of a i case 3 cost of a i case 4

9. Let ∆p=p i (r -i,y) - p i (r -i,x) If M is truthful it must be: ∆p  A (since otherwise when t i =x, i will report y, since in this case her cost will increase by A, and since ∆p>A, her utility will increase!) ∆p ≥ A+k (since otherwise when t i =y, i will report x, since in this case her cost will decrease by A+k, and since ∆p<A+k, this means that the payment’s decrease is less than the cost’s decrease, and so her utility will increase! … but k is strictly positive! Ak xy w i (r -i,y) w i (r -i,x) w i (r -i,r i ) riri Proof (cont’d) Contradiction: g must be monotone!

10. One-parameter mechanisms g: is any monotone algorithm for the underlying one-parameter problem p i (r) = h i (r -i ) + r i w i (r) - ∫ w i (r -i,z) dz 0 riri h i (r -i ): arbitrary function independent of r i For the sake of simplifying the notation: We will write w i (r) in place of w i (g(r)) We will write p i (r) in place of p i (g(r)) Definition: A one-parameter (OP) mechanism is a pair M= such that:

11. Theorem 2 : An OP-mechanism for an OP- problem is truthful. Proof: We show that the utility of a player i can only decrease when i lies Payment returned to i (when she reports r i ) is: p i (r) = h i (r -i ) + r i w i (r) - ∫ w i (r -i,z) dz 0 riri Indipendent of r i We can set h i (r -i )=0 (notice this will produce negative utilities) Truthfulness of OP-mechanisms

12. u i (t i,g(r -i,t i ))= p i (r -i,t i )-v i (t i, g(r -i,t i ))= t i w i (r -i,t i )-∫ w i (r -i,z) dz-t i w i (r -i,t i ) = -∫ w i (r -i,z) dz If i reports x>t i : Her valuation becomes: C = t i w i (r -i,x) Her payment becomes: P= x w i (r -i,x) - ∫ w i (r -i,z) dz  u i (t i,g(r -i,x))= P-C  i is loosing G Proof (cont’d) titi w i (r -i,t i ) x w i (r -i,x) 0 titi C 0 x 0 titi G -P The true utility is equal to this negated area The payment is equal to this negated area w i (r -i,r i )

13. u i (t i,g(r -i,t i ))=-∫ w i (r -i,z) dz If i reports x<t i Her valuation becomes C = t i w i (r -i,x) Her payment becomes P= x w i (r -i,x) - ∫ w i (r -i,z) dz  u i (t i,g(r -i,x))= P-C  i is loosing G Proof (cont’d) titi w i (r -i,t i ) 0 titi G C -P x w i (r -i,x) Player i has no incentive to lie! The payment is equal to this negated area 0 x w i (r -i,r i )

14. On the h i (r -i ) function Once again, we want to guarantee voluntary participation (VP) But when player i reports r i, her payment is: p i (r) = h i (r -i ) + r i w i (r) - ∫ w i (r -i,z) dz 0 riri If we set h i (r -i )= ∫ w i (r -i,z) dz, 0 ∞ p i (r) = r i w i (r) + ∫ w i (r -i,z) dz riri ∞ the payment becomes:  The utility of player i when reporting the true becomes: u i (t i,g(r)) = ∫ w i (r -i,z) dz ≥ 0. titi ∞

15. Summary: VCG vs OP VCG-mechanisms: arbitrary valuation functions and types, but only utilitarian problems OP-mechanisms: arbitrary social- choice function, but only one- parameter types and workloaded valuation functions If a problem is both utilitarian and one-parameter  VCG and OP coincide!

16. A one-parameter mechanism for the private-edge SPT problem

17. The one-parameter SPT problem F: set of spanning tree rooted at s For any T  F f(t,T)= arg min  d T (s,v) =  t e ║ e ║ v e (t e,T)= i.e., v e (t e,T)= t e w e (T), with vVvV e  E(T) tete if e  E(T) 0 otherwise 1 if e  E(T) 0 otherwise w e (T)= Multicast protocol TFTF

18. A corresponding one-parameter mechanism M SPT = g: given the input graph G, the source node s, and the reported types r, compute an SPT S G (s) of G=(V,E,r) by using Dijkstra’s algorithm; p: for every e  E, let a e denote the player owning edge e, and let r(e) be her reported type. Then, the payment for a e is: p e (r)=r(e)w e (r) + ∫ w e (r -e,z) dz r(e) ∞ so that VP is guaranteed.

19. M SPT is truthful M SPT is truthful, since it is an OP-mechanism. Indeed, Dijkstra’s algorithm is monotone, since the workload for a e has always the following shape: 1 Ө e : threshold value where Ө e is the value such that, once fixed r -e : if a e reports at most Ө e, then e is selected in the SPT if a e reports more than Ө e, then e is not selected in the SPT

20. On the payments p e =0, if e is not selected p e = Ө e, if e is selected 1 Ө e : threshold value r(e) p e = r(e) w e (r) + ∫ w e (r -e,z) dz = r(e)+ Ө e - r(e)= Ө e r(e) ∞ p e = r(e) w e (r) + ∫ w e (r -e,z) dz = 0+0 = 0 r(e) ∞

21. An OP-problem is binary demand (BD) if 1. Type owned by each agent a i is a single parameter t i  2. The valuation function of a i is v i (t i,o)= t i w i (o) where the workload w i (o) for a i w.r.t. output o is in {0,1}. When w i (o)=1, we will say that a i is selected in o. A special class of OP-problems

22. An algorithm g for a (minimization) BD-problem is monotone if  agent a i, and for every r -i =(r 1,…,r i-1,r i+1,…,r N ), w i (g(r -i,r i )) has the following shape: Monotonicity for BD problems 1 Ө i (r -i ) riri Ө i (r -i )  {+  }: threshold value Payment for a i becomes: p i (r)= Ө i (r -i )

23. On the threshold values for the SPT problem Let e=(u,v) be an edge in S G (s) (with u closer to s than v) e remains in S G (s) until e is used to reach v from s  Ө e =d G-e (s,v)-d G (s,u) Example 1 1 2 3 2 6 s v u e r(e)=1 1 3-ε 2 3 2 6 s v u e r(e)=3-ε 1 3+ε 2 3 2 6 s v u e r(e)=3+ε  Ө e = 3

24. A trivial solution to find Ө e  e=(u,v)  S G (s) we run the Dijkstra’s algorithm on G-e=(V,E\{e},r) to find d G-e (s,v) Time complexity: k=n-1 edges multiplied by O(m + n log n) time: O(mn + n 2 log n) time The improved solution will cost as much as: O(m + n log n) time

25. Definition of Ө e s d G-e (s,v)= min {d G (s,x)+r(f)+d G (y,v)} f=(x,y)  C(e) x y u v e f

26. Computing d G-e (r,v) (and then Ө e ) is equivalent to finding edge f * such that: Definition of Ө e (cont’d) f * = arg min {d G (s,x)+r(f)+d G (y,v)} f=(x,y)  C(e) = arg min {d G (s,x)+r(f)+d G (y,v)+d G (s,v)} f=(x,y)  C(e) Since d G (s,v) is independent of f call it k(f) Observation: k(f) is a value univocally associated with edge f: it is the same for all the edges of S G (s) forming a cycle with f when f is added to S G (s) = arg min {d G (s,x)+r(f)+d G (s,y)} f=(x,y)  C(e)

27. Definition of k(f) s k(f) = d G (s,x)+r(f)+d G (s,y) x y u v e f

28. Computing the threshold Once again, we build a transmuter (w.r.t. S G (s)), where now sink nodes are labelled with k(f) (instead of w(f) as in the MST case), and we process the transmuter in reverse topological order At the end of the process, every edge e  S G (s) will remain associated with its replacement edge f *, and Ө e = (k(f * )-d G (s,v)) - d G (s,u) Time complexity: O(m  (m,n)), once that d G are given d G-e (s,v )

29. Theorem M SPT can be implemented in O(m + n log n) time. Proof: Time complexity of g: O(m + n log n) (Dijkstra with Fibonacci Heaps) Computing all the payments costs: O(m  (m,n))=O(m + n log n) time Since  (m,n) is constant when m=  (n log log n) Analysis

30. This is the end My only friend, the end… (The Doors, 1967) Thanks for your attention!