The Stackelberg Minimum Spanning Tree Game

The Stackelberg Minimum Spanning Tree Game Jean Cardinal Erik Demaine Samuel Fiorini GwenaëlJoret Stefan Langerman Ilan Newman Oren Weimann G3t R1CH Qu1CK with FTL N3tw0rk1ng!!! Donald Duck Uncle Scrooge Samuel Fiorini GwenaëlJoret Stefan Langerman Ilan Newman Oren Weimann Once upon a time . MIT scientists have invented a revolutionary device to communicate faster than the speed of light. Gyro Gearloose with Randall L. Stephenson Scrooge’s Goal: Infinite profit! Donald’s Goal: Build a network for his nephews Scrooge’s Goal: Infinite profit! But suddenly. a competitor appears! Gladstone Gander (to follow John E. Pepper, Jr. as chairman of Verizon) Example ? 3 1 ? ? 1 2 Gladstone Gander Uncle Scrooge Donald Duck buys MST Example Pricing MST 3 3 1 1 3 3 3 Instance 1 1 2 2 ? 3 revenue = 3 1 ? ? 1 2 1 1 3 1 1 2 1 2 1 1 2 revenue = 4 Assumption: blues have priority over reds of same weight Problem Statement Given a graph G with red and blue edges I Each red edge has a fixed cost c(e) [Gladstone Gander] I The leader [Uncle Scrooge] has to set a price p(e) for each blue edge I The follower [Donald Duck] then computes a MST with the resulting weights Goal: maximize total weight of the blue edges in a MST Context for Stackelberg MST (StackMST) Stackelberg game: I Leader makes one move I Follower makes one move dependent on leader’s move Related work: I “Highway” problems – pricing shortest paths: Labbé, Marcotte, and Savard (Management Science, 1998), Grigoriev, van Hoesel, Kraaij, Uetz, and Bouhtou (WAOA 05), survey by Stan van Hoesel I Bundle pricing: Guruswami, Hartline, Karlin, Kempe, Kenyon, and McSherry (SODA 05), Hartline and Koltun (WADS 05), Grigoriev, van Loon, Sitters, and Uetz (WG 06) I Many other games and mechanisms on spanning trees ∃ cover of size ≤ t ⇐⇒ ∃ pricing giving ≥ (n + t − 1) + 2(m − t) Hardness I Proposition StackMST is NP-hard, even if c(e) ∈ {1, 2} ∀e Proof idea: Reduction from SetCover → sets S1, S2,..., Sm → elements u1, u2,..., un ∃ cover of size ≤ t ⇐⇒ ∃ pricing giving ≥ (n + t − 1) + 2(m − t) Hardness I Proposition StackMST is NP-hard, even if c(e) ∈ {1, 2} ∀e 2 elements Proof idea: sets 1 Reduction from SetCover 2 1 → sets S1, S2,..., Sm 2 1 → elements u , u ,..., u 1 2 n 2 1 2 1 1 ∃ cover of size ≤ t ⇐⇒ ∃ pricing giving ≥ (n + t − 1) + 2(m − t) Hardness I Proposition StackMST is NP-hard, even if c(e) ∈ {1, 2} ∀e 2 elements Proof idea: sets 1 Reduction from SetCover 2 1 → sets S1, S2,..., Sm 2 1 → elements u , u ,..., u 1 2 n 2 1 2 1 1 ∃ cover of size ≤ t ⇐⇒ ∃ pricing giving ≥ (n + t − 1) + 2(m − t) Hardness I Proposition StackMST is NP-hard, even if c(e) ∈ {1, 2} ∀e elements Proof idea: sets 1 Reduction from SetCover 1 1 → sets S1, S2,..., Sm 1 → elements u1, u2,..., un 1 1 1 1 ∃ cover of size ≤ t ⇐⇒ ∃ pricing giving ≥ (n + t − 1) + 2(m − t) Hardness I Proposition StackMST is NP-hard, even if c(e) ∈ {1, 2} ∀e Proof idea: elements sets Reduction from SetCover 1 1 2 1 → sets S1, S2,..., Sm 1 → elements u1, u2,..., un 1 2 1 1 1 Hardness I Proposition StackMST is NP-hard, even if c(e) ∈ {1, 2} ∀e Proof idea: elements sets Reduction from SetCover 1 1 2 1 → sets S1, S2,..., Sm 1 → elements u1, u2,..., un 1 2 1 1 1 ∃ cover of size ≤ t ⇐⇒ ∃ pricing giving ≥ (n + t − 1) + 2(m − t) Hardness II Inapproximability result Proposition StackMST is APX-hard, even if c(e) ∈ {1, 2} ∀e Proof idea: I reduce from VertexCover in graphs with ∆ ≤ 3 I then everything is linear in n I ∃ (1 − )-approx for StackMST ⇒ ∃ (1 + 8)-approx for VertexCover in graphs with ∆ ≤ 3 Bad News for Uncle Scrooge Approximation I Assume that the red costs are c1 < c2 < ··· < ck c − c c − c Let ρ := 1 + 2 1 + ··· + k k−1 c2 ck Z ck 1 c We have: ρ ≤ k and ρ ≤ 1 + dt = 1 + ln k c1 t c1 2 1 0 0 c 1 1 c 2 c 3 2 c 4 3 c 5 Theorem StackMST is ρ-approximable ⇒ min{k, 1 + ln ck }-approximable c1 Analysis: For any price function, revenue = MST − MST ∩ R ≤ MST ∞ − MST 0 k k ∞ 0 X ∞ X 0 ⇒ OPT ≤ MST − MST = ci mi − ci mi i=1 i=1 Where ∞ ∞ I mi := #red edges of cost ci in MST 0 0 I mi := #red edges of cost ci in MST Approximation II The Best-out-of-k algorithm Algorithm Best-out-of-k (uniform pricing): I For i = 1,..., k: assign price ci to all blue edges I Pick i maximizing revenue of the corresponding MST k k ∞ 0 X ∞ X 0 ⇒ OPT ≤ MST − MST = ci mi − ci mi i=1 i=1 Where ∞ ∞ I mi := #red edges of cost ci in MST 0 0 I mi := #red edges of cost ci in MST Approximation II The Best-out-of-k algorithm Algorithm Best-out-of-k (uniform pricing): I For i = 1,..., k: assign price ci to all blue edges I Pick i maximizing revenue of the corresponding MST Analysis: For any price function, revenue = MST − MST ∩ R ≤ MST ∞ − MST 0 Approximation II The Best-out-of-k algorithm Algorithm Best-out-of-k (uniform pricing): I For i = 1,..., k: assign price ci to all blue edges I Pick i maximizing revenue of the corresponding MST Analysis: For any price function, revenue = MST − MST ∩ R ≤ MST ∞ − MST 0 k k ∞ 0 X ∞ X 0 ⇒ OPT ≤ MST − MST = ci mi − ci mi i=1 i=1 Where ∞ ∞ I mi := #red edges of cost ci in MST 0 0 I mi := #red edges of cost ci in MST Approximation II Analysis: wavefront argument bi := #blue edges (of cost ci ) in i-th tree Claim ∞ 0 mi − mi = bi − bi+1 for all i Approximation II Analysis: wavefront argument bi := #blue edges (of cost ci ) in i-th tree Claim ∞ 0 mi − mi = bi − bi+1 for all i Approximation II Analysis: wavefront argument bi := #blue edges (of cost ci ) in i-th tree Claim ∞ 0 mi − mi = bi − bi+1 for all i } Approximation II Analysis: wavefront argument bi := #blue edges (of cost ci ) in i-th tree Claim ∞ 0 mi − mi = bi − bi+1 for all i } Approximation II Analysis: wavefront argument bi := #blue edges (of cost ci ) in i-th tree Claim ∞ 0 mi − mi = bi − bi+1 for all i } Approximation II Analysis: wavefront argument bi := #blue edges (of cost ci ) in i-th tree Claim ∞ 0 mi − mi = bi − bi+1 for all i } Approximation II Analysis: wavefront argument bi := #blue edges (of cost ci ) in i-th tree Claim ∞ 0 mi − mi = bi − bi+1 for all i } Approximation II Analysis: wavefront argument bi := #blue edges (of cost ci ) in i-th tree Claim ∞ 0 mi − mi = bi − bi+1 for all i } Approximation III Analysis: rewriting OPT Denote the revenue of the i-th tree by qi := ci bi From claim, we get ∞ 0 qi qi+1 ∞ 0 qk mi − mi = − for 1 ≤ i ≤ k − 1 and mk − mk = ci ci+1 ck ⇒ Can rewrite upper bound on OPT: k X ∞ 0 q1 q2 q2 q3 qk ci (mi − mi ) = c1 − + c2 − + ··· + ck c1 c2 c2 c3 ck i=1 c2 − c1 ck − ck−1 = q1 + q2 + ··· + qk c2 ck Can tweak the analysis to get: Proposition Best-out-of-k is also a (3 + 2 ln b)-approximation Approximation IV Analysis: final bit Taking j such that qj is maximum, we have OPT Pk c (m∞ − m0) ≤ i=1 i i i qj qj q c − c q c − c q ≤ 1 + 2 1 · 2 + ··· + k k−1 · k qj c2 qj ck qj c − c c − c ≤ 1 + 2 1 + ··· + k k−1 = ρ c2 ck So Best-out-of-k is a ρ-approximation algorithm Approximation IV Analysis: final bit Taking j such that qj is maximum, we have OPT Pk c (m∞ − m0) ≤ i=1 i i i qj qj q c − c q c − c q ≤ 1 + 2 1 · 2 + ··· + k k−1 · k qj c2 qj ck qj c − c c − c ≤ 1 + 2 1 + ··· + k k−1 = ρ c2 ck So Best-out-of-k is a ρ-approximation algorithm Can tweak the analysis to get: Proposition Best-out-of-k is also a (3 + 2 ln b)-approximation And they lived happily ever after. IP formulation (IP) −→relax (LP) Proposition (good news) We have (IP) ≤ (LP) ≤ MST ∞ − MST 0 Proposition (bad news) The integrality gap of (LP) is k on instances with k distinct costs (and with ρ ∼ k) Open questions / outlook Main open question: I Does StackMST admit a constant-factor approximation? Generalizations of StackMST: I Pricing matroids: always ρ-approximable I Stackelberg Steiner trees: also generalizes highway pricing The End.

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