# Unique Sink Orientations of Cubes Motivation and Algorithms Tibor Szabó ETH Zürich (Bernd Gärtner, Ingo Schurr, Emo Welzl)

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Unique Sink Orientations of Cubes Motivation and Algorithms Tibor Szabó ETH Zürich (Bernd Gärtner, Ingo Schurr, Emo Welzl)

Outline of Talk  The Problem  Definition  Examples  Connections to algorithmic problems in Geometry  Linear Programming  Smallest Enclosing Ball Problem  Unique Sink Orientation of Cubes  Properties  Algorithms  Lower bound  (Lots of) open questions

Cubes --- Notation A = {1,2,…,n} = [n] C A is an edge labeled graph : V(C A )=2 A E(C A )={{u,v} : |u  v| =1} λ({u,v}) = a, where {a} = u  v {} {1,2} {2} {1} {3,4} {1,2,3,4} {2,3,4} {1,3,4} {4} {1,2,4} {2,4} {1,4} {3} {1,2,3} {2,3} {1,3} 12 34

Unique Sink Orientation A sink is a vertex without outgoing edges A unique sink orientation (USO) of a cube is an orientation where every face has a unique sink

Find the sink in a USO The USO is given implicitly; it can be accessed through vertex evaluations  Vertex evaluation: returns orientations of all incident edges The Problem

How many vertices have to be evaluated in a USO of an n-cube, until we have evaluated the sink? t(n) worst-case deterministic t(n) worst-case expected for randomized t(1)t(1)= 2= 3/2 ?½½?

? ? ? t(2) ≤ 3

? t(2) ≤ 3 ? ? ?

t(2) = 3 t(2) = 43/20

Further small values t(3) = 5 t̃(3) = 4074633/1369468 t(4) = 7 t̃(4) = ? t(5) = ? [Rote ’01]

WHY ???

Appearences of USOs Orientations of slanted geometric cubes relative to a generic linear function, e.g. Klee-Minty cube (acyclic)

Linear Programming – the straightforward connection MinimumSink acyclic USO

Appearences of USOs Orientations of slanted geometric cubes relative to a generic linear function, e.g. Klee-Minty cube (acyclic) Model for smallest enclosing ball problem (can have cycles as well) [Gärtner et al., ‘01]

a b c Smallest Enclosing Ball Problem n points d-space EXAMPLE

…of a set P of d+1 affinely independent points in d-space. Define orientation on the cube C P : For x  Q  P, Q ← QU{x} iff x  β(Q), where β(Q) is the smallest ball with Q on boundary. This is a USO S is the sink iff β(S) is the smallest ball Smallest enclosing ball enclosing P

a b c a bc abc ac c b ab Smallest enclosing ballUSO of cubes

Appearences of USOs Orientations of slanted geometric cubes relative to a generic linear function, e.g. Klee-Minty cube (acyclic) Model for smallest enclosing ball problem (can have cycles as well) [Gärtner et al., ‘01] Model for linear complementarity problems corresponding to P-matrices (can have cycles) [Stickney-Watson ‘78] Model for general LP [Gärtner-Schurr, ‘03]

Linear Programming – the abstract connection Polynomial algorithm for general USOs might Strongly polynomial algorithm for LP * *

USOs – Related work t(n) ≤ 2 n (???); t(n) ≤ (3/2) n s(n) = (√2) n(1+o(1)) [Aldous ’84] s(n) is the expected number of evaluations the fastest randomized algorithm takes while finding the sink in a cube with an orientation given by a function on the vertices (→ acyclic), which has a unique global sink (no restriction on the faces).

USOs – Specific algorithms RandomEdge can take ((n-1)/2)! steps [Morris ’00] Analysis of RandomEdge for Klee-Minty cube [Gärtner, Henk, Ziegler ’98] RandomEdge on decomposable orientations is O(n 2 ) [Williamson Hoke ’88] RandomFacet t acyc (n) ≤ e 2√n [Gärtner ’95]

An easy first observation t(n) ≤ 2 n-1 + 1 ?? ? ? (?) t(n) ≤ ¾(2 n-1 + 1)

2 n-1 + 1 is best possible For arbitrary orientations

SOMETHING BETTER FOR USOs?

Out-map Out-map s φ of an orientation φ of C n is defined by v → set of labels of outgoing edges incident with v v is a sink iff s φ (v) =  12 3 {2,3} {1,2} {3} {1,2,3} {} {2} {1} {1,3}

Are out-maps of USOs bijections ?????

Lemma: Reversing the orientation of the edges of a fixed label does not affect the unique sink property. a a a a a a a Proof:

Lemma Out-maps of USOs are bijections between the vertex set of C n and 2 [n]. Proof: Suppose u and v have the same out-map value. Reverse all edges of label from s(u)=s(v). This makes both u and v a sink, a contradiction.

Lemma Out-maps of USOs are bijections between the vertex set of C n and 2 [n]. Inherited orientation φ B …of USO φ relative to B  A. φ B defined on C A’, where A’ = A \ B

Inherited Orientation UI …of φ on C A relative to BA.A. The out-map on the sinks of B-labeled faces of C A gives a mapping s’: 2 A ’ → 2 A’, A’ = A \ B. s’ is well-defined s’ is a USO

a bc d B = {b,c}A’ = {a,d} a d d a

Product Algorithm Choose B with |B|=k Perform a search in the (n-k)-cube with the inherited orientation φ B ; takes t(n-k) “hyper”- vertex evaluation Each “hyper”-vertex evaluation is a search for the sink of the corresponding k-cube; takes t(k) vertex evaluations t(n) ≤ t(k) · t(n-k) t(n) ≤ t(k) · t(n-k)

Consequences t(2) = 3 → t(n) ≤ (√3) n ≈ 1.73 n t(3) = 5 → t(n) ≤ (√5) n ≈ 1.71 n t(2) = 43/20 → t(n) ≤ 1.47 n t(3) = 4074633/1369468 → t(n) ≤ 1.44 n 3 [Rote ’01]

Something different… for deterministic

Fibonacci Seesaw Evaluate two arbitrary antipodal vertices v0v0 u0u0 --- Step 0

F G u i-1 v i-1 dim F = i-1 dim G = i-1 b b ? dim F’ = i dim G’ = i uiui F’ G’ t(i-1) evaluations Fibonacci Seesaw --- Step i

→ 0 → 1 → … i → (i+1) … → (n-1) t(n) ≤ 2 + t(0) + t(1) + … + t(n-2) Thus t(n) ≤ F n+2 = O(1.62 n ); where F n is the Fibonacci sequence: F 0 = 0, F 1 = 1, F k+2 = F k+1 + F k. t(4) ≤ 8 < 9 = 2 3 + 1 = 3 · 3 = t(2) · t(2) 2t(0)t(1)t(i)t(n-2)

Is the Fibonacci Seesaw optimal? NO!! The SevenStepsToHeaven Algorithm solves the 4-dimensional problem using at most 7 evaluations. That is optimal; t(4)=7 Combining it with the Fibonacci Seesaw gives: t(n)=O(1.61 n )=o(F n )

Lower bound for t acyc (n)

Tool #1: Product of orientations

Special cases anything (n-1) + 1 = n1 + (n-1) = n

Tool #2: Local change

Important Tool #1 and Tool #2 preserve acyclicity

Strategy against Al We maintain Subset L of labels Set W of requested vertices Orientation φ on C L Partial outmap ξ : W → 2 A

Properties i.|L| = |W| ii.w’ ∩ L ≠ w’’ ∩ L for  w’,w’’  W iii.ξ(w) = φ(w) U ([n] – L) for  w  W

What Al sees at before its fourth request Set of labels = L = {a,b,c} a b c Requested vertices = W = { } Orientation φ on C L CACA

Al’s fourth request Set of labels = L = {a,b,c} a b c Requested vertices = W = { } Orientation φ on C L CACA ? ? ? ? ? ? ?

L := L  {l} l w Update v ∩ L w ∩ L (Tool #1) New request w is “above” at most one old request v in C L Pick l  w  v (otherwise any l  L)

Second phase Choose a translate C’ of C L’ which is empty, i.e. there are no requests in it so far; (2 n-|L’| ≥ n !!!) Tell Al the sink is in C’ and force it solve an (n-log n)-dimensional problem on C’. when |L| = |W| = n-log n

Forcing a recursion by Tool #2 φ on C L C’

Recursion t acyc (n) ≥ n-logn + t acyc (n-logn) i.e. t acyc (n) ≥ n 2 / 2logn

Currently known Deterministic n 2-o(1) ≤ t acyc (n) ≤ t(n) ≤ O(1.61 n ) Randomized t(n) ≤ O(1.44 n ) Randomized acyclic t acyc (n) ≤ e 2√n Log (# of USOs) Θ(2 n log n) [Gärtner ’95] [Matousek ‘01]

Killing RandomFacet F(n) ≤ ½ F(n-1)+ ½ F(n-1)+ ½ n F(0)+½ n F(1)+... ……….+ ½ n F(n-1) A n-1 rand A n-1 Special case of Tool 1: F(n) ≥ e √n

Killing RandomEdge Is harder. One needs Repeated application of Tool 1 and 2 Lots of copies of the Klee-Minty cube

A large class Let ς be the uniform orientation M be a matching of the hypercube Flip the edges of the matching By Tool #2 this is a USO There are a lot of them For this class there is an algorithm finding the sink in FIVE(!!!) steps.

Outlook for Acyclic USOs Exploit acyclicity in order to improve the existing best deterministic algorithm Analyze RandomEdge on AUSO; it could take e c³  n time [Matousek-Szabo, ’04] (it is even worse for general USOs [Morris]; ) No upper bound is known for AUSO. Is the class of orientation constructed by Tool #1 and #2 realizable as an LP?

Outlook for General USOs Provide a faster randomized algorithm, which is conceptually different from the Product Algorithm Prove any lower bound for the expected running time of a randomized algorithm Behaviour of RandomFacet is unresolved (on acyclic USOs the expected running time is e Θ(√n) [Matousek, Gärtner] )

THE END

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