1. Thomas Dueholm Hansen – Aarhus Univ. Uri Zwick – Tel Aviv Univ.
Random-Edge is slower than Random-Facet on abstract cubes
Thomas Dueholm Hansen – Aarhus Univ. Uri Zwick – Tel Aviv Univ.
Hebrew University April 11, 2016

2. Find the highest point in a polytope/polyhedron
Linear Programming
Maximize a linear objective function subject to a set of linear equalities and inequalities
Find the highest point in a polytope/polyhedron

3. The Simplex Algorithm [Dantzig (1947)]
Move up, from vertex to vertex, along edges, until reaching the top.

4. Deterministic pivoting rules
Largest improvement
Largest slope
Dantzig’s rule – Largest modified cost
Bland’s rule – avoids cycling
Lexicographic rule – also avoids cycling
Zadeh’s rule – “least recently entered”
All known to require an exponential number of steps, in the worst-case
[Klee-Minty (1972)] , … , [Amenta-Ziegler (1996)]
[Friedmann (2011)]

5. Randomized pivoting rules
Random-Edge
Choose a random improving edge
Random-Facet
[Kalai (1992)] [Matoušek-Sharir-Welzl (1992)]
To be described shortly
Random-Facet is sub-exponential!
Is Random-Edge sub-exponential???
Is Random-Edge faster than Random-Facet?

6. Random-Facet 2 𝑂 𝑑 ln 𝑛−𝑑 𝑑 2 𝑂 𝑑 ln 𝑛−𝑑 𝑑 [Kalai (1992)] [MSW (1992)]
Fastest known (randomized) pivoting rule
𝑛 = number of inequalities
𝑑 = number of variables = dimension
Assume 𝑛≥2𝑑
2 𝑂 𝑑 ln 𝑛−𝑑 𝑑
[Kalai (1992)] [MSW (1992)]
[Hansen-Z (2015)] (following [Kalai (1992)])
2 𝑂 𝑑 ln 𝑛−𝑑 𝑑

7. Hirsch Conjecture (1957)
The diameter of a d-dimensional, n-faceted polytope is at most n−d
Refuted by [Santos (2010)].
Diameter is still believed to be polynomial.
Quasi-polynomial upper bound [Kalai-Kleitman (1992)] ([Todd (2014)])

8. Abstract objective functions (AOFs)
Acyclic Unique Sink Orientations (AUSOs)
Every face has a unique sink

9. Figure taken from a paper by Gärtner-Henk-Ziegler
Klee-Minty cubes (1972)
Figure taken from a paper by Gärtner-Henk-Ziegler

10. AUSOs of n-cubes 2n facets 2n vertices AUSOs
Williamon-Hoke (1988)
Kalai (1998)
Szabó, Welzl (2001)
Gärtner (2002)
Every subcube has a unique sink
The diameter is exactly n
No diameter issue!

11. Klee-Minty cubes (1972) 𝐾 𝑀 𝑛−1 𝐾 𝑀 𝑛−1
011
111
𝐾 𝑀 𝑛−1
001
101 …
010
110
𝐾 𝑀 𝑛−1
000
100
𝐾 𝑀 𝑛 has a Hamiltonian path (Gray code).
Random-Edge on 𝐾 𝑀 𝑛 takes Θ( 𝑛 2 ), w.h.p. [Balogh-Pemantle (2007)]

12. Turn-based 2-Player 0-Sum Stochastic Games [Shapley ’53] [Gillette ’57] … [Condon ’92]
Total cost
Discounted cost
Limiting average cost
Both players have optimal positional strategies.
Can optimal strategies be found in polynomial time?
Game (two actions per state)  AUSO (of a cube)
Random-Facet is the fastest known algorithm.

13. Random-Facet on the 𝑛-cube [Ludwig (1995)] [Gärtner (2002)]
Start at some vertex 𝑎 of the 𝑛-cube 𝐴. …
𝑖-th coordinate
𝐴 0 𝑖
𝐴 1 𝑖
Split the 𝑛-cube into two (𝑛−1)-cubes 𝐴 0 𝑖 , 𝐴 1 𝑖 along a random coordinate 𝑖. 𝑏 𝑏′ 𝑐 𝑎
Recursively find the sink 𝑏 of the (𝑛−1)-cube 𝐴 𝑗 𝑖 containing 𝑎.
If 𝑏 is not the global sink, move to 𝑏 ′ =𝑏⊕ e 𝑖 .
Exponential?
Sub-exponential!
Recursively find the sink 𝑐 of the (𝑛−1)-cube 𝐴 1−𝑗 𝑖 containing 𝑏′, starting from 𝑏′.
The starting point 𝑏′ of the second recursive call contains valuable information.

14. Random-Facet on the 𝑛-cube [Ludwig (1995)] [Gärtner (2002)]𝑖 𝑛
(After reordering the coordinates according to the hidden order.)
All correct !
Would never be switched !
There is a hidden order of the indices under which the sink returned by the first recursive call correctly fixes the first i bits.
The algorithm does not know the hidden order. But, choosing a coordinate that was fixed has no effect.
Thus, the second recursive call is effectively on an (𝑛−𝑖)-cube.

15. Random-Facet on the 𝑛-cube [Ludwig (1995)] [Gärtner (2002)]𝑖 𝑛
(After reordering the coordinates according to the hidden order.)
All correct !
Would never be switched !
𝑓 𝑛 =𝑓 𝑛−1 + 1 𝑛 𝑖=1 𝑛−1 1+𝑓 𝑛−𝑖
𝑓 𝑛 = 𝑖=0 𝑛 1 𝑖! 𝑛 𝑖 ≤ 𝑖=0 𝑛 𝑛 𝑖 𝑖! 2 = 𝑖=0 𝑛 𝑛 𝑖! 2𝑖 ≤ e 2 𝑛

16. Lower bounds for Random-Edge
2 Ω 𝑛 1/3 for AUSOs [Matoušek-Szabó (2006)]
2 Ω 𝑛 1/4 for LPs [Friedmann-Hansen-Z (2011)]
2 Ω 𝑛 1/2 for AUSOs [here]
The new lower bound is a simplification of the lower bound of Matoušek and Szabó obtained by replacing the Klee-Minty cube, used as a building block, by a path AUSO.
Two main building blocks: Product of AUSOs, Hypersink replacement
Main part of the technical analysis: Random Walk with reshuffles (on a path AUSO)

17. A product (blowup) construction𝐴
𝐴: 0,1 𝑛 → 0,1 𝑛
vertex
outmap
𝐵 𝐱 : 0,1 𝑚 → 0,1 𝑚
𝐶=𝐴× 𝐵 𝐱
𝐶 𝐱,𝐲 =(𝐴 𝐱 , 𝐵 𝐱 (y))
𝐵 000
Adaptation of a slide by Tibor Szabó.

18. Hypersink replacement
Slide by Tibor Szabó.

19. Randomized product + hypersink replacement
vertex
outmap 𝐴
𝐴: 0,1 𝑛 → 0,1 𝑛
𝐵: 0,1 𝑚 → 0,1 𝑚 𝐴
𝑠𝑖𝑛𝑘 𝐵 = 1 𝑚 𝐴
𝐶=𝐴 × 𝑅 𝐵=𝐴× 𝐵 𝐱 𝐴 𝐴′
𝐵 𝐱 obtained from 𝐵 by randomly permuting the indices.
The copy of 𝐴 corresponding to 𝐲= 1 𝑚 is a hypersink. 11
Replace it by a random translation 𝐴′ of 𝐴.
Adaptation of a slide by Tibor Szabó.

20. Random-Edge on a random product [Matoušek-Szabó (2006)]
Each step in 𝐶=𝐴 × 𝑅 𝐵 is either a step in 𝐴, i.e., 𝐱,𝐲 →( 𝐱 ′ ,𝐲), or a step in 𝐵 x , i.e., 𝐱,𝐲 → 𝐱, 𝐲 ′ .
A step 𝐱,𝐲 →( 𝐱 ′ ,𝐲) in 𝐴, does not change 𝐲, but changes the cube 𝐲 is in from 𝐵 𝐱 to 𝐵 𝐱 ′ .
As 𝐵 𝐱 is a random permutation of 𝐵 𝐱 ′ , this corresponds to randomly permuting 𝐲, a reshuffle.
The induced process on 𝐱, until 𝐱=𝑠𝑖𝑛𝑘 𝐴 , is exactly Random-Edge on 𝐴.
The induced process on 𝐲, is a Random Walk with Reshuffles on 𝐵.
Once 𝐲=𝑠𝑖𝑛𝑘 𝐵 , the orientation on 𝐱 changes from 𝐴 to 𝐴′. As 𝐴′ is a random translation of 𝐴′, 𝐲 is a completely random in 𝐴′.
If 𝐱 reaches 𝑠𝑖𝑛𝑘(𝐴) before 𝐲 reaches 𝑠𝑖𝑛𝑘 𝐵 , we get two Random-Edge walks on 𝐴.

21. Random-Edge on a random product [Matoušek-Szabó (2006)]
The probability of reshuffle in each step of Random-Reshuffle on 𝐵 depends on 𝐱. (To be discussed later.)
Lemma: If the probability that Random-Edge on 𝐴 makes less than 𝑇 steps is at most 𝑝, and the probability that Random-Reshuffle on 𝐵 makes less than 𝑇 steps is at most 𝑞, both starting from a random vertex, then the probability that Random-Edge, starting at a random vertex, makes less than 2𝑇 steps on 𝐴 × 𝑅 𝐵 is at most 2𝑝+𝑞.

22. Lower bound Construction [Matoušek-Szabó (2006)]
Let 𝐵 be an 𝑚-AUSO. Compute randomized powers of 𝐵.
𝐴 0 =𝐵 , 𝐴 𝑖 = 𝐴 𝑖−1 × 𝑅 𝐵
𝐴= 𝐴 𝑘 , where 𝑘=𝛾𝑚, for some 𝛾>0.
If Random-Reshuffle on 𝐵 is slow enough, then Random-Edge on 𝐴, which is an 𝛾 𝑚 2 -AUSO, makes at least 2 𝑚 steps, with high probability.
In the last step, we need Random-Reshuffle on 𝐵 to make at least 2 𝛾𝑚 steps, with high probability.
Matoušek-Szabó take 𝐵=𝐾 𝑀 𝑚 to be the Klee-Minty cube.
Random-Reshuffle on Klee-Minty cubes is not slow enough…
Further complications. Lower bound becomes 2 Ω 𝑛 1/3 .
Our modification: Take 𝐵= 𝑃 𝑚 to be the path AUSO.

23. A shortest paths problem gives rise to an AUSO.
A path AUSO
𝑀=𝑛+1 𝑀 𝑀 𝑀 1 1 1 1 2 3 𝑛 1 1 1 1 1
A shortest paths problem gives rise to an AUSO.
A vertex of the AUSO corresponds to a choice of an outgoing edge from each vertex of the graph.
In this particular case, the 𝑖-th bit wants to stay or become a 1 iff all preceding bits are 1.

24. Random-Reshuffle on a path AUSO
𝐱 in 𝐴
𝐲 in 𝐵
𝑎 – outdegree of 𝐱 in 𝐴
𝑖 – number of leading 1s
𝑎≥1
𝑗 – number of non-leading 1s
Outdegree = 𝑗+1
With probability 𝑟= 𝑎 𝑎+𝑗+1 ≥ 1 j+2 , reshuffle 𝐲.
Otherwise, with probability 1 𝑗+1 , change the first 0 to 1,
and probability 𝑗 𝑗+1 , change a random non-leading 1 to 0.
The reshuffle probability 𝑟≥ 1 𝑗+2 chosen by an adversary.

25. Random-Reshuffle on a path AUSO
𝑖 – number of leading 1s
𝐲 in 𝐵
𝑗 – number of non-leading 1s
type = (𝑖,𝑗)
weight = 𝑘=𝑖+𝑗
𝑚− 𝑖+1 𝑗 𝑚 𝑘
A reshuffle on a state of weight 𝑘 creates a state of type (𝑖,𝑗), where 𝑖+𝑗=𝑘, with probability
All states of type (𝑖,𝑗) are obtained with equal probabilities.
The probability of increasing or decreasing 𝑘 depends only on (𝑖,𝑗).
We thus obtain induced random walks on types, and then on weights.

26. Constant drift to the left ⟹ Exponential time to reach 𝑚.
𝑛-cube 0,1 𝑚
Types (𝑖,𝑗)
Weights 𝑘=𝑖+𝑗 1 2 3 𝑚
Constant drift to the left ⟹ Exponential time to reach 𝑚.

27. Induced random walk on the line
Slight complication: From (𝑘,0), the probability of +1 is greater than ½.
Solution: Consider two steps together.
Basic step: Zero or more reshuffles, followed by a step in 𝐵.
Several reshuffles in a row are equivalent to a single reshuffle.
The worst type of weight 𝑘 is (𝑘,0).

28. Induced random walk on the line
Probability of weight increase in one step from (𝑖,𝑗)
𝑟 is the reshuffle probability
𝑢 𝑖 ,𝑗 = 1−𝑟 1 𝑗+1 +𝑟 𝑗 ′ =0 𝑘 𝑚−𝑘+ 𝑗 ′ −1 𝑗′ 𝑚 𝑘 𝑗 ′ +1
= 1−𝑟 1 𝑗+1 +𝑟 𝑚−𝑘 (𝑘+1)(𝑚−𝑘−1) − 1 (𝑚−𝑘−1) 𝑚 𝑘
≤ 1 8 , 8≤𝑘≤𝑚−9

29. Induced random walk on the line
Probability of weight decreasing by 2:
𝑠 1 𝑠 𝑗 ′ =0 𝑘 𝑚−𝑘+ 𝑗 ′ −1 𝑗′ 𝑚 𝑘 𝑗′ 𝑗 ′ 𝑗 ′ −1 𝑗 ′ + 𝑠 𝑠 2 𝑗 ′ + 𝑠 2 𝑗 ′ =0 𝑘 𝑚−𝑘+ 𝑗 ′′ 𝑗′′ 𝑚 𝑘−1 𝑗′′ 𝑗 ′′ 𝑗 ′ −1 𝑗 ′ + 𝑠 𝑠 2 𝑗 ′ + 𝑠 2 𝑗 ′ =0 𝑘 𝑚−𝑘+ 𝑗 ′′ 𝑗′′ 𝑚 𝑘−1 𝑗′′ 𝑗 ′′ +1

30. Concluding remarks and open problems
We presented an 2 Ω 𝑛 lower bound for Random-Edge.
The bound can be slightly improved to 2 Ω 𝑛 log 𝑛 , thus Random-Edge is slower than Random-Facet.
This is the best that can be obtained using hierarchical constructions.
Best upper bound known for Random-Edge is 𝑂(1.8 𝑛 ).
Is Random-Edge sub-exponential?
Is there an algorithm that beats Random-Facet? Is there a polynomial time algorithm?
(The algorithm does not necessarily have to follow a path. It may jump around.)