The document introduces lexicographic reverse search (LRS), an algorithm for enumerating the vertices of a polyhedron given its H-representation. LRS uses a lexicographic pivoting rule within the simplex method and traces all possible paths in reverse to enumerate all vertices. It was developed by Avis and Fukuda to improve upon the original reverse search algorithm. The document outlines the key concepts behind LRS, including lexicographic ordering of vectors, the lexicographic simplex method, and its applications in areas like convex hull problems and linear programming.

1. Introduction to Lexicographic
Reverse Search: lrs
June 29, 2012
Jayant Apte
ASPITRG

2. Outline
●Introduction
●Lexicographic Simplex Algorithm
●Lex-positive and Lex min bases
●The pitfalls in reverse search
●Lexicographic Reverse search
June 29,2012 Jayant Apte. ASPITRG 2

3. Introduction
● Representations of a Polyhedron
● Reverse Search: Some Trivia
● Reverse Search: The high level idea
June 29,2012 Jayant Apte. ASPITRG 3

4. H-Representation of a Polyhedron
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5. V-Representation of a Polyhedron
June 29,2012 Jayant Apte. ASPITRG 5

6. Switching between the two
representations
● H-representation V-representation: The
Vertex Enumeration problem
● Methods:
● Reverse Search, Lexicographic Reverse Search
● Double-description method
● Reverse search is the most robust method
June 29,2012 Jayant Apte. ASPITRG 6

7. Brief history of Reverse Search
● Presented by David Avis and Komei Fukuda in
a 1992 article
“A Pivoting Algorithm for Convex Hulls and
Vertex Enumeration of Arrangements and
Polyhedra"
● A C implementation: rs
● Modified by Avis in 1998 to use lexicographic
pivoting and implemented in rational arithmetic
● A C implementation: lrs
June 29,2012 Jayant Apte. ASPITRG 7

8. Applications
● Finding the vertices and extreme rays given the H-
representation of polyhedron
● Facet enumeration: V-representation to H-
representation
● Enumerating all veritces of Voronoi diagram
● Computing volume of convex hull of set of points
● Estimation of number of vertices of polyhedron
● Nash equilibrium in non-cooperative bi-matrix games
June 29,2012 Jayant Apte. ASPITRG 8

9. Relationship between Simplex Algorithm
and Reverse Search
● Simplex Algorithm is used for solving linear
programs
● Simplex Algorithm starts at any vertex of
polyhedron defined by the constraint set
● Travels along the edges of polyhedron to find
the optimum
● Reverse search runs simplex in reverse
● Traces all possible paths taken by Simplex
algorithm in reverse
June 29,2012 Jayant Apte. ASPITRG 9

10. Relationship between Simplex Algorithm
and Reverse Search
Ref.David Avis, lrs: A Revised Implementation of the Reverse Search Vertex Enumeration
Algorithm
June 29,2012 Jayant Apte. ASPITRG 10

11. Background: Lexicographic Order
June 29,2012 Jayant Apte. ASPITRG 11

12. Vectors
● Comparing two vectors in lexicographic sense
● e.g. [5, 4, 1, 7, 4] [5, 4, 0, 9, 8 ]
● Lex-positive vector
● First non-zero element is positive
● e.g. [0, 4, -1, -7, -4], [5, -4, 1, -7, 4] are lexicographically
positive vectors
● e.g. [0 -4 -1 -7 -4], [-5 - 4 1 -7 4] are lexicographically
negative vectors
● Lex-min vector
● Out of [1 2 3 -5 9],[1 2 0 9 9], [2 3 6 -7 1];
[1 2 0 9 9] is lex-min vector
June 29,2012 Jayant Apte. ASPITRG 12

13. Lexicographic Simplex Method
June 29,2012 Jayant Apte. ASPITRG 13

14. The Linear Programming Problem
Objective Function
Constraint Set
●d-decision variables
●m-constraints
June 29,2012 Jayant Apte. ASPITRG 14

15. Example
June 29,2012 Jayant Apte. ASPITRG 15

16. Nature of Constrained Region
(0.5,0.5,1.5)
(1,1,1)
(0,1,1)
(0,0,1) (1,1,0)
(0,1,0)
(1,0,0)
(0,0,0)
June 29,2012 Jayant Apte. ASPITRG 16

17. Relationship between inequalities and
Vertex: Geometric intuition of Degeneracy
Highlighted inequalities are tight at (0,0,1)
(0,0,1)
June 29,2012 Jayant Apte. ASPITRG 17

18. Relationship between inequalities and
Vertex: Geometric intuition of Degeneracy
(0,0,1)
June 29,2012 Jayant Apte. ASPITRG 18

19. Relationship between Co-Basis and Vertex:
Geometric intuition of Degeneracy
● (0,0,1) is a degenerate vertex
● There are ways of specifying the vertex (0,0,1)
● In general this number are ways of representing a
degenerate vertex, where
n=total number of inequalities that are tight at that vertex
d=number of dimensions
● This number can get very large very quickly.
● One of the places where simplex and by implication
reverse search runs into problems.
June 29,2012 Jayant Apte. ASPITRG 19

20. How to solve this LP?
● Rearrange inequalities to decide a starting
vertex
● Introduce slack variables
● Convert to matrix form
● Convert to dictionary form
June 29,2012 Jayant Apte. ASPITRG 20

21. Starting Simplex at a vertex:
Rearrangement of the inequalities
●Rearrange the inequalities so that last 'd' inequalities
contain initial vertex
(0.5,0.5,1.5)
June 29,2012 Jayant Apte. ASPITRG 21

22. Define Slack Variables
●Convert inequalities to equalities by
introducing slack variables
●Original variables are called decision
variables
June 29,2012 Jayant Apte. ASPITRG 22

23. Our Example
June 29,2012 Jayant Apte. ASPITRG 23

24. Our Example
-1 0 0 1 0 0 0 0 0 0 0 0
0 -1 0 0 1 0 0 0 0 0 0 0
0 0 -1 0 0 1 0 0 0 0 0 0
1 0 0 0 0 0 1 0 0 0 0 0
0 1 0 0 0 0 0 1 0 0 0 0
0 1 1 0 0 0 0 0 1 0 0 0
0 -1 1 0 0 0 0 0 0 1 0 0
1 0 1 0 0 0 0 0 0 0 1 0
-1 0 1 0 0 0 0 0 0 0 0 1
June 29,2012 Jayant Apte. ASPITRG 24

25. Terminology
➢ d decision variables
➢ m constraints
➢ Hence m slack variables
➢ Basis: B: Ordered m-tuple indexing set of m
affinely independent columns of A
➢
➢ Co-basis: N: Ordered d-tuple indexing remaining
d columns
➢ : Sub-matrix of A indexed by B
June 29,2012 Jayant Apte. ASPITRG 25

26. Our example
We define initial basis as
Hence initial co-basis is
-1 0 0 1 0 0 0 0 0 0 0 0
0 -1 0 0 1 0 0 0 0 0 0 0
0 0 -1 0 0 1 0 0 0 0 0 0
1 0 0 0 0 0 1 0 0 0 0 0
0 1 0 0 0 0 0 1 0 0 0 0
0 1 1 0 0 0 0 0 1 0 0 0
0 -1 1 0 0 0 0 0 0 1 0 0
1 0 1 0 0 0 0 0 0 0 1 0
-1 0 1 0 0 0 0 0 0 0 0 1
June 29,2012 Jayant Apte. ASPITRG 26

27. Our example
is invertible
Is a Basic Feasible Solution(BFS)
-1 0 0 1 0 0 0 0 0 0 0 0
0 -1 0 0 1 0 0 0 0 0 0 0
0 0 -1 0 0 1 0 0 0 0 0 0
1 0 0 0 0 0 1 0 0 0 0 0
0 1 0 0 0 0 0 1 0 0 0 0
0 1 1 0 0 0 0 0 1 0 0 0
0 -1 1 0 0 0 0 0 0 1 0 0
1 0 1 0 0 0 0 0 0 0 1 0
-1 0 1 0 0 0 0 0 0 0 0 1
June 29,2012 Jayant Apte. ASPITRG 27

28. Add a cost row
● Add a cost row at the top
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29. Our example
0 1 2 3 4 5 6 7 8 9 10 11 12
1 1 1 1 0 0 0 0 0 0 0 0 0
0 -1 0 0 1 0 0 0 0 0 0 0 0
0 0 -1 0 0 1 0 0 0 0 0 0 0
0 0 0 -1 0 0 1 0 0 0 0 0 0
0 1 0 0 0 0 0 1 0 0 0 0 0
0 0 1 0 0 0 0 0 1 0 0 0 0
0 0 1 1 0 0 0 0 0 1 0 0 0
0 0 -1 1 0 0 0 0 0 0 1 0 0
0 1 0 1 0 0 0 0 0 0 0 1 0
0 -1 0 1 0 0 0 0 0 0 0 0 1
June 29,2012 Jayant Apte. ASPITRG 29

30. Converting Linear program to the
dictionary format
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31. Derive the dictionary from matrix
form
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32. Our example
Cost function value
At current vertex
The vertex
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33. Dictionary Format
Change in Cost
Corresponding to each Cost Corresponding
Co-basic Variable To Dictionary
Basic Indices Co-Basic Indices
Solution Corresponding
To Dictionary
June 29,2012 Jayant Apte. ASPITRG 33

34. Redefine a bit
Basis: B: Ordered m+1-tuple indexing set of m+1 affinely
independent columns of dictionary &
Dictionary is to serve as starting point
Call this the solution of the initial dictionary or the initial
Call this the initial implying that initial basis is identity matrix
June 29,2012 Jayant Apte. ASPITRG 34

35. Our example
0 1 2 3 4 5 6 7 8 9 10 11 12 13
0 1 0 0 0 0 0 0 0 0 0 1 -1.5 -0.5 -2.5 Cost
1 0 1 0 0 0 0 0 0 0 0 0 0.5 -0.5 0.5
2 0 0 1 0 0 0 0 0 0 0 -1 0.5 0.5 0.5 The
3 0 0 0 1 0 0 0 0 0 0 0 0.5 0.5 1.5 vertex
Basis 4 0 0 0 0 1 0 0 0 0 0 0 0.5 -0.5 0.5
5 0 0 0 0 0 1 0 0 0 0 -1 0.5 0.5 0.5
6 0 0 0 0 0 0 1 0 0 0 0 0.5 0.5 1.5
7 0 0 0 0 0 0 0 1 0 0 0 -0.5 0.5 0.5
8 0 0 0 0 0 0 0 0 1 0 1 -0.5 -0.5 0.5
9 0 0 0 0 0 0 0 0 0 1 1 -1 -1 0
June 29,2012 Jayant Apte. ASPITRG 35

36. Simplex Algorithm: The Ingredients
● It travels from one vertex to another, or more specifically one BFS to
another BFS
● For that it uses pivoting operation to obtain the dictionary
corresponding to new BFS.
● A co-basic variable assumes a nonzero value and is said to 'enter'
the basis
● A basic variable 'leaves' the basis
● We get a new dictionary which is 'equivalent' to current dictionary in
algebraic terms, i.e. It carries the same solution
● This new dictionary is obtained by means of 'pivot operation'.
June 29,2012 Jayant Apte. ASPITRG 36

37. Choosing the entering variable
● Choose the entering variable that brings about an increase in cost
One of these is supposed to assume a nonzero value
● Only a negative coefficient at will do as it will mean
that is positive and results in an increase in cost
● If there are many such co-basic variables, choose the one with least
subscript
● If there are no such co-basic variables, current dictionary is optimal and
current BFS is the optimum
June 29,2012 Jayant Apte. ASPITRG 37

38. Choosing entering variable
0 1 2 3 4 5 6 7 8 9 10 11 12 13
0 1 0 0 0 0 0 0 0 0 0 1 -1.5 -0.5 -2.5
1 0 1 0 0 0 0 0 0 0 0 0 0.5 -0.5 0.5
2 0 0 1 0 0 0 0 0 0 0 -1 0.5 0.5 0.5
3 0 0 0 1 0 0 0 0 0 0 0 0.5 0.5 1.5
Basis 4 0 0 0 0 1 0 0 0 0 0 0 0.5 -0.5 0.5
5 0 0 0 0 0 1 0 0 0 0 -1 0.5 0.5 0.5
6 0 0 0 0 0 0 1 0 0 0 0 0.5 0.5 1.5
7 0 0 0 0 0 0 0 1 0 0 0 -0.5 0.5 0.5
8 0 0 0 0 0 0 0 0 1 0 1 -0.5 -0.5 0.5
9 0 0 0 0 0 0 0 0 0 1 1 -1 -1 0
June 29,2012 Jayant Apte. ASPITRG 38

39. Choosing the leaving variable
● Recall lexicographic order
● Construct following matrix:
June 29,2012 Jayant Apte. ASPITRG 39

40. Choosing the leaving variable
● Let and consider the column . If
this column defines an extreme ray.
● Otherwise, let be the index such that is
lexicographically minimum vector of:
● Such a minimum is unique as has full row rank
● This notation can be abbreviated as,
● In case there is no such we set
June 29,2012 Jayant Apte. ASPITRG 40

41. Choosing the Leaving Variable
0 1 2 3 4 5 6 7 8 9 10 11 12 13
0 1 0 0 0 0 0 0 0 0 0 1 -1.5 -0.5 -2.5
1 0 1 0 0 0 0 0 0 0 0 0 0.5 -0.5 0.5
2 0 0 1 0 0 0 0 0 0 0 -1 0.5 0.5 0.5
3 0 0 0 1 0 0 0 0 0 0 0 0.5 0.5 1.5
Basis 4 0 0 0 0 1 0 0 0 0 0 0 0.5 -0.5 0.5
5 0 0 0 0 0 1 0 0 0 0 -1 0.5 0.5 0.5
6 0 0 0 0 0 0 1 0 0 0 0 0.5 0.5 1.5
7 0 0 0 0 0 0 0 1 0 0 0 -0.5 0.5 0.5
8 0 0 0 0 0 0 0 0 1 0 1 -0.5 -0.5 0.5
9 0 0 0 0 0 0 0 0 0 1 1 -1 -1 0
June 29,2012 Jayant Apte. ASPITRG 41

42. Lexicographic choice of leaving
variable
June 29,2012 Jayant Apte. ASPITRG 42

43. Where to get after each
successive iteration?
● Let be current basis inverse and be
the next basis inverse
● and differ only in 1 column
● We can take advantage of this fact and write
● Pre-multiplication by elementary matrix
corresponds to row operations
● Hence, we get new basis inverse by some row
operations
June 29,2012 Jayant Apte. ASPITRG 43

44. The pivot Operation
June 29,2012 Jayant Apte. ASPITRG 44

45. This is how to get after each
successive iteration
● is given as:
June 29,2012 Jayant Apte. ASPITRG 45

46. This is how to get after each
successive iteration
Row multiplication Operation
●Row corresponding to pivot
element is divided by pivot
element
Hence pivot operation automatically yields new basis inverse
June 29,2012 Jayant Apte. ASPITRG 46

47. Result of pivot (5,11)
Basis:0 1 2 3 4 11 6 7 8 9 Co-Basis:10 5 12
0 1 2 3 4 5 6 7 8 9 10 11 12 13
0
1 0 0 0 0 3 0 0 0 0 -2 0 1 -1
1
0 1 0 0 0 -1 0 0 0 0 1 0 -1 0
2
0 0 1 0 0 -1 0 0 0 0 0 0 0 0
3 0 0 0 1 0 -1 0 0 0 0 1 0 0 1
4 0 0 0 0 1 -1 0 0 0 0 1 0 -1 0
Basis
11 0 0 0 0 0 2 0 0 0 0 -2 1 1 1
6 0 0 0 0 0 -1 1 0 0 0 1 0 0 1
7 0 0 0 0 0 1 0 1 0 0 -1 0 1 1
8
0 0 0 0 0 1 0 0 1 0 0 0 0 1
9
0 0 0 0 0 2 0 0 0 1 -1 0 0 1
Identity column
June 29,2012 Jayant Apte. ASPITRG 47

48. Solving Linear Programs:
Simplex Algorithm
June 29,2012 Jayant Apte. ASPITRG 48

49. Iteration 1
You Are Here.
(0.5,0.5,1.5) (1,1,1)
(0,1,1)
(0,0,1) (1,1,0)
(0,1,0)
(1,0,0)
(0,0,0)
June 29,2012 Jayant Apte. ASPITRG 49

50. Iteration 1
0 1 2 3 4 5 6 7 8 9 10 11 12 13
0 1 0 0 0 0 0 0 0 0 0 1 -1.5 -0.5 -2.5 Cost
1 0 1 0 0 0 0 0 0 0 0 0 0.5 -0.5 0.5
2 0 0 1 0 0 0 0 0 0 0 -1 0.5 0.5 0.5 The
3 0 0 0 1 0 0 0 0 0 0 0 0.5 0.5 1.5 vertex
Basis 4 0 0 0 0 1 0 0 0 0 0 0 0.5 -0.5 0.5
5 0 0 0 0 0 1 0 0 0 0 -1 0.5 0.5 0.5
6 0 0 0 0 0 0 1 0 0 0 0 0.5 0.5 1.5
7 0 0 0 0 0 0 0 1 0 0 0 -0.5 0.5 0.5
8 0 0 0 0 0 0 0 0 1 0 1 -0.5 -0.5 0.5
9 0 0 0 0 0 0 0 0 0 1 1 -1 -1 0
June 29,2012 Jayant Apte. ASPITRG 50

51. Pivot(5,11)
June 29,2012 Jayant Apte. ASPITRG 51

52. Iteration 2
You Are Here.
(0.5,0.5,1.5) (1,1,1)
(0,1,1)
(0,0,1)
(1,1,0)
(0,1,0)
(1,0,0)
(0,0,0)
June 29,2012 Jayant Apte. ASPITRG 52

53. Iteration 2
Basis:0 1 2 3 4 11 6 7 8 9 Co-Basis:10 5 12
0 1 2 3 4 5 6 7 8 9 10 11 12 13
0 1 0 0 0 0 3 0 0 0 0 -2 0 1 -1
1 0 1 0 0 0 -1 0 0 0 0 1 0 -1 0
2 0 0 1 0 0 -1 0 0 0 0 0 0 0 0
3 0 0 0 1 0 -1 0 0 0 0 1 0 0 1
4 0 0 0 0 1 -1 0 0 0 0 1 0 -1 0
Basis
11 0 0 0 0 0 2 0 0 0 0 -2 1 1 1
6 0 0 0 0 0 -1 1 0 0 0 1 0 0 1
7 0 0 0 0 0 1 0 1 0 0 -1 0 1 1
8 0 0 0 0 0 1 0 0 1 0 0 0 0 1
9 0 0 0 0 0 2 0 0 0 1 -1 0 0 1
June 29,2012 Jayant Apte. ASPITRG 53

54. pivot(4,10)
June 29,2012 Jayant Apte. ASPITRG 54

55. Iteration 3
You Are Still Here.
Blame degeneracy!
(0.5,0.5,1.5) (1,1,1)
(0,1,1)
(0,0,1)
(1,1,0)
(0,1,0)
(1,0,0)
(0,0,0)
June 29,2012 Jayant Apte. ASPITRG 55

56. Iteration 3
Basis:0 1 2 3 10 11 6 7 8 9 Co-Basis:4 5 12
0 1 2 3 4 5 6 7 8 9 10 11 12 13
0 1 0 0 0 2 1 0 0 0 0 0 0 -1 -1
1 0 1 0 0 -1 0 0 0 0 0 0 0 0 0
2 0 0 1 0 0 -1 0 0 0 0 0 0 0 0
Basis 3 0 0 0 1 -1 0 0 0 0 0 0 0 1 1
10 0 0 0 0 1 -1 0 0 0 0 1 0 -1 0
11 0 0 0 0 2 0 0 0 0 0 0 1 -1 1
6 0 0 0 0 -1 0 1 0 0 0 0 0 1 1
7 0 0 0 0 1 0 0 1 0 0 0 0 0 1
8 0 0 0 0 0 1 0 0 1 0 0 0 0 1
9 0 0 0 0 1 1 0 0 0 1 0 0 -1 1
June 29,2012 Jayant Apte. ASPITRG 56

57. pivot(6,12)
June 29,2012 Jayant Apte. ASPITRG 57

58. Iteration 4
(0.5,0.5,1.5) (1,1,1)
(0,1,1)
(0,0,1) (1,1,0)
You Are Here.
(0,1,0)
(1,0,0)
(0,0,0)
June 29,2012 Jayant Apte. ASPITRG 58

59. Iteration 4
Basis:0 1 2 3 10 11 12 7 8 9 Co-Basis:4 5 6
0 1 2 3 4 5 6 7 8 9 10 11 12 13
0 1 0 0 0 1 1 1 0 0 0 0 0 0 0
1 0 1 0 0 -1 0 0 0 0 0 0 0 0 0
2 0 0 1 0 0 -1 0 0 0 0 0 0 0 0
Basis 3 0 0 0 1 0 0 -1 0 0 0 0 0 0 0
10 0 0 0 0 0 -1 1 0 0 0 1 0 0 1
11 0 0 0 0 1 0 1 0 0 0 0 1 0 2
12 0 0 0 0 -1 0 1 0 0 0 0 0 1 1
7 0 0 0 0 1 0 0 1 0 0 0 0 0 1
8 0 0 0 0 0 1 0 0 1 0 0 0 0 1
9 0 0 0 0 0 1 1 0 0 1 0 0 0 2
June 29,2012 Jayant Apte. ASPITRG 59

60. Observations
● Lexicographic simplex method doesn't travel
through all the bases corresponding to a vertex
● It only visits bases that are lex-positive
● Recall
● A basis is lex-positive if the rows indexed d+1,...,m of D
are lex-positive
●
● Lexicographic pivot rule is reversible!
● is reversible
June 29,2012 Jayant Apte. ASPITRG 60

61. Reverse Search:
The Recipe
● Start with dictionary corresponding to the optimal
vertex
● Ask yourself 'What pivot would have landed me at this
dictionary if i was running simplex?'
● Go to that dictionary by applying reverse pivot to
current dictionary
● Ask the same question again
● Generate the so-called 'reverse search tree'
June 29,2012 Jayant Apte. ASPITRG 61

62. What pivot would have landed me at this dictionary if i
was running simplex?
June 29,2012 Jayant Apte. ASPITRG 62

63. Pitfalls
● Initial basis not being the unique optimum basis
● Degeneracy of other vertices: One vertex
corresponding to many (actually too many)
bases
● Can result in program outputting a vertex more than
once
● Earlier version rs suffered from all these
June 29,2012 Jayant Apte. ASPITRG 63

64. Initial Basis not the unique optimum
basis
● Reverse search must begin at the optimal vertex
corresponding to any cost.
● In rs, one would have to find all possible bases
corresponding to initial vertex and perform reverse search on
them.
● In lrs, we define initial cost in such a way that initial vertex
has a unique Basis
● We define cost as:
● This cost makes the top row of the dictionary
June 29,2012 Jayant Apte. ASPITRG 64

65. Degeneracy of vertices
● Define lex-min bases
●
● Clearly if (i)-(iii) was to be true, we could do
and get that is lexicographically smaller
than
● summarizes the operation
June 29,2012 Jayant Apte. ASPITRG 65

66. Evolution of lrs
All possible Bases A reduced set of bases A reduced set of bases
Very few vertices Called lex-positive Bases Called lex-min Bases
rs |lex-positive BFS>|Vertices|| |Lex-min BFS|=|Vertices|
|BFS|>>|Vertices|
June 29,2012 Jayant Apte. ASPITRG 66

67. Recall these functions
June 29,2012 Jayant Apte. ASPITRG 67

68. Pseudo-code
June 29,2012 Jayant Apte. ASPITRG 68

69. Pseudo-code
●Examine each cobasic column of dictionary by reverse to see if there exists a
lex positive pivot using this column
●Reverse will be executed d times for each dictionary
X Cobasis={a,b,c}
Reverse(a)=p
Cobasis={p,b,c} 1
June 29,2012 Jayant Apte. ASPITRG 69

70. Pseudo-code
X Cobasis={a,b,c}
Reverse(a)=0 Reverse(b)=?
June 29,2012 Jayant Apte. ASPITRG 70

71. Pseudo-code
● When whie loop terminates, return to the parent of current node: backtrack
● Correct j is restored by selectpivot
● Subsequent j=j+1 means next cobasic element of the parent comes under
consideration
Backtrack
X Cobasis={a,b,c}
Reverse(c)=0
Reverse(a)=0
Reverse(b)=0
June 29,2012 Jayant Apte. ASPITRG 71

72. Pseudo-code
Execution continues until all the columns of the starting
●
dictionary are examined
June 29,2012 Jayant Apte. ASPITRG 72

73. Lexicographic Reverse Search on
our example
(0.5,0.5,1.5) (1,1,1)
(0,1,1)
(0,0,1) (1,1,0)
You Are Here.
(0,1,0)
(1,0,0)
(0,0,0)
June 29,2012 Jayant Apte. ASPITRG 73

74. V=(0 0 0)
N=( 10 11 12)
reverse(10)=4 reverse(12)=9
reverse(11)=5 pivot(9,12)
pivot(4,10) V=(0 0 1)
pivot(5,11)
V=(1 0 0) N=(10 11 9)
N=(4 11 12) V=0 1 0)
N=(10 5 12) reverse(11)=6
reverse(4)=0 pivot(6,11) V=(0 1 1)
reverse(12)=8 reverse(10)=4 N=(10 6 9)
pivot(8,12) pivot(4,10)
reverse(12)=6reverse(5)=0 reverse(10)=7
pivot(6,12) pivot(7,10)
V=(1 0 1)
N=(4 11 8) reverse(11)=0 V=(0 0 1) reverse(9)=0
N=(7 11 9)
reverse(11)=5 V=(1 1 0) V=(0 1 1)
pivot(5,11) N=(4 5 12) reverse(9)=0
N=(10 5 6) reverse(8)=0 reverse(12)=0
reverse(11)=8
V=(0.5 0.5 1.5) pivot(8,11) reverse(9)=8
N=(7 8 9) pivot(8,9) reverse(7)=0
V=(1 1 1)
N=(5 6 9) V=(1 0 1)
N=(8 12 9)
reverse(8)=0
reverse(7)=6
pivot(6,7) reverse(9)=0
reverse(5)=0 reverse(6)=0 reverse(9)=0
V=(0.5 0.5 1.5)
N=(6 8 9)
reverse(9)=5
pivot(5,9) reverse(8)=0 reverse(8)=0 reverse(12)=0 reverse(9)=0
reverse(6)=0 V=(1 0 0)
N=(6 8 5)
reverse(6)=0 reverse(8)=0 reverse(5)=0
June 29,2012 Jayant Apte. ASPITRG 74

75. Conclusion
● Lexicographic reverse search is an improved
reverse search for vertex enumeration
● It is robust- not affected by degeneracy
● Yet, it travels more bases than there are
vertices, hence requiring lot of computational
power
June 29,2012 Jayant Apte. ASPITRG 75

76. Starting dictionary
1 0 0 0 0 0 0 0 0 0 1 1 1 0
0 1 0 0 0 0 0 0 0 0 -1 0 0 0
0 0 1 0 0 0 0 0 0 0 0 -1 0 0
0 0 0 1 0 0 0 0 0 0 0 0 -1 0
0 0 0 0 1 0 0 0 0 0 1 0 0 1
0 0 0 0 0 1 0 0 0 0 0 1 0 1
0 0 0 0 0 0 1 0 0 0 0 1 1 2
0 0 0 0 0 0 0 1 0 0 0 -1 1 1
0 0 0 0 0 0 0 0 1 0 1 0 1 2
0 0 0 0 0 0 0 0 0 1 -1 0 1 1
June 29,2012 Jayant Apte. ASPITRG 76

77. Result of pivot(4,10)
(real notation)
1 0 0 0 -1 0 0 0 0 0 0 1 1 -1
0 1 0 0 1 0 0 0 0 0 0 0 0 1
0 0 1 0 0 0 0 0 0 0 0 -1 0 0
0 0 0 1 0 0 0 0 0 0 0 0 -1 0
0 0 0 0 1 0 0 0 0 0 1 0 0 1
0 0 0 0 0 1 0 0 0 0 0 1 0 1
0 0 0 0 0 0 1 0 0 0 0 1 1 2
0 0 0 0 0 0 0 1 0 0 0 -1 1 1
0 0 0 0 -1 0 0 0 1 0 0 0 1 1
0 0 0 0 1 0 0 0 0 1 0 0 1 2
June 29,2012 Jayant Apte. ASPITRG 77

78. Result of pivot(8,12)
1 0 0 0 0 0 0 0 -1 0 0 1 0 -2
0 1 0 0 1 0 0 0 0 0 0 0 0 1
0 0 1 0 0 0 0 0 0 0 0 -1 0 0
0 0 0 1 -1 0 0 0 1 0 0 0 0 1
0 0 0 0 1 0 0 0 0 0 1 0 0 1
0 0 0 0 0 1 0 0 0 0 0 1 0 1
0 0 0 0 1 0 1 0 -1 0 0 1 0 1
0 0 0 0 1 0 0 1 -1 0 0 -1 0 0
0 0 0 0 -1 0 0 0 1 0 0 0 1 1
0 0 0 0 2 0 0 0 -1 1 0 0 0 1
June 29,2012 Jayant Apte. ASPITRG 78

79. Result of pivot(5,11)
1 0 0 0 0 -1 0 0 -1 0 0 0 0 -3
0 1 0 0 1 0 0 0 0 0 0 0 0 1
0 0 1 0 0 1 0 0 0 0 0 0 0 1
0 0 0 1 -1 0 0 0 1 0 0 0 0 1
0 0 0 0 1 0 0 0 0 0 1 0 0 1
0 0 0 0 0 1 0 0 0 0 0 1 0 1
0 0 0 0 1 -1 1 0 -1 0 0 0 0 0
0 0 0 0 1 1 0 1 -1 0 0 0 0 1
0 0 0 0 -1 0 0 0 1 0 0 0 1 1
0 0 0 0 2 0 0 0 -1 1 0 0 0 1
June 29,2012 Jayant Apte. ASPITRG 79

80. Now backtrack all the way to root
June 29,2012 Jayant Apte. ASPITRG 80

81. Result of pivot(5,11)
1 0 0 0 0 -1 0 0 0 0 1 0 1 -1
0 1 0 0 0 0 0 0 0 0 -1 0 0 0
0 0 1 0 0 1 0 0 0 0 0 0 0 1
0 0 0 1 0 0 0 0 0 0 0 0 -1 0
0 0 0 0 1 0 0 0 0 0 1 0 0 1
0 0 0 0 0 1 0 0 0 0 0 1 0 1
0 0 0 0 0 -1 1 0 0 0 0 0 1 1
0 0 0 0 0 1 0 1 0 0 0 0 1 2
0 0 0 0 0 0 0 0 1 0 1 0 1 2
0 0 0 0 0 0 0 0 0 1 -1 0 1 1
June 29,2012 Jayant Apte. ASPITRG 81

82. Result of pivot(4,10)
1 0 0 0 -1 -1 0 0 0 0 0 0 1 -2
0 1 0 0 1 0 0 0 0 0 0 0 0 1
0 0 1 0 0 1 0 0 0 0 0 0 0 1
0 0 0 1 0 0 0 0 0 0 0 0 -1 0
0 0 0 0 1 0 0 0 0 0 1 0 0 1
0 0 0 0 0 1 0 0 0 0 0 1 0 1
0 0 0 0 0 -1 1 0 0 0 0 0 1 1
0 0 0 0 0 1 0 1 0 0 0 0 1 2
0 0 0 0 -1 0 0 0 1 0 0 0 1 1
0 0 0 0 1 0 0 0 0 1 0 0 1 2
June 29,2012 Jayant Apte. ASPITRG 82

83. Backtrack 1 step: pivot(10,4)
1 0 0 0 0 -1 0 0 0 0 1 0 1 -1
0 1 0 0 0 0 0 0 0 0 -1 0 0 0
0 0 1 0 0 1 0 0 0 0 0 0 0 1
0 0 0 1 0 0 0 0 0 0 0 0 -1 0
0 0 0 0 1 0 0 0 0 0 1 0 0 1
0 0 0 0 0 1 0 0 0 0 0 1 0 1
0 0 0 0 0 -1 1 0 0 0 0 0 1 1
0 0 0 0 0 1 0 1 0 0 0 0 1 2
0 0 0 0 0 0 0 0 1 0 1 0 1 2
0 0 0 0 0 0 0 0 0 1 -1 0 1 1
June 29,2012 Jayant Apte. ASPITRG 83

84. Result of pivot(6,12)
1 0 0 0 0 0 -1 0 0 0 1 0 0 -2
0 1 0 0 0 0 0 0 0 0 -1 0 0 0
0 0 1 0 0 1 0 0 0 0 0 0 0 1
0 0 0 1 0 -1 1 0 0 0 0 0 0 1
0 0 0 0 1 0 0 0 0 0 1 0 0 1
0 0 0 0 0 1 0 0 0 0 0 1 0 1
0 0 0 0 0 -1 1 0 0 0 0 0 1 1
0 0 0 0 0 2 -1 1 0 0 0 0 0 1
0 0 0 0 0 1 -1 0 1 0 1 0 0 1
0 0 0 0 0 1 -1 0 0 1 -1 0 0 0
June 29,2012 Jayant Apte. ASPITRG 84

85. Backtrack all the way to root
June 29,2012 Jayant Apte. ASPITRG 85

86. Result of pivot(9,12)
1 0 0 0 0 0 0 0 0 -1 2 1 0 -1
0 1 0 0 0 0 0 0 0 0 -1 0 0 0
0 0 1 0 0 0 0 0 0 0 0 -1 0 0
0 0 0 1 0 0 0 0 0 1 -1 0 0 1
0 0 0 0 1 0 0 0 0 0 1 0 0 1
0 0 0 0 0 1 0 0 0 0 0 1 0 1
0 0 0 0 0 0 1 0 0 -1 1 1 0 1
0 0 0 0 0 0 0 1 0 -1 1 -1 0 0
0 0 0 0 0 0 0 0 1 -1 2 0 0 1
0 0 0 0 0 0 0 0 0 1 -1 0 1 1
June 29,2012 Jayant Apte. ASPITRG 86

87. Result of pivot (7,10)
1 0 0 0 0 0 0 -2 0 1 0 3 0 -1
0 1 0 0 0 0 0 1 0 -1 0 -1 0 0
0 0 1 0 0 0 0 0 0 0 0 -1 0 0
0 0 0 1 0 0 0 1 0 0 0 -1 0 1
0 0 0 0 1 0 0 -1 0 1 0 1 0 1
0 0 0 0 0 1 0 0 0 0 0 1 0 1
0 0 0 0 0 0 1 -1 0 0 0 2 0 1
0 0 0 0 0 0 0 1 0 -1 1 -1 0 0
0 0 0 0 0 0 0 -2 1 1 0 2 0 1
0 0 0 0 0 0 0 1 0 0 0 -1 1 1
June 29,2012 Jayant Apte. ASPITRG 87

88. Result of pivot(8,11)
1 0 0 0 0 0 0 1 -1.5 -0.5 0 0 0 -2.5
0 1 0 0 0 0 0 0 0.5 -0.5 0 0 0 0.5
0 0 1 0 0 0 0 -1 0.5 0.5 0 0 0 0.5
0 0 0 1 0 0 0 0 0.5 0.5 0 0 0 1.5
0 0 0 0 1 0 0 0 -0.5 0.5 0 0 0 0.5
0 0 0 0 0 1 0 1 -0.5 -0.5 0 0 0 0.5
0 0 0 0 0 0 1 1 -1 -1 0 0 0 0
0 0 0 0 0 0 0 0 0.5 -0.5 1 0 0 0.5
0 0 0 0 0 0 0 -1 0.5 0.5 0 1 0 0.5
0 0 0 0 0 0 0 0 0.5 0.5 0 0 1 1.5
June 29,2012 Jayant Apte. ASPITRG 88

89. Result of pivot(6,7)
1 0 0 0 0 0 -1 0 -0.5 0.5 0 0 0 -2.5
0 1 0 0 0 0 0 0 0.5 -0.5 0 0 0 0.5
0 0 1 0 0 0 1 0 -0.5 -0.5 0 0 0 0.5
0 0 0 1 0 0 0 0 0.5 0.5 0 0 0 1.5
0 0 0 0 1 0 0 0 -0.5 0.5 0 0 0 0.5
0 0 0 0 0 1 -1 0 0.5 0.5 0 0 0 0.5
0 0 0 0 0 0 1 1 -1 -1 0 0 0 0
0 0 0 0 0 0 0 0 0.5 -0.5 1 0 0 0.5
0 0 0 0 0 0 1 0 -0.5 -0.5 0 1 0 0.5
0 0 0 0 0 0 0 0 0.5 0.5 0 0 1 1.5
June 29,2012 Jayant Apte. ASPITRG 89

90. Result of pivot(6,10)
1 0 0 0 0 -1 0 0 -1 0 0 0 0 -3
0 1 0 0 0 1 -1 0 1 0 0 0 0 1
0 0 1 0 0 1 0 0 0 0 0 0 0 1
0 0 0 1 0 -1 1 0 0 0 0 0 0 1
0 0 0 0 1 -1 1 0 -1 0 0 0 0 0
0 0 0 0 0 2 -2 0 1 1 0 0 0 1
0 0 0 0 0 2 -1 1 0 0 0 0 0 1
0 0 0 0 0 1 -1 0 1 0 1 0 0 1
0 0 0 0 0 1 0 0 0 0 0 1 0 1
0 0 0 0 0 -1 1 0 0 0 0 0 1 1
June 29,2012 Jayant Apte. ASPITRG 90

91. Backtrack:
pivot(9,5),pivot(7,6),pivot(11,8)
1 0 0 0 0 0 0 -2 0 1 0 3 0 -1
0 1 0 0 0 0 0 1 0 -1 0 -1 0 0
0 0 1 0 0 0 0 0 0 0 0 -1 0 0
0 0 0 1 0 0 0 1 0 0 0 -1 0 1
0 0 0 0 1 0 0 -1 0 1 0 1 0 1
0 0 0 0 0 1 0 0 0 0 0 1 0 1
0 0 0 0 0 0 1 -1 0 0 0 2 0 1
0 0 0 0 0 0 0 1 0 -1 1 -1 0 0
0 0 0 0 0 0 0 -2 1 1 0 2 0 1
0 0 0 0 0 0 0 1 0 0 0 -1 1 1
June 29,2012 Jayant Apte. ASPITRG 91

92. Result of pivot(8,9)
1 0 0 0 0 0 0 0 -1 0 0 1 0 -2
0 1 0 0 0 0 0 -1 1 0 0 1 0 1
0 0 1 0 0 0 0 0 0 0 0 -1 0 0
0 0 0 1 0 0 0 1 0 0 0 -1 0 1
0 0 0 0 1 0 0 1 -1 0 0 -1 0 0
0 0 0 0 0 1 0 0 0 0 0 1 0 1
0 0 0 0 0 0 1 -1 0 0 0 2 0 1
0 0 0 0 0 0 0 -1 1 0 1 1 0 1
0 0 0 0 0 0 0 -2 1 1 0 2 0 1
0 0 0 0 0 0 0 1 0 0 0 -1 1 1
June 29,2012 Jayant Apte. ASPITRG 92

93. Backtrack pivot(9,8),pivot(10,7)
1 0 0 0 0 0 0 0 0 -1 2 1 0 -1
0 1 0 0 0 0 0 0 0 0 -1 0 0 0
0 0 1 0 0 0 0 0 0 0 0 -1 0 0
0 0 0 1 0 0 0 0 0 1 -1 0 0 1
0 0 0 0 1 0 0 0 0 0 1 0 0 1
0 0 0 0 0 1 0 0 0 0 0 1 0 1
0 0 0 0 0 0 1 0 0 -1 1 1 0 1
0 0 0 0 0 0 0 1 0 -1 1 -1 0 0
0 0 0 0 0 0 0 0 1 -1 2 0 0 1
0 0 0 0 0 0 0 0 0 1 -1 0 1 1
June 29,2012 Jayant Apte. ASPITRG 93

94. Result of pivot(6,11)
1 0 0 0 0 0 -1 0 0 0 1 0 0 -2
0 1 0 0 0 0 0 0 0 0 -1 0 0 0
0 0 1 0 0 0 1 0 0 -1 1 0 0 1
0 0 0 1 0 0 0 0 0 1 -1 0 0 1
0 0 0 0 1 0 0 0 0 0 1 0 0 1
0 0 0 0 0 1 -1 0 0 1 -1 0 0 0
0 0 0 0 0 0 1 0 0 -1 1 1 0 1
0 0 0 0 0 0 1 1 0 -2 2 0 0 1
0 0 0 0 0 0 0 0 1 -1 2 0 0 1
0 0 0 0 0 0 0 0 0 1 -1 0 1 1
June 29,2012 Jayant Apte. ASPITRG 94

95. Backtrack to root
June 29,2012 Jayant Apte. ASPITRG 95

96. Proof of uniqueness of initial basis
● Suppose there exists another lex-positive
optimum basis B
● Recall D matrix.
June 29,2012 Jayant Apte. ASPITRG 96

97. Proof of uniqueness of initial basis
B B*={0,...m}
B-B* B*-B N*={m+1,...,m+d}
● Consider pivoting from B* to B. Indices B-B*= must be
brought into basis
● Every pivot will involve subtracting a certain row vector from cost row
● Since B is also optimum
● It follows that is lexicographicallly negative
June 29,2012 Jayant Apte. ASPITRG 97