1. Feasible Combinatorial Matrix Theory
Ariel Fernandez & Michael Soltys
August 29, 2013
Comb Matrix - Soltys MFCS’13 IST Austria Title - 1/21

2. Statistical Archeology: sequence dating (Flinders Petrie, 1899)
900 pre-dynastic Egyptian graves containing 800 representatives of
pottery.
The “graves-versus-varieties” matrix contains vast amount of
information, such as sequential ordering.
[Kendall 1969]
Comb Matrix - Soltys MFCS’13 IST Austria Introduction - 2/21

3. Paleogenomics: DNA sequence organization of ancient living
organisms using similarities and differences between chromosomes
of extant organisms.
[Chauve et al 2008]
Comb Matrix - Soltys MFCS’13 IST Austria Introduction - 3/21

4. Consecutive-ones Property: C1P
Consider a slight relaxation, (k, δ)-C1P: each row has at most k
blocks of 1s and the gap between any two blocks is at most δ.
So (1, 0)-C1P is C1P, and deciding if an A has (k, δ)-C1P is:
polytime for (1, 0)
NP-hard for every k ≥ 2, δ ≥ 1 except (2, 1)
What about (2, 1)? [Patterson 2012]
Comb Matrix - Soltys MFCS’13 IST Austria Introduction - 4/21

5. K¨onig’s Min-Max
0
1 0 1 1
1 0
1
00
0
1
1 0
00
Comb Matrix - Soltys MFCS’13 IST Austria Introduction - 5/21

6. Comb Matrix - Soltys MFCS’13 IST Austria Introduction - 6/21

7. Problems
Cover(A, α):
∀i, j ≤ r(A)(A(i, j) = 1 → α(1, i) = 1 ∨ α(2, j) = 1)
MinCover(A, α):
Cover(A, α) ∧ ∀α ≤ c(α)(Cover(A, α ) → Σα ≥ Σα)
KMM(A, α, β):
MinCover(A, α) ∧ MaxSelect(A, β) → Σα = Σβ
a ΣB
1 formula.
But classical proof is ΠB
2 — is there a ΣB
1 proof?
Comb Matrix - Soltys MFCS’13 IST Austria Problem - 7/21

8. Related Theorems:
Menger’s: size of min cut equals max nr of disjoint s, t-paths
(“Min-Cut Max-Flow”)
Hall’s: ∀k ∈ [n] |Si1 ∪ . . . ∪ Sik
| ≥ k, then there exists a
System of Distinct Representatives.
Dilworth’s: Min nr of chains needed to partition a poset
equals size of max anti-chain of that poset.
Can they all be shown equivalent to KMM with ΣB
0 proofs?
Comb Matrix - Soltys MFCS’13 IST Austria Problem - 8/21

9. ΣB
1 Proof of KMM
Let lA be the min nr of lines necessary to cover A
Let oA be the max selection of ones in A
Comb Matrix - Soltys MFCS’13 IST Austria Main Proof - 9/21

10. 0
1 0 1 1
1 0
1
00
0
1
1 0
00
lA=oA= 3
We want to show with ΣB
1 induction that oA = lA.
Comb Matrix - Soltys MFCS’13 IST Austria Main Proof - 10/21

11. LA uses ΣB
0 induction
LA oA ≤ lA
LA over Z is equivalent to VTC0
And oA ≤ lA follows more or less from the Pigeonhole Principle:
if we can select oA 1s, no two on the same line, then we shall
require at least lA lines to cover those 1s.
Comb Matrix - Soltys MFCS’13 IST Austria Main Proof - 11/21

12. ∃LA oA ≥ lA
Showing oA ≥ lA is more difficult; we use ΣB
1 induction.
Lots of cases, but the interesting case is:
0
where we reduce the general case to the case of the blue matrix
whose cover requires as many lines as rows.
Comb Matrix - Soltys MFCS’13 IST Austria Main Proof - 12/21

13. 00 00 00 0 0 0 0 0 0
Comb Matrix - Soltys MFCS’13 IST Austria Interesting Case - 13/21

14. 10 00 00 0 0 0 0 0 0
Comb Matrix - Soltys MFCS’13 IST Austria Interesting Case - 14/21

15. 0
0 00 00 0 0 0 0 0 01
0 0 0 1 0 0 0 0 0
Comb Matrix - Soltys MFCS’13 IST Austria Interesting Case - 15/21

16. 1
0 00 00 0 0 0 0 0 01
0 0 0 1 0 0 0 0 0
Comb Matrix - Soltys MFCS’13 IST Austria Interesting Case - 16/21

17. 1
0 00 00 0 0 0 0 0 01
0 0 0 1 0 0 0 0 0
Comb Matrix - Soltys MFCS’13 IST Austria Interesting Case - 17/21

18. H
1 1 1
1
1
1
1
1
1
1
1
1
1
1
1
1
A
B
C
D
E
FG
Comb Matrix - Soltys MFCS’13 IST Austria Interesting Case - 18/21

19. J
1 1 1
1
1
1
1
1
1
1
1
1
1
1
1
1
H
1
1
Comb Matrix - Soltys MFCS’13 IST Austria Interesting Case - 19/21

20. ΣB
0 Proof of Equivalence
For example, ΣB
0 proof of Menger → KMM
yx
Left graph has a matching of size k ⇐⇒
Right graph has k disjoint {x, y}-paths
Left graph has a cover of size k ⇐⇒
Right graph has an {x, y}-cut of size k
KMM → Meng
complicated,
[Aharoni 1983]
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21. Some open problems
Frankl’s Theorem: t a positive integer, and m ≤ n(2t −1)
t ; if A
is m × n, and its rows are distinct, then there exists a column
that when deleted, the resulting matrix has at most 2t−1 − 1
pairs of equal rows. (Bondy’s Theorem when t = 1.)
Complexity of decompositions: A = P1 + P2 + · · · + Pn + X;
n boys and n girls, each boy introduced to exactly k girls and
vice versa. Compute a pairing where each boy & girl has been
previously introduced.
Projective Geometry









1 1 1 0 0 0 0
1 0 0 1 1 0 0
1 0 0 0 0 1 1
0 1 0 1 0 1 0
0 1 0 0 1 0 1
0 0 1 1 0 0 1
0 0 1 0 1 1 0









Desargues Thm
Comb Matrix - Soltys MFCS’13 IST Austria Conclusion - 21/21