and other short stories

Zilin Jiang

Arizona State University

October 26, 2024

Joint work with Ron Aharoni, Joseph Briggs, and Ron Holzman

Bárány's colorful Carathéodory theorem

For every family of $n+1$ subsets of $\mathbb{R}^n$, each containing $\vec{a}$ in its convex hull, there exists a rainbow set with the same property.

Definition Given a family $\mathcal{F}$ of subsets $F_1, \dots, F_m$ of $E$ a subset $\{x_1, \dots, x_r\}$ is rainbow if there exist distinct $i_1, \dots, i_r$ such that $x_1 \in F_{i_1}, \dots, x_r \in F_{i_r}$

Recurring theme How large does $\mathcal{F}$ need to be to guarantee a rainbow set satisfying property $\mathcal{P}$ assuming that every member of $\mathcal{F}$ satisfies $\mathcal{P}$?

Recurring theme How large does $\mathcal{F}$ need to be to guarantee a rainbow set satisfying property $\mathcal{P}$ assuming that every member of $\mathcal{F}$ satisfies $\mathcal{P}$?

Drisko's theorem improved by Aharoni and Berger

$2n-1$ matchings of size $n$ in a bipartite graph have a rainbow matching of size $n$

Aharoni, Holzman and J.

$rn - r + 1$ fractional matchings of size $n$ in an $r$-partite hypergraph have a rainbow fractional matching of size $n$

Recurring theme How large does $\mathcal{F}$ need to be to guarantee a rainbow set satisfying property $\mathcal{P}$ assuming that every member of $\mathcal{F}$ satisfies $\mathcal{P}$?

Odd cycles: $\mathcal{P} =$ "being an odd cycle in a fixed $K_n$"

How many odd cycles in a fixed $K_n$ are needed to guarantee a rainbow odd cycle?

$n = 3$? Three odd cycles suffice

What's the answer in general?

How many odd cycles in a fixed $K_n$ are needed to guarantee a rainbow odd cycle?

Answer $n$ odd cycles suffice

Proof Take a maximal rainbow forest

At least one odd cycle is not used

This odd cycle is fully contained in a rainbow tree

Some edge doesn't respect bipartition of rainbow tree

Add that edge to the rainbow tree

How many odd cycles in a fixed $K_n$ are needed to guarantee a rainbow odd cycle?

Answer $n$ odd cycles suffice

Example $n-1$ identical cycles of length $n$ have no rainbow cycle

When $n$ is odd: $n$ odd cycles are necessary

When $n$ is even: $n-1$ odd cycles are necessary

Question Improvement for even $n$?

For every family of $n-1$ odd cycles in $K_n$, if no rainbow odd cycle exists, then ...

Think about $n = 5$...

Definition A family $\mathcal{O}$ of cycles is a pruned cactus if all the cycles are identical to a cycle on $\lvert \mathcal{O} \rvert + 1$ vertices, or $\mathcal{O}$ can be partitioned into two pruned cacti $\mathcal{O}_1$ and $\mathcal{O}_2$ such that $\cup \mathcal{O}_1$ and $\cup \mathcal{O}_2$ share exactly one vertex

Pruned cactus

Each cycle of length $n$ repeats $n-1$ times

# cycles $=$ # vertices $- 1$

No rainbow cycle

Aharoni, Briggs, Holzman, and J.

For every family of $n-1$ odd cycles in $K_n$, if there exists no rainbow odd cycle, then the family is a pruned cactus.

In a pruned cactus, (a) each cycle of length $n$ repeats $n-1$ times, and (b) # cycles $=$ # vertices $- 1$

Pruned cactus of odd cycles has even number of cycles

Corollary When $n$ is even, $n-1$ odd cycles suffice

Aharoni, Briggs, Holzman, and J.

For every family $\mathcal{O}$ of $n$ odd cycles in $K_{n+1}$, if no rainbow odd cycle, then $\mathcal{O}$ is a pruned cactus.

Proof: Break into 3 cases.

- There exists $\mathcal{K} \subsetneq \mathcal{O}$ such that $v(\cup \mathcal{K}) \le \lvert \mathcal{K} \rvert + 1$
- Every odd cycle in $\mathcal{O}$ is Hamiltonian
- For every $\mathcal{K} \subsetneq \mathcal{O}$, $v(\cup \mathcal{K}) > \lvert \mathcal{K} \rvert + 1$, and some odd cycle in $\mathcal{O}$ is not Hamiltonian

Case 1: There exists $\mathcal{K} \subsetneq \mathcal{O}$ such that $v(\cup \mathcal{K}) \le \lvert \mathcal{K} \rvert + 1$

Proof sketch "induction helps"

By induction, $\mathcal{K}$ is a pruned cactus

Without loss of generality, $\mathcal{K}$ consists of identical odd cycles $O$

Each odd cycle $O_i$ in $\mathcal{O}\setminus\mathcal{K}$ is partitioned by $V(O)$ into arcs, one of which is odd, say $P_i$

After contracting $V(O)$, odd arc $P_i$ becomes odd cycle

(a) rainbow cycle exists, or (b) new odd cycles form a pruned cactus

Case 2: Every odd cycle in $\mathcal{O}$ is Hamiltonian

Proof sketch

Let $S$ be a rainbow star of maximum size

As in Case 1, contract the leaves of $S$ and apply induction

Proof of Case 3 For every $\mathcal{K} \subsetneq \mathcal{O}$, $v(\cup \mathcal{K}) > \lvert \mathcal{K} \rvert + 1$, and some odd cycle in $\mathcal{O}$ is not Hamiltonian

Suppose $\mathcal{O} = \{O_1, \dots, O_n\}$ and $v \not\in O_n$

Consider $\mathcal{O}' = \{O_1-v, \dots, O_{n-1} - v\}$

They are connected subgraphs, and

for every $\mathcal{K}' \subseteq \mathcal{O}'$, $v(\cup \mathcal{K}') \ge \lvert \mathcal{K}' \rvert + 1$

Rado's matroid theorem

Given a matroid with ground set $E$, for every subsets $E_1, \dots, E_m$ of $E$, a rainbow independent set of size $m$ exists if and only if $\mathrm{rank}(\bigcup_{i\in I} E_i) \ge |I|$ for every $I \subseteq [m]$

Corollary

For every connected subgraphs $E_1, \dots, E_m$ in $K_{m+1}$ a rainbow spanning tree exists if and only if $v(\bigcup_{i\in I} E_i) \ge |I| + 1$ for every $I \subseteq [m]$

Proof of Case 3 For every $\mathcal{K} \subsetneq \mathcal{O}$, $v(\cup \mathcal{K}) > \lvert \mathcal{K} \rvert + 1$, and some odd cycle in $\mathcal{O}$ is not Hamiltonian

Suppose $\mathcal{O} = \{O_1, \dots, O_n\}$ and $v \not\in O_n$

Consider $\mathcal{O}' = \{O_1-v, \dots, O_{n-1} - v\}$

They are connected subgraphs, and

for every $\mathcal{K}' \subseteq \mathcal{O}'$, $v(\cup \mathcal{K}') \ge \lvert \mathcal{K}' \rvert + 1$

Rainbow spanning tree exists

Odd cycle $O_n$ is fully contained in this rainbow tree ...

Proposition For every family of $n$ cycles in $K_n$, there exists a rainbow cycle

Observation For a pruned cactus with $n-1$ cycles (on $n$ vertices), there exists no rainbow cycle.

For every family of $n-1$ cycles in $K_n$, if there exists no rainbow cycle, then ...

Definition A family $\mathcal{O}$ of cycles is a saguaro if $\mathcal{O}$ is a pruned cactus, or $\mathcal{O}$ can be partitioned into two saguaros $\mathcal{O}_1$ and $\mathcal{O}_2$ and a single cycle $O$ such that $\cup \mathcal{O}_1$ and $\cup \mathcal{O}_2$ share no vertex, and $O$ is an even cycle that alternates between $\cup \mathcal{O}_1$ and $\cup \mathcal{O}_2$

Aharoni, Briggs, Holzman, and J.

For every family of $n-1$ cycles in $K_n$, if there exists no rainbow cycle, then the family is a saguaro.

Zichao Dong and Zijian Xu

For every family of $\lfloor{(6n-1)/5}\rfloor$ even cycles in $K_n$, there exists a rainbow even cycle

Example there exist 6 squares on 6 vertices without rainbow even cycles

Glue a new copy at one vertex to the union of the previous ones

Györi, independently Frankl, Füredi and Simonyi, independently Goorevitch and Holzman

More than $n^2/8$ distinct triangles in $K_n$ have a rainbow triangle

Observation Take $n/4$ disjoint pairs of vertices, and connecting each pair to each of the remaining $n/2$ vertices with a triangle

How many cycles of length $a\mathbb{Z}+b$ are needed in $K_n$ to guarantee a rainbow cycle of length $a\mathbb{Z}+b$?

Theorem $2n-1$ matchings of size $n$ in a bipartite graph have a rainbow matching of size $n$

Conjecture $2n$ matchings of size $n$ ~~in a bipartite graph~~ have a rainbow matching of size $n$

Zilin Jiang

Arizona State University

zilinj@asu.edu

Arizona State University

zilinj@asu.edu