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 $\mathcal{P}$?

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

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

Proof: Break into 3 cases.

1. There exists $\mathcal{K} \subsetneq \mathcal{O}$ such that $v(\cup \mathcal{K}) \le \lvert \mathcal{K} \rvert + 1$
2. Every odd cycle in $\mathcal{O}$ is Hamiltonian
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

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

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 ...

Rainbow odd cycles

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 $\cup \mathcal{O}_2$

Rainbow even cycles

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

Rainbow triangles

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