Subfield: Combinatorics


Question: In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group.

It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make.

| ARMLton |  |
| :--- | :--- |
| Resident | Dishes |
| Paul | pie, turkey |
| Arnold | pie, salad |
| Kelly | salad, broth |


| ARMLville |  |
| :--- | :--- |
| Resident | Dishes |
| Sally | steak, calzones |
| Ross | calzones, pancakes |
| David | steak, pancakes |

The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}.

A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad.

Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$.

In order to avoid the embarrassing situation where two people bring the same dish to a group dinner, if two people know how to make a common dish, they are forbidden from participating in the same group meeting. Formally, a group assignment on $T$ is a function $f: T \rightarrow\{1,2, \ldots, k\}$, satisfying the condition that if $f\left(r_{i}\right)=f\left(r_{j}\right)$ for $i \neq j$, then $r_{i}$ and $r_{j}$ do not know any of the same recipes. The group number of a town $T$, denoted $\operatorname{gr}(T)$, is the least positive integer $k$ for which there exists a group assignment on $T$.

For example, consider once again the town of ARMLton. A valid group assignment would be $f($ Paul $)=f($ Kelly $)=1$ and $f($ Arnold $)=2$. The function which gives the value 1 to each resident of ARMLton is not a group assignment, because Paul and Arnold must be assigned to different groups.


For a dish $D$, a resident is called a $D$-chef if he or she knows how to make the dish $D$. Define $\operatorname{chef}_{T}(D)$ to be the set of residents in $T$ who are $D$-chefs. For example, in ARMLville, David is a steak-chef and a pancakes-chef. Further, $\operatorname{chef}_{\text {ARMLville }}($ steak $)=\{$ Sally, David $\}$.


If $\operatorname{gr}(T)=\left|\operatorname{chef}_{T}(D)\right|$ for some $D \in \operatorname{dish}(T)$, then $T$ is called homogeneous. If $\operatorname{gr}(T)>\left|\operatorname{chef}_{T}(D)\right|$ for each dish $D \in \operatorname{dish}(T)$, then $T$ is called heterogeneous. For example, ARMLton is homogeneous, because $\operatorname{gr}($ ARMLton $)=2$ and exactly two chefs make pie, but ARMLville is heterogeneous, because even though each dish is only cooked by two chefs, $\operatorname{gr}($ ARMLville $)=3$.


A resident cycle is a sequence of distinct residents $r_{1}, \ldots, r_{n}$ such that for each $1 \leq i \leq n-1$, the residents $r_{i}$ and $r_{i+1}$ know how to make a common dish, residents $r_{n}$ and $r_{1}$ know how to make a common dish, and no other pair of residents $r_{i}$ and $r_{j}, 1 \leq i, j \leq n$ know how to make a common dish. Two resident cycles are indistinguishable if they contain the same residents (in any order), and distinguishable otherwise. For example, if $r_{1}, r_{2}, r_{3}, r_{4}$ is a resident cycle, then $r_{2}, r_{1}, r_{4}, r_{3}$ and $r_{3}, r_{2}, r_{1}, r_{4}$ are indistinguishable resident cycles.
Let $T$ be a town such that, for each $D \in \operatorname{dish}(T),\left|\operatorname{chef}_{T}(D)\right|=3$.

Prove that if $T$ contains a resident cycle such that for every dish $D \in \operatorname{dish}(T)$, there exists a chef in the cycle that can prepare $D$, then $\operatorname{gr}(T)=3$.

Solution:
First we prove that for any town $T$ and any $D \in \operatorname{dish}(T), \operatorname{gr}(T) \geq\left|\operatorname{chef}_{T}(D)\right|$.

Let $D \in \operatorname{dish}(T)$. Suppose that $f$ is a valid group assignment on $T$. Then for $r, s \in \operatorname{chef}_{T}(D)$, if $r \neq s$, it follows that $f(r) \neq f(s)$. Hence there must be at least $\left|\operatorname{chef}_{T}(D)\right|$ distinct groups in the range of $f$, i.e., $\operatorname{gr}(T) \geq\left|\operatorname{chef}_{T}(D)\right|$.

By the conclusion above, it must be the case that $\operatorname{gr}(T) \geq 3$. Let $C=\left\{r_{1}, r_{2}, \ldots, r_{n}\right\}$ denote a resident cycle such that for every dish $D \in \operatorname{dish}(T)$, there exists a chef in $C$ that can prepare $D$. Each resident is a chef for two dishes, and every dish can be made by two residents in $C$ (although by three in $T$ ). Thus the number of residents in the resident cycle $C$ is equal to $|\operatorname{dish}(T)|$, which was proved to be even in the previous part.

Define a group assignment by setting

$$
f(r)= \begin{cases}1 & \text { if } r \notin C \\ 2 & \text { if } r=r_{i} \text { and } i \text { is even } \\ 3 & \text { if } r=r_{i} \text { and } i \text { is odd. }\end{cases}
$$

For any $D \in \operatorname{dish}(T)$, there are exactly three $D$-chefs, and exactly two of them belong to the resident cycle $C$. Hence exactly one of the $D$-chefs $r$ will have $f(r)=1$. The remaining two $D$-chefs will be $r_{i}$ and $r_{i+1}$ for some $i$, or $r_{1}$ and $r_{n}$. In either case, the group assignment $f$ will assign one of them to 2 and the other to 3 . Thus any two residents who make a common dish will be assigned different groups by $f$, so $f$ is a valid group assignment, proving that $\operatorname{gr}(T)=3$.