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.
For each odd positive integer $d \geq 3$, prove that $\operatorname{gr}\left(\mathcal{F}_{d}\right)=d$.

Solution:
Fix $D \in \operatorname{dish}\left(\mathcal{F}_{d}\right)$. Then for every other dish $D^{\prime} \in \operatorname{dish}\left(\mathcal{F}_{d}\right)$, there is exactly one chef who makes both $D$ and $D^{\prime}$, hence $\left|\operatorname{chef}_{\mathcal{F}_{d}}(D)\right|=d-1$, which is even because $d$ is odd. Thus for each $D \in \operatorname{dish}\left(\mathcal{F}_{d}\right)$, $\left|\operatorname{chef}_{\mathcal{F}_{d}}(D)\right|$ is even. Because $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$ is odd and $\left|\operatorname{chef}_{\mathcal{F}_{d}}(D)\right|=d-1$ for every dish in $\mathcal{F}_{d}$, problem 12c applies, hence $\operatorname{gr}\left(\mathcal{F}_{d}\right)>d-1$.

Label the dishes $D_{1}, D_{2}, \ldots, D_{d}$, and label the residents $r_{i, j}$ for $1 \leq i<j \leq d$ so that $r_{i, j}$ is a $D_{i}$-chef and a $D_{j}$-chef. Define $f: \mathcal{F}_{d} \rightarrow\{0,1, \ldots, d-1\}$ by letting $f\left(r_{i, j}\right) \equiv i+j \bmod d$.

Suppose that $f\left(r_{i, j}\right)=f\left(r_{k, \ell}\right)$, so $i+j \equiv k+\ell \bmod d$. Then $r_{i, j}$ and $r_{k, \ell}$ are assigned to the same group, which is a problem if they are different residents but are chefs for the same dish. This overlap occurs if and only if one of $i$ and $j$ is equal to one of $k$ and $\ell$. If $i=k$, then $j \equiv \ell \bmod d$. As $j$ and $\ell$ are both between 1 and $d$, the only way they could be congruent modulo $d$ is if they were in fact equal. That is, $r_{i, j}$ is the same resident as $r_{k, \ell}$. The other three cases $(i=\ell, j=k$, and $j=\ell$ ) are analogous. Thus $f$ is a valid group assignment, proving that $\operatorname{gr}\left(\mathcal{F}_{d}\right) \leq d$. Therefore $\operatorname{gr}\left(\mathcal{F}_{d}\right)=d$.