A correspondent writes:

A few times, over the past few weeks, I've seen a particular friend of mine, and he's had a game with him, a game called Spot It. It takes the form of a bunch of cards, each of which has eight different symbols on it; the game mechanic depends on the fact that each pair of cards has exactly one symbol in common.

[...]

It turns out there are 57 distinct symbols. 42 of them appear on eight cards each, 14 appear on seven cards each, and one appears on six. This looks suspiciously like "57 symbols, 57 cards, each card has eight symbols, each symbol appears on eight cards" with two cards missing.

[...]

It's trivial to deduce what the 'missing' cards are, and, with them added, it is then a 57-card deck with the same property; that any two cards have exactly one symbol in common.

Then I noticed that 57 equals 57. That is, that the number of cards in the 'complete' deck equals the number of distinct symbols. Then, I drew up a 57x57 matrix, where one axis is card identity and the other is symbol identity. It looks symmetric enough, and, on checking, it turns out that the "one bit in common" property holds in the other dimension too (which means, mapped back into cards, that, for every pair of symbols, there is exactly one card that has both of them).

Then a few questions (paraphrased):

1. Why $57$; is there anything special about this number?
2. If we choose a different number of symbols per card, can we always construct a card game like this? How many cards would there be?
3. Are there always the same number of cards and symbols in a complete game? If a game has fewer cards than symbols, is it always possible to complete it?
4. Is it possible to add more cards even if none are "missing"?

I added a question of my own.

1. Is the incidence matrix actually symmetric (for some permutation of rows and columns)? If so, do all such card games have symmetric incidence matrices?

I found question E interesting because it implies that there is a bijection between cards and symbols with nice properties;* in particular it relates these card games to certain regular graphs of diameter $2$. More on this later.

Here's an example of what we're talking about. The triangle game has cards $a,b,c$ and symbols $\alpha,\beta,\gamma$:

 $a$ $b$ $c$ $\beta,\gamma$ $\alpha,\gamma$ $\alpha,\beta$

There are $3$ cards and $3$ symbols, each card has $2$ symbols, each symbol appears on $2$ cards, each pair of cards has one symbol in common, and each pair of symbols appears together on one card. There is a pleasing duality between cards and symbols, and we can write down the dual of this game:

 $\alpha$ $\beta$ $\gamma$ $b,c$ $a,c$ $a,b$

In the dual game, the card $\alpha$ has symbols $b,c$ because those were the cards with the symbol $\alpha$ in the original game. Notice that in this case, the dual is essentially the same game: there's a bijection $$\varphi = \{~a \mapsto \alpha,~b \mapsto \beta,~c \mapsto \gamma~\}$$ such that a card $x$ has a symbol $\varphi(y)$ if and only if the card $\varphi(x)$ has the symbol $y$ in the dual game. This relates to the fact that the incidence matrix is symmetric:

 $\begin{matrix} \alpha&\beta&\gamma \end{matrix}$ $\begin{matrix} a \\ b \\ c \end{matrix}$ $\begin{pmatrix} 0&1&1 \\ 1&0&1 \\ 1&1&0 \end{pmatrix}$

It's usually a good idea to come up with some notation.

Definition. An incidence structure is $(C,S,\in)$ where $C$ and $S$ are disjoint finite non-empty sets, and $\in$ is a relation with domain $C$ and range $S$.**

Interpret $C$ as the set of cards, $S$ as the set of symbols, and $x \in \sigma$ as "card $x$ has symbol $\sigma$". The bit about domains and ranges means there are no "blank cards" or "phantom symbols" - every card has a symbol, and every symbol appears on some card.

Definition. For $x : C$, $S_x = \{~\sigma : S~|~x \in \sigma~\}$, and for $\sigma : S$, $C_{\sigma} = \{~x : C~|~x \in \sigma~\}$.

$S_x$ is the set of symbols on a card $x$, and $C_\sigma$ is the set of cards with symbol $\sigma$.

Definition. A Card & Symbol Game (or CSG) is an incidence structure such that

$\newcommand\prop{[\textsf{#1}]}\prop{UNIFORM}$ For some $n \ge 1$, every card has $n+1$ symbols (i.e. $\exists n \ge 1 \bullet \forall x : C \bullet \#S_x = n+1$)

$\prop{1-COMMON}$ Any two distinct cards have a unique common symbol (i.e. $\forall x,y : C~|~x \ne y \bullet \#(S_x \cap S_y) = 1$)

Call this $\prop{CSG} \equiv \prop{UNIFORM} \land \prop{1-COMMON}$.

The use of $n+1$ rather than $n$ is because mathematicians like to go back and change their notation in order to make things easier later, once they know what will make things easier.

Some more definitions will be useful for our investigation:

Definition. An incidence structure is square if there are the same number of cards and symbols (i.e. $\#C = \#S$). Call this $\prop{SQUARE}$.

Definition. An incidence structure is regular if the number of symbols on each card equals the number of cards with each symbol, and this number is at least $2$. Call this $\prop{REGULAR}$.

Definition. Incidence structures $\mathcal{G} = (C,S,\in)$ and $\mathcal{G}' = (C',S',\in')$ are isomorphic (or $\mathcal{G} \cong \mathcal{G}'$) if there are bijections $\varphi : C \to C'$ and $\psi : S \to S'$ such that $x \in \sigma$ if and only if $\varphi(x) \mathrel{\in'} \psi(\sigma)$.

Obviously if $\mathcal{G}$ satisfies a property $\prop{P}$ and $\mathcal{G} \cong \mathcal{G}'$, then $\mathcal{G}'$ also satisfies $\prop{P}$.

We'd like to talk about duals:

Definition. The dual of an incidence structure $\mathcal{G} = (C,S,\in)$ is $\mathcal{G}^\star = (S,C,\ni)$, where $\sigma \ni x$ if and only if $x \in \sigma$.

Definition. $\mathcal{G}$ is self-dual if $\mathcal{G} \cong \mathcal{G}^\star$. Call this $\prop{SELF-DUAL}$.

Obviously $\mathcal{G}^{\star\star} = \mathcal{G}$. It will be useful to have a notation for dual properties:

Meta-definition. For a property $\prop{P}$, an incidence structure $\mathcal{G}$ satisfies $\newcommand\propd{[\textsf{#1}^\star]}\propd{P}$ if and only if $\mathcal{G}^\star$ satisfies $\prop{P}$. For example,

$\propd{UNIFORM}$ For some $n \ge 1$, every symbol appears on $n+1$ cards

$\propd{1-COMMON}$ Any two distinct symbols appear together on a unique card

Obviously $[\mathsf{P}^{\star\star}] \equiv \prop{P}$. Due to the symmetry between cards and symbols in their definitions, $\prop{SQUARE}$, $\prop{REGULAR}$ and $\prop{SELF-DUAL}$ are each equivalent to their dual properties.

We'll be interested in finding more logical connections between these properties; let's write $\newcommand\infer{\vDash} \prop{P} \infer \prop{Q}$ to mean that $\prop{Q}$ can be inferred from $\prop{P}$.

Proposition. $\prop{REGULAR} \equiv \prop{SQUARE} \land \prop{UNIFORM} \land \propd{UNIFORM}$.

Proof. Obviously $\prop{REGULAR} \equiv \prop{UNIFORM} \land \propd{UNIFORM}$ with the same $n$. Consider $\in$ as a set of pairs $x \mapsto \sigma$. Then $\#{\in} = \#C \times \#S_x$ for each $x : C$, and also $\#{\in} = \#S \times \#C_\sigma$ for each $\sigma : S$. Since $\#C_\sigma = \#S_x$, we have $\#C = \#S$.

Conversely, suppose $\#C = \#S$, then $\#C \times \#S_x = \#{\in} = \#S \times \#C_\sigma$ for each $x:C$, $\sigma:S$, so $\#C_\sigma = \#S_x$. $\square$

So if $\mathcal{G}$ is square, uniform and dual-uniform, then the values of $n$ are the same. This is not necessarily the case if $\mathcal{G}$ is not square: for example, the incidence structure $\mathcal{B}_3$ defined by the incidence matrix $$\begin{pmatrix} 0&0&0&~&1&1&1 \\ 0&0&1&~&1&1&0 \\ 0&1&0&~&1&0&1 \\ 0&1&1&~&1&0&0 \\ 1&0&0&~&0&1&1 \\ 1&0&1&~&0&1&0 \\ 1&1&0&~&0&0&1 \\ 1&1&1&~&0&0&0 \end{pmatrix}$$ is uniform with each $\#S_x = 3$ and dual-uniform with each $\#C_\sigma = 4$. (Exercise: generalise this construction to $\mathcal{B}_n$ with $2^n$ cards, $2n$ symbols, each card having $n$ symbols and each symbol appearing on $2^{n-1}$ cards.) This isn't a CSG, as it doesn't satisfy $\prop{1-COMMON}$; another construction gives $\mathcal{C}_4$ with incidence matrix $$\begin{pmatrix} 1&1&1&0&0&0 \\ 1&0&0&1&1&0 \\ 0&1&0&1&0&1 \\ 0&0&1&0&1&1 \end{pmatrix}$$ which generalises to $\mathcal{C}_n$ with $n$ cards, $\frac{1}{2}n(n-1)$ symbols, each card having $n-1$ symbols and each symbol appearing on $2$ cards. (The details of this construction are left as another exercise.) This is a CSG satisfying $\propd{UNIFORM}$ but not $\propd{1-COMMON}$.

Some second-order theorems:

Meta-theorem. $(\prop{P} \infer \prop{Q}) \infer (\propd{P} \infer \propd{Q})$.

Proof. Suppose there is an incidence structure $\mathcal{G}$ satisfying $\propd{P} \land \neg \propd{Q}$. Then $\mathcal{G}^\star$ satisfies $\prop{P} \land \neg \prop{Q}$, a contradiction. $\square$

Meta-theorem. $\prop{SELF-DUAL} \land \prop{P} \infer \propd{P}$.

Proof. If $\mathcal{G}$ satisfies $\prop{P}$ then $\mathcal{G}^\star \cong \mathcal{G}$ satisfies $\propd{P}$. $\square$

These will be useful later.

We turn our attention now to projective geometry. The reason for this will be immediately clear:

Definition. A finite projective plane (or FPP) is an incidence structure $(P,L,\in)$ where $P$ is a set of points, $L$ is a set of lines, $p \in \lambda$ means "point $p$ is on line $\lambda$", and

$\prop{1-COMMON}$ Any two distinct points are on a unique common line

$\propd{1-COMMON}$ Any two distinct lines meet at a unique common point

$\prop{NON-DEGENERATE}$ There are four distinct points such that no three are on any line

Call this $\prop{FPP} \equiv \prop{1-COMMON} \land \propd{1-COMMON} \land \prop{NON-DEGENERATE}$.

Definition. $P_\lambda$ is the set of points on a line $\lambda$, and $L_p$ is the set of lines through a point $p$, as before.

The words "point" and "line" are deliberately undefined - we can interpret them in a geometric sense (if we're generous about how straight a line has to be): This is the Fano plane, the smallest FPP. But we could instead define "point" to mean card, and "line" to mean symbol, and then we can draw the Fano plane as a card game diagram:

 $a$ $b$ $c$ $d$ $e$ $f$ $g$ $\alpha,\beta,\gamma$ $\alpha,\delta,\epsilon$ $\alpha,\zeta,\eta$ $\beta,\delta,\eta$ $\beta,\epsilon,\zeta$ $\gamma,\epsilon,\eta$ $\gamma,\delta,\zeta$

It's easy to check that the Fano plane is a CSG, and isomorphic to its dual (which is also an FPP),

 $\alpha$ $\beta$ $\gamma$ $\delta$ $\epsilon$ $\zeta$ $\eta$ $a,b,c$ $a,d,e$ $a,f,g$ $b,d,g$ $b,e,f$ $c,e,g$ $c,d,f$

via the isomorphism $\varphi = \{~a \mapsto \alpha,~b \mapsto \beta,~\ldots,~g \mapsto \eta~\}$ and $\psi = \varphi^{-1}$.

In fact, every FPP is a CSG, and its dual is also an FPP. Let's prove this.

Theorem. The dual of an FPP is an FPP (i.e. $\prop{FPP} \infer \propd{FPP}$).

Proof. We only need to check $\propd{NON-DEGENERATE}$.

Choose $p,q,r,s : P$ as in $\prop{NON-DEGENERATE}$. By $\prop{1-COMMON}$, there are lines $\lambda,\mu,\nu,\xi : L$ joining $p$ to $q$, $q$ to $r$, $r$ to $s$ and $s$ to $p$ respectively. Suppose, without loss of generality, that $\lambda,\mu,\nu$ meet at a single point: by $\propd{1-COMMON}$ the only point on both $\lambda$ and $\mu$ is $q$, and so $\nu$ contains all three of $q,r,s$, a contradiction. $\square$

By a meta-theorem, it follows that $(\prop{FPP} \infer \prop{P}) \infer (\prop{FPP} \infer \propd{P})$. I.e., if every FPP satisfies some property $\prop{P}$, then every FPP also satisfies $\propd{P}$.

Lemma. Every line in an FPP contains at least $3$ points.

Proof. Choose $p,q,r,s : P$ as in $\prop{NON-DEGENERATE}$, and without loss of generality let $p \notin \lambda$ and $q \notin \lambda$. By $\prop{1-COMMON}$ there are $\mu,\nu,\xi : L_p$ with $q \in \mu$, $r \in \nu$ and $s \in \xi$, which are distinct by $\prop{NON-DEGENERATE}$.

Then by $\propd{1-COMMON}$ there are $t,u,v : P_\lambda$ with $t \in \mu$, $u \in \nu$ and $v \in \xi$ which are distinct as each pair of $\mu,\nu,\xi$ already meet at $p$. $\square$

By the dual lemma, every point lies on at least $3$ lines.

Lemma. For any two points in an FPP, there is a line containing neither.

Proof. By $\prop{1-COMMON}$ choose $\mu : L$ with $p \in \mu$ and $q \in \mu$. By the lemma, there is a point $r \in P_\mu - p - q$. By $\prop{NON-DEGENERATE}$ there is a point $s : P - r$ with $s \notin \mu$, so choose $\lambda : L$ with $r \in \lambda$ and $s \in \lambda$. By $\propd{1-COMMON}$, $\lambda$ and $\mu$ only meet at $r$, so $p \notin \lambda$ and $q \notin \lambda$. $\square$

By the dual lemma, for any two lines, there is a point on neither.

Theorem. Every FPP is a CSG (i.e. $\prop{FPP} \infer \prop{CSG}$).

Proof. We have $\prop{1-COMMON}$ already, so we only need to check $\prop{UNIFORM}$: that every point is on the same number of lines.

We will show that given any point $p : P$ and line $\lambda : L$ with $p \notin \lambda$, the number of lines through $p$ equals the number of points on $\lambda$. Given $\mu : L_p$, $\mu \ne \lambda$ so by $\propd{1-COMMON}$ there is a unique point $q : P_\lambda$ with $q \in \mu$. Conversely, given $q : P_\lambda$, by $\prop{1-COMMON}$ there is a unique $\mu : L_p$ with $q \in \mu$. This gives a bijection $: L_p \to P_\lambda$.

Then given distinct points $p,q : P$, by the lemma there is a line $\lambda : L$ with $p \notin \lambda$ and $q \notin \lambda$, so $\#L_p = \#P_\lambda = \#L_q$. $\square$

Notice that the proof also gives us $\prop{FPP} \infer \prop{REGULAR}$, as $\#L_p = \#P_\lambda$.

The problem right now is that the dual of a CSG is not necessarily a CSG; $\mathcal{C}_n^\star$ satisfies $\prop{UNIFORM}$ but not $\prop{1-COMMON}$, for example. Worse still, call this one $\mathcal{L}_5$:

 $a$ $b$ $c$ $d$ $e$ $\alpha,\Omega$ $\beta,\Omega$ $\gamma,\Omega$ $\delta,\Omega$ $\epsilon,\Omega$

This CSG is uninteresting because the common symbol is always $\Omega$. Nonetheless, it's a CSG by our definition, and the dual game

 $\Omega$ $\alpha$ $\beta$ $\gamma$ $\delta$ $\epsilon$ $a,b,c,d,e$ $a$ $b$ $c$ $d$ $e$

satisfies neither $\prop{UNIFORM}$ nor $\prop{1-COMMON}$. Of course we could change our definition of CSGs to include $\propd{UNIFORM}$ and $\propd{1-COMMON}$, so that the dual of a CSG is always a CSG. But this is unsatisfying and may be too strict: it would disallow "incomplete" CSGs, as deleting cards from a CSG preserves $\prop{UNIFORM}$ and $\prop{1-COMMON}$ but neither $\propd{UNIFORM}$ nor $\propd{1-COMMON}$.

A better approach would be to try to isolate what it means for a CSG to have "missing" cards, and then relate this to "completeness". We'll consider a CSG to be "complete" if it satisfies $\propd{UNIFORM}$ and $\propd{1-COMMON}$, i.e. every symbol appears on the same number of cards, and every pair of symbols appears together, as in the completed original game. Conveniently, this is exactly $\propd{CSG}$.

Definition. A CSG is complete if its dual is a CSG. Call this $\prop{COMPLETE} \equiv \prop{CSG} \land \propd{CSG}$.

Definition. An extension of an incidence structure $\mathcal{G} = (C,S,\in)$ is an incidence structure $\mathcal{G}' = (C',S',\in')$ for which $C \subseteq C'$, $S \subseteq S'$, and $\in$ is the restriction of $\in'$ to $C$ and $S$ (i.e. ${\in} \subseteq {\in'}$ as sets of pairs). The extension is non-trivial if $\mathcal{G}' \ne \mathcal{G}$.

An expansion is an extension such that $\in$ is the restriction of $\in'$ to $C$. Write $\mathcal{G} \le \mathcal{G}'$ for an expansion.

A completion is an expansion which is a complete CSG.

Proposition. If $\mathcal{G} \le \mathcal{G}'$ is a non-trivial expansion, then $C \ne C'$.

Proof. Suppose $C = C'$. Then $\in$ is the restriction of $\in'$ to its own domain, so ${\in} = {\in'}$ and hence $S' =$ range of ${\in'} =$ range of ${\in} = S$, and the expansion is trivial; a contradiction. $\square$

Definition. An incidence structure is CSG-expansible if it has a non-trivial expansion which is a CSG. Call this $\prop{CSG-EXPANSIBLE}$.

An incidence structure is completable if it has a completion. Call this $\prop{COMPLETABLE}$.

So extensions and expansions may add more cards and more symbols, but an extension might add new symbols to existing cards, whereas an expansion cannot change any existing cards. An expansion is trivial if and only if it adds no cards. A completion need not be non-trivial, so $\prop{COMPLETE} \infer \prop{COMPLETABLE}$.

Now let's try to answer some of the questions.

Theorem. $\prop{1-COMMON} \land \prop{REGULAR} \infer \#C = n^2 + n + 1$.

Proof. Fix a card $x : C$. Given any other card $y : C - x$, by $\prop{1-COMMON}$ there is a unique $\sigma : S_x$ for which $y$ is in $C_{\sigma} - x$. Therefore $\{x\}$ and the $C_{\sigma} - x$ form a partition of $C$.

By $\prop{REGULAR}$, $\# S_x = n+1$, and each $\# (C_{\sigma} - x) = n$. Therefore $\#C = 1 + n(n+1)$. $\square$

This answers question A: $57 = 7^2 + 7 + 1$.

Theorem. An incidence structure is a complete CSG if and only if it is either an FPP or a triangle.

Proof. Since complete CSGs satisfy $\prop{1-COMMON}$ and $\propd{1-COMMON}$, we only need to consider complete CSGs which don't satisfy $\prop{NON-DEGENERATE}$. These are degenerate finite projective planes, all of which are classified:

• The empty set. This is not an incidence structure by our definition.
• A single line, with any number of points. This fails $\prop{UNIFORM}$, as we need $n \ge 1$.
• As before, but one of the points is on any number of additional lines. This still fails $\prop{UNIFORM}$.
• A single point, with any number of lines. This fails $\propd{UNIFORM}$, as we need $n \ge 1$.
• As before, but one of the lines contains any number of additional points. This still fails $\propd{UNIFORM}$.
• A line $\mu$ with $C_{\mu} = \{p_1,\ldots,p_{n+1}\}$, and an additional point $q$ with $S_q = \{\lambda_1,\ldots,\lambda_{n+1}\}$, where each $C_{\lambda_i} = \{p_i, q\}$. This satisfies $\prop{UNIFORM}$ only when $n=1$, giving the triangle.

Conversely, $\prop{FPP} \infer \prop{CSG}$ and by a meta-theorem $\prop{FPP} \infer \propd{FPP} \infer \propd{CSG}$, so every FPP is a complete CSG, as is the triangle. $\square$

This answers the first part of question B: all FPPs with $n \le 10$ are known, and there is no FPP with $n=6$, so there is no complete CSG with $7$ symbols per card, for example. $n$ is called the order of the FPP.

Corollary. Every complete CSG is regular.

Proof. $\prop{FPP} \infer \prop{REGULAR}$, and the triangle is regular. $\square$

This answers the second part of question B: if there is a complete CSG with $n+1$ symbols per card, then it has $n^2 + n + 1$ cards.

Corollary. Every complete CSG is square.***

Proof. $\prop{REGULAR} \infer \prop{SQUARE}$. $\square$

This answers question C: yes, a complete CSG must have equal numbers of cards and symbols; and no, there are CSGs which are not completable. For example, a CSG with $1$ card and $7$ symbols is not completable, though it has "missing cards" in the sense that it is CSG-expansible; however, it has a CSG-expansion which is not CSG-expansible.**** In contrast, $\mathcal{L}_n$ is not completable for $n \ge 3$, but $\mathcal{L}_n \le \mathcal{G}$ is a non-trivial CSG-expansion exactly when $\mathcal{G} \cong \mathcal{L}_{n'}$ for some $n' > n$; so however many cards are added, with whatever new symbols, more can still be added!

Some things are known about completability: for example, Dow showed in 1982 that, in other words, any CSG with $\#S = n^2 + n + 1$ and $\#C > n^2 - 2\sqrt{n+3} + 6$ is completable, and if $\#C > n^2 - n + 1$ then the completion is unique (up to isomorphism).

Theorem. $\prop{COMPLETE} \infer \neg \prop{CSG-EXPANSIBLE}$.

Proof. Let $\mathcal{G} \le \mathcal{G}'$ be a non-trivial expansion, and fix $x : C$ and $y : C' \setminus C$. Let $\chi : S$ be the common symbol between $x$ and $y$. There is another symbol $\sigma : S_x - \chi$, and since $G$ satisfies $\propd{UNIFORM}$ there is another card $w : C_\sigma - x$.

Suppose $G'$ satisfies $\prop{1-COMMON}$. Let $\tau : S$ be the unique common symbol between $w$ and $y$. Since $\mathcal{G}$ satisfies $\propd{1-COMMON}$ there is a card $z : C_\chi - x$ with $z \in \tau$. Then $y$ and $z$ have two common symbols $\chi$ and $\tau$, a contradiction. Therefore $\mathcal{G}'$ is not a CSG. $\square$

This answers question D: no, if the CSG is complete then we can't add more cards and still have a CSG.

Question E asks if an incidence matrix is equal to its own transpose. This is apparently a stronger condition than $\prop{SELF-DUAL}$, which only requires the incidence matrix to be some permutation of its transpose.

Definition. An incidence structure is self-polar if there is a bijection $\varphi : C \to S$ such that $x \in \varphi(y)$ if and only $y \in \varphi(x)$. Call this $\prop{SELF-POLAR}$.

Proposition. $\prop{SELF-POLAR} \infer \prop{SELF-DUAL}$.

Proof. $\psi = \varphi^{-1}$ will do. $\square$

Proposition. Every self-dual CSG is complete.

Proof. By meta-theorem, $\prop{SELF-DUAL} \land \prop{CSG} \infer \propd{CSG}$. $\square$

FPPs can be systematically constructed using vector spaces over finite fields, yielding FPPs where $n$ can be any prime power. (It's an open problem whether there are any FPPs where $n$ is not a prime power.) FPPs constructed this way are always self-polar; FPPs constructed by other methods are not necessarily self-dual, and apparently it's another open problem as of 2009 whether every self-dual FPP is self-polar.

To answer question E: the particular card game introduced above is a complete CSG with $8$ symbols per card, so it is isomorphic to the unique FPP with $n=7$, which is self-polar. However, there are complete CSGs which are not self-dual, the smallest two of which are isomorphic to a dual pair of FPPs with $n=9$, so they have $91$ cards each with $10$ symbols.

For general incidence structures, $\prop{SELF-POLAR}$ is strictly stronger than $\prop{SELF-DUAL}$. In 2003, Brouwer et al. found the smallest possible self-dual incidence structure which is not self-polar:

 $a$ $b$ $c$ $d$ $e$ $f$ $g$ $\alpha,\beta,\gamma,\epsilon$ $\alpha,\beta,\delta,\zeta$ $\alpha,\zeta,\eta$ $\beta,\epsilon,\eta$ $\alpha,\gamma$ $\beta,\delta$ $\gamma,\delta$

There are $7$ others of the same size, up to isomorphism.

Returning to the start of this post, our correspondent also noted that (in other words) there is a one-to-one correspondence between CSGs and constant-weight codes for which each codeword has Hamming weight $n+1$ and each pair has Hamming distance $2n$. We've shown that complete CSGs are almost exactly the same thing as finite projective planes; another closely related area is the study of symmetric block designs, which (for $\lambda = 1$) are exactly the same thing as complete CSGs, although this result is a theorem rather than by definition.

I said we'd come back to regular graphs with diameter $2$. If the incidence matrix of a CSG is symmetric (i.e. the CSG is self-polar), then we can interpret it as the adjacency matrix of a graph in which $x,y:C$ are joined by an edge if and only if $x \in \varphi(y)$. In this context,

• $\prop{UNIFORM}$ means the graph is $n+1$-regular, i.e. each vertex has $n+1$ edges;
• $\prop{1-COMMON}$ means each pair of distinct vertices $x,y:C$ are joined by a unique route of length $2$;
• $\prop{SELF-POLAR}$ means the graph is undirected.

For $x \in \varphi(x)$, the graph has an edge from $x$ to itself, and self-edges may be used to form routes.***** This gives a correspondence between self-polar CSGs and graphs of this form. Here's the graph corresponding to the Fano plane: However, the graph produced depends on which polarity $\varphi$ is used. Automorphisms of the graph correspond to conjugates of $\varphi$, but in general the polarities may not form a single conjugacy class in the group of automorphisms of the FPP.

Another thing you can do with CSGs is colour the edges of complete graphs such that the monochrome sets are themselves complete graphs. The colour of an edge joining two cards is given by their unique common symbol. Here's the Fano plane again, with the symbols $\alpha,\beta,\ldots,\eta$ in order of hue: Here each monochrome set is complete on $3$ vertices, because the Fano plane is dual-uniform; and each pair of colours meets at one vertex, because the Fano plane is dual-1-common. So that's nice.

*Combinatorics would greatly benefit from a theorem which says "if two sets appear near each other, have the same cardinality, and this cardinality is an unusually mundane number like $57$, then there is a natural bijection with some nice properties which explains why this has happened."

**Combinatorists may recognise that the usual definition of incidence structure does not require that everything has some incidence. The relation symbol is $\in$ because we can use a set-theoretic model where $S \subseteq \mathbb{P} C$. Astute combinatorists may recognise that the usual definition of a uniform incidence structure is the dual of the definition given here.

***Ryser's theorem gives the converse: every square CSG is complete!

****This follows because Ryser's theorem gives a strict bound $\#C < n^2 + n + 1$, so it cannot be expanded indefinitely.

*****The friendship theorem implies that the only such graph with no self-edges is the triangle. This is a corollary of the fact that every polarity $\varphi$ of an FPP has an absolute point $x:P$ such that $x \in \varphi(x)$.