Algorithm to create "Spot It"/"Dobble" cards, but with two common images from any two cards In the ordinary "Spot It"/"Dobble" game, there is exactly one common image between any two cards. At the moment I am trying to modify the algorithms already available on the internet so that any two cards have two, three, etc. elements in common.
Unfortunately, I have failed and am now looking for ideas on how such an algorithm could look in concrete terms. Do you have any ideas for such an algorithm?

 A: One can generate such a deck with a symmetric block design, which consists of

*

*a set $\mathcal S$ (of symbols)

*a collection $\mathcal C$ of subsets of $\mathcal S$ called blocks---in our case, cards---such that there are as many cards as there are symbols, and each $2$-subset of symbols occurs on equally many, say, $\lambda$, cards.

This concept is a special case of the concept of combinatorial design.
If we denote by

*

*$v$ the number of symbols, equivalently the number of cards, and

*$k$ the number of cards in which any given symbol appears---which for symmetric designs is the same as the number of symbols on each card---

then $$\lambda (v - 1) = k (k - 1) ,$$
and we call a design with given parameters a $(v, k, \lambda)$-design. The order of a symmetric design is $k - \lambda$.
It turns out that every two distinct cards have $\lambda$ symbols in common, so $(v, k, 1)$-designs (with $k > 2$) correspond to projective planes and therefore to the Spot-It like games outlined in MJD's blog post mentioned in the comments, and the question is asking about the construction of $(v, k, \lambda)$-designs with $\lambda > 1$.
There are lots of constructions for producing these:
Example 1 (faces of $n$-simplices) One easy construction is taking a set of $n$ symbols and making $n$ cards, each of which omits one of the symbols, so that each pair of cards intersects in $n - 2$ symbols, giving a $(n, n - 1, n - 2)$ design, which thus has order $1$. (We can think of such a design as an $n$-simplex, the cards as the codimension-$1$ faces of the simplex, and the sets of symbols common to two given cards as the codimension-$2$ face at which the corresponding faces meet.)
Example 2 (complementary designs) Given any $(v, k, \lambda)$-design, we can produce a $(v, v - k, \lambda + v - 2 k)$-design by replacing the set of symbols on each card with the complement of that set. The new design is called the complement of the old one and the two designs have the same orders.
Example 3 (Hadamard designs) For any Hadamard matrix of size $(4 m) \times (4 m)$, we can construct a $(4 m - 1, 2 m - 1, m - 1)$-design (called a Hadamard $2$-design), which thus has order $m$: By rearranging rows and columns of the Hadamard matrix, we can arranging for the first row and first column to have all entries $+1$. Delete that row and column, that is, consider the lower-right $(4 m - 1) \times (4 m - 1)$ minor $\Lambda$. Then, for each $(i, j)$ the card $C_j$ includes the $i$th symbol if and only if $\Lambda_{ij} = -1$.
Example 4 (Grassmannians of hyperplanes over finite fields) For any finite field $\Bbb F_q$ and any dimension $d \geq 2$, let the symbols and cards be the lines and hyperplanes, respectively, in $\Bbb F_q^{d + 1}$, and declare that a symbol $\ell$ is on a card $H$ if and only if $\ell \subset H$. This defines a $\left(\frac{q^{d + 1} - 1}{q - 1}, \frac{q^d - 1}{q - 1}, \frac{q^{d - 1} - 1}{q - 1}\right)$-design, which thus has order $q^{d - 1}$. The special case $d = 2$ yields projective planes.
A design with $\lambda = 2$---that is, a deck so that any two distinct cards have exactly two symbols is common---is called a biplane. A biplane of order $m$ corresponds to a deck with $\frac{1}{2} (m^2 + 3 m + 4)$ symbols and as many cards.

*

*The only biplane of order $0$ corresponds to a trivial deck of $2$ cards, each of which have the same $2$ symbols:
$$01, \qquad 01$$


*The only biplane of order $1$ is given by taking $n = 4$ in Example 1:
$$012, \qquad 013, \qquad 023, \qquad 123$$


*The only biplane of order $2$ is the complement (Example 2) of the Fano plane, the smallest of the projective planes, i.e., the projective plane $\Bbb P(\Bbb F_2^3)$ corresponding to the $2$-element field $\Bbb F_2$:
$$0136, \qquad 0145, \qquad 0235, \qquad 0246, \qquad 1234, \qquad 1256, \qquad 3456 .$$


*The only biplane of order $3$ arises from the unique Hadamard design (Example 3) with $m = 3$. Since $11$ is prime, we can apply the Paley construction to produce a $12 \times 12$ Hadamard matrix and hence an $11$-card deck:
\begin{multline}
01247, \quad
0136\textrm{A}, \quad
0259\textrm{A}, \quad
03789, \quad
04568, \quad
12358, \\
1489\textrm{A}, \quad
15679, \quad
23469, \quad
2678\textrm{A}, \quad
3457\textrm{A} .
\end{multline}
This configuration is called the Paley biplane, and it corresponds to the exceptional inclusion $PSL(2, 5) \hookrightarrow PSL(2, 11)$.


*There are several distinct biplanes of order $4$, including one arising from the classic Kummer configuration, which corresponds to a deck of $16$ cards and as many symbols.


*The Bruck-Ryser-Chowla Theorem implies that there are no biplanes of order $5$ or $6$.
I haven't seen the term before, but one might as call a design with $\lambda = 3$ a triplane; a triplane of order $m$ corresponds to a deck with $\frac{1}{3} (m^2 + 5 m + 9)$ symbols and as many cards, so a triplane cannot have order equivalent to $2 \pmod 3$.

*

*There is a triplane of order $0$ analogous to the trivial biplane.

*Taking $n = 5$ in Example $1$ gives a triplane of order $1$.

*The dual of the Paley biplane is a triplane of order $3$.

*Taking $q = 2, d = 3$ in Example 4 gives a triplane of order $4$, and it corresponds to a deck of $15$ cards and as many symbols.

A: For anyone who's interested: I used the advice posted in this comment:
https://stackoverflow.com/questions/65780880/algorithm-to-create-spot-it-dobble-cards-but-with-two-common-images-from-an#comment116311020_65780880
So let's assume I take two decks of cards with completely different images.
A1     B1
A2     B2
A3     B3
...    ...
An     Bn

Now, all those cards get shuffled
A23     B50
A4      B12
A18     B17
...     ....

and combined into one large card.
A23 + B50
A4  + B12
A18 + B17
...   ....

This will result in two images matching between each pair of cards (one of set A and one of set B)
