Now that you have read the instructions to play EvenQuads, we will show you some of the mathematics (courtesy of Dr. Lauren Rose and Jeffrey Pereira) hidden behind this game.

We know that an EvenQuads deck consists of 64 cards. We can check this by just counting the cards in a deck. But, without counting the cards, why 64? The cards have three types of attributes: the symbol featured in the card, the color of the symbol, and the number of symbols. Notice there are three attributes per card and each attribute has four states. Hence, the number of cards must be \(4 \times 4 \times 4 = 64\).

Suppose we started with the following three cards

How many different cards can complete this to a Quad? There are three different colors, so the fourth card must be yellow. There are two spirals and one double square, so using the two-and-two rule, the fourth card must be a double square. These three cards each have one symbol, so the fourth card must also have exactly one symbol. Hence, the only card that completes these three cards to a Quad must be:

In fact, this property holds no matter which three cards we start with! There is always a unique fourth card that will complete a Quad.

**Why?** Let’s look at just one attribute:

**If all three cards have the same state**, the fourth card must also be in that state.

**If all three cards have different states,** the fourth card must have a state different from the other three.

**If the three cards do not have all the same state or all different states**, then there must be one state that appears exactly twice and another state that appears exactly once. The only way to satisfy the Quad rules is to have a two-and-two Quad, so the state appearing exactly once must also be on the fourth card.

In all cases, the fourth state of the attribute is determined.

**Repeating this argument for all three attributes,** there is a unique way to complete the given three cards to a Quad. Q.E.D

In order to analyze EvenQuads algebraically, we will represent each attribute using arithmetic modulo \(2\). That is, our numbers will only be \(0\) and \(1\), and the way we will add them up will be:

\(0+0=0\) \(0+1=1\) \(1+0=1\) \(1+1=0\)We will represent each of the four states of the three attributes of EvenQuads as binary strings (corresponding to elements of \(\mathbb{Z}_2 \times \mathbb{Z}_2\)).

This is all summarized in the following table:

Number | Symbol | Color | Representation |

1 | Spiral | Green | \(01\) |

2 | Polyhedron | Blue | \(10\) |

3 | Square | Pink | \(11\) |

4 | Circle | Yellow | \(00\) |

Because each card has three attributes, we can represent a card as a vector (number, symbol, color) =\((n, l, c) \in (\mathbb{Z}_2 \times \mathbb{Z}_2)^3\) ,

where the values for \(n\), \(l\) and \(c\) are obtained from the table above.

We get that, for example, \((00,10,11)\) represents the card featuring four Pink Polyhedra.

**Theorem**: Cards A, B, C and D form a Quad if and only if \(A+B+C+D = (00, 00, 00)\) using componentwise binary addition.

**Example :** The Quad {(1, Polyhedron, Blue), (1, Polyhedron, Pink), (1, Circle, Green), (1, Circle,Yellow)}

can be represented as a subset of \((\mathbb{Z}_2 \times \mathbb{Z}_2)^3\) as follows

\({(01, 10, 10), (01, 10, 11), (01, 00, 01), (01, 00, 00)}\).

We see \((01, 10, 10) + (01, 10, 11) + (01, 00, 01) + (01, 00, 00) = (00,00,00)\) which indicates that the theorem is satisfied.

Similarly, the Quad

is also \({(01, 01, 01), (00, 11, 11), (11, 11, 00), (10, 01, 10)}\); the sum of these vectors also yields \((00,00,00)\), as expected.

**Proof of Theorem. **

(\(\Rightarrow\)) Assume that \(A, B, C, D\) form a Quad. For each of the three attributes, we must show that the sum of the states on the cards is \(00\). There are three cases for the states of a given feature \(\mathcal{F}\) in the four cards.

Case 1: All four states are the same. In this case, our sum is of the form \(a+a+a+a\), which will be equal to \(00\), as the operation is addition modulo \(2\).

Case 2: All states are different. In this case the sum is \(00+10+01+11=00\) .

Case 3: There are two states, each on two cards. Suppose the states are \(a\) and \(b\). Then their sum is \(a+a+b+b\), which is also \(00\), as all computations are performed modulo \(2\).

Since the states of each attribute add up to \(00\), we have that \(A+B+C+D= (00,00,00)\).

(\(\Leftarrow\)) Assume that \(A+B+C+D= (00,00,00)\).

It follows that the sum of the states of each feature add up to \(00\).

Suppose that \(A, B, C, D\) do not form a Quad. Let the states of an attribute \(\mathcal{F}\) be \(a, b, c, d\). If the states are not all the same, all different or half and half, then the only possibilities are that states occur asymmetrically, as either \(aaab\) or \(aabc\). In either case, since the operations are performed modulo \(2\), we would get that \(a+a=00\), and thus we will end up with either \(a+a+a+b=a+b\) or \(a+a+b+c=b+c\). However, in \(\mathbb{Z}_2 \times \mathbb{Z}_2\), every element is its own (additive) inverse. Hence, both \(a+b=00\) and \(b+c=00\) yield contradictions. It follows that only the possibilities discussed in the (\(\Rightarrow\)) direction of this proof yield Quads. Q.E.D

Publications that reference EvenQuads and offer additional mathematical ideas:

*Toward a Combinatorial Theory of SET and Related Card Games*, Jonathan Schneider, https://arxiv.org/abs/2010.01882*How many cards should you lay out in a game of EvenQuads?: A detailed study of 2-caps in AG(n,2)-*

Julia Crager, Felicia Flores, Timothy E. Goldberg, Lauren L. Rose, Daniel Rose-Levine, Darrion Thornburgh, Raphael Walker https://arxiv.org/abs/2212.05353