A look at generalized perfect shuffles

When handed an unshuffled deck of cards, most of us respond in the same way. We attempt (however feebly) to do a so-called perfect shuffle. A perfect shuffle splits a deck of cards into two equal stacks and then perfectly interlaces the cards from the two stacks one after the other. Only experienced gamblers and magicians can perform perfect shuffles reliably, and yet the mathematics behind perfect shuffles has a rich history, including everything from mathematical card tricks to sophisticated research.

There are two ways that the interlacing of a perfect shuffle can be done. The out shuffle (denoted $O$) is the instance in which the original top card remains at the top, while the in shuffle (denoted $I$) is the instance in which the original top card becomes the second card, lying underneath the top card of the second stack. See Figure 1 for an example. Perfect shuffles are permutations of the set of $2n$ cards, so the group generated by the in shuffle and out shuffle $\langle I,O\rangle$ is a subgroup of the symmetric group $S_{2n}$. In 1983, Persi Diaconis, Ron Graham, and William Kantor determined the group $\langle I,O\rangle$ for a deck of $2n$ cards for all $n$ [4]. (The development of this paper by Diaconis et al. was nicely described in narrative form in Science [8].)

One might expect that the order of these shuffle groups $\langle I,O\rangle$ would grow with the size of the deck, but this is not the case. Compare, for example, decks with 30 cards and 32 cards, respectively. With 30 cards, the group generated by in and out shuffles has order $15!\cdot 2^{14}$, which is over 21 quadrillion. And yet, with 2 more cards (32 cards total), the shuffle group has order $160$. (As we will see later, when the deck size is a power of 2, the shuffle group is always relatively small.)

Several colorful variations on perfect shuffles have since been studied, including flip shuffles, horseshoe shuffles, Monge shuffles, and milk shuffles, to name a few [2, 3, 4, 9]. At the conclusion of their paper, Diaconis et al. mentioned another natural variation – the so-called generalized perfect shuffles. We describe these shuffles here.

Suppose one has a deck of $mn$ cards. A generalized perfect shuffle (also called an $m$-shuffle) is performed first by dividing the cards into $m$ equal stacks (the first stack contains the first $n$ cards, the second stack contains the next $n$ cards, and so on). Place these $m$ stacks side by side in order from left to right. A priori, there are $m!$ different patterns which one can use to interlace the cards from these $m$ stacks to reassemble the deck. We consider two interlacing patterns which directly generalize the standard in and out perfect shuffles:

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  Out $m$-shuffle (denoted $O_{m}$). Pick up the cards from the $m$ stacks from left to right. In this case, the original top card will remain the top card when the shuffle is complete.

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  In $m$-shuffle (denoted $I_{m}$). Pick up the cards from the $m$ stacks from right to left. In this case, the original top card will now be the $m^{th}$ card when the shuffle is complete.

See Figure 2 for examples of 3-shuffles. With this perspective, the standard perfect shuffles studied by Diaconis et al. are a special case, being 2-shuffles.

A variety of results have already been discovered about generalized perfect shuffles (or, $m$-shuffles, as we will call them). The group generated by an $m$-shuffle and the simple cut was investigated by Morris and Hartwig [12]. They found the group in most cases to be either the full symmetric group of the deck or the alternating group of the deck. Working even more generally, others have considered allowing all $m!$ interlacing patterns of the $m$ stacks rather than restricting to the in and out $m$-shuffles. The groups generated in this broader context have been determined for several infinite families [1, 10].

Here in this work, we return to the original question posed by Diaconis et al. [4] and which was posed again in [10]. Namely, we seek to find the group generated by in and out $m$-shuffles $\langle I_{m},O_{m}\rangle$. The solution to the analogous question for the group $\langle I,O\rangle$ of standard perfect shuffles was a mathematical and computational tour de force requiring many cases. We expect the situation with generalized shuffles to be similarly complex. Here in this paper, we restrict to decks of size $m^{k}$ for some $m,k>1$. With this card deck, we consider $m^{y}$-shuffles for some $y<k$.

A natural starting place is to consider 2-shuffles on decks of size $2^{k}$. This special case fits into the setting of standard perfect shuffles and was studied by Diaconis et al. They proved the following.

Broadening this result, we find the structure of the $m^{y}$-shuffle group on a deck of $m^{k}$ cards, for all possible values of $m,k$, and $y$. The result nicely generalizes the previous special case.

Certainly a few examples will clarify this result, and so we discuss examples throughout. Note that the order and the structure of these shuffle groups in Theorem 2 are independent of the value of $m$.

We begin by reviewing some of the beautiful mathematics underlying standard perfect shuffles (see [4] for the proofs). We only fit in a sampling of results here; further interesting patterns (and mathematical card tricks to impress your friends) are plentiful and provide ample opportunity for reading and investigation [5, 6, 11, 13, 14].

Suppose we have a deck of $2n$ cards. Index each card by its distance from the top of the deck from $0$ to $2n-1$. Simple formulas give the location of each card after performing a standard perfect shuffle. The out shuffle fixes the cards with index $0$ and $2n-1$. And, for any card with index $i<2n-1$, the card’s position after an out shuffle $O$ is as follows:

It follows, then, that the order of the out shuffle is given by the order of $2$ in $\mathbb{Z}_{2n-1}$. An in shuffle $I$ moves the card with index $i$ as follows:

From this, one can show that the order of the in shuffle is the order of $2$ in $\mathbb{Z}_{2n+1}$.

For example, consider a deck of 52 cards. One needs to do only 8 out shuffles to bring the deck back to its original order, since 2 has order 8 in $\mathbb{Z}_{51}$. On the other hand, one must do 52 in shuffles before the deck returns to its original order, as 2 has order 52 in $\mathbb{Z}_{53}$.

Perfect shuffles preserve several types of symmetry in a deck of cards. Consider a pair of cards which are each located the same distance from the center of the deck. After an in or out shuffle, these two cards will both move to positions which are again the same distance from the center of the deck. Because of this, we say that in and out shuffles preserve central symmetry.

The set of all permutations in $S_{2n}$ that preserve central symmetry form a subgroup of $S_{2n}$, which we denote by $B_{n}$. Observe that $B_{n}$ has order $n!~{}\!2^{n}$, and so it follows that the shuffle group $\langle I,O\rangle$ on $2n$ cards has order at most $n!~{}\!2^{n}$. In fact, the group $\langle I,O\rangle$ is isomorphic to $B_{n}$ if and only if $n\equiv 2\pmod{4}$ and $n>6$. A deck of 52 cards $(n=26$) is one example of this. Save for a few exceptional cases, the remaining shuffle groups $\langle I,O\rangle$ can be described as a kernel (or an intersection of kernels) of different homomorphisms from $B_{n}$ to $\mathbb{Z}_{2}$ (see [4] for details). The significant exceptional case is when the number of cards is a power of 2 (say, $2^{k}$). In this case, the shuffle group has order $2^{k}k$ and has structure $\mathbb{Z}_{2}^{k}\rtimes\mathbb{Z}_{k}$, as stated in Theorem 1.

Let’s consider a specific example to illustrate Theorem 1. Begin with just 4 cards. A priori, we might be able to reach any of $24$ different card arrangements by shuffling these 4 cards. But Theorem 1 tells us that we only reach 8 card arrangements with standard perfect shuffles, and the shuffle group is isomorphic to $(\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{2}$. (This group is more commonly recognized as the dihedral group of a square, $D_{4}$.)

This group’s Cayley graph nicely illustrates the relationships among the card arrangements and shuffles. Let us consider each of these 8 possible card arrangements to be vertices on a graph. If one card arrangement can be obtained from another by an in shuffle or out shuffle, we draw a directed edge between the two associated vertices. In this example, the resulting graph is a cube. See Figure 3.

Now we expand our perspective to consider a deck of $mn$ cards, together with in and out $m$-shuffles (denoted $I_{m}$ and $O_{m}$, respectively). We begin by studying the basic properties of these shuffles. The following lemma (also proved in [10]) provides the critical information of where cards end up after an $m$-shuffle.

With these basic facts in place, we turn our focus to the special case where the size of the deck of cards is a power of $m$, say, $m^{k}$. As before, we index each card in the deck by its distance from the top of the deck from $0$ to $m^{k}-1$. The index $i$ on each card can be naturally expressed in base-$m$ as a $k$-tuple of digits: $(x_{1},x_{2},\ldots,x_{k})$, where

In Lemma 3, we discussed the effect of in and out $m$-shuffles on a card’s index, but now we describe $m$-shuffles’ effects on the base-$m$ index of a card.

At last, we are ready to determine the structure of the group of $m^{y}$-shuffles on a deck of $m^{k}$ cards for all possible $m$, $k$, and $y$. Alongside these discoveries, we give concrete examples of these groups. The tools used here are all standard group theory techniques.

We proceed in three cases, each of which is proved separately: Case (1) $y$ is odd and relatively prime to $k$ (Theorem 6), Case (2) $y$ is even and relatively prime to $k$ (Theorem 7), and Case (3) $y$ and $k$ are not relatively prime (Corollary 8). Together, these three results prove Theorem 2 stated in the introduction.

Lest we lose the forest for the trees, let’s look at an example that illustrates the theorem we have just proved. Suppose we have $m^{2}$ cards, for some natural number $m>1$. Theorem 6 tells us that we are only able to reach a total of 8 card orderings with $m$-shuffles on these $m^{2}$ cards. The shuffle group is $(\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{2}$. (This group is more commonly known as the dihedral group of the square, $D_{4}$.) The associated Cayley graph illustrates the relationship among the shuffles. In this case, the graph is a cube, as shown in Figure 4. Observe that the group is no different from the special case of 2-shuffles on a deck of 4 cards which we discussed earlier. Indeed, the group structure is independent of the value of $m$.

We now consider the case where $y$ is even and relatively prime to $k$. The theorem and proof are reminiscent of the previous case when $y$ is odd, with a few necessary twists.

As an example of the above theorem, suppose that we have $m^{3}$ cards for some natural number $m>1$. If we do in and out $m^{2}$-shuffles on the deck, we are only able to reach 12 card orderings regardless of the value of $m$. By Theorem 7, the shuffle group is $(\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes_{\phi}\mathbb{Z}_{3}$, where $\phi(1)(a_{1},a_{2})=(a_{2},a_{1}+a_{2})$. This group is isomorphic to the alternating group $A_{4}$. The associated Cayley graph takes the shape of the cuboctahedron as in Figure 5.

We conclude by considering the case where $k$ and $y$ are not relatively prime. This case follows as a corollary of the previous two results and, together with Theorems 6 and 7, completes the proof of Theorem 2.

A natural next step in this investigation, of course, is to understand the situation for deck sizes which are not equal to $m^{k}$. We wrote a computer program to determine the order of $m$-shuffle groups for small deck sizes. See Figure 6 for this information.

Another natural question returns to the idea of symmetry and shuffle groups. As mentioned previously, the standard perfect shuffle group $\langle I,O\rangle$ for a deck of $2n$ cards cannot be larger than $2^{n}n!$, since all elements in the group preserve central symmetry. Similarly, the $m$-shuffle groups $\langle I_{m},O_{m}\rangle$ on a deck of $2n$ (or $2n+1$) cards also preserve central symmetry and hence can be no larger than $2^{n}n!$, as well [10]. In their work on standard shuffles, Diaconis et al. described precisely the values of $n$ for which the standard shuffle group on $2n$ cards reaches this maximum value [4]. We ask whether such a result can be generalized? For example, consider a deck of $6$ cards. The maximum order possible is $48$, which is achieved with the 3-shuffle group. With a deck of 12 cards, however, the maximum order is 46,080, which is not realized by any $m$-shuffle group. What is the pattern?

Acknowledgements: We thank Kent Morrison for his insightful comments on an early draft of this paper.