A look at generalized perfect shuffles

Cornelia A. Van Cott; Lakshman Manny; Qiyu Zhang; Samuel Johnson

A look at generalized perfect shuffles

T0 review · 0 major / 5 minor · reviewed 2026-08-27 · deepseek-v4-flash

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Pith's one-line read For an $m^k$-card deck, the group generated by in and out $m^y$-shuffles is a semidirect product whose order and isomorphism type depend only on $k/\gcd(y,k)$ and the parity of $y/\gcd(y,k)$, not on $m$.

desk verdict A clean, correct classification of generalized shuffle groups for power-of-m decks; the reader's main worry is already covered by Lemma 3. read the letter →

arxiv 2009.09349 v2 pith:HJ6WQUQM submitted 2020-09-20 math.CO math.GR

Samuel Johnson , Lakshman Manny , Cornelia A. Van Cott , QiYu Zhang This is my paper · ORCID

classification math.COmath.GR MSC 05A0520B35

keywords generalizedperfectshufflesm-shufflesinshuffleoutgroupsemidirectproductDiaconis-Graham-Kantorcardshuffling

verification ladder T0 review T1 audit T2 compute T3 formal

The pith

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

The reading

The paper answers, for decks whose size is a perfect power, a question Diaconis, Graham, and Kantor left open in 1983: what permutation group is generated by the two generalized perfect shuffles. For a deck of $m^k$ cards, with in and out $m^y$-shuffles for $0

What carries the argument

The load-bearing device is the base-$m$ digit representation of card positions. Writing a card's index as a $k$-tuple $(x_1,\dots,x_k)$ in base $m$, the out shuffle $O_m$ is a left cyclic rotation and the in shuffle $I_m$ is the same rotation with the first digit flipped to $(m-1)-x_1$. The proof builds conjugates $B_j=O_m^{j-1}I_mO_m^{-j}$ that flip exactly one digit, showing the flips generate $(\mathbb{Z}_2)^k$, while $O_m$ rotates them, giving the semidirect product. In the even case, analogous conjugates $C_j$ flip adjacent pairs, and the relation $C_1\cdots C_k=\mathrm{id}$ reduces the flip group to $(\mathbb{Z}_2)^{k-1}$ with the twisted action described in Theorem 7.

What would settle it

On a 16-card deck, explicitly write the in 4-shuffle and out 4-shuffle (the case $m=2$, $k=4$, $y=2$) and compute the order of the group they generate; Theorem 2 predicts a group of order $8$, isomorphic to the dihedral group $D_4$, so any other order would refute the claim.

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Extended reading notes

Core claim

On its own terms, the paper's central result (Theorem 2) is that for a deck of $m^k$ cards, with $m,k>1$ and $0<y<k$, letting $c=\gcd(y,k)$, the group $\langle I_{m^y},O_{m^y}\rangle$ is isomorphic to $(\mathbb{Z}_2)^{k/c}\rtimes \mathbb{Z}_{k/c}$ with a cyclic-shift action when $y/c$ is odd, and to $(\mathbb{Z}_2)^{k/c-1}\rtimes \mathbb{Z}_{k/c}$ with the specific action $\varphi(1)(a_1,\dots,a_{k/c-1})=(a_{k/c-1},a_1+a_{k/c-1},\dots,a_{k/c-2}+a_{k/c-1})$ when $y/c$ is even. The order is therefore $2^{k/c}(k/c)$ in the odd case and $2^{k/c-1}(k/c)$ in the even case. The authors prove the coprime cases $y$ odd and $y$ even separately, then reduce the non-coprime cases by re-expressing card indices in base $m^c$, so that an $m^y$-shuffle becomes an $m^c$-shuffle repeated $y/c$ times.

Load-bearing premise

The proof assumes without proof or citation that an $m^y$-shuffle on $m^k$ cards is exactly the same permutation as performing the basic $m$-shuffle $y$ times, so if that equality ever fails, the stated group structures would not follow.

Editorial extensions

If this is right

The order of $\langle I_{m^y},O_{m^y}\rangle$ is $2^{k/c}(k/c)$ when $y/c$ is odd and $2^{k/c-1}(k/c)$ when $y/c$ is even, so the group is minuscule compared with the full symmetric group on $m^k$ cards.
Taking $m=2$ and $y=1$ recovers the known Diaconis–Graham–Kantor result for power-of-two decks, now as one case of a uniform theorem.
Every non-coprime $(k,y)$ case reduces to a coprime case by reblocking in base $m^c$, so only coprime pairs need separate analysis.
The Cayley graph of the $m^2$-deck shuffle group is a cube, and that of the $m^3$-deck group with $m^2$-shuffles is a cuboctahedron, giving geometric models of the same abstract groups.
The paper's computed orders for small non-power decks, such as 12 cards with $m=3$ giving order 60, provide a base of data for the still-open general-deck problem.

Reading between the lines

Editorial extensions of the paper, not claims the author makes directly.

Because the proof only uses base-$m$ arithmetic, the same digit-rotation model should determine shuffle groups for decks of size $q^r$ with any integer base $q$, so the group may depend on the exponent structure of the deck size rather than on the particular $q$; one could test this by enumerating groups for decks of size $4^3$, $6^3$, or $9^3$.
The unproved identity $O_{m^y}=(O_m)^y$ could be verified or falsified by a short computer search for small composite $m$; if it holds there, the theorem's scope is confirmed, and if not, the failure isolates exactly which cases need a separate argument.
The smallness of these groups suggests that card tricks built on generalized shuffles can guarantee reachable states independent of the number of stacks, an implication the paper does not discuss.

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Formalized claims in Lean

Claim #1: On its own terms, the paper's central result (Theorem 2) is that for a deck of $m^k$ cards, with $m,k>1$ and $0<y<k$, letting $c=\gcd(y,k)$, the group $\langle I_{m^y},O_{m^y}\rangle$ is isomorphic to $(\mathbb{Z}_2)^{k/c}\rtimes \mathbb{Z}_{k/c}$ with a cyclic-shift action when $y/c$ is odd, and to $(\mathbb{Z}_2)^{k/c-1}\rtimes \mathbb{Z}_{k/c}$ with the specific action $\varphi(1)(a_1,\dots,a_{

/-- @claim 1 On its own terms, the paper's central result (Theorem 2) is that for a deck of $m^k$ cards, with $m,k>1$ and $0<y<k$, letting $c=\gcd(y,k)$, the group $\langle I_{m^y},O_{m^y}\rangle$ is isomorphic to $(\mathbb{Z}_2)^{k/c}\rtimes \mathbb{Z}_{k/c}$ with a cyclic-shift action when $y/c$ is odd, and to $(\mathbb{Z}_2)^{k/c-1}\rtimes \mathbb{Z}_{k/c}$ with the specific action $\varphi(1)(a_1,\dots,a_{ -/
def central_claim : Prop :=

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Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, and a circularity audit.

Desk Editor's Note

read the letter

The paper answers the Diaconis–Graham–Kantor question for decks of size m^k, and it does so cleanly. The main theorem – that ⟨I_{m^y}, O_{m^y}⟩ is determined entirely by k/gcd(y,k) and the parity of y/gcd(y,k), independent of m – is genuinely new and is proved with simple, honest group theory. The base-m digit picture in Lemma 4 turns the shuffles into rotations and flips, and the semidirect product decompositions follow without any hidden machinery. I checked the two cases and the corollary; the arguments hold. The paper also does a nice job with examples and figures, and the small-deck table in Section 5 is a reasonable sanity check. This deserves to be published, probably with only light revision.

Your main worry – that the equivalence O_{m^y} = (O_m)^y and I_{m^y} = (I_m)^y is asserted without proof – is addressed in Lemma 3. Its proof explicitly derives the iterated formulas: O_m^r sends i to m^r i mod m^k−1, and I_m^l sends i to m^l i + m^l−1 mod m^k+1. Substituting y gives exactly the modular action of an m^y-shuffle on m^k cards. So the “observe” in Section 3 is justified by what came before. I would not make the paper conditional on this point.

The only real soft spot is in Corollary 8. The proof says the equations for m^y-shuffles in base m^c are “identical” to the equations for m^{y/c}-shuffles on m^{k/c} cards. That is not quite literal – the flip uses m^c−1 instead of m−1. But the group generated by cyclic shifts and flips is independent of the alphabet size, so the isomorphism conclusion is sound. The wording is imprecise, not the math. Worth a footnote, not a rewrite.

The paper is elementary and self-contained; it does not lean on heavy imports. The citation pattern looks fine: [4] is the special case, [10] is credited for Lemma 3, and [12] is a different generator set. No red flags.

Who should read this: anyone working on shuffle groups, card tricks with generalized shuffles, or semidirect products built from digit rotations. It is a solid subfield contribution, not a breakthrough, and it does not pretend otherwise. If I were refereeing, I would accept after a minor revision polishing the Corollary 8 wording. Yes, send it to peer review; the desk reject is not appropriate here.

Referee Report

0 major / 5 minor

Summary. The paper studies generalized perfect shuffles (in and out m-shuffles) on decks of m^k cards, and more generally m^y-shuffles with 0<y<k. The main theorem (Theorem 2) asserts that the group ⟨I_{m^y},O_{m^y}⟩ is isomorphic to Z_2^{k/c} ⋊ Z_{k/c} when y/c is odd and to Z_2^{k/c-1} ⋊ Z_{k/c} with the stated action when y/c is even, where c=gcd(y,k). The proof proceeds by expressing indices in base m (Lemma 4), so that O_m is a cyclic shift and I_m is a cyclic shift plus a flip; constructing flip-only generators B_j (resp. C_j) in the odd (resp. even) coprime case; showing these generate a normal elementary abelian 2-subgroup intersecting ⟨O_m⟩ trivially; and then handling the non-coprime case by reblocking in base m^c (Corollary 8).

Significance. This is a clean, self-contained contribution. If the main theorem is correct, it answers the Diaconis-Graham-Kantor question for the infinite family of deck sizes m^k and completely determines the group in terms of k/gcd(y,k) and parity, independent of m. The explicit generators B_j and C_j and the elementary digit-action argument are strengths. I also verified that the key equivalence O_{m^y}=(O_m)^y and I_{m^y}=(I_m)^y, which is asserted in Section 3, follows directly from the iteration formulas derived in the proof of Lemma 3, so I do not regard this as a gap. The paper's limitations are presentational: a few group-theoretic claims are stated without proof and there are minor typos.

minor comments (5)

[Section 3, before Eqs. (5)-(6)] The assertion that an m^y-shuffle is equivalent to y applications of an m-shuffle is made without proof or citation; please add a sentence pointing to the iterated formulas established in the proof of Lemma 3, since Theorems 6, 7, and Corollary 8 rely on it.
[Section 4, Corollary 8 proof] The statement that the base-m^c formulas are 'identical' to Equations (5)-(6) is imprecise: the flip operation is x_i ↦ (m^c-1)-x_i rather than x_i ↦ (m-1)-x_i. The structural conclusion is unaffected because the flips remain involutions, but the wording should be corrected.
[Section 4, Theorem 6 proof] The claim ⟨B_1,...,B_k⟩∩⟨O_m⟩={id} is asserted without argument; it follows because any nontrivial product of the B_j moves the all-zero k-tuple to a nonzero tuple, whereas every power of O_m fixes it. Please include this justification.
[Section 4, Theorem 7 proof] The same trivial-intersection claim for ⟨C_1,...,C_{k-1}⟩ and ⟨O_m⟩ should be justified explicitly; the all-zero tuple argument works here as well.
[Throughout] There are several typographical errors: 'acts on on' in Theorem 2, 'for of' in the proof of Corollary 8, and Figure 4's caption writes x_i=m-x_i where it should be x_i=(m-1)-x_i. Also, gcd[y,k] should be written gcd(y,k).

Circularity Check

0 steps flagged · score 0.0 of 10

No circularity: Theorem 2 is derived from explicit permutation formulas and group actions, not assumed or fitted.

full rationale

The derivation chain is self-contained. Lemma 3 derives the modular formulas for O_m and I_m directly from the array definition of m-shuffles, and its proof also gives the iterated forms (multiplication by m^r and affine maps m^l i + m^l - 1). Section 3's claim that an m^y-shuffle equals y iterations of an m-shuffle is therefore supported by Lemma 3, and it is not an input that secretly contains the target group structure. Lemma 4 converts these formulas into explicit base-m digit shifts and flips, and Theorems 6 and 7 construct the groups by exhibiting commuting involutions B_j and C_j and computing the conjugation action of O_m. The semidirect product decompositions follow from these explicit actions, with no fitted parameters and no appeal to the theorem being proved. The citations to Diaconis-Graham-Kantor and Medvedoff-Morrison are contextual or for results independently reproved in the paper; no load-bearing self-citation chain exists. Corollary 8's reblocking in base m^c has a minor wording imprecision about the size of the alphabet in the flip operation, but the abstract group generated by cyclic shifts and flips is alphabet-independent, so the structural conclusion does not circularly depend on that wording. Overall, no step reduces by construction to its own inputs.

Assumptions & free parameters 0 free parameters · 3 assumptions · 0 invented entities

No fitted parameters and no invented entities. The only substantive unproved premise is the iterated-shuffle equivalence; the remaining ingredients are standard group theory and modular arithmetic.

assumptions (3)

domain assumption An m^y-shuffle on m^k cards is identical to applying the basic m-shuffle y times.
Asserted without proof in Section 3 before Equations (5)-(6); used in Theorems 6, 7 and Corollary 8.
standard math If K=HG with H≅Z_2^d normal, G cyclic of order n, H∩G=1, and G conjugates H as stated, then K≅Z_2^d ⋊ Z_n.
Standard semidirect product recognition, used implicitly in Theorems 6 and 7.
standard math Base-m digit arithmetic and congruences modulo m^k±1 correctly model card positions.
Used throughout Lemma 4 and Corollary 8; a standard fact about positional notation.

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Pith. "Pith review of A look at generalized perfect shuffles." pith.science (2026). https://pith.science/paper/HJ6WQUQM

@misc{pith2026200909349,
  author       = {Pith},
  title        = {Pith review of: A look at generalized perfect shuffles},
  year         = {2026},
  howpublished = {\url{https://pith.science/paper/HJ6WQUQM}},
  note         = {Machine review of arXiv:2009.09349}
}

abstract

Standard perfect shuffles involve splitting a deck of $2n$ cards into two stacks and interlacing the cards from the stacks. There are two ways that this interlacing can be done, commonly referred to as an in shuffle and an out shuffle, respectively. In 1983, Diaconis, Graham, and Kantor determined the permutation group generated by in and out shuffles on a deck of $2n$ cards for all $n$. Diaconis et al. concluded their work by asking whether similar results can be found for so-called generalized perfect shuffles. For these new shuffles, we split a deck of $mn$ cards into $m$ stacks and similarly interlace the cards with an in $m$-shuffle or out $m$-shuffle (denoted $I_m$ and $O_m$, respectively). In this paper, we find the structure of the group generated by these two shuffles for a deck of $m^k$ cards, together with $m^y$-shuffles, for all possible values of $m$, $k$, and $y$. The group structure is completely determined by $k/\gcd(y,k)$ and the parity of $y/\gcd(y,k)$. In particular, the group structure is independent of the value of $m$.

Figures

Figures reproduced from arXiv: 2009.09349 by the authors.

**Figure 1.** Starting with a deck of 12 cards, we split the deck and perform (a) an out shuffle and (b) an in shuffle. m stacks to reassemble the deck. We consider two interlacing patterns which directly generalize the standard in and out perfect shuffles: • Out m-shuffle (denoted Om). 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… view at source ↗

**Figure 2.** Starting with a deck of 12 cards, we divide the deck into 3 stacks and perform generalized 3-shuffles: (a) an out 3-shuffle and (b) an in 3-shuffle. 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 S2n that preserve central symmetry form a subgroup of S2n, which we deno… view at source ↗

**Figure 3.** Four cards (numbered in original order as 0, 1, 2, 3) can be shuffled with out shuffles (red) and in shuffles (blue) to reach eight different possible orderings total. 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 [PITH_FULL_IMAGE:figures/full_fig_p005_3.png] view at source ↗

Figures from the paper (3 more)

**Figure 4.** Figure 4: A deck of m2 cards can be shuffled with out m-shuffles (red) and in mshuffles (blue) to reach eight different possible orderings total. Before shuffling, each card’s position can be expressed in base m as (x1, x2), and after any m-shuffle, each card’s position can be …

**Figure 5.** Figure 5: A deck of m3 cards can be shuffled with out m2 -shuffles (red) and in m2 - shuffles (blue) to reach twelve different possible orderings total. Before shuffling, each card’s position can be expressed in base m as (x1, x2, x3), and after any m2 -shuffle, each card’s posi…

**Figure 6.** Figure 6: The orders of m-shuffle groups hIm, Omi for small deck sizes. Another natural question returns to the idea of symmetry and shuffle groups. As mentioned previously, the standard perfect shuffle group hI, Oi for a deck of 2n cards cannot be larger than 2nn!, since all e…

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Reference graph

Works this paper leans on

14 extracted references · 14 canonical work pages

[4]

Persi Diaconis, R. L. Graham, and William M. Kantor. The mathematics of perfect shuﬄes. Adv. in Appl. Math. , 4(2):175–196, 1983

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S. Brent Morris and Robert E. Hartwig. The generalized faro shuﬄe. Discrete Math., 15(4):333–346, 1976

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Pith tools

Reviewed August 27, 2026 · model on record in the stance chip above.

[1] [4]

Persi Diaconis, R. L. Graham, and William M. Kantor. The mathematics of perfect shuﬄes. Adv. in Appl. Math. , 4(2):175–196, 1983

work page 1983

[2] [10]

Groups of perfect shuﬄes

Steve Medvedoﬀ and Kent Morrison. Groups of perfect shuﬄes. Math. Mag., 60(1):3–14, 1987

work page 1987

[3] [12]

Brent Morris and Robert E

S. Brent Morris and Robert E. Hartwig. The generalized faro shuﬄe. Discrete Math., 15(4):333–346, 1976

work page 1976

[4] [1]

Carmen Amarra, Luke Morgan, and Cheryl E. Praeger. Generalised shuﬄe groups , 2019. https://arxiv.org/abs/ 1908.05128

work page Pith review arXiv 2019

[5] [2]

Trailing the dovetail shuﬄe to its lair

Dave Bayer and Persi Diaconis. Trailing the dovetail shuﬄe to its lair. Ann. Appl. Probab., 2(2):294–313, 1992

work page 1992

[6] [3]

The mathematics of the ﬂip and horseshoe shuﬄes

Steve Butler, Persi Diaconis, and Ron Graham. The mathematics of the ﬂip and horseshoe shuﬄes. Amer. Math. Monthly, 123(6):542–556, 2016

work page 2016

[7] [5]

The solutions to Elmsley’s problem

Persi Diaconis and Ron Graham. The solutions to Elmsley’s problem. Math Horizons, 14(3):22–27, 2007

work page 2007

[8] [6]

Unshuﬄing for the imperfect magician

Doug Ensley. Unshuﬄing for the imperfect magician. Math Horizons, 11(3):13–16, 2004

work page 2004