ADVANCES IN APPLIED MATHEMATICS
ARTICLE NO.
21, 205]227 Ž1998.
AM980597
The Cycling of Partitions and Compositions under
Repeated Shifts
Jerrold R. Griggs*, † and Chih-Chang Ho†
Department of Mathematics, Uni¨ ersity of South Carolina, Columbia,
South Carolina 29208
Received March 28, 1998; accepted May 5, 1998
In ‘‘Bulgarian Solitaire,’’ a player divides a deck of n cards into piles. Each move
consists of taking a card from each pile to form a single new pile. One is concerned
only with how many piles there are of each size. Starting from any division into
piles, one always reaches some cycle of partitions of n. Brandt proved that for
n s 1 q 2 q ??? qk, the cycle is just the single partition into piles of distinct sizes
1, 2, . . . , k. Let D B Ž n. denote the maximum number of moves required to reach a
cycle. Igusa and Etienne proved that D B Ž n. F k 2 y k whenever n F 1 q 2
q ??? qk, and equality holds when n s 1 q 2 q ??? qk. We present a simple new
derivation of these facts. We improve the bound to D B Ž n. F k 2 y 2 k y 1, whenever n - 1 q 2 q ??? qk with k G 4. We present a lower bound for D B Ž n. that is
likely to be the actual value. We introduce a new version of the game, Carolina
Solitaire, in which the piles are kept in order, so we work with compositions rather
than partitions. Many analogous results can be obtained. Q 1998 Academic Press
1. INTRODUCTION
In the early 1980s, an article by Martin Gardner brought widespread
attention to a card game that had attracted the curiosity of some European mathematicians. Called Bulgarian Solitaire, the game works as follows: First divide a finite deck of n cards into piles. A move consists of
removing one card from each pile and forming a new pile. The operation is
repeated over and over. We pay attention only to how many piles there are
of each positive size, ignoring the locations of the piles.
Thus, Bulgarian Solitaire can be viewed as a way of changing one
partition into another. For instance, the partition into k parts of distinct
* Research supported in part by NSA Grant MSP MDA904-95H1024.
†
Research supported in part by NSF Grant DMS-97-01211.
205
0196-8858r98 $25.00
Copyright Q 1998 by Academic Press
All rights of reproduction in any form reserved.
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GRIGGS AND HO
sizes from 1 to k is preserved under the operation. Clearly, for any n, any
start eventually leads to a cycle of partitions, since there are only finitely
many partitions of n altogether. What is striking in playing the game is
that, starting from any partition of a deck of size n s 1 q 2 q ??? qk
cards, one always arrives eventually at this stable partition into sizes 1
through k. This effect is all the more dramatic in that it seems to take
quite a while in some cases with only moderately large k, say k s 5.
We do not know yet the reason for the appellation ‘‘Bulgarian.’’ However, we were introduced to an ordered variation of the game by a
Bulgarian visitor, Andrey Andreev. In his game, which we shall call
Carolina Solitaire, one also maintains an ordering of the nonempty piles.
Say we begin with n cards divided into a row of piles of sizes c1 , . . . , c r ) 0,
Ý i c i s n. One card is removed from each pile, and these r cards are then
placed in a pile ahead of the others. Any exhausted pile Žsize 0. is ignored;
only nonempty piles are considered. For a triangular number n, say
n s 1 q 2 q ??? qk, this game also appears to arrive at a stable division,
with piles of sizes k, k y 1, . . . , 1.
After proving this fact for Carolina Solitaire, as well as deriving bounds
on how soon the cycling begins for arbitrary n, we asked experts for
references to related work, and our search finally led us to the literature
on Bulgarian Solitaire. Our methods are easily adapted to this simpler
unordered game. In fact, we obtain improved bounds on the maximum
number of shifts needed for any game on n cards to cycle. We shall also
mention work on other variationŽs. of Bulgarian Solitaire, most notably
one called Montreal Solitaire.
Let us now define the games formally, present some examples and
calculations, and describe the plan of the paper. For a positive integer n,
we say l s Ž l1 , l2 , . . . . is a partition of n, and we write l & n, provided
the nonnegative integers l1 G l 2 G ??? add up to n. If l s ) l sq1 s 0, we
say l has s parts, and we often drop the zeroes, writing l s Ž l1 , l2 , . . . , l s ..
The shift operation B on l is the partition of n, denoted BŽ l., obtained
by decreasing each part l i by one, inserting a part s, and discarding any
zero parts. So B Ž i. Ž l. denotes the partition obtained by successively
applying the shift operation B to l a total of i times. Starting with a
partition l, we describe Bulgarian Solitaire by repeatedly applying the
shift operation to obtain the sequence of partitions
l , B Ž l . , B Ž2. Ž l . , . . . .
For a couple of simple examples, we note that l s Ž2, 1, 1, 1, 1. & 6 gives
the sequence Ž2, 1, 1, 1, 1., Ž5, 1., Ž4, 2., Ž3, 2, 1., Ž3, 2, 1., . . . , while l s Ž6, 1.
& 7 yields Ž6, 1., Ž5, 2., Ž4, 2, 1., Ž3, 3, 1., Ž3, 2, 2., Ž3, 2, 1, 1., Ž4, 2, 1., . . . . The
first example is fixed at the partition Ž3, 2, 1. after three steps, while the
second example reaches a cycle after two steps.
CYCLING UNDER REPEATED SHIFTS
207
We say a partition m & n is B-cyclic if B Ž i. Ž m . s m for some i ) 0.
Brandt w3x characterized all B-cyclic partitions for Bulgarian Solitaire. In
particular, when n is a triangular number, 1 q 2 q ??? qk, he proved that
Ž k, k y 1, . . . , 2, 1. is the unique B-cyclic partition of n, one that we end
up with, no matter where we start. Note that if n is not triangular, there is
no fixed partition under B.
To measure how long it takes for Bulgarian Solitaire to cycle, for l & n
we let d B Ž l. denote the smallest integer i G 0 such that B Ž i. Ž l. is
B-cyclic. Let D B Ž n. [ max d B Ž l.: l & n4 , so that for any l & n, D B Ž n.
steps reach a cycle. Trivially, D B Ž1. s D B Ž2. s 0 and D B Ž3. s 2. The
values of D B Ž n., 4 F n F 36, are displayed in Fig. 1, which we worked out
by computer. The data are arranged using the representation n s 1 q 2
q ??? qŽ k y 1. q r, 1 F r F k.
Brandt w3x conjectured that D B Ž n. s k 2 y k for triangular n, n s 1 q 2
q ??? qk. Hobby and Knuth w7x confirmed this for k F 10 by computer. It
was first proven by Igusa w8x and Etienne w5x, apparently independently at
about the same time, although Etienne’s proof was published much later.
ŽNote that Etienne’s paper is noted as having been received in 1984.. In
1987 Bentz w2x also gave a different proof. In Section 2 we present our
proof of Brandt’s result characterizing all B-cyclic partitions. This leads to
our simple, new proof of the value of D B Ž n. for triangular numbers n,
given in Section 3.
Igusa and Etienne went on to obtain the upper bound D B Ž n. F k 2 y k
for general n, represented as 1 q 2 q ??? qŽ k y 1. q r, 1 F r F k. It is by
no means sharp for every n Žsee Fig. 1.. Igusa w8x mentioned that this
bound follows easily from Brandt’s conjecture using a comparison theorem
by Akin and Davis w1, Theorem 3Žc.x. We will describe this comparison
theorem and show the implication in Section 4. We then prove a better
general bound ŽTheorem 4.4., which is D B Ž n. F k 2 y 2 k y 1, when 1 F r
- k and k G 4. ŽCases n s 2, 5 violate the bound.. We also present a
FIG. 1.
D B Ž n. for n s 4, 5, . . . , 36.
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GRIGGS AND HO
lower bound for D B Ž n. that we suspect is the correct value in general,
although a proof of this claim has eluded us.
In Section 5 we consider Carolina Solitaire, the ordered version of
Bulgarian Solitaire. Analogously with d B Ž l. and D B Ž n. for Bulgarian
Solitaire, we introduce d C Ž l. and D C Ž n. for Carolina Solitaire. We show
that, for triangular n s 1 q 2 q ??? qk, D C Ž n. s k 2 y 1. For other n of
the form n s 1 q 2 q ??? qŽ k y 1. q r, 1 F r - k, we prove D C Ž n. F
k 2 y k y 2, provided that k G 4. We also present a lower bound for
D C Ž n. that we conjecture is the correct value in general.
Other variations of Bulgarian Solitaire have been introduced: Yeh and
Servedio w9, 10x studied a variation on circular compositions; Cannings and
Haigh w4x investigated ‘‘Montreal Solitaire,’’ in which game the rule from l
to BŽ l. is changed when an exhausted pile Žsize 0. in l occurs. Akin and
Davis w1x also introduced ‘‘Austrian Solitaire.’’ In this game, a special pile
called the bank is reserved. Each move consists of taking a card from each
pile into the bank, and then generating some new piles from the bank by a
certain rule.
2. CHARACTERIZING CYCLIC PARTITIONS
We begin by describing B-cyclic partitions of n, a result discovered by
Brandt w3x. Etienne w5x as well as Akin and Davis w1x also gave proofs.
Although Etienne proved the result for triangular n only, the idea in his
proof is simple and can be applied to general n. The approach presented
here is similar to Etienne’s and we include it for the reader’s benefit.
THEOREM 2.1 w3x. Let n s 1 q 2 q ??? qŽ k y 1. q r, 1 F r F k. Then
l & n is B-cyclic if and only if l has the form
Ž k y 1 q d ky 1 , k y 2 q d ky2 , . . . , 2 q d 2 , 1 q d 1 , d 0 . ,
1
where each d i is 0 or 1 and Ý ky
is0 d i s r.
COROLLARY 2.2. If n s 1 q 2 q ??? qk, then Ž k, k y 1, . . . , 2, 1. is the
unique B-cyclic partition of n.
In order to prove the theorem, we introduce a natural array representation of a partition l. We are particularly interested in how repeatedly
shifting l affects the position of the ones in the array.
For a partition l s Ž l1 , l2 , . . . , l s . & n, we associate a Ž0, 1.-array
`
Ml s w m i j x i , js1 ,
where m i j s
½
1,
if j F s and i F l j ,
0,
otherwise.
The columns of Ml correspond to the parts of l.
CYCLING UNDER REPEATED SHIFTS
209
We notice that M B Ž l. can be obtained directly from Ml by a shifting
process on a Ž0, 1.-array. In later sections, such a shifting process can help
us evaluate d B Ž l. easily for some particular partitions l. We describe this
shifting process as follows: For a Ž0, 1.-array M, we say that the wth
diagonal of M consists of entries m i j , where i q j y 1 s w. Assume
l s Ž l1 , l2 , . . . , l s . & n and its associated array Ml is given.
Step 1: Diagonally circular shifting.
For each diagonal of Ml, say
¨ 1 , ¨ 2 , . . . , and ¨ w are the entries Žlisted from left to right. in this diagonal,
we replace them by ¨ w , ¨ 1 , ¨ 2 , . . . , and ¨ wy1 , respectively. Then we obtain
a new array; denote it MlX . Note that the numbers of ones in columns of
MlX are s, l1 y 1, l 2 y 1, . . . , l s y 1, 0, 0, 0, . . . . If s G l1 y 1 then MlX is
the array M B Ž l. . If s - l1 y 1 then we need an extra step to obtain M B Ž l. .
Step 2: Left shifting. We remove all zero entries in the first column of
MlX and shift each entry at the Ž i, j .-position, i G s q 1, j G 2, to the
Ž i, j y 1.-position. The new array is M B Ž l. .
Figure 2 shows two examples of the shifting process described above.
Proof of Theorem 2.1. Ž¥. Assume that l has the stated form. The
array Ml has all ones on the first k y 1 diagonals and all zeroes beyond
diagonal k. Each shift just shifts entries on diagonal k. Thus, B Ž k . Ž l. s l,
and l is B-cyclic.
Ž«. In the shifting process, we notice that Step 1 keeps every entry
on the same diagonal, while Step 2 always brings the entry 1 at the
Ž s q 1, 2.-position to a diagonal with smaller index. Thus, for any B-cyclic
partition m , the shifting process from Mm to M B Ž m . should involve Step 1
only.
FIG. 2. A shifting process on the arrays.
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GRIGGS AND HO
Now assume that l is B-cyclic and l cannot be expressed in the stated
form. Then for some w, there is a 0 in the wth diagonal and a 1 in the
Ž w q 1.th diagonal. Since the series of shifting processes from Ml to
M B Ž l . to M B Ž2. Ž l . , and so on involves Step 1 Ždiagonally circular shifting.
only, and since the integers w and w q 1 are relatively prime, we will
reach Žin at most w 2 y w y 1 steps. an array Mm which has 0 as Ž1, w .-entry and 1 as Ž w q 1, 1.-entry. ŽFigure 3 shows an example for such a series
of shifting processes.. Then the shifting process from Mm to M B Ž m .
involves Step 2, a contradiction.
It is easy to see from Theorem 2.1 that for a given n with r s k, k y 1,
or 1, starting from any partition of n and repeatedly applying the shift
operation B always reaches a unique cycle of partition of n.
FIG. 3. A series of shifting processes on the associated array of l s Ž2, 2, 1, 1..
CYCLING UNDER REPEATED SHIFTS
211
For general n, Brandt w3x also showed that the number of cycles for
Bulgarian Solitaire is
1
Ý
k
d <gcdŽ k , r .
f Ž d.
krd
,
rrd
ž /
where f Ž d . is the Euler f-function. The derivation of this formula was
explained in detail later by Akin and Davis w1x.
3. EVALUATING D B Ž n. FOR TRIANGULAR NUMBERS n
The partition l s Ž2, 2, 1, 1. in Fig. 3 gives D B Ž3. G 3 2 y 3. In general,
we have
THEOREM 3.1 w1, 2, 5x.
If n s 1 q 2 q ??? qk, then D B Ž n. G k 2 y k.
Proof. It suffices to present l & n such that d B Ž l. s k 2 y k. Let
l s Ž l1 , l2 , . . . , l kq1 ., where l1 s k y 1, l i s k y i q 1 for i s
2, 3, . . . , k, and l kq 1 s 1. By repeatedly applying the shifting process
Ždescribed in the previous section. to the arrays Ml, M B Ž l. , . . . , it is easy
to see that
B Ž k . Ž l . s Ž l1 , l2 , . . . , l ky1 , l k q 1 . ,
B Ž2 k . Ž l . s Ž l1 , l2 , . . . , l ky1 q 1, l k . ,
..
.
B ŽŽ ky2. k . Ž l . s Ž l1 , l2 , l 3 q 1, l 4 , l5 , . . . , l k . ,
B ŽŽ ky1. ky1. Ž l . s Ž l1 q 2, l2 , l 3 , . . . , l ky1 . ,
B ŽŽ ky1. k . Ž l . s Ž k, k y 1, k y 2, . . . , 2, 1 . .
Therefore d B Ž l. s k 2 y k.
Before we prove in Theorem 3.7 that, for triangular n, the lower bound
k 2 y k is the actual value of D B Ž n., we need some notation and lemmas.
Let us start with the example for n s 15: The 176 partitions of 15 can be
arranged in a tree illustrated in Fig. 4 Žfrom w3x. so that the vertices
correspond to the partitions and going down corresponds to applying B.
ŽThus, the root of the tree is the partition Ž5, 4, 3, 2, 1... In this figure, each
vertex is labeled with the number of parts for its corresponding partition.
For a partition l & n, we associate a sequence seq B Ž l. s ² c1 , c 2 , . . . :,
where c i is the number of parts in B Ž iy1. Ž l.. For instance, the top left
vertex in Fig. 4 Žcorresponding to the partition l s Ž4, 4, 3, 2, 1, 1. & 15.
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GRIGGS AND HO
FIG. 4. A tree for the partitions of 15.
has the associated sequence seq B Ž l. s ²6, 5, 5, 5, 4, 5, 6, 5, 5, 4, 5, 5, 6, 5,
4, 5, 5, 5, 6, 4, 5, 5, 5, 5, 5, . . . :.
In Fig. 4, we observe that the pattern ‘‘x y 1, x, x, . . . , x, x q 1’’ appears
quite often in most associated sequences. It will play a pivotal role in our
process for evaluating D B Ž n.. To study this pattern, we associate each
partition with a diagram: Given a partition l s Ž l1 , l2 , . . . , l s . & n, let
seq B Ž l. s ² c1 , c 2 , . . . : be its associated sequence. Then diagram B Ž l. is
defined to be the diagram shown in Fig. 5, where each l i-column Žresp.,
c i-column. has l i Žresp., c i . ones on the top and has infinitely many zeroes
on all other entries. For example, if l s Ž4, 2, 2, 2. then diagram B Ž l. is
shown in Fig. 6Ža..
We notice that there are c i ones in the row corresponding to c i .
Furthermore, diagram B Ž l. can be regarded as bookkeeping for the shift
CYCLING UNDER REPEATED SHIFTS
213
FIG. 5. diagram B Ž l..
operation B on l, since we can easily obtain B Ž i. Ž l. from diagram B Ž l.
for any integer i. For example, the rectangle below c 3 Žsee Fig. 6Žb.. gives
B Ž3. Ž l. s Ž4, 3, 2, 1. and the rectangle below c6 gives B Ž6. Ž l. s Ž4, 3, 2, 1..
PROPOSITION 3.2.
Then
Assume n G 1, l & n, and seq B Ž l. s ² c1 , c 2 , . . . :.
Ž1. c iq1 F c i q 1 for i G 1;
Ž2. for n s 1 q 2 q ??? qŽ k y 1. q r, where 1 F r F k, the sum of
any k consecuti¨ e terms in the sequence is at most k Ž k y 1. q r;
FIG. 6. Ža. diagram B Ž l. for l s Ž4, 2, 2, 2..
Žb. The rectangle below c 3 .
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GRIGGS AND HO
Ž3. Ž the sandwich property . if i - j and c i - x - c j , then there exist
integers p and q such that i F p - q F j, q G p q 2, and Ž c p , c pq1 , . . . , c q .
s Ž x y 1, x, x, . . . , x, x q 1..
Proof. Ž1. follows from the definition of the shift operation B. To
prove Ž2., we see for i ) 0 in diagram B Ž l. that c iq1 q c iq2 q ??? qc iqk is
the number of 1s in the k rows labelled by c iq1 , c iq2 , . . . . These come
from B Ž i. Ž l. Žin the rectangle below c i ., or from the staircase consisting of
j y 1 squares to the right of c iqj , 1 F j F k. Thus, c iq1 q ??? qc iqk F n q
1 q 2 q ??? qŽ k y 1. s k Ž k y 1. q r.
For Ž3., choose p as large as possible such that i F p - j and c p F x y 1.
We next require a series of lemmas that rely on Proposition 3.2 and
diagram B Ž l..
LEMMA 3.3.
Then
Let l & n s 1 q 2 q ??? qk and seq B Ž l. s ² c1 , c 2 , . . . :.
Ž1. d B Ž l. s t, where ² c t , c tq1 , . . . : s ² k y 1, k, k, . . . :;
Ž2. if d B Ž l. s t G k q 1 then at least one of the following holds:
Ži. Ž c p , c pq1 , . . . , c q . s Ž k y 1, k, k, . . . , k, k q 1. for some p, q
with t y k F p - q F t y 1 and q G p q 2;
Žii. Ž c p , c pq1 , . . . , c q . s Ž k y 2, k y 1, k y 1, . . . , k y 1, k . for
some p, q with t y k q 1 F p - q F t q 1 and q G p q 2.
Proof. Ž1. By Theorem 2.1, we have d B Ž l. s t where seq B Ž l. s
² c1 , . . . , c t , k, k, k, . . . : and c t / k. Then Proposition 3.2 forces c t s k y 1.
Ž2. For i ) j, e i, j denote the entry in the c i-row and the c j-column
of diagram B Ž l.. Then e t, ty1 s 1. Since c t s k y 1, we have e t, i s 0 for
some i, t y k F i F t y 2. We choose such i as large as possible.
Then e t, iq1 s 1, and, since c tq1 s k s c t q 1, comparing rows for c t and
c tq1 in diagram B Ž l., we have e tq1, iq1 s 1 as well. The column below c iq1
then has ones down at least as far as the row for c tq1. Proposition 3.2Ž1.
gives c i G c iq1 y 1, which forces all entries in the c i-column above the 0 at
e t, i to be ones. Therefore, c i s t y i y 1 and c iq1 s t y i.
If i G t y k q 1, then c i F k y 2, and Žii. holds by the sandwich property. Else, we have i s t y k, c tyk s k y 1, and c tykq1 s k. If c j F k y 2
for some j, t y k q 2 F j F t y 1, Žii. holds by the sandwich property,
while if c j G k q 1 for some such j, Ži. holds. Else, suppose for contradiction that k y 1 F c j F k, for j s t y k q 2, t y k q 3, . . . , t y 1: Since
i s t y k, the row for c t in diagram B Ž l. consists of k y 1 ones followed
by all zeroes. The row for c tq1 is then k ones followed by all zeroes. There
are k ones at the tops of columns c tq1 , c tq2 , . . . . Hence, in the rectangle
CYCLING UNDER REPEATED SHIFTS
215
below c ty1 , we have just k y 1 ones in the first row, followed by k y j q 1
ones in row j, 2 F j F k. Rows below that are all zero. In total, we have
n y 1 ones in the rectangle, which is a contradiction since this rectangle
represents, after reordering the columns if necessary, a partition of n.
LEMMA 3.4. Let l s Ž l1 , l2 , . . . , l s . & n G 1 and seq B Ž l. s
² c1 , c 2 , . . . :. If, for some p, k,
Ž c p , . . . , c pqk . s Ž k y 2, k y 1, k y 1, . . . , k y 1, k . ,
then p q k F n q 1.
Proof. Let e i, j Žresp., f i, j . denote the entry in the c i-row and the
c j-column Žresp., l j-column. of diagram B Ž l.. Then we have e pqky2, p s 1
and e pq ky1, p s 0, since c p s k y 2. By the given condition, in the c pqk-row
we also have e pq k, j s 1 for j s p q 1, p q 2, . . . , p q k y 1. Note that in
addition to these k y 1 entries of 1, there is exactly one more 1 in the
c pq k-row, since c pqk s k.
If f pq k, j s 1 for some j, 1 F j F s, then p q k F l j F n. Else, e pqk, j s
1 for some j, 1 F j F p y 1. In fact, j s 1, for, otherwise, we have
e pq k, jy1 s 0, and, since c pqk s k s c pqky1 q 1, comparing rows for c pqk
and c pq ky1 in diagram B Ž l., we also have e pqky1, jy1 s 0. Similarly, since
c pq ky1 s c pqky2 , e pqky2, p s 1, and e pqky1, p s 0, and comparing rows
for c pq ky1 and c pqky2 in diagram B Ž l., we have e pqky2, jy1 s 0. Then
c j ) c jy1 q 1, a contradiction. So j s 1, and then p q k y 1 F c1 F n.
LEMMA 3.5. Let l & n G 1 and seq B Ž l. s ² c1 , c 2 , . . . :. If for some
integers x, p, q, q G p q 3, and
Ž c p , c pq1 , . . . , c q . s Ž x y 1, x, x, . . . , x, x q 1 . ,
then either p F x or
Ž c p9 , c pXq1 , . . . , c q9 . s Ž xX y 1, x9, xX , . . . , xX , xX q 1 .
for some integers xX , pX , qX with xX F x, 2 F qX y pX - q y p, and p y x F pX
- qX F p q 1.
Proof. We assume p G x q 1 and prove the existence of xX , pX , and qX .
For i ) j, let e i, j denote the entry in the c i-row and the c j-column of
diagram B Ž l.. Then e p, py1 s 1. Since c p s x y 1, we have e p, p9 s 0 for
some p9, p y x F p9 F p y 2. We choose such p9 as large as possible.
Then e p, j s 1 for j s pX q 1, pX q 2, . . . , p y 1, and, since c pq1 s x s c p
q 1, comparing rows for c pq 1 and c p in diagram B Ž l., we also have
e pq 1, j s 1 for j s pX q 1, pX q 2, . . . , p. The column below c p9q1 then has
ones down at least as far as the row at c pq 1. Proposition 3.2Ž1. gives
c p9 G c p9q1 y 1, which forces all entries above the zero at e p, p9 to be ones.
Therefore c p9 s p y p9 y 1, c p9q1 s p y p9, and e pq2, p9q1 s 0.
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GRIGGS AND HO
Since c pq 2 s x s c pq1 , comparing rows for c pq2 and c pq1 in
diagram B Ž l., we have e pq2, j s 1 for j s p9 q 2, p9 q 3, . . . , p q 1. If
e pq 3, p9q2 s 1 then c p9q2 s p y p9 q 1, and we are finished. So we may
assume e pq 3, p9q2 s 0 and c p9q2 s p y p9.
Since c pq 3 s x s c pq2 , comparing rows for c pq3 and c pq2 in
diagram B Ž l., we have e pq3, j s 1 for j s p9 q 3, p9 q 4, . . . , p q 2. If
e pq 4, p9q3 s 1 then c p9q3 s p y p9 q 1, and we are finished. So we may
assume e pq 4, p9q3 s 0 and c p9q3 s p y p9. Continue this process.
So we may assume e qy 1, p9qqypy2 s 0, c p9qqypy2 s p y p9, and e qy1, j
s 1 for j s p9 q q y p y 1, p9 q q y p, . . . , q y 2. Since c q s x q 1 s
c qy 1 q 1, comparing rows for c q and c qy1 in diagram B Ž l., we have
e q, p9qqypy1 s 1, and hence c p9qqypy1 s p y p9 q 1. This completes the
proof.
LEMMA 3.6.
Let l & n G 1 and seq B Ž l. s ² c1 , c 2 , . . . :. If
Ž c p , c pq1 , c pq2 . s Ž x y 1, x, x q 1 . ,
then p F x.
Proof. For i ) j, let e i, j denote the entry in the c i-row and the
c j-column of diagram B Ž l.. Then e p, py1 s 1. Assume the contrary, i.e.,
p ) x. Then c p s x y 1 - p y 1 implies e p, i s 0 for some i, 1 F i F p y 2.
Choose such i as large as possible. Then e p, iq1 s 1. Since c pq1 s x s
c p q 1 and c pq2 s x q 1 s c pq1 q 1, comparing rows for c p , c pq1 , and
c pq 2 in diagram B Ž l., we also have e pq2, iq1 s e pq1, iq1 s 1. Thus c iq1 )
c i q 1, a contradiction.
We are now in a position to prove
THEOREM 3.7 w5, 8x.
If n s 1 q 2 q ??? qk, then
D B Ž n . s k 2 y k.
Proof. Note that from Theorem 3.1 we have D B Ž n. G k 2 y k. Now we
prove D B Ž n. F k 2 y k. Let l & n. It suffices to show d B Ž l. F k 2 y k.
We may assume k G 3, since it is easy to verify that d B Ž l. F k 2 y k for
k s 1, 2. Further, we may also assume d B Ž l. G k q 1. ŽOtherwise, d B Ž l.
F k - k 2 y k, since k G 3.. Let seq B Ž l. s ² c1 , c 2 , . . . :, where
² c t , c tq1 , . . . : s ² k y 1, k, k, k, . . . :. Then d B Ž l. s t and at least one of
Ži., Žii. of Lemma 3.3 holds.
If Žii. holds with q s p q k, we have p s t y k q 1 and q s t q 1. Note
that Lemma 3.4 gives p q k F n q 1. Then d B Ž l. s t s p q k y 1 F n
F k 2 y k, since k G 3.
Else, Ži. or Žii. holds and q F p q k y 1. By Lemmas 3.5 and 3.6,
seq B Ž l. has at most k 2 y 2 k y 1 terms before the c p-term. Then d B Ž l.
s t F Ž p y 1. q k q 1 F Ž k 2 y 2 k y 1. q k q 1 F k 2 y k.
CYCLING UNDER REPEATED SHIFTS
217
We observe that, for k s 2 Žresp., k s 3., l s Ž1, 1, 1. Žresp., l s
Ž2, 2, 1, 1.. is the only partition of n s 1 q 2 q ??? qk achieving d B Ž l. s
k 2 y k. For k G 4, we have
THEOREM 3.8. Let k G 4 and n s 1 q 2 q ??? qk. If l & n with d B Ž l.
s k 2 y k, then the associated array Ml s w m i j x`i, js1 satisfies the following
conditions Ž illustrated in Fig. 7.:
Ž1.
Ž2.
Ž3.
m i j s 1 for j F k q 3 y 2 i;
m i j s 0 for j G 2 k q 1 y 2 i;
<Ž i, j .: j s k q 4 y 2 i, m i j s 04< F 2.
Proof. If k G 4 and d B Ž l. s k 2 y k, the proof of Theorem 3.7 implies
Ži. of Lemma 3.3 holds and q s p q k y 1. Further, we have seq B Ž l. s
² c1 , c 2 , . . . :, where Ž c k , c kq1 , c kq2 . s Ž k y 1, k, k q 1. and Ž c 2 k , c 2 kq1 ,
c 2 kq2 , c 2 kq3 . s Ž k y 1, k, k, k q 1..
For i ) j, let e i, j denote the entry in the c i-row and the c j-column of
diagram B Ž l.. Then e kq2, kq1 s 1 and e kq2, k s 1. We also have e kq2, i s 1
for i s 1, 2, . . . , k y 1. ŽOtherwise, we choose i as large as possible such
that 1 F i F k y 1 and e kq 2, i s 0. By comparing rows for c kq2 , c kq1 , and
c k in diagram B Ž l., we have e k, i s e kq1, i s 0, and hence c iq1 ) c i q 1, a
contradiction.. Therefore, c i G k q 2 y i for i s 1, 2, . . . , k q 2, and hence
condition Ž1. holds.
FIG. 7. Conditions for Theorem 3.8. Among the ?Ž k q 3.r2@ entries marked by an
asterisk, at most two are zeroes.
218
GRIGGS AND HO
We also have e2 k, i s 1 for i s k q 1, k q 2, . . . , 2 k y 1. ŽOtherwise, we
choose i as large as possible such that k q 3 F i F 2 k y 2 and e2 k, i s 0.
Then e2 k, iq1 s 1. We note that e2 kq1, kq1 s 1 and e2 kq2, kq1 s 0, since
c kq 1 s k. By comparing rows for c 2 k , c2 kq1 , and c 2 kq2 in diagram B Ž l., we
have e2 kq2, iq1 s e2 kq1, iq1 s 1, and hence c iq1 ) c i q 1, a contradiction..
Therefore, e2 k, i s 0 for i s 1, 2, . . . , k, and condition Ž2. holds.
Since e2 k, kq3 s 1, by comparing rows for c 2 k , c 2 kq1 , c 2 kq2 , and c 2 kq3 in
diagram B Ž l., we have e2 kq3, kq3 s e2 kq2, kq3 s e2 kq1, kq3 s 1, and hence
2
Ž .
c kq 3 G k. So we have Ý kq
js1 e kq3, j G k, and hence condition 3 holds.
We have checked by computer that the converse of this theorem is true
for k s 4, 5, 6, 7. ŽIn particular, the converse with condition Ž3. removed is
also true for k s 4, 5, 6.. When k s 8, there are 1276 partitions satisfying
all three conditions in Theorem 3.8, but 9 partitions among them have
d B Ž l. s 20 / k 2 y k. So the three conditions given in Theorem 3.8 are
necessary, but not sufficient, for the associated array of extremal partitions.
4. EVALUATING BOUNDS ON D B Ž n. FOR
NONTRIANGULAR NUMBERS n
For any nontriangular n, we can write n s 1 q 2 q ??? qŽ k y 1. q r,
where 1 F r - k. In this section we first prove D B Ž n. F k 2 y k based on
Theorem 3.7 and a comparison theorem from w1x. Then we improve this
bound to k 2 y 2 k y 1 for k G 4. We also present a lower bound that is
likely the actual value.
Let L s D`is1 L i , where L i denotes the set containing all partitions of
i. For l s Ž l1 , l2 , . . . . and m s Ž m 1 , m 2 , . . . . in L, we define a partial
ordering Ži.e., a reflexive, antisymmetric and transitive relation. by
l F m m li F mi
for i s 1, 2, . . . .
Then we have the following theorem due to Akin and Davis:
THEOREM 4.1 w1, Thm. 3Žc.x.
BŽ m ..
If l, m g L and l F m , then BŽ l. F
Proof. Let x 1 , x 2 , . . . and y 1 , y 2 , . . . Žnot necessarily in nonincreasing
order. be the parts of two partitions x and y, respectively. We observe that
if x i F yi for i G 1, then x F y. ŽTo verify this, we may assume further,
without loss of generality, that the x i ’s are already in nonincreasing order.
If y 1 - y 2 , we interchange y 1 and y 2 ; then x i F yi for i G 1 still holds. If
y 2 - y 3 , we interchange y 2 and y 3 ; then x i F yi for i G 1 still holds.
Continuing this process, eventually we can arrange all yi ’s in nonincreasing
order and x i F yi for i G 1 still holds. Therefore x F y..
CYCLING UNDER REPEATED SHIFTS
219
Now we assume l, m g L and l F m. Let l s Ž l1 , l 2 , . . . . and m s
Ž m 1 , m 2 , . . . ., where l1 G ??? G l s ) l sq1 s 0, m 1 G ??? G m t ) m tq1 s
0, and l i F m i for i G 1. Then the parts of BŽ l. are x 1 s s, x 2 s l1 y 1,
x 3 s l2 y 1, . . . , x sq1 s l s y 1, x sq2 s 0, x sq3 s 0, . . . . Similarly, the
parts of BŽ m . are y 1 s t, y 2 s m 1 y 1, y 3 s m 2 y 1, . . . , ytq1 s m t y 1,
ytq2 s 0, ytq3 s 0, . . . . Since x i F yi for i G 1, we have BŽ l. F BŽ m . by
the observation above.
Combining this theorem with Theorem 3.7, we can show
THEOREM 4.2 w5, 8x.
Let n s 1 q 2 q ??? qŽ k y 1. q r, 1 F r - k. Then
D B Ž n . F k 2 y k.
Proof. Let l & n. It suffices to show d B Ž l. F k 2 y k. We choose m , n
such that m & Ž1 q 2 q ??? qŽ k y 1.., n & Ž1 q 2 q ??? qk ., and m F l
2
2
F n . By Theorem 4.1, we have that B Ž k yk . Ž m . F B Ž k yk . Ž l. F
Ž k 2yk . Ž .
Ž k 2yk . Ž .
B
n . Note that Theorem 3.7 gives B
m s Ž k y 1, k y
2
2
2, . . . , 2, 1. and B Ž k yk . Ž n . s Ž k, k y 1, . . . , 2, 1.. Thus B Ž k yk . Ž l. has the
form stated in Theorem 2.1 and d B Ž l. F k 2 y k.
We prove an analogue of Lemma 3.3 for nontriangular n. Applying it
along with Lemmas 3.4, 3.5, and 3.6 leads to an improved upper bound on
D B Ž n. for nontriangular n.
LEMMA 4.3. Let l & n s 1 q 2 q ??? qk q r, 1 F r - k, and seq B Ž l.
s ² c1 , c 2 , . . . :.
Ž1. If Ž c t , . . . , c tqky1 . s Ž c tqk , . . . , c tq2 ky1 . s ??? , i.e., c tqi s c tqiqk
for i G 0, then d B Ž l. F t y 1. ŽThus we can find d B Ž l. by choosing such t
as small as possible..
Ž2 . If c t s k y 1, c tq 1 s k, c tq i s c tq iq k for i G 0, and
Ž c tyk , . . . , c ty1 . / Ž c t , . . . , c tqky1 ., then at least one of Ži., Žii. of Lemma 3.3
holds.
Proof. Ž1. If c tqi s c tqiqk for i G 0, then each c tqi is the number of
parts for some B-cyclic partition, and hence Theorem 2.1 gives k y 1 F
c tqi F k for i G 0. In particular, we note that the rows for
c t , c tq1 , . . . , c tqky1 in diagram B Ž l. each consists of k y 1 or k ones. Then
B Ž ty1. Ž l. has the stated form of Theorem 2.1, since it is represented by
the rectangle below c ty1 , after reordering the columns if necessary. Therefore, B Ž ty1. Ž l. is B-cyclic, and hence d B Ž l. F t y 1.
Ž2. The proof is similar to that of Lemma 3.3Ž2.. In the last case, we
suppose for contradiction that c tyk s k y 1, c tykq1 s k, and k y 1 F c j
F k, for j s t y k q 2, t y k q 3, . . . , t y 1. Similar to the proof of Ž1.,
220
GRIGGS AND HO
then the rectangle below c tyky1 in diagram B Ž l. gives that B Ž tyky1. Ž l. is
B-cyclic. Thus, by Theorem 2.1, B Ž tyky1qi. Ž l. s B Ž ty1qi. Ž l. for i G 0.
Counting the number of parts in each partition, we have Ž c tyk , . . . , c ty1 . s
Ž c t , . . . , c tqky1 ., a contradiction.
THEOREM 4.4.
Let n be nontriangular,
n s 1 q 2 q ??? q Ž k y 1 . q r , 1 F r - k.
Then
Ž1.
Ž2.
D B Ž n. F k 2 y 2 k y 1 for k G 4;
furthermore, equality holds when k G 4 and r s k y 1.
Proof. Ž1. We may assume k G 5, since we have D B Ž n. F 7 for k s 4
from Fig. 1. Let l & n. It suffices to show d B Ž l. F k 2 y 2 k y 1. Let
seq B Ž l. s ² c1 , c 2 , . . . :. Let B Ž ty1. Ž l. be B-cyclic for some t G 1. Then
Theorem 2.1 gives c tqi s c tqiqk and k y 1 F c tqi F k for i G 0. Further,
we may assume c t s k y 1 and c tq1 s k, since r / k. We choose such t as
small as possible. If t F k, then d B Ž l. F t y 1 F k y 1 - k 2 y 2 k y 1,
since k G 5. So we may assume t G k q 1 and Ž c tyk , . . . , c ty1 . /
Ž c t , . . . , c tqky1 .. Then, by Lemma 4.3, at least one of Ži., Žii. of Lemma 3.3
holds.
If Žii. holds with q s p q k, we have Ž p, q . s Ž t y k q 1, t q 1.. Note
that Lemma 3.4 gives p q k F n q 1. Then d B Ž l. F t y 1 s p q k y 2
F n y 1 - k 2 y 2 k y 1, since k G 5.
Else, if Ži. or Žii. holds and q F p q k y 2, by Lemmas 3.5 and 3.6,
seq B Ž l. has at most k 2 y 3k y 1 terms before the c p-term. Then d B Ž l.
F t y 1 F Ž p y 1. q k F Ž k 2 y 3k y 1. q k s k 2 y 2 k y 1.
Else, suppose for contradiction that Ži. or Žii. holds with q s p q k y 1:
if Ži. holds with q s p q k y 1, we have Ž p, q . s Ž t y k, t y 1. and c ty1 s
k q 1, which is a contradiction, since, by Proposition 3.2Ž2., we have
c ty1 F k; else, Žii. holds with q s p q k y 1, we have Ž p, q . s Ž t y k q
2, t q 1. and c tykq2 s k y 2, which is also a contradiction, since, by
comparing rows for c tq1 and c t in diagram B Ž l., we have c tykq2 / k y 2.
Ž2. Assume k G 4 and r s k y 1. By Ž1., it suffices to present l & n
such that d B Ž l. s k 2 y 2 k y 1. Let l s Ž l1 , l2 , . . . , l kq1 ., where l1 s k
y 1, l2 s k y 2, l i s k y i q 1 for i s 3, 4, . . . , k, and l kq1 s 1. Imitating the proof of Theorem 3.1, we can show that d B Ž l. s k 2 y 2 k y 1.
So far we have obtained an upper bound for D B Ž n. in Theorem 4.4.
Next we will find a lower bound for D B Ž n. and conjecture this lower
bound is the actual value of D B Ž n..
221
CYCLING UNDER REPEATED SHIFTS
LOWER BOUND THEOREM 4.5.
1 F r F k,
For n s 1 q 2 q ??? qŽ k y 1. q r, G 3,
¡Ž k y 3 y r . k q r q 2,
for 1 F r -
~
for r s
D B Ž n . G n y k q 1,
¢Ž r y 2. k q r ,
for
ky1
2
ky1
2
kq1
2
or
,
kq1
2
,
- r F k.
Proof. It suffices to present l & n such that d B Ž l. is the stated lower
bound on D B Ž n.. We consider three cases Žassociated arrays for extremal
partitions in Cases Ž1. and Ž3. are illustrated in Fig. 8.:
Case Ž1.. 1 F r - ?Ž k y 1.r2@. Let l s Ž l1 , l2 , . . . , l k ., where l1 s
k y 2, l i s k y i for i s 2, 3, . . . , k y r y 1, and l i s k y i q 1 for i s
k y r, k y r q 1, . . . , k. Imitating the proof of Theorem 3.1, we can show
that d B Ž l. s Ž k y 3 y r . k q r q 2.
Case Ž2.. r s ?Ž k y 1.r2@ or ?Ž k q 1.r2@. Let l s Ž1, 1, . . . , 1. with
n parts of 1. By applying induction on i, we can show that
B Ž1q1q2q3q? ? ?qi. Ž l. s Ž n y Ž1 q 2 q ??? qi ., i, i y 1, i y 2, . . . , 2, 1. for i
s 1, 2, . . . , k y 2. Thus d B Ž l. s 1 q 1 q 2 q 3 q ??? qŽ k y 2. q Ž r y 1.
s n y k q 1.
FIG. 8. Associated arrays for two extremal partitions in Theorem 4.5.
222
GRIGGS AND HO
Case Ž3.. ?Ž k q 1.r2@ - r F k. Let l s Ž l1 , l 2 , . . . , l kq1 ., where l i
s k y i for i s 1, 2, . . . , k y r q 1, l i s k y i q 1 for i s k y r q 2,
k y r q 3, . . . , k, and l kq 1 s 1. Imitating the proof of Theorem 3.1, we
can show that d B Ž l. s Ž r y 2. k q r.
CONJECTURE 4.6 w5x.
Ž1.
Assume n G 3.
If
n g 1 q 2 q ??? q Ž K y 1 . , 1 q 2 q ??? q Ž K y 1 . q
Ky2
2
,
then D B Ž n. strictly decreases from Ž K y 1. 2 y Ž K y 1. to uŽ K 2 y 2 K .r2v.
Ž2. If
n g 1 q 2 q ??? q Ž K y 1 . q
Ky2
2
, 1 q 2 q ??? q Ž K y 1 . q K ,
then D B Ž n. strictly increases from uŽ K 2 y 2 K .r2v to K 2 y K.
Actually Etienne gave K 2 y K instead of Ž K y 1. 2 y Ž K y 1. in Ž1. and
gave Ž K q 1. 2 y Ž K q 1. instead of K 2 y K in Ž2.. Both of these seem to
be mistakes.
According to w5x, this conjecture was confirmed for n s 3, 4, 5, . . . , 55.
Here we suggest a stronger conjecture that is confirmed for n s
3, 4, 5, . . . , 36 Žsee Fig. 1., and also for r s k y 1, k.
CONJECTURE 4.7.
‘‘D B Ž n. s .’’
In Theorem 4.5 ‘‘D B Ž n. G ’’ can be replaced with
5. CAROLINA SOLITAIRE: A VARIATION OF
BULGARIAN SOLITAIRE
Let us now define Carolina Solitaire formally, and then derive analogues
of some of the results for Bulgarian Solitaire. For a positive integer n, we
say l s Ž l1 , l2 , . . . . is a composition of n, and we write l * n, provided
that the positive integers l1 , l2 , . . . , l s Žnot necessarily in nonincreasing
order. add up to n and l i s 0 for i G s q 1. We say such l has s
Žpositive. parts, and we often drop the zeroes, writing l s Ž l1 , l2 , . . . , l s ..
The shift operation C on l is the composition of n, denoted C Ž l.,
obtained by decreasing each part l i by one, inserting a part s s as the first
part, and discarding any zero parts. So C Ž i. Ž l. denotes the composition
obtained by successively applying the shift operation C to l a total of i
times. Starting with a composition l, we describe Carolina Solitaire by
CYCLING UNDER REPEATED SHIFTS
223
repeatedly applying the shift operation C to obtain the sequence of
compositions
l , C Ž l . , C Ž2. Ž l . , . . . .
For a couple of simple examples, we note that l s Ž2, 1, 1, 1, 1. * 6 gives
the sequence Ž2, 1, 1, 1, 1., Ž5, 1., Ž2, 4., Ž2, 1, 3., Ž3, 1, 2., Ž3, 2, 1., Ž3, 2, 1., . . . ,
while l s Ž6, 1. * 7 yields Ž6, 1., Ž2, 5., Ž2, 1, 4., Ž3, 1, 3., Ž3, 2, 2., Ž3, 2, 1, 1.,
Ž4, 2, 1., Ž3, 3, 1., Ž3, 2, 2., Ž3, 2, 1, 1., Ž4, 2, 1., . . . . The first example is fixed
at the composition Ž3, 2, 1. after five steps, while the second example
reaches a cycle after four steps.
We say a composition m s Ž m 1 , m 2 , . . . , m s . * n is a permutation of a
composition l s Ž l1 , l2 , . . . , l s . * n if the parts m i of m are a permutation of the parts l i of l. We notice the connection between Bulgarian
Solitaire and Carolina Solitaire: If m * n is a permutation of l & n then,
for i ) 0, C Ž i. Ž m . * n is a permutation of B Ž i. Ž l. & n.
Similar to B-cyclic partition, d B Ž l., and D B Ž n. in Bulgarian Solitaire,
we can define Žresp.. C-cyclic composition, d C Ž l., and D C Ž n. in Carolina
Solitaire: We say a composition m * n is C-cyclic if C Ž i. Ž m . s m for some
i ) 0. We will prove in Theorem 5.1 that l * n is C-cyclic if and only if
l & n is B-cyclic. To measure how long it takes for Carolina Solitaire to
cycle, for l * n we let d C Ž l. denote the smallest integer i G 0 such that
C Ž i. Ž l. is C-cyclic. Let D C Ž n. [ max d C Ž l.: l * n4 , so that for any
l * n, D C Ž n. steps reach a cycle. Trivially, D C Ž1. s D C Ž2. s 0 and D C Ž3.
s 3. The values of D C Ž n., 4 F n F 36, are displayed in Fig. 9, which we
worked out by computer. The data are arranged using the representation
n s 1 q 2 q ??? qŽ k y 1. q r, 1 F r F k.
We will show in Theorem 5.5 that D C Ž n. s k 2 y 1 whenever n s 1 q 2
q ??? qk. When n is a nontriangular number, we will prove in Theorem
5.8 that D C Ž n. F k 2 y k y 2 for k G 4. We also present in Theorem 5.9 a
lower bound for D C Ž n. that we conjecture is the correct value in general.
FIG. 9.
D C Ž n. for n s 4, 5, . . . , 36.
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GRIGGS AND HO
Now we start with the characterization of C-cyclic compositions. Similarly to the associated Ž0, 1.-array for a partition, we can define the
associated Ž0, 1.-array for a composition l s Ž l1 , l2 , . . . , l s . * n:
`
Ml s w m i j x i , js1 ,
where m i j s
½
1,
if j F s and i F l j ,
0,
otherwise.
The columns of Ml correspond to the parts of l.
We notice that MC Ž l. can be obtained directly from Ml by a shifting
process on a Ž0, 1.-array: Assume l s Ž l1 , l2 , . . . , l s . * n and its associated array Ml is given.
Step 1: Diagonally circular shifting. This is the same as Step 1 for
Bulgarian Solitaire described in Section 2. After Step 1, we obtain a new
array; denote it MlX . Note that the numbers of ones in columns of MlX are
s, l1 y 1, l 2 y 1, . . . , l s y 1, 0, 0, 0, . . . . If l i y 1 s 0 and l iq1 y 1 ) 0
for some i, then we need an extra step to obtain MC Ž l. . Otherwise, MlX is
the array MC Ž l . .
Step 2: Left shifting. Whenever there is some integer j such that the
jth column of MlX is an all-zero column and the Ž j q 1.th column of MlX is
not an all-zero column, we remove the jth column and shift each column
with larger index one column to the left. We repeat this shifting until there
does not exist such an integer j.
Using the shifting process described above and imitating the proof of
Theorem 2.1, we can show
THEOREM 5.1. Let n s 1 q 2 q ??? qŽ k y 1. q r, 1 F r F k. Then l *
n is C-cyclic if and only if l has the form
Ž k y 1 q d ky 1 , k y 2 q d ky2 , . . . , 2 q d 2 , 1 q d 1 , d 0 . ,
1
where each d i is 0 or 1 and Ý ky
is0 d i s r.
COROLLARY 5.2. If n s 1 q 2 q ??? qk, then Ž k, k y 1, . . . , 2, 1. is the
unique C-cyclic composition of n.
By Theorems 2.1 and 5.1, l * n is C-cyclic if and only if l & n is
B-cyclic. So the number of cycles for Carolina Solitaire is the same as that
for Bulgarian Solitaire.
Next we shall prove D C Ž n. s k 2 y 1 for triangular n.
PROPOSITION 5.3. Let n s 1 q 2 q ??? qk and l * n. Then l is a
permutation of Ž k, k y 1, . . . , 2, 1. if and only if d C Ž l. F k y 1.
CYCLING UNDER REPEATED SHIFTS
225
Proof. If l is a permutation of Ž k, k y 1, . . . , 1., it is easy to see that
C Ž ky1. Ž l. s Ž k, k y 1, . . . , 1. and d C Ž l. F k y 1. If l is not a permutation of Ž k, k y 1, . . . , 1., then there exists m s C Ž i. Ž l., i G 0, such that m
is not a permutation of Ž k, k y 1, . . . , 1. and C Ž m . is a permutation of
Ž k, k y 1, . . . , 1.. We note that m must be a permutation of Ž k q 1, k y
1, k y 2, k y 3, . . . , 2., and hence C Ž ky1. Ž m . s Ž k, k y 1, . . . , 3, 1, 2. is not
C-cyclic. Therefore, d C Ž m . G k and d C Ž l. G k.
Combining Proposition 5.3 with Theorem 2.1, we have the following
corollary that describes the relation between Carolina Solitaire and Bulgarian Solitaire when n is a triangular number.
Let n s 1 q 2 q ??? qk and let l * n be a permuta-
COROLLARY 5.4.
tion of l9 & n.
Ž1.
Ž2.
d C Ž l. F k y 1 if and only if d B Ž l9. s 0.
If d C Ž l. G k then d C Ž l. y d B Ž l9. s k y 1.
Combining Corollary 5.4 with Theorem 3.7, we can prove:
THEOREM 5.5.
If n s 1 q 2 q ??? qk, then
D C Ž n . s k 2 y 1.
Now we turn our attention to the value of D C Ž n. for nontriangular n.
PROPOSITION 5.6.
Let l * n s 1 q 2 q ??? qŽ k y 1. q r, 1 F r - k.
Ž1. If l is a permutation of some C-cyclic composition of n, then
Ž
d C l. F k y 1.
Ž2. If l is not a permutation of any C-cyclic composition of n, then
d C Ž l. G k y 1.
Proof. Ž1. Assume l is a permutation of m * n, where m has the
stated form of Theorem 2.1. Then we note that C Ž ky1. Ž l. also has the
stated form of Theorem 2.1. Thus C Ž ky1. Ž l. is C-cyclic and d C Ž l. F k y 1.
Ž2. It is easy to verify the statement for k F 3. So we may assume
k G 4. If l is not a permutation of any C-cyclic composition of n, then
there exists m s C Ž i. Ž l., i G 0, such that m is not a permutation of any
C-cyclic composition and C Ž m . is a permutation of some C-cyclic composition. By Theorem 2.1, we note that m must be a permutation of
Ž k q d ky 1 , k y 2 q d ky3 , k y 3 q d ky4 , . . . , 3 q d 2 , 2 q d 1 , d 0 q d ky2 .,
where each d i is 0 or 1, d 0 F d ky2 , and Ž d ky1 , d ky2 . / Ž0, 1.. Then
C Ž ky2. Ž m . does not have the stated form of Theorem 2.1. So d C Ž m . G k y
1, and hence d C Ž l. G k y 1.
226
GRIGGS AND HO
We note that from the condition ‘‘d C Ž l. s k y 1’’ we cannot determine
whether l is a permutation of some C-cyclic composition of n. For
example, when k s 4, l s Ž3, 2, 4. * 9 with d C Ž l. s 3 is a permutation of
the C-cyclic composition Ž4, 3, 2.; however, l s Ž5, 4. * 9 with d C Ž l. s 3
is not a permutation of any C-cyclic composition.
Similarly to Corollary 5.4 for triangular n, we have the following
corollary that describes the relation between Carolina Solitaire and Bulgarian Solitaire when n is not a triangular number:
COROLLARY 5.7. Let n s 1 q 2 q ??? qŽ k y 1. q r, 1 F r - k, and let
l * n be a permutation of l9 & n.
Ž1.
If d C Ž l. F k y 2 then d B Ž l9. s 0.
Ž2.
If d C Ž l. G k y 1 then d C Ž l. y d B Ž l9. s k y 2 or k y 1.
Proof. Ž1. follows immediately from Proposition 5.6Ž2. and we can
prove Ž2. by applying induction on d C Ž l.. ŽFor the induction basis, we note
that if d C Ž l. s k y 1, then d B Ž l9. s 0 or 1 by Proposition 5.6..
We are now in a position to prove
THEOREM 5.8.
1 F r - k. Then
Let n be nontriangular, n s 1 q 2 q ??? qŽ k y 1. q r,
Ž1.
D C Ž n. F k 2 y k y 2 for k G 4;
Ž2.
furthermore, equality holds when k G 4 and r s k y 1.
Proof. Ž1. follows from Theorem 4.4 and Proposition 5.7. To prove Ž2.,
it suffices to present l * n such that d C Ž l. s k 2 y k y 2. Let l s
Ž l1 , l 2 , . . . , l kq1 ., where l1 s k y 1, l2 s k y 2, l i s k y i q 1, for i s
3, 4, . . . , k, and l kq 1 s 1. Imitating the proof of Theorem 3.1, we can show
that d C Ž l. s k 2 y k y 2.
Using the results for D B Ž n. and considering the relations between
d C Ž l. and d B Ž l., we have proven Theorems 5.5 and 5.8 for D C Ž n.. We
can also prove these two theorems in another way: Note that we have
Lemmas 3.3, 3.4, 3.5, 3.6, and 4.3 for Bulgarian Solitaire. If we consider
analogous lemmas for Carolina Solitaire, then we can prove Theorems 5.5
and 5.8 directly.
Similarly to Theorem 4.5 for D B Ž n., we have the following ‘‘lower bound
theorem’’ for D C Ž n.:
227
CYCLING UNDER REPEATED SHIFTS
LOWER BOUND THEOREM 5.9.
1 F r F k,
¡Ž k y 2 y r . k q r ,
DC
Ž n . G~
For n s 1 q 2 q ??? qŽ k y 1. q r G 3,
for 1 F r -
n y 1,
for
n,
for
¢Ž r y 1. k q r y 1,
for
ky1
2
ky1
2
kq1
2
ky1
2
FrF
FrF
,
kq1
2
kq1
2
and r s 1,
and r G 2,
- r F k.
Proof. Similarly to the proof of Theorem 4.5, we consider three cases:
1 F r - ?Ž k y 1.r2@, ?Ž k y 1.r2@ F r F ?Ž k q 1.r2@, and ?Ž k q 1.r2@ - r
F k. For each case, we take the same extremal partition Žcomposition. as
that in the proof of Theorem 4.5.
We conclude this paper by giving the following conjecture that is
confirmed for n s 3, 4, . . . , 36 Žsee Fig. 9., and also for r s k y 1, k.
CONJECTURE 5.10.
‘‘D C Ž n. s .’’
In Theorem 5.9 ‘‘D C Ž n. G ’’ can be replaced with
ACKNOWLEDGMENTS
We are grateful to Andrey Andreev for introducing us to Bulgarian Solitaire and Carolina
Solitaire. We thank Ron Graham for pointing us to the literature on Bulgarian Solitaire.
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