VARIATIONS OF THE 15

A thesis submitted to the Kent State University Honors College in partial fulfillment of the requirements for University Honors

by

Lisa Rose Hendrixson

May, 2011

Thesis written by

Lisa Rose Hendrixson

Approved by

, Advisor

, Chair, Department of Mathematics

Accepted by

, Dean, Honors College

ii

TABLE OF CONTENTS

LIST OF FIGURES...... iv

ACKNOWLEDGEMENTS ...... v

CHAPTERS

I INTRODUCTION ...... 1

II THE HISTORY OF THE 15 PUZZLE ...... 3

III MATHEMATICS AND THE 15 PUZZLE ...... 6

IV VARIATIONS OF THE 15 PUZZLE ...... 14

V CONCLUSION ...... 22

BIBLIOGRAPHY ...... 23

iii LIST OF FIGURES

• Figure 1.The 15 Puzzle ...... 2

• Figure 2.Pictoral representation of the above ...... 8

• Figure 3.The permutation multiplied by the transposition (7, 8) ...... 9

• Figure 4.Two odd length cycles inside the 15 Puzzle ...... 10

• Figure 5.The two cycles, put together...... 11

• Figure 6.Permutting 3 blocks cyclically...... 12

• Figure 7. of the 15 Puzzle...... 13

• Figure 8.Starting position for the first variation...... 15

• Figure 9.Creating a single transposition inside the variation...... 16

• Figure 10.Showing the switch of the blank spaces...... 16

• Figure 11.Puzzle with a fixed block...... 17

• Figure 12.Bipartite graph of a puzzle wth a glued-down block...... 21

iv ACKNOWLEDGEMENTS

I would like to thank Dr. Donald White for all his help and support during the process of writing this thesis. Without his dedication, it would not have been possible.

Also, I would like to thank Dr. Mark Lewis, Dr. Elizabeth Mann, and Dr. Sara Newman for their willingness to serve on my defense committee and for all of their helpful comments and support along the way.

v INTRODUCTION

The 15 Puzzle: to look at, it does not look like much. Just a tray, with 15 little numbered blocks. Admittedly, these blocks come out of the tray easily, but this is only exciting for as long as it takes to find the piece that fell out. It is difficult to imagine that such a thing would excite masses of people to a fervor equivalent to that of a group of mid- dle school girls at a boyband concert. Who would have guessed the mathematical insight lurking beneath the simple design? The answer is, perhaps, too simple: a mathematician would indeed find the hidden complexities.

The 15 Puzzle is a device that few know much about in the present time. It has lost its intrigue to newer and more complicated that have taken center stage as both toys and catalysts for mathematical thought. Most people fail to understand what a 15 Puzzle is without a somewhat longwinded explanation. Essentially, the 15 Puzzle is a small tray with 15 numbered blocks that can be pulled out of the tray and rearranged in any manner that the player desires. The desired position is when the blocks are arranged as in Figure

1, with the 1 block in the top left corner and the rest of the blocks continuing in numerical order left to right and then top to bottom. From there, the blocks can be removed from the tray and replaced in any order the player wants. It should be stated that taking the pieces out and putting them back in the solved position is cheating.

The goal of this thesis is to educate the reader about the 15 Puzzle, including its

1 2

definition, the somewhat murky history, and the mathematics behind it.

Figure 1: The 15 Puzzle CHAPTER ONE

THE HISTORY OF A PHENOMENON

The origins of the 15 Puzzle are shrowded in mystery. The most popular story is that the 15 Puzzle was invented by Sam Loyd, an American in the mid 19th Century. Loyd was a famous and prolific puzzle maker, who, sadly, was particularly lax when it came to giving credit to actual inventors. Regardless, he became known as “America’s Greatest

Puzzlist,” with some help from the work of Henry Dudeney, his British counterpart [12].

Loyd began playing chess at the age of 14, and within two years became famous for related problems. This led to a job as Problem Editor for the American Chess Journal and the writing of a book entitled Chess Strategy. However, this book has a copyright de- screpancy: while the date says 1878, Alain White wrote in his book about Loyd that Chess

Strategy was not published until 1881. If this is to be believed, then Loyd would have been working on the book while the 15 Puzzle craze was going on, a not impossible feat, but perhaps improbable. At any rate, Loyd’s name was never mentioned in connection with the 15 Puzzle during the puzzle’s peak of popularity. Instead, Loyd first mentioned having invented the puzzle a decade after the craze had ended. After his death, Loyd’s son, Sam

Loyd, Jr., took up his father’s business and even the credit for some of his father’s puzzles.

What he did not do, however, is give his father, or himself, any credit concerning the 15

Puzzle [12].

3 4

Another theory on the origin of the 15 Puzzle is that a child in a deaf-mute school in Hartford, Connecticut invented the puzzle. No evidence to support this claim has ever been found [12].

A third theory about the invention of the 15 Puzzle is that Noyes Chapman devel- oped the puzzle in 1879. Chapman was a New York postmaster. Chapman did apply for a patent for a puzzle he called “Block Solitare Puzzle.” However, the application was re- jected because of the similarity to a puzzle developed more than a year prior by Ernest U.

Kinsey. Kinsey had a 6 × 6 puzzle with lettered blocks, instead of numbered ones. Despite this, it seems that Chapman gave puzzles to Anna and James Belden. James Belden was an ex-mayor of Syracuse, New York and a business associate of Noyes Chapman’s son, Frank

Chapman. From there, Mr. and Mrs. Belden took the puzzle to the community of Watch

Hill, Rhode Island, where copies were manufactured based on the originals from Noyes

Chapman. It is unknown how the puzzle made its way from Watch Hill to Hartford, or further still to Boston, where Matthias Rice began manufacturing and selling the puzzle in earnest [12].

Rice began making a verson of the puzzle he referred to as “The Gem Puzzle” in

December 1879 in his woodworking shop in Boston, Massachusetts. He sold the puzzles for 75 cents, and the idea was picked up by other manufacturers when the puzzle proved to be popular. The craze spread outward, west to the rest of America and east into Europe,

Asia, and Australia, where it remained popular until as late as June 1880. It was hypothe- sized that players would go mad from working the puzzle, although doctors discredit that 5

notion as those who went insane had demonstrated evidence of the condition prior to expo- sure to the 15 Puzzle. The trouble, of course, stemmed from the fact that the puzzle is not always solvable, an assumption that the average player made. Rewards were offered for a solution to the problem of switching the 14 block and the 15 block. It took a rigorous proof using group theory to show that these people need not have bothered: switching the 14 and

15 blocks creates a puzzle which is unsolvable. This proof and another, more recent proof will be demonstrated in the next chapter [12].

The 15 Puzzle opened up a market for similar puzzles, many with letters, or dif- ferent dimensions, etc. A three-dimensional puzzle was even constructed and patented by

Charles I. Rice in 1889. The 15 Puzzle left its impact on the world, making way for the

Rubik’s Cube craze of the 1980s. But for those who prefer a simple pastime that is easily

fit into one’s pocket, the 15 Puzzle is still available for sale. For the technologically savvy, computer programs and internet sites offer the puzzle, although only in solvable configura- tions. And, naturally, mathematicians, being an obsessive lot, still play with the puzzle and the mathematical thoughts it provokes [12]. CHAPTER TWO

MATHEMATICS AND THE 15 PUZZLE

It should come as little surprise that, while the general population obsessed over the

15 Puzzle, so too did mathematicians. The puzzle is solved when the blocks are in numer- ical order, left to right, and top to bottom. The first proofs of possible (solvable) positions and impossible (unsolvable) positions appeared in early 1880. The proof most often stated as the first is that of William Woolsey Johnson concerning the fact that the odd are impossible. A permutation is the product of transpositions. A transposition is when two blocks, from anywhere in the puzzle are switched, while leaving the rest alone, often by taking those two blocks straight up out of the tray and switching them. Johnson’s article was published in the December 1879 issue of the American Journal of Mathematics [14], but was not actually released until April 1880 because the journal was running late at the time. The American Journal of Mathematics published Johnson’s proof alongside another proof by William E. Storey, who proved that all even permutations are solvable positions.

Even permutations are those that are the product of an even number of transpositions, and odd permutations are those that are the product of an odd number of transpositions [12].

In general, mathematics from the 1800s is difficult to read and understand. How- ever, Jerry Slocum and Dic Sonneveld have given a brief sketch of Johnson’s proof in their book entitled The 15 Puzzle:How It Drove the World Crazy [12]. In their proof, they use

6 7

the example permutation

(1, 9)(2, 13, 10, 7, 15, 6, 4, 8, 16, 5, 11, 3)(12)(14) =

(1, 9)(2, 13, 10, 7, 15, 6, 4, 8, 16, 5, 11, 3), which uses 16 to represent the empty space. The above notation is standard notation for a permutation, where the 1 goes to the 9 and vice versa, while the 2 goes to 13, 13 goes to 10, 10 goes to 7, and so on until we get to the 3, which goes back to 2. Note that we can leave out the 12 and 14 because they remain in their solved positions in the mixed-up puzzle [12].

Johnson first splits the frame of the 15 Puzzle into 16 cells, numbered from left to right and top to bottom. The 15 blocks are then placed into the frame in any way that the player wishes. This creates a permutation that would be required to put the blocks back into the solved position. From there, the permutation of the blocks mentioned above is represented cyclically [12].

Next, Johnson considers what would happen if any two blocks were switched. If the two switched pieces are in different cycles, the cycles will be joined together into a single cycle. If the two blocks are in the same cycle, the switch will result in two sep- arate cycles. So, let the switch be between 7 and 8. Then, the permutation becomes

(1, 9)(2, 13, 10, 8, 16, 5, 11, 3)(4, 7, 15, 6), by multipying the original permutation by the transposition (7, 8), as is demonstated in Figure 3 [12].

Then, Johnson defines permutations as either even or odd depending on whether they are the product of an even or odd number of transpositions. In the original example, 8

Figure 2: Pictoral representation of the above permutation.

(1, 9)(2, 13, 10, 7, 15, 6, 4, 8, 16, 5, 11, 3) is even since it has 12 transpositions, the single transposition out front and the 11 in the longer cycle since the longer cycle has length 12, as well as the two identity cycles (12) and (14). In comparison, the permutation after 7 and 8 have been switched is odd, since it is the product of three odd permutations and the two identity cycles. Thus, to get from the orginal example to the modified one, we need an odd permutation, since the sum of an even number and an odd number is again odd [12].

Finally, Johnson points out that legal moves consist of switching the , or in this case the 16 block, with any one of the blocks that are adjacent to it. Now, for the blank space to begin and end its journey in the lower right-hand corner, it has no choice but 9

Figure 3: The permutation multiplied by the transposition (7, 8)

to have moved an even number of times. But this points out that it is impossible to get from

the original (even) example to the modified (odd) example. Thus, Johnson has proved that

it is impossible to get from an even position, created by permuting the blocks an even num-

ber of times, to an odd position, created by permuting the blocks an odd number of times.

Since the solved position, represented by (1), is an even permutation, it is impossible to solve a puzzle that is in an odd position [12].

After this first rigorous proof of the 15 Puzzle, there were many rehashings of the problem. Some were more complicated and others were more simple and concise. One of these proofs for the original 15 Puzzle was completed by Edward L. Spitznagel, Jr. and published in 1967. The proof centers around the path taken by the empty space as the num- 10

bered tiles are moved around the puzzle [13].

Spitznagel splits his proof into two main pieces, the unattainable configurations and the attainable ones, where configurations refers to the positions of the tiles in relation to one another [13].

The unattainable configurations are those that require an odd permutation of the blocks to return to the solved position, i.e., picking them up out of the tray and placing them in their solved positions. Spitznagel makes this point with the example in Figure 4.

A combination of the paths in Figure 4 is also presented in Figure 5, a reproduction from

Figure 4: Two odd length cycles inside the 15 Puzzle.

Spitznagel’s article.

Permutations that are caused by any of the above simple circuits are cyclic, and the

product of an even number of transpositions. Also, the circuits of the blank space have odd 11

Figure 5: The two cycles, put together. length, switching the blocks an even number of times. This makes them even permutations.

Therefore, any product of these permutations will be even, making it impossible to ever get an odd permutation [13].

The proof of attainable positions for the 15 Puzzle that Spitznagel gives is some-

what based on the fact that A15 has no proper normal subgroups. This property of An will be used later to prove possible configurations in variations of the 15 Puzzle [13].

The first step of the proof is to show that any three blocks that are lined up either horizontally or vertically can be permuted cyclically, while the remainder of the puzzle can be returned to its original position. Spitznagel uses the illustration in Figure 6 to make his point.

It can be observed that the top three blocks have been permuted cyclically, going 12

Figure 6: Permuting 3 blocks cyclicly.

from (a, b, c) to (b, c, a), while the bottom row returns to its original position. The next step

is to show that no matter where the three blocks occur in the puzzle, as long as they are

lined up, the same effect can be produced [13].

However, it is important to notice that these moves are 3-cycles, meaning they are

permutations of length 3. Since 3 is an odd number, the 3-cycles are actually even per-

mutations. Thus, the attainable configurations must contain all the 3-cycles. The group of

attainable positions is a subgroup of A15. Spitznagel uses the cycle (1, 2, 3) in particular.

He also states that since every conjugate of (1, 2, 3) is in this subgroup, and this subgroup

is normal in A15, then the subgroup must be A15 itself. Thus, the attainable permutations of the 15 Puzzle are represented by A15 [13]. This is the most common way of showing that An is simple, i.e., it has no proper, normal subgroups. 13

Shortly after Spitznagel’s proof was published, Richard M. Wilson proved his the- orem linking the 15 Puzzle to graph theory [16]. The trick is to put the puzzle in the solved position, and create a graph with each block or the empty space as a vertex and edges be- tween adjacent blocks or spaces. Wilson said that for puzzles of this nature, the possible positions were represented by An if a bipartite graph could be formed, and Sn if it was impossible to create such a bipartite graph. A bipartite graph is a graph where the vertices can be split into two disjoint sets such that no vertex is connected with an edge to another vertex in the same set.

From Figure 7, it can be seen that the possible positions are represented by A15, as

Figure 7: Bipartite graph of the 15 Puzzle.

has been proven before.

Now that the solvablity of even permutations and nonsolvability of odd permuta-

tions inside the 15 Puzzle has been demonstrated in several different ways, the next natural

question that arises is this: what about variations of the 15 Puzzle? The next chapter is

dedicated to a few of the many variations that have been toyed with over the last 130 years. CHAPTER THREE

VARIATIONS OF THE 15 PUZZLE

From the previous chapter, we have seen in several proofs that the possible moves inside the 15 Puzzle are represented by the collection of even permutations in S15, namely

A15. If we have any rectangular puzzle, with dimensions m×n and mn−1 blocks, then the possible moves are represented by Amn−1 [9]. The question now becomes, does this hold if we change the puzzle slightly? We will use a few of the techniques used by the experts mentioned previously to answer this question.

We want to discern whether the possible moves in the puzzle displayed in Figure

8 are represented by A15, S15, or yet another group. First, let moves be the act of sliding a block into the empty space, and let transpositions be a switch of any two blocks, leaving the rest alone. We know that if we leave off the top blank space, we will get all the permutations in A15. So, certainly, we have at least A15. It remains to be seen that we will also have S15, and we can show that we have the entire group if we can produce a single transposition of blocks in the puzzle, i.e., if we can switch some pair of blocks and leave the remaining blocks alone. Adding a single transposition to A15 generates S15 because A15 is a maximal subgroup of Sn. This can be done with the puzzle in question, using the steps in Figure 9.

Thus, we can observe that a single transposition has been created, in particular,

(1, 2), since those two pieces have been switched and all the other blocks retain their posi-

14 15

Figure 8: Starting position for the first variation. tions. Therefore, it can be stated with confidence that the possible moves on this puzzle are represented by the permutations in S15, and the puzzle is solvable, regardless of the starting position.

Of course, there is a reason why this is possible. The puzzle actually contains two blank spaces. If we could distinguish between them, we would see that they actually switch places with one another, as seen in Figure 10.

Thus, we have actually generated two transpostions, but since the blank spaces are essentially identical, the player does not notice and the puzzle is solvable for all starting points, i.e., the entire group S15 is possible inside the 15 Puzzle.

Now, we consider a different puzzle entirely, which can be seen in Figure 11. Let 16

Figure 9: Creating a single transposition inside the variation.

Figure 10: Showing the switch of the blank spaces. 17

the orange square be a block that is free to move, but must return to the same position in which it began.

Figure 11: Puzzle with a fixed block.

We want to see if the possible moves in this puzzle are represented by S14, A14, or some other group of permutations. To do this, we shall continue in a more general way, and then use the general statement to say something about the possible moves.

Theorem 1. Let G = An be the group of even permutations acting on the set S =

{1, 2, 3, ..., n}. If H = StabG(a), where a is a member of S, then H is isomorphic to

An−1.

Before we continue on the proof, we should list the results that will be used. It 18

should be mentioned that some of these results depend on H from Theorem 1 being a sub- group of Sn and |Sn : H| = 2. Proofs of the first three lemmas appear in [2].

Lemma 1. The group An is a simple group for n ≥ 5.

Lemma 2. If H is a subgroup of G and |G : H| = 2, then H is a normal subgroup in G.

Lemma 3. Let G be a finite group acting on S.

a. The orbits of S (under the action of G) partition S.

b. For any x ∈ S, |Gx| = [G : Gx].

|H||K| Lemma 4. If H and K are subgroups of G, then |HK| = |H∩K| .

Proof. Suppose we have a group G, with subgroups H and K. Define S to be the set

of left cosets of K, i.e., S = {gK|g ∈ G}. Then, define a of H on S by setting a · (xK) = axK for all a ∈ H and x ∈ G. If we consider the orbit O(K) of the coset of K, we have that O(K) = {hK|h ∈ H}. Also, |O(K)| = [H : StabH(K)]. Since

StabH(K) = {a ∈ H|aK = K} = H ∩ K, note that |StabH(K)| = |H ∩ K|. Now, there are

|H| ` = |H ∩ K| distinct cosets since the orbits of K partition of the set, and we shall denote the distinct

cosets h1K, h2K . . . , h`K. If we take the disjoint union of them, we get that

` [ HK = hiK. i=1 19

Finally, since each coset is disjoint from the other cosets and they each have size |K|,

` X |HK| = |K| = `|K|. i=1

Thus, |H||K| |HK| = . |H ∩ K|

Lemma 5. If H is a subgroup of Sn and H is not contained in An, then Sn = HAn.

Proof. Suppose H 6⊆ An. Since An ⊆ Sn, we know that AnH ⊆ Sn. However, An is

a maximal subgroup of Sn because [Sn : An] = 2. This means that if H 6⊆ An, then

An ( HAn and so HAn = Sn.

Lemma 6. If H is a subgroup of Sn and H is not contained in An, then |An : H ∩An| = 2.

This is a direct result of the two previous lemmas.

∼ Now, we will use these results to prove that H = An−1.

Proof. (Proof of Theorem 1) First, note that Sn and An are groups of permutations of elements of the set S, where a is any element of S. Now, we know that G = An acts transitively on S from [2], i.e., for each x, y ∈ S, there exists g ∈ G such that gx = y.

Thus, the size of the orbit of a is |O(a)| = n, which is the size of S. Now because

|A | |O(a)| = |A : H| = n , n |H|

we can say |A | |A | 1 n! 1 |H| = n = n = = (n − 1)! = |A |. |O(a)| n 2 n 2 n−1 20

Thus, |H| = |An−1|. So, both H and An−1 are in Sn−1 and, because they have the same size,

|Sn−1 : H| = |Sn−1 : An−1| = 2.

This means that both H and An−1 are normal in Sn−1. Also, An−1 is simple for n ≥ 6 and so has no proper nontrivial normal subgroups. If H ⊆ An−1, it is clear that H = An−1, since |H| = |An−1|. So, assume that H 6⊆ An−1. Because both H and An−1 are normal in Sn−1, and An−1 is maximal in Sn−1, it is possible to say that Sn−1 = An−1H. Finally, since H and An−1 are subgroups of Sn−1, and all of these groups are finite,

|An−1||H| |Sn−1| = |An−1H| = and |Sn−1| = 2|An−1|. |An−1 ∩ H|

Putting these two equations together yields

|An−1||H| = 2|An−1|, |An−1 ∩ H| |A | n−1 = 2, |An−1 ∩ H|

and

|An−1 : An−1 ∩ H| = 2.

However, this says that An−1 ∩ H is normal in An−1, which is a contradiction since An−1

∼ is simple. Thus, An−1 = H.

Therefore, in the puzzle we are considering, if we stabilize the sixth block in the

original 15 Puzzle, the possible moves are represented by the permutations in A14. It should

be noted that the choice of the fixed block in completely arbitrary and does not affect the 21

proof at all.

Finally, consider the same puzzle, but with the sixth block being unable to move from its position. The simplest way to consider such a puzzle is to use Richard Wilson’s

Theorem regarding block puzzles such as the 15 Puzzle [16]. In simplest terms, the theo- rem states that the possible moves are represented by S14 if the previously defined graph is

not bipartite and represented by A14 if the graph is bipartite. Figure 12 is made by drawing lines between blocks that are adjacent in the solved position, and shows that a bipartite graph can be obtained from the puzzle.

Thus, by Wilson’s Theorem, the possible moves are represented by the permuta-

Figure 12: Bipartite graph of a puzzle with a glued-down block.

tions in the group A14.

These variations are just a few of the many posibilities that could be studied using

various areas of mathematics such as group theory and graph theory. CONCLUSION

Throughout this discussion of the 15 Puzzle, we have covered a great deal of mate- rial. We began with the history of the 15 Puzzle, delved into the mathematics of those who came before, and added some small insights to various deviations of the puzzle. However, this is far from the end of the 15 Puzzle.

The 15 Puzzle has excited mathematicians for years. Its life as a mathematical ob- ject has stretched from 1879 to the present day. Mathematicians continue to publish papers on the 15 Puzzle, as well as use it as an example for more complicated and intricate math- ematical thoughts.

Perhaps even more exciting is the fact that theoretical mathematics is not the only area that uses the 15 Puzzle. Psychologists use it to conduct experiments on memory and cognition [17]. Also, because of the parity in the puzzle, it is possible to use it in simple cryptography [11].

Thus, though the puzzle may be a quiet sort of object that causes little com- motion anymore, it is far from dead. Its applications are numerous and varied, and give cause to study the 15 Puzzle for many years to come.

22 BIBLIOGRAPHY

[1] A.F. Archer, A modern treatment of the 15 puzzle, The Amer. Math. Monthly 106

(1999), 793-799.

[2] John A. Beachy and William D. Blair, Abstract Algebra, Third Edition, Waveland

Press, 2006.

[3] A.L. Davies, Rotating the fifteen puzzle, Math. Gazette 54 (1970), 237-240.

[4] A. Fink and R. Guy, Rick’s tricky six puzzle: S5 sits specially in S6, Math. Mag. 82

(2009), 83-102.

[5] L.E. Horden, Sliding Piece Puzzles, Oxford Univ. Press, 1986.

[6] S. Hurd and D. Trautman, The knight’s tour on the 15-puzzle, Math. Mag. 66 (1993),

159-166.

[7] D. Joyner, Adventures in Group Theory: Rubik’s Cube, Merlin’s machine, and other

mathematical toys, 2nd Edition, Johns Hopkins University Press, Baltimore, 2008.

[8] S. Kim, The 15 puzzle, Discover, Vol. 27, Issue 7, 2006, (72-78).

[9] H. Leibeck, Some generalizations on the 14-15 puzzle, Math. Mag. 44 (1971), 185-

189.

[10] H.V. Mallison, An array of squares, Math. Gazette 24 (1940), 119-121.

23 24

[11] T. Mizuki, Y. Kugimoto, and H. Sone, Secure Multiparty Computations Using the

15 Puzzle, Combinatorial Optimization and Applications: First International Con-

ference, COCOA 2007, Xi’an, China August 14-16, 2007: proceedings, 255-266.

[12] Jerry Slocum and Dic Sonneveld, The 15 Puzzle: How it Drove the World Crazy. The

Puzzle that Started the Craze of 1880. How America’s Greatest Puzzle Designer, Sam

Loyd, Fooled Everyone for 115 Years, Slocum Puzzle Foundation, Beverly Hills, CA,

2006.

[13] E.L. Spitznagel, Jr., A new look at the 15-Puzzle, Math. Mag. 40 (1967), 171-174.

[14] W.E. Storey, Notes on the 15 Puzzle, II, Amer. J. Math. 2 (1879), 399-404.

[15] S.K. Stein, Mathematics: the Man-Made Universe, W. H. Freeman and Company,

San Francisco, CA, 1963.

[16] Richard M. Wilson, Graph Puzzles, homotopy, and the alternating group, J. Combin.

Theory Ser. B 16 (1974), 86-96.

[17] P. Zygmunt, Solving combinatorial problems: the 15 puzzle, Memory and Cognition

33 (2005), 1069-1084.