Auslander, Joel

Benjamin, Arthur T.

Wilkerson, Daniel S.

Scholarship@Claremont

Claremont CollegesScholarship @ ClaremontAll HMC Faculty Publications and Research HMC Faculty Scholarship2-1-1993Optimal LeapfroggingJoel AuslanderHarvey Mudd CollegeArthur T. BenjaminHarvey Mudd CollegeDaniel S. WilkersonHarvey Mudd CollegeThis Article is brought to you for free and open access by the HMC Faculty Scholarship at Scholarship @ Claremont. It has been accepted for inclusionin All HMC Faculty Publications and Research by an authorized administrator of Scholarship @ Claremont. For more information, please contactscholarship@cuc.claremont.edu.Recommended CitationAuslander, Joel, Arthur T. Benjamin and Daniel S. Wilkerson. "Optimal Leapfrogging." Mathematics Magazine, Vol. 66, No. 1,pp.14-19, February, 1993.Optimal LeapfroggingAuthor(s): Joel Auslander, Arthur T. Benjamin and Daniel Shawcross WilkersonSource: Mathematics Magazine, Vol. 66, No. 1 (Feb., 1993), pp. 14-19Published by: Mathematical Association of AmericaStable URL: http://www.jstor.org/stable/2690465 .Accessed: 10/06/2013 19:06Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .http://www.jstor.org/page/info/about/policies/terms.jsp .JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range ofcontent in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact support@jstor.org. .Mathematical Association of America is collaborating with JSTOR to digitize, preserve and extend access toMathematics Magazine.http://www.jstor.org This content downloaded from 134.173.130.46 on Mon, 10 Jun 2013 19:06:37 PMAll use subject to JSTOR Terms and Conditions14 MATHEMATICS MAGAZINE Optimal Leapfrogging JOEL AUSLANDER Harvard University Cambridge, MA 02138 ARTHUR T. BENJAMIN Harvey Mudd College Claremont, CA 91711 DANIEL SHAWCROSS WILKERSON University of.California Berkeley, CA 94720 This article arose from trying to determine the fastest way of moving checkers from the lower left-hand corner of a Go board to the upper right-hand corner, with no opponent in the way. The pieces move in "Chinese checkers" fashion by shifting or jumping in a way we soon illustrate and later describe precisely. Our goal is to move our pieces from a prescribed origin position to a prescribed destination position in as few turns as possible. We regard our Go board as a subset of the integer-point lattice, Z2. Suppose we have four indistinguishable checkers initially situated at the points (0,0), (1,0), (0, 1), and (1, 1), and we wish to move them to the points (17,17), (17,18), (18,17), and (18,18). See FIGURE 1. One might begin by maneuvering into the snake configuration {(O, 0), (1, 1), (2,2), (3,3)). (This can be done in four moves: for example, a hop, (0, 1) > (2, 1); a shift, (2, 1) -- (2,2); a two-hop jump (1, 0) -- (1, 2) -> (3,2); and a shift (3,2) (3,3).) Then apply the following three-move procedure: Shift the bottom piece (at (0,0)) to the right (to (1,0)), then triple-hop that piece (to (3,4)), then shift that piece to the right (to (4,4)). The pieces end up in the snake configuration (starting at (1, 1)), and the same three-move procedure can be applied. If we apply this procedure 15 times, we reach {(15,15),(16,16),(17,17),(18,18)). Four moves later, we will have reached our destination using 4 + (15 X 3) + 4 = 53 moves alto- gether. However, a much faster trajectory exists. In one move, hop the piece at (1,0) to (1,2) reaching the serpent configuration {(0, 0), (0, 1), (1, 1), (1,2)). See FIGURE 2. Then apply the following two-move procedure: Double-hop the bottom piece (from (0,0) to (2,2)), then double-hop the left-most piece (from (0, 1) to (2,3)). Once again, the pieces end up in the "serpent configuration" (starting at (1, 1)), and the same two-move procedure can be applied. If we apply this procedure 16 times, we reach {(16,16), (16,17), (17,17), (17,18)). Two moves later, we will have reached our desti- nation using only 35 moves. The second trajectory is faster because the serpent configuration requires only two moves to translate itself in the direction (1, 1), a feat requiring three moves by the snake configuration. In this article, we characterize the speediest configurations for the above game (played in n dimensions) and thereby prove that the second trajectory is in fact optimal. Precisely, we consider the problem of efficiently moving a collection of p indistin- guishable pieces over the integer lattice Zn. The movement rules are analogous to those of Chinese checkers, and are as follows. At all times, pieces occupy distinct points in Zn. At each move, exactly one piece is displaced. If a piece is situated at the point x = (x1 ... . X) E Zn and, for some i E {1, . . ., n), the point x + ei is unoccupied (where ei is the i-th unit vector), then the piece may shift there; similarly for x - e. This content downloaded from 134.173.130.46 on Mon, 10 Jun 2013 19:06:37 PMAll use subject to JSTOR Terms and ConditionsVOL. 66, NO. 1, FEBRUARY 1993 15 If x + e, is occupied, but x + 2ei is not, then the piece can hop over the occupant of x + ei to arrive at x + 2ei, where it may either remain or hop over another adjacent piece, etc. (Similarly for a hop over x - ei to x - 2ei. A move consists of either a shift or a jump (a sequence of one or more hops by a single piece). 18-- - - - - - - -_ _ _ _ _ _ _ _ - - - Goal 3 2 1 0 l 0 12 3 18 FIGURE 1 Solitaire Chinese checkers on a Go board. FIGURE 2 Serpent movement. We begin with some definitions. A placement of pieces is a size p subset of Zn, usually denoted by X = {x1,.. )x ). We define the centroid of a placement X to be c(X) = - xu 1~ ~ ~ ~~ = which is a vector in - Zn. For placements X and Y, the displacement from X to Y is p defined by n d(X, Y) _ _(c_(y) -c (X)) where ci(Y) - ci(X) is the i-th component of c(Y) - c(X). Loosely, displacement measures the distance between placements, where the directions ei are viewed as positive directions. Note that d can be negative, or nontrivially zero. For m ? 1, an in-move trajectory X0, X1).. )XM is a sequence of placements where X.+ 1 is reach- able from Xv, in a single move. The speed of an in-move trajectory from X to Y is defiGnedl as d_(X Y) /m. This content downloaded from 134.173.130.46 on Mon, 10 Jun 2013 19:06:37 PMAll use subject to JSTOR Terms and Conditions1 6 MATHEMATICS MAGAZINE THEOREM 1. Any trajectory speed is < 2 - 2/p, where p ? 2 is the number of pieces. When p = 1, the speed is bounded by 1. Proof. From the definition, the speed of the trajectory X0, X1, .. ., X77 is the average of the speeds of its m moves. Each of these is of the form d(X;, X1V+)/1. If p = 1, the move is a shift and the displacement is 1, provided the shift is positive. If p > 2, the move is either a shift or a jump of at most p - 1 hops. The centroid is displaced by at most (2 p - 2)/p, the maximum being attained by p - 1 positive hops. Note that this maximum speed is not sustainable: There are various configurations in which a "long-jump" of p - 1 hops can be made, but the next move can be another long-jump only in the rather trivial one-dimensional case with p = 2. This is a special case of our next theorem, that a "repeatable" trajectory has speed at most 1, which we will call the speed of light. We say that placement Y is a translate of X if there exists a E Z " such that Y = X + a, i.e., {y1 . ., yp} = {xl +a, ... , xP + a}. Such placements X and Y are said to be represented by the same configuration. For x E Z', define lxii to be Ein= x , and for all integers M, let the border M be {x E Zn: lix II = M}. Define the tail and head of a placement X as t(X) = miniix1Ji, h(X) = maxilx,jil. it U THEOREM 2. Let Y be a translate of X. Then any trajectory from X to Y has speed at most 1. Proof. Suppose Y = X + a for some a E Z7l. For ease of notation, assume that Yi = xi+ a, i = 1,. . I p, whence n n p d(X,Y)= (ci(Y) - ci(X))= - a ai lall. i= 1 i=1 u=1 Next we observe that the tail (and the head) cannot increase by more than 1 after each move. Therefore, since t(Y) = min%,1II(xi + a)II = minP=111xill + Ilaill = t(X) + Ilall, it follows that the number of moves m needed for a trajectory from X to Y is at least lall = d(X, Y). If d(X, Y) < 0, then since m> 1, the trajectory has nonpositive speed. Otherwise, since m ? d(X, Y), its speed is d(X, Y)/m < 1. A placement X is called a speed-of-light placement if there exists a nonzero vector a e Zn and a speed one trajectory (called a speed-of-light trajectory) from X to X + a. In FIGURE 3, we illustrate speed-of-light configurations for the two-dimen- sional case (where p = 1, 2, and 4, respectively). In fact, the next theorem demon- strates that these are the only such configurations intwo dimensions, and essentially the only ones for higher dimensions, too. THEOREM 3. Thefollowing are speed-of-light configurations: The atom {x} (when p = 1), the frog {x,x+ei} 1 <i <n (when p =2), and the serpent {x, x + ei, x + ei + ej, x + 2ei + ej} 1 <i =j < n (when p =4). No other speed-of-light configurations exist. The first part of the theorem is straightforward. The atom can translate itself (in the direction eP) by shifting itself from {x} to {x + etj, a speed one maneuver when This content downloaded from 134.173.130.46 on Mon, 10 Jun 2013 19:06:37 PMAll use subject to JSTOR Terms and ConditionsVOL. 66, NO. 1, FEBRUARY 1993 17 p = 1. The frog {x, x + ei) translates itself (in the direction ei) in a single hop to {x + ei, x + 2ei), a speed-one maneuver when p = 2. When p = 4, the serpent per- forms two consecutive double-hops to go from {x, x + ei, x + ei + ej, x + 2ei + ej) to {x + ei + ej, x + 2ei + ej, x + 2ei + 2ej, x + 3ei + 2ej), translating itself in the direc- tion ei + ei in two moves. We establish the second part of the theorem by proving a series of necessary conditions that must be satisfied by speed-of-light objects. LEMMA 1. Every move in a speed-of-light trajectory must simultaneously increase the values of the back border and the front border. Hence a speed-of-light placement X contains a unique piece on border t(X), and a speed-of-light move must "jump" that piece to a point on border h(X) + 1. Proof. As argued in proving Theorem 2, an m-move trajectory from X to X + a has speed hlall/m, where m llall. Note that t(X + a) = t(X) + Ilall, h(X + a) = h(X) + Ilall. We observe (as in the proof of Theorem 2) that the functions t and h cannot increase by more than 1 each move. Hence, in order for m = Ilall, we must simultaneously increase the values of both borders each move. LEMMA 2. Given a speed-of-light placement X and t(X) < M < h(X), there is at most one occupied point x E X with lix II = M. Proof. Suppose, to the contrary, that more than one piece is situated on border M. By Lemma 1, the first M - t(X) moves of the trajectory involve moving pieces from borders with values less than M to borders with values greater than M, after which our new back border has value M. But then this border has more than one piece, contradicting Lemma 1. LEMMA 3. When p ? 2, every move in a speed-of-light trajectory is a jump. Proof. Since p 2 2, we have h(X) > t(X) by Lemma 1. Since a shift can not take a back border piece beyond border t(X) + 1 < h(X), it can not expand the front border, as required. Notice that in a speed-of-light trajectory, for a piece on border M to make "forward progress," it must hop over a piece on border M + 1 and land on border M + 2. It follows from Lemma 3 that every speed-of-light placement X has at least one piece on every border between t(X) and h(X). Thus, we have LEMMA 4. Every speed-of-light placement X must have exactly one piece on each border between border t(X) and border h(X). Consequently, h(X) = t(X) + p - 1. LEMMA 5. If X is a speed-of-light placement with p 2 2 pieces, then p must be even. Proof. By Lemmas 1 and 4, the first move must jump a piece from border t(X) to border t(X) + p. Since a jump changes the border value by an even number, p must be even. 0 @~00 0 0 0.0 Atom Frogs Serpents FIGURE 3 Speed-of-light configurations. This content downloaded from 134.173.130.46 on Mon, 10 Jun 2013 19:06:37 PMAll use subject to JSTOR Terms and Conditions18 MATHEMATICS MAGAZINE Proof of Theorem 3. By Lemma 3, when p = 2, all speed-of-light configurations must be of the form {x, x + ei1 for some 1 < i < n, i.e., the frogs. By Lemma 5, p = 3. We can restrict our attention to the case where p ? 4. It remains to prove that the only possible remaining speed-of-light configurations are of the serpent variety. Imagine that we have a speed-of-light trajectory that makes one move every second. Suppose the front border piece of our speed-of-light placement X is presently (t = 0) situated at x E Zn, with lix II = M. We shall focus our attention on only those points in Zn that occupy borders of value M or higher. Thus at t = 0, all that we see is a single piece, situated at x (see FIGURE 4). When t = 1, the front border's value has increased to M + 1. The new front border piece must have made its final hop over x to land on the point x + ei for some 1 < i < n. See FIGURE 5. I I _ t - 1- xT\ I I IL M FIGURE 4 What we see when t = 0. I I I I \::+\I I I I I x x+ei I I] I I M+ I M FIGURE 5 What we see when t = 1. When t = 2, the front border's value has increased to M + 2. The piece that landed there had to make its final hop over x + ei. Hence the piece on border M + 2 must be situated at x + ei + e1 for some 1 <j < n. Observe that i Oj, for otherwise the piece at x would have hopped over x + ei to point x + 2ei thus leaving no piece with border value M, contradicting Lemma 4. Thus, at t = 2, we see three pieces, situated at x, x + ei, x + ei + ej, where 1 <i j< n. (See FIGURE 6.) When t = 3, a piece is jumped to the new front border M + 3 and, since it had to hop over the piece at x+ ei + ej, it must end up at x+ ei + e1 + ek for some 1 < k < n. We know that k j by the same argument as i =#j above. we now show that, in fact, k = i. The new front border piece, before it made its final hop over x+ ei + e. to x+ ei + e.+ ek, must have been at the point x+ ei + e -ek on border M + 1. But how did it get there? It had to hop over the sole piece on border M, situated at x. But this requires ei + e- ek to be a unit vector, which is only possible when k = i or k =j. And since k #j, we have k = i. Hence at t = 3, we see four pieces, situated at x, x + ei, x + ei + ej, and x + 2ei + ej, as in FIGURE 7. When t = 4, the back border piece, wherever it is, jumps to border M + 4, landing on x + 2ei + e. + ek for some 1 < k < n. By the argument in the preceding paragraph, the hop over the piece at x + 2ei + ei had to come from the point x + 2ei reached by a hop over x + ei, and so that hop came from x. Hence the jump originated at x. Thus p = 4, and our current configuration (as well as our original one) must have been a serpent. This content downloaded from 134.173.130.46 on Mon, 10 Jun 2013 19:06:37 PMAll use subject to JSTOR Terms and ConditionsVOL. 66, NO. 1, FEBRUARY 1993 19 I X I _ x ,+ e \ e\ x x+ei I1 \ M+2 M+ 1 M FIGURE 6 What we see when t = 2. I ~~~~~~~x + 2e, + ej x + e2 e ie+ \ I Xx+e,+ej I x x + e M+3 M FIGURE 7 What we see when t = 3. Returning to the problem at the outset of the article, we see that the 35-move trajectory must be optimal because the original configuration is translated a distance of 34 units, and the square configuration is not a speed-of-light configuration. As a consequence of this theorem, we see that no speed-of-light configurations exist when the number of pieces is three or greater than four. In these cases, it is easy to create configurations that are translatable with speed 2/3 (e.g., the snake configura- tion with p pieces). The question of whether the speed 2/3 is optimal for the two- (and higher-) dimensional problem, when p = 3 or p > 4, remains open. More specifically, it remains unknown how to optimally translate six or nine pieces (arranged in a triangle or square) from the lower left-hand corner to the upper right-hand corner of the Go board. Acknowledgments. We dedicate this paper to the Summer Studies in Mathematics program at Hamp- shire College, directed by Professor David Kelly, where this research took place. HCSSiM is a mathematics program for high-ability high school students. During the 1986 program, Benjamin led a week-long, open-ended course, where students explored properties of jumping games. Two students, Auslander and Wilkerson, proved the two-dimensional version of this theorem and presented it at an MAA section meeting in Baltimore in 1986. We also thank Professor Alan J. Goldman of Johns Hopkins University for suggesting this problem to us, Greg Levin for assistance with the figures, and the anonymous referees for many valuable suggestions. REFERENCE Readers may also enjoy Wheels, Life and Other Mathematical Amusements by Martin Gardner (W. H. Freeman and Co., 1983), particularly the chapters analyzing the games of "Halma" and "Life." This content downloaded from 134.173.130.46 on Mon, 10 Jun 2013 19:06:37 PMAll use subject to JSTOR Terms and Conditions

Optimal Leapfrogging

Keck Graduate Institute

Claremont Colleges
Scholarship @ Claremont
All HMC Faculty Publications and Research HMC Faculty Scholarship
2-1-1993
Optimal Leapfrogging
Joel Auslander
Harvey Mudd College
Arthur T. Benjamin
Harvey Mudd College
Daniel S. Wilkerson
Harvey Mudd College
This Article is brought to you for free and open access by the HMC Faculty Scholarship at Scholarship @ Claremont. It has been accepted for inclusion
in All HMC Faculty Publications and Research by an authorized administrator of Scholarship @ Claremont. For more information, please contact
scholarship@cuc.claremont.edu.
Recommended Citation
Auslander, Joel, Arthur T. Benjamin and Daniel S. Wilkerson. "Optimal Leapfrogging." Mathematics Magazine, Vol. 66, No. 1,
pp.14-19, February, 1993.
Optimal Leapfrogging
Author(s): Joel Auslander, Arthur T. Benjamin and Daniel Shawcross Wilkerson
Source: Mathematics Magazine, Vol. 66, No. 1 (Feb., 1993), pp. 14-19
Published by: Mathematical Association of America
Stable URL: http://www.jstor.org/stable/2690465 .
Accessed: 10/06/2013 19:06
Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .
http://www.jstor.org/page/info/about/policies/terms.jsp
 .
JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of
content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms
of scholarship. For more information about JSTOR, please contact support@jstor.org.
 .
Mathematical Association of America is collaborating with JSTOR to digitize, preserve and extend access to
Mathematics Magazine.
http://www.jstor.org 
This content downloaded from 134.173.130.46 on Mon, 10 Jun 2013 19:06:37 PM
All use subject to JSTOR Terms and Conditions
14 MATHEMATICS MAGAZINE 
Optimal Leapfrogging 
JOEL AUSLANDER 
Harvard University 
Cambridge, MA 02138 
ARTHUR T. BENJAMIN 
Harvey Mudd College 
Claremont, CA 91711 
DANIEL SHAWCROSS WILKERSON 
University of.California 
Berkeley, CA 94720 
This article arose from trying to determine the fastest way of moving checkers from 
the lower left-hand corner of a Go board to the upper right-hand corner, with no 
opponent in the way. The pieces move in "Chinese checkers" fashion by shifting or 
jumping in a way we soon illustrate and later describe precisely. Our goal is to move 
our pieces from a prescribed origin position to a prescribed destination position in as 
few turns as possible. 
We regard our Go board as a subset of the integer-point lattice, Z2. Suppose we 
have four indistinguishable checkers initially situated at the points (0,0), (1,0), (0, 1), 
and (1, 1), and we wish to move them to the points (17,17), (17,18), (18,17), and 
(18,18). See FIGURE 1. One might begin by maneuvering into the snake configuration 
{(O, 0), (1, 1), (2,2), (3,3)). (This can be done in four moves: for example, a hop, 
(0, 1) > (2, 1); a shift, (2, 1) -- (2,2); a two-hop jump (1, 0) -- (1, 2) -> (3,2); and a shift 
(3,2) (3,3).) Then apply the following three-move procedure: Shift the bottom 
piece (at (0,0)) to the right (to (1,0)), then triple-hop that piece (to (3,4)), then shift 
that piece to the right (to (4,4)). The pieces end up in the snake configuration 
(starting at (1, 1)), and the same three-move procedure can be applied. If we apply 
this procedure 15 times, we reach {(15,15),(16,16),(17,17),(18,18)). Four moves 
later, we will have reached our destination using 4 + (15 X 3) + 4 = 53 moves alto- 
gether. However, a much faster trajectory exists. In one move, hop the piece at (1,0) 
to (1,2) reaching the serpent configuration {(0, 0), (0, 1), (1, 1), (1,2)). See FIGURE 2. 
Then apply the following two-move procedure: Double-hop the bottom piece (from 
(0,0) to (2,2)), then double-hop the left-most piece (from (0, 1) to (2,3)). Once again, 
the pieces end up in the "serpent configuration" (starting at (1, 1)), and the same 
two-move procedure can be applied. If we apply this procedure 16 times, we reach 
{(16,16), (16,17), (17,17), (17,18)). Two moves later, we will have reached our desti- 
nation using only 35 moves. The second trajectory is faster because the serpent 
configuration requires only two moves to translate itself in the direction (1, 1), a feat 
requiring three moves by the snake configuration. In this article, we characterize the 
speediest configurations for the above game (played in n dimensions) and thereby 
prove that the second trajectory is in fact optimal. 
Precisely, we consider the problem of efficiently moving a collection of p indistin- 
guishable pieces over the integer lattice Zn. The movement rules are analogous to 
those of Chinese checkers, and are as follows. At all times, pieces occupy distinct 
points in Zn. At each move, exactly one piece is displaced. If a piece is situated at the 
point x = (x1 ... . X) E Zn and, for some i E {1, . . ., n), the point x + ei is unoccupied 
(where ei is the i-th unit vector), then the piece may shift there; similarly for x - e. 
This content downloaded from 134.173.130.46 on Mon, 10 Jun 2013 19:06:37 PM
All use subject to JSTOR Terms and Conditions
VOL. 66, NO. 1, FEBRUARY 1993 15 
If x + e, is occupied, but x + 2ei is not, then the piece can hop over the occupant of 
x + ei to arrive at x + 2ei, where it may either remain or hop over another adjacent 
piece, etc. (Similarly for a hop over x - ei to x - 2ei. A move consists of either a shift 
or a jump (a sequence of one or more hops by a single piece). 
18-- - - - - - - 
-_ _ _ _ _ _ _ _ - - - Goal 
3 
2 
1 
0 l 
0 12 3 18 
FIGURE 1 
Solitaire Chinese checkers on a Go board. 
FIGURE 2 
Serpent movement. 
We begin with some definitions. A placement of pieces is a size p subset of Zn, 
usually denoted by X = {x1,.. )x ). We define the centroid of a placement X to be 
c(X) = - xu 
1~ ~ ~ ~~ = 
which is a vector in - Zn. For placements X and Y, the displacement from X to Y is p 
defined by 
n 
d(X, Y) _ _(c_(y) -c (X)) 
where ci(Y) - ci(X) is the i-th component of c(Y) - c(X). Loosely, displacement 
measures the distance between placements, where the directions ei are viewed as 
positive directions. Note that d can be negative, or nontrivially zero. For m ? 1, an 
in-move trajectory X0, X1).. )XM is a sequence of placements where X.+ 1 is reach- 
able from Xv, in a single move. The speed of an in-move trajectory from X to Y is 
defiGnedl as d_(X Y) /m. 
This content downloaded from 134.173.130.46 on Mon, 10 Jun 2013 19:06:37 PM
All use subject to JSTOR Terms and Conditions
1 6 MATHEMATICS MAGAZINE 
THEOREM 1. Any trajectory speed is < 2 - 2/p, where p ? 2 is the number of 
pieces. When p = 1, the speed is bounded by 1. 
Proof. From the definition, the speed of the trajectory X0, X1, .. ., X77 is the 
average of the speeds of its m moves. Each of these is of the form d(X;, X1V+)/1. If 
p = 1, the move is a shift and the displacement is 1, provided the shift is positive. If 
p > 2, the move is either a shift or a jump of at most p - 1 hops. The centroid is 
displaced by at most (2 p - 2)/p, the maximum being attained by p - 1 positive hops. 
Note that this maximum speed is not sustainable: There are various configurations 
in which a "long-jump" of p - 1 hops can be made, but the next move can be another 
long-jump only in the rather trivial one-dimensional case with p = 2. This is a special 
case of our next theorem, that a "repeatable" trajectory has speed at most 1, which we 
will call the speed of light. 
We say that placement Y is a translate of X if there exists a E Z " such that 
Y = X + a, i.e., {y1 . ., yp} = {xl +a, ... , xP + a}. Such placements X and Y are said to 
be represented by the same configuration. For x E Z', define lxii to be Ein= x , and 
for all integers M, let the border M be {x E Zn: lix II = M}. Define the tail and head 
of a placement X as 
t(X) = miniix1Ji, h(X) = maxilx,jil. 
it U 
THEOREM 2. Let Y be a translate of X. Then any trajectory from X to Y has speed at 
most 1. 
Proof. Suppose Y = X + a for some a E Z7l. For ease of notation, assume that 
Yi = xi+ a, i = 1,. . I p, whence 
n n p 
d(X,Y)= (ci(Y) - ci(X))= - a ai lall. 
i= 1 i=1 u=1 
Next we observe that the tail (and the head) cannot increase by more than 1 after 
each move. Therefore, since t(Y) = min%,1II(xi + a)II = minP=111xill + Ilaill = 
t(X) + Ilall, it follows that the number of moves m needed for a trajectory from X to 
Y is at least lall = d(X, Y). If d(X, Y) < 0, then since m> 1, the trajectory has 
nonpositive speed. Otherwise, since m ? d(X, Y), its speed is d(X, Y)/m < 1. 
A placement X is called a speed-of-light placement if there exists a nonzero vector 
a e Zn and a speed one trajectory (called a speed-of-light trajectory) from X to 
X + a. In FIGURE 3, we illustrate speed-of-light configurations for the two-dimen- 
sional case (where p = 1, 2, and 4, respectively). In fact, the next theorem demon- 
strates that these are the only such configurations intwo dimensions, and essentially 
the only ones for higher dimensions, too. 
THEOREM 3. Thefollowing are speed-of-light configurations: 
The atom {x} (when p = 1), 
the frog {x,x+ei} 1 <i <n (when p =2), and 
the serpent {x, x + ei, x + ei + ej, x + 2ei + ej} 1 <i =j < n (when p =4). 
No other speed-of-light configurations exist. 
The first part of the theorem is straightforward. The atom can translate itself (in 
the direction eP) by shifting itself from {x} to {x + etj, a speed one maneuver when 
This content downloaded from 134.173.130.46 on Mon, 10 Jun 2013 19:06:37 PM
All use subject to JSTOR Terms and Conditions
VOL. 66, NO. 1, FEBRUARY 1993 17 
p = 1. The frog {x, x + ei) translates itself (in the direction ei) in a single hop to 
{x + ei, x + 2ei), a speed-one maneuver when p = 2. When p = 4, the serpent per- 
forms two consecutive double-hops to go from {x, x + ei, x + ei + ej, x + 2ei + ej) to 
{x + ei + ej, x + 2ei + ej, x + 2ei + 2ej, x + 3ei + 2ej), translating itself in the direc- 
tion ei + ei in two moves. 
We establish the second part of the theorem by proving a series of necessary 
conditions that must be satisfied by speed-of-light objects. 
LEMMA 1. Every move in a speed-of-light trajectory must simultaneously increase 
the values of the back border and the front border. Hence a speed-of-light placement X 
contains a unique piece on border t(X), and a speed-of-light move must "jump" that 
piece to a point on border h(X) + 1. 
Proof. As argued in proving Theorem 2, an m-move trajectory from X to X + a 
has speed hlall/m, where m llall. Note that t(X + a) = t(X) + Ilall, h(X + a) = 
h(X) + Ilall. We observe (as in the proof of Theorem 2) that the functions t and h 
cannot increase by more than 1 each move. Hence, in order for m = Ilall, we must 
simultaneously increase the values of both borders each move. 
LEMMA 2. Given a speed-of-light placement X and t(X) < M < h(X), there is at 
most one occupied point x E X with lix II = M. 
Proof. Suppose, to the contrary, that more than one piece is situated on border M. 
By Lemma 1, the first M - t(X) moves of the trajectory involve moving pieces from 
borders with values less than M to borders with values greater than M, after which 
our new back border has value M. But then this border has more than one piece, 
contradicting Lemma 1. 
LEMMA 3. When p ? 2, every move in a speed-of-light trajectory is a jump. 
Proof. Since p 2 2, we have h(X) > t(X) by Lemma 1. Since a shift can not take a 
back border piece beyond border t(X) + 1 < h(X), it can not expand the front 
border, as required. 
Notice that in a speed-of-light trajectory, for a piece on border M to make "forward 
progress," it must hop over a piece on border M + 1 and land on border M + 2. It 
follows from Lemma 3 that every speed-of-light placement X has at least one piece 
on every border between t(X) and h(X). 
Thus, we have 
LEMMA 4. Every speed-of-light placement X must have exactly one piece on each 
border between border t(X) and border h(X). Consequently, h(X) = t(X) + p - 1. 
LEMMA 5. If X is a speed-of-light placement with p 2 2 pieces, then p must be even. 
Proof. By Lemmas 1 and 4, the first move must jump a piece from border t(X) to 
border t(X) + p. Since a jump changes the border value by an even number, p must 
be even. 
0 @~00 0 0 0.0 
Atom Frogs Serpents 
FIGURE 3 
Speed-of-light configurations. 
This content downloaded from 134.173.130.46 on Mon, 10 Jun 2013 19:06:37 PM
All use subject to JSTOR Terms and Conditions
18 MATHEMATICS MAGAZINE 
Proof of Theorem 3. By Lemma 3, when p = 2, all speed-of-light configurations must 
be of the form {x, x + ei1 for some 1 < i < n, i.e., the frogs. By Lemma 5, p = 3. We 
can restrict our attention to the case where p ? 4. It remains to prove that the only 
possible remaining speed-of-light configurations are of the serpent variety. Imagine 
that we have a speed-of-light trajectory that makes one move every second. Suppose 
the front border piece of our speed-of-light placement X is presently (t = 0) situated 
at x E Zn, with lix II = M. We shall focus our attention on only those points in Zn that 
occupy borders of value M or higher. Thus at t = 0, all that we see is a single piece, 
situated at x (see FIGURE 4). 
When t = 1, the front border's value has increased to M + 1. The new front border 
piece must have made its final hop over x to land on the point x + ei for some 
1 < i < n. See FIGURE 5. 
I I 
_ t - 1- 
xT\ I 
I IL 
M 
FIGURE 4 
What we see when t = 0. 
I I I 
I \::+\I I 
I I I 
x x+ei 
I I] 
I 
I 
M+ I 
M 
FIGURE 5 
What we see when t = 1. 
When t = 2, the front border's value has increased to M + 2. The piece that landed 
there had to make its final hop over x + ei. Hence the piece on border M + 2 must be 
situated at x + ei + e1 for some 1 <j < n. Observe that i Oj, for otherwise the piece 
at x would have hopped over x + ei to point x + 2ei thus leaving no piece with border 
value M, contradicting Lemma 4. 
Thus, at t = 2, we see three pieces, situated at x, x + ei, x + ei + ej, where 
1 <i j< n. (See FIGURE 6.) 
When t = 3, a piece is jumped to the new front border M + 3 and, since it had to 
hop over the piece at x+ ei + ej, it must end up at x+ ei + e1 + ek for some 
1 < k < n. We know that k j by the same argument as i =#j above. we now show 
that, in fact, k = i. The new front border piece, before it made its final hop over 
x+ ei + e. to x+ ei + e.+ ek, must have been at the point x+ ei + e -ek on 
border M + 1. But how did it get there? It had to hop over the sole piece on border 
M, situated at x. But this requires ei + e- ek to be a unit vector, which is only 
possible when k = i or k =j. And since k #j, we have k = i. Hence at t = 3, we see 
four pieces, situated at x, x + ei, x + ei + ej, and x + 2ei + ej, as in FIGURE 7. 
When t = 4, the back border piece, wherever it is, jumps to border M + 4, landing 
on x + 2ei + e. + ek for some 1 < k < n. By the argument in the preceding paragraph, 
the hop over the piece at x + 2ei + ei had to come from the point x + 2ei reached by 
a hop over x + ei, and so that hop came from x. Hence the jump originated at x. Thus 
p = 4, and our current configuration (as well as our original one) must have been a 
serpent. 
This content downloaded from 134.173.130.46 on Mon, 10 Jun 2013 19:06:37 PM
All use subject to JSTOR Terms and Conditions
VOL. 66, NO. 1, FEBRUARY 1993 19 
I X 
I _ x ,+ e \ e\ 
x x+ei 
I1 \ M+2 
M+ 1 
M 
FIGURE 6 
What we see when t = 2. 
I ~~~~~~~x + 2e, + ej 
x + e2 e ie+ 
\ I Xx+e,+ej I 
x x + e M+3 
M 
FIGURE 7 
What we see when t = 3. 
Returning to the problem at the outset of the article, we see that the 35-move 
trajectory must be optimal because the original configuration is translated a distance 
of 34 units, and the square configuration is not a speed-of-light configuration. 
As a consequence of this theorem, we see that no speed-of-light configurations exist 
when the number of pieces is three or greater than four. In these cases, it is easy to 
create configurations that are translatable with speed 2/3 (e.g., the snake configura- 
tion with p pieces). The question of whether the speed 2/3 is optimal for the two- 
(and higher-) dimensional problem, when p = 3 or p > 4, remains open. More 
specifically, it remains unknown how to optimally translate six or nine pieces 
(arranged in a triangle or square) from the lower left-hand corner to the upper 
right-hand corner of the Go board. 
Acknowledgments. We dedicate this paper to the Summer Studies in Mathematics program at Hamp- 
shire College, directed by Professor David Kelly, where this research took place. HCSSiM is a mathematics 
program for high-ability high school students. During the 1986 program, Benjamin led a week-long, 
open-ended course, where students explored properties of jumping games. Two students, Auslander and 
Wilkerson, proved the two-dimensional version of this theorem and presented it at an MAA section 
meeting in Baltimore in 1986. We also thank Professor Alan J. Goldman of Johns Hopkins University for 
suggesting this problem to us, Greg Levin for assistance with the figures, and the anonymous referees for 
many valuable suggestions. 
REFERENCE 
Readers may also enjoy Wheels, Life and Other Mathematical Amusements by Martin Gardner (W. H. 
Freeman and Co., 1983), particularly the chapters analyzing the games of "Halma" and "Life." 
This content downloaded from 134.173.130.46 on Mon, 10 Jun 2013 19:06:37 PM
All use subject to JSTOR Terms and Conditions


English

http://scholarship.claremont.edu/cgi/viewcontent.cgi?article=1530&context=hmc_fac_pub

Optimal Leapfrogging

Abstract

Similar works

Full text

Available Versions

Scholarship@Claremont

Keck Graduate Institute