Gardner M. Sixth Book of Mathematical Diversions from Scientific American

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Mathematical

Games

this Lvas the first appearance of the hexa-

gon in print. Karl Fabel sent

letter

he had received

from hlartin Kiihl, of Han-

over, Germany, showing that Kiihl, too, had

independently discovered the hexagon

(about

1940)

but had not published it.

feature story by Karl Abraham on

Clifford

\T7.

hdams' discovery of the pat-

tern appeared in the Philadelphia

Evening

Rulletirl,

July

19,

1963,

page

18;

a follow-up

story giving the solution (which readers

had been asked to find) appeared in the

July

issue.

Answers

The combinatorial questio~ls are answered

as follows:

Four socks guarantee a

n~atchi~~g pair.

A dollar can be changed in

202

distinct

ways. For

full solution, using recursive

computation, see the last two pages of

George

Polya's

HOZC

Sul~e

It;

Second

edition; New York: Doubleday,

1957.

strip of five stamps, blank on both

sides, can be folded in

distinct ways

[see Figure

191.

(If the stamps are printed

on one side, you might think the number of

ways would double, but it increases only

25.

Why?)

The

problern of finding

formula for

stamps remains unsolved, but recursive

procedures by which the number of

differ-

19.

Answer

the

stamp-folding problem

Combinatorial

Theory

ent folds can be calculated easily with

computers have

been developed. This prob-

lem was first posed

by S.

Ulam.

1961

Mark

Wells, using a computer at Los

Alamos Scientific Laboratory,

found the

number of distinct foldings for six, seven,

eight, and

nine stamps to be

38,

120, 353,

and 1,148 respectively. More recent re-

sults will be found in

Map-folding

Problem,"

Lunnon, in

Mathe-

matics of' Computation,

January, 1968;

and in "Folding a Strip of Stamps," by

John

Koehler, in

Journal

Comhinato-

rial Theory,

September,

1968.

The problern

of finding a nonrecursive formula is more

difficult,

and also unsolved, if one asks for

different ways of folding square sheets into

a packet of unit squares.

To change the magic hexagon to a hexa-

gon with

22 as the sum of each three-cell

row, 42 as the sum of each four-cell row,

and 62 as the sum of each five-cell row,

replace the number in each cell with the

difference between that number and 20.

The problem of placing stamps with

values of one, two, three, four, and five

cents in a four-by-four square, with no two

stamps of the

same value in any row, col-

umn, or diagonal (including the

smaller

diagonals), can be answered with a maxi-

mum value of

cents

[see Figure

20,

which

shows

one of many solutions].

This is

probably two cents more than most readers

were able to achieve if they used four

fours and left two cells empty. The trick is

to use

only three four-cent stamps. "The

reader will probably find, when he sees

the solution," wrote

Heriry Dudeney in

20.

solution to the

stamp-placing

problem

Anluseme~lts in Matllei?zatics

(Problem 308),

"that, like the stamps themselves, he is

licked."

Donald

Knuth found that Dudeney's

solution could be enlarged to a remarkable

five-by-five square, satisfying all the con-

ditions and giving

the maxinlum total value

of 75 cents. Simply

add

top row of

5, and a right border (reading down-

ward) of

Each value appears

five times in the

square. This is equivalent

to the

problern of superin~posing five solu-

tions to the problem of the non-attacking

queens

the order-fj board. (See Chapter

16 of my

Unexpected Hanging;

New York:

Simon and Schuster,

1969.)

Mathematical Games

References

Coi~z/~inc~toriczl AII(I~~.F~s.

Percy

*\.

\lachlahon.

Cambridge: Cambridge University Press,

1'01.

1915; Vol.

1916. (Reprint

vols.

in one]. Bronx,

N.Y.: Chelsea Publishing Co.,

1060.)

Ail Introductior~ to Corrzbinutoriczl Anulysis.

John Kiordan. New k'ork: John \Viley and

Sons, 1958.

Cornbinutoricrl Jlathen~afics.

Herbert John Ry-

ser. New York: John \Viley and Sons,

1963.

Al~plied Co~nbil~(ltori(~l .Il~~~l~e~nutic~.

Frank

Harary. New York:

John

LViley and Sons, 1964.

Conzbitlcltorinl Theory.

5Iarshall Hall, Jr. Wal-

tharn, hlass.: Blaisdell Publishing Co., 1967.

Eleinentary ComOiizatorial Ancl1~~si.s.

\Iartin

Eisen. New York: Gordon and Breach, 1969.

Bouncing Balls in

Polygons and Polyhedrons

THROUGHOUT

recorded history the bounc-

ing ball has been indispensable equip-

ment for a dazzling variety of indoor and

outdoor sports. Games exploiting it range

from the child's simple bouncing of a rub-

ber ball ("One, two, three O'Lary

.")

sports such as tennis, handball, and bil-

liards in which the ability to judge angles

of incidence and reflection is essential to

a player's skill.

The balls

shine round and clear, quick blobs

of color on faultless Felds,

where rapid vengeance rolls

and clicks, returns

or poorly judged, deflects

to pass and spend itselfin motion

rebounding gingerly from cushions

. . .

Herman

Spector,

"B.A. (Billiard Academy)"

Mathematicians and physicists are no-

toriously fond of pool and billiards. It is

easy to understand why. The gingerly re-

bounds within faultless fields can be pre-

cisely calculated. Lewis Carroll, who taught

mathematics at the University of Oxford,

enjoyed playing billiards, particularly on

circular

table he had made for himself.

much prized collector's item is the first

edition of a two-page leaflet, published by

Carroll in

1890

and never reprinted, that

explains his rules for this game.

Hundreds of recreational problems con-

cern the rebounds of elastic balls within

perimeters of various shapes. Consider,

for example, the following old puzzle:

You have two vessels with respective ca-

pacities of

and

pints. Beside you is a

large tub of water. Using only the two ves-

sels (and excluding all dodges such as mark-

ing the containers or tilting them to obtain

fractional amounts), how can you measure

exactly two pints?

The question can be answered by trial

and error or by applying variou: algebraic

procedures. What has all this to do with

bouncing balls? Surprisingly, liquid-mea-

suring puzzles of this type can be solved

Mathematical Games

easily by graphing the paths of balls bounc-

ing inside

rllotnl?oidal

tables! (The method,

using what topologists call a "directed

graph," was first explained

hl.

Tweedie in

The Alatlzemuticcrl Gazette

July,

1930.)

The cushions of such tables

are

best drawn on isometric graph paper:

paper with

lattice of equilateral triangles.

In this case the sides of the table are

and

units

[see

Figure

811.

Readings on the

horizontal axis represent the amount of

water in

tlle 11-pint vessel at any time and

readings on

tlle vertical axis tell how much

water is in the 7-pint vessel.

To use the graph, imagine a ball at point

the lower left corner. It travels to the

right along the base of the rhomboid until

it strikes the right-hand cushion at a point

labeled

on the base line: the 11-pint

vessel has been filled

and the 7-pint con-

tainer remains empty. After bouncing

off

the right-hand cushion the ball travels up

and to the left until it hits the top cushion

at point

on the horizontal co-ordinate

22.

Graph

Tartaglia's puzzle

and on the seventh line on the side co-

ordinate. This plot indicates that 7 pints

have been transferred from the 11-vessel

to the 7-vessel, leaving

pints in the

larger vessel.

If you continue to follow the bouncing

ball until it strikes a point marked

keep-

ing a record of each step, you will obtain the

18-step answer shown below the graph.

Slanting arrows indicate that water is poured

from one vessel into another. The vertical

arrows show either that the 7-vessel is being

emptied into the tub or that the 11-vessel is

being filled.

Is this the shortest answer? No; an alter-

native procedure is to begin by filling the

7-vessel. This is graphed by starting the ball

at the

point and rolling it

along the

table's left side. If the reader traces the

ball's

~ath until it strikes a

point, keeping

a record of the steps, he will find that his

ball computer bounces out a solution in

steps -the minimum.

LVith a little ingenuity one can devise

ball-bounce computers for any liquid-pour-

ii~g puzzle in which no more than three

vessels are involved. Consider the oldest of

all three-vessel problems, which goes back

Nicola Fontana, the sixteenth-century

Italian mathematician who called himself

Tartaglia ("The Stammerer"). A11 eight-pint

vessel is filled with

water. By means of two

empty vessels that

hold five and three pints

respectively, divide the eight pints evenly

between the two larger vessels.

The graph

for this problem is shown in Figure

22.

Here

the

eight-pint vessel is represented by

line

paralleling a main diagonal of the rhomboid.

The ball begins as before in the

corner. It

is easy to trace a path that computes the

minimum solution, which requires seven

operations.

23.

Graph for vessels

of volumes

and

\Vllen the two smaller vessels have no

common divisor and the

thircl vessel is

equal to or greater

than the sum of the

smaller vessels, it is possible to measure

out any whole

rlu~llber from 1 to the capac-

ity of the middle-sized vessel. For example,

with vessels of

15, 16-, and 31-pint capac-

ities one can measure

any quantity from

16. This is not possible if the two smaller

vessels have a

comnlon divisor.

graph for

vessels of

4,6, 10 will not bounce the ball to

any odd number, and vessels of

3,9, 12 will

measure only the quantities 3, 6, 9.

(In

both

cases only

nlultiples of the coinrnon divisor

can be measured.) If the largest vessel is

smoller

than the sum of the other two, there

are further limitations. For example, vessels

7,9,

12 require that

corner of the rhom-

boidal graph be sliced off

[see Figure

231.

The bouncing ball will measure any quan-

tity

from

except

,Altholigll

and 9

have

common divisor, the smallness of

the third vessel makes it impossible to

ol~tain

When the largest vessel is

larger

than the

sun1 of the other two, the graph continues

to be applicable. The reader may

enjoy

applying it to the followirlg variation of

Tartaglia's problem, as posed by Sam Loyd

on page 304 of his

fanlous

Cyclo),edicl

Puzzles.

(This is one of the puzzles for

which the

Cyclopedic1

fails to furnish an

answer, a fact that may explain why the puz-

zle has never been reprinted.)

Some

U.S.

soldiers nlanaged to "capture"

a 10-gallon keg of beer. "They naturally

sampled

part of it," writes Loyd, making

use of 3-gallon and 5-gallon containers. The

rest of the beer was carried back to camp in

three equal portions-one

the keg and

the other two in the two containers. How

much did they

drink and how did they mea-

sure the remainder into three equal (non-

zero) parts? The best solution is the one

with the fewest steps for the entire pro-

cedure. Each step, including the drinking

operation, involves an integral number of

gallons, and it is assumed that no beer is

wasted by being tossed out.

You

lnay find it entertaining to experi-

ment

with vessels of various sizes, using the

ball computer to explore all that can be

done with them. For more information

about the technique, including its exten-

sion to four vessels by

means of tetrahedral

graphs, the interested reader is referred to

the book by T.

O'Beirne listed at the

close of this chapter.

different type of ball-bouncing prob-

lem is that of finding cyclic paths

along

which a ball can bounce forever inside a

polygon, always tracing the same path and

hitting each side

only once in each cycle.

Such problems can be solved by using the

powerful technique of mirror reflection.

table in the shape of

square provides a

simple example. Figure

shows a square

reflected along three different sides, and the

colored

line is its only cyclic path with seg-

ments of equal length. Folding the four

squares into a unit-square packet trans-

forms the straight line

into the cyclic path.

At this

point two interesting questions

arise. Are there cyclic paths with equal seg-

ments inside the solid analogues of the

square and equilateral triangle: the cube

and tetrahedron? The ball is

assunled to be

an idealized elastic particle (or a light ray

24. Equal-segment path in a square

inside

solid with interior mirror surfaces),

taking straight paths

in zero gravity and

bouncing off the sides in the usual

manner:

with equal angles of incidence and reflec-

tion on a plane perpendicular to the side

against which it bounces.

The ball nlust

strike each face only once during the cycle

and travel the

same distance between each

consecutive pair of bounces. (Striking an

edge or corner is not regarded as striking

the faces meeting at that edge or corner;

otherwise the cube problem would be

solved

by a ball moving back and forth be-

tween two diagonally opposite corners.)

Warren Weaver, in one of his

many arti-

cles on Lewis Carroll, has disclosed that the

cube problem is found among Carroll's

un-

published and mathematical notes. It is the

25.

Solution to Lewis Carroll's

cube-and-ball problem

sort of problem that would appeal to the

inventor of circular billiards. Actually the

notion of playing billiards inside a cubical

"table" is not as farfetched as it might seem.

With gigantic space stations perhaps only a

few decades away it takes no great prophetic

ability to foresee a variety of three-dimen-

sional sports that will take advantage of

zero gravity. Pool adapts neatly to a rectan-

gular room with cushion walls, floor and

ceiling, corner pockets, and balls numbered

from

that are initially arranged in

tetrahedral formation. Of course, there

would be difficulties. Air resistance offers

much less friction than the felt surface of a

pool table does. If the tetrahedron were

broken by a fast cue ball, entropy would

increase at a rapid rate. It would be hard to

keep out of the way of balls flying about in

random directions like the molecules of a

gas in thermal equilibrium!

But back to Carroll's problem. The reflec-

tion technique used with squares can be

applied to cubes. Five reflections are re-

quired and the colored line in Figure

traces the desired path. It is one of four

Bouncing

Balls

different paths, identical in shape, that

solve the problem. (If all six faces of the

cube are ruled into nine smaller squares,

each path touches every face at one corner

of the central square.) Figure 26 shows a

cardboard model that demonstrates the

path after the six cubes have been "folded"

into one another. The cord is held in place

by passing loops through small holes and

securing them on the outside with pegs

made of wood. If you think of the cube as

being formed of 27 smaller cubes, you will

see that every segment of the path is a diag-

onal of a small cube. Each segment there-

fore has a length of

1/*

on a unit cube. The

path's total length is

2s.

26.

Model showing path inside a cube

As far as I know, Hugo Steinhaus was the

first to find this path. (See his

One Hun-

dred Problems in Elementary Mathematics;

New York: Basic Books, 1964, Problem

33.

The book is a translation of the 1958 Polish

edition.) The solution was later redis-

covered by Roger

Hayward, who published

it in

Recreational Mathematics Magazine,

June, 1962. The shape of the path, he writes,

is known to organic chemists as a

"chair-

shaped hexagon." It occurs often in car-

bon compounds, such as cyclohexane, in

which six carbon atoms are single-bonded

in a ring with other atoms attached outside

the ring. "It is interesting to note," writes

Oliver of the Hewlett-Packard Com-

pany in

Palo Alto, California, "that the path

appears as a

2 rectangle in all projec-

tions of the cube taken perpendicular to a

face, as a rhombus in three of the isometric

projections taken parallel to a diagonal of

the cube, and as a regular hexagon in the

fourth isometric view. A queer figure, but

that's the way the ball bounces!"

A similar cyclic path inside a tetrahedron

was discovered by John

Conway and

later, independently, by

Hayward in 1962.

It is easy to reflect a tetrahedron three times

[see Figure

271

and find a cyclic path that

touches each side once. The difficult trick

is to find a cyclic path with equal segments.

One is shown

the colored line. There are

three such paths, all alike, touching each

face of the solid at one corner of a small

equilateral triangle in the center of the face.

The side of this small triangle is a tenth of

the edge of a tetrahedron with an edge of

Each segment of the ball's path has a length