2. Although nearly every submission we receive at Math Horizons comes in electronically, last spring a short story called
“Superstrings and Thelma” arrived in our old-fashioned mailbox with a simple note attached: “Is this short story something
you can use? I wrote the math column in Scientific American for 25 years. If my piece is not right for Math Horizons, there is
no need to send it back. All best, Martin.”
Just seeing the name on the page caused a palpable stir. Martin Gardner was well into his nineties, yet here we were, in
possession of a charming piece of original fiction that had come directly from the typewriter of the most celebrated voice in
recreational mathematics. What is the proper editorial response to someone whose life’s work arguably made the existence
of a magazine like Math Horizons a possibility? Sadly, we did not get the chance to craft a reply.
As a means of giving our newer readers a broader picture of Gardner’s influence, we decided to run the short story alongside
a report from the ninth biennial “Gathering for Gardner” held in March of this year. Then, on May 22, Martin Gardner passed
away, and our plans evolved into fashioning an appropriate tribute. We’ve added a previously published piece by Gardner
that is more typical of his mathematical writing and selected several companion articles where his influence is unmistakable.
Special thanks go to former MH editor Don Albers for penning an opening essay on short notice.
In one way or another, all of us who love mathematics have received something from Martin Gardner over the years—and
we have no plans to send it back.
Bruce Torrence and Stephen Abbott
Co-Editors, Math Horizons
Dr. Stephens’s Living Legacy
Dear Math Horizons,
I recently received a copy of your February 2010 issue with
the article titled “The Best Undergraduate Math Program
You’ve Never Heard Of” by Reuben Hersh. My godfather,
Sylvester Reese (pictured, far right), sent me the article
because I had always known my father, Delegate Howard P.
Rawlings (pictured, far left) was a protégé of Dr. Clarence
Stephens when he attended Morgan. In fact, Dr. Stephens
had such a profound impact on my father that my younger
sister Stephanie was named after him. Now we know why.
Four years ago, I came to work at a community college where
the students are likely similar to those Dr. Stephens taught at
Morgan . . . poorly prepared, black, with low confidence. I
have become frustrated by our inability to provide students
with the nurturing environment to learn and EXCEL in math. I
clearly got this from my father. Because of your article, I now
understand how my father learned this from Dr. Stephens.
Thank you for highlighting Dr. Stephens’s work at Morgan
and Potsdam! Know his work was furthered by his students
like Delegate Rawlings, former chair of the Maryland State
Appropriations Committee and advocate for higher
education; his namesake, Mayor of the City of Baltimore
Stephanie Rawlings-Blake; and me.
Sincerely,
Lisa Rawlings
To Our Readers:
Letters:
Rusczyk Is Right On
Dear Editors,
In the “Conversation with Richard Rusczyk” (February 2010),
Mr. Rusczyk presents one of the best-stated arguments for
both the why and the how of the teaching and learning of
math that I have ever read. His criticism of the “standard
curriculum” seems more pointed at elementary school than
high school math, at least in my experience, but his concise
answer to the question “What should we train our students
for?” was both compelling and inspiring.
Bob Nortillo
Niles West High School
3. 5 Martin Gardner, 1914–2010: Magical Man of Numbers and Letters
Don Albers The founding editor of Math Horizons offers some personal reflections
on the life and impact of the undisputed champion of popular mathematics.
6 Talkative Eve / Superstrings and Thelma
Martin Gardner A rerun of a 1996 Math Horizons piece followed by a previously
unpublished short story provide a glimpse into Martin Gardner's creative endeavors.
9 Gathering for Gardner
Bruce Torrence The author finds himself in the middle of a puzzler’s paradise where
mathematics is king, probability rules, and brevity is not only the soul of wit but a way
to get rich.
13 The View from Here: Confronting Analysis
Tina Rapke An unfiltered firsthand account of one student’s
struggle to stay rational in her attempts to understand the real numbers.
16 Infinite Bottles of Beer on the Wall
Donald Byrd A classic ditty for long car rides goes transfinite
in the service of illustrating the principles of cardinal arithmetic.
18 Congo Bongo
Hsin-Po Wang A mathematical jungle drumming challenge has legendary
problem-solvers Jacob Ecco and Justin Scarlet standing on their heads—hopefully
at the same time.
22 Port-and-Sweep Solitaire
Jacob Siehler If you have exhausted all the challenges of traditional peg solitaire, a
new variation called “Port-and-Sweep” will take you in some imaginary new directions.
26 The Mathematics behind Lifesaving Kidney Exchange Programs
Olivia M. Carducci What happens when a willing organ donor turns out to be
incompatible with a loved one’s blood type? The tools of graph theory may have
vital implications for the evolving formulation of a national kidney exchange network.
30 The Playground
The Math Horizons problems section, edited by Derek Smith
34 Aftermath: Facebook and Texting vs. Textbooks and Faces
Susan D’Agostino
What is really happening on the laptops of note-taking students during class—and to what effect?
p22
SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS 3
p18
Contents
p7
p9
p26
p13
5. Science, a brilliant debunking of
pseudoscience, remains a best-seller
58 years after publication. His
Annotated Alice has sold over 400,000
copies, and the first printing of the
most recent revision sold out a few
days after publication. His works on
philosophy, especially the philosophy of
science, continue to be very popular.
Over the course of his life, Gardner
wrote more than 70 books, of which
half are still in print.
As a young editor, I was greatly
impressed by his kindness, generosity,
and gentle manner. We talked frequent-
ly on the phone, and finally in 2000 I
met him in person for the first time at
his home in Hendersonville, North Car-
olina. For most of the day we chatted in
his library about his life and books. I
was also treated to a demonstration of
some of his magic tricks and a tour of
his extensive files. By late afternoon, he
said that it was time to go upstairs and
join our wives in the living room where
he proudly demonstrated his ability to
prepare martinis.
I knew Martin Gardner for 32 of his 96
years. As he was an inspiration to the
end, I regret that it could not have
been longer.
DOI: 10.4169/194762110X525511
a high school student
and wanted to attend
Caltech, but at that time
(1932), Caltech required
two years of liberal arts
at a college before
transferring. So he went
to the University of
Chicago, where he
became hooked on phi-
losophy. After service in
World War II, his writing
career began to blos-
som with a position at
Humpty Dumpty’s Mag-
azine for Little Children.
In 1957 he began writing
his monthly column for
Scientific American. Most mortals find
the idea of writing a significant column
on mathematics every month for 25
years to be mind-boggling.
Although many of his columns dealt
with topics from computer science,
Gardner always composed at a
typewriter, and the manuscripts that he
sent to me when I was editor of Math
Horizons and the College Mathematics
Journal contained his revisions in
pencil. Given that he was the author, I
would have been happy if he had
submitted his manuscripts written with
crayons on lined paper. It’s very
pleasing that today the MAA is the
publisher of nearly all of his books on
mathematics and a CD containing the
15 books of all his Scientific American
columns.
Mathematics, however, was only one
of Gardner’s interests. He wrote
extensively about close-up magic and
counted some of the best-known
magicians among his friends. His book
Fads and Fallacies in the Name of
SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS 5
M
ay 22 Martin Gardner died. For
25 of his 95 years, he wrote
“Mathematical Games and
Recreations,” a monthly column for
Scientific American magazine, which
was without a doubt the most popular
and influential column on mathematics
that has ever existed. His columns
inspired thousands of readers to delve
more deeply into the large world of
mathematics that he loved to explore
and explain. His readers included ama-
teur and professional mathematicians
and a host of people from many other
fields. Among his correspondents were
many famous mathematicians and sci-
entists, including John Horton Conway,
Persi Diaconis, Ron Graham, Douglas
Hofstadter, Richard Guy, Don Knuth,
Sol Golomb, and Roger Penrose. His
extensive network of friends often
enabled him to bring new develop-
ments in mathematics in clear English
to his legions of fans.
What is astounding is that he never
took a single college course in mathe-
matics. He excelled in mathematics as
Don Albers
Martin Gardner, 1914-2010
Magical Man of Numbers and Letters
About the author: Don Albers is
currently the editorial director of the
books program at the MAA. He is also
the founder of Math Horizons, serving
as its first editor from 1993 through
1998. He is co-editor of a number of
books, including Mathematical People:
Profiles and Interviews, which contains
an in-depth conversation with Martin
Gardner.
email: dalbers@maa.org
6. 6 SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS
DID. Because EVE must be smaller
than 303, E is 1 or 2. Of the 14
possibilities (121, 141, … , 292) only
242 produces a decimal fitting
.TALKTALK…, in which all the digits
differ from those in EVE and DID.
The unique answer is
242/303 = .798679867986… If EVE/DID
is not assumed to be in lowest terms,
there is one other solution, 212/606 =
.349834983498…, proving as Joseph
Madachy has remarked, that EVE
double-talked.
Three Squares
There are many ways to prove that
angle C in the figure is the sum of
angles A and B. Here is one (see Figure
2). Construct the squares indicated by
red lines. Angle B equals angle D
because they are corresponding angles
of similar right triangles. Since angles A
and D add to angle C, B can be substi-
tuted for D, and it follows immediately
that C is the sum of A and B.
This little problem produced a flood of
letters from readers who sent dozens of
other proofs. Scores of correspondents
avoided construction lines by making
the diagonals equal to the square roots
of 2, 5, and 10, then using ratios to find
two similar triangles from which the
desired proof would follow. Others
generalized the problem in unusual
ways.
You have six weights. One pair is red,
one pair white, one pair blue. In each
pair, one weight is a trifle heavier than
the other but otherwise appears to be
exactly like its mate. The three heavier
weights (one of each color) all weigh
the same. This is also true of the three
lighter weights.
In two separate weighings on a balance
scale, how can you identify which is the
heavier weight of each pair?
ANSWERS
Talkative Eve
As stated earlier, to obtain the simplest
fraction equal to a decimal of n repeat-
ed digits, put the repeating period over
n 9’s and reduce to its lowest terms. In
this instance TALK/9,999, reduced to
its lowest terms, must equal EVE/DID.
Consequently, DID is a factor of 9,999.
Only three such factors fit DID: 101,
303, 909.
If DID = 101, then EVE/101 =
TALK/9,999, and EVE = TALK/99.
Rearranging terms, TALK = (99)(EVE).
EVE cannot be 101 (since we assumed
101 to be DID), and anything larger
than 101, when multiplied by 99, has a
five-digit product. And so DID = 101 is
ruled out.
If DID = 909, then EVE/909 =
TALK/9,999, and EVE = TALK/11.
Rearranging terms,
TALK = (11)(EVE). In
that case, the last
digit of TALK would
have to be E. Since
it is not E, 909 is
also ruled out.
Only 303 remains
as a possibility for
Talkative Eve
T
his cryptarithm (or alphametic,
as some puzzlers prefer to call
them) is an old one of unknown
origin, surely one of the best and, I
hope, unfamiliar to most readers:
The same letters stand for the same
digits, zero included. The fraction
EVE/DID has been reduced to its
lowest terms. Its decimal form has a
repeating period of four digits. The
solution is unique. To solve it, recall that
the standard way to obtain the simplest
fraction equivalent to a decimal of n
repeating digits is to put the repeating
period over n 9’s and reduce the
fraction to its lowest terms.
Three Squares
Using only elementary geometry (not
even trigonometry), prove that angle C
in Figure 1 equals the sum of angles A
and B.
I am grateful to Lyber Katz for this
charmingly simple problem. He writes
that as a child he went to school in
Moscow, where the problem was given
to his fourth-grade geometry class for
extra credit to those who solved it.
“The number of blind alleys the
problem leads to,” he adds, “is
extraordinary.”
Red, White, and Blue Weights
Problems involving weights and
balance scales have been popular
during the past few decades. Here is an
unusual one invented by Paul Curry,
who is well known in conjuring circles
as an amateur magician.
EVE
DID
TALKTALKTALK= . ...
Talkative Eve
Martin Gardner
This article first appeared in the April 1996 issue of Math Horizons.
Figure 1. Prove that angle A plus angle B equals angle C.
7. “That’s me,” I said.
She smiled and held out a hand. “I’m
Thelma O’Keefe. We were in the same
algebra 101 class.”
We shook hands.
“You won’t remember me,” she said. “I
was fat in those days, and shy, and not
very pretty.”
“That’s hard to believe,” I said. “You
look gorgeous now.”
“Well, thank you, kind sir,” she said,
SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS 7
Charles Trigg published 54 different
proofs in the Journal of Recreational
Mathematics Vol. 4, April 1971, pages
90–99. A proof using paper cutting, by
Ali R. Amir-Moéz, appeared in the same
journal, Vol. 5, Winter 1973, pages 8–9.
For other proofs, see Roger North’s
contribution to The Mathematical
Gazette, December 1973, pages
334–36, and its continuation in the
same journal, October 1974, pages
212–15. For a generalization of the
problem to a row of n squares, see
Trigg’s “Geometrical Proof of a Result
of Lehmer’s,” in The Fibonacci
Quarterly, Vol. 11, December 1973,
pages 539-40.
Red, White, and Blue Weights
One way to solve the problem of six
weights—two red, two white, and two
blue—is first to balance a red and a
white weight against a blue and a white
weight.
If the scales balance, you know there
are a heavy and a light weight on each
pan. Remove both colored weights,
leaving the white weights, one on each
side. This establishes which white
weight is the heavier. At the
same time it tells you which
of the other two weights
used before (one red, one
blue) is heavy and which is
light. This in turn tells you
which is heavy and which is
light in the red-blue pair not
yet used.
If the scales do not balance
on the first weighing, you
know that the white weight
on the side that went down
must be the heavier of the
two whites, but you are still
in the dark about the red
and the blue. Weigh the
original red against the
mate of the original blue (or
the original blue against the mate of the
original red). As C. B. Chandler (who
sent this simple solution) put it, the
result of the second weighing, plus the
memory of which side was heavier in
the first weighing, is now sufficient to
identify the six weights.
For readers who liked working on this
problem, Ben Braude, a New York City
dentist and amateur magician, devised
the following variation. The six weights
are alike in all respects (including color)
except that three are heavy and three
light. The heavy weights weigh the
same and the light weights weigh the
same. Identify each in three separate
weighings on a balance scale.
DOI: 10.4169/194762110X524684
Figure 2. Construction for proof of the three-square
theorem.
S
everal years ago I was a
graduate student at the
University of Chicago. I was
working on my doctorate in physics,
about possible ways to test superstring
theory, when my brother in Tulsa died
suddenly from a heart attack. Both
parents had earlier passed away. After
the funeral I drove past my past,
marveling at the enormous changes
that had taken place since I grew up
there. I drove by the red brick building,
now an enormous warehouse, that had
once been Tulsa Central High. My
grades in history, Latin, and English lit
were low, but I was good in math and
had a great physics teacher. He was
mainly responsible for my majoring in
physics after a scholarship took me to
the University of Chicago.
While I was having dinner at a popular
restaurant on the corner of Main and
Sixth streets, the waitress stared at me
with a look of surprise. “Are you
Michael Brown?”
Short
Fiction
Superstrings and Thelma
Martin Gardner
8. smiling. “You were a whiz at algebra.
Do you remember when you caught Mr.
Miller in a mistake he made on the
blackboard, and how embarrassed he
was?”
“I remember. He was a miserable
teacher. I think he hated math.”
“I know I hated it,” Thelma said.
“I’m sorry to hear that. Math can be
exciting and beautiful when it’s taught
properly.”
After Thelma brought my receipt and
credit card, I said, “Any chance I could
see you after work? Maybe you could
steer me to a late night bar where we
could chat about old times?”
“I’m free at eleven,” she said.
I followed Thelma’s car to a pleasant
little bar on the outskirts of town, near
where she lived. She was divorced, she
told me, and had a boy of ten who was
probably asleep in her apartment. The
bar served only beer. She said she
didn’t drink anything with more alcohol
than beer. Her ex, she added, had an
alcohol problem. I didn’t press her for
details. Instead, I fear I talked too much
about myself, and
mainly about
superstrings.
I did my best to
explain that strings
were inconceivably tiny loops, like
rubber bands, that vibrated at different
rates. Their frequencies generated all
the properties of the fundamental
particles, such as electrons and quarks.
The simplest string vibration produces
gravitons, conjectured particles that
carry gravity waves.
“You have nice dark eyes,” she
interrupted.
“Thanks,” I said. “Your eyes aren’t so
bad either.”
I tried hard to explain how a famous
physicist named Ed Witten had
generalized strings to what he called
M-theory. The M stands for
membranes, or branes for
short. A superstring is a
brane of one dimension.
Other branes have high-
er dimensions. Our uni-
verse, I said, has ten or
eleven dimensions, of
which six or seven are
squeezed into
compact tiny
spheres that are
attached to every
point in our space-
time.
“I didn’t under-
stand a word you
spoke,” Thelma
said. “It sounds
nutty to me. Do
you believe all
that?”
“Mostly. I think strings are for real,
but I’m not so sure about Witten’s
membranes.”
“Is everything made of strings?”
Thelma asked.
“Everything.”
“And what are the strings made of?”
“Nothing. They’re just pure
mathematical structures.”
“If the universe isn’t made of anything,”
she said, “how come it exists?”
“Good question. Nobody knows.”
“Well, maybe God knows,” she said.
Outside the bar, standing by our cars,
Thelma invited me to her apartment for
some coffee.
“No,” I said. “I really can’t stay another
minute. I have to be up early to catch a
plane to Chicago. It was great getting
to know you.”
“Will I see you again?”
“It would be a pleasure,” I said.
Like a fool I failed to ask for her
address and phone number. We shook
hands. She said goodbye, then startled
me with a quick kiss on the mouth.
Almost a year drifted by. My thesis
was published as a
book by the
University of Chica-
go Press. My
suggestions for test-
ing string theory were taken seriously
by most stringers. There was hope that
some of the tests might actually be
made by a new atom cruncher under
construction in Switzerland. There were
vague rumors about a Nobel prize.
I was unable to get Thelma out of my
head. I kept thinking of her wonderful
smile, and how good she smelled. It
wasn’t perfume. Was it her hair? I
thought about her more than I thought
about superstrings!
The University of Oklahoma, at
Norman, hired me as an assistant
professor. A suburb of Oklahoma City,
Norman is only a few hours drive from
Tulsa.
8 SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS
Photograph courtesy of Adeline C., http://www.flickr.com/photos/adollinn/
I was unable to get Thelma out of my head. I kept thinking
of her wonderful smile, and how good she smelled.
9. None of the waitresses at the
restaurant where Thelma had worked
knew what had happened to her. She
left her job six months before, and they
hadn’t heard from her since.
No Thelma O’Keefe was in the Tulsa
phone book. I drove back to Norman
feeling sad and frustrated. Should I hire
a detective? The Norman yellow book
had a long list of “investigators” and
two detective agencies.
I was planning to call one of the
agencies when my telephone rang. It
was Thelma!
“I heard you were asking about me,”
she said.
“Yes. How did you get my phone
number?”
“It’s on the Internet. How are strings?”
“Not so good. It didn’t predict dark
matter. It didn’t predict dark energy. It
even failed to pass one of my tests.
Lots of stringers are starting to have
doubts, including me.”
“If we meet again,” said Thelma, “don’t
tell me about it.”
Further Reading
Two recent books attacking string/M
theory as pseudoscience are Not Even
Wrong, by mathematician Peter Woit
(Basic Books, 2006), and The Trouble
with Physics, by Lee Smolin (Mariner
Books, 2007). See Chapter 18, “Is
String Theory in Trouble?” in my book,
The Jinn from Hyperspace (Prometheus
Books, 2008).
Editor's note: A provocative and
widely discussed article in which
string theory is used to suggest that
gravity is not a fundamental force, but
is rather a consequence of entropy, is
Erik Verlinde's “On the Origin of
Gravity and the Laws of Newton”
(2010), which can be found at
http://arxiv.org/abs/1001.0785.
DOI: 10.4169/194762110X525557
SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS 9
Bruce Torrence
Gathering for Gardner
Photographs courtesy of Bruce Torrence
I
t’s a chaotic scene in the lobby of the
Ritz-Carlton. There’s a dental con-
vention getting under way, and as
travel-weary orthodontic professionals
trickle into the packed room, they
encounter a bewildering display. Every
tabletop holds a collection of fascinat-
ing objects—puzzles of every imagina-
ble shape and design—around which
magicians, mathematicians, and puzzle
masters gather to discuss their latest
inventions. The conversations are
animated, and there is a tangible
feeling that something important is
unfolding. And indeed there is:
unbeknownst to the dentists, the ninth
Gathering for Gardner has just begun.
Atlanta has been host to this unlikely
convention of tinkerers and free
thinkers for almost two decades. They
assemble to pay homage to Martin
Gardner, the prolific and magnetic
author whose interests spanned the
seemingly disparate disciplines of
mathematics, puzzles, magic, and the
spirited debunking of pseudoscience.
This invitation-only affair attracts lumi-
naries in all of these fields, and despite
their obvious differences, there is a fer-
tile and dynamic common ground and
a deep mutual respect among the
participants.
What Has Tuesday
Got to Do with It?
The organization of the conference is
simple: there is one grand conference
room with a single stage. Speakers give
brief presentations in turn and are
awarded a dollar coin for each minute
that they finish ahead of their allotted
time—an innovative and surprisingly
inexpensive management tactic. Only
one speaker really cashed in: Gary
Foshee, a mechanical puzzle collector
10. the week and that births are independ-
ent of one another. With these assump-
tions in hand, the puzzle succumbs
easily to elementary probability theory.
If one denotes a single birth by a two-
tuple such as (boy, Tuesday) or (girl,
Sunday), then there are 14 equally likely
scenarios for a single birth. Foshee has
two children, and with no other infor-
mation, it follows that there are 142 =
196 equally likely poss bilities for two
children. But at least one of his children
is a (boy, Tuesday), and some simple
counting reveals that just 27 of the 196
outcomes satisfy this criterion. To see
this, simply note that there are 14
cases where the Tuesday boy is the
first born, and 14 where he is the sec-
ond born, and subtract the single case
that was counted twice. Among these
27 equally likely poss bilities, how many
include two boys? Exactly 13—there
are seven with (boy, Tuesday) as the
first child, and seven with (boy, Tues-
day) as the second child, from which
we subtract the one we counted twice.
Hence, the answer to Foshee’s riddle is
13/27, close to, but not exactly, 1/2.
Of course, the riddle leads naturally to
a host of other questions. Does “one is
a boy born on a Tuesday” mean that
the other was not born on a Tuesday? I
don’t read it this way. After all, the
question makes clear that there is a
poss bility that the second child is a
boy, so why shouldn’t the child be
allowed to arrive on a Tuesday also?
More glaring is the counterintuitive
nature of the result itself. Why should
something like the day of the week
affect the outcome? In thinking about
this, it’s important to understand that
had the day of the week not been
mentioned, the answer would be 1/3,
not 1/2, for only one of the three
equally likely gender scenarios BB, BG,
GB yields two boys. It’s all about
counting the possible outcomes. If you
are still wondering what Tuesday has to
do with it, you may wish to consult the
Further Reading section at the end of
the article.
illusion and perception, and on and on.
The topics bounced so completely from
one idea to the next that the effect was
both intoxicating and refreshing. Spon-
taneous discussion arose in and out of
the conference room. “What does
Tuesday have to do with it?”
Foshee’s puzzle proved a perfect
catalyst for discussion, and it is both
fun and instructive to reason through it.
One first has to make a few assump-
tions, and most of the people I spoke
with agreed that it’s best to keep it
simple. Let’s assume that there are no
multiple births, and that any single birth
is equally likely to be a boy or a girl.
Assume further that births are uniformly
distributed among the seven days of
and designer, faced the audience and
spoke slowly, “I have two children. One
is a boy born on a Tuesday. What is the
probability I have two boys?” After a
pause, he continued. “The first thing
you think is ‘What has Tuesday got to
do with it?’ Well, it has everything to do
with it.” And with that, he stepped
down (and collected a hefty stack of
coins from the organizers).
Other talks focused on space-filling
curves and genome folding, origami
mazes, Lewis Carroll’s mathematics,
dancing tessellations, psychological
explanations for children’s magic
theory, the history of Rubik’s cube, the
relation between computer hacking and
invention, self-replicating machines,
10 SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS
11. Just as Foshee’s riddle yields to
elementary probability theory, the
vast majority of questions and puzzles
posed at the gathering can be
approached using basic principles.
Each question has some clever twist,
and of course this is the hallmark of a
good brainteaser. Not all yield so easily,
however. Bill Gosper, discoverer of the
“glider gun” in Conway’s game of life
and considered by many to be the
founder of the hacker community,
proudly shared his “Dozenegger”
puzzle. This is a physical puzzle in
which twelve circles, each a different
size meticulously laser-cut from a sheet
of acrylic, must be snuggly packed (but
not forced) into a larger elliptical cavity.
(You can find it on his website—see the
Further Reading section.)
Magic and illusion played a big part in
the proceedings. Individual presenta-
tions in this area fell squarely in the field
of cognitive psychology, highlighting
peculiarities in human perception. A
cube with a corner removed can be
interpreted at least three ways (a large
cube with a smaller cube cut out of it; a
large cube with a smaller cube jutting
out from it; and a large three-sided
“room” with a cube sitting in its back
corner). Verses of Led Zepplin’s
“Stairway to Heaven,” played
backwards can be made to say pretty
much whatever you want provided the
listener reads those words as it plays.
Cinematic scenes in which actors
changed costumes out of frame and
returned in their new garb went
unnoticed by the entire audience (until
we were prompted to look for the
change in a second showing).
Performance artists and magicians also
took to the stage and presented
stunning illusions with amazing skill.
Attendees were treated to close-up
magic after hours by some of the best
in the business. The net effect was one
of fascination with a subtly disturbing
aftertaste. Everyone came away with a
similar feeling: how is it that we can be
deceived so easily? Seeing and
believing will never again be the same.
Abstract Structures
Another highlight of the gathering was
an afternoon dedicated to socializing
and sculpture building. Attendees were
invited to participate in the construction
and installation of several mathematical
sculptures at the home of Tom
Rodgers, one of the main organizers of
the event. On the lighter side, literally,
Vi Hart led a group that created
geometric balloon art. At the other
extreme, Chaim Goodman-Strauss
collaborated on a weighty steel
sculpture that suggested a space-filling
curve packed neatly into a cube. Other
sculptures were created using materials
such as aluminum, wood, bamboo, and
plastic. Under the direction of their
designers—George Hart, Carlo Séquin,
Akio Hizume, Rinus Roelofs, among
others—it was a tour de force of master
craftsmen at the top of their game. The
collaborative enterprise also
emphasized to the pure mathemati-
cians among us some of the practical
difficulties that arise when an abstract
idea is realized as a solid structure.
Every sculpture presented unique
challenges, and their successful
resolution led to a deep sense of
fulfillment as the day came to a close.
Back in the grand conference room, the
onslaught of ideas continued. Princeton
mathematician John H. Conway gave a
spirited presentation on the arithmetic
SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS 11
12. of lexicographic
codes, or lexicodes
as he calls them.
Pure mathematics
par excellence! Later,
Scott Morris and
Bruce Oberg gave
back-to-back
presentations
heralding and
bemoaning,
respectively, the
number nine (this
being the ninth
Gathering for
Gardner). Pure nonsense, with a nod to
Gardner’s fictitious numerologist and
polymath, Dr. Matrix.
As the barrage of topics flowed from
the podium, what at first seemed a
disparate hodgepodge of individually
fascinating ideas began to gel into a
coherent form. There was mathematics,
there were puzzles, there was sleight of
hand and deception, and there was
mathematical artwork—the realization
of abstract ideas into concrete form.
There is beauty in all these, of course.
But together the topics celebrate noth-
ing less than our capacity to reason,
and they demonstrate the supremacy
of that endeavor. It is reason, after all,
that enables us to solve mathematical
problems and puzzles, and it is reason
that enables us to test our often flawed
perceptions and arrive at the truth.
Collectively, the topics also pay
homage to Martin Gardner himself, as
they are precisely the themes that
arose again and again in his writing. In
the end, it is deeply inspiring to witness
the extent to which his legacy lives on.
Further Reading
For more on Foshee’s Tuesday birthday
problem, see Andrew Gelman’s
Statistical Modeling, Causal Inference,
and Social Science blog entry for May
27, 2010, titled “Hype about conditional
probability puzzles” at
http://www.stat.columbia.edu/~co
ok/movabletype/archives/2010/05/
hype about cond.html.
To feel the sting of Bill Gosper’s
twelve-circle problem, download a
(free) playable computer version of the
puzzle at
http://demonstrations.wolfram.co
m/TheTroublesomeTwelveCircleProb
lem/.
View the “Dozenegger” at
http://gosper.org/Dozenegger3.pdf.
To test your skill at spotting a sleight of
hand, check out “The Color-Changing
Card Trick” at
http://www.youtube.com/watch?v=v
oAntzB7EwE.
The Ambiguous Corner Cube was first
discussed in C. L. Strong’s “The
Amateur Scientist” column in the
November 1974 issue of Scientific
American, p. 126. A template for
constructing a physical model can be
found at
http://www.bu.edu/lite/inkjet
science/pdfs/ProjectLITECornerCu
beThirdCut.pdf.
A brief introduction to Conway’s
lexicode theorem is available at
http://www.dpmms.cam.ac.uk/semin
ars/Kuwait/abstracts/L25.pdf,
and a more complete treatment can be
found in MASS Selecta: Teaching and
Learning Advanced Undergraduate
Mathematics, edited by S. Katok, A.
Sossinsky, and S. Tabachnikov (AMS,
2003).
For an introduction to the mysterious
Dr. Matrix, see Martin Gardner’s The
Magic Numbers of Dr. Matrix
(Promethius Books, 1985). You might
also enjoy the “Ask Dr. Matrix” online
tool at
http://www.iread.it/ask matrix.php.
DOI: 10.4169/194762110X525593
12 SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS
About the author: Bruce Torrence is
the Garnett Professor of Mathematics
at Randolph-Macon College, and a
co-editor of Math Horizons.
email: btorrenc@rmc.edu
13. SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS 13
W
hen I think of my experience
learning analysis, there is a
mixture of emotions. There
are great feelings of accomplishment
and perseverance that are seeded in
feelings of anxiousness, discourage-
ment, and
inadequacy. Before
analysis, I found
math enjoyable and
interesting. Never
before had it left a
bad taste in my mouth. I took math
courses because I found them easy
and they were good GPA boosters.
Analysis would change all of that. It
was unlike any other math I had ever
seen—to be honest, I wasn’t even sure
if it was math.
As I walked into the first lecture of my
first analysis class, I was excited but
had those first-day butterflies: Will the
course be interesting? Will I catch on
quickly? Will the professor be
impressed with my abilities? I had
taken a course from this professor
before, so that was not causing me too
much worry, but the text, Principles of
Mathematical Analysis, by Walter
Rudin, didn’t look like anything I had
seen before.
Then the professor walked in and
began to talk about Dedekind cuts and
other obscure-sounding mathematical
notions—density, orderings, countabili-
ty. It didn’t take long for me to realize I
was in way over my head. I had jumped
into the deep end, and I was drowning
quickly. Before this class, I never
understood my friends’ reactions when
I told them that I was majoring in math-
ematics: “I’m not a math person,” “It’s
too hard,” “I have math dyslexia.”
Now here I was, reconsidering my own
choice of major,
about to give up on
my hopes of
becoming a mathe-
matician. It was as
if I was taking the
math ride; it was fun, it was good, but
now I had to get off the bus. I had
reached the end of my abilities.
Not being the type of person to give up
easily, and with graduation so close, I
knew I had no choice but to complete
the course. And I did—with hard work
and extraordinary effort. I’ve since gone
on to graduate school and even passed
a candidacy examine in analysis! I am
Confronting AnalysisTina Rapke
It was as if I was taking the math ride; it was fun,
it was good, but now I had to get off the bus.
The View from Here
14. 14 SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS
sharing my story because I believe that
my initial reaction to analysis is not
unique, and the strategies I used to
make it through this experience could
be helpful to others.
Irrational Thoughts
My struggles began with the very first
assignment. Looking back at my notes
from that term, I see that we were
asked to show that the set of positive
rationals with squares greater than 2
has no least element. I stumbled
through the proof. It was clear that I
was trying to mimic a similar proof from
the class notes. Did I really understand
this argument? Apparently not! The
difficulties I was having are glaringly
obvious when I look at the second
assignment. There we were asked to
do the following problem from page 44
of Rudin:
“Regard Q, the set of all rational
numbers, as a metric space, with
d(p, q) = | p – q |. Let E be the set of all
p ʦ Q such that 2 < p2 < 3. Show that
E is closed and bounded in Q but that
E is not compact.”
The first line in my response was, “Let s
be the least rational such that s2 > 2.”
Wow, that’s amazing because I had
just proved that no such least element
exists! This demonstrates how
confused I really was during the
course—proving something one day
and then claiming the opposite on the
very next assignment.
I remember a conversation with the
professor after receiving a poor mark
on the assignment. We talked about
how I thought I could arrange the
rationals in order. Part of my confusion
came with the introduction of the
concept of “countability.” I knew that a
set was countable if there was a one-
to-one correspondence between the
set and the natural numbers, and we
had learned that the rationals were a
countable set, which meant that they
could be arranged in a sequence:
r1, r2, r3, …. Now, given any collection
of natural numbers, it is always possi-
ble to go through and pick out the
smallest one, so it seemed to me that
given a rational number rn on my list, I
could go through the list of remaining
rationals and let rn+1 be the “next
largest one.” At the professor’s request,
I experimented with trying to name the
“next largest” rational, and as you can
guess, I wasn’t able to do so.
I remember thinking in terms of
“thickness,” which turned out to be a
temporary stand-in for density. Given a
rational, I could not name the next one.
I related this to the same property of
the real numbers, so the rational
numbers and the real numbers were
similar in terms of “thickness.” But then
this blurred into my understanding of
cardinality and suggested to me that
the rationals had a cardinality close to
that of the reals. But the real numbers
are uncountable and the rationals are
countable. Why?
Looking back at all the course
assignments, I remember feeling tense
and anxious. My stomach was
constantly tied in knots. I wanted to run
away and pretend the assignments
never existed. I felt uneasy because I
knew I wouldn’t be able to complete
them. I didn’t understand what was
really going on. If I didn’t grasp the
concepts, how could I complete the
assignments?
Making Progress
Rudin’s definition says a set is
countable if there is a one-to-one
correspondence with the natural
numbers, which for me meant that to
show A is countable I had to find an
explicit function from the naturals to the
set A. But Rudin explains that the
rationals are countable in a less direct
manner. He first provides the following
result: Let A be a countable set, and let
Bn be the set of all n-tuples
(a1, a2, …, an) where each ak ʦ A. Then
Bn is countable. After establishing this,
he notes that we can apply this
theorem with n = 2 by observing that
the rational numbers are of the form
a/b where a and b are integers.
This is all well and good, but when
first learning about countability, I was
confused because I wanted to see a
correspondence with the natural
numbers via some explicit function.
Many years later, I found the following
theorem in Principles in Real Analysis
by Aliprantis: For an infinite set A the
following statements are equivalent:
15. SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS 15
(i) A is countable; (ii) There exists a
subset B ʕ N (where N is the set of
natural numbers) and a function
f : B A that is onto; (iii) There exists a
function g: A N that is
one-to-one.
Currently, I like justifying that the
rationals are countable by using the
following steps:
1. First consider the set P of positive
rationals.
2. Let B = {2n3m : n, m ʦ N} which is
a subset of N.
3. Construct the function f : B P
by letting f(2n3m) = n/m.
4. Appeal to the above theorem.
5. Finally, bring in the negative
rationals by noting that the union
of two countable sets is
countable.
My Story Now
I completed my first analysis course,
did fairly well, and received a
scholarship as a consequence. I’ve
written my Ph.D. candidacy exam in
analysis and passed. One of my
favorite memories of studying was one
night when analysis crept into my
dreams. I woke up in a panicky cold
sweat. In my dream I was being chased
by some analysis monster. My only
defense was to use the “blancmange
function” (a continuous but nowhere
differentiable function discussed in
Hairer and Wanner’s book Analysis by
Its History) as a boomerang. I took it as
a good sign at the time that analysis
concepts were finding their way into my
subconscious.
While studying for my candidacy exam,
I used several different textbooks. I
can’t say with certainty that my initial
misconceptions with density and
countability would have been resolved
by seeing the theorems from
Aliprantis’s Principles in Real Analysis
sooner, but this book fundamentally
changed my understanding of these
ideas—which brings up an important
point about textbooks: Before encoun-
tering analysis, I never consulted differ-
ent sources. If you’re having trouble in
math, you should consider finding
alternate textbooks and other
resources. I like Aliprantis and
Burkinshaw because of the way that
they lay things out and because they
have a very conversational style. They
use the word “we” a lot and make it
seem that the proofs are a team effort.
Understanding Analysis by Stephen
Abbott is another good resource. It is
also very well laid out, and the author
doesn’t give things up; he makes you
work, but he guides you and gives
many strong hints. I really enjoy the
discussions at the beginning of each
section, which include some history
and motivation. I also like Analysis by
Its History because it provides some
well-known counterexamples, and I like
learning through history. As for a first
text in analysis, I recommend A Friendly
Introduction to Analysis by Witold
Kosmala because it has many figures
and examples. However, over all of
these, I find Rudin to be the most
useful reference text. Rudin goes
straight to the point with very few fillers.
I like to think that Rudin provides a
good struggle—you need to read
between the lines, fill in the gaps, and
try really hard.
Since that first course, I’ve gone on to
take many other analysis courses,
including measure theory and
functional analysis, and I’ve done quite
well. It took hard work, but I’ve come to
appreciate the beauty and proof
techniques of analysis, and I now find it
fascinating.
In the end, if you’re having trouble with
analysis and find yourself in this story,
YOU ARE NOT ALONE! I’ve heard
many other successful graduate
students say that they had similar
issues when they first confronted
analysis. So keep with it. Some
struggles reap big rewards!
DOI: 10.4169/194762110X525566
→
→
→
The blancmange function
About the author: Tina Rapke is a
Ph.D. candidate at the University of
Calgary. She is pursuing an interdisci-
plinary degree in mathematics and
education.
email: tkrapke@ucalgary.ca
Photographs courtesy of Stephen Abbott
16. 16 SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS
A
classic summer-camp song—at least that’s
the way I think of it—goes like this:
100 bottles of beer on the wall,
100 bottles of beer;
If one of those bottles should happen to fall,
99 bottles of beer on the wall.
99 bottles of beer on the wall,
99 bottles of beer;
If one of those bottles should happen to fall,
98 bottles of beer on the wall.
(etc.)
And so on until you get to zero. The idea, of
course, is to entertain kids on a long trip with a
long song. For a longer trip, make a longer song
by starting with 200 or 300 bottles. Some years
ago, the following playful variation—for really
long trips!—occurred to me:
Infinite bottles of beer on the wall,
Infinite bottles of beer;
If one of those bottles should happen to fall,
Infinite bottles of beer on the wall.
Infinite bottles of beer on the wall,
Infinite bottles of beer;
If one of those bottles should happen to fall,
Infinite bottles of beer on the wall.
(etc.)
And then I thought, why stop with this basic fact
about transfinite arithmetic? So I started creating
other versions and, once word got out, began
receiving different verses from friends. A sampling
of this collection appears below where most, but
by no means all, are by me. The vast majority are
mathematically inspired, and collectively they
actually carry some pedagogical value. I have no doubt many
more entertaining and illuminating variations are possible, so
please pass along your contributions.
Basic transfinite version:
Infinite bottles of beer on the wall,
Infinite bottles of beer;
If finite bottles should happen to fall,
Infinite bottles of beer on the wall.
Infinite bottles of beer on the wall,
Infinite bottles of beer;
If finite bottles should happen to fall,
Infinite bottles of beer on the wall.
(etc.)
Larger infinity version:
Uncountable bottles of beer on the wall,
Uncountable bottles of beer;
If countable bottles should happen to fall,
Uncountable bottles of beer on the wall.
General transfinite version:
Aleph-n bottles of beer on the wall,
Aleph-n bottles of beer;
If, where m < n, Aleph-m bottles should happen to fall,
Aleph-n bottles of beer on the wall.
(etc.)
Indeterminate version:
Infinite bottles of beer on the wall,
Infinite bottles of beer;
If infinite bottles should happen to fall,
Indeterminate bottles of beer on the wall.
(The End)
Geometric progression version:
2100 bottles of beer on the wall,
2100 bottles of beer;
Donald Byrd
Infinite Bottles of Beer
A Cantorial Approach to Cantorian
Arithmetic and other Mathematical Melodies
CartoonbyJohnJohnson
17. SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS 17
If half of those bottles should happen to fall,
299 bottles of beer on the wall.
299 bottles of beer on the wall,
299 bottles of beer;
If half of those bottles should happen to fall,
298 bottles of beer on the wall.
…
(optional ending)
One bottle of beer on the wall,
One bottle of beer;
If half of those bottles should happen to fall,
Half of a bottle of beer on the wall.
(etc.)
Euler’s identity version:
100 bottles of beer on the wall,
100 bottles of beer;
If e2πi of those bottles should happen to fall,
99 bottles of beer on the wall.
(etc.)
Differential version:
100 bottles of beer on the wall,
100 bottles of beer;
If dbeer/dt = –1,
99 bottles of beer on the wall.
(etc.)
Infinitesimal version:
100 bottles of beer on the wall,
100 bottles of beer;
If db bottles should happen to fall,
100 bottles of beer on the wall.
(etc.)
Identity version:
100 bottles of beer on the wall,
100 bottles of beer;
If none of those bottles should happen to fall,
100 bottles of beer on the wall.
(etc.)
Topological dimension version:
100 bottles of beer in a square,
100 bottles of beer;
If one of the dimensions should happen to fall,
10 bottles of beer in a line.
10 bottles of beer in a line,
10 bottles of beer;
If one of those dimensions should happen to fall,
One bottle of beer on the wall.
(The End)
Fractal dimension version:
1 bottle of beer on the wall,
1 bottle of beer;
If the middle third of that bottle should happen to fall,
(1/3 + 1/3) bottles of beer on the wall.
(1/3 + 1/3) bottles of beer on the wall,
(1/3 + 1/3) bottles of beer;
If each middle third of those bottles should fall,
(1/9 + 1/9 + 1/9 + 1/9) bottles of beer on the wall.
(etc.)
Russell’s Paradox version:
Four lines of lyrics on the wall,
Four lines of lyrics;
If one of the lines should happen to fall,
(The End)
Base 2 version:
100 bottles of beer on the wall,
100 bottles of beer;
If one of those bottles should happen to fall,
11 bottles of beer on the wall.
(etc.)
Continuum hypothesis version:
Beth-1 bottles of beer on the wall,
Beth-1 bottles of beer;
Take Aleph-1 down, and pass them around:
There may be no bottles of beer on the wall.
(The End)
DOI: 10.4169/194762110X524657
About the author: Donald Byrd is Woodrow Wilson
Indiana Teaching Fellow and Adjunct Associate Professor
of Informatics and Music at Indiana University. Lacking
the discipline to work hard at mathematics, he took up
computing as the path of least resistance,
but he still enjoys the ideas of math greatly.
email: donbyrd@indiana.edu
18. 18 SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS
This document was considered so valuable that it was put
inside the treasure chest for safekeeping. It was a lot safer
than was originally thought.
The sponsor of the expedition was definitely not pleased with
the current state of affairs, so she hired a team of mathemati-
cians to try to open it. The task was seemingly hopeless, and
many gave up, until Dr. Jacob Ecco arrived with his sidekick,
Professor Justin Scarlet.
“My dear professor,” said Dr. Ecco, “we must approach this
challenge systematically. What is the most basic principle in
problem solving?”
“Downsizing,” replied the professor, who was well versed
with the methods of Dr. Ecco. “Suppose there is only one
bongo drum. Then the treasure chest will open
automatically.”
“What if there are two bongo drums?” asked Dr. Ecco.
Hsin-Po Wang
Congo Bongo
Illustration courtesy of Greg Nemec
Dedicated to the memory of Martin Gardner
A
n expedition into Congo uncovered a treasure chest
in the shape of a regular octagon. At each corner was
a bongo drum. A scroll attached to the chest, written
in French, explained that there was a genie inside each
bongo drum. A genie is either standing upright or doing a
handstand. One may strike a number of bongo drums at the
same time. When a bongo drum is struck, the genie inside
will change its posture from right side up to upside down, or
vice versa. The treasure chest will open if and only if all
genies are right side up, or all are upside down. However,
each time some bongo drums are hit the treasure chest will
spin rapidly on its vertical axis. As the bongo drums are all
identical in appearance, after the rotation it is impossible to
tell which of them had just been hit.
Unfortunately, the scroll did not record the exact procedure
by which the treasure chest might be opened. However, it
mentioned that such a procedure had been documented.
19. SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS 19
“Well, if the treasure chest is not already open, hitting either
of the bongo drums will do it.”
“Excellent! As usual, my dear friend, you have provided me
with the inspiration that leads to the solution.”
“How?” cried Professor Scarlet. “I have made but the most
trivial observation.”
“Nevertheless, a solution to the case with two bongo drums
leads to a solution to the case with four bongo drums, which
in turn leads to a solution to the case with eight bongo
drums,” Dr. Ecco proclaimed.
“I can smell that you are going to use induction, but l still do
not see the connection from one case to the next,” replied
Professor Scarlet.
“Be patient. Obviously, we will be treating pairs of bongo
drums as a single bongo drum. How can we identify pairs of
bongo drums? Adjacency is unsatisfactory because after the
treasure chest spins around, it is impossible to tell whether a
bongo drum is to be paired with its neighbor to the left or its
neighbor to the right.”
“The obvious solution is to identify opposite pairs of bongo
drums,” pronounced Professor Scarlet.
“What insight!” said Dr. Ecco with a slight smile.
“This is fine if the genies in each opposite pair of bongo
drums are both right side up or both upside down. But there
is no guarantee that they start off like that,” his companion
pointed out.
“True, but there is a way to make them end up that way. Let 0
or 1 indicate whether a genie is right side up or upside down,
and consider a four-sided chest. If we are really lucky, the ini-
tial state may already be (0,0,0,0) or (1,1,1,1).”
“You may be good enough to be that lucky, but that never
works with me,” complained Professor Scarlet.
“I am not counting on luck. I am just singling out the most
favorable scenario. We may regard (0,0,0,0) and (1,1,1,1) as
the same state. Call it state 0. It is also referred to as an
absorbing state, in that once we enter it we do not leave (for
the problem is then solved). How many other states are
there?” asked Dr. Ecco.
Professor Scarlet paused. “Well, we have three other states,
namely, (0,1,1,1), (0,0,1,1), and (0,0,0,1).”
“You are correct as far as the number of other states is
concerned, but wrong about their composition,” Dr. Ecco
said. “By symmetry, we may consider (0,1,1,1) and (0,0,0,1)
as the same state. We call it state 3. Your (0,0,1,1) is state 2,
which is not the same as (0,1,0,1). This I call state 1.”
“Yes!” exclaimed Professor Scarlet excitedly. “In state 1, the
genies in each opposite pair of bongo drums are both right
side up or both upside down. In the case with two bongo
drums, we hit one of them. So here we hit an opposite pair of
bongo drums, and the treasure chest will open.”
“We will call hitting an opposite pair of bongo drums opera-
tion A, and as you keenly observed, performing this opera-
tion from state 1 will unlock the chest. What happens when I
perform operation A if we are in state 2 or state 3?”
“Let me see,” Professor Scarlet said. “Ah, we will remain in
the same state as before.”
“How can I move from state 2 to state 0 or 1?” Dr. Ecco
asked.
“Well, we may hit an adjacent pair of bongo drums. I
suppose this will be operation B. If we hit both zeros, or both
ones, we will be in state 0 immediately. If we hit one 0 and
one 1, we will be in state 1.”
“Indeed, you can check that state 3 is left unchanged by
operation B, just as it was unchanged by operation A,” Dr.
Ecco agreed.
“The rest is easy now. We perform operation C by striking
just one bongo drum. This will change state 3 into state 0,
state 1, or state 2,” Professor Scarlet said triumphantly.
“Not so fast,” Dr. Ecco cauntioned. “After operation B, we
may be in state 1, and if we rush into operation C now, state
1 will become state 3, and we will be going in circles.”
“I was hasty,” admitted Professor Scarlet. “We must perform
operation A once more to clear state 1 before taking care of
state 3. After operation C, we will clear the lower states with
the sequence ABA again. So the overall procedure for the
four-sided chest is ABACABA.”
“Excellent. Here, l have drawn a state transition diagram for
you,” Dr. Ecco said.
Here is the same drawing, with an emphasis on the structure
of the transition operations:
20. 20 SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS
“That explains everything clearly!” said the professor. “Now
I am confident we are onto something that will crack the
eight-sided treasure chest. Let’s see. We wish to treat each
pair of opposing bongo drums as a single entity. As before,
the simplest states are those in which the genies in each
opposite pair of bongo drums are both right side up or both
upside down. Here is my drawing of this part of the state
transition diagram. It is really the same as yours, except that
operation A means hitting every other pair of opposing
bongo drums; operation B means hitting any two adjacent
pairs of opposing bongo drums; operation C means hitting
any pair of opposing bongo drums. By performing the
sequence ABACABA when starting from any such simple
state, the treasure chest will open.”
“Excellent, my dear professor,” exclaimed Dr. Ecco. “These
states together form an expanded absorbing state in the
overall diagram below. It is the box marked 4, where the box
marked m contains all states with m matching opposite pairs,
for 0 ≤ m ≤ 4.”
“This diagram looks suspiciously like the ones before, except
that states 4 and 0 are no longer equivalent. Why do you
have a state 2 and a state 2’? And what are the operations D,
E, F, and G?”
“The states with 2 matching pairs are classified according to
whether those matching pairs are alternating or adjacent,” Dr.
Ecco said. “The former are grouped under state 2 while the
latter are grouped under state 2’. Operation D hits any four
adjacent bongo drums. Operation E hits any two bongo
drums separated by one other drum. Operation F hits any
two adjacent bongo drums. Operation G hits any one bongo
drum.”
“Let me see. If we denote the sequence ABACABA by X,
then the sequence that will open the treasure chest is
XDXEXDXFXDXEXDXGXDXEXDXFXDXEXDX,” summed up
Professor Scarlet.
“We keep repeating X so that whenever we enter state 4, we
will not return to another state. Whatever the initial state of
the treasure chest may be, it will open by the end of this
sequence.”
The sponsor of the expedition was delighted with their
analysis. She hired a team of Congo natives to strike the
bongo drums in the prescribed fashion. Lo and behold, the
treasure chest opened right before her eyes. Unable to
contain her excitement, she reached inside, but found only
another scroll. It was written in Kituba, but the most
prominent line read something like
XDXEXDXFXDXEXDXGXDXEXDXFXDXEXDX.
Later, back in his New York apartment, Dr. Ecco said to
Professor Scarlet, “That was hilarious, wasn’t it? By the way,
have you thought of the obvious question?”
“Yes, l have,” replied the professor. “For what number of
bongo drums can such a treasure chest be opened? From
our analysis, the general procedure when the number is a
power of two is clear. We treat each opposing pair as a single
entity, thereby reducing to the preceding case. Then we
progressively move all states into the expanded absorbing
state. But what happens if the number of drums is not a
power of two?”
21. “Then unlocking the chest is not always possible,” Dr. Ecco
replied. “For in this case the number of bongo drums has an
odd prime factor p. Suppose the pagan god Hemba is having
fun with us. He will choose p evenly spaced bongo drums
and make sure that the genies inside are not all right side up
and not all upside down. Now ignore all other bongo drums
except these p. The two types of bongo drums are not equal
in number since p is odd. In order for us to succeed, we must
strike precisely the bongo drums of one type. But Hemba will
spin the treasure chest so that the bongo drums we plan to
hit include at least one from the opposite group. This way, he
can keep us from opening the treasure chest forever.”
See problem 249 in the Playground on page 30 for more fun
with these treasure chests.
Further Reading
This problem was posed in the Senior A-Level paper in the
2009 Fall Round of the International Mathematics
Tournament of the Towns. It is related to a problem posed in
Martin Gardner’s famous “Mathematical Games” column in
Scientific American (Feb. 1979), which is reproduced in his
anthology Fractal Music, Hypercards and More Mathematical
Recreations (W.H. Freeman, 1992). Gardner’s version of the
problem is also treated in three other sources:
Ted Lewis and Steve Willard, The rotating table, Mathematics
Magazine, 53 (1980): 174–179.
William Laaser and Lyle Ramshaw, Probing the rotating table,
in The Mathematical Gardner, edited by David Klarner,
(Wadsworth, 1981) 288–307.
Albert Stanger, Variations on the rotating table problem,
Journal of Recreational Mathematics, 19 (1987): 307–308
and 20 (1988): 312–314.
Jacob Ecco and Justin Scarlet are fictitional characters
created by Dennis Shasha of the Courant Institute, New York.
He is a leading creator of mathematical puzzles. Shasha has
written a number of puzzle books featuring these clever
detectives. A good place to start is his book The Puzzling
Adventures of Doctor Ecco (Dover, 1998).
DOI: 10.4169/194762110X525601
About the author: Hsin-Po Wang is enrolled in Taipei
Municipal Jianguo High School. He won a gold medal in the
2009 International Mathematical Olympiad and several
awards in the International Mathematics Tournament of the
Towns. He has also participated in science fair competitions
in Taiwan and in the United States.
email: billy820621@yahoo.com.tw
Cartoon courtesy of xkcd.com
xkcd
SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS 21
22. 22 SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS
with the peg solitaire “problem space” sets in, and one
abandons the game, abandons solving individual puzzles in
favor of proving theorems, or seeks a twist on the rules to
make the game new again. Opting for the latter, I propose a
variation that I call “Port-and-Sweep Solitaire” (PaSS). PaSS
is also played on a grid, and the First Rule of PaSS is that
each square may hold 0, 1, or 2 counters—no more, no less.
I usually play on my computer, but checkers on a checker-
board work just as well (and are pleasingly tactile). PaSS
permits two types of move:
1. Sweep move: Add to four consecutive
squares on the board, and
2. Port move: Add to three consecutive squares
on the board.
Both moves are subject to the First Rule, and as in OPS,
they may be performed in all four directions. Lacking several
hundred years of tradition, PaSS does not yet boast a single
defining problem, but it does admit a wealth of varied and
interesting ones. Figure 1 is a small, but typical, puzzle: we’re
asked to reduce a given arrangement of sixteen counters on
a 5 ϫ 5 board to a single counter in the center. Up to
symmetry, there are just three opening moves possible: port
a 2 to the center, port a 2 to one of the corners, or sweep
three 1’s to the edge.
Figure 2 shows a partial solution to this problem, using the
third option to start. See if you can carry it to completion.
This is a very forgiving problem; there are many ways to
complete the solution I’ve begun, and my opening isn’t too
special. Any two moves at the start can be carried to a suc-
cessful conclusion. But if you arrive at one of the positions in
Figure 3 on your third move, you will inevitably get stuck with
no legal moves and at least three counters on the board.
Figure 4 offers three more problems on the 5 ϫ 5 board,
which are somewhat less forgiving. In one case you can go
Figure 1. Left: A small puzzle in its initial state.
Right: The desired final state.
Jacob Siehler
Port-and-Sweep Solitaire
Image courtesy of Cristóbal Vila - etereaestudios.com
H
ow does this happen? I just wanted a nice game
where I didn’t have to count higher than two, and I
ended up dealing with imaginary numbers. But let me
back up: I’ve been a little obsessed with a puzzle lately, and I
would like to explain what’s puzzling me and how the square
root of –1 can sneak in where you least expect it. The puzzle
in question is a relative of the classic peg solitaire game
pictured above, and before I introduce it properly, we can
have a quick brush-up on the traditional version. You could
write a whole book on the subject [1] but as you probably
know, the game is quite simple: it is played on a grid, where
each square may hold a peg (or a marble, or some other
marker). The only allowable move is to jump one peg over an
adjacent peg into an empty square, removing the jumped
peg from the board:
This can be done in any direction on the grid—up, down, left,
or right. An equivalent way of looking at the move is that it
consists of adding to the “peg count” of three
consecutive squares on the grid—subject to the rule that a
square can’t contain more than one peg, or a negative
number of pegs.
The best-known problem in ordinary peg solitaire (OPS) asks
the player to reduce an almost-full board of 33 squares to a
single peg in the center square (see photo), but of course
other problems are possible—easier problems to warm up
on, or fresh challenges for those who have solved the classic
configuration [8]. Eventually, though, a feeling of familiarity
23. SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS 23
astray on the first move! Despite the modest length of the
problems and the small board size, I think you’ll find their
solutions satisfying. See Problem 250 in the Playground on
page 30.
Sorry, You Can’t Get There from Here
Trying to reduce the problem in Figure 5 to a single counter
will be less satisfying, however, and after a certain number of
failed attempts, you will begin to suspect that it can’t be
done. How can you prove that no solution exists, without
examining every possible sequence of moves?
Figure 4. Three more games to try on a 5 ϫ 5 board.
An elegant way is to use an invariant—a function whose
value is determined by the state of the board and will not
vary when we change the board by legal moves. If such a
function can be found that takes one value on the pristine
state of the problem and a different value on the target state,
then no sequence of legal moves can ever connect the two.
PaSS has a natural affinity for mod 3 arithmetic, and we can
take advantage of that to define a valuable pair of invariants
for the game. These invariants will assign to each board a
value in a charming and petite arithmetic universe of just nine
elements; namely, the set
F9 = {0, 1, 2, i, 1 + i, 2 + i, 2i, 1 + 2i, 2 + 2i}.
F9 is a discrete, mod 3 analogue of the complex number
system. Elements have the form a + bi; we can refer to a as
the real part and b as the imaginary part. Addition uses the
simple rule (a + bi) + (c + di) = (a + c) + (b + d)i, reducing both
parts mod 3 to stay in the range {0, 1, 2}. To multiply, treat i
as a square root of –1, but in the mod 3 world, –1 is the same
as +2, so you can avoid minus signs by declaring i2 = 2. That
leads to the peculiar, but totally legitimate, multiplication
formula
(a + bi) + (c + di) = (ac + 2bd) + (ad + bc)i.
F9 is an abelian group under addition, and the nonzero
elements form an abelian group under multiplication—that is,
F9 is a field of nine elements. We don’t need much of the
multiplicative structure for the present purposes, but note
that i(2i) = 1, that is, i–1 = 2i.
Now, coordinatize your board in the usual Cartesian manner,
letting the lower left square be (0,0). Let S(a,b) denote the
number of counters on square (a,b), and define
where all arithmetic—sums, products, and powers—takes
place in F9, so each of the sums above, no matter how
lengthy, will reduce to one of the nine elements.
Suppose a position SЈ is derived from S by a sweep
to the right, taking one counter from each of
(a, b), (a + 1, b), (a + 2, b) and adding two to (a + 3, b). Then
since the sum in parentheses is zero in F9. Similar
calculations show that the value of π is unchanged by
sweeps in the other three directions. And if SЈ is derived from
S by porting a 2 from (a, b) to (a + 2, b), then
π π
π
( ') ( )
( ) (
S S i i
S i
a b a b
a b
= − +
= +
+ + +
+
2
1
2
++
=
i
S
2
)
( ),π
π π( ') ( )S S i i i ia b a b a b a b
= − − − ++ + + + + + +1 2 3
2
= − + + +
=
+
π
π
( ) ( )
( ),
S i i i i
S
a b
1 2 3
π μ( ) ( , ) ( ) ( , )
( , ) (
S S a b i S S a b ia b
a b
a b
a
= =+
∑ and
, )
,
b
∑
Figure 2. A promising opening sequence of moves.
Figure 3. Dead-end configurations.
Figure 5. An impossible game.
24. 24 SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS
and likewise for ports in other directions. This shows that the
value of π remains unchanged under all legal moves in the
game, and it’s routine to check that has the same property.
Thus, the values of π and separate the possible positions
on a given board into 81 distinct classes, and no play
between positions in different classes is ever possible. A
5 ϫ 5 board with a single counter in the center has π = 1 and
= 1.
What of the board in Figure 5? Its invariants are π = 2 + i and
= 1 + 2i, so it can never be played to a 1 in the center. In
fact, it could never be played to a single 1 or single 2 on any
square, since no single term in the sums for π and can
contribute both a real and imaginary part.
I should admit that I did not have to overexert myself to
discover π and , because I knew of an analogous pair of
invariants for ordinary peg solitaire, described by de Bruijn
[6], using mod 2 arithmetic and a field of four elements. PaSS
can therefore be seen as a plausible answer to the analogy
problem, “What game is to the field of nine elements as peg
solitaire is to the field of four elements?”
More Problems: Four in the Corners
Back to problems you can solve: Figure 6 offers a few
additional problems on a 6 ϫ 6 board. Boards with even
sides have no center square, so as an attractive, symmetric
alternative, these are designed to play down to four 1’s, one
in each of the four corners.
How big is the “problem space” of potentially interesting
four-corners puzzles? Not very big, if we’re picky about what
we consider an interesting problem. I prefer problems with
a lot of symmetry, and there are 39 = 19,683 six-by-six
boards that have 90-degree rotational symmetry (including
the four-corners configuration itself, and some dull things like
the all-zero board). Taking the π and invariants into account
Figure 6. Some four-corners games, where the winning
configuration has a 1 in each corner.
reduces the possibilities by a factor of 34, so at most
35 = 729 of these could plausibly be played to four corners.
Among those, a large number share the disappointing
property exhibited by Figure 7: one can easily reduce the red
cells to a single 1 in the upper left corner, using ports alone,
without affecting the rest of the board.
Due to symmetry, this can be repeated three more times,
verbatim, to solve the problem. Boring! A port-and-sweep
problem should require both ports and sweeps, I’m sure you
agree. Eliminating the boring problems leaves a list of 273
candidates to be investigated—some that can’t be solved,
some too small or too obvious to be interesting, some too
large to be fun, and just a few in the Goldilocks region:
tantalizing, but elusive. Of course, if we’re open to problems
with less symmetry, there will be many more possibilities, and
there are many tactics yet to be discovered which will aid in
their construction and solution.
As a parting shot, the 7 ϫ 7 board in Figure 8 was generated
by a computer working backward from a single 1 in the
center and trying to obtain a symmetric configuration (in this
case, just a reflective symmetry). I did not have the computer
save its steps, so I know the problem can be solved, but I
don’t know how! I believe it’s large enough to vex attempts to
solve it by sheer computing power—but perhaps a reader will
discover a more ingenious, tactical approach. See Problem
250 in the Playground on page 30.
Mutatis Mutandis
Ordinary peg solitaire has been studied extensively, and while
Beasley’s book [1] is the most comprehensive single work,
other stimulating articles can be found. Anyone who is keen
to explore will find that most of the problems and techniques
that have been developed for OPS can be successfully,
Figure 8. Can you play this to a single 1 in the center?
Figure 7. A four-corners game where no sweeping is needed.
25. SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS 25
ahem, ported over to PaSS—but there are surprises to be
found in the process. For example, “resource counts” can be
constructed which provide an additional tool for proving
certain PaSS problems unsolvable, but they are governed by
more restrictive inequalities than in OPS. PaSS seems to be
more complicated than OPS on a one-dimensional (1 ϫ n)
board. And an investigation of the “Solitaire Army” problem
([2], [4]) will show that PaSS counters have much more
forward mobility than their ordinary peg counterparts. That
problem might lead you to wonder, “What is to PaSS as the
Golden Ratio is to OPS?”—but I don’t believe any works of
art will be inspired by the answer.
Now, I’m going to have just one more try at the game in
Figure 8…
Further Reading
[1] John Beasley, The Ins and Outs of Peg Solitaire, Oxford
University Press, 1985.
[2] George I. Bell, George Bell’s Peg Solitaire Page,
http://home.comcast.net/
~gibell/pegsolitaire/index.html.
[3] George I. Bell, A fresh look at peg solitaire, Mathematics
Magazine, 80(1) 2007, 16–28.
[4] Elwyn R. Berlekamp, John H. Conway, and Richard K.
Guy, Winning Ways for Your Mathematical Plays, volume 4,
2nd edition, AK Peters, 2004. See chapter 23, pages
803–842.
[5] Arie Bialostocki, An application of elementary group
theory to central solitaire. The College Mathematics Journal,
29(3) 1998, 208–212.
[6] N. G. de Bruijn, A solitaire game and its relation to a finite
field, Journal of Recreational Mathematics, 5 1972, 133–137.
[7] Christopher Moore and David Eppstein, One-dimensional
peg solitaire, and duotaire, In Richard J. Nowakowski, editor,
More Games of No Chance, Cambridge University Press,
2002.
[8] Jacob A. Siehler, Problems in peg solitaire,
http://demonstrations.wolfram.com/ProblemsInPegSo
litaire/.
DOI: 10.4169/194762110X525575
About the author: Jacob Siehler is an assistant professor of
mathematics at Washington and Lee University, who grew
up reading Martin Gardner and who continues to grow by
reading Martin Gardner.
email: siehlerj@wlu.edu
26. 26 SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS
identify “multiple patient exchanges”
like the one described for these two
couples. The doctors get all the glory,
but mathematics makes the exchange
work.
Background
The only cure for end-stage renal failure
is a kidney transplant. Dialysis can
extend a patient’s life while the
patient waits for a kidney to
become available but
does nothing to cure
the disease. Accord-
ing to the National
Kidney
Foundation, as of
April 2007 there
were 70,870
patients on the
United Network
for Organ Shar-
ing (UNOS) wait
list for a kidney
transplant. In 2006,
18,016 patients
received a kidney
transplant while 3,916
died waiting for a kidney to
become available. Without an
increase in the number of kidneys
available for transplant, the UNOS wait
list will continue to grow.
One potential source of kidneys is living
donors. A living donor is a person who
is willing to donate one of his kidneys.
People are born with two kidneys, but it
is possible to live a long, healthy, and
body. The two couples do not know
each other and they have different
doctors, but they will soon transform
each other’s lives. It turns out that Jim
can donate his kidney to Bob and Mary
can donate her kidney to Kathy. The
two families are beneficiaries of a
kidney exchange clearinghouse.
A kidney exchange clearinghouse
gathers information about incompatible
patient-donor pairs and attempts to
M
ary sighs deeply. She is sitting
next to her husband, Bob,
who has fallen asleep. Bob is
hooked up to a dialysis machine as he
has been three times a week for the
last year. Mary thinks about all the
traveling she and Bob had planned to
do in their retirement years. That’s all
on hold now; Bob can’t be away
from the hospital for more than
three days at a time. He
needs a kidney transplant.
Mary would be happy
to donate one of her
kidneys, but they
are not compatible
with Bob’s blood
type. Bob is on
the list for a
cadaver trans-
plant, but the
wait for someone
with his blood
type averages
three years, and he
must get much sicker
to advance to the top
of the list. In the mean-
time, their lives are on hold.
Nearby, Kathy holds her breath as
she waits for the doctor to tell her the
results of her test. She is about to find
out if her husband, Jim, can donate a
kidney to her. The doctor says “positive
crossmatch.” Kathy and Jim are not
sure what a positive crossmatch is, but
they know it means Jim’s kidney
cannot be transplanted into Kathy’s
Olivia M. Carducci
Three kidneys that would not otherwise be available for transplant have
been made available, and three people’s lives have been saved.
The Mathematics behind
Lifesaving Kidney Exchange Programs
27. SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS 27
active life with only one. In 2006, more
than a third of all transplanted kidneys
came from living donors. Unfortunately,
significant numbers of willing donors
are unable to donate due to blood or
tissue incompatibilities. This is the
impetus for the creation of kidney
exchange programs.
In a kidney exchange, a group of
donors who are unable to give a kidney
directly to their loved ones donate
instead to another member of the
group. For example, consider three
donor-recipient pairs: (Da, Ra), (Db, Rb),
and (Dc, Rc). If Da donates a kidney to
Rb, and Db’s kidney goes to Rc, and
Dc’s kidney goes to Ra, then this is a
three-way kidney exchange. Three
kidneys that would not otherwise be
available for transplant have been
made available, and three people’s lives
have been saved. There are several
regional kidney exchange programs,
most notably in New England, Ohio,
and Baltimore. As we will see,
mathematics, and graph theory in
particular, is behind the algorithms that
are used to identify potential
exchanges.
The Top Trading Cycles
and Chains Algorithm
One of the first algorithms utilized for
exchanges is the Top Trading Cycles
and Chains (TTCC) algorithm.
Consider a set of donor-recipient pairs
(Di, Ri). Donor Di will often be referred
to as kidney Di and recipient Ri will
often be referred to as patient Ri.
Assume that kidney Di is incompatible
with recipient Ri, so that an exchange
of some sort is required. We will use w
to represent the cadaver wait list. Each
patient ranks the available kidneys (and
possibly the wait list w), with the most
preferred kidney ranked first. For
example, consider a kidney exchange
problem with eight pairs (D1, R1), …,
(D8, R8), and with preferences as
follows:
Note that some lists end with w while
others end with the recipient’s donor
kidney. A recipient whose list ends with
w is willing to have his donor’s kidney
transplanted in exchange for the
recipient being moved near the top of
the wait list. A recipient whose list ends
with his donor is only interested in an
exchange that matches him with a
living donor. If the recipient is matched
with his donor’s kidney, the recipient
does not receive a kidney in this round
of the exchange.
The TTCC algorithm consists of several
rounds, or stages. At each stage, the
exchange problem will be modeled as a
directed graph. Recall that a directed
graph consists of a set of points, called
vertices, and a set of ordered pairs of
vertices, called edges. At the end of
each stage, a subset of the available
kidneys will be assigned so the
exchange problem for the next stage
will be smaller.
Consider a directed graph with a vertex
for each patient Ri, each donor Di, and
the cadaver wait list w. There is an
edge from Di to Ri and from Ri to the
patient’s most preferred kidney (or to
w).
TTCC Algorithm
Step 1: Formulate the directed graph
with vertices for all available donor-
recipient pairs.
Step 2: If there are no cycles in the
graph, skip to step 3. Otherwise,
identify all cycles, and carry out the
corresponding exchanges. (If edge
(Ri, Dj) is in the cycle, then patient Ri
receives kidney Dj. If i = j, then patient
Ri does not receive a kidney in this
exchange.) Remove all donor-recipient
pairs that were in a cycle from the list of
available donor-recipient pairs. If the
available donor-recipient list is empty,
stop; otherwise, go to step 1.
Step 3: All patients are in a chain that
ends with w, but patients can be in
more than one such chain. Choose a
chain ending with (Rp, w). If edge
(Ri, Dj) is in the chosen chain, then
patient Ri receives kidney Dj, and Rp
will receive priority on the wait list.
Remove the donor-recipient pairs in
the chain from the list of available
donor-recipient pairs, but the kidney at
the start of the chain is still available. If
the available donor-recipient list is
empty, stop; otherwise, go to step 1.
Let’s return to the example above. For
step 1 in the initial stage, we construct
the directed graph as shown in Figure
1. (We don’t show vertex w as no
edges are incident with it.)
Figure 1. Directed graph for the initial stage of step 1.
R D D w R D D D w
R D D D w R D w
R D D w R
1 4 5 5 3 1 7
2 1 3 7 6 7
3 4 5
: :
: :
: 7 3 7
4 8 2 8 2 8
:
: :
D D
R D D w R D D
28. Step 2: The only cycle is
(D1, R1, D4, R4, D8, R8, D2, R2, D1).
Make the corresponding assignments,
and remove the donor-recipient pairs
from the list of available pairs.
Step 1: The new graph is shown in
Figure 2. Note that the edge (R3, D5)
has been added since kidney D4 is no
longer available.
Step 2: The only cycle is
(D5, R5, D3, R3, D5). Make the
corresponding assignments, and adjust
the list of available donor-recipient
pairs.
Step 1: The new graph is shown in
Figure 3.
Step 2: Again, there is one cycle,
(D7, R7, D7). This assignment means
patient 7 does not receive a kidney in
this run of the algorithm.
Step 1: The new graph is shown in
Figure 4.
Step 2: Go to step 3.
Step 3: The only path is (D6, R6, w),
which means patient 6 is given priority
on the cadaver queue and kidney 6 is
made available to a patient on the
cadaver queue. Stop.
The final assignments are:
and D6 is available to the cadaver
queue.
The cycle found in stage 1 illustrates
the main problem with the TTCC
algorithm: this cycle involves four
patients and kidneys. The logistics of
coordinating four transplants makes
this impractical. Each transplant
requires two operating rooms and their
staffs. Some incompatibilities between
recipients and donors can only be
identified by mixing their blood. Only 10
percent of pairwise matches will exhibit
such incompatibilities, but with a cycle
of three donor-recipient pairs, such an
incompatibility will occur 27 percent of
the time, and with a cycle of four pairs
such an incompatibility will occur 47
percent of the time. The TTCC
algorithm can generate arbitrarily long
cycles.
Pairwise Exchanges
Exchanges that involve only two
donor-recipient pairs (like the one in the
introduction) will be called pairwise
exchanges. Consider an undirected
graph G with a vertex for each donor-
recipient pair and an edge between two
vertices if each of the two donor
kidneys is compatible with the recipient
from the other vertex. In this case, all
compatible kidneys are viewed as
equally desirable. This may seem
problematic, but since success rates
are comparable for all compatible live
kidney donations, many transplant
surgeons consider it reasonable.
Let V denote the vertices of G and E
the edges. A matching is a subset of
E such that each vertex in V is in at
most one edge of . Each edge of
represents a pairwise exchange. An
optimal set of pairwise exchanges is a
matching containing the most edges,
called a maximum cardinality matching.
A patient whose vertex is not included
in the maximum cardinality matching
does not receive a donor kidney in this
exchange.
An augmenting path for is a path
from a vertex not in an edge of to
another vertex not in an edge of
whose edges are alternately not in
and in . An augmenting path can be
used to increase the number of edges
in a matching by one. The following
theorem can be found in most under-
graduate graph theory texts.
Theorem: A matching in a graph G is
a maximum cardinality matching if and
only if there is not an augmenting path
for in G.
Maximum Cardinality Matching
Algorithm:
Step 1: Formulate the graph G with a
vertex for each donor-recipient pair and
an edge between compatible pairs. Set
the matching = л.
Step 2: If there is no augmenting path
in G, then output the maximum
cardinality matching . Otherwise, find
an augmenting path P in G.
Step 3: Form the new matching by
reversing the roles of the edges in P;
edges in P ʝ are removed from
and edges in P – are added to . Go
to step 2.
As an example, consider seven donor-
recipient pairs, (D1, R1), (D2, R2), …,
(D7, R7) represented by vertices v1, v2,
…, v7, respectively, with the recipients’
compatible donors as listed below.
28 SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS
R R R R R R R R
D D D D D w D D
1 2 3 4 5 6 7 8
4 1 5 8 3 7 2
R D D D R D D
R D D D D R D D D
R D D
1 2 4 7 5 4 6
2 1 3 4 7 6 4 5 7
3 2 4
: :
: :
: D R D D D D D
R D D D
7 7 1 2 3 4 6
4 3 5 6
:
:
Figure 2. Directed graph for the second
stage of step 1.
Figure 3. Directed graph for the third
stage of step 1.
Figure 4. Directed graph for the final
stage of step 1.
29. Since D2 appears on R1’s list and D1
appears on R2’s list, there is an edge
from v1 to v2; although D4 appears on
R1’s list, D1 does not appear on R4’s
list, so there is no edge between v1 and
v4.
Step 1: Formulate the graph
representing the compatibilities. This is
shown in Figure 5.
Step 2: Augmenting path: (1, 2). (Any
edge could be chosen.)
Step 3: = {(1, 2)}
Step 2: Augmenting path: (3, 2), (2, 1),
(1, 7)
Step 3: = {(3, 2), (1, 7)}
Step 2: Augmenting path: (4, 3), (3, 2),
(2, 1), (1, 7), (7, 6)
Step 3: = {(4, 3), (2, 1), (7, 6)}
Step 2: There are no augmenting paths
left, so the matching {(4, 3), (2, 1), (7, 6)}
is a maximum cardinality matching.
Clearly there are other matchings that
also include three edges, for example
{(1, 7), (3, 4), (5, 6)}, but no matching
includes more than three edges. The
medical community can prioritize
patients for living-donor transplants
similar to the prioritizing on the cadaver
waiting list, and the matching that
includes the highest priority patients
can be chosen as the exchange to be
carried out.
Obviously, considering only pairwise
exchanges avoids the issue of long
cycles. However, if we apply the
pairwise exchange algorithm to the
example used to illustrate the TTCC
algorithm, we see that there is only one
pairwise exchange possible: patients 1
and 5. Permitting longer cycles clearly
has the potential to help more patients.
Evidence from simulations suggests
that permitting three-way exchanges
helps significantly more patients than
allowing only pairwise exchanges, but
that any gains from permitting longer
cycles are minimal.
Current Work
Integer programming has been used to
generate pairwise and three-way
exchanges. Large integer programming
problems can require an unreasonable
amount of time to solve, even on the
fastest computers, so this solution was
limited to regional exchanges. Recently,
Tuomas Sandholm of Carnegie Mellon
University has developed a method of
breaking down the large integer
programming problem into manageable
pieces. His approach is currently being
tested on a regional level, but the hope
is that one day there will be a national
database for living donors similar to the
cadaver queue.
How does the story end? It remains to
be seen. The effort to organize a fair
and efficient kidney exchange at the
national level is ongoing. One thing is
certain: The algorithm that determines
how the kidneys are allocated is
essential. The donors, recipients, and
doctors must have faith that the
allocation is fair, or the system will fail.
Mathematics is key to this lifesaving
endeavor.
Acknowledgment
The author would like to thank Anthony
Nance for many helpful discussions
and much encouragement on this
project.
Further Reading
A compendium of information on organ
donation is published online by the
National Kidney Foundation at
www.kidney.org/news/newsroom/fs
new/25factsorgdon&trans.cfm. See
www.unos.org for an up-to-the-minute
count on the number of people waiting
for some type of transplant. For
information on the New England Kidney
Exchange Program, see Alvin E. Roth,
Tayfun Sönmez, and M. Utku Ünver, “A
Kidney Exchange Clearing house in
New England,” American Economic
Review, Papers and Proceedings, 95
(2), May 2005, 376–380. A good source
of information on pairwise matchings is
Alvin E. Roth, Sönmez Tayfun, and M.
Utku Ünver, “Pairwise Kidney
Exchange,” Journal of Economic
Theory, 125, 2005, pp. 151–188. To
better understand why three-way
exchanges may be optimal, see Susan
L. Saidman et al., “Increasing the
Opportunity of Live Kidney Donation by
Matching for Two- and Three-Way
Exchanges,” Transplantation, 81 (5),
March 15, 2006, pp. 773–782. The
integer programming approach being
developed at Carnegie Mellon
University is outlined in an article
appearing online at
www.carnegiemellontoday.com/arti
cle.asp?aid=514.
DOI: 10.4169/194762110X525539
SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS 29
About the author: Olivia Carducci
teaches all levels of mathematics at
East Stroudsburg University. She is the
proud mother of four children, both a
Boy Scout and Girl Scout leader, and a
soccer coach.
email: ocarducci@po box.esu.edu
Figure 5: Graph for step 1.
30. THE SANDBOX
In this section, we highlight problems that anyone can
play with, regardless of mathematical background. But just
because these problems are easy to approach doesn’t
necessarily mean that they are easy to solve!
This issue’s Sandbox problem comes from Joseph Ward, a
high school student at Silver Creek Academy (AZ). Joseph
was playing around with pencils one day when he discovered
that some numbers of pencils could be stacked lengthwise
to form the lowest k > 1 rows of an equilateral triangle when
viewed from the end. Two different ways to stack n = 9
pencils are shown below:
He also found that, as hard as he tried, there were some
numbers of pencils, such as n = 8, that could not be stacked
in this way. Problem 248, Pencil Stack, asks you to deter-
mine the positive integers n for which there is at least one
way to form the lowest k > 1 rows of an equilateral triangle.
THE ZIP-LINE
This section offers problems with connections to articles that
appear in this issue. Not all of the problems in this section
require you to read the corresponding articles, but doing so
can never hurt, of course.
The first Zip-Line problem is motivated by the article “Congo
Bongo” on pages 18–21. As described on the first page of
the article, we attempt to open a treasure chest in the shape
of a regular n-gon with a bongo drum at each corner. Inside
each drum there is a genie standing either upright or on its
hands, and striking any subset of the drums at the same time
will cause all of the genies inside of those drums to invert
themselves. The treasure chest will open if you can devise a
30 SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS
sequence of drum strikes that leaves all n genies oriented in
the same way; the complicating factors are that you can’t see
the genies’ orientations inside the drums and that the chest
rapidly spins a random amount between each set of simulta-
neous drum strikes.
The end of the article explains why no sequence of drum
strikes is guaranteed to open an n-drum chest unless n is a
power of 2. However, for other n it may be possible to have
a pretty good chance of opening the chest during a certain
sequence of drum strikes, assuming that the spinning
between strikes is truly random. This leads us to Problem
249, Congo Bongo not so Wrongo: What is the smallest
number of drum strikes needed on a three-drum treasure
chest to give you at least a 90 percent chance of opening
it? (As a separate further challenge, which may require
computer assistance, try to answer the same question for
a five-drum chest.)
The other Zip-Line problem, Problem 250, is simple to state:
Win at Port-and-Sweep! The rules of the game are
described in the first paragraphs of “Port-and-Sweep
Solitaire” on pages 22–25: Each square of the board must
have a value of 0, 1, or 2 at all times; a “port” move subtracts
2 from one cell and adds 1 to a cell distance 2 away; a
“sweep” move subtracts 1 from three consecutive cells and
adds 2 to a cell next in line. The goal is to end with a board
where every cell is 0 except for a single 1 in the middle.
(a) Start with the 3rd board in Figure 4 from the article:
(b) Start with the challenge board in Figure 8:
Welcome to the Playground! Playground rules are posted at the bottom of page 33, except for the most important one: Have fun!
THE PLAYGROUND!Derek Smith
31. the per-class mean class size is
and the per-student mean class size is
Since all of the numerators and denominators are positive,
cross multiplying or dividing shows that S ≥ C if and only if
At this point, the group from Taylor invoked the famous
Cauchy-Schwartz inequality,
which holds for all positive quantities x1, x2, …, xk and y1, y2,
…, yk. Simply letting xi = 1 and yi = ci for each i then turns
the Cauchy-Schwartz inequality into exactly what we wanted
to show!
The companion problem to Problem 240 asked if the
per-class standard deviation for class size must always be
at least as large as the per-student standard deviation for
class size; for the example above, we get (sample) standard
deviations of and respectively. The answer is no:
check this for 5 students distributed into one class of size 2
and three classes of size 1.
September’s second Playground problem came from Tuan Le
of Fairmont High School (CA). Problem 241, As Easy as
a, b, c, was another inequality: show that
is greater than or equal to
when k = 1 and a, b, and c are non-negative real numbers.
As a further challenge problem, Tuan asked you to find the
largest value of k for which this relationship is guaranteed to
hold.
No correct “by hand” solutions were received by the
submission deadline; there were two incorrect submissions.
However, Stan Wagon of Macalester College offers
computer-based solutions to both the regular problem
(without the need for the non-negativity condition on a, b,
and c) and the challenge problem in the form of a
Mathematica notebook which can be found at
http://www.maa.org/mathhorizons/supplemental.htm.
R k a b b c c a= − − −(( )( )( ))2
L a a b a c b b a b c c c a c b= − − + − − + − −4 4 4
( )( ) ( )( ) ( )( )
x y x yi
i
k
i
i
k
i i
i
k
2
1
2
1 1= = =
∑ ∑ ∑
⎛
⎝
⎜
⎞
⎠
⎟
⎛
⎝
⎜
⎞
⎠
⎟ ≥
⎛
⎝
⎜
⎞
⎠
⎟⎟
2
,
k c c c c c ck k( ) ( ) .1
2
2
2 2
1 2
2
+ + + ≥ + + +
S
c c c
c c c
k
k
=
+ + +
+ + +
1
2
2
2 2
1 2
.
C
c c c
k
k
=
+ + +1 2
,
THE JUNGLE GYM
Any type of problem may appear in the Jungle Gym—climb
on!
Matthew McMullen of Otterbein University (OH) provides
the Jungle Gym problem, Twice Sliced. Polya’s Pizzeria
produces pizzas that are perfectly round, 12 inches in
diameter, and cut into six congruent slices. You and a friend
decide to share a slice by making a straight cut that divides
the slice into equal areas, and you would like to consider
cuts besides the “boring” cut that goes through the midpoint
of the crust. Part (a) of Problem 251 asks you to describe all
poss ble cuts, like cut a in the figure below, that don’t go
through the curved outer crust of the slice, while part (b) asks
you to describe all of the cuts (like b) that do.
FEBRUARY WRAP-UP
Larry Lesser of the University of Texas at El Paso led off
February’s Sandbox by asking you to compare two types of
averages: if a 9-student school consists of two 2-student
classrooms and one 5-student classroom, then you can
check that the mean class size on a per-class basis is 3,
while the mean class size on a per-student basis is
33/9 = 11/3. Problem 240, Meaningful Means, asks if
11/3 > 3 is just one case of a more general phenomenon: if n
students are distributed among k non-empty classes, must
the per-student mean class size always be at least as large
as the per-class mean class size?
Correct solutions were received from student Xie Jun
(University of Michigan–Shanghai Jiao Tong University Joint
Institute) and a student group consisting of Seth Foote,
Megan Frantz, and Audrey Nice (Taylor University, IN), as well
as from Allen Fuller (Gordon College). Also, Alexander
Bogomolny’s solution to this problem can be found at his
“Cut The Knot” website.
Here, we highlight the solution given by the group from
Taylor. Let ci represent the number of students in class i.
Then
n c c ck= + + +1 2 ,
SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS 31
3 2 5. ,
