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FRAGMENTS
Contents Interior fragments
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The Politics before the Politics: Census 2010, Reapportionment, and Redistricting Tommy Ratliff and Karen Saxe The mathematics is anything but black and white in the once-a-decade ritual to decide how the U.S. House of Representatives parcels out its seats.
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The Quest for God’s Number Rik van Grol On which day did God create Rubik’s Cube, and did it come with an instruction manual?
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A Colorful Proof of Pick’s Theorem Jack E. Graver and Yvette A. Monachino A color-coded matching algorithm is the key to a clever new proof of an old friend.
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A Conversation with Corey Greenspan Stephen Abbott A graduating senior’s journey into the irrational world of π leads to
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fame, fortune, and the discovery of a cool new rock band.
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Alumni Profiles: Love Actuary Nicole Belmonte An aspiring actuarial fellow reveals her secret to professional success: All you need is (to) love (math).
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Paging Dr. Freud! Stephen Walk More than Freudian analysis is required to get to the roots of this puzzling dream.
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Have an Abacus—and Own a Piece of Mathematical History Eli Maor The ancient art of counting on the suan pan is being preserved by the author, who shares his fondness for the form and function of these early analog computers.
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Painting by Parametric Curves and Van Gogh’s Starry Night Stephen Lovett, Matthew Arildsen, Jon Jones, Anna Larson, Rebecca Russ
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For this professor and his students, exploring the artistic possibilities of plotting parametrically makes a strong impressionism.
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The Playground! The Math Horizons problems section, edited by Derek Smith
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Aftermath: Does the Master’s Degree in Mathematics Get Too Little Respect? Carl Cowen The former president of the MAA weighs in on “the most marketable degree in the mathematical sciences.” NOVEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS
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Math Horizons is for undergraduates and others who are interested in mathematics. Its purpose is to expand both the career and intellectual horizons of students. STEPHEN ABBOTT Middlebury College BRUCE TORRENCE Randolph-Macon College Editors LOIS M. BARON Managing Editor KARA KELLER Art Director DEREK SMITH Lafayette College Problems Editor Math Horizons (ISSN 1072-4117) is published four times a year (September, November, February, and April) by the Mathematical Association of America, 1529 Eighteenth Street, NW, Washington, DC 20036. November 2010, Volume XVIII, Issue 2. Periodicals postage paid at Washington, DC, and additional mailing offices. Annual subscription rates are $29.00 for MAA members, $38.00 for nonmembers, and $49.00 for libraries. Bulk subscriptions sent to a single address are encouraged. The minimum order is 20 copies ($200.00); additional subscriptions may be ordered in units of 10. To order, call (800) 331-1622. For advertising information, call (877) 622-2373. Printed in the United States of America. Copyright ©2010. The Mathematical Association of America. POSTMASTER: Send address changes to Math Horizons, MAA Service Center, P.O. Box 90973, Washington, DC 20090-1112.
EDITORIAL BOARD Ezra Brown Virginia Tech Edward Burger Williams College Nathan Carter Bentley University Timothy Chartier Davidson College Katherine Crowley Washington & Lee University William Dunham Muhlenberg College Joseph Gallian University of Minnesota Duluth Deanna Haunsperger Carleton College Dan Kalman American University Stephen Kennedy Carleton College Mark McClure University of North Carolina at Asheville
John Allen Paulos Temple University Loren Pitt University of Virginia Tommy Ratliff Wheaton College Adrian Rice Randolph-Macon College Marian Robbins California Polytechnic State University Karen Saxe Macalester College Peter Schumer Middlebury College Laura Taalman James Madison University James Tanton St. Mark's School Tom Tucker Colgate University Robin Wilson The Open University
Instructions for Authors Math Horizons is intended primarily for undergraduates interested in mathematics. Thus, while we especially value and desire to publish high-quality exposition of beautiful mathematics, we also wish to publish lively articles about the culture of mathematics. We interpret this quite broadly—we welcome stories of mathematical people, the history of an idea or circle of ideas, applications, fiction, folklore, traditions, institutions, humor, puzzles, games, book reviews, student math club activities, and career opportunities and advice. Manuscripts may be submitted electronically to Editors Stephen Abbott (
[email protected]) and Bruce Torrence (
[email protected]). If submitting by mail, please send two copies to Bruce Torrence, Math Department, Randolph-Macon College, P.O. Box 5005, Ashland, VA 23005-5505. Subscription Inquiries email:
[email protected] Web: www.maa.org Fax: (301) 206-9789 Call: (800) 331-1622 or (301) 617-7800 Write: Math Horizons, MAA Service Center, P.O. Box 90973, Washington, DC 20090-1112 THE MATHEMATICAL ASSOCIATION OF AMERICA 1529 Eighteenth Street, NW Washington, DC 20036
On the cover: Flag Cube, by Lucas Garron. Lucas is an undergraduate at Stanford University and an avid speed cuber. The image was made in POV-Ray. Each of the 96 Rubik’s cubes is shown in a legitimate position. Hence, with enough time and enough Rubik’s cubes, anyone could build a real version of this image. 4
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O, IC! ACROSS 1. Dispose of documents 6. Bitmap font file type 9. October birthstone 13. Stowed a rope 15. Logical negative of iff 16. Apprehends 17. Little llama 18. Square grid of points? 20. Property? 22. Awaiting further action 23. Trait desirous of proofs 25. Mathematical publisher 28. King of Spain 29. Wellington film workshop 33. One whose stay is extended 37. Straw-colored 39. Self-help bestseller? 41. Element of a power set 42. Patronize the ETS 43. Fit like gears 44. What knot theorists do 45. Wee little one 46. Funds raised 51. Univ of Chicago color 54. VA tourist attraction? 58. Motherly love? 60. Luminous phenomena 61. Demolish 62. Beloved math game 63. Gives desired result 64. Secluded valley 65. XXIX times XIX 66. Hamper
DOWN 1. Lion King villian 2. Gap (in a proof) 3. Performs a discontinuous operation on a surface 4. Overjoy 5. Model plane feature 6. Candy with a dispenser 7. Ant or bee society
by Sam Vandervelde 1
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8. Money to Bern 9. Surjective 10. Jack and Jillʼs objective 11. Generic quadrilateral name 12. Englandʼs financial center 14. April 15, 1707 for instance 19. Ubiquitous article 21. Snowy bird 24. Requiring oxygen 25. Gorge 26. X-men character 27. Thirteen per deck 29. 2u/ t2 = c2 2u object 30. Have being 31. —— one (slim odds) 32. Anxiety, German style 34. Spoonʼs running partner 35. df/dx = kx, for example
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36. Mouse medic 37. What pigs canʼt do 38. Ignited 40. Bohemian Rhapsody group 44. It is commonly removed 46. Email protocol 47. Like a sphere 48. Electronic marketplace 49. Celtic VIP 50. Yes —— Bob 51. Word after corn or piece 52. Opposite of subtractz 53. Nothing to a Parisian 55. Cabbage Patch, for one 56. USDA top rating 57. Trivial 58. Part of MSRIʼs URL 59. $1M proof organization
For the solution, visit http://www.maa.org/mathhorizons/supplemental.htm.
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The Politics before the Politics: Census 2010, Reapportionment, and Redistricting Tommy Ratliff and Karen Saxe s you are certainly aware, 2010 is a census year in the United States. The Constitution requires a national census once every 10 years, and the first took place in 1790 to determine the number of seats each of the then 15 states would have in the U.S. House of Representatives. At that time there were 3,231,533 people living in the United States (slaves not included), and there were 105 representatives to serve this population.
A
The current House size, 435 members, has been unchanged since 1911, except for a temporary increase to 437 at the time of admission of Alaska and Hawaii as states. An apportionment has been made on the basis of each census since 1790, except following the census of 1920 (keep reading!). While the primary reason for the decennial census is apportionment, it also was created to gain a better understanding of where people lived and to establish patterns of settlement. Nowadays, census data also help determine the allocation of federal funds for community services, such as school lunch programs, and new construction, such as highways and hospitals. Every year, more than $400 billion is distributed to local, state, and tribal governments based on census data. According to the U.S. Census Bureau, cost estimates for the 2010 census are more than $14 billion. Overall, there is a stunning amount of money spent in running the census, and at stake due to it.
Hamilton versus Jefferson For the first 150 years of U.S. history, there was great controversy over the apportionment of congressional seats to the states. Indeed, the very first bill to be vetoed was Washington’s veto of Alexander Hamilton’s apportionment bill. Hamilton’s method awarded one fewer seat to Virginia (Washington’s home state) than would have a rival method proposed by Thomas Jefferson. Hamilton’s method is an example of a quota method. You first choose the size of the House of Representatives, which
we will call h, and then calculate the so-called standard quota (or just quota) for each state, which is simply its proportion of the population times h. You give each state the whole number contained in its quota and then assign any remaining seats to the states with the largest remainders. To see this in action, consider the very simple nation that consists of four states as shown with a House size of h = 54.
Messing with Texas Each state usually redraws its congressional districts once a decade, following the census. However, the state of Texas modified its district boundaries twice following the 2000 census. In 2001, the State House of Representatives was controlled by the Democratic Party and the State Senate was controlled by the Republican Party. Since the two bodies could not agree on a redistricting plan, a three-judge panel redistricted the state to incorporate two new seats while mostly following the previous boundaries, which were drawn in 1990 when Democrats controlled both chambers in the legislature. When the Republicans gained control of the State House of Representatives in 2003, the legislature took to redrawing the congressional districts, which the Democrats strongly opposed. Not only did these lawmakers refuse to appear at the state capital, which prevented the legislature from having a quorum—the minimum number required to conduct state business—but on two occasions up to 50 Democratic lawmakers fled to Oklahoma or New Mexico to be out of the jurisdiction of the state of Texas! A compromise was reached in October of 2003, but the redistricting had very real consequences for both Texas and the U.S. House of Representatives: the Texas congressional delegation went from a 17 to 15 advantage for the Democrats in the 2002 elections to a 21 to 11 advantage for the Republicans in 2004.
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Population
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% Pop
Quota
Round Down
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39.47 %
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38,000
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Hamilton Apportionment After calculating the quota for each state and the initial allocation of seats, two seats remain to be allocated. Since states C and D have the highest remainder, each is given an additional seat, resulting in the final apportionment. Besides the fact that Jefferson felt that Hamilton’s method disadvantaged Virginia in 1790, Jefferson objected to Hamilton’s method because the apportionment is not based upon using a consistent district size across all states.
A : 13,000
C : 4,000
18 / 19
6/5
D : 6,000 9/8
B : 15,000 21 / 22
Figure 1. Our small nation with each state’s population and Hamilton (red) / Jefferson (green) apportionments. Jefferson proposed a divisor method. You choose the size of the House h and then pick a divisor x that serves as the ideal district size. You divide each state’s population by x to find the quotient and round down to the nearest integer to determine the state’s apportionment. If need be, you adjust the divisor x until there are exactly h seats apportioned. If we apply this method to our small nation with a House size of h = 54 and divisor x = 680 (found after some experimentation), we can see that we get a different apportionment, where both large states gain a seat compared to using Hamilton’s method. It is perhaps not surprising that Jefferson’s method exhibits a bias in favor of large states since it always ignores the
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NOVEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS
fractional part of the quotient, which is potentially more significant with a smaller quotient.
You could overcome this bias by not 54 always rounding down. John Quincy Adams proposed a divisor method where you always round up, which exhibits a small-state bias. Daniel Webster proposed the divisor method where you round as normal depending on whether the remainder is less than 0.5 or greater than or equal to 0.5. But there is nothing special about choosing the arithmetic mean of n and n + 1 as the dividing point.
Paradoxical Twists Our first example would seem to argue in favor of using Hamilton’s method, but there are definitely some issues that can arise with it. For example, suppose that we expand the House for our small nation from size h = 54 to h = 55. If you recalculate the Hamilton apportionment (do it!), you will find that both states A and B gain a seat, but state D loses a seat! This is known as the Alabama paradox and was first identified following the 1880 census, when Alabama would receive eight representatives in a 299-seat House, but only seven representatives in a 300-seat House, using Hamilton’s method. This means that Hamilton’s method is not monotone with respect to the House size h. Perhaps more disturbing than the Alabama paradox is the population paradox that was identified in 1910. In this case, the House size stays fixed, but the state populations increase. It is possible that state B can grow both faster and by a larger absolute amount than state D, but B loses a seat to D. You can check that this occurs with our small nation with a House size of h = 55 when states A and C grow by 10 percent, state B grows by 5 percent, and state D grows by 4 percent.
Jefferson Apportionment
State
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4,000
5.88
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6,000
8.82
8
Total
38,000
Jefferson Apportionment
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A disturbing feature of this paradox is that it can provide an incentive for a state to underreport its population. Balinski and Young prove that the only methods that avoid the population paradox are the divisor methods (see the Further Reading section); this is the good news about divisor methods. The bad news is that they do not necessarily adhere to the quota rule; that is, after determining the standard quota for each state and determining for each the lower quota n and upper quota n + 1 and doing the apportionment using a divisor method, a state could end up with neither n nor n + 1 representatives! In fact, as Balinski and Young show in their book’s appendix B, this occurs when applying Jefferson’s method to the 2000 census: California’s quota is 52.447, but Jefferson’s method would apportion 55 seats to the state. To define a divisor method, we just need to specify the dividing point in the interval [n, n + 1] that determines when we round up or round down. One of the most interesting is the Hill-Huntington method, which initially does not sound like a divisor method at all. Joseph Hill was chief statistician at the U.S. Census Bureau. Mathematician Edward Huntington taught at Williams and Harvard, and served as MAA president, AAAS president, and AMS vice president! Following their method, you allocate the seats so that no transfer of any one seat between two states can reduce the percentage difference in representation between those two states. It is not at all clear that this process of trading seats would ever end, but this is in fact equivalent to a divisor method where the dividing point between n and n + 1 is the geometric mean n(n + 1). In 1911 Congress passed a bill apportioning 433 seats, with the stipulation that the territories of Arizona and New Mexico would each receive one seat if admitted as states before the next census. This is the heritage of the House of Representatives being set at 435 members. Every value between 390 and 440 was calculated using Webster’s method, and 433 was chosen because it resulted in no state losing a seat. The migration from rural to urban areas changed the population distribution so much that rural states effectively blocked any reapportionment after the 1920 census. At this stage, there was a lack of understanding, to some degree, of the subtleties involved with apportionment methods. Do certain methods favor small states and others favor large states? Is there fundamental bias in a given apportionment method? Though no reapportionment took place following the 1920 census, a National Academy report supported the Hill-Huntington method.
Apportionment Matters Apportionment determines not only the number of seats in the House of Representatives, but also the number of votes in the Electoral College, which affects the influence of each state in electing the president. Each state receives a number of votes in the Electoral College equal to its total representation in Congress: the number of seats in the House of Representatives plus two for the state’s two senators. The method of apportionment can have very real consequences as demonstrated in the 1876 U.S. presidential election. After the 1870 census, there were initially 283 seats apportioned (this number was chosen because Hamilton’s and Webster’s methods agreed for this size House), but then nine seats were added a few months later, and these were distributed to states in a way that did not agree with either Hamilton’s or Webster’s method. Although Tilden defeated Hayes in the popular vote, there was a dispute over the electoral votes in three states, and Hayes eventually defeated Tilden 185 to 184 in the Electoral College using the ad hoc method of apportionment. However, if Hamilton’s method had been used, then Tilden would have won the election!
To avoid future failures to apportion, a bill was passed in 1929 stating that the president will send to Congress the results of the census and the apportionment of the 435 seats computed in three ways: using (1) the method used in the preceding apportionment, (2) Webster’s method, and (3) the Hill-Huntington method. If Congress does not then reapportion itself, the reapportionment will be based on (1). As it turned out, in 1930 the Hill-Huntington and Webster’s method agreed. In 1940, the Hill-Huntington method gave an extra seat to the Democrats, so Congress selected it, and it has remained in use ever since.
District Dynamics For the time being, the method of apportionment and House size seem fairly stable. But the fun is far from over! After each reapportionment, the redistricting process begins anew, whereby each state is re-carved in order to determine the geographic district that each member of Congress will represent. Redistricting is done at the state level, but there are three general principles that guide redistricting in all states. First, states are to strive for population equity between districts. It is easy to decide if a state has adhered to this; for example, the 13 districts in New Jersey contained (according to 2000 census numbers) either 647,257 or 647,258 people. Pretty good!
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In practice, there are districts that are not contiguous (like New Jersey’s Thirteenth) and many that are unarguably noncompact (like Arizona’s Second and Illinois’s Fourth).
“Gerrymandering” is not new. This 1812 cartoon shows the Massachusetts districts drawn to favor the incumbent Democratic-Republican Party candidates of Governor Elbridge Gerry over the Federalists.
Congressional District 13 nationalatlas.gov
ey
West New York
Congressional District
13
Union City
95
Ne
TM
k Tp
Union
Hudson
Ga rd en
Essex
w
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280
Je rs
Pkwy
Bergen
County
Jersey City
78
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5
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Union
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Perth Amboy 9
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nationalatlas.gov
CO
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Congressional District
2 Mohave Coconino
Kingman
Bullhead City
Navajo
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CALIFORNIA
Lake Havasu City Yavapai
La Paz
NEW MEXICO
Mohave
Surprise
10
17
County
2 1
Peoria Phoenix
150
violated. Find a congressional district map of your home state (at http://www.nationalatlas.gov/government.html); are there any noncontiguous districts?
45 6 3 7
Maricopa Goodyear 0
TM
8
Arizona (8 Districts)
300 MIles
Congressional District 4 nati onal atlas. gov
94
Congressional District
4
L a ke Michigan
TM
Cook
County
16 5-10 4 14 13 1-3 11
294
Stone Park
17 Cook Cicero
8
Chicago
Lake Michigan
16 10
6 90 94
14
5 7 3
13
4 1 2
0
5
10 Miles
11
19 12
9 34
55
15
18
290
IND.
Illinois (19 Districts)
The average population per district in the entire country was 646,946 people after the most recent redistricting. But since we require districts to lie within states, the variation of average district populations between states is large—the district with the most people is Montana’s single district of 905,306, and the district with the fewest is Wyoming’s single district, with 495,304 people. Second, districts are to be contiguous; that is, they are to be path connected. Again, it is easy to see if this criterion is
8
The third requirement is that, generally speaking, districts need to be drawn in compact shapes. Most states describe their compactness requirements in vague terms, but some states include specific measures in their redistricting laws. Iowa, for example, gives two such measures in its 2000 Legislative Guide to Redistricting. Compactness was introduced, along with contiguity, by federal law in the early 1900s and started playing a much larger role in the 1960s, in the context of racial discrimination.
NOVEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS
By all measures, circles are compact. Extremely jagged edges and skinny appendages are features of low compactness, and they are the hallmarks of gerrymandered districts. Gerrymandering—the process of creating a district to achieve political goals—has been with us almost since the beginning. So how can we measure compactness? These measures can usually be categorized in one of three ways: as an area measure, a perimeter measure, or a population measure. As an example, you might find the smallest circle containing the district and take the ratio of the district’s area to that of the circle. This ratio is always between 0 and 1 (the closer to 1,
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the more compact the district). This is known as the Roeck test (1961) and is an area measure.
The population measure of Hofeller and Grofman (1990) takes the ratio of district population to that of the convex hull of the district. Note that the measure—using all three of these specific tests—of a circular district is 1. Finally! Representatives have been elected, and their districts have been determined; let the politics begin!
Further Reading A good place to begin is the book Fair Representation: Meeting the Ideal of One Man, One Vote (second edition), by Michel Balinski and H. Peyton Young (Brookings Institution Press, 2001). There is a wealth of information on the current census at http://2010.census.gov/2010census/index.php. A brief history of apportionment of the U.S. House of Representatives can be found at http://www.census.gov/population/www/censusdata/ apportionment/history.html. Writing in The New Yorker, Jeffery Toobin’s “Drawing the Line” (March 6, 2006) discusses gerrymandering. It looks specifically at Tom DeLay’s redistricting of Texas in 2003, in which Republicanheld districts went from a two-seat disadvantage to a 10-seat advantage with respect to districts held by Democrats! The upcoming redistricting process is explored in depth, and the least compact of the current congressional districts are revealed at http://www.redistrictingthenation.com. To bring the redistricting process to life, you may enjoy playing the Redistricting Game at http://www.redistrictinggame.org. About the authors: Tommy Ratliff is professor of mathematics at Wheaton College in Massachusetts, and Karen Saxe is professor and chair of mathematics at Macalester College in Minnesota. They both teach courses on electoral system design and voting theory. Collaborating on this article has been a great way to share their interests in the many ways in which mathematics rears its head in the political arena. email:
[email protected] [email protected]
STRONG GRADUATE DEGREES
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The Cox test (1927) is one of the older perimeter measures. You simply take the ratio of district area to that of a circle with the same perimeter. It is amusing to note that the Cox measure of Illinois’s Fourth District is approximately 0.0368. The perimeter of this district is about 116 miles, and the area is approximately 39.43317 square miles; a circle with the same perimeter is much bigger—about 1,070.8 square miles!
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The Quest for God’s Number Rik van Grol ubik’s Cube triggered one of the largest puzzle crazes in the world. The small mechanical puzzle, invented by Ernõ Rubik in Hungary, has sold more than 350 million copies. Although it has existed since 1974, the popularity of the cube skyrocketed around 1980 when the cube was introduced outside of Hungary. In the early days, simply solving the puzzle was the main issue, especially because no solution books were available and there was no Internet. But solving the puzzle in the shortest amount of time was also hot news. In the early 1980s the best times were on the order of 24 seconds.
R
11 seconds. The foundation for a fast solving time is a good algorithm, and so the search for efficient algorithms has been an important area of study since the early 1980s. The ultimate goal is to discover the best method of all—the optimal solution algorithm—which has been dubbed God’s algorithm. God’s algorithm is the procedure to bring back Rubik’s Cube from any random position to its solved state in the minimum number of steps.
The maximum of all minimally needed number of steps is referred to as God’s number. This number can be defined in several ways. The By the end of the most common is in 1980s, the craze was Determining God’s number by independently terms of the number starting to ebb, but of face turns required, improving the upper and lower bounds was a in certain groups the but it can also be puzzle remained very quest that lasted for three decades—but it has measured as the much alive, and now finally come to an end. In July 2010 the upper number of quarter Rubik’s Cube is turns. Whereas a and lower bounds met at the number 20. making a comeback. quarter turn is either a Solving the cube the positive or negative 90° turn, a face turn can be either of quickest—speed cubing—is currently a major activity. The these or a 180° turn. A 180° face turn is equal to two quarter fastest times are under 8 seconds with the average around
10
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turns. Earlier this year, after decades of gradual progress, it was determined that God’s number is 20 face turns. Thus, if God’s algorithm were used to solve the cube, no starting position would ever require more than 20 face turns. Apart from determining God’s number, another major question has been to find out whether God’s algorithm is an elegant sequence of moves that can be “easily” memorized, or if, instead, God’s algorithm amounts to a short procedure with giant lookup tables. If it’s the latter case, then no one will ever be able to learn how to solve Rubik’s Cube in the minimal number of moves (read on to learn why this is true). Still, even if God’s algorithm has no practical purpose, it is interesting to know what God’s number is.
Playing God (with Small Numbers) If you start with a solved cube and ask someone to make a few turns, you will (after some practice) be able to return it to the solved state in a minimum number of moves as long as the initial number of scrambling moves is not too large. With fewer than four scrambling moves, it is easy; with four, it becomes tricky. With five, it is simply hard. Some experts can handle six or even seven scrambling moves, but any more and it is essentially impossible to solve the cube in the minimal number of moves. Generally speaking, most algorithms take between 50 and 100 moves. What if you were to randomly turn the cube 1,000 times? Will it take 1,000 moves to get it back? No, it still takes most algorithms 50 to 100 moves because the algorithms are designed to work from any starting position.
Starting Small: The 222 Cube Unlike the classic 333 Rubik’s Cube, the 222 cube has been completely analyzed. God’s number is 11 in face turns and 14 in quarter turns. To find these values, all possible configurations of the 222 cube were cataloged, and for each of these configurations, the minimal number of turns needed to reach the solved state was determined. This brute force approach was possible because there are “only” about 3.7 million configurations to study. To calculate the number of configurations for the 222 cube, we start with the observation that the eight corner cubies (as they are called) can be permuted in 8! ways. For any such permutation, each corner cubie can be oriented in
three ways, leading to 38 possible orientations. However, given the orientation of seven corners, the orientation of the eighth is determined by the puzzle mechanism, so the corners really have 37 orientations. As the orientation of the whole cube is not fixed in space (any one of the eight corners can be placed in, say, the top-front-right position, and once it is placed there, the entire cube can be rotated so that any one of three faces is on top), the total number of permutations needs to be divided by 8 3 = 24. Hence, the total number of positions is
8 ! × 37 = 7 ! × 36 = 3, 674, 160. 8×3 There are only 2,644 positions for which 11 face turns are required to solve the puzzle. Assuming all configurations have the same likelihood of being a starting position, the average number of face turns required to solve the puzzle is 9. Likewise, there are only 276 positions from which 14 quarter turns are required, and on average, 11 quarter turns are required to solve the puzzle.
A Leap in Complexity: The 333 Cube Until recently, God’s number for the 333 cube was not known. From the late 1970s until now the search area has been limited by two numbers: the lower bound and the upper bound. The lower bound Glow is determined by proving that there are positions that require at least Glow turns. The high upper bound G is determined by proving that no position requires more than Ghigh turns. So, how many configurations are there? With 8 corner cubies and 12 edge cubies, there are 8!12!38212 different patterns, but not all patterns are possible: • With 8 corners there are 8! corner permutations, and with 12 edges there are 12! edge permutations. However, because it is impossible to interchange two edge cubies without also interchanging 2 corner cubies, the total number of permutations should be divided by 2. • Turning of corner cubies (keeping their position, but cycling the colors on their three faces) needs to be done in pairs—only 7 corner cubies can be turned freely.
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• Flipping of edge cubies (keeping their position, but switching the colors of their two faces) needs to be done in pairs—only 11 edge cubies can be flipped freely. Because of the six center pieces, the orientation of the cube is fixed in space, so the number of permutations should not be divided by 24 as with the 222 cube. Hence, the number of positions of the 333 cube is
8 ! × 12 ! 38 212 × × = 43, 252, 003, 274, 489, 856, 000 2 3 2 ≈ 4.3 × 1019. This is astronomically bigger than 3,674,160 for the 222 cube, and it made searching the entire space computationally impossible. For instance, if every one of the 350 million cubes ever sold were put in a new position every second, it would take more than 3,900 years for them to collectively hit every possible position (with no pair of cubes ever sharing a common configuration). Or to put it another way: if a computer were capable of determining the fewest number of moves required to solve the cube for 1,000 different starting positions each second, it would take more than a billion years of computing time to get through every configuration. Determining God’s number by independently improving the upper and lower bounds was a quest that lasted for three decades—but it has finally come to an end. In July 2010 the upper and lower bounds met at the number 20.
Raising the Lower Bound Using counting arguments, it can be shown that there exist positions requiring at least 18 moves to solve. To see this, one counts the number of distinct positions achievable from the solved state using at most 17 moves. It turns out that this number is smaller than 4.31019. This simple argument was made in the late 1970s (see Singmaster’s book in the Further Reading section), and the result was not improved upon for many years. Note that this is not a constructive proof; it does not specify a concrete position that requires 18 moves. At some point, it was suggested that the so-called superflip would be such a position. The superflip is a state of the cube where all 12
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the cubies are in their correct position with the corner cubies orientated correctly, but where each edge cubie is flipped (orientated the wrong way). It took until 1992 for a solution for the superflip with 20 face turns to be found, by Dik Winter. In 1995 Michael Reid proved that this solution was minimal, and thus a new lower bound for God’s number was found. Also in 1995, a solution for the superflip in 24 quarter turns was found by Michael Reid, and it was later proved to be minimal by Jerry Bryan. In 1998 Michael Reid found a new position requiring more than 24 quarter turns to solve. The position, which he calls the superflip composed with four spots, requires 26 quarter turns. This put the lower bound at 20 face turns or 26 quarter turns.
Lowering the Upper Bound Finding an upper bound requires a different kind of reasoning. Perhaps the first concrete value for an upper bound was the 277 moves mentioned by David Singmaster in early 1979. He simply counted the maximum number of moves required by his cube-solving algorithm. Later, Singmaster reported that Elwyn Berlekamp, John Conway, and Richard Guy had come up with a different algorithm that took at most 160 moves. Soon after, Conway’s Cambridge Cubists reported that the cube could be restored in at most 94 moves. Again, this reflected the maximum number of moves required by a specific algorithm. A significant breakthrough was made by Morwen Thistlethwaite. Where algorithms up to that point attacked the problem by putting various cubies in place and performing subsequent moves that left them in place, he approached the problem by gradually restricting the types of moves that could be executed. Understanding this method requires a brief introduction to the cube group. As we work with the cube, let’s agree to keep the overall orientation of the cube fixed in space. This means that the center cubies on each face will never change. We may then label the faces Left, Right, Front, Back, Up, and Down. The cube group is composed of all possible combinations of successive face turns, where two such combinations are equal if and only if they result in the same cube configuration. We denote the clockwise quarter turns of the faces by L, R, F, B, U, and D, and use concatenation as the group operation. For instance, the product FR denotes a quarter turn of the front face followed by a quarter turn of the right face, while F2 denotes a half turn of the front face. The group identity, I, is no move at all. So, for example, F4 = I. Thistlethwaite divided the cube group G0 into the following nested chain of subgroups:
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G0 = 〈 L, R, F, B, U , D 〉 G1 = 〈 L, R, F, B, U 2 , D 2 〉 G2 = 〈 L, R, F 2 , B 2 , U 2 , D 2 〉 G3 = 〈 L2 , R 2 , F 2 , B 2 , U 2 , D 2 〉 G4 = { I } Thistlethwaite’s algorithm works by first performing a few moves that result in a configuration that no longer requires quarter turns of the U and D faces (although it still requires half turns). From this point on, only moves in the subgroup G1 are needed. A few additional G1 moves puts the cube into a position where quarter turns are no longer needed for the F and B faces, and so on. The algorithm uses large lookup tables at each step, and while it is not practical for humans, the successive sets of moves per subgroup are small enough to allow computer analysis. Initially, Thistlethwaite showed that any configuration could be solved in at most 85 moves. In January 1980 he improved his strategy to yield a maximum of 80 moves. Later that same year, he reduced the number to 63, and then again to 52. After this rush of activity, progress stalled for several years. In 1989 Hans Kloosterman reported an algorithm that required at most 44 moves, which he later improved to 42. In 1992 Herbert Kociemba improved Thistlethwaite’s algorithm by reducing it to a two-phase algorithm requiring only the subgroups G0, G2, and G4. Using Kociemba’s ideas, Michael Reid announced in 1995 that he had improved the upper bound to 29 face turns. At about this time, Richard Korf introduced a new approach. Instead of using a fixed algorithm, his strategy simultaneously searched for a solution along three different lines of attack. On average, his algorithm appeared to solve the cube in 18 moves. There was, however, no worst-case analysis, and so the upper bound held still at 29. Things turned quiet again until 2006, when Silviu Radu initiated a new countdown by reducing the upper bound to 27. The next year, Gene Cooperman brought it down to 26. With the lower bound of 20 face turns in sight, Tomas Rokicki entered the picture, reducing the upper bound to 25 in March 2008. Working with John Welborn, he had it down to 24 by April, then 23 in May, and 22 by August of the following year. Finally this past July, Rokicki announced an upper bound of 20, the established value of the lower bound and therefore the long-sought-after value of God’s number. Rokicki worked with Kociemba, as well as mathematician Morley Davidson and Google engineer John Dethridge. The team used symmetry arguments to significantly reduce the
search space and then managed to partition the space of all configurations that remained into pieces small enough to fit onto a modern computer. They then made use of the enormous computing resources available through Google. Had the entire problem been done on a single good desktop PC, they report it would have taken about 35 years to complete, but by farming out different pieces to a large number of computers, the team was able to complete the calculation in just a few weeks. It is a remarkable achievement.
Epilogue Now that the quest for God’s number for the classic 333 cube has come to an end, it is time to look ahead. Is there an elegant version of God’s algorithm—one that a human could implement? And what is God’s number for the 444, 555, 666, and 777 cubes, all of which are now on the market? Finding God’s number for such puzzles is a challenge that should last far into the future. The number of positions for the 777 cube reaches 2.010160.
Further Reading Two influential early books are David Singmaster’s Notes on Rubik’s Magic Cube (Enslow Publishers, 1981), which is both readable and has many technical details; and Patrick Bossert’s You Can Do the Cube (Puffin Books, 1981), a how-to guide written when the author was 12 years old that sold more than 1.5 million copies. Another classic, complete with the requisite group theory, is Inside Rubik’s Cube and Beyond, by Christoph Bandelow (Birkhäuser, 1982). On the web, an excellent how-to guide with several links to other sources can be found at Japp’s Puzzle Page: http://www.jaapsch.net/puzzles/. A brief history of the quest for God’s number can be found on Tom Rokicki’s site, http://www.cube20.org. About the author: Rik van Grol is editor of Cubism For Fun, the English newsletter of the Dutch Cube Club (see http://CFF.helm.lu). He lives in the Netherlands. email:
[email protected] DOI: 10.4169/194762110X535961
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A Colorful Proof of Pick’s Theorem Jack E. Graver and Yvette A. Monachino he standard lattice is the set of all points in the plane with integer coordinates. A polygon whose vertices are points of the lattice is called a lattice polygon. Pick’s Theorem can be used to find the area of any lattice polygon.
T
Theorem. (G. Pick, 1899) Let P be a lattice polygon with I lattice points in its interior and B lattice points on its boundary. If A denotes the area of P, then A = I + B/2 – 1. (See figure 1.)
I=6
FRAGMENTS
Interior fragments
Side fragments
B = 10
Ordinary fragments
A = 6 + 5 - 1 = 10 Vertex fragments
Figure 2. usual lattice—it’s the dual lattice that divides the plane into unit squares such that each unit square has a lattice point at its center. An edge is defined as a “side” of a unit square; when we use the term side, we are speaking about a side of the polygon. The sides of the polygon cut some of the Figure 1. squares into pieces, and we call those pieces that lie inside the polygon fragments. Fragments are categorized by the There are many published proofs of this result, but almost all location of the lattice point. Fragments containing a lattice of them are constructed in the same basic way. They begin point in their interior are called interior fragments. If the lattice by showing that Pick’s formula gives the area for some set of point in the basic shapes; fragment is the set of all The published proofs are all constructed in the same basic way. located on triangles, the the set of right boundary of the polygon but is not at a vertex, then it is triangles and rectangles, and the set of triangles of area 1/2 defined to be a side fragment. A fragment containing a vertex are the three most commonly used sets. Once it is proven of the polygon is called a vertex fragment, and a fragment that the area of these basic shapes is given by Pick’s formuthat contains no lattice point is an ordinary fragment. la, one shows that all lattice polygons can be constructed by Examples of these fragments are labeled in figure 2. adding and subtracting these shapes, and that the construction process respects Pick’s formula. The paper by Grünbaum and Shephard (see the Further Reading section) includes a good set of references to these proofs. We offer a new proof in hopes that it will contribute to a deeper understanding of this important theorem. Before beginning the proof, we introduce a few concepts and definitions. The lattice that we are working with is not the
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A key observation is that the area of the polygon is the sum of the areas of its fragments. Another concept crucial to our proof is that of matching fragments. Two fragments are matched if they share a boundary segment under a 180-degree rotation about the center of a side of the polygon. (See figure 3.) Because we
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are rotating around the center of a side, vertex fragments are matched to other vertex fragments. Furthermore, each vertex fragment is matched to its two neighboring vertex fragments when the vertices are ordered cyclically around the polygon. In his paper “A ‘Natural’ Approach to Pick’s Theorem,” Gordon Haigh introduced the dual lattice and used matching to prove Pick’s formula for right triangles. But he then completed his proof of Pick’s Theorem in the usual manner.
MATCHING FRAGMENTS
contains exactly one nonordinary fragment. Each color class containing an interior fragment can be assembled into a full square, contributing I square units to the total area. Each color class containing a side fragment can be assembled into a half square. Hence, the side fragments contribute S/2 to the total area. Each vertex fragment is actually a sector of a square—a sector of a square being a region of the square between two rays from the center lattice point. The collection of vertex sectors can then be arranged to cover a square V/2 – 1 times. Taking all of the fragments together, we see that the total area is I + S/2 + V/2 – 1 = I + B/2 – 1. Figure 4 illustrates this process for our example. All that is left is to fill in a few details.
5 vertex fragments cover a square cove squa e 3/2 times
a
c
COLOR CLASSES ASSEMBLED
a&b a,b &c
a
a,b,c &d d
a,b,c, d&e
6 interior fragments
Figure 3. Pick’s Theorem makes a distinction between lattice points in the interior of the polygonal region and those on the boundary. Focusing on the boundary lattice points, we let V denote the number of vertex boundary points of the polygon and use S = B – V for the remaining number of side boundary points. Thus, Pick’s formula becomes A = I + S/2 + V/2 – 1, where, as above, I denotes the number of interior points. The final concept we need for our proof is that of a coloring scheme for the fragments. The fragments containing lattice points are assigned distinct colors c1, c2, …, cI + S + V. Now we just have the ordinary fragments left to color. We do this by the following process: if an uncolored fragment is matched to a colored fragment, then it adopts the color of the latter. The process can be continued until all the fragments are colored. Of course, we must show that each fragment is assigned a color and that the assignments are independent of the order in which the fragments are colored. With these concepts in hand, the proof is easy to outline. First, we color the vertex, side, and interior fragments with distinct colors. Next, the ordinary fragments are matched and colored, with the result being that each color class
e
5 side fragments
b
Figure 4.
Checking the Details To prove Pick’s Theorem, we have to show three things: (1) every fragment receives a unique color; (2) the fragments in a particular color class add up to a full square, a half square, or a sector of a square; and (3) the vertex sectors cover a square (V/2 – 1) times. To do this, we need a better understanding of the half turns that match up fragments. Let’s first see why a half turn about the center of a side will always map the dual lattice onto itself. A side of our lattice polygon is a segment connecting two lattice points, so there are only a few possibilities for the coordinates of the center point of this segment. It could be another lattice point (if the “rise” and “run” of our segment are both even integers) or a vertex in the dual lattice (if the “rise” and “run” are both odd). Or it could fall on a midpoint of an edge (if the “rise” and NOVEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS
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“run” have different parity). In every case, it’s straightforward to see that a 180-degree rotation around the center point not only takes the dual lattice onto itself, but also maps the boundary component along the side of a fragment onto a corresponding “matched” boundary component on the same side.
t s r
x
t
x
s r
Figure 5. Once the initial color assignments are made, only the ordinary fragments remain to be colored. To show that our algorithm gives a unique coloring of each ordinary fragment, we pick a point within an arbitrary ordinary fragment (such as point r or point t in figure 5) and draw the directed line segment from that point to the lattice point of the square that contains it. This segment must cross a side of the polygon and, in fact, may do so several times.
continue to fit with the new matched fragment, inside a single lattice square, and without any overlap. This implies that the process can’t involve the same fragment twice (or else it would be periodic) and must eventually end after a finite number of steps at an interior or side fragment. When this occurs, all the fragments encountered en route inherit the color of the terminating fragment. To show that the union of the fragments of a given color is a full square for an interior fragment, we reverse the above construction. Consider an interior fragment and select any point within the square containing it. This time, consider a directed segment from the center lattice point heading out to the selected point. Now it is possible to follow that segment—using successive half turns whenever a side of the polygon is encountered—until we arrive at a fragment containing (the image of) the selected point. The fragments encountered along the way must all be the same color, and as before, the rotated images of the fragments will fit into a single lattice square without overlap. Because the selected point was an arbitrary point in the original square, we conclude that the fragments of that color class can be assembled into a complete square. If we start with a side fragment, a similar argument applies, provided our arbitrary point is chosen in such a way that the directed segment from the center lattice point passes through the interior of the side fragment. Following the same logic, we find that the fragments of this color class can be arranged into precisely half of a square. Finally, recall that vertex fragments are always sectors of a square, and the sum of the interior angles of any polygon is (V – 2) x 180⬚ = (V/2 – 1) x 360⬚. Hence, when added together, the vertex fragments cover a square (V/2 – 1) times.
Further Reading Starting from our selected point, consider the first side of P encountered along this segment, and note that this side is independent of our selected point. After a half turn about the midpoint of this side, the rotated image of the ordinary fragment “fits” into a square in the dual lattice. There is a natural image for the directed line segment, which still terminates at a lattice point. Continuing in the same direction, the rotated directed line segment must enter a new fragment, which is interior, side, or ordinary. If it is interior or side, we assign our original fragment the color of this fragment. If it is ordinary, we continue along our directed line segment until we (necessarily) encounter a new side of our polygon and repeat the matching process. Because each successive half turn uses a different side from the one preceding it, the images of previous fragments will 16
B. Grünbaum and G. C. Shephard, “Pick’s Theorem,” American Mathematical Monthly 100(2) 1993, 150–161. G. Haigh, “A ‘Natural’ Approach to Pick’s Theorem,” Mathematical Gazette, 64 1980, 173–177. About the authors: Jack Graver is a professor of mathematics at Syracuse University working in both combinatorics and graph theory. Yvette Monachino is completing a master’s degree in both mathematics and mathematics education at Syracuse University. The two have been giving workshops for math teachers in the Syracuse City Schools under Title II B Mathematics Science Partnership grant. This paper arose out of one of those workshops. email:
[email protected] [email protected] DOI: 10.4169/194762110X535998
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3643678925903600113305305488204665213841469519415116094330572703657595919
A Conversation with Corey Greenspan Stephen Abbott
3.14159265358979323846264338327950288419716939937510582097494459230781640628620
CG: That was all I knew. I kept memorizing up to the day of the competition, and that happens to be where I ended up.
CG: It depends on what the numbers are. The two groups 7950 and 1971, for whatever reason, I looked at them and I saw years.
MH: I assume there are national competitions for memorizing π?
MH: 7950 looks like a PIN, not a year!
CG: I’m sure—there are people who know thousands of digits. MH: Do you know the world record? CG: No. I think it’s 14,000, or probably more by now. MH: What’s your best guess? CG: I’ll go with 14,000.
n March 15 last spring (the actual Pi Day fell on a Sunday), senior Corey Greenspan of Southern New Hampshire University dominated his school’s memorizing π competition, reproducing 419 correct digits. Math Horizons caught up with the recent graduate to find out what motivated him and what tricks he used to pull off this cognitive feat.
O
Math Horizons: Was this your first competition of this sort? Corey Greenspan: It was my second time. The department held the competition two years ago, and I came in third or fourth. MH: And how many digits did you get that time? CG: It was something like 117. MH: And that wasn’t enough. CG: It was only about half enough. MH: And so you had to improve your technique this time around to get 419. Did you mess up on 420, or was that all you memorized?
MH: All right, I’ll pick 14,001 and we’ll see who is closer. [Note: Lu Chao officially recited 67,890 digits, but the unofficial world record seems to be 100,000 digits by Akira Haraguchi of Japan.] OK, so what method did you use? When I looked up your phone number on the computer, I had to write it down so I wouldn’t forget it on the way to dial the phone. CG: I actually can’t remember phone numbers very well either. MH: What? CG: (laughing) It’s funny, because that is one of the strategies that people use. My problem with that method is, how do you remember which phone number comes next? If I remember five phone numbers, I still don’t know what order to put them in.
CG: I can’t explain it. I saw it and it just looked like a year. You know what I mean? MH: Hey, you are the champion, so it’s OK. CG: I guess I am a little strange. MH: OK, but we’re talking 419 digits so that’s like 100 different years you have to remember. CG: Well, the whole thing wasn’t years. Another example was a group of about 15 numbers which I noticed contained three 62s. I just memorized the short groups of numbers in between, and once I got the first number, it would all trigger itself. Instead of memorizing it all at once, it was more like I would just work through it each time I got there. MH: Like memorizing the first word of a poem, and the rest just comes out. CG: Exactly. Another thing I did was to put the numbers in subgroups of about 20 or 25 digits because remembering the end of a group was easier than thinking of the digits as all being in the middle of one long string. Once you got to the end of a group, you just have to remember how to start the next group. MH: And how do you do that?
MH: Fair enough. CG: I used a lot of different strategies. Using the same thing over and over would get old and a little tedious. I usually just used whatever came to me the first time I looked at a set of numbers. So, for instance, I thought of certain groups of numbers as years. MH: How many numbers do you consider at one time?
CG: I numbered the groups using the first number. The fourth group of numbers actually started with a four, and the fifth group of numbers started with a five. MH: Cool idea. You’re making me think I might be able to do this. How many people actually competed? CG: Not as many as last time
89 9862 8034825342117067982148086513282306647093844609550582231725359408128481 18
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MH: Having spent all that time immersed in the decimal expansion, do you have a spiritual connection to π now—some kind of cosmic enlightenment for what it means to be irrational? CG: I don’t think so. MH: 419 digits and you haven’t been mathematically transformed in some way? CG: Not really. Sorry to disappoint you. MH: Did it at least improve your dating life? CG: (laughs) No, but I do have a girlfriend. MH: Well, that’s good. Did she help you study at all? CG: When she saw the numbers 303 she told me about a band called 3OH!3, and that’s how I actually remembered those three numbers. We’d always look at each other and laugh when I got to that part. MH: But here’s what I still don’t quite get: 419 total digits minus these three digits, minus the eight digits in the two “years” you mentioned—I mean, we still have a long way to go! CG: It was a long list of tricks—there were some patches where a high density of repeated digits made things easier. In another section, the first three digits of my phone number showed up. I also did a lot of visualizing—the number 8462 I always saw as a graph on a vertical number line where you start at the 8, go halfway down to the 4, then up to the 6 and down to the 2. MH: Amazing. So what did you win? CG: I got $100 and two pies. I had originally asked for cucumber pies, but common sense prevailed and I got blueberry. MH: What’s next for you, competitionwise?
CG: They did invite me back next year. MH: OK, but I am going to recommend they pick a new number, like 2 to keep it fair. CG: I’ve seen online competitions for memorizing e. MH: I’ll challenge you to a digits of 1/3 contest. CG: There are competitions for everything, but maybe not that one. MH: What about reproducing digits of π while riding a unicycle? CG: You know, I thought of trying to recite the digits of π while solving a Rubik’s Cube, but I never did it. MH: Ouch. Any final words of advice for future π-memorizers out there? Was it worth the effort? CG: I think it was—and not just for the prize. The key is to never stop studying. It’s been only a few months since the competition, and I can probably only remember the first group or two. MH: How many can you do right now? Say them out loud, and I’ll check you on the computer. CG: 3.1415926535897932384626433832. That’s all I got. MH: Awesome! The next digit is a 7 by the way. CG: It’s the 7950 I think, actually. MH: Hey, you’re right! That is definitely your PIN. CG: You can try it.
Further Amusements The astute reader will observe that the first 419 digits of π have been placed around the border of this article. To see something truly out of the ordinary, check out http://www.youtube.com/watch?v=b UGjUCHSKLM. DOI: 10.4169/194762110X535970
8 10 975665933446 12847564823378678316 52712019091456485669234603486104543266482133936072602
141273724587006606315588174881520920962829254091715364 unfortunately, but I was one of the first to go, and so I think my score might have scared some folks off.
49
Greenspan
481117450284102701938521105559644622948954930381964428
Overstimulated
Cartoon courtesy of xkcd.com
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Alumni Profiles
loveactuary Nicole Belmonte
bout six years ago I was a freshman in college, and all I knew about my future career was that I wanted it to involve mathematics. Soon a professor told me about a career path I’d never heard of before: the actuarial profession. Actuaries seemed to have everything I was looking for—a job based in math and statistics, exceptionally high job satisfaction, prestige within the field with paychecks to match, and regular work hours. I was sold. The standard avenue to receiving actuarial credentials is a series of exams, and so I decided to get that process started early. My sophomore year, I sat for my first exam and passed, and that marked my official decision to strive for an actuarial career.
A
The next exam I sat for, I was not so lucky. From that failing grade I learned 20
a couple of important lessons: these exams were going to require serious time and effort, and I would need to use all available materials to help me get through them. The exam I failed was on probability, and since I had taken some probability classes, I sat for it with virtually no extra studying. I never had to study very hard for math tests, but actuarial exams are on an entirely different level. I’ve since learned that the generally accepted rule is that each hour of the exam requires 100 hours of studying. The shortest exam is two-and-a-half hours long; the rest are up to four hours. For property casualty insurance there are nine exams. You are number people, so I don’t have to tell you what an undertaking this is. It’s said that becoming an actuarial fellow is comparable to earning a doctorate.
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With a workload like that, organizing my time and sticking to a plan has been crucial. I’ve had the most success when I’ve created a study schedule and started on the material about three months before the exam. Some people plan out the number of hours per week they want to put in; others break it down to the day. I’ve found the more detailed my schedule, the more likely I am to stick to it. With a weekly goal, it’s easy to keep putting off studying day by day as other plans arise. It’s amazing how productive I can be in other aspects of my life when I should be studying—my house is clean, my laundry is done, my meals are prepared in advance. A daily schedule makes it a little harder to procrastinate. The minimal preparation I did for the failed exam was all from one textbook.
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There are multiple resources available to people studying for actuarial exams—particularly the first few—and it was definitely a mistake not to take advantage of them. I was fortunate in the fact that the Mathematical Sciences Department at Bentley University offered preparatory classes for the first two actuarial exams. Having a professor around who knew the material and could answer my questions was invaluable. The courses also helped keep my studying on pace—I couldn’t fall behind if I wanted to keep my grades intact. After taking the prep course for the probability exam, I not only passed but got the highest grade possible on it. This exam success, combined with some internship experience, eventually led to a full-time offer in an actuarial rotational program at a major insurance company in the fall of my senior year.
have a bit of a reputation for being, well, dull. I once was told that the definition of an actuary is “someone who is too boring to be an accountant.” Personally, I think even an actuary could come up with a better joke than that. But times are changing and an actuary has to be more than exceptionally good at math. Interviewers are looking for people with personality—people who can not only do the math but also explain it to coworkers. While I’ve been asked in interviews what I would consider when deciding whether to increase rates in a state, I’ve also been asked how I would pitch that idea to my CEO. I’ve had to give writing samples. One interviewer actually asked, “If you were a Harry
allows some study time at work, and it pays for me to attend courses and seminars offered by actuarial organizations to help prepare for all the exams. There are also dozens of other people studying for exams with me, whereas at school I was the only one staying in on During my senior year, I Down the road, my options are even more open. Many weekends to study opted to take an independof the upper-level managers in my company come from when exam time was ent study course to prepare actuarial backgrounds—even our CEO is an actuary. approaching. I for the third exam. While this suppose misery really meant I had to teach myself does love company, although the Potter character, who would you be most of the concepts, it freed up my reward of passing—in the form of both and why?” Actuaries today are expectschedule to allow for more study time. personal accomplishment and a pay ed to be very well-rounded people, and It also provided me with a professor to raise—makes the studying much less we have to be able to demonstrate that help with the topics I found difficult. It miserable. in a two-hour interview. was especially great to have someone around who could relate to what I was The exams are the only major downside Now that I’m working, exams pose going through (we would allocate at of the career so far. I am working for a different challenges. The material, and least 10 minutes at the start of each company that offers a rotational therefore the study technique, is very meeting to complain to each other program for people like me still taking different. The lower-level exams test about the notation), and with her help I exams, meaning that before I’m various mathematical subjects, while was able to pass my third actuarial assigned a permanent position, I will the upper-level exams are more exam. Actually, the professor who have three temporary positions in focused on the work actuaries do. I’m worked with me on this independent different areas of the company. The also required to take two exams each study was the same professor who told most traditional actuarial work involves year instead of the one per year I was me about being an actuary in the first pricing insurance or setting loss taking in college. Study time is now in place, so I owe her many thanks for reserves, but my first rotation involved addition to a full-time job, and there helping me get where I am today. working with a team that monitors a aren’t professors guiding me along the variety of trends, both internal and way—but the lessons I learned in Landing a job did not come without external to the company. I got a great college still hold. I now know to put in challenges, of course. The actuarial overview of the industry and saw how the hours and take advantage of any profession is gaining popularity, so actuarial work ties into the business materials I can get my hands on. My many qualified people are interviewing side of things. job pays for my study guides and for positions. Now, I know actuaries NOVEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS
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One of my ongoing projects was to monitor an indicator of the average model year and MSRP (manufacturer suggested retail price) of vehicles we insure. We looked at trends over time, and our results were used to help project expected premium and losses for the following year. This one metric reflected a variety of things, both specific to our business and about the economy in general. When gas prices soared at the end of 2008, the general public made a shift to compact cars instead of more expensive (higher MSRP) gas guzzlers, so we saw a steep drop in the average value of the indicator. Then, due to the recession, consumers purchased used cars instead of new cars or just held onto their old vehicles longer, and this could be seen clearly in our average model year data. With the “Cash for Clunkers” initiative last year, we saw a spike in the indicator that flattened out when the program ended. It’s been interesting to see things you hear about in the news directly affect our data and to learn how to incorporate those realizations
into financial planning and strategic decisions.
Paging Dr. Freud!
about to sting. Instead, the bee took out a stopwatch and a stethoscope and checked my PULSE.
Stephen Walk
I had the most disturbing dream last night. Let me tell you about it… saw a giant QUARTER rolling through my ATTIC. It made a mess there, but then crashed through the window and rolled out onto a FARM, where it knocked over a MULE. The mule picked up an AX and started to chop the quarter into pieces, but he was chased off by a convocation of ravenous EAGLES. They swooped at me, too, but I opened a package of photos and fought off the eagles with the sharp edge of a photo-NEGATIVE. This attracted the attention of a giant BEE that zoomed at me as if it was
I
There are so many questions to be answered when determining rates and reserves for insurance companies, and actuaries are called upon to answer many of those questions. We’re expected to understand all different aspects of the industry, from pricing and reserving, to claims settling and marketing, to finance and relevant laws and regulations. Because of this, we are well respected, and significant weight is given to actuarial opinion. I’ve recently begun my second rotation, where I’m seeing another side of the business and another way to apply my skills. Whereas my first rotation was a monitoring role, this rotation combines monitoring with a more typical actuarial role—pricing insurance—where I can apply what I’ve learned from exams. Once I’ve completed this rotation, there are still many new options for my next role. I could learn reserving, another traditional aspect of the business; or go
I knocked Photo by Max Halberstadt, 1922 the bee off Sigmund Freud my arm with an OAR, and it landed in the middle of a group of coal MINERS who were lined up in a SQUARE formation. From the sky plummeted RUDOLPH the Red-Nosed Reindeer, flattening all of the miners. On Rudolph’s back was Aunt BEE from the Andy Griffith Show! She looked very SCARED because she
to a position in predictive modeling, which just recently emerged in the insurance industry; or take on a role that I don’t even know exists at this point. Down the road, my options are even more open. Many of the upperlevel managers in my company come from actuarial backgrounds—even our CEO is an actuary—and outside my company, other actuaries are working in consulting or other financial fields. The rumors I heard six years ago are true—my budding mathematical career is everything I hoped it would be. About the author: Nicole Belmonte is a senior actuarial analyst at Liberty Mutual Insurance Co. in Boston. She received a degree in mathematics from Bentley University in 2008 and is now working toward her associateship and fellowship with the Casualty Actuarial Society. email: nicole.belmonte@ libertymutual.com DOI: 10.4169/194762110X535952
and Rudolph had just flown through a flock of mynah birds. Then I saw the MYNAHS do a low flyover just above Aunt Bee’s head and fly out to sea, but the sea looked very weird—it was all light brown and furry!! I then realized that the FURRY SEA was contained in a giant martini glass. My favorite bartender, OLIVER, reached over to put a giant olive into the glass (that’s why we call him “Oliver”). Then he leaned back and collapsed, exhausted, onto TWO HAYstacks… And then I woke up.
Can someone out there interpret my dream? Does it reveal my future—or some deep, dark secret of my psyche? Help! See page 33 for the answer DOI: 10.4169/194762110X12863022878093
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Have an Abacus—and Own a Piece of Mathematical History Eli Maor
In almost every home throughout the length and breadth of the Celestial Land [China], the abacus is found. C. C. Woo The Fundamental Operations of Bead Arithmetic e all remember the cute little abacus we used to play with as children. We might even have been taught how to count on it and spell out the names of the first few numbers, using the abacus’s beads as a visual, tangible aid. But in the Orient, the abacus had actually been used as a computing device for hundreds of years, until the electronic hand-held calculator brought about its demise.
W
The abacus, or counting board, was already known in antiquity. In its most primitive form it was just a tray or flat surface on which pebbles—calculi in Latin—were placed, counted, and sorted into groups of units, fives, tens, and so on. Indeed, the word abacus may have come from the Greek abax, a tray or flat pan (some scholars believe the word comes from the Hebrew avak, dust—a reference to a flat surface on which a layer of sand or dust was strewn and calculations done with the operator’s fingers).
Figure 1. The Salamis tablet. 24
Later, grooves were cut into the surface to facilitate the movement of the pebbles, which were gradually replaced by counters—little coins or button-like objects. A further improvement came when the grooves were assigned different denominations—the rightmost groove for units, the one to its left for fives, the next one for tens, followed by fifties, hundreds, and so on—a primitive sort
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of a placevalue system. Figure 1 shows one of the oldest abaci to come to us, a marble slab found on the Greek island of Salamis; its exact date is unknown. When the Hindu-Arabic numeration system was introduced to Europe in the Figure 2. A contest between an abacist and Middle Ages, it an algorist, 1503. was at first met with a great deal of skepticism. Those few who were numerically literate—mainly merchants and scholars—were accustomed to doing calculations with Roman numerals. They found the Hindu numerals strange and confusing, especially the mysterious zero, a symbol that stood for the empty slot—for “nothing”—yet when appended at the end a number had the effect of increasing it tenfold. This was simply beyond the comprehension of most people, who preferred to do their calculations on the counting board where numbers could be visualized as concrete, physical objects. Yet slowly, chiefly through the efforts of Mohammed ibn Musa al-Khwarizmi (ca. 780–850) and Leonardo Fibonacci (ca. 1170–1230), the new system gained acceptance. A famous engraving, taken from a book titled Margarita Philosophica by Gregorius Reisch (Freiburg, 1503), depicts an allegorical contest between an “abacist” and an “algorist,” the latter doing his sums with Hindu-Arabic numerals (see figure 2). The judge is Miss Arithmetica, and the expressions on the faces of the contestants make it clear who is the winner. It was a contest between the analog computer, which uses physical objects to represent numbers, and its digital counterpart: the “Old Math” versus the “New.”
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The Long Road to China From Europe merchants traveling the famed Silk Road carried the abacus to China. Called a suan pan, the Chinese version consists of a rectangular wooden frame with a cross bar along its length and a number of bamboo rods on which beads can be moved up and down.
which four are already in the “up” position. What to do? Instead of adding seven earth beads, subtract three earth beads, and add one earth bead to the next column. This is precisely the “carry” operation we learned in school, except that it is done mechanically on the abacus. The following calculation makes this clear: 4 + 7 = 4 + (10 – 3) = (4 + 10) – 3 = 14 – 3 = 11.
Each rod has two beads above the cross bar and five below it. The beads above the bar are called heaven beads, and those below it earth beads. Each heaven bead is worth five earth beads. A bead on any given rod has a value tenfold the same bead on the rod to its right. Thus (counting from right to left), the rods stand for units, tens, hundreds, and so on. The number of rods can vary from 9 to 21 and on occasion even 27 (it is usually an odd number, but I’ve seen abaci with an even number of rods, too). Of course, the more rods there are, the larger the numbers the device can handle.
Now the tens column shows 9 (the original 8 plus the 1 that was carried over). To this column you now add 1, which gives the tens column a total value of 10. So you clear this column by moving all its beads away from the bar, and carry over 1 to the next column, the hundreds. Finally, adding 3 to this column, you get the result 901. The same procedure also works in reverse: to subtract, say, 7 from 25, add three beads to the units column and subtract (“borrow”) one bead from the tens column: 25 – 7 = 25 – (10 – 3) = (25 – 10) + 3 = 15 + 3 = 18. Of course, if you are a beginner, it will take you a while to do this mechanically; but a skilled abacist can do these operations—and, indeed, much more complex ones, including multiplication, division, and square root extraction—with lightning speed, adding or subtracting the complement of a number (that is, 10 minus the number) and moving a bead in the next column up or down as required. This fits well with the Oriental tradition, where memorization and rote learning are taught to children starting at a young age.
From Old Math to New Figure 3. The number 584 represented on a suan pan. To operate the suan pan, place it in front of you on a table, with the rods running away from you (that is, turned 90 degrees from the usual way we think of an abacus). In its “neutral” position, all beads are pushed away from the cross bar. Once a bead is moved toward the bar, it assumes a value determined by the rod that it is on.
From China the suan pan made its way to Japan, where it became known as the soroban. The Japanese made two modifications to the Chinese abacus: they replaced the two heaven beads on each rod with a single bead and, in 1930, reduced the earth beads on each rod from five to four (see figure 4). This considerably simplified operating the abacus because it removed the step of clearing the rods when it was time to exchange them for a bead of the next highest denomination. This version of the soroban became a national
Suppose you want to add 317 to 584. Start by “clearing” the abacus; that is, move all beads away from the cross bar. To enter 584, begin with the rightmost column and move four earth beads up to the cross bar. Then go to the next column and move one heaven bead down and three earth beads up to the bar. Now go to the third column and move one heaven bead down. Your abacus now shows the number 584 (see figure 3). To add 317, start again with the rightmost column, where you must move up seven earth beads to join the four beads already there. Alas, only five earth beads are available, of
Figure 4. Two modern sorobans.
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standard, and practically every Japanese child was taught its use in school. Indeed, a person applying for a job in the civil service, such as accounting or engineering, had to take a government-administered exam for speed and proficiency in using the soroban. Even the dawn of the electronic age did not diminish the Japanese passion for their soroban. In a famous contest between a Japanese postal clerk and an American GI, held in Tokyo in 1946 before a large audience, the two contestants were given 50 problems in addition, subtraction, multiplication, and division. The Japanese clerk, using a soroban, won in three of the four categories; only in multiplication did the American, who used a desktop electronic calculator, win out. As with the allegorical contest between the medieval abacist and algorist, the real contest of 1946 symbolized the transition from the Old Math to the New.
Figure 7. Three miniature abaci.
Figure 8. A Japanese writing box with abacus. Figure 5. A decorative abacus.
Collecting abaci can be fun—but be warned: it can be addictive! Until recently you could find them at flea markets or antique stores, but now you can get them on eBay, often for as little as $5. They come in all sizes, they don’t need a battery, and they can be nice home decorations (see figures 5, 6, 7, and 8). Over the years I’ve acquired some 80 of them, but finding the space to store them (not to mention the disapproving glances from my wife) increasingly limits my passion. Abaci of various kinds can also be found in public places. Figures 9 and 10 show two abaci in Israel, the first in a park in the town of Ra’anana, the second in the lobby of a central bank near Tel Aviv—the largest abacus I’ve yet seen.
Figure 6. The author with the largest abacus in his collection.
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But except for die-hard enthusiasts, the abacus is fast disappearing, following in the wake of the logarithmic slide rule and the numerical tables that once were part of every
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All photographs courtesy of Eli Maor
Figure 11. A Casio dual abacus/calculator. Figure 9. Abacus in a park in Israel.
around? Whichever the case, it is another sign of the transition from the Old Math to the New. About the author: Eli Maor is the author of e: The Story of a Number, The Pythagorean Theorem: A 4,000-Year History, Trigonometric Delights, To Infinity and Beyond, and Venus in Transit. He teaches the history of mathematics at Loyola University in Chicago. email:
[email protected] DOI: 10.4169/194762110X536005 Research topic: Moduli Spaces of Riemann Surfaces
Education Theme: Making Mathematical Connections
A three-week summer program for graduate students undergraduate students mathematics researchers undergraduate faculty secondary school teachers math education researchers
IAS/Park City Mathematics Institute (PCMI) July 3-23, 2011 Park City, Utah
Figure 10. Abacus on display in a bank near Tel Aviv, Israel. algebra and trigonometry textbook. Perhaps the most poignant sign of its decline is a hybrid abacus/calculator produced in the 1970s by Casio (see figure 11). It was meant to let someone do his or her calculation on the abacus, and then recheck it on the calculator. Or perhaps it was the other way
Organizers: Benson Farb, University of Chicago; Richard Hain, Duke University; and Eduard Looijenga, Universiteit Utrecht. Graduate Summer School Lecturers: Carel Faber, Stockholm; Soren Galatius, Standford; Ursula Hamenstadt, Bonn; Makoto Matsumoto, Tokyo; Martin Möller, Bonn; Andrew Putman, Rice University; and Natalie Wahl, Copenhagen. Clay Senior Scholars in Residence: Joseph Harris, Harvard University, and Dennis P. Sullivan, SUNY Stonybrook. Other Organizers: Undergraduate Summer School and Undergraduate Faculty Program: Aaron Bertram, University of Utah; and Andrew Bernoff, Harvey Mudd College. Secondary School Teachers Program: Gail Burrill, Michigan State University; Carol Hattan, Vancouver, WA; and James King, University of Washington. Applications: pcmi.ias.edu Deadline: January 31, 2011 IAS/Park City Mathematics Institute Institute for Advanced Study, Princeton, NJ 08540 Financial Support Available
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Painting by Parametric Curves and Van Gogh’s Starry Night Stephen Lovett, Matthew Arildsen, Jon Jones, Anna Larson, Rebecca Russ athematics is sometimes described as the study of patterns, and so it is not surprising that mathematics enjoys a close connection with art. The following story, which comes from an investigative project given in a second-semester calculus class, explores a possibly new connection between these two fields.
M
Mathematics and Art Many of us have marveled at the beauty and detail of the Mandelbrot set, admired Escher’s eerie use of symmetry, and enjoyed playing with a computer algebra system’s rendering of a complicated surface. In math departments everywhere, faculty and students proudly hang posters displaying beautiful pictures that arise from mathematics: Klein bottles, adjacency graphs, or Thomas Banchoff’s flat torus in a 3-sphere. Capturing the internal beauty of mathematics through art is what Keith Devlin calls “making the invisible visible.”
© Museum of Modern Art
Vincent van Gogh’s famous Starry Night.
The above instances are examples of mathematics generating art, but there is another dimension (pun intended) to the connection—mathematics at the service of art.
JPEG file format came into vogue, using ever more technical mathematics to digitally encode images. Thinking about pixelated JPEG images calls to mind Seurat’s pointillism and ultimately the larger impressionist movement, which brings us around to our main focus—the broad brush strokes of postimpressionist Vincent van Gogh.
Celtic and Islamic works of art are known for their elegant use of tilings and polygonal symmetry patterns. Renaissance art saw the use of projective geometry to create the illusion Painting by Parametric Curves of perspective. During the American postwar era, the Postimpressionism includes the careful, minute detail of a paint-by-number painting technique became popular. technique like pointillism, but also the sweeping brush Art critics would turn over in their graves to find Leonardo strokes of Vincent van Gogh. Rather than focusing on pixels, da Vinci paired with the slogan “Everyman a Rembrandt.” we can model Van Gogh’s stylistic approach by supplying However, the paint-by-number method is in essence parametric curves a structural that plot the brush precursor to the strokes, effectively Rather than focusing on individual pixels, we bitmap file format tracing the tip of used to store can model impressionistic techniques by supplying the brush as it images digitally. In glides across the parametric curves that plot the brush strokes. a bitmap image, canvas. With the each pixel is use of a computer algebra system, we can select colors, assigned a hexadecimal code, and each code corresponds curve thickness, and parametric equations that fully describe to a unique color. Only a few years after the bitmap, the the characteristics of a single stroke.
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To demonstrate this new approach, we will use a well-known work of art, Van Gogh’s masterpiece Starry Night, because of its iconic, gracefully familiar curves. Aside from some of the details in the village, one notices that Van Gogh uses a small number of distinct styles of brush strokes, as if his hand liked a certain motion and so repeated it. Our goal, therefore, is to represent the various predominant brush strokes in the painting. We begin with the central stroke in the sky. We propose
( x(t ), y(t )) = (12
t − 6t + 15.5, − 2.5 sin(t + 2 ) + 18
)
where t 僆 [0.15, 8.5]. This follows the shape of a sine curve before beginning to curl back under itself. While y (t) is simply a scaled and shifted sine function, the x(t) function changes direction when the derivative x⬘(t) = 0; i.e., when t = 1, and this causes it to curl back. These parametric equations give the stroke:
20 19 18 17 16 5
10
15
20
Here we chose the RGB code for the color as (0.8, 0.9, 1). This color code corresponds to 80 percent red, 90 percent green, and 100 percent blue. For systems where RGB colors are given as integers between 0 and 255, our color choice would correspond to the triple (204, 230, 255). For a second stroke of the sky, we start from a figure-eight curve (cos(t), sin(2t)), a simple example of a Lissajous curve. We scale it as needed and plot it only over a useful portion of its natural domain. Hence, we use
( x(t ), y(t )) = ( −5 cos(t ) + 22, − 1.5 sin(2t ) + 14.8 ) where t 僆 [-π/3, 4π/3], and we get: 16.0 15.5
trees are common in this region. With this identification, we choose a dark green color with (R, G, B) = (0, 0.15, 0). For the foreground, we first model the tall outline of the cypress. We use the pair of equations
⎛
( x(t ), y(t )) = ⎜ 8 − 4 − sin(2t − 10), 22 − ⎝
where t 僆 [-3π, 3π]. The sine term varies between –1 and 1, and because the domain for the parametric curve is between –3π and 3π, the subtraction of t/4 in the x-coordinate makes the outline span from approximately –3 to 3. Adding 8 shifts the span from approximately 5 to 11. The y-coordinate, on the other hand, begins near 0 and moves up to 22 before returning to its starting point. This is because a squared term is subtracted from 22, and the domain is centered at the origin. Combining these components, the graph forms a wavy line that moves up from 0 to 22 and then comes back down.
14.5 14.0
t2 ⎞ 4 ⎟⎠
20
15
10
5
5
6
7
8
9
10
11
For a second piece of the cypress tree, we try to imitate Van Gogh’s wrist movement and draw a brush stroke that is reminiscent of a flame. To create this effect, we use the parametric curve
( x(t ), y(t )) = (1.8 cos(t ) + 11.8, 0.18(t 2 − 4t ) + 1) where t 僆 [0, 3π]. As with the tall piece in the foreground, by pairing a cosine function for the x-component with a quadratic in the y-component, we produce the effect of an oscillating curve that comes down toward the x-axis and then curves back up. The minimum occurs when y⬘(t) = 0; that is, when t = 2. 4
15.0
t
10
8
6
3
13.5 18
20
22
24
26
There is wide debate about the nature of the dark feature to the left and in the foreground of the painting. The Museum of Modern Art in New York identifies it as a cypress tree. This assertion has considerable merit since Starry Night depicts the village of Saint-Rémy in southern France, and cypress
4
2
2
1
14.5
15.0
15.5
10
11
12
13
16.0
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To give more definition to the cypress in the foreground, we add a third wavy stroke that gives the same wispy effect that makes this dark foreground element so mysterious. We employ one more cosine function paired with a quadratic:
⎛
t2
⎞
( x(t ), y(t )) = ⎜ cos(t ) + 15, 27 − 1⎟ ⎝
20.0
19.5
⎠
where t 僆 [3π/2, 4π].
19.0
10
18.5
8 6 4 2
18.0 10
20
30
40
50
To draw the horizon as shown above, we use a wavy curve that moves generally upward, similar to a parabola opening along the x-axis. However, to create the rolling hills effect, we add a sine function to the x-component. We choose
10.0
10.5
11.0
11.5
12.0
12.5
( x(t ), y(t )) = (1.67 sin(4t ) + 5t + 63, − 0.075t 2 + 12 ) where t 僆 [-4π, -2π/3]. To obtain a rich blue where the sky touches the hills, we use (R, G, B) = (0, 0, 0.45). The last elements we add to our version of Starry Night are the stars. For these, the linear spirals so familiar from polar graphs are our inspiration. Converting the polar equation = a to parametric equations and shifting, we obtain:
( x(t ), y(t )) = ( at cos(t ) + x0 , at sin(t ) + y0 ) where (x0, y0) is the center of the spiral and t 僆 [0, -2πn] with n being the desired number of loops. The constant a determines the radius of the spiral. In terms of colors for the stars, (R, G, B) = (1, 1, 0) gives a basic yellow while (R, G, B) = (1, 0.8, 0) mixes in less green and so produces a more orange tint.
Re-creating Starry Night As Van Gogh once stood before his canvas, brushes in hand, overlooking Saint-Rémy-de-Provence one evening during his stay at the sanitarium, so we now sit before a computer one afternoon in a math department office. Poised to press the “enter” key on our keyboard, it’s not clear whether we are at the cusp of grandeur or taking a small step toward insanity. Nevertheless … click. Before the art critics weigh in, we should keep in mind that we have only “painted” nine brush strokes, and it’s safe to assume Van Gogh used a few more than that. Still, with just
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these few strokes—and perhaps a pair of calculus-colored glasses—some essence of Starry Night has been captured in our simple equations. Admittedly, it takes a good bit of tinkering to find the equations that follow the artist’s line, but at least we have the comfort of knowing our brush strokes are digitally saved. We can always change colors, enlarge, tweak, or delete any stroke we feel is out of place. With this flexibility we hope that we will be more inclined to deal with our artistic angst by removing errant brush strokes rather than removing an earlobe.
About the authors: Stephen Lovett is associate professor of mathematics at Wheaton College. Matthew Arildsen (political science), Jon Jones (physics), Anna Larson (mathematics), and Rebecca Russ (engineering) were members of Lovett’s fall 2009 calculus II class. email:
[email protected] DOI: 10.4169/194762110X535989
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THE PLAYGROUND! Derek Smith
Welcome to the Playground! Playground rules are posted at the bottom of page 33, except for the most important one: Have fun!
THE SANDBOX
THE ZIP-LINE
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 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.
November’s Sandbox problem, An Alternate Solution, is communicated by Mike Develin (D.E. Shaw) and Ariel Levavi (UC–San Diego). A group of n prisoners is waiting in a group holding area for the following sequence of events to take place:
November’s Zip-Line problem takes a census of your powers of geometry. Near the end of the article on “The Politics before the Politics: Census 2010, Reapportionment, and Redistricting” on pages 5–9, the authors describe several ways to measure the “compactness” of a planar region R. One is based on the Roeck test, which computes the ratio of the area of R to the area of the smallest circle containing R. Another is the Cox test, which computes the ratio of the area of R to the area of the circle with the same perimeter as R.
(1) The prisoners will soon be isolated in individual cells. (2) Their captor will secretly assign a distinct rational number to each of the prisoners. (3) The captor will slide under the door of each prisoner’s cell a list telling which rational number is assigned to each of the n – 1 other prisoners.
Problem 253, Rating Ratios, asks: if all you know about R is that it is a quadrilateral, can you tell which is larger, the Roeck ratio or the Cox ratio?
(4) After a few minutes, each prisoner must slide back under the door a single sheet of paper with either a 0 or a 1 written on it.
THE JUNGLE GYM
(5) The captor will then take all of the 0s and 1s and line them up in order according to the least-to-greatest order of the rational numbers assigned to the prisoners. The prisoners will all be released if the sequence of 0s and 1s alternates, either as 0, 1, 0, 1, ... or as 1, 0, 1, 0, ... Otherwise, all of the prisoners will remain incarcerated until one of them proves or disproves the Riemann Hypothesis. When n = 2, prisoners A and B can guarantee their release by having A write 0 and B write 1 on their sheets of paper, since regardless of the rational numbers assigned to each of them the resulting sequence will either be 0, 1 (if A’s rational number is smaller) or 1, 0. Problem 252 asks about the case when n = 3: what’s the best strategy that a group of three prisoners should agree to while waiting in the holding area? As a challenge problem, ask the same question for n greater than 3.
Any type of problem may appear in the Jungle Gym—climb on! Problem 254, Chewsing Integers, is offered by Gary Gordon (Lafayette College). First, we need a few quick definitions that generalize the standard formula for binomial coefficients. If a1 < a2 < … is a strictly increasing sequence of positive integers and m is a positive integer, define m? to be
m ? = a1a2 am , the product of the first m numbers in the sequence. Also, define 0? = 1. Finally, if n and k are integers such that 0 ≤ k ≤ n, define “n chews k” as
n n? = . k k ?(n − k )? You can check that “n chews k” is a generalization of the binomial coefficient “n choose k”: the two concepts are
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equal when the given sequence is 1 < 2 < 3 < … because then m? = m!
This problem is similar to a problem from the 2004 William Lowell Putnam Mathematics Competition, where basketball star Shanille O’Keal shoots free throws.
Suppose that you are told two things: (1) The sequence a1 < a2 < … is an arithmetic sequence, which means that it is of the form x < x + y < x + 2y < … (2) “n chews k” is an integer for all non-negative integers k and n such that 0 ≤ k ≤ n. What can you conclude about the sequence?
Problem 246, Hex on You, remains open, with a hint, as described in the next section.
APRIL WRAP-UP The first problem in April’s Playground was The Mighty Mathletes by David Seppala-Holzman (St. Joseph’s College). At one point in the baseball season, the Mathletes had a winning percentage of 50 percent. Then they went on a tear, so that later in the season they had a winning percentage of 90 percent. Problem 244 asked if there are any other specific winning percentages that the Mathletes must have hit during the season.
The first Zip-Line problem in April’s Playground concerned an explorer wishing to travel as far as possible into the desert with her camel Ralph. Ralph must eat bananas while he walks, and he consumes them at a steady rate of one banana per mile. Problem 245, Going Bananas, asked how far they could travel from their starting point if they had an initial stockpile of 4,000 bananas, with the restriction that Ralph can carry at most 1,000 bananas at any time. The duo can get to a distance of just over 1,676 miles away from the starting point, as discovered by student Joseph Ward, the University of Minnesota–Duluth Math Club, the Northwestern University Problem Solving Group, and the student group of Kim Moorehead, Ben Nelson, and Chris Vroon (Taylor University). As with the solution to a similar problem in James Tanton’s article “A Dozen Harmonious Problems” in April’s issue, the key is to establish stockpiles of bananas along the route. All four correct solutions used stockpiles at the positions A, B, and C shown below. Start A
Student Joseph Ward (Silver Creek Academy High School) and the student group of Jennie Clinton, Bill Solyst, and Danielle Urbanowicz (Taylor University), as well as Michael Faleski and Dmitry Fleischman, found that the set of guaranteed winning fractions (wins to total games played) must be a subset of {1/2, 2/3, 3/4, …, 9/10}. A team that loses one of its first two games and then wins the next eight games will achieve every fraction in this set and no others. Both Joseph and the group from Taylor went on to show that each of those fractions must be hit, regardless of the way the team goes from 1/2 to 9/10. Suppose that a team currently has a wins out of b games, and suppose that the fraction of wins satisfies a/b < n/(n + 1) for some positive integer n. If the team’s fraction could skip over n/(n + 1) with a win in the next game, then we would also know that n/(n + 1) < (a + 1)/(b +1). The first inequality implies that an + a < bn while the second implies that bn < an + a + 1. Combining these, we are required to have the strict inequalities
an + a < bn < an + a + 1 for integers a, b, and n, an impossibility since the terms on each end differ by only 1.
1000/5
B
C 1000/3
Finish 1000
1000/7
Each segment of length 1000/(2k + 1) is traversed a total of 2k + 1 times: k + 1 times left-to-right and k times right-to-left. Three of the solutions hit the labeled points in the order
SASASASABABABCBCF. For each segment XY moving toward the right, Ralph is always loaded with 1,000 bananas at X. For XY moving to the left, the number of bananas deposited at X should ensure that the duo arrives at Y carrying no bananas. The fourth solution used a different route of travel among the labeled points—can you find it? One might complain that the solutions just given require messy banana fractions: for instance, according to the travel route displayed above, exactly 5,000/7 bananas should first be deposited at A. We leave open the question of how far the duo can travel into the desert if only whole bananas may be deposited or packed between travel segments.
April’s second Zip-Line problem was the final problem mentioned in the “Aftermath” column on the Intermediate
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Value Theorem by Bruce Peterson. Suppose a planar set has the property that the maximum distance between any two points in the set is 1. Problem 246, Hex On You, asks you to find the side length of the smallest regular hexagon that is guaranteed to contain it. This proved to be the most challenging of April’s problems: there were several solutions submitted with the correct numerical answer—the side length needs to be at most 1/ 3 —but none of the solutions provided an appropriate proof. The most common misstep was the claim that if S is any set such that the maximum distance between two points in S is 1, then S can be contained in a circle of diameter 1. As a counterexample, let S be an equilateral triangle of side length 1:
Problem 247 asked if there must be a point P in the interior of the triangle such that
∠PAB =
∠PBC ∠PCA = . 2 3
Student Joseph Ward proved that, yes, such a point P must exist. His argument asks you to follow, as the value of increases from 0, the path of the point P that is at the intersection of the rays leaving A and C. When = 0, P is at A. As increases from 0, all the rays rotate in a clockwise manner until one of the rays (corresponding to the minimum m of the angle measures A, B/2, and C/3) first reaches the other side of the angle where it is based. If m = A, then the path of P will end at C; if m = C/3, then the path of P will end on the side BC; and if m = B/2, then the path of P may end in the interior of the triangle. In all three cases, you can check that there must be an angle
(0, m) such that P lies on the ray emanating from B, giving the desired point P. Why? As Joseph points out, this could have been Problem 246!
SUBMISSION & CONTACT INFORMATION
We’ll leave the problem open with the following hint: you just might want to use the topic of the “Aftermath” column that generated the problem!
“FIVE MORE MINUTES, KIDS!” The final problem from April was As Easy As 1-2-3 by Christopher Thron (Texas A&M University–Central Texas). Suppose that a triangle has vertices A, B, and C labeled clockwise as shown in the picture below.
The deadline for submitting solutions to problems in this issue is January 10, 2011.
B
A
α
2α
P?
The Playground features problems for students at the undergraduate and (challenging) high school levels. All problems and/or solutions may be submitted to Derek Smith, Mathematics Department, Lafayette College, Easton, PA 18042. Electronic submissions (preferred) may also be sent to
[email protected]. Please include your name, email address, and school affiliation, and indicate if you are a student. If a problem has multiple parts, solutions for individual parts will be accepted. Unless otherwise stated, problems have been solved by their proposers.
DOI: 10.4169/194762110X536023
Answer for the Paging Dr. Freud! riddle on page 22: The “dream,” when visualized in detail and mentally rehearsed a few times, is a mnemonic for the quadratic formula. (Read aloud the words in small caps: “Quarter-attic farm-mule: ax eagles...”)
3α
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Aftermath
Does the Master’s Degree in Mathematics Get Too Little Respect? Carl Cowen f you think about the history of science, mathematics sits in a unique position: everything that has ever been true in mathematics is still true! We no longer believe that the elements are Earth, Air, Fire, and Water, for example, but Euclid’s description of geometry in the plane is still correct. Modern physics rests on developments from the late 19th century onward, with recognition that Newton’s discoveries provide a working foundation. Modern chemistry is largely a 20th-century science, and molecular biology starts with the discovery of the role of DNA in the mid-20th century. A fundamental difference between undergraduate education in mathematics and that of the other sciences is that we (mostly) take students to the early 20th century or so while the other sciences take students to the research forefront.
I
As an example, a few years ago I taught a course on computational neuroscience for juniors and seniors with a mathematical background or a biological background (prerequisites: two semesters of calculus for biology students; differential equations for math students; and at least junior standing in a mathematics, statistics, engineering, or biological sciences major. Note that no biology prerequisites were asked of the math students). During the semester, we read a research paper from 1988. The math students were astonished: they mostly had never seen a research paper, or if they had, they had never seen one that new! The biology students were also astonished: they had seen many research papers, but they had never seen one that old! Thus, our science colleagues have a quite different perspective on undergraduate and graduate education than we do. A Ph.D. in chemistry at Purdue University requires two (two!) classroom courses, and the rest is research. A Ph.D. in mathematics usually includes 10 to 15 classroom courses! My own opinion is that the study for the master of science degree is the most intensive learning experience in the mathematical sciences. Much more mathematics is learned than at the undergraduate level because the study is so much deeper, and more is learned than at the Ph.D. level because there the learning is specialized and research focused. Thus, first and foremost, I regard the M.S. as the time when students acquire a broad and deep understanding of mathematics. Further, most of the master’s program is devoted to studying late 19th-, 20th-, and 21st-century mathematics. Indeed, an 34
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M.S. program should put a student close (say 1950s–1970s era) to the research forefront in at least one area. Most M.S. programs include Ph.D. qualifier courses. This is fundamental, broad, and deep material in comparison to undergraduate work. As a profession, we put too little emphasis on the M.S. and give it too little respect. We should be encouraging many more of our undergraduate students to go to graduate school and get an M.S. degree. Mathematics faculty are good at encouraging the “best” students to go to graduate school, but we should be encouraging the top third of our students to go on—they are surely qualified for the experience and would benefit greatly from the added education. Moreover, the job surveys I’m familiar with suggest that the M.S. is the most marketable degree in the mathematical sciences. This is a consequence, I believe, of the fact that M.S. students know much more mathematics than undergraduates and are less likely than Ph.D.s to be “distracted” by research interests (in the minds of those who are looking for mathematical expertise in filling job openings). There are several important career paths for M.S. degrees. The M.S. in statistics is the professional degree for a statistician. As I understand it, except for specialized areas such as the pharmaceutical industry where the Ph.D. is preferred, most “working” statisticians have an M.S. in applied statistics or biostatistics. The two-year college faculty member in mathematics is usually expected to have a “plain vanilla” M.S. in mathematics with enough statistics background to be able to teach beginning statistics courses. Both of these career paths are full of opportunities! Aftermath essays are intended to be editorials and do not necessarily reflect the views of the MAA. To respond, go to Aftermath at www.maa.org/mathhorizons. About the author: Carl Cowen is professor of mathematical sciences and director of the Actuarial Science Program at Indiana University–Purdue University at Indianapolis. He is a former president of the MAA. email:
[email protected] DOI: 10.4169/194762110X536032
