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Mathematical Idol 2010

April 2010

The following effect was inspired by a well-known card trick, and while it does use cards of a sort—on which appear names of famous mathematicians—it has otherwise been taken some distance from its origins. (More on that, and a version with a regular deck of cards, later.)

Stand before an audience, your assistant by your side. "Hello, and welcome to Mathematical Idol 2010! Thanks for coming. Hundreds of top mathematicians from down through the ages have worked hard to be remembered here today, observing patterns, formulating definitions, finding examples and counterexamples, and proving theorems, as well as providing inspiration for countless others in the field." Pick up and briefly display a small stack of index cards.

"Our goal is to pick one of these as Mathematical Idol 2010. We believe that audience participation is important, and in a moment, I'll ask for your help in selecting the overall winner! First prize is  nights at Hilbert's Hotel (with a guarantee of never having to move room), dinner with Paul Erdös at the Caffeine Conjectures Café, an indefinite publishing contract with a continuum of the profession's leading journals, and a lifetime membership of the Mathematical Association of America." Pause for applause.

"After numerous preliminary Gauss/Jordan elimination rounds—I'm afraid Jordan was eliminated early on—we are down to just eight contestants: in alphabetical order they are Archimedes, Euclid, Euler, Gauss, Hilbert, Newton, Pythagoras and Riemann. Please give them a big hand!"

Display eight index cards, each bearing the name of one of these mathematicians, and say, "I need a volunteer to decide how to narrow this list down to three." Somebody will come forward. Thank that person, and continue.

"In a moment, while my assistant here steps out of the room, you'll be given these cards, and asked to pick the top three! You can deliberate carefully, weighing up their various accomplishments and contributions to mathematical knowledge, or you can pick them at random, it's up to you. Once you have reached your decision, I'll then play the role of the Mathematical Idol judges, and decide on the ranking of those finalists. Only I will know who the overall winner is." Pause, to allow that to sink in. 

"You'll then ask my assistant to rejoin us, whereupon you'll reveal to her the third and second place runners-up, no more. That means the winner could be any of the other six contestants. How could anybody know who the winner is, based only on that information? One would have as much chance as rolling a die and getting a six. Yet, my assistant will soon announce the correct identity of Mathematical Idol 2010!"

For instance, as we will see below, if Riemann gets third place, and Newton second, then Gauss wins overall.

 

Your assistant can pull this off no matter which three finalists are chosen; it has nothing to do with the relative talents of the mathematicians, or expectations of specific rankings. It can even be done remotely, using a speakerphone both for volunteer-to-assistant communication and for the final announcement of the winner; a quite effective method, and an easy one, as it allows for the subtle introduction of at least one (honest) cheat sheet.

Here's the set-up in more detail. Introduce your mathematically savvy assistant—or somebody with an off-the-chart good memoryand in due course have her step outside. Hand out eight index cards, on which are written the names of the mathematicians, one per card. The volunteer looks through them, and hands three of them back to you. The other five are put away. You deliberate for a moment, perhaps making idle chit chat (or having the volunteer read out some biographical information on these three mathematicians), as you do the required calculations in your head. You then display the finalists in a face-down row, and clearly indicate the precise ranking. Cover the card which represents Mathematical Idol 2010 with a cup or glass.

Point to the covered card, declaring, "That's the overall winner, you'll find out who it is presently. Please step outside and bring my assistant back in." Once your assistant has rejoined the gathering you, have the third and second place runner-up cards turned over, revealing their names. Sure enough, your assistant soon names the overall winner. Finally, have the card under the cup or glass retrieved, to confirm the accuracy of her announcement, namely, the identity of Mathematical Idol 2010!

Hyper Shift Incidence

It's all based on this little-known fact:

If three items are randomly chosen from eight known ones, then you can arrange two of the three in a row to reveal to a savvy assistant the identity of the third one.

"Known" here means known to both you and your assistant. "Savvy" implies somebody who (like you) can do non-trivial mental gymnastics; though possession of a prodigious memory would also do the trick, as experience bears out (keep reading).

It's an easy exercise to figure out ways to do this if the choices are made from six items, and such strategies can be modified to work for seven, but if eight are allowed, it's quite a bit trickier. There's no hope for more than eight (can you see why?).

This is a special case of a more general result which is often associated with a famous two-person card trick from sixty years ago. What's crucial is the ranking of those three finalists, which you determine, after performing some mental calculations explained below. From the third and second place rankings alone, your assistant can indeed figure out who the overall winner must be. Once again, it's a case of mathematical, not physical or verbal, communication!

(For a more magical performance, one can utilize a twist on the "magician's choice" to give the illusion that you play no role in the ordering of the finalists.)

Let's take a closer look at the numbers. You start with any one of the 8!/(3!5!) possible selections of 3 objects from 8, which we'll see determines one specific way (of the 3! available) to rank them. Meanwhile, your assistant learns of one of the 8 x 7 possible runners-up arrangements (for third and second places). She is expected to figure out the overall winner from that information alone, and she succeeds, essentially because 8!/(3!5!) ≤ 8 x 7 and 8!/(3!5!) ≥ 8 x 7.

The trick is to come up with a logical (or memorable) one-to-one correspondence between the 56 unordered sets of three finalists and the 56 ordered runners-up arrangements, so that both of you can perform this live in real time. There are many ways to to do this, which we encourage interested readers to pause and consider before going any further.

The method which we now explain readily extends to larger numbers; indeed, it's the simplest nontrivial case of an algorithm due to Elwyn Berlekamp, which has gets rediscovered frequently and has appeared in print several times in recent years in connection with the optimal version of the aforementioned two-person card trick (references are provided later).

The general result says: For a deck of (up to) n! + n - 1 cards, if we chose any n of them at random, it is possible to arrange some n - 1 of them in a row in such a way as to indicate the identity of the remaining one. The famous Fitch Cheney card trick (published in 1950) is the case n = 5 for a standard 52 card deck; extending this to a 124 card deck requires an altogether more sophisticated approach. It's easy to see that this upper limit on the deck size cannot be exceeded.

Our focus here is on the case n = 3. We follow the treatment found in a delightful online Maple worksheet by John Cosgrave, of St. Patrick's College, Dublin.

Let's identify Archimedes, Euclid, Euler, Gauss, Hilbert, Newton, Pythagoras, and Riemann with 0, 1, 2, 3, 4, 5, 6 and 7, respectively, and suppose that c0 < c1 < c2 are chosen from {0, 1, ..., 7}. The winning label w = c0 + c1 + c2 (mod 3) determines the winner cw (namely, Mathematical Idol 2010). You convey this fact to your assistant by arranging the remaining two cj's in the appropriate order. Note that cw ≥ w. In fact,

cw - w is in {0, 1, ..., 5} and gives the position of cw if those cj's which are smaller than it are eliminated, so that its position is "bumped back" a value or two if necessary.

Next, you must compute the sum s of the two runners-up. Since s = w - cw (mod 3), then the non-negative number cw - w is congruent to -s (mod 3). Your assistant can figure out cw - w (mod 3)—and hence s (mod 3)for herself, as she knows the two runners-up. By observing the order of these finalists, she can deduce the exact value of cw - w, and hence the winner cw.

By calculating -s (mod 3), your assistant narrows down the value of cw - w to one of two possibilities, differing by 3. That's where the order of the two runners-up comes in: you use ascending order to indicate the first possibility (0, 1 or 2), and descending for the second (3, 4 or 5). Hence your assistant can determine cw - wexactly. Finally, the number of runners-up which are less than cw - w communicates w, which in turn tells your assistant how much to bump up cw - w by to find Mathematical Idol 2010.

For example, suppose that the three finalists you are handed are Archimedes, Euler and Pythagoras, corresponding to 0, 2 and 6. Since w = 0 + 2 + 6 = 2 (mod 3), then Pythagoras is Mathematical Idol 2010, corresponding to c2 = 6. This fact must be coded by the correct ordering of the two runners-up Archimedes and Euler, corresponding to 0 and 2. The sum of these is s = 0 + 2 = 2 (mod 3), and cw - w = 1 (mod 3) as expected. The negative of the sum s reveals 1 (mod 3) to your assistant, and if the two finalists in question are displayed in descending order 2, 0 (i.e., Euler in third position, Archimedes in second), that communicates the fact thatcw - w = 1 + 3 = 4 (as opposed to 1). Finally, since both of the runners-up are visibly less than 4, your assistant can infer that w = 2cw - w = 4 and c2 = 4 + 2 = 6; Pythagoras must be Mathematical Idol 2010.

Let's look a second example, this time from your assistant's perspective only. Suppose that she sees runners-up Euclid and Pythagoras (corresponding to 1 and 6) in that order. She computes s = 1 + 6 = 1 (mod 3), deducing that cw - w = -s = 2 (mod 3). So she knows that either cw - w = 2 or 5. Since the two runners-up are in ascending order, she infers that cw - w = 2. Exactly one of the runners-up is less than 2, so she concludes that w = 1, and finally bumps up by 1. Gauss must be Mathematical Idol 2010!

Live performance may be rendered less stressful by the availability of discretely placed customized cheat sheets, one for you and another for your assistant. (If the speakerphone option is used, your assistant can read hers freely and hence no longer needs to be highly trained or have an amazing memory.) Your cheat sheet, as illustrated below, is alphabetized in the left column by finalists, showing the overall winner (in italics), whereas the right column shows the exact order in which the runners-up must be revealed to your assistant.

YOUR PERSPECTIVE  
Finalists (winner in italics) Runners-up in order
Archimedes, Euclid, Euler Euclid, Euler
Archimedes, Euclid, Gauss Archimedes, Gauss
Archimedes, Euclid, Hilbert Archimedes, Euclid
Archimedes, Euclid, Newton Euclid, Newton
Archimedes, Euclid, Pythagoras Archimedes, Pythagoras
Archimedes, Euclid, Riemann Euclid, Archimedes
Archimedes, Euler, Gauss Archimedes, Euler
Archimedes, Euler, Hilbert Euler, Hilbert
Archimedes, Euler, Newton Archimedes, Newton
Archimedes, Euler, Pythagoras Euler, Archimedes
Archimedes, Euler, Riemann Euler, Riemann
Archimedes, Gauss, Hilbert Archimedes, Hilbert
Archimedes, Gauss, Newton Gauss, Archimedes
Archimedes, Gauss, Pythagoras Gauss, Pythagoras
Archimedes, Gauss, Riemann Archimedes, Riemann
Archimedes, Hilbert, Newton Hilbert, Newton
Archimedes, Hilbert, Pythagoras Pythagoras, Archimedes
Archimedes, Hilbert, Riemann Hilbert, Archimedes
Archimedes, Newton, Pythagoras Newton, Archimedes
Archimedes, Newton, Riemann Newton, Riemann
Archimedes, Pythagoras, Riemann Riemann, Archimedes
Euclid, Euler, Gauss Euler, Gauss
Euclid, Euler, Hilbert Euclid, Hilbert
Euclid, Euler, Newton Euler, Euclid
Euclid, Euler, Pythagoras Euler, Pythagoras
Euclid, Euler, Riemann Euclid, Riemann
Euclid, Gauss, Hilbert Euclid, Gauss
Euclid, Gauss, Newton Gauss, Newton
Euclid, Gauss, Pythagoras Euclid, Pythagoras
Euclid, Gauss, Riemann Gauss, Euclid
Euclid, Hilbert, Newton Newton, Euclid
Euclid, Hilbert, Pythagoras Hilbert, Euclid
Euclid, Hilbert, Riemann Hilbert, Riemann
Euclid, Newton, Pythagoras Newton, Pythagoras
Euclid, Newton, Riemann Riemann, Euclid
Euclid, Pythagoras, Riemann Pythagoras, Euclid
Euler, Gauss, Hilbert Gauss, Hilbert
Euler, Gauss, Newton Euler, Newton
Euler, Gauss, Pythagoras Gauss, Euler
Euler, Gauss, Riemann Gauss, Riemann
Euler, Hilbert, Newton Hilbert, Euler
Euler, Hilbert, Pythagoras Hilbert, Pythagoras
Euler, Hilbert, Riemann Riemann, Euler
Euler, Newton, Pythagoras Pythagoras, Euler
Euler, Newton, Riemann Newton, Euler
Euler, Pythagoras, Riemann Pythagoras, Riemann
Gauss, Hilbert, Newton Newton, Hilbert
Gauss, Hilbert, Pythagoras Pythagoras, Gauss
Gauss, Hilbert, Riemann Hilbert, Gauss
Gauss, Newton, Pythagoras Newton, Gauss
Gauss, Newton, Riemann Riemann, Newton
Gauss, Pythagoras, Riemann Riemann, Gauss
Hilbert, Newton, Pythagoras Pythagoras, Newton
Hilbert, Newton, Riemann Riemann, Hilbert
Hilbert, Pythagoras, Riemann Pythagoras, Hilbert
Newton, Pythagoras, Riemann Riemann, Pythagoras

The second table shows your assistant's cheat sheet, which is alphabetized in the right column by the runners-up (in the order in which they are revealed to her), whereas the left column shows all three finalists, with the winner in italics. Your assistant's cheat sheet has the same 56 rows as yours, it's just that they have been re-ordered for ease of lookup, being alphabetized by the entries in the second column, as opposed to the first. 
 

  ASSISTANT'S PERSPECTIVE
Finalists (winner in italics) Runners-up in order
Archimedes, Euclid, Hilbert Archimedes, Euclid
Archimedes, Euler, Gauss Archimedes, Euler
Archimedes, Euclid, Gauss Archimedes, Gauss
Archimedes, Gauss, Hilbert Archimedes, Hilbert
Archimedes, Euler, Newton Archimedes, Newton
Archimedes, Euclid, Pythagoras Archimedes, Pythagoras
Archimedes, Gauss, Riemann Archimedes, Riemann
Archimedes, Euclid, Riemann Euclid, Archimedes
Archimedes, Euclid, Euler Euclid, Euler
Euclid, Gauss, Hilbert Euclid, Gauss
Euclid, Euler, Hilbert Euclid, Hilbert
Archimedes, Euclid, Newton Euclid, Newton
Euclid, Gauss, Pythagoras Euclid, Pythagoras
Euclid, Euler, Riemann Euclid, Riemann
Archimedes, Euler, Pythagoras Euler, Archimedes
Euclid, Euler, Newton Euler, Euclid
Euclid, Euler, Gauss Euler, Gauss
Archimedes, Euler, Hilbert Euler, Hilbert
Euler, Gauss, Newton Euler, Newton
Euclid, Euler, Pythagoras Euler, Pythagoras
Archimedes, Euler, Riemann Euler, Riemann
Archimedes, Gauss, Newton Gauss, Archimedes
Euclid, Gauss, Riemann Gauss, Euclid
Euler, Gauss, Pythagoras Gauss, Euler
Euler, Gauss, Hilbert Gauss, Hilbert
Euclid, Gauss, Newton Gauss, Newton
Archimedes, Gauss, Pythagoras Gauss, Pythagoras
Euler, Gauss, Riemann Gauss, Riemann
Archimedes, Hilbert, Riemann Hilbert, Archimedes
Euclid, Hilbert, Pythagoras Hilbert, Euclid
Euler, Hilbert, Newton Hilbert, Euler
Gauss, Hilbert, Riemann Hilbert, Gauss
Archimedes, Hilbert, Newton Hilbert, Newton
Euler, Hilbert, Pythagoras Hilbert, Pythagoras
Euclid, Hilbert, Riemann Hilbert, Riemann
Archimedes, Newton, Pythagoras Newton, Archimedes
Euclid, Hilbert, Newton Newton, Euclid
Euler, Newton, Riemann Newton, Euler
Gauss, Newton, Pythagoras Newton, Gauss
Gauss, Hilbert, Newton Newton, Hilbert
Euclid, Newton, Pythagoras Newton, Pythagoras
Archimedes, Newton, Riemann Newton, Riemann
Archimedes, Hilbert, Pythagoras Pythagoras, Archimedes
Euclid, Pythagoras, Riemann Pythagoras, Euclid
Euler, Newton, Pythagoras Pythagoras, Euler
Gauss, Hilbert, Pythagoras Pythagoras, Gauss
Hilbert, Pythagoras, Riemann Pythagoras, Hilbert
Hilbert, Newton, Pythagoras Pythagoras, Newton
Euler, Pythagoras, Riemann Pythagoras, Riemann
Archimedes, Pythagoras, Riemann Riemann, Archimedes
Euclid, Newton, Riemann Riemann, Euclid
Euler, Hilbert, Riemann Riemann, Euler
Gauss, Pythagoras, Riemann Riemann, Gauss
Hilbert, Newton, Riemann Riemann, Hilbert
Gauss, Newton, Riemann Riemann, Newton
Newton, Pythagoras, Riemann Riemann, Pythagoras

One performance option is to include biographical snippits on the index cards, some of which could be read out along the way, such as:

Lenny Euler had over 900 releases to his name, some of them still influential today. Who could forget "The Seven Bridges of Konigsberg," "Euler's Identity Crisis," "The Infinite Product of Your Love," "Blinded by the Light" or the posthumous "Master of Everything." While he used e and mail a lot, it never occurred to him to try e-mail.

Or perhaps,

Bernie Riemann had one brother and four sistersfamily was always integral to his life. In college, he "already sang like a canary," according to one of his teachers, Moritz Stern. His lifelong ambition was to understand the music of the primes in terms of the elusive Catherine Zeta Jones function. He claimed that: "half the battle is keeping it real," but neither he nor anyone else has been able to prove this to date.

Or perhaps not.

Kiddy Chaperone

Here's a version with a regular deck of cards. Have somebody shuffle and deal out eight cards face up. On an index card write down the names of those dealt, as you ask the volunteer to carefully chose five cards for a poker hand. The idea, you stress, is to get a good hand "without making all of the selections obvious." Those five cards are then turned face down and set aside. 

Display two of the three cards which were not selected face up, banishing the third one from view. Your assistant joins the proceedings for the first, having previously seen or heard nothing, and is handed your list of all eight cards. She sizes up the situation—and the volunteer—for a moment, and soon correctly identifies all five cards in the chosen poker hand. 

The big deal here is that it appears that only two of the unused cards have been identified, merely narrowing down the poker hand to one of six possibilities. However, she can tell what the third unused card is, and hence announce the complete poker hand with 100% accuracy, provided that: (1) she knows the identities of all eight cards, and (2) both of you are in agreement on a specific ordering (0 to 7, as before) for these cards. The list which you make (in advance of the poker hand selection), and later hand over, takes care of both of these requirements. You may opt to jot down the cards according to some pre-arranged full deck order (e.g., CHaSeD) or not. Smaller and less conspicuous numerical (0 to 7) versions of the cheat sheets can be used. 

As noted in Better Poker Hands Guaranteed (the June 2006 Card Colm), there is an 89% chance of getting at least "one pair" among eight random cards. Assuming that the volunteer's poker hand includes all such desirable cards, one can play up your assistant's unerring ability to deduce the identity of all five cards selected based on "reading the volunteer's mind." This effect can of course be repeated.

Reenforced Lies

"Mathematical Idol" can be found—in a two-stage, extended form, in which all eight contestants are ranked—in Mathematical Wizardry for a Gardner (edited by Ed Pegg Jr, Alan H. Schoen & Tom Rodgers), A.K. Peters, April 2009. 

For the grand finale of the 2005 season of Mathematical Idol, during the banquet at the MAA Seaway Section Meeting, at SUNY Geneseo, NY, that October, a non-mathematical assistant's apparently photographic memory of her cheat sheet resulted in a flawless performance. 

In addition to the two weblinks provided earlier for the classic Fitch Cheney trick and its generalization, here are two fine resources: (1) Michael Kleber, "The Best Card Trick," Mathematical Intelligencer24, 1, 9--11, Winter 2002, and (2) Shai Simonson & Tara Holm, "Using a Card Trick to Teach Discrete Mathematics," PRIMUSXIII, 3, 248--269, September 2003. 

"Hyper shift incidence" is an anagram of "Fitch Cheney inspired," "reenforced lies" is an anagram of "Idol references," and "kiddy chaperone" is an anagram of "dicey poker hand." 

Dummy View - NOT TO BE DELETED