# Two Summer Difference Certainties

June 2009

Disclaimer: what we describe here can only be done with a genuinely shuffled deck by those who can casually yet confidently glance through the card faces seeking five which enjoy a special relationship to one another, whose identities are then memorized instantly, while the performer chats nonchalantly to the audience about how random a shuffled deck is. A much easier version, in which the performer does all of the (ahem) shuffling, is also suggested before we sign off.

Consider the following effect:

From an audience-shuffled deck, deal five cards in a face-down row.
Ask for two volunteers. Have the first one mix those cards thoroughly
before offering any two of them to the other one; the mixer takes the
remaining cards. The volunteers each scan their card faces, sum
those values, then look at the other person's cards, and compare
sums. The person with the larger sum announces by how much that
one exceeds the other sum, or else one volunteer indicates that the
sums are equal. For instance, one person may say, "My total is
three larger than hers." Based on this information alone, you, the
mathemagician, correctly identify each of the cards being held.

Yes, you really can name all values and suits! This effect, which is done more or less as described above with an unprepared deck, is based on an extension of a probabilistic version of the key concept explored in Additional Certainties, the February 2008 Card Colm. It requires some quick thinking and fast memory work, and can even be repeated if you are feeling brave.

A more straightforward, and definitely less stressful, version, using a partially rigged deck, does not bear repetition; nevertheless, it's discussed briefly at the end below.

## Richest 2-Sums

Let's begin by studying small sets of positive integers {a1, a, ... , ak} with the property that all possible k(k-1)/2 unordered pairs have unique sums. We say that such 2-sum rich sets are 2-sum'r, pronounced "two summer." (In the literature these called Sidon sets or sequences). Of course, the ai's must be distinct, and there is no harm is assuming that they are also listed in ascending order.

Examples include: {1, 2, 3, 5}, {1, 2, 3, 7, 100, 2009}, {1, 2, 4, 6, 10}, {1, 2, 5, 7, 12}, {1, 3, 4, 5, 9}, and {4, 5, 7, 100}.

Clearly, any a1 < a2 < a3 yields a 2-sum'r set of size three. Given any collection of four or more numbers, the 2-sums a + b and c + d are equal if and only if a - c = d - b. Hence, coming up with 2-sum'r sets is equivalent to coming up with sets which avoid equal differences for distinct pairs. (This connects Sidon sets with so-called Golomb Rulers).

Hence, we can be sure that {1, 2, 5, 7} is 2-sum'r---without computing a single sum---since it's obvious that no two differences are the same. Similarly, {1, 2, 4, 7} is 2-sum'r, despite the fact that 7 - 4 = 4 - 1, as is {1, 2, 4, 7, 10, 14}. On the other hand, {1, 2, 3, 6, 11, 13} is not 2-sum'r, since 13 - 11 = 3 - 1 (corresponding to 13 + 1 = 3 + 11).

Suppose {a1, a, ... , ak} is a 2-sum'r set of size k. How can we extend it, by appending ak+1, to yield a 2-sum'r set of size k + 1? What is needed here is that all new 2-sums ak+1 + ai (for i < k + 1) be distinct from any 2-sums that arose earlier. One way to guarantee this is to make sure that the smallest possible new 2-sum exceeds the largest earlier 2-sum. Hence, a sufficient condition is that ak+1 + a1 > ak-1 + ak, or equivalently ak+1 > ak-1 + ak - a1. If a1 = 1, we can take ak+1 = ak-1 + ak.

In this way, {1, 2, 3} can be extended to {1, 2, 3, 5}, and {1, 2, 4} can be extended to {1, 2, 4, 6}. The condition above is sufficient, but by no means necessary, e.g., while {1, 2, 5, 7} can be extended to {1, 2, 5, 7, 12} as above, it can also be extended to the "tighter" {1, 2, 5, 7, 9}.

## Getting Lucky (Again) with a Shuffled Deck

We claim, without proof, that if the card faces in a shuffled deck are scanned,

There a high probability of finding a run of five adjacent cards whose values are 2-sum rich.

For instance, here is such a run:

We have found in practice that a good way to locate such a run is to search for the Aces and see what values are nearby. In the rare cases where this does not yield the desired result, repeat searching for the Twos. The idea is to "accept" the next two lowest values within some five-card run containing an Ace or a Two, and just worry about whether the addition (so to speak) of the two other cards in the run leads to a 2-sum'r set. If you don't have the kind of luck we claim is very likely, settle for a four-card 2-sum'r run instead.

## What a Difference Two Summers Make

To perform the trick as advertized, hand the deck out for shuffling. Take it back, and fan the card faces while commenting on how different combinations of cards have the same sums, and how difficult it would be to determine which cards contributed to a given total. Say, "In a moment I'm going to try to divine all of the cards two of you pick, based on even less information than that."

What you are actually doing while blathering on like this is calmly scanning the faces seeking five adjacent 2-sum'r cards. It's okay to let the audience see some card faces, as long as they are not the ones you're about to zero in on!

Let's assume that the five cards shown have been located. Surreptitiously cut these to the top and do some riffle shuffles which do not disturb them, all the while chatting casually. The key now is to sear into your brain the card total (26 here) as well as a clear image in ascending order as follows:

Some people find it easier to remember the value sequence {1, 3, 5, 8, 9} and the suit order (either as CDHCH or ♣ ) separately.

In general, you've memorized the top (four or) five card values and suits, and the resulting total t. Once those cards are dealt face-down to the table, and further mixed, pay attention to which spectator gets two cards and which gets three. Each now computes a sum, s1 and s2; assume for the sake of argument that the first is larger than the second.

You are told the difference d = s1 - s2, and you already know t = s1 + s2. Addition and division by 2 yields s1; if that corresponds to the sum of two values, you can quickly determine what they are, if it corresponds to the sum of three, just subtract it from t and work with s2 first instead.

For instance, in the case of the five cards above, with total value 26, if the first spectator has two cards and reports that his sum exceeds the other spectator's sum by eight, then since (26 + 8)/2 = 17, you can quickly deduce that he must have an 8 and a 9, and assuming that you haven't forgotten the suits or the other card values, you can now conclude the effect as desired.

## Remove Email? Yikes!

If you're not confident that your brain is up to speed to master the quick search and memory work required above, you could forego handing out the deck for genuine shuffling, and instead just use five fixed 2-sum'r cards. Plant those on top of the deck, and maintain them there through some riffle shuffles. For instance, you could use the following cards, which coincidentally also have total 26.

This basically takes us back to Additional Certainties, the February 2008 Card Colm, with one difference---literally---rather than a sum, conveying multiple card identities.

Colm Mulcahy (colm@spelman.edu) completed his PhD at Cornell in 1985, under Alex F.T.W. Rosenberg. He has been in the department of mathematics at Spelman College since 1988, and writing Card Colms---the only MAA columns to actively encourage lying on a regular basis---bi-monthly since October 2004. For more on mathematical card tricks, including a guide to topics explored in previous Card Colms, see http://www5.spelman.edu/~colm/cards.html.

Colm is delivering the MAA Lecture for Students (on mathematical card magic) at MAA MathFest in Portland, Oregon, at 1pm on Thursday, 6th August.

"Richest 2-Sums" is an anagram for "2-Sum Rich Sets" and "Remove Email? Yikes!" is an anagram of "Memory Like A Sieve?!"