# Card Colm (by Colm Mulcahy)

August 2007

Dedicated with great fondness to the memory of friend, physicist and cook extraordinaire
Ron Keith, of Emporia, Kansas, who slipped away while this column was being written.

## Sixy Alpha Omegas

How can one arrange cards in a circle---or numbered beads in a bracelet---so that if any one is picked at random, its value is some constant, say 6? Hint: the circle would be very small if we only used cards from a single deck.

What about a bracelet for which any pair of adjacent beads yields the same total, such as 13? (We trust this question does not leave the reader at 6s and 7s.) A moment's reflection reveals that we must have alternating values, like 3 and 10 (or 6 and 7, etc.). The bracelet can have any (even) number of beads.

What about a bracelet for which any three adjacent beads yields the same total? It turns out that we must have beads whose values repeat in cycles of length three, and a similar conclusion holds if we want any adjacent k beads to yield the same sum. Packets of cards, repeating in such cycles, retain some structure after a single riffle shuffle; these were considered in the Gilbreath shuffle Card Colm in August 2006 (and a year before that in the special case k = 2).

Let's explore in a different direction. A bracelet in which single beads alternate between two values is, well, just that. We are back to examples like [5 8 5 8 5 8 .... 5 8] (even length). Are there bracelets for which the sum of adjacent pairs of beads alternates between two values, say n and n+1 for some n? If we try with n = 10 and starting beads 5 and 6, we get the following runaway example which never closes back on itself:

[5 6 4 7 3 8 2 9 1 10 0 11 -1 12 -2 13 .... ]

Alas, the same thing happens if we seek alternating sums n and n+k for some k > 1. While these do have magical potential, they are not pursued here.

How about a bracelet for which any three adjacent beads yields a total which is always n or n+k for some n, k? That can be arranged! Once again, we'll restrict attention to the case k = 1. Here are examples with n = 7 and 14, having triple sums of 7, 8 and 14, 15, respectively:

 3 2 2 1 1 4
 8 2 5 4 1 9

It's easy to show that bracelets which alternately force triple sums of n or n+1 must basically have 6 beads (or be repeats of such bracelets, e.g., with 12 or 18 beads). Upon reflection (or rotation, or both), they must be of the form:

 a b n+1-a-b n-a-b b-1 a+1

for some a and b. Since any a, b work, there is essentially a two parameter family of possibilities. A few of these bracelets consist of two beads repeated three times, e.g., in the case of the smallest positive example, which is [1 2 1 2 1 2] (with n = 4). To simplify things below, we mostly restrict attention to positive beads. E.g., if n = 10, there are examples such as [1 5 4 2 4 5] and [2 4 4 3 3 5]; note that in both cases, several beads are repeated.

If we use cards, whose values range from 1 (Ace) to 13 (King), there are only finitely many such bracelets, and the same holds if we follow the convention, established in Card Colm in February 2007, that black cards have the usual values and red ones have corresponding negative values (Jokers can be used to represent zero). These "negative connotations" are worth bearing in mind in all that follows, although we leave it to the reader to generate examples of these.

## The Sixiest One

Are there examples with distinct positive beads? If so, then the smallest possible value for n would be 10, since 1 + 2 + 3 + 4 + 5 + 6 = 21, and it's easy to see that the alternating triple sums of any such bracelet must be 10 and 11. This leads us to the most perfectly pleasing example:

 1 6 4 3 5 2

This is a remarkable circle, for the following six reasons:
• It uses each of six consecutive numbers.

• These numbers trace out a regular geometric pattern when taken in numerical order.

• Triple sums centered on even numbers are constant (10 in this case).

• Triple sums centered on odd numbers are constant (11 in this case).

• The numbers depicted have the αΩ (alpha Omega) property: read in the shape of a capital Omega, they are in alphabetical order.

• It's possible to count out and discard the cards so that they are revealed in numerical order.
Assume that the cards are arranged in a packet in alphabetical order, 5, 4, 1, 6, 3, 2, with the 5 on top. Move 1 card from top to bottom for each letter in the words "count it" thus bringing the 4 to the top. Now move 1 card from top to bottom and set the next card aside (it's the Ace). Next, move 2 cards from top to bottom and set the next card aside (it's the 2). Third, move 3 cards from top to bottom and set the next card aside (it's the 3). Keep going until only one remains; the cards come off the packet in order 1, 2, 3, 4, 5, 6!

This curiosity is language independent. For a related spelling effect in English, as well as additional comments on the order above, see "One of Six and Another of Half a Dozen" in the May 2007 issue of Word Ways, The Journal of Recreational Linguistics. Versions in over 40 languages, from Albanian to Welsh, have been explored too; the Icelandic results are particularly nice!

## C++

Another example that evokes positive feelings is [8 5 4 7 6 3], which can be obtained from the perfect [5 4 1 6 3 2] circle by simply adding 2 to each bead and rotating a little:

 4 7 5 6 8 3

This is also quite remarkable, for six similar reasons:
• It too uses each of six consecutive numbers.

• These six numbers trace out the same kind of regular geometric pattern as before.

• Triple sums centered on even numbers are constant (16 in this case).

• Triple sums centered on odd numbers are constant (17 in this case).

• Amazingly, the numbers depicted also have the αΩ (alpha Omega) property: read in the shape of a capitol Omega, they are in alphabetical order.

• A modified counting out and discarding reveals the cards in reverse numerical order.
The last item above involves some non-mathematical hokey pokey, but may be worth considering.

Start with the cards in alphabetical order 8, 5, 4, 7, 6, 3, with the 8 on top. This can be achieved by starting with the cards fanned in numerical order from left to right, moving the 6 and 7 so that they go between the 3 and 4, and then counting out the cards face down, to reverse their order. Move 1 card from top to bottom for each letter in the words "count it out" thus bringing the 6 to the top. Now move 8 cards from top to bottom and set the next card aside (it's the 8). Next, move 7 cards from top to bottom and set the next card aside (it's the 7). Here comes a slick move: separate the remaining cards with your hands to reveal two pairs, wave them around a bit without showing their faces, and rejoin them in reverse order, so it's now 5, 4, 6, 3 from the top. Move 6 cards from top to bottom and set the next card aside (it's the 6). There's another slick move: take the top two cards in one hand and the remaining one in the other hand, wave them around and rejoin them in reverse order. You may now count out the 5, 4 and 3 without any further ado.

## Putting Our Bracelets On the Table

 n, n+1 # of distinct bracelets bracelet representatives 4, 5 1 (0) [1 2 1 2 1 2] (period 2) 5, 6 1 (0) [1 2 2 2 1 3] 6, 7 2 (0) [1 2 3 2 1 4], [1 3 2 2 2 3] 7, 8 3 (0) [1 2 4 2 1 5], [1 3 3 2 2 4], [2 3 2 3 2 3] (period 2) 8, 9 4 (0) [1 2 5 2 1 6], [1 3 4 2 2 5], [1 4 3 2 3 4], [2 3 3 3 2 4] 9, 10 5 (0) [1 2 6 2 1 7], [1 3 5 2 2 6], [1 4 4 2 3 5], [2 3 4 3 2 5], [2 4 3 3 3 4] 10, 11 7 (1) [1 2 7 2 1 8], [1 3 6 2 2 7], [1 4 5 2 3 6], [1 5 4 2 4 5], [2 3 5 3 2 6], [2 4 4 3 3 5], [3 4 3 4 3 4] (period 2) 11, 12 8 (1) [1 2 8 2 1 9], [1 3 7 2 2 8], [1 4 6 2 3 7], [1 5 5 2 4 6], [2 3 6 3 2 7], [2 4 5 3 3 6], [2 5 4 3 4 5], [3 4 4 4 3 5], 12, 13 10 (2) [1 2 9 2 1 10], [1 3 8 2 2 9], [1 4 7 2 3 8], [1 5 6 2 4 7], [1 6 5 2 5 6], [2 3 7 3 2 8], [2 4 6 3 3 7], [2 5 5 3 4 6], [3 4 5 4 3 6], [3 5 4 4 4 5] 13, 14 12 (3) [1 2 10 2 1 11], [1 3 9 2 2 10], [1 4 8 2 3 9], [1 5 7 2 4 8], [1 6 6 2 5 7], [2 3 8 3 2 9], [2 4 7 3 3 8], [2 5 6 3 4 7], [2 6 5 3 5 6], [3 4 6 4 3 7], [3 5 5 4 4 6], [4 5 4 5 4 5] (period 2) 14, 15 14 (4) [1 2 11 2 1 12], [1 3 10 2 2 11], [1 4 9 2 3 10], [1 5 8 2 4 9], [1 6 7 2 5 8], [1 7 6 2 6 7], [2 3 9 3 2 10], [2 4 8 3 3 9], [2 5 7 3 4 8], [2 6 6 3 5 7], [3 4 7 4 3 8], [3 5 6 4 4 7], [3 6 5 4 5 6], [4 5 5 5 4 6] 15, 16 16 (5) [1 2 12 2 1 13], [1 3 11 2 2 12], [1 4 10 2 3 11], [1 5 9 2 4 10], [1 6 8 2 5 9], [1 7 7 2 6 8], [2 3 10 3 2 11], [2 4 9 3 3 10], [2 5 8 3 4 9], [2 6 7 3 5 8], [2 7 6 3 6 7], [3 4 8 4 3 9], [3 5 7 4 4 8], [3 6 6 4 5 7], [4 5 6 5 4 7], [4 6 5 5 5 6] 16, 17 19 (7) [1 2 13 2 1 14], [1 3 12 2 2 13], [1 4 11 2 3 12], [1 5 10 2 4 11], [1 6 9 2 5 10], [1 7 8 2 6 9], [1 8 7 2 7 8], [2 3 11 3 2 12], [2 4 10 3 3 11],[2 5 9 3 4 10], [2 6 8 3 5 9], [2 7 7 3 6 8], [3 4 9 4 3 10], [3 5 8 4 4 9], [3 6 7 4 5 8], [3 7 6 4 6 7], [4 5 7 5 4 8], [4 6 6 5 5 7], [5 6 5 6 5 6] (period 2) 17, 18 21 (8) [1 2 14 2 1 15], [1 3 13 2 2 14], [1 4 12 2 3 13], [1 5 11 2 4 12], [1 6 10 2 5 11], [1 7 9 2 6 10], [1 8 8 2 7 9], [2 3 12 3 2 13], [2 4 11 3 3 12], [2 5 10 3 4 11], [2 6 9 3 5 10], [2 7 8 3 6 9], [2 8 7 3 7 8], [3 4 10 4 3 11], [3 5 9 4 4 10], [3 6 8 4 5 9], [3 7 7 4 6 8], [4 5 8 5 4 9], [4 6 7 5 5 8], [4 7 6 5 6 7], [5 6 6 6 5 7]
(The 12,13 and 16,17 rows were corrected on 25 June 2009)

This table tracks the n, n+1 bracelet representatives with positive beads, up as far as n = 17 (by which stage numbers like 14 and 15, which are harder to accommodate with a pack of cards, start appearing). The non-repeating ones are listed in italics, with the number of these given parenthetically after the counts in the second column.

Buried above are the αΩ (alpha Omega) examples seen earlier, namely, [5 4 1 6 3 2] = [1 6 3 2 5 4] and [8 5 4 7 6 3] = [3 6 7 4 5 8]. All of the representative bracelets are given in "numerical order"---if we imagine their beads concatenated---then the sums obtained by adding the first three beads in each bracelet as listed are always n (as opposed to n + 1).

The occasional period 2 examples are noted. It can be seen that the perfectly sixy example encountered above is unique: there is basically only one way to arrange 1, 2, ...., 6 in a circle to get alternating sums of 10 and 11. Moreover, given any collection of six distinct positive numbers occurring in any of the examples tabulated, there is a unique way to arrange them to get the desired triple sums.

Further generalizations suggest themselves, and some of these will appear in the October Card Colm.

It's time to take a look at selected card applications of all of the above.

## Magic Charms

We now present ideas for exploiting the kind of bracelets seen above. Some of these need to be fleshed out to turn them into engaging performances. We refer readers to last February, April, and June for details on a range of numerical predictions and forces. While we've used the term "beads" up to now for the numbers that make up our bracelets, let's switch to the more natural "charms"---which seems entirely fitting in the context of magic.

A standard magician's way of exploiting the abilty to force, say, either 12 or 13 on a spectator, is to have a known card thirteen deep in a deck with its name written ahead of time on a prediction slip. If 13 is forced, have the thirteenth card down turned over, whereas if 12 is forced, you ask that this number of cards be removed, pause to review how fair and random the method of number selection was---while trying to keep a straight face---and request that "the next card" be turned over. Either way, you come out looking very magical indeed.

We suggest placing the predicted card at position nineteen (from the top), with the first six cards having values [1 3 8 2 2 9] or [1 4 7 2 3 8], or any of the other seven possibilities listed in the table above for forcing 12 or 13. Now, casually overhand shuffle making sure that only the bottom half of the deck is in fact disturbed. Deal out the top six cards into a circle, and then shuffle the rest of the cards while retaining the top stock. Have a spectator point to any of the six cards on display, and have it and its immediate neighbours on each side picked up. Remove the three remaining cards and bury them one by one in the bottom of the stack of retained cards. Have the selected cards summed, and proceed as above to thunderous applause.

Suppose you wish to force 14 or 15 (and hence, effectively, 15, as seen above!). Arrange twenty-four cards to consist of these four bracelet values stacked on top of one another in any order, using random suits: [1 2 11 2 1 12], [2 5 7 3 4 8], [2 6 6 3 5 7] and [3 4 7 4 3 8]. Aces, Jacks and Queens are used for 1, 11 and 12, respectively. We have not used any value more than four times, so this is indeed possible using a single deck of cards. Deal into four circles of six cards each, face-up, and have one circle selected by a spectator. Any card and its two neighbours are selected. Jumble all of the cards not being used before proceeding, so that no evidence is left behind. The remaining three cards are summed, and the rest is automatic.

Instead of dealing cards into circles, packets of six could be presented, with the spectator asked to choose any one---perhaps after you point out how different they all are. The chosen packet could then be cut as often as is desired. At the conclusion of this, have the top three cards and bottom three cards separated and ask that one of these groups be discarded.

It is possible to arrange 1, 2, 3, ...., 7, 8 in a circle in alphabetical order, and force a spectator to eliminate either the 1 and 2 or the 7 and 8. This leaves one of the two αΩ (alpha Omega) bracelets studied above, hence either 10 or 11 or 16 or 17 can be forced. If a known card is in position 11 from the top of a packet of size 27, all is well. If 10 or 11 is determined, proceed in the usual way. Otherwise, have the cards examined with the packet flipped over, so that the faces are visible. Counting to the 17th card from the face, or removing 16 cards, leads to the desired card.

Here's one way to get a spectator to make the right elimination choices, starting with 8, 5, 4, 1, 7, 6, 3, 2 in a clockwise circle (possibly face down), bearing in mind that 1 is across from 2, and 7 is across from 8. Explain that the spectator will eliminate two cards: one picked at random and the other the one across from that. Ask the spectator to point to (not pick) one card. There is a 50% chance it will belong to one of the pairs of interest. If so, say, "Okay, we'll eliminate this card and the one across from it." If the 4 is selected indead, say, "Count four clockwise from that number, the next card is the 2, which we now eliminate along with the one opposite 1." If the 8, 5 or 6 is selected, say "Count that number anti-clockwise, starting there, and eliminate the next card and the one opposite it." All cases are now covered. Tighten up the circle by closing the two gaps: it should look like a sixy αΩ.

Colm Mulcahy ([email protected]) has been in the department of mathematics at Spelman College since 1988. He has never, as a rule, worn bracelets, and he doesn't intend starting now. He hopes to cure himself of his recent addiction to galactic embers (and acetic gamblers) by the end of of 2007, but suspects that this will not happen before November. For more on mathematical card tricks, including a guide to topics explored in previous Card Colms, see http://www5.spelman.edu/~colm/cards.html.

Past Columns