Dealing with some maths Hayden Tronnolone School of Mathematical Sciences University of Adelaide August 20th, 2012
To call a spade a spade First, some dealing... Hayden Tronnolone (University of Adelaide) Dealing with some maths August 20th, 2012 2 / 17
To call a spade a spade First, some dealing... This trick is called The Royal Hummer. It uses a technique called a Hummer shuffle. Hayden Tronnolone (University of Adelaide) Dealing with some maths August 20th, 2012 2 / 17
I heart Hummer shuffles Definition (Hummer shuffle) Given a deck of cards, a Hummer shuffle is performed by 1 flipping the top two cards, and then 2 cutting the deck at any point. Introduced by Bob Hummer (1905 1981). To understand how this is used, we first consider a simple trick that uses the Hummer shuffle. Hayden Tronnolone (University of Adelaide) Dealing with some maths August 20th, 2012 3 / 17
Ten-Card Hummer 1 Take a deck of ten cards, all face-down. 2 Perform any number of Hummer shuffles. 3 Reverse the cards in the even positions. This will leave exactly five cards face-up. This was introduced by Hummer in his manuscript Face-up/Face-down Mysteries (1942). To understand why this works, we need to consider what properties of the deck are preserved by a Hummer shuffle and the limits on the achievable arrangements of cards. Hayden Tronnolone (University of Adelaide) Dealing with some maths August 20th, 2012 4 / 17
Card-inality Consider a deck of 2n cards (we won t discuss odd numbers). How many ways can we arrange these cards (order and orientation)? The 2n cards can be arranged in 2n! ways. Each card can be face-up or face-down, so for each ordering there are 2 2n ways to orientate the cards. This gives possible arrangements. 2 2n 2n! For the ten-card deck n = 5, so there are 3, 715, 891, 200 possible arrangements. For 52 cards there are 2.7 10 34. Hayden Tronnolone (University of Adelaide) Dealing with some maths August 20th, 2012 5 / 17
Theorems are ace But, when using Hummer shuffles, there is a limitation: Theorem (Even Odd) Take a deck of 2n cards, all face-down. After any number of Hummer shuffles the number of face-up cards at even positions equals the number of face-up cards at odd positions. The Ten-Card Hummer works because of this theorem. Suppose there are a cards face-up at even positions, so there are 10 a face-down. This must also hold for the odd positions. Flipping every even card leaves 10 a face-up. In total there are (10 a) + a = 10 cards face up. We can prove this theorem by induction. Hayden Tronnolone (University of Adelaide) Dealing with some maths August 20th, 2012 6 / 17
So are proofs To begin with this is trivially true. Suppose we have performed some number of Hummer shuffles and the result is still true. Flipping the top two cards preserves the equality. Why? Hayden Tronnolone (University of Adelaide) Dealing with some maths August 20th, 2012 7 / 17
Flipping two cards Hayden Tronnolone (University of Adelaide) Dealing with some maths August 20th, 2012 8 / 17
Cutting Cutting one card just swaps the odd and even cards. Cutting an even number doesn t change the odd and even cards. Cutting an odd number is like cutting one card and then cutting an even number of cards. So, the theorem is proved....... Hayden Tronnolone (University of Adelaide) Dealing with some maths August 20th, 2012 9 / 17
Underwhelmed? Join the club Under the assumptions of the even odd theorem, the orientation of 2n 1 cards decides the orientation of the last. So, there are at most 2 2n 1 2! possible arrangements. Not much better! Theorem (Hummer constraint) Take a deck of 2n cards, all face-down, and apply any number of Hummer shuffles. Suppose card j is at position k. Define p(j) to be 1 if j is face-up and 0 otherwise. Then s(k) = j + k + p(j) (mod 2) is the same at every position k. Hayden Tronnolone (University of Adelaide) Dealing with some maths August 20th, 2012 10 / 17
A quick proof is a good proof Proof. Initially, card j is at position j and is face-down, so at every position s(k) = 2j 0 (mod 2). Suppose it is true after some number of Hummer shuffles. Flipping the top two cards increases the score associated with the first card by 1, but either adds or removes 1 due to the flip, so the parity of s(2) is unchanged. The same holds for s(1). Cutting an even number of cards doesn t change the parity of s(k). Cutting one card changes the parity of every s(k). Cutting an odd number is equivalent to cutting one card and then cutting an even number of cards. Hayden Tronnolone (University of Adelaide) Dealing with some maths August 20th, 2012 11 / 17
The diamond-standard bound It is possible to use Hummer shuffles to put the 2n cards in any order. However, the Hummer constraint theorem means that for any given order, there are only two possible sets of orientations. So, there are only 2 2n! arrangements. Hayden Tronnolone (University of Adelaide) Dealing with some maths August 20th, 2012 12 / 17
On the flip side What if we instead start with the cards in order but with alternating orientations? (Exercise: can this arrangement be reached using Hummer shuffles on a face-down deck?) Theorem (Alternating deck) Consider a deck of cards that initially has alternating orientations. The alternating pattern is preserved by Hummer shuffles. Hayden Tronnolone (University of Adelaide) Dealing with some maths August 20th, 2012 13 / 17
Another proof?! (Part 1) Flipping the top two cards. Hayden Tronnolone (University of Adelaide) Dealing with some maths August 20th, 2012 14 / 17
Another proof?! (Part 2) Cutting one card (even case also holds)....... Hayden Tronnolone (University of Adelaide) Dealing with some maths August 20th, 2012 15 / 17
Finally... We are now in a position to explain the Royal Hummer. Hayden Tronnolone (University of Adelaide) Dealing with some maths August 20th, 2012 16 / 17
Finally... We are now in a position to explain the Royal Hummer. I m not going to. This is an exercise. Along with: Hayden Tronnolone (University of Adelaide) Dealing with some maths August 20th, 2012 16 / 17
Finally... We are now in a position to explain the Royal Hummer. I m not going to. This is an exercise. Along with: 1 Why don t these tricks work with an odd number of cards? 2 What about flipping more than two cards? Hayden Tronnolone (University of Adelaide) Dealing with some maths August 20th, 2012 16 / 17
Reference Hayden Tronnolone (University of Adelaide) Dealing with some maths August 20th, 2012 17 / 17