Batch 2 - Class 99 - Graphene Trampolines

Pre-Class Problem:

Attendance: Anisha, Smiti, Tishyaa, Liza, Khushi, Arnav, Anishka, Anshi

Class Notes:
Richard Smalley (1943-2005) was awarded the Nobel Prize in Chemistry for discovering that he could make tiny soccer balls out of carbon atoms. In 2005, he started to shrink, and became so small that he could bounce on tiny trampolines that make up graphene. Richard Smalley spent his time making graphene trampoline puzzles.

Here is one. Smalley can move on the trampoline basis roll of dice. He can not go back to a cell he has already visited.

Lets try another trampoline. We will try a few different starting points, and then color a cell yellow, green or red basis whether starting out from that cell can lead to both success and failure, only success or only failure. Try this one out and color all cells basis success and failure possible there.
          


Solve for a few more. Older students, find the probability of success for the yellow cells.

          

Card Magic!

Note that this trick does not work all the time. It works about 85% of the time.

Trick:  The victim chooses a number (say between 1 and 10), and then you start at the beginning of a shuffled deck (or better, a shuffled pair of decks) and you deal the cards face up, slowly, one at a time. As you deal, the victim silently counts the cards until his number of cards is dealt. In other words, if he chose 5, he will silently count the cards you deal as 1, 2, 3, 4, 5. When his card (the fifth in this example) is reached, he looks at the rank of that card which will be from ace (= 1) to king. If the card is a face card (jack, queen or king), count its rank as 5 (This is not what we did in class). As soon as he sees that card, the rank is his new number, and he begins counting from one again beginning with the next card you turn up. Repeat this process, and each time he reaches his count, that card’s rank is his new number. Finally, as he is counting up the deck will run out, and he will not have gotten to his number. Whatever card he was using as his latest number is his final card. At this point, you tell him what that card is.

Discussion: The reason this works is that as you’re both going through the deck, if you ever happen to hit the same card as the victim, from then on you will both be locked into the same “path” through the rest of the deck and will certainly arrive at the same final card. You can make a mathematical estimate as follows. The “average” advance will be by: (1 + 2 + 3 + · · · + 9 + 10 + 5 + 5 + 5)/13 = 5.38 cards, so you will take, on average, about 52/5.38, or about 8 steps. The density of the victim’s cards is about 1/5.38 = .186, so there’s a 1 − .186 = .814 chance of a miss. But you have to miss all 8 times, and this will occur about (.814)8 = .174 so you’ll win about .836 of the time.


This has mathematical application in cryptology to solve for discrete logarithms. Its called Pollard's Kangaroo method!

Homework
              

References:
     http://mathpickle.com/project/graphene-trampoline-logic-probability/
     http://mathpickle.com/wp-content/uploads/2016/06/Graphene-Trampolene-Nano.pdf
     http://www.geometer.org/mathcircles/CardTricks.pdf
     Mathematical Puzzles, Geoffrey Mott-Smith