Categories > Card Magic > Fixing Elmsley's "The Obedient Faro"

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 Posted 1598483701 Reply with quote  #1 Someone posted on the Magic Café re: an item in The Collected Works of Alex Elmsley Volume II. It’s “The Obedient Faro” on p. 346. The person was asking about the math on p. 348 that is supposed to tell you how to determine the starting position of the target card and then which sequence of OO, OI, IO or II faros to perform.Here O = OUT Faro; I = IN Faro. So IO means an IN followed by an OUT.The description on p. 348 is incorrect. I showed the poster how to fix things and then thought that faro shufflers on TMF might be interested. The premise is this: How to you get a card to any position in the deck using 2 faro shuffles?The answer involves two things – 1. The starting point in the deck of the target card and 2. The sequence of In and Out Faros needed.The good news is that you’ll always do 2 faro shuffles based on binary math. In binary (base 2) 0 = 00; 1 = 01; 2 = 10 and 3 = 11. It’s because the 2nd “place value” in binary is 2 instead of 10. 45 means 4 tens and 5 ones. In binary 10 means 1 two and 0 ones. 11 means 1 two and 1 one which equals 3.Here’s what to do:1. Decide where you want the target card to end up. Call that N.2. Determine how many cards will be ABOVE the target card in that position. That’s N-1. So if the target position is 36, there will be 35 cards above the target when it’s in position.Now the math:3. Calculate N-1 divided by 4. There will be a quotient and a remainder. For example if N-1 is 35 as in the example above, division by 4 would give you 8 with remainder 3. Let’s call 8 Q for quotient and 3 R for remainder. These two values tell you exactly what to do.4. Position the target card at Q + 1 and then use R in binary form to determine In and Out Faros.Example: Position desired is 36 (N) so we divide N-1 (35) by 4 getting Q = 8 and R = 3. Place the target card, say ace of spades, at Q+1 or 9th from the top and then perform the binary for 3 which is II. That’s two In faros.Example 2: Position desired is 23. Divide 23 -1 (22) by 4 getting Q = 5 and R = 2. So position AS at Q+1 =6 from the top and perform binary 2 (IO) an IN followed by an OUT. That will put the AS at 23.Remember: R = 0 do OO; R = 1 do OI; R = 2 do IO; R = 3 do IINot super practical but interesting.Mike 0 1 1

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 Posted 1598496464 Reply with quote  #2 So is it just a beautiful coincidence of language that I looks like 1 and O looks like 0? This would still work if binary 2 was @#, right? We just happen to have letters that resemble the digits, which is quite beautiful really, even symmetrical. __________________When you come to a fork in the road, take it! 0 0 0

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 Posted 1598498892 Reply with quote  #3 Very nice description Mike.  I just tried it, and it works!It will be fun to practice this and see what routine can be created.  i.e. selected card, and how, when, why to get it to a specific starting spot after knowing a target location.  Also, how, what determines the target location.I don't have "The Collected Works of Alex Elmsley Volume II", so I am not familiar with that routine, but I follow the math you laid out here.  I like it!Tom __________________"How can I help you do your job better?"-T. Kracker 0 0 0

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 Posted 1598535150 Reply with quote  #4 I from a mem deck you can get the card into position using a technique like that used by Jason Ladanye and Darwin O. After you do the two faros, I think that dealing the cards into four FU piles will allow you to get back to stack. There will be runs of cards in stack order that have to be extracted. There might be an effect that makes dealing the FU piles logical. I don't think I'd blow the stack just to position a named card at a named number with two faros, though.M 0 0 0

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 Posted 1598565918 Reply with quote  #5 Thanks, Mike. I don't remember understanding this very well when I encountered it in the Elmsley book, but it made sense as you explained it. 0 0 0

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 Posted 1598603766 Reply with quote  #6 Quote:Originally Posted by TheAmazingStanley So is it just a beautiful coincidence of language that I looks like 1 and O looks like 0? This would still work if binary 2 was @#, right? We just happen to have letters that resemble the digits, which is quite beautiful really, even symmetrical. Certainly, and also Elmsley's experience as a computer programmer.See the quote below from the section "Mathematics of the weave shuffle" page 311."While investigating the effect of combined in- and out-shuffles, I fell into the practice of abbreviating them as 'I' and 'O'. This led me to the discovery of a fortunate coincidence, for I noticed that my sequences of 'I's and 'O's could be read as 1s and 0s; and these could be manipulated with binary arithmetic to yield useful instructions for shuffle sequences." __________________So much to do, so little time !! 0 0 0
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