

glowball Special user Nashville TN 653 Posts 
Easier Harding stack deck and method.
See further below for Method D. I think it's pretty good. But first I want to talk just a little bit about my three failures IE Method A, Method B, Method C which were all disasters (either the deck randomness was totally unacceptable or the formulas would not work). I liked the randomness look of the Harding physical stack but wanted to come up with an easier way to determine the theoretical position in the imaginary new deck order stack (did not like subtracting 13 or 26 or 39). Method A: I tried to get away from the four groups of 13 in the imaginary stack and use four groups of 10 (each group of 10 would be for one suit and I started at position 10 and ended on position 49) and filled in the other positions with the face cards. I spent a lot of time on this. This made the calculations for the number cards very easy but the deck looked horrible because the formula threw all the like value cards together. So I abandoned method A. Method B: to try and get the same value cards away from each other I added the pip number of points to my formula (and still trying to salvage the four groups of 10 concept). But this formula caused the physical deck to have card sequences 9x, 8x, 7x, 6x, 5x, 4x, 3x, 2x. This broke the suit sequence up (a good thing) but the sequential value strings of cards would make the spectators notice. So I abandoned method B. Method C: use the value of the card (1 thru 13) times it's pip points/lobes thus say The King of Clubs would be 13 times 3 equals 39, the 8 of clubs would be 8 times 3 equals 24. But I started seeing all kinds of overlaps where two different cards would calculate to the same position. So I abandoned method C. Method D: the very good method (I think). I'll bet dollars to donuts somebody else has already done this. I wonder how well it would work if the imaginary deck was composed of four aces, followed by the four twos, followed by the four threes etc? Using shcd let's say the eight of hearts was named then the position would be: 7 times constant 4 = 28 plus the two lobes on the heart equals 30 then flip to become position 03. This may work and give a random looking deck similar to the normal Harding deck. Note that calculations using this method D would be far easier than the standard Harding method and deck. Below is Method D imaginary deck also showing it's physical position after its name: 01 AS 10 02 AH 20 03 AC 30 04 AD 40 05 2S 50 06 2H 6 fixed rule breaker 07 2C 7 fixed rule breaker 08 2D 80 5+5 is 35 09 3S 90 5+5 is 45 10 3H 01 11 3C 11 12 3D 21 13 4S 31 14 4H 41 15 4C 51 16 4D 61 5+5 is 16 17 5S 71 5+5 is 26 18 5H 81 5+5 is 36 19 5C 91 5+5 is 46 20 5D 02 21 6S 12 22 6H 22 23 6C 32 24 6D 42 25 7S 52 26 7H 62 5+5 is 17 27 7C 72 5+5 is 27 28 7D 82 5+5 is 37 29 8S 92 5+5 is 47 30 8H 03 31 8C 13 32 8D 23 33 9S 33 34 9H 43 35 9C 53 5+5 is 08 36 9D 63 5+5 is 18 37 10S 73 5+5 is 28 38 10H 83 5+5 is 38 39 10C 93 5+5 is 48 40 10D 04 41 JS 14 42 JH 24 43 JC 34 44 JD 44 45 QS 54 5+5 is 09 46 QH 64 5+5 is 19 47 QC 74 5+5 is 29 48 QD 84 5+5 is 39 49 KS 94 5+5 is 49 50 KH 05 51 KC 15 52 KD 25 end of harding imaginary Method D ndo Method D Harding physical stack: 3H, 5D, 8H, 10D, KH, 2H, 2C, 9C, QS, AS, 3C, 6S, 8C, JS, KC, 4D, 7H, 9D, QH, 5D, 3D, 6H, 8D, JH, KD, 5S, 7C, 10S, QC, AC, 4S, 6C, 9S, JC, 2D, 5H, 7D, 10H, QD, AD, 4H, 6D, 9H, JD, 3S, 5C, 8S, 10C, KS, 2S, 5C, 7S End of method D physical Harding stack. I think this one is very good and super simple, well, going from card name to number is easy. Card named. Formula to get position: you always multiply 4 times 1 less than the target value and then add the pip point/lobe count and then flip the two digits and then if greater than 52 do the 5+5. Card position named. Formula to get card name: I'll do a quick example on one number. Let's say 37: flip it to become 73 (because it's greater than 52 we do the 5+5 making it 28). Then divide by 4 giving quotient of 7 and remainder of 0. Note that a remainder of 1 means spades, remainder of 2 means hearts, remainder of 3 means clubs, remainder of 4 (zero treated as 4) means diamonds (using the remainder to determine the suit via pip points/lobes ie: shcd). Therefore the target card for 37 is the 7D. We got the seven from the quotient and we got the suit from the remainder. A more concise way of saying the above is: 1. flip the number 2. if greater than 52 then 5+5 3. divide by 4 4. quotient for value, remainder for suit Note that for flipping purposes: stated numbers lower than 10 are mentally treated as having a single zero in front of them IE the number 8 is 08. Then when it gets flipped it becomes 80. 
Nikodemus Special user 830 Posts 
Hi glowball,
I am pretty sure I understand the problem you are trying to solve, but just to be sure... The Harding system is a very simple rule to convert one card location to another. Therefore we can use it to generate a randomlooking stack from an ordered sequence. The ordered arrangement we start with is assumed to be AK for each of the four suits. [This is new deck order for many brands, but NOT Bicycle. But this doesn't matter  it's just a theoretical starting point.] The problem is  do we actually know the positions of all the cards in that simple ordering? Personally (like you I presume) I don't. It's easy for the first 13 cards; but for the rest I would need to subtract 13, 26 or 39 as you mentioned. Your solution is to imagine the cards start in a different sequence  all the Aces, then the 2's, etc. You then provide a formula to calculate the position of a named card in this order, followed by the Harding transformation to give its position in the "random" stack. You presumably believe this is an improvement because you think your formula is easier than subtracting 13/26/39. But my honest opinion is that your formula is not really an improvement. It is just a different flavour of the same basic approach. You are still using TWO calculations to derive the location of a card  (1) the formula to give its position in the "imaginary" ordered sequence. (2) the Harding transformation. To me this will always be too slow. It makes much more sense just to properly memorise a stack. Alternatively use a system that allows you to calculate the "random" location more directly. (Doug Dyment has done a LOT of work on this subject). My opinion is the Harding system is a great formula IF (and only if) you already know the first sequence perfectly. Then it is a simple calculation to create a different derived sequence. So if you already know the AK sequence perfectly (or can calculate it instantly) it is great. Otherwise, as I stated, you just end up with a twostage calculation. Interestingly this means the Harding system could be used with a fully memorised stack, to easily generate an alternative stack, if needed. 
glowball Special user Nashville TN 653 Posts 
Nickodemus, good points.
Here is where I'm coming from: Wanted: simple algorithmic stack that a few of my magician friends can do acaan with one fellow magician as a shill when I am not present (I normally let them use my Aronson deck with me as a shill and they get to be the star doing the trick for a waiter or waitress). The randomness of the deck is not super important thus Si Stebbins is okay. Don't need any builtin features. I strictly want a simple algorithmic stack (my magician friends won't memorize 52 cards). The ease of learning and using the algorithm outweighs just about everything else. I'm 79 years old and experienced (20 years) Aronson stack acaan user and am looking for the easiest algorithmic stack (that has passible randomness) for my magician friends to use when I'm gone. I'm also having fun doing the research and coming up with my own ideas that I want to share. Hope this explains why I'm focusing on algorithmic acaan stacks. Thanks much for your thoughts. glowball. PS: later, when I have time I'll discuss the issue of ease of calculating standard Harding vs Method D. 
hcs Elite user Germany, Magdeburg 477 Posts 
Quote: You win!
On Apr 8, 2022, glowball wrote: ... Please pass me an email. HCS 
landmark Inner circle within a triangle 5153 Posts 
Glowball, I don't understand the 5+5 adjustment. What does that consist of? I don't see it in your post.
Click here to get Gerald Deutsch's Perverse Magic: The First Sixteen Years
All proceeds to Open Heart Magic charity. 
Claudio Inner circle Europe 1757 Posts 
Quote:
I don't understand the 5+5 adjustment. What does that consist of? I don't see it in your post. 08 2D 80 5+5 is 35 because 08 > 80 > 52, therefore 08  5 = 03; flip and add: 30 + 5 = 35; 09 3S 90 5+5 is 45 because 09 > 90 > 52, therefore 09  5 = 04; flip and add: 40 + 5 = 45. 
landmark Inner circle within a triangle 5153 Posts 
Got it. Thanks.
Click here to get Gerald Deutsch's Perverse Magic: The First Sixteen Years
All proceeds to Open Heart Magic charity. 
glowball Special user Nashville TN 653 Posts 
Oops.
In my first big post above about Method D the calculation to go from card to number is correct, however the calculation to go from number to card is missing a step. Immediately after the divide by 4 step there should be a step inserted to add one to the quotient if the remainder is other than zero (meaning do this only if spades, hearts, clubs). The below is the correct algorithm to go from number to card: Spectator says number, magician (or confederate) calculates card: 1. flip the number. 2. if flipped greater than 52 then 5+5. 3. divide by constant 4. 4. If remainder NOT zero then add 1 to quotient. 5. use quotient for value, remainder for suit (use SHoCkeD but zero remainder (instead of 4) means diamonds). 
Nikodemus Special user 830 Posts 
Hi glowball,
Slightly tangential to your original question  When you are involved, you don't necessarily need to be the stooge. You could let your friends be the stooge if you taught them a secret signalling system. 
glowball Special user Nashville TN 653 Posts 
Method D Harding physical stack:
3H, 5D, 8H, 10D, KH, 2H, 2C, 9C, QS, AS, 3C, 6S, 8C, JS, KC, 4D, 7H, 9D, QH, 5D, 3D, 6H, 8D, JH, KD, 5S, 7C, 10S, QC, AC, 4S, 6C, 9S, JC, 2D, 5H, 7D, 10H, QD, AD, 4H, 6D, 9H, JD, 3S, 5C, 8S, 10C, KS, 2S, 5C, 7S I am responding to Nicodemus post that my Method D is perhaps no better than the standard Harding. The standard Harding requires the mental imagery of the spread of suits and using that mental imagery calculating the theoretical position of a named card (or named number). then the flipping and the 5 + 5 is applied if necessary. My Method D eliminates the mental imagery and eliminates the figuring where the target card is amongst the imaginary spread. The Method D formula takes the magician straight to the wanted result. Examples of the mental process of standard Harding versus method D (using CHaSeD for the Harding but use SHoCkeD for method D); Example A: spectator names 9H: Standard Harding: 1. imagine a spread of the deck and notice that hearts is in second group (CHaSeD suits). 2. add 13 (because hearts) to the 9 giving 22. Now ready for the flip and possible 5+5. Method D: 1. multiply constant 4 times 8 (8 is one less than target 9) giving 32. 2. add the two lobes on the heart 32 +2 = 34 Now ready for the flip and possible 5+5. Notice that the "multiply by the constant 4" eliminates the necessity of an imaginary spread and eliminates the need to figure whether to use 13, 26, or 39 and eliminates the addition of this number to the target number. This "multiply by 4" takes the magician directly near the wanted number so just have to add the lobes and then flip. Both methods have to deal with the 5+5 issue so that's a wash when comparing the two methods. I contend that Method D is more direct and quicker than standard Harding when going from card name to card position. I might agree with Nicodemus (that Method D is no better than standard Harding) when it comes to going from position to card (spectator names a number from 1 to 52). Mod Stack by evanthx published here on the Magic CafÃ© in 2006. https://themagiccafe.com/forums/viewtopi......&start=0 My Method D stack is very similar to the Mod Stack (older than my Method D) but the Mod Stack starts the aces in a different position and treats diamonds as zero whereas Method D treats diamonds as lobe count 4 allowing a direct addition of the lobe count (Spades point=1, Hearts=2, Clubs=3, Diamonds=4). The Mod Stack also addresses next card calculation whereas my method D stack is strictly focused on any card and any number. Alan Shaxon's version (also older? than Method D and Mod Stack) maybe similar and there are others (see references by ddyment in the above link). Method D is nothing new. 
glowball Special user Nashville TN 653 Posts 
Oops, had 2 typos on the stack itself. On the physical deck position 20 I erroneously had a second 5D, this is wrong It should be AH in position 20. Also position 51 should be 4C not 5C.
Below is the corrected Method D physical stack: 3H, 5D, 8H, 10D, KH, 2H, 2C, 9C, QS, AS, 3C, 6S, 8C, JS, KC, 4D, 7H, 9D, QH, AH, 3D, 6H, 8D, JH, KD, 5S, 7C, 10S, QC, AC, 4S, 6C, 9S, JC, 2D, 5H, 7D, 10H, QD, AD, 4H, 6D, 9H, JD, 3S, 5C, 8S, 10C, KS, 2S, 4C, 7S End of method D physical Harding stack. 
glowball Special user Nashville TN 653 Posts 
After physically putting this Method D deck together I see that it is a horrible looking deck because of very long runs of red cards and then very long runs of black cards. This was very hard to see on paper because the red and black do not stand out on paper/computer screen (it's all black "ink").
If the hearts and spades positions are switched this should make a pretty good looking deck. I will try that and if it looks decent I will post the new physical deck. The formulas will not change except that spades will have a lobe count of two and the hearts will have a lobe count of one. 
Claudio Inner circle Europe 1757 Posts 
Quote:
On Apr 15, 2022, glowball wrote: Hereâ€™s the red ones in bold: 3H, 5D, 8H, 10D, KH, 2H, 2C, 9C, QS, AS, 3C, 6S, 8C, JS, KC, 4D, 7H, 9D, QH, AH, 3D, 6H, 8D, JH, KD, 5S, 7C, 10S, QC, AC, 4S, 6C, 9S, JC, 2D, 5H, 7D, 10H, QD, AD, 4H, 6D, 9H, JD, 3S, 5C, 8S, 10C, KS, 2S, 4C, 7S You're right, it does not look good. 
glowball Special user Nashville TN 653 Posts 
I'm going to create a new thread for method G which will be cleaned up and will have a good mix of red and black cards.
I used method D as the base for method G but made the following changes / improvements: 1. started the imaginary stack with the four kings (This jogged everything down in the imaginary deck by four which allowed the direct multiplication against the actual card value not one less than the card value as method D did). Note that this also changed the "number to card" formula to not add one instead to subtract one from the quotient if the card is a diamond. 2. switched the positions of the spades and the hearts to get a good mixture of reds and blacks throughout the physical deck (but this means the lobe count calculations must treat hearts as 1 and spades as 2). 3. Kings are to be treated as value zero not value thirteen.. 4. the 06 and 07 cards are AC and QD 5. the QC is in position 25 (in Method G there were three queens in a row at positions 5, 6, 7 which looked really bad so I had to break it up and swap the queen of clubs with the Ace of Clubs). Because of this there are three cards to memorize instead of two: AC at position 6 QD at position 7 QC at position 25 Note that method D and method G used the imaginary deck only for me to create the deck and create its formulas. Unlike the standard Harding the magician does not directly use the imaginary deck because my algorithms take care of finding the wanted values. 
glowball Special user Nashville TN 653 Posts 
Don't try to use this method D it is totally bad instead use Method G see the below thread:
https://www.themagiccafe.com/forums/view......orum=205 
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