

glowball Special user Nashville TN 662 Posts 
Seven Queens mod 4 acaan calculation
I have posted more details to this under secret sessions and some info under shuffled not stirred but I wanted you math guys to have some visibility to this and perhaps you have knowledge of this mod 4 technique being used on a tetradistic stack before? Thanks. The main essence of this calculation is: Add the card value and it's suit then mod 4 that number thus you should know the 0, 13, 26, 39 factor. Then just add in the value of the named card's paired twin card. I have been looking for a nonmemorized stack that meets the following three criteria (for my magician friends that do not wish to memorize a 52 card deck): My (glowball) Criteria for a nonmem deck: 1. Has reasonable randomness look ie: better than a Si Stebbins. 2. Can do an acaan (any card) calculation a little easier and faster than the Harding stack. 3. Can fairly quickly calculate what the next card is from a known card identity. Off the top of my head the stack that comes immediately to mind is the Osterlind Breakthrough stack which is quite close to meeting my three criteria however I would like the calculations to be a little bit easier than the Breakthrough stack. A tetradistic stack is fine as long as it meets the three criteria, especially criteria 1. I recently came up with this system (won't be surprised if someone else has done this long ago) that I believe meets the three criteria. The unique part of this system is a special way of using the mod 4 calculation. This is a tetradistic arrangement with Harry Riser type pairs where the paired values increment by one from Ace to two to three etc thru King (ie: 1 thru 13) and different suits (somewhat like Eight Kings stack). Nothing new here. But here is where my mod 4 technique comes into play. Making mod 4 work: Thought process and creating this stack: Hopefully the first group of 13 cards could all mod 4 to a result of the same value and then the second group of 13 could mod 4 to a different value (all the same mod 4 value within the group) and so on for the third group of 13 and then for the fourth group of 13. If this could be done then it would be a simple matter of associating one of four starting factors (0, 13, 26, 39) with each group and then adding the paired card value and bingo we have the position of the spectator stated card. Pretty easy. Let's say the following is the sequence of the 13 random looking cards but their suits have not yet been determined for our ideal stack. 7, Q, 8, K, 10, 9, A, 3, 6, 5, J, 2, 4 I did a lot of experimenting and found the simple solution of adding the suit value to the card value to come up with a number to apply the mod 4. I use SHoCkeD for the values of the suits: Spades are 1 (Spade has one point) Hearts are 2 (Hearts have two lobes) Clubs are 3 (clubs have three lobes) Diamonds are 4 (Diamonds have four points) Let's say that our four groups of 13 cards that we call them: Group 0 Group 1 Group 2 Group 3 For purposes of this post the term "mod 4" will simply mean the remainder value when dividing some number by the constant 4. I'll explain why group 0 is listed first but for now let's see how we can make every card in a group have the same mod 4 result. For this explanation I will arbitrarily start with "group 3" (the physical last group) and ask the question "what must the suits be for each card in that group to make the mod 4 result equal 3". With the card values in mind: 7, Q, 8, K, 10, 9, A, 3, 6, 5, J, 2, 4 The below is just a description of how this deck was created and may seem complicated, in actual practice the acaan calculations are much much simpler and I will explain later. How the deck was created: Let's start with the first card in "group 3" which is a seven (all groups in this stack start with seven) and ask what must it's suit be? in order for the seven plus it's suit value to mod 4 result to 3? Well our next target value above seven that will mod 4 to a result of three is the target value 11. Therefore how much must we add to the seven to get to 11? The answer is obviously 4. And 4 equates to diamonds. Therefore in the deck we are constructing group 3 (which is physically the last group) must start with the Seven of Diamonds. The second card in "group 3" is a queen and to figure out its suit in order to mod 4 to a result of three we need to raise the value of 12 (Queen's value) to 15 which is an increment of three. Three equates to clubs, therefore the second card in "group 3" must be the Queen of Clubs. The third card is an eight and to figure out its suit in order to mod 4 to a result of three we need to raise the value of 8 to 11 which is an increment of three which equates to clubs. Therefore the third card in "group 3" must be the eight of clubs. I'll do one more just to illustrate the principle. The fourth card is a king and to figure out its suit in order to mod 4 to a result of three we must raise its value from 13 to 15 which means two more notches up the ladder. Value 2 equates to hearts therefore the fourth card of "group 3" is the king of hearts. I think you get the idea (using the appropriate suit for each card value to make it mod 4 to the same value as the other cards in that same group). The next to last group of 13 cards I'm calling "group 2" which means all the cards in that group of 13 will have a mod 4 result of 2 (because I made sure that each card had the appropriate suit to make its value plus suit be 2 above the nearest multiple of four, note that the multiples of four that we will be concerned with are: 0, 4, 8, 12, 16) The second group of 13 cards I'm calling "group 1" which means all the cards in that group of 13 cards will have a mod 4 result of 1 (because I made sure that each card had the appropriate suit to make its value plus suit be 1 above the nearest multiple of four). The first group of 13 cards I'm calling "group 0" which means all the cards in that group of 13 cards will have a mod 4 result of 0 (because I made sure that each card had the appropriate suit to make its value plus suit be 4 above the nearest multiple of four which means the remainder when dividing by 4 will equal zero). I have already done all the calculations and embedded them in to the following physical stack therefore at performance time your task is much simpler. One more major element in the calculation is the FACTOR that is associated with the mod 4 result: Result Factor 0 equals 0 1 equals 13 2 equals 26 3 equals 39 At performance time after a spectator names a card the mental calculation of the position of that card within the deck is quite quick: Add the card value and it's suit and mod 4 that number thus you should know the above factor. Then just add in the value of the named card's paired twin card. Easy peasy. This uses the Harry Riser twins pairs technique. The pairs pattern is mostly my own in order to have a good color mixture of suits and still meet the needs of a hidden 1 thru 13 arrangement. Paired twins: Ace and 7 (have a sharp angle at top) 2 and Queen (2 headed Queen) 3 and 8 (the 3 looks like half of an eight) 4 and King (the K and 4 have 4 corners) 5 and 10 (five and dime store) 6 and 9 (same symbol upside down) Jack stands alone. Card to position calculation Example: A spectator names the two of clubs: Magician or a shill mentally calculates: 2C 2 plus 3 equals five which mod 4's to 1. Thus the factor is 13. The paired value of a two is a queen which is 12. 13 + 12 = 25 thus the two of clubs is at position 25. The reason I made the first 13 cards group 0 is so that the other three groups will have their mod 4 result numbers be the first digit of their factors. Example group 2 factor is 26. Group three factor is 39. Group 1 factor is 13. Group zero factor is zero. This makes the determination of the factor very quick and easy. Note that if the mod 4 result is 0 then the position of the card is super simple it's just the value of the paired twin. Seven Queens V1 acaan Stack: 7S, QD, 8D, KC, 10H, 9C, AC, 3S, 6H, 5C, JS, 2H, 4D 7H, QS, 8S, KD, 10C, 9D, AD, 3H, 6C, 5D, JH, 2C, 4S 7C, QH, 8H, KS, 10D, 9S, AS, 3C, 6D, 5S, JC, 2D, 4H 7D, QC, 8C, KH, 10S, 9H, AH, 3D, 6S, 5H, JD, 2S, 4C To get the position of a named card: Do a mod 4 on the card value plus it's SHCD suit value in order to get the 0 or 13 or 26 or 39 factor then add the paired twin card value. https://www.themagiccafe.com/forums/view......forum=37 Click on the above link in secret sessions for more details. 
glowball Special user Nashville TN 662 Posts 
In some of my other posts about the Seven Queens stack and the Jackknife stack I refer to the first physical group of 13 cards as being group 4 whereas in this post I refer to the first physical group of 13 cards as group zero because I have defined the mod 4 function here as simply being the remainder when dividing a number by 4.
The above fact only comes into play when the card's value plus it's suit equals 4 or 8 or 12 or 16. In those cases the FACTOR is zero. I had run into computer functions that did modulus calculations that would never have zero as a result and would use the modulus number instead of zero if the modulus number divided into the target number an even number of times (zero remainder) and was thinking in those terms when I made the aforementioned posts. In those other posts I should have defined mod 4 as simply being the remainder when dividing by 4 and allowing zero to be one of the possible remainders. This would eliminate having to use the term "group 4" in those other posts because "group 0" works perfectly as the name for the first physical group of 13 cards. 
glowball Special user Nashville TN 662 Posts 
The above link into secret sessions is not working.
The below link should work: https://www.themagiccafe.com/forums/view......forum=37 Above goes into secret sessions. 
glowball Special user Nashville TN 662 Posts 
Next card SUIT new rules:
Instead of Napoleon battlefield using the NEXT card value and looking at prior card suit, just base the rules on the PRIOR card value and PRIOR card suit to get the next card suit (and use John F Kennedy White House for the next card suit memory methods). This makes the calculation a little easier when doing a trick where the deck is cut and the magician glimpses the bottom card and needs to know what the top card is. The results are exactly the same using this new method, it's just the new memory method for the next card suit is based entirely on the prior card value and suit in order to get the next card SUIT (the next card VALUE still uses the "Seven Queens ate ..." story memory method). Next suit: Forward one SHoCkeD: 3, 6, 10, J Backward one SHoCkeD: 7, 8, K Same color different shape: A, 2, 4, 5 Exact same suit: Q, 9 "Next card suit" White House memory methods: 10 days JFK (Jack) has Marilyn Monroe (36) as a guest at the White House and they go FORWARD for a walk SHoCkeD. He has a 78yearold King visitor who exits the White House by taking a BACKWARD step SHoCkeD. There are young (1, 2, 4, 5 yr olds) visitors in line that each has a guest following them of the same color but different shape. The Queen (Jacqueline Kennedy) is dressed to the nines but never changes her suit. Note that this next card suit memory method is NOT used when doing the acaan trick, it is used when doing a trick where the deck is cut and the magician glimpses the bottom card and quickly mentally uses the "Seven Queens ate ..." to know the VALUE of the top card and the White House story to know the SUIT of the top card. It is also used when revealing a series of several cards that were dealt to spectators after the magician knows the value of one of the cards. 7S, QD, 8D, KC, 10H, 9C, AC, 3S, 6H, 5C, JS, 2H, 4D 7H, QS, 8S, KD, 10C, 9D, AD, 3H, 6C, 5D, JH, 2C, 4S 7C, QH, 8H, KS, 10D, 9S, AS, 3C, 6D, 5S, JC, 2D, 4H 7D, QC, 8C, KH, 10S, 9H, AH, 3D, 6S, 5H, JD, 2S, 4C 
glowball Special user Nashville TN 662 Posts 
See The thread below for how to quickly calculate the top card when the bottom card is glimpsed by the magician:
https://www.themagiccafe.com/forums/view......orum=205 Note that this 13 stories method can also be used when revealing a series of six cards dealt. Above thread is in shuffled not stirred. 
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