Shave and a haircut (base 6)

This whole post is about what we magicians can do with just 2 binary bits (and rounding the top two corners of some playing cards so we can know what certain cards are when any six cards are consecutively dealt from a memorized stack).

I apologize for the pun "Two Bits" but like Roger rabbit I just could not resist:

https://youtu.be/6ds6w7SkHyw

Background:
I have tried to come up with a marking scheme that could be used on any stack using a blank back deck of playing cards (a spectator would cut the deck and deal six playing cards to other spectators. The objective of this marking scheme would identify the value of at least one of those playing cards but not it's suit.

The magician will simply ask the one spectator with the marked card what their suit is: "I'm getting a good mental reading on everybody except for your card I'm having trouble with the suit, please just tell me the suit not the value of the card, just the suit".

Since this is a memorized deck once the magician knows the value and suit of one card the magician now knows what all 6 cards are.

I have come up with a few successful schemes that I have published on the Café, however each of those schemes has a negative feature(s) that I would like to improve.

For example one of my schemes rounded (widened) all 4 corners of all 52 cards in the deck. This scheme uses the binary method of adding 1, 2, 4, and 8. Each of the four corners of the card represents that binary value (if a corner was widened then the magician just added up those corners on a card to get its value, simple).

The problem with this scheme is that if a spectator re-orients the card 180 degrees and the magician does not realize that is what happened then the magician will mentally calculate the wrong value for the card. Also when the deck is being handled the magician must really watch the orientation of the deck and it's very easy to get confused.

Things can easily go wrong, therefore to prevent this orientation problem I did the following (after properly rounding the corners of the cards) I put a big red dot on the bottom face of all 52 cards (not on its back!) so that when doing the trick the magician tells the spectators to hold their card with one hand with their thumb on the red dot.

I have used the humorous patter: "this is a thumbprint recognition device that will transfer your thoughts". Instead of patter about a thumbprint I have also used the patter: "the red dot is an activator button to transfer your thoughts so hold it firmly with your thumb". By utilizing the red dot this way during the performance: this insured the spectators proper orientation of their card. However another problem arose because the spectator's hand could obscure one of the bottom corners.

Top two corners:
Because of the above issues I decided to try a scheme of using just the top two corners to encode a value (and round the diagonally opposite corners the same way so there will not be any orientation problem thus eliminating the need of the red dot). However, this reduces the number of "bits" from 4 to 2 (four "bits" can encode a value up to 16 which is plenty because the maximum needed is 13 for a king).

By reducing the observed corners to two this means there is only two "bits" ("shave and a haircut", ha, ha). Using two binary "bits" can normally encode 4 decimal values ie: 0, 1, 2, 3 which is not nearly enough to identify all the playing cards therefore I changed the scheme to identify just certain cards throughout the stack (I call them spy cards because they are the only cards with their corners rounded).

Normally two binary bits can encode the values 0, 1, 2, 3 but in this situation we can only encode the values 1, 2, 3 (we cannot encode the value "0" on a spy card because the non-spy cards will all have the value "0" giving a false signal).

With all this in mind I came up with a very specific scheme just for the Tamariz Mnemonica stack and then did the same for the Aronson stack. I'm calling this the "specific value spy cards method". The below link is in secret sessions:

https://www.themagiccafe.com/forums/view......forum=37

Note that the above link using the "specific value spy cards method" is extremely easy and requires no mathematical calculations.

However using the above "specific value spy cards method" means that any other memorized stack will require a lot of analysis to come up with which cards to make spy cards to mark with rounded corners (as few as possible values and spread somewhat evenly throughout the stack).

Therefore instead of the above, I want to develop a generic encoding method for ANY memorized stack using just two bits (two corners). Note that on a spy card we can round (widen) the upper right corner only, the upper left corner only, or both the upper right corner and the upper left corner (and of course their diagonally opposite corners, but the magician only looks at the top two corners). That is three values that can be encoded ie: 1, 2, and 3. This is not nearly enough information to have a GENERIC scheme for ANY stack, therefore we need two spy cards side by side and I think we can have a generic scheme to identify the position (within the stack) of the second spy card.

Note that this generic scheme does not require the magician to ask the name of the suit of any card nor does it require an equivoque of any kind.

Here is my first attempt at such a scheme showing where the spy cards are located and their widened corner value:

xxxx11xxxx12xxxx13xxxx21xxxx22xxxx23xxxx31xxxx32xx33
Note that the x represents a non-spy card.

The generic (works on any mem stack) algorithm:
The FIRST of the two spy card pairs: the magician mentally thinks of the marked card value (of 1, 2, 3) then subtracts one from it then multiplies by 18.
The SECOND marked card value (of 1, 2, 3) is directly multiplied by 6 and the result is added to the above.
This gives the position of the second marked card of the pair. Because this is a memorized deck the magician now knows what all six cards are.

Example 1: magician sees following on the back of six cards: xxx12x
Note that the x means an unmarked card.
Mentally calculate the algorithm as follows:
Always subtract 1 from the first number of the pair: ie: in this case 1 minus 1 equals zero then that result times 18 equals 0. Next the second number of the pair is multiplied by 6 therefore in this case that result is 2 times 6 which equals 12.
The 12 is added to the zero giving a final result of 12 which is the position of the second card of the pair.

Example 2: magician sees following on the back of six cards: x31xxx
Mentally calculate the algorithm as follows:
Always subtract 1 from the first number of the pair: ie: in this case 3 minus 1 equals 2 then that result times 18 equals 36. Next the second number of the pair is multiplied by 6 therefore in this case that result is 1 times 6 which equals 6.
The 6 is added to the 36 giving a final result of 42 which is the position of the second card of the pair.

Once the magician knows the position of a card in the memorized stack then all six cards are known.

This scheme lets the magician have only 18 marked cards. That's the good news, the bad news is that there are eight places (out of 52) where the spectator can cut the deck where there is one marked card as the first card of the six and one marked card in the 6th position and positions two through five are unmarked cards. A magician cannot easily use the algorithm in those 8 cases.
Let me see if there's an easy memory trick for each of those eight cases. See below little list: Note that the first card of the six cards is in position:

52: 3xxxx1
6: 1xxxx1
12: 2xxxx1
18: 3xxxx2
24: 1xxxx2
30: 2xxxx2
36: 3xxxx3
42: 1xxxx3

Note that position 48 will never be a problem because it has 33 close enough to it that either 32 (position 48) or 33 (position 52) will be fully in view. Note that the two marked cards of 33 indicate (via the algorithm) the value 54 but of course the magician knows to mentally convert that to 52.

One way to overcome this nuisance is to mentally bring the first and last card values together in your mind for example 2xxxx2 becomes simply 22 and then have a mnemonic story for it. Below are the 8 situations and possible mnemonic stories to go with each one:

52: 3xxxx1 31 yr old pilot of B-52 bomber
6: 1xxxx1 11 yr old should not play with 6 gun
12: 2xxxx1 21 yr old stands on head reversed
18: 3xxxx2 32 Jim Brown got 18 yd per carry
24: 1xxxx2 12 eggs are doubled to 24
30: 2xxxx2 22 rifle and 30 ought
36: 3xxxx3 33 inch is almost 1 yard (36 inch)
42: 1xxxx3 13 is unlucky for universe (42)

Note that in the Hitchhiker's Guide to the Galaxy the number 42 was the meaning for life the universe and everything. Note that Jim Brown (greatest running back ever) wore number 32 but he didn't average 18 yd per carry, this is just a memory trick (he did average an incredible 5.2 yards per carry for entire career).

Reminder: If the magician sees two marked cards (rounded corner cards) together then the magician mentally uses those two cards.
When the magician does not see two marked cards together and sees that the first and the last card among the six cards are marked (rounded corners) he abandons the algorithm and simply uses the above mnemonic stories to know the position of the first marked card among the six.

This generic solution requires 18 cards to have their corners rounded (widened) and should work well with any memorized stack. I think using a blank back deck of cards gets the spectators minds off of the possibility of marked cards.

A way to eliminate the need for the 8 mnemonic stories is to deal seven cards instead of six cards because that way the magician is assured that there will always be two spy cards together among the seven cards. Therefore when seven cards are dealt the magician can always use the algorithm by mentally focusing on the two spy cards that are together.

Note that on some cell phone viewers the below got cut off to the right. So here it is again but specified on two lines so that it should be easily seen if you are using a cell phone on the Café.

Spy card placement (any stack) and marked value:
xxxx11xxxx12xxxx13xxxx21xxxx22xxxx23
xxxx31xxxx32xx33
Note that the x represents a non-spy card.