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The Magic Cafe Forum Index » » Magical equations » » Aronson bits morph to P3 bits (deck) (2 Likes) Printer Friendly Version

glowball
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From a physical Aronson deck this is how quickly and accurately to physically create a wrapping 52 card 26-26 (1s and 0s) DeBruijn deck (assuming Hearts and Diamonds are binary "1"'s). Note that a spectator can initially cut this deck as many times as they wish.

To perform the Leo boudreau's "Eyes closed if you have a red card" trick with this wrapping DeBruijn deck you will also need a full crib sheet. Therefore during performance after each of the six spectators have their card in hand (and you know the octal pattern of the six cards by looking at the six spectators' eyes) I openly use a funny titled book named "ESP Made Easy" to look up the octal code crib sheet info.

This gets a chuckle from the audience because they think I'm pulling their leg to get a laugh with such a funny titled book. There is no such book, I just bought a thin dollar book from Goodwill and created and printed my own simple title page and glued it on the front and glued the full crib sheet info on page 50.

Below is the eight step procedure to convert an Aronson deck to this P3 DeBruijn 26-26 (1s and 0s) deck from ThomasB.

Way ahead of performance time starting with an Aronson deck:
Move the Ace of Spades to after the 2D
Move the 3S and AD to immediately after the AS
Move the AH and 8S up 3 places after the 8c (before the 7S)
Move 3D thru 10D (9 cards) and insert them after the QS (before the 6D)
Move 8h, 9c, qs together after the JH
Move the JH and 8H back to after the 10c
Move the JD after the 3C
Move position 52 (9D) to after the 9H

This deck is slightly better than some other DeBruijn's because it does not have six zeros nor six ones in a row. This means with this deck you will always have at least one black card and one red card among the six chosen cards, this is a good thing because with some other decks if they happen to cut to where there is 6 black cards or 6 red cards this might tip the trick a little bit.

This physical Deck is now:
Js Kc 5c 2h 9s 3h 6c 8d Ac 10s 5h 2d As 3s Ad Kd 7d 8c Ah 8s 7s 5S Qd Jc 9c Qs 10c Jh 8H 4s 10h 6h 3c Jd 2s 9h 9d Ks 6s 4c 3d 7h Qh 5d 7c 4h Kh 4d 10d 6d Qc 2c

Keep this TomasB P3 (I'm just calling it P3 to have a name) deck in your card-case so you are ready to perform.

SORTED (for fast human lookup) Crib Sheet for TomasB's P3 DeBruijn:
01=8s 7s 5S Qd Jc 9c
02=2c Js Kc 5c 2h 9s
03=9c Qs 10c Jh 8H 4s
04=Qc 2c Js Kc 5c 2h
05=Js Kc 5c 2h 9s 3h
06=Jc 9c Qs 10c Jh 8H
07=Ks 6s 4c 3d 7h Qh
10=6d Qc 2c Js Kc 5c
12=Ah 8s 7s 5S Qd Jc
13=8d Ac 10s 5h 2d As
13=9d Ks 6s 4c 3d 7h
14=Qd Jc 9c Qs 10c Jh
17=2d As 3s Ad Kd 7d
20=5S Qd Jc 9c Qs 10c
21=9s 3h 6c 8d Ac 10s
23=3c Jd 2s 9h 9d Ks
24=8c Ah 8s 7s 5S Qd
25=5c 2h 9s 3h 6c 8d
26=6c 8d Ac 10s 5h 2d
30=10d 6d Qc 2c Js Kc
31=Kd 7d 8c Ah 8s 7s
33=Jh 8H 4s 10h 6h 3c
34=9h 9d Ks 6s 4c 3d
35=10h 6h 3c Jd 2s 9h
36=5h 2d As 3s Ad Kd
37=Qh 5d 7c 4h Kh 4d
40=7s 5S Qd Jc 9c Qs
41=Ac 10s 5h 2d As 3s
42=Kc 5c 2h 9s 3h 6c
43=As 3s Ad Kd 7d 8c
45=Qs 10c Jh 8H 4s 10h
47=6s 4c 3d 7h Qh 5d
50=7d 8c Ah 8s 7s 5S
51=Jd 2s 9h 9d Ks 6s
52=2h 9s 3h 6c 8d Ac
54=3h 6c 8d Ac 10s 5h
55=8H 4s 10h 6h 3c Jd
56=6h 3c Jd 2s 9h 9d
57=5d 7c 4h Kh 4d 10d
60=2s 9h 9d Ks 6s 4c
62=4s 10h 6h 3c Jd 2s
63=4c 3d 7h Qh 5d 7c
64=10s 5h 2d As 3s Ad
65=3s Ad Kd 7d 8c Ah
66=10c Jh 8H 4s 10h 6h
67=7c 4h Kh 4d 10d 6d
70=4d 10d 6d Qc 2c Js
71=Kh 4d 10d 6d Qc 2c
72=Ad Kd 7d 8c Ah 8s
73=4h Kh 4d 10d 6d Qc
75=3d 7h Qh 5d 7c 4h
76=7h Qh 5d 7c 4h Kh

Your Aronson routines will then NOT work with this deck but you will have a true 52 bit wrapping DeBruijn deck where red cards are binary "1"s deck that you can do Leo Boudreau's "Eyes closed if you have a red card" trick with 6 spectators and they can initially cut this full real deck as many times as they want.

For those interested here are the P3 DeBruijn bits:
0001010100110011101000100001101101011000111101111100

Note: originally TomasB gave me a bunch of 26-26 DeBruijns and I thought the P1 design was the best at crudely approximating the Aronson and NOT having six "1" bits nor six "0" bits in a row. First I modified it slightly by moving the first two bits to the end and called it P2. Then I modified P2 to flip all the bits because the original Aronson setup more closely accommodated 3 zero bits (black cards) at the beginning and throughout so I named this the P3 DeBruijn.

Many thanks to TomasB for the DeBruijn's and many thanks to Claudio for that idea to use octal codes on the crib sheet.
glowball
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Reminder that I'm using low bits first on the octal code technique.
Example:
After the six spectators have their cards (hidden from you) and you (the magician using the Leo Boudreau technique) say "concentrate on your card" pause "I'm having trouble with the red cards, if you have Hearts or Diamonds close your eyes and concentrate extra hard":

First Spectator: eyes closed
Second Spectator: eyes closed
Third Spectator: eyes open

Fourth Spectator: eyes open
Fifth Spectator: eyes open
Sixth Spectator: eyes closed

From the above pattern you secretly mentally create two octal binary digits from 110 001 which is 3 and 4 ie: 34 on the crib sheet is:
34=9h 9d Ks 6s 4c 3d
therefore you know that the first spectator has the 9h, second spectator has 9d, third spectator has Ks, fourth spectator has 6s, fifth spectator has 4c, sixth spectator has the 3d.
glowball
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Correction: the 9d is octal 16 not 13.

Corrected:
SORTED (for fast human lookup) Crib Sheet
for TomasB's P3 DeBruijn:

01=8s 7s 5S Qd Jc 9c
02=2c Js Kc 5c 2h 9s
03=9c Qs 10c Jh 8h 4s
04=Qc 2c Js Kc 5c 2h
05=Js Kc 5c 2h 9s 3h
06=Jc 9c Qs 10c Jh 8h
07=Ks 6s 4c 3d 7h Qh
10=6d Qc 2c Js Kc 5c
12=Ah 8s 7s 5S Qd Jc
13=8d Ac 10s 5h 2d As
14=Qd Jc 9c Qs 10c Jh
16=9d Ks 6s 4c 3d 7h
17=2d As 3s Ad Kd 7d
20=5S Qd Jc 9c Qs 10c
21=9s 3h 6c 8d Ac 10s
23=3c Jd 2s 9h 9d Ks
24=8c Ah 8s 7s 5S Qd
25=5c 2h 9s 3h 6c 8d
26=6c 8d Ac 10s 5h 2d
30=10d 6d Qc 2c Js Kc
31=Kd 7d 8c Ah 8s 7s
33=Jh 8h 4s 10h 6h 3c
34=9h 9d Ks 6s 4c 3d
35=10h 6h 3c Jd 2s 9h
36=5h 2d As 3s Ad Kd
37=Qh 5d 7c 4h Kh 4d
40=7s 5S Qd Jc 9c Qs
41=Ac 10s 5h 2d As 3s
42=Kc 5c 2h 9s 3h 6c
43=As 3s Ad Kd 7d 8c
45=Qs 10c Jh 8h 4s 10h
47=6s 4c 3d 7h Qh 5d
50=7d 8c Ah 8s 7s 5S
51=Jd 2s 9h 9d Ks 6s
52=2h 9s 3h 6c 8d Ac
54=3h 6c 8d Ac 10s 5h
55=8H 4s 10h 6h 3c Jd
56=6h 3c Jd 2s 9h 9d
57=5d 7c 4h Kh 4d 10d
60=2s 9h 9d Ks 6s 4c
62=4s 10h 6h 3c Jd 2s
63=4c 3d 7h Qh 5d 7c
64=10s 5h 2d As 3s Ad
65=3s Ad Kd 7d 8c Ah
66=10c Jh 8h 4s 10h 6h
67=7c 4h Kh 4d 10d 6d
70=4d 10d 6d Qc 2c Js
71=Kh 4d 10d 6d Qc 2c
72=Ad Kd 7d 8c Ah 8s
73=4h Kh 4d 10d 6d Qc
75=3d 7h Qh 5d 7c 4h
76=7h Qh 5d 7c 4h Kh

glowball yelnif
alea
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Quote:
On Dec 6, 2018, glowball wrote:
Reminder that I'm using low bits first on the octal code technique.
Example:
After the six spectators have their cards (hidden from you) and you (the magician using the Leo Boudreau technique) say "concentrate on your card" pause "I'm having trouble with the red cards, if you have Hearts or Diamonds close your eyes and concentrate extra hard":

First Spectator: eyes closed
Second Spectator: eyes closed
Third Spectator: eyes open

Fourth Spectator: eyes open
Fifth Spectator: eyes open
Sixth Spectator: eyes closed

From the above pattern you secretly mentally create two octal binary digits from 110 001 which is 3 and 4 ie: 34 on the crib sheet is:
34=9h 9d Ks 6s 4c 3d
therefore you know that the first spectator has the 9h, second spectator has 9d, third spectator has Ks, fourth spectator has 6s, fifth spectator has 4c, sixth spectator has the 3d.


First of all for congrats to you (and Thomas) for these amazing threads on the de Bruijn sequences (homework left by Diaconis-Graham in their M.M. ten years ago....).

May you clarify how did you get "3", "4" in the example? Thanks
glowball
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Alea, in my above example:

First Spectator: eyes closed
Second Spectator: eyes closed
Third Spectator: eyes open

Fourth Spectator: eyes open
Fifth Spectator: eyes open
Sixth Spectator: eyes closed

The above mentally becomes 110 001 but I am using reverse order binary bits meaning the low order bits come first (mathematicians always show the low order bits to the far right where I am showing the low order bits to the far left in this thread). Therefore 110 translates to 1 + 2 + 0 = 3. The other number 001 translates to 0 + 0 + 4 = 4 (0 on ones position, 0 and the twos position, 1 in the fours position).

Take a look at my most recent (February 2022) thread that uses only four chosen cards (no duplicates) that is similar to this trick and uses a stack in its pristine sequence.

https://www.themagiccafe.com/forums/view......forum=99

Hope This helps. Larfin glowball.
alea
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Quote:
On Feb 16, 2022, glowball wrote:
Alea, in my above example:
....

The above mentally becomes 110 001 but I am using reverse order binary bits meaning the low order bits come first (mathematicians always show the low order bits to the far right where I am showing the low order bits to the far left in this thread). Therefore 110 translates to 1 + 2 + 0 = 3. The other number 001 translates to 0 + 0 + 4 = 4 (0 on ones position, 0 and the twos position, 1 in the fours position).
....
Hope This helps. Larfin glowball.


Oh my God, you’re right... the reverse order of course... : I had the answers in front of my eyes!! I spent the entire afternoon reading (and uncoding!) all the Bruijn correlated threads ( mainly yours), so much boatload to need really a rest in my binary research!
Thanks, you are truly an awesome Bruijn -sleuth!
alea
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Quote:
On Feb 16, 2022, glowball wrote:
Alea, in my above example:
....

The above mentally becomes 110 001 but I am using reverse order binary bits meaning the low order bits come first (mathematicians always show the low order bits to the far right where I am showing the low order bits to the far left in this thread). Therefore 110 translates to 1 + 2 + 0 = 3. The other number 001 translates to 0 + 0 + 4 = 4 (0 on ones position, 0 and the twos position, 1 in the fours position).
....
Hope This helps. Larfin glowball.


Oh my God, you’re right... the reverse order of course... : I had the answers in front of my eyes!! I spent the entire afternoon reading (and uncoding!) all the Bruijn correlated threads ( mainly yours), so much boatload to need really a rest in my binary research!
Thanks, you are truly an awesome Bruijn -sleuth!
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