Editted from normal PA {104}. When there is a pair of same suit: a single object is used to represent that 2 ranks notation. Total objects:{169}. No, user found. Antelex is willing to try once he can finish with cards in one minute.
Antelex had tried the concept, it works but like previous mentioned, without seeing all the cards, the pa boost system bugs. [relevant link]
The img/deck column is only a reference number because of grouping and other reasons, the number might vary in different person.
I guess I’d consider PAO a system of only 17, not 52, images. Maybe that’s not what you want to represent with that table though.
Anyway, to me it is one newly composed image per location. Yes with information from three different Ps, As, and Os… but in the end composed into just one new image
I am not sure how others think, some would say it is actually 52 images, but some says it is only 17. But I can only fill a digit in that box. I had changed the subtitle, is it clear now?
Added a new field ‘undefined card system’. If you have any similar thought please post in here. I will constantly update this table if I encounter any new relevant content.
Actually, it requires 51 images and last 1 card left.
But if you say 17 images that means you consider like this
Every single image is combination of 3 images .
At last you can say that it require 17 images but actually didn’t.
Then I might add a remark there, the img/deck is only a reference numbers, bc of grouping and other reasons, they might vary in each person. Thanks for your positive feedback.
base case: no suit pair = normal PA
case 1: a suited pair happens = an obj used in the same loci → OA
case 2: consecutive suit pair = two objects connected in a loci → OO
The core idea of PA-boost is to make use of ALL PEGS in traditional PAO{300}. Apart from traditional PA {104} pegs, there remains 300-104=196 pegs, take O{100} and create O’{69} = 169 for suit pair cards.
Now, there are 96 pegs, 48 for P, another 48 for A. By finding the most common pattern (prove by logic and simulator), we can convert the remaining pegs as SHORTCUTS!
You might want to remove the remove the 3-card option. Obviously, nobody has images for 52*51*50=132,600 card combinations memorized. Even using a block system alongside it, you’d still have to memorize \frac{132,600}{2}=66,300 images.
Just saying, because people new to it always think that they are using a 3-card system with their PAO; when in fact, it’s just a once card system that let’s you build a compound image of three of your pre-memorized images in a particular location.
Updated the table, and all the thoughts in my mind has already posted in here.
I am grateful to all your votes, your vote can contribute the card memorizing industry in two way. Newcomers: better card system planning. Geniuses: creating a innovative system.
About the update, please refer to this old post. Pseudo 2-card system (Nov '17) - #14 by Antelex