Brainstrorming a 3-Card or 4-Card Mnemonic System

General Memory Chat
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Do you think a three card mnemonic system is possible?

A two card system requires 2,704 images (or even 2,652, since there aren’t duplicate cards in a single deck).

A three-card system would require 140,608 images (52x52x52), unless someone could figure out how to compress the data more efficiently.

Can anyone think of a way to do it?

What if suits were separated from card values and memorized separately?

There are four suits. That means 64 possible combinations of suits between the three cards (4x4x4).

Then you would have three cards with 13 possible values. That would make 2,197 images (13x13x13). So, 3 cards would be represented by 2 images. This still is less efficient than the Ben System, which uses 2 images for 4 cards, but is more efficient than having 140,608 images.

Is there anyone out there that can figure out how to make it even more efficient and still use 10,000 images or fewer? The Ben System can place a deck of cards in nine loci. Simon Reinhard’s number system uses 10,000 images, so it’s known that at least those two things can be accomplished.

I mentioned 4-card system in the title, because if the suits are separated from the card values, there are only 256 combinations of the suits with four cards. Maybe there is a way to compress the card values. 13x13x13x13 = 28,561 so it isn’t efficient to just use the unmodified card values.

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So it means that if we can compress 4 card per 1 image then the system will reduce alot, right?
I see some good point from my language, vietnamese, which is latin based. Normally, every image is built by 2 words, such as: triangular = tam giác
So if we apply the rule as per from Ben:

  1. For the first word
  • the first consonant = combination of suits. Ex: club/diamond - t
  • The vowel = rank/number of the first card. Ex: Ace = `a’
  • second consonant = rank/number of the second card. Ex: 3 = m
    ==> we have a word “tam”, which means ace of club & 3 of diamond
  1. For the second word
    Similarly, the first consonant = combination of suits: diamond/spade - gi/j
  • The vowel = rank/number of the first card. Ex: Ace = `a’
  • Second consonant = rank/number of the second card. Ex: J = c
    ==> we have a word “giac”, which means ace of diamond & Jack of spade

Both these words “tam giác” means triangular and has this image. Plz advise how many images for this system? :slight_smile:

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There are a number of ways to do this. All of them require noting the order of suits, as you have observed. A system for compressing data in three cards could be to have 1300 images and 64 unique states of affairs, or descriptors, such as “it is raining.” Then just read the digits, 4 of hearts 2 of clubs 9 of hearts as a rainbow that is being rained on, for instance. The extra 300 images are for J Q K, but I’m sure you’ve thought of this. If you wanted to compress four cards, you could use the same method, but with 256 states of affairs, and 13 secondary descriptors, such as material or color, which would be determined by the first card. The next three would be read in the same way, and one could sidestep having to memorize more than 1556 items.

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I have descriptors like this that are useful only in certain situations, such as flushes or two or more repeating card values. In these instances, I am “lucky” and change the way those cards are encoded from the ‘regular’ PAO rules that I use.

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Right now I’m trying to finish my 2704 system and with this system it’s possible to expand it to 140,608 without creating a new system but it would just take forever.
I thought about a way to put 10 cards at a locus but I don’t think I’ll use it
2704 Images + Action
Image + Action + Image + Action + Image = 3 images
and the actions are gonna have to be simple
some people think PAO for cards have 156 images but I think some actions don’t count as an image.
pushing,punching,kicking,throwing ← these are just ways to associate person&object
swinging golf club, throwing a baseball ← same but requires an additional image
Person / throw / object = 2
Person / throw a baseball / object = 3
so 2704 with action can put 10 images at a locus but it will take away creativity.
I know it sounds useless but I’m just throwing out an idea.

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For me, being forced to make a creative association is always more memorable. And always takes longer.

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Well, it’s not a 3 or four-card system, but in the hope of reviving this very cool thread, here’s a potential variation on the typical loci-saving 3img/loci that is commonly used by 2-carders. Because PAO works for many people, it’s safe to say that for these people, there is time-saving benefit to be found in encoding information in the form of the relationship of objects, e.g. “action.” The Actions that are commonly used require either persons or robustly anthropomorphized objects as a subject. 2704 PAO might be pretty cool, but who dares to be the first to find out? A relatively simple alternative to creating 2704 distinct actions could be to create 1352 tableau, each focusing on two objects, as in the Sistine Chapel, and contain 2 of the 2704 objects in each tableau, using a 2-block system like JM does. During Memorization, the two subjects of the tableau are determined by four cards, the tableau itself is determined by two cards, and the lack of 1352 tableau is made up for by assigning the primary and secondary subjects within them depending on which block the tableau card is from.

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One image for every ordering of 52 cards would make 52! (approx 10^67) images, that’s more images than there are grains of sand on a trillion trillion trillion earths (10^21 x 10^27 = 10^48). (10,000 images isn’t that bad is it?)

So if our goal is to have the least amount of images encode a deck of cards, then we have some serious constraints ahead given our current methods.

A die (dice) that has X equally probable and distinguishable outcomes must have at least X sides. This, in practice, shows how we have to carry every “other possible” outcome around to every loci.

As above, our perfect die would have 52! (10^67) sides, which is obviously not possible.

A system that encodes 4 cards per IMAGE can be likened to having 4 dice with 52 sides each. To encode a deck of cards you would need to roll each one 13 times. [The number of possible ways these 4 dice can come up after 13 rolls is actually greater than 52! (it equals 52^52), but we only “need” (must) carry as much baggage as is the number of possible out comes of the 4 dice (52^4)]. In doing this we basically shrink our deck of cards to 13 decks with 4 cards per deck.

A PAO system turns this 52^4 above into 52x4. Each loci has an IMAGE that is actually composed of 4 images. The IMAGE is created on the spot and so you can drop the baggage of carrying around all the possible outcomes.

If suits are encoded into the location of the loci themselves then you would only need 13x4 = 52 images to encode four cards per loci.

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Spam ^^^^ Josh, could you ban him?

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Sorry about that. I deleted the comment and blocked the user.

EDIT: just deleted the other comment too (upon request).

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Roughly speaking, each card holds about log2 (52) = 5.7 bits of information (neglecting optimizations that a single deck will have no duplicate cards). So the number of images you need in a system increases exponentially as you demand more “efficiency” assuming you are defining efficiency as using fewer images per card.

It plays out something like this. Somewhere around 11k images, you actually get 2 bits of information extra per image or Ben system (which is at 11.4 bits per image), which would allow one to memorize a deck in 22 images rather than the Ben system 26. (Exactly two bits more than Ben system is 4 * 52 * 52 = 10816). (I’m trying stay close to Josh’s 10,000 image mark that he mentioned in the OP as being a reachable goal.) So with all that effort, maybe you could save 1 locus per deck if storing 3 images per locus like Ben. Let’s call this imaginary system an 11k system. At 13.4 bits per image, you save effectively another image for 6 or so Ben system images. It just doesn’t seem promising to me. Wouldn’t it be easier to learn to store 4 images per locus instead of 3? Or just rehearse your journeys better?

Maybe I’m wrong. At the top levels, maybe saving a locus per deck would be a key differentiator. It was amazing to read how Zoomy encodes his last few cards in a speed deckafter putting down the deck.

Images per deck Number of images in system Log2(images in system)
52 52 5.700439718 (Boris or 52 images, 52 loci, also PAO if you count P, A, and O as separate images)
51 57 5.832890014
50 61 5.930737338
49 67 6.06608919
48 73 6.189824559
47 80 6.321928095
46 88 6.459431619
45 97 6.599912842
44 107 6.741466986
43 119 6.894817763
42 134 7.06608919
41 151 7.238404739
40 171 7.417852515
39 195 7.607330314
38 223 7.8008999
37 259 8.016808288
36 302 8.238404739
35 355 8.471675214
34 422 8.721099189
33 506 8.982993575
32 615 9.2644426
31 756 9.562242424
30 943 9.881113961
29 1194 10.22158712
28 1538 10.58683979
27 2018 10.97871046
26 2704 11.40087944 (Ben System)
25 3710 11.85720347
24 5225 12.35121532
23 7580 12.88798213
22 11377 13.47383256 Sweet spot?
21 17749 14.11545012
20 28948 14.82117606
19 49709 15.60121946
18 90645 16.46793982
17 177400 17.43664648
16 377582 18.52643046
15 888816 19.76152526
14 2364432 21.17306222
13 7311616 22.80175887 (Imaginary unique image per 4 cards)

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Your post makes absolutely no sense to me. Don’t worry about explaining it further, but are you sure you’re right about being able to compress the data? Kudos if you really did figure that out. But you’re right - carrying it out is not a good idea unless you have a 10,000 object system like Simon does. In that case it might be worth moving forward, but the returns are so small compared to the 33% increase you would get with digits by moving to a 4-digit over a 3-digit system, (which means that as long as you don’t lose pace by more than 25% percent, it’s worth it…or is it 33%?) Actually, even if your pace fell beneath that threshold it could still be worth it to reduce the number of loci needed so greatly. But you wouldn’t get that with cards.

****No. That method is extremely tempting, 13 loci per 2 decks, but if it weren’t so tricky to do, top MA’s would definitely be doing it. It’s just like using a PAOX method instead of PAO. No one seems to be able to make good use of 4 cards per locus with a PAO variation, and it certainly is not easier. I’m like the rest - 3/locus is fastest for me, even though it’s easier to move at a slightly increased speed with 2/locus, recall declines because the reason 2/locus is faster is that you can spend less time linking. More loci with worse links = worse recall.

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I guess that was confusing. Let me provide one concrete implementation, though I don’t think you will like it.

First, create 4 * 52 * 52 images, i.e., 4 Ben System’s worth. PM me when you are done and I’ll buy you a beer. :slight_smile:

Let’s call these four groups of 2704 images each BenA, BenB, BenC, BenD. It is important that they are distinct and that upon seeing an image in your mind, you very quickly know whether it’s in BenA, BenB, BenC, or BenD.

You will be processing cards in groups of 7.

Look at the first card in the deck. We’ll call this the master card. This card is going to tell you how to encode the next 6 six cards, but you may forget this card after the next 6 are encoded, because you will be able to reconstruct it later.

The suit of the first card will tell you how to encode the next two cards.

spade = BenA
heart = BenB
club = BenC
diamond = BenD

Encode cards 2 and 3 as above.

Look at the rank of the 1st card. Convert to a number in the range 1-13. Call this number R. This will first you how to encode cards 4 and 5, as follows:

1-4 = BenA
5-8 = BenB
9-12 = BenC
13 = BenD

Finally, take find the remainder of R divided by 4. This will tell you how to encode cards 6 and 7, as follows:

0 = BenA
1 = BenB
2 = BenC
3 = BenD

Draw the next card, repeat. You pay attention to what you are doing in 7 card chunks, the 1st card in each chunk being a kind of master card.

So when you are done memorizing, of course you have to get your images back into cards. This is easy for cards 2-7, 9-14, etc. but remember I’ve said you can forget the master card in each group of 7? You can get it back by working backwards. Cards 2 and 3 will tell you the suit of master card 1. Cards 4 and 5 will tell whether the rank is in the range 1-4, 5-8, etc. and cards 6 and 7 will help you resolve the precise rank of the master card.

This system would allow one to encode 7 cards in 3 images, packing slightly better than Ben. But you need 4 Ben systems worth of images to do it and you have to learn how to work in groups of 7. Improvements in the use of the master card are possible to make it easier to use, but that’s about the gist of it.

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So this master card idea is really just a way of taking the information encoded by the master card, which is 6 bits (2^6 = 64, enough to specific a specific card in a 52-card deck) and scattering it across images that come after that so that you don’t have to have an image for the master card itself. You can scatter those bits any way you want, as long as there are six of them.

If you had even more images, you could do better, of course. At 8 Ben systems worth (8 * 52 * 52 = 21632 images), it is possible to do even better, because you’d be able to process 5 cards at a time (1 master, plus 2 pairs of regular cards), so efficiency of 5 cards in 2 images. An example of master card use for that would be:

Look at suit of master card and whether rank is even or odd, and that tells you how to encode cards 2 and 3:

spade even = BenA
spade odd = BenB
heart even = BenC
heart odd = BenD
club even = BenE
club odd = BenF
diamond even = BenG
diamond odd = BenH

divide the rank of the master card by two and discard any remainder. The result will tell you which system to use for cards 4 and 5:

0 = BenA
1 = BenB
2 = BenC
3 = BenD
4 = BenE
5 = BenF
6 = BenG

(You never encode cards 4 and 5 with BenH because there isn’t a rank above 13 (King)).

Basically same idea as I mentioned before. Upon decoding, the image for cards 2 and 3 will tell you the suit and the parity of the master card, and the image for cards 4 and 5 will give you the exact rank…

A similar idea can be applied for decimal digits. Let’s say you normally use a 3-digit system but you can’t imagine getting to 10,000 images, so a full 4-digit system is out of reach. You can only get to 4,000. Not bad, though. Here’s one way to take advantage of it, by observing that having 4,000 images is like having 2 extra bits per image to work with over having just 1,000, and that a decimal digit can be completely specified by 4 bits (because 2^4 = 16 is bigger than 10.) You scatter those four bits of a “master digit” among the next 6 digits.

method:

Convert the master digit to binary:

0 = 0000
1 = 0001
2 = 0010
3 = 0011
4 = 0100
5 = 0101
6 = 0110
7 = 0111
8 = 1000
9 = 1001

Okay, time to encode digits 2, 3, 4. Which system to use? Look at the 1st 2 bits above:

00 = BenA
01 = BenB
10 = BenC

(BenD will not be used for digits 2-4. It’s not needed for decimal digits.)

Next, look the last 2 digits of the binary representation of the master digit. This says how to encode digits 5, 6, 7 in your sequence:

00 = BenA
01 = BenB
10 = BenC
11 = BenD

Now you can forget the master digit. You will process digits 7 at a time, but you get 7 digits in 2 images instead of just 6.

BTW, even at 2,000 images you could get something, but it doesn’t save much. You’d have to distribute the 4 bits of the master digit across the next 3 * 4 = 12 digits, so you’d only be able to forget 1 digit in 13, which likely isn’t worth it.

For a 2-digit system, let say you had 400 images before you gave up on repeating Ben. I’ll call them SmallA, SmallB, SmallC, and SmallD, 100 images each.

Master digit binary conversion as above, generally same as above except you process 5 digits at a time and discard the master, so that’s 5 digits in 2 images rather than the 4:2 ratio one would get would only 100 images. Since in competition the rows are a multiple of 5, this might be worth considering. But if you have 400 images, it might be really tempting to put in the extra work to make a 1000 image system.

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Dude, that’s brilliant. for 13 values, use 4 groups of images to distribute the numeric data among 2 images. God, that is a seriously clever idea. It would be a terrible nightmare to create and implement, but given a decade of work, it seems like a person really could save time. A person should allow ten years or so to learn to plqy the piano the way they want…You would have to weigh it against the continuous learning curve of a simpler system, though :frowning: Even if it weren’t advantageous in practice, it’s at least a pleasure to think about!

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Here’s another idea if you already have 1000 images for 3-digit numbers and also a 3x3 binary system (nine bits).

You can get 3 cards in 2 images this way. The 1st image comes from your digit system. It is simply:

D1 = Rank1 mod 10
D2 = Rank2 mod 10
D3 = Rank3 mod 10

So J = 1, Q = 2, K = 3.

So 2-3-2 just becomes 232. Rank(Q) = 12, so Q-3-2 also becomes 232 (as does 2-K-2 or 2-K-Q, etc. your 1st images loses some rank information, but we’ll fix that up with the binary images.)

The binary digits for your next image can be done card by card, so it’s simple. Each card will have 3 binary digits:

Binary Digit 1 = 1 if card is face card, 0 otherwise (this puts back the info lost in the decimal image)
Binary Digit 2 = 1 if card is red, 0 if black
Binary Digit 3 = 1 if card is diamond or spade, 0 otherwise (maybe you think diamonds and spades are both pointy like ‘1’ and clubs and hearts are more rounded like ‘0’.

Examples:

Example 1:

2​:spades: 3​:heart: 2​:clubs::

1st image: Decimal image (from your 1000 images for digits): 232

2nd image: Binary image (from your 3x3 binary digit system):

001
010
000

Example 2:

Q​:spades: 3​:heart: K​:diamonds::

1st image: Decimal image (from your 1000 images for digits): 233

2nd image: Binary image (from your 3x3 binary digit system):

101
000
111

Hard parts about this:

  1. Prerequisites that you already have a) a 3-digit decimal system (like Ben) and b) at least a 9 binary digit binary system.

  2. Looking at 3 cards at once.

  3. If using a 3x3, “seeing” the 9 binary digits from that are shown above would require massive amounts of practice.

  4. Reconstruction. Fortunately the reconstruction can be done 1 card at a time.

Advantages: Two images, 3 cards.

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Quick way to make 3 card system:

You just need 52 distinguishable

  1. O - objects
  2. S - skins
  3. P - faces/heads
The shape of the object remains, but it has corresponding skin and a place where you put the face/head. For example: pear + hedhehog's skin + Jim Carrey = a pear that has spikes protruding out everywhere, and on top it has Jim Carrey's funny face. Later this hybrid acts as a whole when it performs the A.

Some of the skins you can even take from the objects themselves (e.g. one ‘object’ could have been a hedgehog). And they can also be simply patterns, e.g. spirals or stripes,or compound like sand (a sandy/stripy pear). They can also have movement (spinning spiral, flowing sand). Important is that they are distinguishable from each other.

I think it is more memorable than object or person alone, because

  1. you spend some time for combining the three cards (and thus you see the combination more time)
  2. it has both unique shape (thanks to object) and is also emotional (thanks to face)

I composed my system in two days (took a number from PAO, assigned it to card. P+O+skin/detail from the O if there was a good one, then built the resulting hybrid person that does its A). Finding memorable skins wasn’t that hard, but putting the head into right place, then creating the arms and legs and finally seeing the resulting hybrid acting as a whole took time (the hybrid must be able to perform the A-s). If someone will use this system for competing, then for greater speed I think it’s necessarry to play all the 2704 OS combinations through in one’s mind, but putting the head to the place meant for it shouldn’t be tricky.

I’ll practise it and post my progression (atm I’m too slow recalling the POSA associated with the card). What I use now is 6 cards per locus - OSPAPS. OSP does the A to (predetermined) detail of its locus, the detail has P head that reacts emotionally and then the detail’s skin changes into S. Oh, and I have different color for each suit (if possible the O and S of a card are colored respectively).

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I know I’m going sound pretty basic, but can’t you just use the PAO and Dominic system. You’ll have 13 stage palace with PAO. Then have 4 locations and then the other two suits use the Dominic system.

Ex: K(hearts), 7(spades), J(clubs)

K= King sitting on a throne 7= Samurai unsheathing a sword J= Johnny Knoxville riding in a shopping cart

For first suit indicates location: Spades= Ocean Clubs= Forest Diamonds= Casino Hearts= My house

For the second and third suit use Dominic system to recall a person.

So the image would yield= A king unsheathing a shopping cart at my house with Cousin Skeeter (SC=63)

Idk just a thought

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52 X 51 is correct for pair-card-image.It’s 52X51X50 for 3-card-image.i just thought about it last night and found this post right now.im a very beginner (since may 2020).ill be here with a less-effort method.Thank You

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