Binary Number Memorization Systems
Memorizing binary data (e.g., 01110100010100100100…) is even easier than memorizing decimal numbers, because a larger number of digits can be encoded in a single location. These techniques work on any type of binary data, including zeros and ones or the order of black and red in a deck of cards.
These methods are very easy to learn, but probably aren’t the most efficient for memory competitions. If you want to memorize binary numbers for memory competitions, see the section below on Complex Binary Memorization Systems.
For three binary digits, there are eight possibilities:
- 000 = None = N
- 111 = All = A
- 001 = Bottom = B
- 100 = Top = T
- 011 = Lower = L
- 110 = Upper = U
- 101 = Outer = O
- 010 = Middle = M
110-101 becomes U and O. Make a mnemonic image with the letters U and O, for example, “unicorn” and “opossum”. Place those images in the first location in your Memory Palace, and then move on to the next six digits.
See this blog post for a full tutorial: Learn how to memorize the order of black & red in a deck of cards
This is an interesting system that was mentioned in Gary Lanier’s memory journal in a post titled, Binary Numbers You Can See. In this system, mnemonic images are created from the shapes of the binary numbers:
- 000 = the Three Stooges
- 001 = glasses staring at a wall
- 011 = bowling ball, 2-pin split
- 111 = three fence posts
- 110 = a bull’s two horns goring a matador’s face
- 100 = putter, golf ball, cup
- 010 = a canon with a wheel on each side
- 101 = a soccer ball making a goal
See also Number Shape System.
- 000 = 0
- 001 = 1
- 010 = 2
- 011 = 3
- 100 = 4
- 101 = 5
- 110 = 6
- 111 = 7
We recommend that you learn the correct way to count in binary. If you aren’t familiar with it, check out this page.
If you have a one-digit system like the Number Shape System and your mnemonic image for 4 is a flag, then the binary number 100 would be represented by a flag.
If you have a two-digit system like the Major System or Dominic System, and the mnemonic image for 45 is a werewolf, then the binary number 100-101 would be represented by a werewolf. If you are placing two images per locus in memory palaces, you can encode 12 binary digits per locus. If you are placing three images per locus, like in a [PAO system](/wiki/Person-Action-Object (PAO) System/), you can encode 18 binary digits per locus.
If you have a three-digit system, you could encode nine binary digits in a single image, and between 18 and 27 binary digits per locus.
A swan looks kind of a like the number 2, which is 010 in binary.
If binary numbers are combined into 3x3 grids of nine digits each, it only requires 512 images. The grids can be read from top to bottom, converting each row into a decimal digit.
For example, the following grid could use the same mnemonic image as 065, if you have a three-digit decimal number memorization system:
000 110 101
The reason for 065 is because:
- 000 in binary is 0 in decimal
- 110 in binary is 6 in decimal
- 101 in binary is 5 in decimal
The advantage to binary grids over something like the [PAO system](/wiki/Person-Action-Object (PAO) System/) is that your images will exactly fit the 30 columns in a competition row. I.e., 30 digit rows divided by 3 digits width per image is 10 images per row. In contrast, a 2-digit PAO system that encodes 18 binary digits per PAO set will have PAO sets that overlap the lines. The binary grid system will however only use 24 out of the 25 rows on a competition sheet (8x3). You can skip the last row, or use a different system on it.
The potential to miss 90 points for missing one image is one disadvantage to this system. This occurs because you take numbers from 3 rows for one image.
The differences in efficiency between this and the Ben System are that the grids only encode nine digits per image while the Ben System encodes 10. The grids require 512 images and the Ben System requires 1,024.
The Ben System can encode 30 binary digits per location, which is the exact number of digits in a row of binary numbers at memory competitions. The Ben System for binary numbers requires 1,024 mnemonic images and is significantly more complex that the simpler methods above. The technique is listed in the binary numbers section of the Ben System wiki page.
For people who are interested in competition memorization, here is a table that compares the efficiency of different binary number memorization systems. The columns are the type of system, the rows are the number of images per location in a memory palace, and the table cells contain the number of binary digits that can be stored at each locus. For example, if you use a 2-digit Major System and place three images per locus (e.g., PAO), you can encode 18 binary digits per locus.
|Lanier System||1-digit system||2-digit system||3-digit system||3x3 Grids||Ben System|
|1 image per location||3||3||6||9||9||10|
|2 images per location||6||6||12||18||18||20|
|3 images per location||9||9||18||27||27||30|
The number of images required also varies greatly:
- a 9-digit system (3-3-3) requires 512 images (888)
- a 10-digit system (4-3-3) requires 1,024 images (1688)
- an 11-digit system (4-4-3) requires 2,048 images (16168)
- a 12-digit system (4-4-4) requires 4,096 images (161616)
- How to Build a Memory Palace
- List of Memory Techniques
- Does Photographic Memory Exist?
- Memory Palace Alternatives
- Memory Palace Alternatives