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10,000 Image Memory Systems

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For an exam of many revision facts, it would be nice to have many hundreds of imaginary places at which to visualise the facts; but it would take time to learn those places themselves as well as taking time to learn them in sequence.

I tried a few ways to make a system of many images which occur in sequence. These places would be the sequential order in which facts need to be recalled - like the BLOKES system but with a bigger set of sequential images.

One image location system idea is to use more than 5 images for each of the town 100 shops. If I had 10 images of LIbraries or of parts of a LIbrary, and I learned them in order then the 100 place town system could represent 1000 places.

With a big sequential image set, you want the images to be sufficiently unique so that your recall does not confuse one image with another. However, that is challenging!

See the 'Picture Frame 10000' article for an ambitious idea for building a 10,000 location system.

As for the items which someone could imagine at mnemonic imagined places, could there be a 10,000 item system where any one item represents a 4 digit number?

I compromised on that and I think that a 5000 item system would be a more efficient idea. If I wanted to represent the number 5001 then I would use the image I have for 0001 and use the context around it to imply that 5000 needs to be added to it.

I also thought that a 10 digit number might be represented by a 1/1000 person doing a 1/1000 action on a 4 digit number person. If I have two sets of 1000 people then my choice of set determines if my 1/5000 person needs to have a + 5000 added on.

Below and in the Maths.. article, I write more about what maths answers can be cryptically stored in those images: in the hair/hat and colour; and the clothes outfit and colours.

I actually have all the pieces I need to make that system - I'm just lacking spare time!

For people memorising cards, a 52 x 52 system could be represented mostly by using a subset of the people from the 50 x 100 = 5000 system.



In the image below, you see how the 0-24 hair and hats from the Male 1000 system; and the 0-24 hair and hats of the Women 1000 system can be used to represent numbers 00-49. This required making male SH be a shaven head with a single Mohican central section, male CH be a 'comb over' from the man's right to left of a few strands of hair; and a braided hair 'SH' image for a woman; and a dangling cork hat for a 'CH' woman image.

P and Q are not in the 0-24 nor the 25-49 range, of themselves.

Colours 0 to 9 can be put on the end of the 2 digit hair style to make a 3 digit number from 000 to 499.

Now consider colours 0-9 and colours 10 to 19. If the colour of the hair is in the range 10-19 then add 500 to the three digit number.

In this way, a coloured hair style can mean a number from 000 to 999. The answer to multiplication problems is a 4 digit answer. eg. 5 x 8 = 0040 . What the hairstyle would represent is the 004; and you would have to work out the final 0 by using normal maths. That sounds silly for 5 x 8 but it makes sense for 55 x 88: you know the final digit is the same final digit as 5 x 8 : the '0' of 40. Put another way, the rightmost digit of a multiplication problem is the easiest to work out because it has the least steps to solve.

With a male outfit, a male outfit colour, and the top outfit digit and the lower half outfit digit form a three digit number also.

I have 1000 male face images (facing at about 30 degrees); and the idea is that different hair styles can make person 000 also be person 1000, 2000, 3000 ,4000 but with different hair. However, whenever I multiply by 0 in a calculation, the answer is 0 and that means that I need a rule to vary the hairstyle but for the person doing maths to know that 0 is the answer; and to not interpret the hairstyle as a number. If the first two digits of a 0000 to 49999 number are 00 (pronounced as syllable 'BO') or '01' ('BI') then use hair and clothing based on a different set of rules: pi digits. So person 00 x 00 is the 141592 of 3.141592 : 141 as the hair and 592 as the clothes. There is also simple maths if the last syllable of a name word is BI (the 01 syllable); someone with the name 'CHEBO' would be multiplication maths using 00 ('BO') and so its answer is 0; someone with the name 'DOBI' would be a multiplication by 01 ('BI'); so the answer is the 'DO'part. The first syllable being a "BO" or "BI" could be approached like that too since the answer returned is too simple to bother memorising.

00 x 01 is going to be the next 6 digits of pi: 653589 expressed as hair 653-500 [see the 500 rule above giving 153]; and 589 for clothes.

In about 40 cases, a face and its hair will be duplicating another person's face and hair. I would use Facegen faces to produce about 40 people to represent them:

N1 N2 Male hair or hat and colour Coloured clothes DDDD DDD
20 00 000 000 2000 000
30 00 000 000 3000 000
40 00 000 000 4000 000
11 00 000 000 1100 100
21 00 000 000 2100 100
31 00 000 000 3100 100
41 00 000 000 4100 100
41 12 049 109 4112 112
12 00 000 000 1200 200
22 00 000 000 2200 200
32 00 000 000 3200 200
42 00 000 000 4200 200
13 00 000 000 1300 300
23 00 000 000 2300 300
33 00 000 000 3300 300
43 00 000 000 4300 300
14 00 000 000 1400 400
24 00 000 000 2400 400
34 00 000 000 3400 400
44 00 000 000 4400 400
15 00 000 000 1500 500
25 00 000 000 2500 500
35 00 000 000 3500 500
45 00 000 000 4500 500
16 00 000 000 1600 600
26 00 000 000 2600 600
36 00 000 000 3600 600
46 00 000 000 4600 600
17 00 000 000 1700 700
27 00 000 000 2700 700
37 00 000 000 3700 700
47 00 000 000 4700 700
18 00 000 000 1800 800
28 00 000 000 2800 800
38 00 000 000 3800 800
48 00 000 000 4800 800
19 00 000 000 1900 900
29 00 000 000 2900 900
39 00 000 000 3900 900
49 00 000 000 4900 900

The first 1000 people's data details: 0000 to 0999

1000-1999 people's data details.

2000-2999 people's data details.

3000-3999 people's data details.

4000-4999 people's data details.

The images of people give the first 3 digits of a 4 digit answer.

45 x 76 is 3420 but it would just give you the 342 of the left part of the answer.

Even 8 x 5 would be served up an answer of 004 since 0040 requires the 4th digit!

Below is a table of syllable endings and the kind of 4th digit you get when multiplying with them. But you have to exclude any calculation with BO or BI since they are easy to solve (as mentioned above).

Also, I present an alphabetical order. So DEHA (an E and an A) would be looked for down the table as A followed by E:

Note: More O and I 'syllable ending' logic is presented in a table further down this article.

Digit 1 Digit 2 Answer
Examples:
A A 4

A A 9 7 x 7 (hair or hat of ) HAZA, TAZA (clothes of HAZA)
A E 6

A E 1 7 x 3 or 3 x 7 KALE, LEKA, PAGE
A U 8

A U 3 7 x 9 or 9 x 7 CHACHU, HUHA
E E 4

E E 9 3 x 3 JELE
E U 2

E U 7 3 x 9 or 9 x 3 WUYE
U U 6

U U 1 9 x 9 HUPU

The A E pattern normally gives the answer 6 as the fourth digit but 7 x 3 or 3 x 7 is the exception; and DEHA is a 7 times a 7. Look up DE and HA in the table below to see what I mean:

Digits Syllable Ends in 3 or 7 or 9
3 BE 3
7 CHA 7
9 CHU 9
13 DE 3
17 FA 7
19 FU 9
23 GE 3
27 HA 7
29 HU 9
33 JE 3
37 KA 7
39 KU 9
43 LE 3
47 MA 7
49 MU 9
53 NE 3
57 PA 7
59 PU 9
63 RE 3
67 SA 7
69 SU 9
73 SHE 3
77 TA 7
79 TU 9
83 VE 3
87 WA 7
89 WU 9
93 YE 3
97 ZA 7
99 ZU 9

By knowing the table of answers and knowing rote the syllables which end in 3, 7 or 9, I can quickly decide what the fourth digit is, with practice!

Because DEHA is the same as saying, "What is 13 x 27, the real world answer is 0351 but the image of a person (the coloured hair or hat) gives 035. I get the 1 by considering the EA in the table [alphabetically: AE] and noting that a 3 and 7 invoke the exception: the end digit is 1 rather than 6.

An additional awareness of which syllables have digits ending in 1, 5 and 6 will help with maths involving syllables that end in O or I [and are not BO or BI]:

[Again, alphabetical sorting so that, for instance, FIJU (16 x 34) is IU and JUFI is arranged to be IU also]

Multiply This By This Answer End Digit
A I 7
A I 2
A O 5
A O 0
E I 8
E I 3 3 x 1, 1 x 3
E O 5 3 x 5, 5 x 3
E O 0
I I 6 if there is a 6 in either part of the question: 1 x 6 , 6 x 1, 6 x 6
I I 1
I O 0
I O 5 1 x 5, 5 x 0
I U 9 1 x 9, 9 x 1
I U 4
O O 0
O O 5 5 x 5
O U 0
O U 5 5 x 9, 9 x 5

I also gave the people surnames to do some additions maths.

What do the bald heads of these people look like before hair or hats are added? See my article about Face Generating Software.