Toward More Efficient Number Mnemonics

By Keith, Mike | Word Ways, August 2009 | Go to article overview

Toward More Efficient Number Mnemonics


Keith, Mike, Word Ways


As pointed out by Ross Eckler in the Nov 2008 Word Ways ("Mnemonics for Number Sequences", p. 297), the well-known type of mnemonic which uses the length of successive words to represent a sequence of decimal digits (with a ten-letter word for each occurrence of zero) is not particularly efficient. If the sequence of digits being represented has a uniform probability distribution (as is the case for the digits of [pi] and e, for example) then in the long run this scheme will have an "inefficiency ratio", defined as (total number of letters used)/(number of digits represented), of 5.5. This is pretty far away from 1, which could be considered ideal in some sense.

Here is a poem that captures the first 134 digits of [pi] with somewhat better efficiency.

   Darkness: heavy, dull, silky but somehow grotesque,
   Interred within a frozen cell
   Lined in macabre skin-framed pallets.

   He wipes blood, old as an ache,
   Pokes at a fly in disgust.
   Chimes urge religious locals to vows of love,
   Knees showing a form of loyalty he privately lacks.

   Cramped, heavily bound, he heaves
   A calm death-mocking word,
   Pushing closer to tears.
   Quiet scrapes over roads by shrill axles die,
   Soundless as a scarecrow.

   Voices of bygone folks flow swiftly
   Over grave, granite, or greenwood,
   Dying in a December sky.

To extract the digits of [pi] from this text, follow these simple rules:

(1) Take each word of three or more letters from the text, in order.

(2) Extract two digits from each word ("first digit" and "second digit") like this:

* Calculate the word's score in the game of Scrabble by adding up the score of the individual letters. The fight-most digit of the word score is the first digit.

* Add up the numerical values of the letters (using A=1, B=2, C=3, etc.). The right-most digit of this sum is the second digit.

Here is how this works out for the first few words of the poem:

Word         Letter sum (= second digit)   Letter sum (= second digit)

DARKNESS     4+1+18+11+14+5+19+19 = 91     4+1+18+11+14+5+19+19 = 91
HEAVY        8+5+1+22+25 = 61              8+5+1+22+25 = 61
DULL         4+21+12+12 = 49               4+21+12+12 = 49
SILKY        19+9+12+11+25 = 76            19+9+12+11+25 = 76

with the underlined right-most digits making 3,1,4,1,5,9,2,6, the first eight digits of [pi]. …

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