533 Rate this article:
No rating

Base 60 encoding of positive floating point numbers in IDL

Atle Borsholm

Here is an example of representing numbers efficiently using a restricted set of symbols. I am using a set of 60 symbols (or characters) to encode floating point numbers as strings of any selected length. The longer the strings are, the more precise the numbers will potentially be.


Here is an example of a representation, this is restricted to positive numbers, in order to keep the example short.

IDL> a=[14.33, 3.1415, 12345]
IDL> a
       14.330000       3.1415000       12345.000
IDL> base60(a)
IDL> base60(a, precision=8)
IDL> base60(base60(a)) - a
 -4.5533356836102712e-006 -4.6258149324351905e-006    -0.016666666666424135
IDL> base60(base60(a, precision=8)) - a
 -9.2104102122902987e-012 -4.6052051061451493e-013 -7.7159711509011686e-008
In this example, it can be seen that the 5-digit representations are not as close to the original numbers as the 8-digit representations.
The code example for the base60 function is listed below.
; Converts from a numeric type to a base 60 representation
; Converts from a base 60 string to a floating point representation
; PRECISION is only used to determine how many symbols to use when encoding,
; and is ignored for decoding.
function Base60, input, precision=precision
  compile_opt idl2,logical_predicate
  ; set default precision of 5 digits for encoding only
  if ~keyword_set(precision) then precision = 5
  ; base 60 symbology
  symbols = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ!@#$%^&*'
  base = strlen(symbols)
  ; fast conversion from symbol to value
  lut = bytarr(256)
  lut[byte(symbols)] = bindgen(base)
  if isa(input, /string) then begin
    ; convert from base60 string to float
    ; find exponent first
    scale = replicate(double(base),n_elements(input)) ^ $
      (lut[byte(strmid(input,0,1))] - base/2)
    res = dblarr(n_elements(input))
    for i=max(strlen(input))-1,1,-1 do begin
      dig = lut[byte(strmid(input,i,1))]
      res += dig
      res /= base
    res *= scale
  endif else begin
    ; convert from float to base60 strings
    ; encode exponent(scale) first
    ex = intarr(n_elements(input))
    arr = input
    dbase = double(base)
    repeat begin
      dec = fix(arr ge 1)
      ex += dec
      arr *= dbase ^ (-dec)
      inc = fix(arr lt 1/dbase)
      ex -= inc
      arr *= dbase ^ inc
    endrep until array_equal(arr lt 1 and arr ge 1/dbase,1b)
    if max(ex) ge base/2 || min(ex) lt -base/2 then begin
      message, 'Number is outside representable range'
    bsym = byte(symbols)
    res = string(bsym[reform(ex+base/2,1,n_elements(ex))])
    for i=1,precision-1 do begin
      arr *= base
      fl = floor(arr)
      arr -= fl
      res += string(bsym[reform(fl,1,n_elements(fl))])
  return, res

Please login or register to post comments.