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Base 60 encoding of positive floating point numbers in IDL

Thursday, February 02, 2017

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)
FotV*
FdiDx
HdzS*
IDL> base60(a, precision=8)
FotV**aO
FdiDx*^c
HdzS****
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
endfor
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'
endif
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))])
endfor
endelse
return, res
end

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