Haskell logbase error

Question

I am trying to calculate the length of an Integer in Haskell, using the fact that the length is equal to truncate (log10(x)+1).

Using Integers I created:

len :: Integer -> Integer
len i = toInteger (truncate (logBase 10 (fromIntegral i)) + 1)

Unfortunately, not all numbers get the correct length. I tried a few different cases and found that:

logBase 10 10         = 1.0
logBase 10 100        = 2.0
logBase 10 1000       = 2.9999..6
logBase 10 10000      = 4.0
logBase 10 100000     = 5.0
logBase 10 1000000    = 5.9999999

Is there a reason why logBase 10 1000 doesn't return 3.0? How do I get the correct log-value for 1000 in base 10?

Columbous

logBase is defined as log y / log x; the division is probably the culprit, as while the log will be correct (with regard to roundings), the division of them doesn't have to be. – Osteogenesis

@BartekBanachewicz So there is no way to get the correct result? Tried using (log 1000) / (log 10), but still 2.99996. Guess that the log-values might be rounded down for 1000 and up for 10, thus it will be slightly smaller than 3. – Columbous

Suppose log is not necessary, you could avoid floating point operation with repeatedly divide by 10 (and higher power of 10). – Hittite

@Columbous The easiest way to get the correct result could be to use numbers package, and call logBase 10 1000 :: BigFloat Prec50, then round that into float or double. Easiest, but not necessarily the best. – Osteogenesis

If you don't need Double Floating precision then use Float type instead and it seems fine. Such as logBase 10 (1000 :: Float) would return 3.0 or functionally logBase 10 . (fromInteger :: Integer -> Float) $ 1000 would do the same. – Layton

Answer 1

There is an integer log base function in GHC modules which has type Integer -> Integer -> Int#.

Example usage:

{-# LANGUAGE MagicHash #-}

import Control.Monad
import GHC.Integer.Logarithms ( integerLogBase# )
import GHC.Exts (Int(..))

main = do
  forM_ [(1::Int)..20] $ \n -> do
    let a = 10^n-1
        la = I# (integerLogBase# 10 a)
        b = 10^n
        lb = I# (integerLogBase# 10 b)
    putStrLn $ show a ++ " -> " ++ show la
    putStrLn $ show b ++ " -> " ++ show lb

Output:

9 -> 0
10 -> 1
99 -> 1
100 -> 2
999 -> 2
1000 -> 3
9999 -> 3
10000 -> 4
99999 -> 4
100000 -> 5
999999 -> 5
1000000 -> 6
9999999 -> 6
...
9999999999999999999 -> 18
10000000000000000000 -> 19
99999999999999999999 -> 19
100000000000000000000 -> 20

Chocolate

Answer 2

If you don't need Double Floating precision then use Float type instead and it seems fine. Such as logBase 10 (1000 :: Float) would return 3.0 or functionally logBase 10 . (fromInteger :: Integer -> Float) $ 1000 would do the same.

As per your code toInteger seems redundant since truncate :: (Integral b, RealFrac a) => a -> b already does that job. So may simply do like

len :: Integer -> Integer
len = (+1) . truncate . logBase 10 . (fromIntegral :: Integer -> Float)

This will work correctly up until 9999987.

Layton

Answer 3

Can't you just go

amountOfDigits = length . show

(if you only interested in amount of digits in traditional base 10)

Demetriusdemeyer

This would fail for numbers with leading zeros. – Englishism

Answer 4

len :: Integer -> Integer
len x
  | x <= 1 = 1 -- doesn't handle negative numbers
  | otherwise = ceiling (logBase 10 (fromIntegral x :: Float))

Englishism