Base58Check encoding: Difference between revisions

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===version 2===
===version 2===
does what the others do
does what the others do
<lang>/* REXX */
<lang rexx>/* REXX */
s="123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz"
s="123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz"
Numeric Digits 1000
Numeric Digits 1000

Revision as of 11:02, 14 August 2017


The popular encoding of small and medium-sized checksums is base16, that is more compact than usual base10 and is human readable... For checksums resulting in hash digests bigger than ~100 bits, the base16 is too long: base58 is shorter and (when using good alphabet) preserves secure human readability. The most popular alphabet of base58 is the variant used in bitcoin address (see Bitcoin/address validation), so it is the "default base58 alphabet".

Write a program that takes a checksum (resultant hash digest) integer binary representation as argument, and converts (encode it) into base58 with the standard Bitcoin alphabet — which uses an alphabet of the characters 0 .. 9, A ..Z, a .. z, but without the four characters 0, O, I and l.

The reference algorithm is at the Bitcoin's Base58Check page.

Haskell

<lang haskell>import Numeric (showIntAtBase)

chars = "123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz"

base58Encode :: Integer -> String base58Encode n = showIntAtBase 58 (chars !!) n ""

main :: IO () main = mapM_ (putStrLn . base58Encode)

       [25420294593250030202636073700053352635053786165627414518,
        0x61,
        0x626262,
        0x636363,
        0x73696d706c792061206c6f6e6720737472696e67,
        0x516b6fcd0f,
        0xbf4f89001e670274dd,
        0x572e4794,
        0xecac89cad93923c02321,
        0x10c8511e]</lang>
Output:
6UwLL9Risc3QfPqBUvKofHmBQ7wMtjvM
2g
a3gV
aPEr
2cFupjhnEsSn59qHXstmK2ffpLv2
ABnLTmg
3SEo3LWLoPntC
3EFU7m
EJDM8drfXA6uyA
Rt5zm

Perl 6

<lang perl6>sub encode_Base58 ( Int $x ) {

   constant @codes = <
         1 2 3 4 5 6 7 8 9
       A B C D E F G H   J K L M N   P Q R S T U V W X Y Z
       a b c d e f g h i j k   m n o p q r s t u v w x y z
   >;
   return @codes[ $x.polymod( 58 xx * ) ].join.flip;

}

my @tests =

   25420294593250030202636073700053352635053786165627414518 => '6UwLL9Risc3QfPqBUvKofHmBQ7wMtjvM',
   0x61                    => '2g',
   0x626262                => 'a3gV',
   0x636363                => 'aPEr',
   0x73696d706c792061206c6f6e6720737472696e67 => '2cFupjhnEsSn59qHXstmK2ffpLv2',
   0x516b6fcd0f            => 'ABnLTmg',
   0xbf4f89001e670274dd    => '3SEo3LWLoPntC',
   0x572e4794              => '3EFU7m',
   0xecac89cad93923c02321  => 'EJDM8drfXA6uyA',
   0x10c8511e              => 'Rt5zm',

use Test; for @tests {

   is encode_Base58(.key), .value, "{.key} encodes to {.value}";

} </lang>

REXX

version 1

Following the description in https://www.anintegratedworld.com/how-to-manually-calculate-base58check-encoding/ I get the result expected there. Apart for the leading 1 the program works also for the inputs shown above. <lang rexx>/* REXX */ s="123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz" Numeric Digits 100 k='00010966776006953D5567439E5E39F86A0D273BEED61967F6'x n=c2d(k) o= Do Until n=0

 rem=n//58
 n=n%58
 o=o||substr(s,rem+1,1)
 End

o=o||substr(s,1,1) Say reverse(o)</lang>

Output:
16UwLL9Risc3QfPqBUvKofHmBQ7wMtjvM

version 2

does what the others do <lang rexx>/* REXX */ s="123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz" Numeric Digits 1000 cnt_ok=0 Call test 'N',25420294593250030202636073700053352635053786165627414518,,

            '6UwLL9Risc3QfPqBUvKofHmBQ7wMtjvM'

Call test 'X','61'x ,'2g' Call test 'X','626262'x ,'a3gV' Call test 'X','636363'x ,'aPEr' Call test 'X','73696d706c792061206c6f6e6720737472696e67'x,,

             '2cFupjhnEsSn59qHXstmK2ffpLv2'

Call test 'X','516b6fcd0f'x ,'ABnLTmg' Call test 'X','bf4f89001e670274dd'x ,'3SEo3LWLoPntC' Call test 'X','572e4794'x ,'3EFU7m' Call test 'X','ecac89cad93923c02321'x ,'EJDM8drfXA6uyA' Call test 'X','10c8511e'x ,'Rt5zm' Call test 'X','10c8511e'x ,'check_error_handlimng' Say cnt_ok 'tests ok' Exit test:

 Parse Arg how,k,res
 If how='X' Then
   k=c2d(k)
 o=
 Do Until k=0
   rem=k//58
   k=k%58
   o=o||substr(s,rem+1,1)
   End
 o=reverse(o)
 If o=res Then cnt_ok+=1
 Else Do
   Say 'expected:' res
   Say 'found   :' o
   End
 Return</lang>
Output:
expected: check_error_handlimng
found   : Rt5zm
10 tests ok