SEDOL
From Rosetta Code
Programming Task
This is a programming task. It lays out a problem which Rosetta Code users are encouraged to solve, using languages they know.
710889 B0YBKJ 406566 B0YBLH 228276 B0YBKL 557910 B0YBKR 585284 B0YBKTProduce this output:
7108899 B0YBKJ7 4065663 B0YBLH2 2282765 B0YBKL9 5579107 B0YBKR5 5852842 B0YBKT7
Contents |
[edit] Ada
with Ada.Text_IO; use Ada.Text_IO; procedure Test_SEDOL is subtype SEDOL_String is String (1..6); type SEDOL_Sum is range 0..9; function Check (SEDOL : SEDOL_String) return SEDOL_Sum is Weight : constant array (SEDOL_String'Range) of Integer := (1,3,1,7,3,9); Sum : Integer := 0; Item : Integer; begin for Index in SEDOL'Range loop Item := Character'Pos (SEDOL (Index)); case Item is when Character'Pos ('0')..Character'Pos ('9') => Item := Item - Character'Pos ('0'); when Character'Pos ('B')..Character'Pos ('D') | Character'Pos ('F')..Character'Pos ('H') | Character'Pos ('J')..Character'Pos ('N') | Character'Pos ('P')..Character'Pos ('T') | Character'Pos ('V')..Character'Pos ('Z') => Item := Item - Character'Pos ('A') + 10; when others => raise Constraint_Error; end case; Sum := Sum + Item * Weight (Index); end loop; return SEDOL_Sum ((-Sum) mod 10); end Check; Test : constant array (1..10) of SEDOL_String := ( "710889", "B0YBKJ", "406566", "B0YBLH", "228276", "B0YBKL", "557910", "B0YBKR", "585284", "B0YBKT" ); begin for Index in Test'Range loop Put_Line (Test (Index) & Character'Val (Character'Pos ('0') + Check (Test (Index)))); end loop; end Test_SEDOL;
The function Check raises Constraint_Error upon an invalid input. The calculated sum is trimmed using (-sum) mod 10, which is mathematically equivalent to (10 - (sum mod 10)) mod 10.
Sample output:
7108899 B0YBKJ7 4065663 B0YBLH2 2282765 B0YBKL9 5579107 B0YBKR5 5852842 B0YBKT7
[edit] BASIC
Works with: QuickBasic version 4.5
DECLARE FUNCTION getSedolCheckDigit! (str AS STRING) DO INPUT a$ PRINT a$ + STR$(getSedolCheckDigit(a$)) LOOP WHILE a$ <> "" FUNCTION getSedolCheckDigit (str AS STRING) IF LEN(str) <> 6 THEN PRINT "Six chars only please" EXIT FUNCTION END IF str = UCASE$(str) DIM mult(6) AS INTEGER mult(1) = 1: mult(2) = 3: mult(3) = 1 mult(4) = 7: mult(5) = 3: mult(6) = 9 total = 0 FOR i = 1 TO 6 s$ = MID$(str, i, 1) IF s$ = "A" OR s$ = "E" OR s$ = "I" OR s$ = "O" OR s$ = "U" THEN PRINT "No vowels" EXIT FUNCTION END IF IF ASC(s$) >= 48 AND ASC(s$) <= 57 THEN total = total + VAL(s$) * mult(i) ELSE total = total + (ASC(s$) - 55) * mult(i) END IF NEXT i getSedolCheckDigit = (10 - (total MOD 10)) MOD 10 END FUNCTION
[edit] Forth
create weight 1 , 3 , 1 , 7 , 3 , 9 ,
: char>num ( '0-9A-Z' -- 0..35 )
dup [char] 9 > 7 and - [char] 0 - ;
: check+ ( sedol -- sedol' )
6 <> abort" wrong SEDOL length"
0 ( sum )
6 0 do
over I + c@ char>num
weight I cells + @ *
+
loop
10 mod 10 swap - 10 mod [char] 0 +
over 6 + c! 7 ;
: sedol" [char] " parse check+ type ; sedol" 710889" 7108899 ok sedol" B0YBKJ" B0YBKJ7 ok sedol" 406566" 4065663 ok sedol" B0YBLH" B0YBLH2 ok sedol" 228276" 2282765 ok sedol" B0YBKL" B0YBKL9 ok sedol" 557910" 5579107 ok sedol" B0YBKR" B0YBKR5 ok sedol" 585284" 5852842 ok sedol" B0YBKT" B0YBKT7 ok
[edit] Fortran
Works with: Fortran version 90 and later
MODULE SEDOL_CHECK
IMPLICIT NONE
CONTAINS
FUNCTION Checkdigit(c)
CHARACTER :: Checkdigit
CHARACTER(6), INTENT(IN) :: c
CHARACTER(36) :: alpha = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"
INTEGER, DIMENSION(6) :: weights = (/ 1, 3, 1, 7, 3, 9 /), temp
INTEGER :: i, n
DO i = 1, 6
temp(i) = INDEX(alpha, c(i:i)) - 1
END DO
temp = temp * weights
n = MOD(10 - (MOD(SUM(temp), 10)), 10)
Checkdigit = ACHAR(n + 48)
END FUNCTION Checkdigit
END MODULE SEDOL_CHECK
PROGRAM SEDOLTEST
USE SEDOL_CHECK
IMPLICIT NONE
CHARACTER(31) :: valid = "0123456789BCDFGHJKLMNPQRSTVWXYZ"
CHARACTER(6) :: codes(10) = (/ "710889", "B0YBKJ", "406566", "B0YBLH", "228276" , &
"B0YBKL", "557910", "B0YBKR", "585284", "B0YBKT" /)
CHARACTER(7) :: sedol
INTEGER :: i, invalid
DO i = 1, 10
invalid = VERIFY(codes(i), valid)
IF (invalid == 0) THEN
sedol = codes(i)
sedol(7:7) = Checkdigit(codes(i))
ELSE
sedol = "INVALID"
END IF
WRITE(*, "(2A9)") codes(i), sedol
END DO
END PROGRAM SEDOLTEST
Output
710889 7108899 B0YBKJ B0YBKJ7 406566 4065663 B0YBLH B0YBLH2 228276 2282765 B0YBKL B0YBKL9 557910 5579107 B0YBKR B0YBKR5 585284 5852842 B0YBKT B0YBKT7
[edit] J
There are several ways to perform this in J. This most closely follows the algorithmic description at Wikipedia:
sn =. '0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ' ac0 =: (, 10 | 1 3 1 7 3 9 +/@:* -)&.(sn i. |:)
However, so J is so concise, having written the above, it becomes clear that the negation (-) is unneccsary.
The fundamental operation is the linear combination (+/@:*) and neither argument is "special". In particular, the coefficients are just another array participating in the calculation, and there's no reason we can't modify them as easily as the input array. Having this insight, it is obvious that manipulating the coefficients, rather than the input array, will be more efficient (because the coefficients are fixed at small size, while the input array can be arbitrarily large).
Which leads us to this more efficient formulation:
ac1 =: (, 10 | (10 - 1 3 1 7 3 9) +/@:* ])&.(sn i. |:)
which reduces to:
ac1 =: (, 10 | 9 7 9 3 7 1 +/@:* ])&.(sn i. |:)
Which is just as concise as ac0, but faster.
Following this train of thought, our array thinking leads us to realize that even the modulous isn't neccesary. The number of SEDOL numbers is finite, as is the number of coefficients; therefore the number of possible linear combinations of these is finite. In fact, there are only 841 possible outcomes. This is a small number, and can be efficiently stored as a lookup table (even better, since the outcomes will be mod 10, they are restricted to the digits 0-9, and they repeat).
Which leads us to:
ac2 =. (,"1 0 (841 $ '0987654321') {~ 1 3 1 7 3 9 +/ .*~ sn i. ])
Which is more than twice as fast as even the optimized formulation (ac1), though it is slightly longer.
[edit] Java
import java.util.Scanner; public class SEDOL{ public static void main(String[] args){ Scanner sc = new Scanner(System.in); while(sc.hasNext()){ String sedol = sc.next(); System.out.println(sedol + getSedolCheckDigit(sedol)); } } public static int getSedolCheckDigit(String str){ if(str.length() != 6){ System.err.println("Six chars only please"); return -1; } str = str.toUpperCase(); int[] mult = {1, 3, 1, 7, 3, 9}; int total = 0; for(int i = 0;i < 6; i++){ char s = str.charAt(i); if(s == 'A' || s == 'E' || s == 'I' || s == 'O' ||s == 'U'){ System.err.println("No vowels"); return -1; } total += (Character.isDigit(s) ? Character.digit(s, 10) : s - 55) * mult[i]; } return (10 - (total % 10)) % 10; } }
[edit] Perl
This program reads from standard input.
sub sum {my $n = 0; $n += $_ foreach @_; return $n;} sub zip {my $f = shift; my @a = @{shift()}; my @b = @{shift()}; my @result = (); push(@result, $f->(shift @a, shift @b)) while @a and @b; return @result;} sub char_to_v {my $c = shift; $c =~ /[A-Z]/ and $c = ord($c) - ord('A') + 10; return $c;} my @weights = (1, 3, 1, 7, 3, 9); sub sedol {my $s = shift; my @vs = map {char_to_v $_} split //, $s; my $checksum = sum (zip sub {$_[0] * $_[1]}, \@vs, \@weights); my $check_digit = (10 - $checksum % 10) % 10; return $s . $check_digit;} while (<>) {chomp; print sedol($_), "\n";}
[edit] Python
import string # constants sedolchars = string.digits + string.ascii_uppercase sedol2value = dict((ch, n) for n,ch in enumerate(sedolchars)) for ch in 'AEIOU': del sedol2value[ch] sedolchars = sorted(sedol2value.keys()) sedolweight = [1,3,1,7,3,9,1] def checksum(sedol): tmp = sum(sedol2value[ch] * sedolweight[n] for n,ch in enumerate(sedol[:6]) ) return sedolchars[ (10 - (tmp % 10)) % 10] for sedol in ''' 710889 B0YBKJ 406566 B0YBLH 228276 B0YBKL 557910 B0YBKR 585284 B0YBKT '''.split(): print sedol + checksum(sedol)
Categories: Programming Tasks | Encryption | Ada | BASIC | Forth | Fortran | J | Java | Perl | Python
