SEDOLs
You are encouraged to solve this task according to the task description, using any language you may know.
For each number list of 6-digit SEDOLs, calculate and append the checksum digit. That is, given this input:
710889 B0YBKJ 406566 B0YBLH 228276 B0YBKL 557910 B0YBKR 585284 B0YBKT
Produce this output:
7108899 B0YBKJ7 4065663 B0YBLH2 2282765 B0YBKL9 5579107 B0YBKR5 5852842 B0YBKT7
Ada
<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; </ada> 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
BASIC
<qbasic>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</qbasic>
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
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.
Java
<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) ? Integer.valueOf(""+s) : s - 55) * mult[i]; } return (10 - (total % 10)) % 10; } }</java>
Python
<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)
</python>