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27210131191041119112111115115101115115101115
31161041015115112101971142111102710010111511
61051101215104111108100115311610410141029711
61012111102311610410151191111141081002105110
3104105115510497110100115

 No Solution Yet Submitted by Juggler Rating: 3.0000 (3 votes)

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 Indeed, decryption verified. Comment 7 of 7 |

DATA     27210131191041119112111115115101115115101115
DATA     31161041015115112101971142111102710010111511
DATA     61051101215104111108100115311610410141029711
DATA     61012111102311610410151191111141081002105110
DATA     3104105115510497110100115

FOR i = 1 TO 5
s\$ = s\$ + LTRIM\$(RTRIM\$(sn\$))
NEXT

DO
l = VAL(LEFT\$(s\$, 1))
s\$ = MID\$(s\$, 2)
FOR i = 1 TO l
IF LEFT\$(s\$, 1) < "3" THEN
a\$ = CHR\$(VAL(LEFT\$(s\$, 3)))
s\$ = MID\$(s\$, 4)
ELSE
a\$ = CHR\$(VAL(LEFT\$(s\$, 2)))
s\$ = MID\$(s\$, 3)
END IF
PRINT a\$;
NEXT
PRINT " ";
LOOP UNTIL s\$ = ""

produces

He who possesses the spear of destiny holds the fate of the world in his hands

I different variation would have deleted all the bolded digits and replaced all but the first with "32", and a "46" could have been placed at the end, as all meaningful 2-digit codes begin with 3 or higher, and 2-digit codes with 1 or 2.

 Posted by Charlie on 2005-09-28 20:30:28

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