...Ordinal numbers (or ordinal numerals) are words representing position or rank in a sequential order.
Your task is, using 2 separate programs (of which can be made from 2 different languages), to output the ordinal sequence from first to
nth. You will be outputting the full word
second as opposed to
The challenge of ordinal numbers has been brought up before, particularly in this entry. In this challenge, ordinals are merely a vehicle to facilitate the unique conditions detailed below.
You must make a program that, when given the input of
n must output anything.
n will always be a positive, non-zero integer no larger than 999.
Valid output includes but is not limited to:
- Any output to
- Creation of files / folders / etc
- A graphical interface or images of any kind
You must make a program that uses the output of part 1's program to output a sequence of ordinal numbers, starting from 1 (first), up to whatever
n was parsed in part 1.
- The total bytes for part 2 must not exceed the total bytes for part 1 (less than or equal to).
- Not case sensitive.
- Output must contain only the ordinal sequence (only characters a-Z) and whitespace (newlines allowed).
- Can be output to any source, so long as it is visible either during or after execution.
- Program does not need to terminate so long as its output is correct.
- Output is not required to have any grammar, but may optionally include it (hyphens, commas, "ands", etc).
nine hundred ninety ninthis just as acceptable as
nine hundred and ninety-ninth.
n is 8
FIRST SECOND THIRD FOURTH FIFTH SIXTH SEVENTH EIGHTH
The hierarchy of win conditions is:
- The lowest number of bytes in part 1
- The lowest number of bytes in part 2
Entry #1 | Part 1 = 32 bytes, Part 2 = 22 bytes Entry #2 | Part 1 = 31 bytes, part 2 = 30 bytes Entry #2 wins - Part 1 contains 31 bytes vs 32 bytes --- Entry #1 | Part 1 = 21 bytes, Part 2 = 33 bytes Entry #2 | Part 1 = 80 bytes, Part 2 = 70 bytes Entry #2 wins - Entry #1 disqualified (Part 2 contains more bytes than Part 1) --- Entry #1 | Part 1 = 50 bytes, Part 2 = 49 bytes Entry #2 | Part 1 = 50 bytes, Part 2 = 50 bytes Entry #1 wins - Part 1 is equal, Part 2 contains 49 bytes vs 50 bytes