Check whether letters of word are in alphabetical order
CJam, 8 bytes
lel_$_&=
Here is a test harness for all examples in the challenge. This returns 0
or 1
(which are falsy and truthy, respectively, in CJam).
And here is a script to filter the word list in the question (takes a few seconds to run). You'll have to copy the word list into the input field manually, because it's too long for a permalink.
Explanation
l "Read input.";
el "Convert to lower case.";
_$ "Get a copy and sort it.";
_& "Remove duplicates (by computing the set intersection with itself).";
= "Check for equality with original (lower case) word.";
Regex (any flavor), 55 bytes
Some people don't consider regex to be a programming language, but it's been used before, and it's not close to being the shortest.
^a?b?c?d?e?f?g?h?i?j?k?l?m?n?o?p?q?r?s?t?u?v?w?x?y?z?$
I've added one byte for the i
(case-insensitive) flag. This is very straightforward and might be shorter to generate on the fly.
If regex alone are not allowed, you can use this 56-byte Retina program suggested by Martin Büttner:
i`^a?b?c?d?e?f?g?h?i?j?k?l?m?n?o?p?q?r?s?t?u?v?w?x?y?z?$
Running this on the wordlist linked above yielded 10 6-letter words in alphabetical order.
["abhors", "almost", "begins", "begirt", "bijoux", "biopsy", "chimps", "chinos", "chintz", "ghosty"]
Python 3, 44 bytes
*s,=input().lower()
print(sorted(set(s))==s)
A simple approach - check uniqueness, check sortedness.