Alternating chars and nested parenthesesin (e)grep

I'm looking for a regex that finds all words in a list that do not have characters next to each other that are the same. (this is an exercise)

So abcdef is printed, but aabcdef is not.

I tried both

egrep "^((.)[^\1])*$"

and egrep "^((.)[^\2])*$" words but, other than being not sure which one would be right, they don't work.

I know I can go egrep -v "(.)\1", but i want to use the regex in an OR structure with some other ones, so that's not possible.

For those in开发者_开发百科terested, the full exercise is to find all words that have exactly two character pairs, so aacbb and aabbd are matched, but abcd and aabbcc are not.

Thanks,

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egrep is deprecated. use grep -E

eg

echo "aacbb" | grep -E "(\w)\1"

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Is egrep a requirement? Or can we switch to something more powerful like perl,python etc?

A think a negative look ahead assertion would work here:

#!/usr/bin/env python

import re

test1 = "abcdef"
test2 = "aabcdef"
test3 = "abbcdef"

r = re.compile(r"^(?:(.)(?!\1))*$")

assert r.match(test1) is not None
assert r.match(test2) is None
assert r.match(test3) is None

I guess the two groups version can be made by combining three of these expressions with ones that do match pairs.