How to use a posix BRE or a posix ERE regex to match a line that doesn't contain 834 at any place in string?

I want to find grep '!<pattern>' file which will be equal to grep -v '<pattern>' file.

It is theoretical question on regex, that's why I don't want to use -v flag.

I figured out how to list lines, that doesn't contain one character: ^[^8]*$, for my case I also tried ^$|^.$|^..$|^([^8]..|.[^3].|..[^4])*$, but that's not working as expected.

  • 2
    – mikeserv
    Commented Nov 16, 2015 at 19:20
  • @mikeserv, not working either :( Commented Nov 16, 2015 at 20:05
  • yeah. think i stuck in one too many asterisks. the (8(...)*)* will match 8 then fall back to the [^8]* when one of the [^3]|3[^4] stuff matches. you have to something very like that - but expanded for all possibilities. regex is math - its why * 0 is a place-holder. negation is perfectly valid regex - you should not try to avoid doing it.
    – mikeserv
    Commented Nov 16, 2015 at 20:07
  • @mikeserv, Anyway, why my pattern is not working? Shouldn't it check each character from beginning of line to be not a part of 834? Commented Nov 17, 2015 at 16:27

1 Answer 1


The ^([^8]*(8([^38]|3[^4])+)*)+$ pattern should do the trick, thanks to mikeserv, who pointed it out to me.

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