Homework & Coursework Questions

Thanks @bakunin
I think this

Code:

 grep -v  '^(..)+$'  filename

will work although this will also take spaces into consideration ,as I am not aware if those were to be ignored or not
Please let me know if I have missed something or there are any other more simpler approaches as I will be very pleased to learn about it too

I think spaces are characters like any other. The important point in such problems is not to do it this way or that way (like including the spaces or not), but to be aware of the limitations of ones solution. As you are aware that your solution only works under a narrowly defined set of definitions your take is impeccable.

One little problem remains, though: you seem to have developed this with a GNU-grep. GNU-grep - like other regexp-based GNU programs (i.e. sed) uses GNU-BREs (Basic Regular Expressions), which are slightly different from POSIX-BREs. In this case, you specifically use the "+" quantor, which means "1 or more". In POSIX-BREs this quantor doesn't exist, only the "*", which means "0 or more". If you want to be portable across different Unix-flavours you should write the regexp this way:

Code:

/^..\(..\)*$/

which does the same as yours.

Second, it is debatable if 0 is an even number. This is a matter of definition, though, so if you define 0 to be even you should replace "+" with "*".

I hope this helps.

bakunin

Code:

grep '^\(..\)*.$' file

will be portable.

Yes, I would say grep '^\(..\)*.$' file the best solution.

Code:

$ cat file
1

12
123
12345
123456

Code:

$ grep '^\(..\)*.$' file
1
123
12345