Grep Pattern and the next word

Question

I have a text file with lines like this (only the last two strings matter):

1 sometext Pattern
2 sometext Pattern  5Asda5}  
3 sometext Pattern asd2-asd  
4 sometext Pattern a-d
5 sometext Pattern   foobar  
6 sometext Pattern {asd  
7 Pattern Pattern something 123asd  
8 Pattern sometext asd    
9 Pattern 5h{1,2}b

I need to find each line with "Pattern" as last word. A word is defined in this case as string with letters and hyphens. There can be leading spaces or spaces following a string (like this: Pattern\t\t\t\t\t$)

My take at this is

egrep  '(\bPattern\b +[^a-zA-Z\-]{2,} *$)|(\bPattern\b *$)' file.txt 

to find all lines with Pattern followed by space(s) and a non-"word" at the end or with Pattern at the end. But the first part does not work as intended. I need another way to find the non-words.

Could you show me a way to grep line 1,2,3,6,9 but not the rest? I like to stay away from awk as we did not cover it in class.

Answer 1

Note that \b, in grep implementations that support it, is for the transition between a word character (alnum and underscore) and non-word character (or vise versa). So \bPattern\b would match in foo-Pattern.bar for instance.

If you're looking for a blank-delimited Pattern word, you can't use \b.

Here, it seems you're looking for a blank-delimited Pattern that is followed by a list of zero or more blank-delimited strings that have characters other than letters and -, so:

non_word='([^[:blank:]]*[^[:blank:][:alpha:]-][^[:blank:]]*)'
grep -E "(^|[[:blank:]])Pattern([[:blank:]]+$non_word)*[[:blank:]]*$"