Remove duplicate words and just print lines in which this occurs - regex

Question

I have a challenge to look in a file if a sentence contains 2 identical consecutive words. If so, you print the word; otherwise, you don't print the sentence.
Example:
abc2 1 def2 3 abc2
F4
--------------
dea 123 123 zy45
12 12
abc cd abc cd
xyz%$#! xyz%$#! kk
xyzxyz
abc h h h h
After running the program the output will be:
dea 123 zy45
12
xyz%$#! kk
abc h h h
3
This is what I have so far:
sed '/\([^\([^ ]\+\)[ ]\+\1]\)/d' F4 >|tmp
I got this so far but this is only separating between the sentences that have the double word and sentences that don't.

Answer 1

Your sed expression was quite accurate. However, it needed some mangling to make it work:
$ sed -nr 's/\b(\S+)\s+\1(\s|$)/\1/p' file
dea 123 zy45
12
xyz%$#! kk
abc h h h
The idea is the one you already implemented: match a given word with [^ ] and see if you match it again with \1. What I added is all of this to be replaced with \1 so the repeated block disappears.
Instead of [^ ] it is also useful to use \S and instead of [ ], \s. Note also the usage of \b as a word boundary to prevent false positives like fedorqui qui and the usage of \1(\s|$) to prevent other false positives like hello helloa (thanks WalterA for the examples!). Note the usage of \s|$ to match either a space or the end of the line; \b matches any not-word character, which makes it not useful for the case with xyz%$#! kk.
To prevent all lines to be printed, we use sed -n. This way, we just print (with p) those that go through the regular expression that was defined.
Note the usage of -r to get rid of all those escaping to capture groups. Without it, the command would be:
sed -n 's/\b\([^ ]\+\)[ ]\+\1/\1/p' file
Let's test it with a more comprehensive input:
$ cat a
abc2 1 def2 3 abc2
F4
--------------
dea 123 123 zy45
12 12
abc cd abc cd
xyz%$#! xyz%$#! kk
xyzxyz
fedorqui qui
hello helloa
abc h h h h
$ sed -nr 's/\b(\S+)\s+\1(\s|$)/\1/p' a
dea 123zy45
12
xyz%$#!kk
abc hh h

Answer 2

I was looking for a sed solution that seemed to be easy. perhaps in this case awk is better (F4 is the inputfile):
awk '{
for (i=2; i<=NF; i++) {
if ($(i-1)==$i) {
$i="";
printf("%s\n", $0);
break;
}
}
}' F4
I am not complete happy with this solution, since it will leave a double FieldSep in $0 after deleting the doubled word, but literally the OP did not see that a space or tab should be deleted too.