removing just the middle of the file with sed - sed

Question

I want to use sed as a part of the pipeline to preserve just 10 first and 10 last lines of its input. It would not be working on physical files, but just reading from STDIN and outputting to STDOUT. The amount of data in stream is bigger than machine RAM (or its disk space), so it needs to relatively efficient. It also must work in stream mode, without creating temporary files (no writeable filesystems).
Extra bonus if it could display one line instead of all of the middle it deleted:
for example, if I had input lines containing numbers from 1 to 100000, I would need it to output (line with literal <cut> text would be nice, but is optional):
1
2
3
4
5
6
7
8
9
10
<cut>
99991
99992
99993
99994
99995
99996
99997
99998
99999
100000
The best I've come up with is to have it output first 10 lines, and last one 1 line with:
yes ' ' | head -n 100000 |nl | \
sed -e '$q;11,$d'`
which outputs
1
2
3
4
5
6
7
8
9
10
100000
but I need it to output more context (10 lines instead of just 1) at the end of data too.
Update: length of the input stream is unknown and will vary, 100000 above is just an example.
Update: as noted in the question and the tag, I need it in sed, not awk, perl or other programming languages in which it is more easy to accomplish (that requirement, along with no tmp files, is due to fact it is embedded system with limited commands and resources available)
Update: if the input is less then that 10+10 lines, it should ideally just print the whole input

Answer 1

You can try following command:
sed -n 'H; 1,10 { p; b }; g; s/\n[^\n]*//; h; $ { s/\n/<cut>\n/; p }'
sed has two blocks to save content, pattern space and hold space. The first one is used to parse current line, and the second one can be used as a backup. The approach is to save in hold space the last ten lines processed.
H saves every line to hold space, g recover hold space, then remove oldest line and save again to hold space, and in last line ($) print adding your magic word in front of it.
The whole command:
yes ' ' | head -n 100000 |nl|\
sed -n 'H; 1,10 { p; b }; g; s/\n[^\n]*//; h; $ { s/\n/<cut>\n/; p }'
Yields:
1
2
3
4
5
6
7
8
9
10
<cut>
99991
99992
99993
99994
99995
99996
99997
99998
99999
100000
And said that, follow the advice from Ed Morton, because awk is simpler and easier to debug or modify some weeks later.
UPDATE:
You can append to hold space after first ten lines and check if there are more than 10 newline characters in it before removing oldest as FIFO structure:
sed -n '1,10 { p; b }; H; g; /\(\n[^\n]\+\)\{11\}/ s/\n[^\n]*//; h; $ { s/^\n//; p }'
Now it's more challenging to know where to add the <cut> string in the edge case of 20 input lines, but I will leave it as an exercise for you.

Answer 2

sed is for simple substitutions on a single line, that is all. For anything else, including this task, you should be using awk:
$ cat tst.awk
BEGIN { beg=(beg?beg:3); end=(end?end:3) }
NR<=beg
{ rec[(NR-1)%end+1] = $0 }
END {
print "<cut>"
for (i=1;i<=end;i++) {
print rec[(NR+i-1)%end+1]
}
}
$ seq 10 | awk -f tst.awk
1
2
3
<cut>
8
9
10
$ seq 10 | awk -v beg=2 -v end=4 -f tst.awk
1
2
<cut>
7
8
9
10
I see you've added a "it has to be sed" requirement to your question but I'll leave this answer here for future readers looking for a sensible way to perform the task.

Answer 3

This might work for you (GNU sed):
sed '1,10b;:a;$!{N;s/\n/&/10;Ta;D};i\<cut>' file
Print the first 10 lines as normal. Collect the next 11 lines and if it is not the end of file, delete the first of them and repeat always maintaining the last 10 lines. At the end of the file, insert a line containing <cut> and print the remaining 10 lines.