Regex explanation needed - regex

Question

My student guideline contains this example.
I need find all -tation in transportation but for example not deportation.
I use regex with lookbehind assertion
/(?<=transpor)tation/
Also there is an equivalent
/tation(?<=transportation)/
It says first regex
is more efficiency because it hasn't backtracking - engine doesn't
check tation second time.
Ok. It is good.
And there is obscure phrase for me
Usually it would either match the text the first time or use a
lookahead to check that it match the second pattern instead of
backtracking through it a second time with a lookbehind.
I can't understand both hints (in particular about lookahead assertion).(before or and after or)
I believe my riddle (puzzle) will be clear to native English speaker as a minimum.
Thanks.

Answer 1

I wouldn't give much thought to it; the statements seem to be based on assumptions about the internal logic of regular expression engines which aren't universally valid.
To shed some light on this, I looked at the debugging info of Perl's regular expression compiler.
Code:
use re 'debug';
/(?<=transpor)tation/;
/tation(?<=transportation)/;
Output:
Compiling REx `(?<=transpor)tation'
size 11 Got 92 bytes for offset annotations.
first at 1
1: IFMATCH[-8](8)
3: EXACT <transpor>(6)
6: SUCCEED(0)
7: TAIL(8)
8: EXACT <tation>(11)
11: END(0)
anchored `tation' at 0 (checking anchored) minlen 6
Offsets: [11]
12[8] 0[0] 5[8] 0[0] 0[0] 12[0] 12[0] 14[6] 0[0] 0[0] 20[0]
Compiling REx `tation(?<=transportation)'
size 13 Got 108 bytes for offset annotations.
first at 1
1: EXACT <tation>(4)
4: IFMATCH[-14](13)
6: EXACT <transportation>(11)
11: SUCCEED(0)
12: TAIL(13)
13: END(0)
anchored `tation' at 0 (checking anchored) minlen 6
Offsets: [13]
1[6] 0[0] 0[0] 24[14] 0[0] 11[14] 0[0] 0[0] 0[0] 0[0] 24[0] 24[0] 26[0]
Freeing REx: `"(?<=transpor)tation"'
Freeing REx: `"tation(?<=transportation)"'
Here, the compiled expressions don't seem to be substantially different.