Suppose recursion for output of Galois LFSR is $u_1=u_3=u_4=0$
What kind of taps are here, and initial state? How to uncover this information?
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In general, you're missing both the polynomial (taps) and some symbols, so there is no unique solution.
The LFSR is determined by its generating polynomial and thus its degree/length/taps. If the LFSR is 5 bits or more in length you're out of luck, you need 5 equations so 5 bits of information for any hope of solving this.
If it is 4 bits long, and primitive, you're in general out of luck again in the absence of any extra information, since all loadings, and in particular both states $$(s_1,s_2,s_3,s_4)=(0,s_2,0,0)$$ with $s_2$ arbitrary are possible.
However, if $s_2=0$ this will give a trivial LFSR of all zero output, so if you're not in this setting you can assign $s_2=1$ and proceed, if the LFSR is 4 bits long. Then you might try all possible primitive LFSRs (there are 2 of length 4) and have 2 candidates.
If it is 3 bits or less, again, if the all zero state can be ruled out, you can proceed by trying all primitive LFSRs and in fact now you have 4 bits of info (4 symbols after ruling out all zero state) for a $L=3$ so you should be able to determine the LFSR uniquely if it is primitive (there are 2 here).
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$\begingroup$ Suppose any $s_2$ works. Ive been thinking, if initial configuration is, lets say $1,0,0,0,0,0$, and tap is to 5h bit (0 which is after 1 in initial state) It should produce 4 consequtive zeroes, and given any $s_2$ works, should correspond to initial requirement $u_1=u_3=u_4$ Do I miss anything? $\endgroup$ Commented Sep 4, 2015 at 5:09