TITLE: Effective way of finding generators on the curve and the rank conjecture
QUESTION [5 upvotes]: Hello everyone, 
I have never heard of a polynomial time running algorithm that finds the generators of elliptic curves efficiently. I do know that Nagell-Lutz theorem is useful in computing the torsion part in $$E(\mathbb{Q})=\mathbb{Z}^{\phi} \oplus E_{\rm{Torsion }}(Q).$$ So what about the count of $\phi$ and the effective generation of points. 
I do know 2-descent , that doesn't not work perfectly always. Regarding the 3-descent , only some curves having CM are said to pass through it. But are there any latest advancements in the area of computing the generators of the curve, if so please give some references. 
I also read the rank conjecture, for expressing the rank of the underlying abelian group $E(\mathbb{Q})$ in terms of the order of vanishing of the taylor expansion of associated $L$-function. So I again got stuck. 
Why do one go for a rank, if one has a point that has infinite order ? . 

Rank 1-curves usually mean that they have a point of infinite order on them that can be used to generate all other points by successive chord and tangent methods.
Rank 2- Curves have 2 such points of infinite order that can be used to generate all other points. 

So my question is why should  one bother about $n$ points ( of infinite order ) if we have one point of infinite order. To express an analogous statement, suppose think that you are given a secret map that will lead you to some treasure. After the hard journey, you at-last said " Eureka" and you have found a " Machine X" . Its written on the Machine X, that it will produce dollar notes, as many as you want. So its limit is infinity. You can extract infinite number of dollar notes from the machine. You took that machine and packed it and again saw the secret-map. There are in fact some other markings on the map that will lead you to a place that contains the same Machine X. 
Do you again go to those places searching for another Machine-X if you have a Machine-X that will produce infinite amount of money. I think the analogy is clear. So if we already have a point that produces infinite points why to again bother about other points that give rise to infinite points. 
So if I am right, does the task of finding $\phi$ number of generators reduce to the task of finding one generator ? . That will make the above equation look like this $$E(\mathbb{Q})=\mathbb{Z} \oplus E_{\rm{Torsion }}(Q).$$
( Which is nothing but the Rank-1 situation ).
Thank you.

REPLY [6 votes]: We do not currently know an effective algorithm to compute the rank of an elliptic curve or to compute generators for its Mordell-Weil group. One can "do descents by day and search for points by night", and in practice, the process will stop. BTW, people do 3 and 4 (and maybe even 5) descents these days, and they're not restricted to CM curves. Finally, I'll mention that if $E(\mathbb{Q})$ has rank 1, then one can use Heegner points to find a rational point of infinite order, or one can use the value of $L'(E,1)$ (sort of as described in Keller's answer) to do a very efficient search. (This last algorithm is in a paper of mine). However, neither method is practical if the conductor $N$ of the curve is too large. Roughly, for both methods one needs to compute the value of a series that doesn't begin to converge until you take $O(\sqrt{N})$ terms.