Suppose that we have a knot $K$ with 40 crossings which is not a cable knot or an alternating knot.
Then, what is the easiest way to compute Ozsváth-Szabó's invariant $\tau(K)$?
Are there any softwares which compute $\tau(K)$ if we insert Gauss code or PD code of $K$?
For Rasmussen's $s$-invariant, Knottheory package works very nicely for this purpose.
In principle, a computer can compute $\tau$ for any knot, using grid diagrams (or variants thereof); Baldwin and Gillam computed $\tau$ for knots up to 11 crossings a while ago. However, this is not practical. Unlike with Khovanov homology, there are no programs that can compute even $\widehat{HFK}$ for "large" knots (say, more than 20 crossings). I have heard that Zoltán Szabó has a program that can compute $\widehat{HFK}$ for large knots (maybe around 50 crossings) in a matter of minutes; I don't think he has made it public yet, though, and I don't know if it computes $\tau$.
There are a couple of tricks you can use, that sometimes help:
Plamenevskaya gave a lower bound on $\tau$ using the self-linking number of any representative of $K$; more precisely, if $T$ is a transverse representative of $K$, then $$ {\rm sl}\,T \le 2\tau(K) - 1.$$ The inequality is in fact an equality if $T$ is the closure of a quasi-positive braid (i.e. the closure of a braid that's a product of conjugates of positive generators). Of course you can use it for both $K$ and its inverse, and see what you get. To generate some random Legendrian/transverse representatives of $K$ you can use Gridlink.
Another lower bound is given in terms of the unknotting number (I believe this is due to Ozsváth and Szabó): $$ u(K) \le \tau(K),$$ and in fact there are more refined versions of this with signed unknotting numbers (for which I'm afraid I have no references, but I can try and dig them up).
Of course, you can always try and simplify your knot using concordances, since $\tau$ is a concordance invariant (and then combine it with the previous tricks).
Upper bounds depend a lot on the kind of information you have. Plamenevskaya's inequality also happens to give an upper bound (since it gives a lower bound on $\tau$ and $-\tau$); the Seifert and slice genus are obvious candidates, but the latter is usually quite hard to get your hands on.
EDIT: added something about computing $\widehat{HFK}$ and the relationship between $\tau$ and the self-linking number.
This is to elaborate on the comment I made above about Livingston's paper.
Livingston also proves (Theorem 4) that if a knot is null-homologous in a fiber surface of a torus knot and it bounds a quasipositive surface, then $\tau(K)=g(K)=g_4(K)$, where $g(K)$ is the three-genus and $g_4(K)$ is the four-genus. As pointed out by Hedden, this is equivalent to saying that if $K$ is strongly quasipositive, then $\tau=g(K)=g_4(K)$. Hedden also proves that for fibered knots the converse is true, i.e. $\tau(K)=g(K)=g_4(K)$ implies $K$ is strongly quasipositive.
So if you have some kind of Floer-homology-free way of figuring out whether your knot is fibered and (if it is) what its fiber surface is like (i.e. what's the three-genus? and is it strongly quasipositive?), that might be used to tell you $\tau(K)$.
As for figuring out the genera and fiberedness of your knot: in Gabai's paper ``Detecting fibered links in $S^3$" he gives a general strategy (using his sutured manifold theory) for figuring out whether any link in $S^3$ is fibered and what the fiber surface is like. Gabai classified the fiberedness and genera of pretzel knots and Hirasawa and Murasugi used Gabai's approach to do the same for Montesinos knots. I think you might be able to find similar info about arborescent knots. Apparently Stoimenow also has a paper classifying the strong-quasipositivity and fiberedness of closed 3-braids as well.
In that paper of Hedden's mentioned above he also connects the dots to contact topology --- in particular if your knot is fibered and the corresponding open book decomposition supports a tight contact structure, Hedden tells you that $\tau(K)=g(K)=g_4(K)$ as well. In all these approaches you need to be able to determine genus, which is not easy.
