Sasha's argument is pretty technological. You can really do this almost by hand, though.

Let $A$ be the algebra in question, let $r$ be its Jacobson radical (that is, the subspace of strictly upper triangular matrices), and let $E$ be the subalgebra of the diagonal matrices in $A$ (which is a complement to $r$) Notice that both $A$ and $r$ are $E$-bimodules.

The algebra $A$ has a projective resolution as a bimodule of the form $A\otimes_E r^{\otimes_E *}\otimes_E A$ which looks exactly like the Hochschild resolution but the inner copies of $A$ have been replaced by $r$, and all tensor products involved are over $E$ and not over the base field; the differentials in the complex have exactly the same formula as the usual Hochschild differential. This can be checked easily —it is a nice exercise— but you can find the details in a nice paper by Claude Cibils on square-zero algebras, if I recall correctly. (This is like the reduced Hochschild resolution, but instead of killing the copy of $k$ inside $A$, we kill the whole of $E$; almost anything useful that one wants to do equires that we be aware of this complex, so it is important to keep it at hand)

Now, $HH_*(A)$ is $Tor^{A^e}_*(A,A)$, so it can be computed as the homology of the complex obtained from $A\otimes_E r^{\otimes_E *}\otimes_E A$ by applying the functor $A\otimes_{A^e}(\mathord-)$. You should explicitly describe this complex: its only non-zero term is the $0$th one, so its homology is very, very easy to compute!

(The same thing can be done for every triangular algebra, that is, every algebra whose ordinary quiver is acyclic)

Sasha's argument is pretty technological. You can really do this almost by hand, though.

Let $A$ be the algebra in question, let $r$ be its Jacobson radical (that is, the subspace of strictly upper triangular matrices), and let $E$ be the subalgebra of the diagonal matrices in $A$ (which is a complement to $r$) Notice that both $A$ and $r$ are $E$-bimodules.

The algebra $A$ has a projective resolution as a bimodule of the form $A\otimes_E r^{\otimes_E *}\otimes_E A$ which looks exactly like the Hochschild resolution but the inner copies of $A$ have been replaced by $r$, and all tensor products involved are over $E$ and not over the base field; the differentials in the complex have exactly the same formula as the usual Hochschild differential. This can be checked easily —it is a nice exercise— but you can find the details in a nice paper by Claude Cibils on square-zero algebras, if I recall correctly. (This is like the reduced Hochschild resolution, but instead of killing the copy of $k$ inside $A$, we kill the whole of $E$; almost anything useful that one wants to do equires that we be aware of this complex, so it is important to keep it at hand)

Now, $HH_*(A)$ is $Tor^{A^e}_*(A,A)$, so it can be computed as the homology of the complex obtained from $A\otimes_E r^{\otimes_E *}\otimes_E A$ by applying the functor $A\otimes_{A^e}(\mathord-)$. You should explicitly describe this complex: its only non-zero term is the $0$th one, so its homology is very, very easy to compute!

(The same thing can be done for every triangular algebra, that is, every algebra whose ordinary quiver is acyclic)

There are various programs people have written to compute Hochschild homology and cohomology; for example, I understand that Ed Green and his students have written code to do non-comm. Groebner bases on quotients of paths algebras and, probably, to compute (co)homology, and there are others (I have written code to handle very special cases, for example) I am not aware of any other approach apart from «try to be smart about n.c. Groebner bases, work hard for a minimal resolution, and then just do linear algebra.» The first step is pretty well understood for quotients of path algebras, say; the second one can be don algorithmically, I think; the third one is of course very well understood. In practice, interesting examples tend to result in huuuuge computations, and the result is somewhat unenlightening.

Sasha's argument is pretty technological. You can really do this almost by hand, though.

Let $A$ be the algebra in question, let $r$ be its Jacobson radical (that is, the subspace of strictly upper triangular matrices), and let $E$ be the subalgebra of the diagonal matrices in $A$ (which is a complement to $r$) Notice that both $A$ and $r$ are $E$-bimodules.

The algebra $A$ has a projective resolution as a bimodule of the form $A\otimes_E r^{\otimes_E *}\otimes_E A$ which looks exactly like the Hochschild resolution but the inner copies of $A$ have been replaced by $r$, and all tensor products involved are over $E$ and not over the base field; the differentials in the complex have exactly the same formula as the usual Hochschild differential. This can be checked easily —it is a nice exercise— but you can find the details in a nice paper by Claude Cibils on square-zero algebras, if I recall correctly.

Now, $HH_*(A)$ is $Tor^{A^e}_*(A,A)$, so it can be computed as the homology of the complex obtained from $A\otimes_E r^{\otimes_E *}\otimes_E A$ by applying the functor $A\otimes_{A^e}(\mathord-)$. You should explicitly describe this complex: its only non-zero term is the $0$th one, so its homology is very, very easy to compute!

(The same thing can be done for every triangular algebra, that is, every algebra whose ordinary quiver is acyclic)

Sasha's argument is pretty technological. You can really do this almost by hand, though.

Let $A$ be the algebra in question, let $r$ be its Jacobson radical (that is, the subspace of strictly upper triangular matrices), and let $E$ be the subalgebra of the diagonal matrices in $A$ (which is a complement to $r$) Notice that both $A$ and $r$ are $E$-bimodules.

The algebra $A$ has a projective resolution as a bimodule of the form $A\otimes_E r^{\otimes_E *}\otimes_E A$ which looks exactly like the Hochschild resolution but the inner copies of $A$ have been replaced by $r$, and all tensor products involved are over $E$ and not over the base field; the differentials in the complex have exactly the same formula as the usual Hochschild differential. This can be checked easily —it is a nice exercise— but you can find the details in a nice paper by Claude Cibils on square-zero algebras, if I recall correctly. (This is like the reduced Hochschild resolution, but instead of killing the copy of $k$ inside $A$, we kill the whole of $E$; almost anything useful that one wants to do equires that we be aware of this complex, so it is important to keep it at hand)

Now, $HH_*(A)$ is $Tor^{A^e}_*(A,A)$, so it can be computed as the homology of the complex obtained from $A\otimes_E r^{\otimes_E *}\otimes_E A$ by applying the functor $A\otimes_{A^e}(\mathord-)$. You should explicitly describe this complex: its only non-zero term is the $0$th one, so its homology is very, very easy to compute!

(The same thing can be done for every triangular algebra, that is, every algebra whose ordinary quiver is acyclic)