failure of $\square(\kappa)$ at an inaccessible $\kappa$ How can we force the failure of $\square(\kappa)$ at an inaccessible $\kappa$, where 
$\square(\kappa)$ is defined as follows: There is a sequence $(C_i:i< \kappa)$ such that:
(1) $C_{i+1} = \{i\}$ and $C_i$  is closed and cofinal in $i$ if $i$ is a limit
ordinal.
(2) If $i$ is a limit point of $C_j$, then $C_i = C_j \cap i$.
(3) There is no club $C$ (a subset of $\kappa$) such that for all limit points $i$ in
$C$ the equality $C_i= C \cap i$ holds.
Update. By a recent result of Magidor and Vaananen ( On Löwenheim-Skolem-Tarski numbers for extensions of first order logic), it is consistent, relative to the existence of a supercompact cardinal, that $\square(\kappa)$ fails at the least strongly inaccessible cardinal. 
Is it possible to reduce the consistency strength to a weakly compact cardinal or even a measurable cardinal?
 A: The consistency strength of the failure of $\square(\kappa)$ for non-weakly Mahlo inaccessible cardinal $\kappa$ (in particular, for the first strongly inaccessible) is higher than weakly compact. For example it implies that $0^\#$ exists.
http://www.jstor.org/stable/27590333
We know that global square holds in $L$, i.e. there is a sequence $\langle C_\alpha | \alpha \text{ is singular limit ordinal} \rangle$ with the properties: $\forall \alpha,\, \text{otp } C_\alpha < \alpha$ and for every accumulation point $\beta$ of $C_\alpha$, $C_\alpha \cap \beta = C_\beta$. 
In $V$, there is a club $D\subset \kappa$ through the singular cardinals below $\kappa$. If $0^\#$ doesn't exists, the covering theorem will imply that those cardinals are singular in $L$ as well, and therefore $C_\alpha$ is defined for every $\alpha \in D$.
Now we can define a $\square(\kappa)$ sequence, $\langle E_\alpha | \alpha < \kappa \rangle$ in $V$ as follow:


*

*For $\alpha \in \text{acc } D$, $E_\alpha = C_\alpha \cap D$ (unless this set is bounded below $\alpha$. In this case, $\text{cf }\alpha = \omega$ and we can choose $E_\alpha$ to be arbitrary $\omega$ sequence cofinal in $\alpha$). 

*For $\alpha \notin \text{acc }D$, $E_\alpha = \alpha \setminus (\max D\cap \alpha)$


It's easy to see that $\langle E_\alpha | \alpha < \kappa\rangle$ is coherent (since the global square is coherent) and thread-less (since every thread meets $D$ club many times, but $\text{otp }E_\alpha < \alpha$ for every $\alpha \in D$).
Edit: As Mohammad noted, we can get better lower bound by using more appropriate inner model. We need to violate global square on the singular cardinals. By $\square$ On the singular cardinals (Theorem 1), we need at least many measurable cardinals of uncountable Mitchell order in order to get a model without global square on the singular cardinals. 
A: In general, one cannot force the failure of $\square(\kappa)$ at a fixed cardinal $\kappa$. Indeed, if $\kappa$ is any regular uncountable cardinal which is not weakly compact in $L$, then there is a nontrivial $\square(\kappa)$ sequence which is moreover constructible. The fact that $\kappa$ is not weakly compact in $L$ cannot be destroyed by forcing. On the other hand, $\square(\kappa)$ always fails at a weakly compact cardinal.
