TQBF PSPACE-COMPLETE : Why this algorithm is exponential but Savitch's not? So this is a question pertaining to the proof for $PSPACE-COMPLETE$ (for TQBF for example). The idea is to first prove the $L$ $is$  $PSPACE$(easy part) and next is to prove $PSPACE-COMPLETE$. The latter requires demonstrating an algorithm which computes the L in polynomial space. This is usually achieved by having recursive calls such that is re-used. 

In TBQF proof, the equation $\phi_{i+1}(A,B)$= $\exists Z [\phi_{i+1}(A,Z) 
 \land \phi_{i+1}(Z,B) ]$ ($Z$ is mid-point )is default recursive relation for computing TBQF truth. In any standard proof, it is said that $\phi_{i+1}(A,B)$ is computed two times and for $m$ nodes, this formula explodes hence, other recursive-relation should be used to bound.

However in Savitch's proof, the recursive relation is $Path(a,b,t)$ = $Path(a,mid,t-1)$ AND $Path(mid,b,t-1)$ accepts then ACCEPT. In proof, it is stated that this relation reuses-spaces.

My Question is Why in TBQF relation space explodes while in Path, it is reused? Both of these relations looks more or less same to me because both refers to i-1 instances and will need space to store them?.
In the proof of the $\textit{PSPACE-}$completeness of TQBF, we need to construct a formula of some specific type. The recursive midpoint algorithm is easy to implement (and Savitch's Theorem requires finding the $\textit{algorithm}$), while the same idea, if performed in a similar manner, results in an exponential formula (and the completeness proof requires the $\textit{formula}$). 

More generally, suppose there's some language $\textit{L}$, an optimal algorithm $A$ deciding $L$, and the smallest formula (maybe quantified) $\varphi$, s.t. $\varphi(x) = 1 \Leftrightarrow x \in L$. For me, it's not very clear why $A\in \textit{PSPACE}$ would imply that  $\varphi$ is small (I only see it as a consequence of the $\textit{PSPACE}-$completeness theorem)