Suppose the logarithm of the price follows a standard Brownian bridge from $O$ to $C$ hitting high (maximum) of $H$ and low (minimum) of $L$ on the way. The paths can be constructed with the application of the reflection principle.
Take first the simpler task of constructing Brownian paths with OHC property. We start with a Brownian bridge connecting the opening price $O$ at opening time $t=0$ with price $2H-C$ at closing time $t=1$. Amongst the paths constructed, delete the ones cross below $H$ from above after having crossed above $H$ from below for the first time. For each of the remaining path, reflect the part beyond the stopping time of first crossing $H$ around $H$.
The original task can be accomplished by repeatedly and carefully applying the exact same principle of reflection. The algorithm is a bit more complicated though. We divide all the path into disjoint subsets be sequence of hitting time of $H$ and $L$ between the opening point $O$ (let's set the price of $O$ at $0$ and starts at time $0$ and ends at time $1$) and closing point $C$. In the following description of the algorithm, I am going to sacrifice rigour for sake of descriptive simplicity --- until someone asks questions and ask me to filling th details. A path possessing the required property will start from $0$ and alternatingly hit $H$ and $L$ then end at $C$. Let $h_k$ be the stopping time of the path hitting price $H$ for the $k$'th time after the path hits $L$. So between $O$ and $C$, the set of hitting time sequences in order of occurrence is $\{(h_1,l_1),(h_1,l_1,h_2),(h_1,l_1,h_2,l_2),...\}$ union with $\{(l_1,h_1),(l_1,h_1,l_2),(l_1,h_1,l_2,h_2),...\}$.
The paths generating each hitting time sequence correspond to a subset of Brownian bridges emanating from $0$ and ending at different price points $p$ at time $t=1$ with density proportional to $e^{-p^2}$. Let $k$ run through all natural numbers. For $(l_1,h_1,...,l_k,h_k)$, the Brownian bridge ends at $p=C-2k(H-L)$; for sequence $(l_1,h_1,...,l_k)$ it ends at $p=-C-2k(H-L)+2H$; for $(h_1,l_1,h_2,l_2,...,h_k,l_k)$, it ends at $p=C+2k(H-L)$; for $(h_1,l_1,h_2,l_2,...,h_k)$, it ends at $p=-C+2(k-1)(H-L)+2H$. Amongst all the Brownian bridge paths thus constructed any path hitting any price line in the set $B=\{H+i(H-L): i\in \mathbf Z\}$ for consecutively the second time is eliminated.
Now fold or reflect all thus constructed along the price lines of $\{H+i(H-L): i\in \mathbf Z\}$. The paths thus formed are those required.