# Computing partition function of Potts model on a "thin" 2D grid

I am looking for an exact way of evaluating the partition function of the $$q$$-state Potts model on a $$2D$$ grid of size $$N \times H$$:

$$\mathcal{Z}(\beta) = \sum_{\left(\sigma_{i, j}\right)_{\substack{1 \leq i \leq N\\1 \leq j \leq H}} \in \left\{1,\,\ldots,\,q\right\}^{N \times H}}\exp\left(\beta\left\{\sum_{\substack{1 \leq i < N\\1 \leq j \leq H}}\mathbf{1}\left[\sigma_{i, j} = \sigma_{i + 1, j}\right] + \sum_{\substack{1 \leq i \leq N\\1 \leq j < H}}\mathbf{1}\left[\sigma_{i, j} = \sigma_{i, j + 1}\right]\right\}\right)$$

More specifically, I am interested in the regime where $$N$$ is arbitrarily large but $$H$$ is held constant. The requirement for the exact analytic methods are the following:

• It should have complexity polynomial in $$N$$.
• It should have complexity independent of $$q$$.
• However, let us say the complexity in $$h$$ does not matter in first approximation (it may be exponential or worse).

As a non-expert in analytic methods of statistical physics, I have currently come up with the following:

• Apply the naive transfer matrix method along the large, size $$N$$, dimension. This involves matrices of size $$q^H \times q^H$$, hence does not satisfy the requirement of independence in $$q$$.
• By exploiting the symmetries of the transfer matrix, I could reduce that to matrices of size at most $$2^HB_H^2$$, where $$B_H$$ (Bell number) is the number of partitions of a set of $$H$$. This is now independent of the number of states $$q$$, though quite huge as a function of $$H$$!

Is is the best one can do indeed? As a non-expert, I have good reasons to doubt that and I would appreciate if more knowledgeable people could share their thoughts!