A 3-by-3 unknot grid has tilde homology of rank four

The marking permutation sends the $O$ in row $0$ to the $X$ in column $1$ and then to the $O$ in row $1$; it continues from row $1$ through column $2$ to row $2$, and from row $2$ through column $0$ back to row $0$. Thus the grid has one component. In its planar projection the oriented path has vertices $(0.5,0.5),(1.5,0.5),(1.5,1.5),(2.5,1.5),(2.5,2.5),(0.5,2.5)$ and returns to $(0.5,0.5)$. These six axis-parallel edges form a simple polygon and meet only at consecutive endpoints, so the projection has no crossings and is an unknot diagram.

The six grid states, in lexicographic order as row-to-column permutations, are $012,021,102,120,201,210$. The only cells containing neither marking are $(r,c)=(0,2),(1,0),(2,1)$; each is isolated from the other two by marked cells across every side, including the torus identifications. Every marking-free rectangle made of more than one cell contains a pair of side-adjacent cells and hence a marking. Therefore the only possible tilde rectangles are these three single-cell rectangles.

For state $201$, the SW and NE corners of cell $(r,c)=(0,2)$ lie in that state and the other two corners give state $021$; cell $(1,0)$ similarly gives $201\to 210$, and cell $(2,1)$ gives $201\to 102$, with the row interval in the last cell wrapping around the torus. No other state has the SW/NE pair of corners of any of these empty cells. Thus the differential matrix, with rows and columns ordered $012,021,102,120,201,210$ and rows indexing targets, has rows $000000,000010,000010,000000,000000,000010.$ It sends only the basis vector for state $201$ to the sum of the three distinct target basis vectors, so its rank is one.

By [F3] this is a chain complex. Rank-nullity gives $\dim \ker \mathrm{\partial }=6-1=5$ and $\dim \mathrm{im}\mathrm{\partial }=1$; hence the homology has dimension $5-1=4$.