The 2-by-2 unknot grid has tilde homology of rank two

Starting at the bottom-left $O$, the oriented route follows its row to the bottom-right $X$, then follows that column to the top-right $O$, then that row to the top-left $X$, then that column back to the starting $O$. Thus the link has one component. Its planar projection has no transverse self-intersection: the four segments meet only at these marked turning points, and smoothing the four corners gives a simple closed planar curve, which bounds a disk. Therefore it is an unknot.

There are exactly two grid states: $\mathbf{x}=\{(0,0),(1,1)\}$ and $\mathbf{y}=\{(0,1),(1,0)\}$, with coordinates read modulo two. The rectangles from $\mathbf{x}$ to $\mathbf{y}$ are the bottom-left and top-right cells, each containing one $O$; the rectangles from $\mathbf{y}$ to $\mathbf{x}$ are the bottom-right and top-left cells, each containing one $X$. No such rectangle is counted in tilde mode, and there are no rectangles from a state to itself. Hence the tilde differential is zero.

The tilde chain group has its two grid states as a basis, and its differential is zero by step 1.2. The chain condition is supplied by [F3], so its homology equals this two-dimensional chain group and has rank two.