 Solution

Put $R=\mathbb{Z}[U]$, with $\mathrm{deg}U=-2$. In genus one, an intersection point is itself a generator of the Heegaard Floer chain complex. Thus the free $R$ basis isThere are infinitely many integral generators $U^{m}j$, $m\ge 0$, but these seven points are the module generators.

The obstruction to a Whitney disk class is the alpha-and-beta joining cycle in $H_1(Y)$. Identify the opposite sides of the square and straighten the two small fingers of the alpha curve. Its algebraic intersection with beta is three; the remaining intersections, one in each residue class, are represented by $1,4,5$. The four small bigons connect $1$ to $2$ to $3$, and $5$ to $6$ to $7$. A joining cycle between consecutive remaining strands is a meridian, which generates $H_1(Y)=\mathbb{Z}/3$. Consequently the equivalence classes are exactlyEquivalently, these are the three Spin-c structures represented in the Heegaard diagram.

Use the domain convention ${\partial }_{\alpha }D=y-x$. The positive bigons run from $2$ to $1$, from $2$ to $3$, from $5$ to $6$, and from $7$ to $6$. Their Maslov indices are one: their Euler measure is $1/2$, and the two endpoint averages contribute $1/4$ each. Write their basepoint multiplicities as $a,b,c,d$, respectively. The relative grading in Heegaard Floer homology is thereforeThe reproduced figure has no visible marked $z$. If $z$ is outside all four small bigons, as in the usual intended diagram, $a=b=c=d=0$. One may normalize the relative Heegaard Floer gradings asEach class has its own arbitrary additive constant, and $\mathrm{gr}(U^{m}j)=\mathrm{gr}(j)-2m$. If the basepoint is put inside one of these disjoint bigons, exactly its multiplicity is one; the preceding boxed difference formulas give the answer without an unstated basepoint assumption.