Ways to complete the grid puzzle

I want to complete the cross figure shown here. I would like all eleven answers to be prime numbers and none to have a leading zero. Also I don’t want any digit to appear more than once in any row or column or on either of the two long diagonals.

complete-the-grid-puzzle

Given the starting values shown, how many different ways could I complete the grid?

Reveal Solution

We use $1d$ for “1 down”, etc., and introduce notation as shown here:

complete-the-grid-solution-key

Evidently $\{{\sf C},{\sf J},{\sf M}\}=\{1,3,7\}$ and we rule out 5 of the possible six permutations as follows:

If $({\sf C},{\sf J},{\sf M})=(1,3,7)$ or $({\sf C},{\sf J},{\sf M})=(1,7,3)$ or $({\sf C},{\sf J},{\sf M})=(7,3,1)$ then it quickly follows that $6a=691$ and ${\sf D}=6$ . This implies that ${\sf E}\in\{1,7\}$ and this clashes with either $3a$ or $9d$ and these cases fail.
If $({\sf C},{\sf J},{\sf M})=(3,7,1)$ then ${\sf I=1}$ and there is no way to choose ${\sf F}$ .
If $({\sf C},{\sf J},{\sf M})=(7,1,3)$ then ${\sf F}=4$ and ${\sf G}\in\{1,9\}$ which are numbers that neighbour ${\sf G}$ .

We are left with $({\sf C},{\sf J},{\sf M})=(3,1,7)$ and the grid has been narrowed down to the summary on the right here:

complete-the-grid-solution

To finish the job we look at cases according to $5d$ . There are 4 possibilities:

If $5d=271$ then every square is uniquely defined except ${\sf A}\in\{5,8\}$ .
So there are two possible grids here.
If then and we can now consider two subcases according to
1. If $9d=37$ then $2d=61$ and $7d=67$ . The grid is unique in this case.
2. If $9d=97$ then $2d\in\{31,61\}$ and $7d\in\{31,61\}$ . Four different grids here.
If $5d=571$ then the grid is unique except ${\sf A}\in\{2,8\}$ .
If $5d=631$ then the grid is uniquely defined except for ${\sf H}\in\{3,6\}$ .