Beware the Mine Squares!

There are a total of 16 cells. The 3 by 3 grid of numbers, each number indicates the total bombs for the four cells surrounding it. The corner numbers each give the total of the four cells around them. There is no overlap, and they account for every cell. Add up the four corner numbers (3 + 1 + 1 + 3). That should be the answer if the numbers are possible. The only possibility I am leaving off is an existence proof. It is possible but pointless for the numbers to indicate an impossible configuration that could never be satisfied by all the numbers. Putting some faith in the problem writer, I did one quick mental check that made me believe it was possible, but I did not write it down and may have lacked rigor.

Consider the 3-square bomb at either corner, 3 out of 4 nodes must have the bombs, leaving one free.

Now we choose the free node to be at the big square's corner, the adjacent 2-squares will be automatically occupied with 2 bombs each. However, since the center square's number is 3, at least one more bomb must be placed in either territory, making it contradicted. Thus, the free node can't be at the far corner.

Then, for generality, if the free node is the one shared between the lower right 3-square and the center 3-square territory, that will leave the rest of the nodes of the center to have bombs and consequently making the adjacent 2-squares at middle upper and left middle to be occupied. However, the 3-square at the upper left will still need a place to plant bomb on either of the 2-squares, making it again contradicted.

Hence, the free node must be in other corner, and that will lead to these combinations:

With reflection, there are $2$ possible ways to plant the bombs according to the numbers, resulting in $\boxed{8}$ bombs in total.

The central $3$ can be done in only two ways (the other two being effectively the same due to symmetry of the problem).

The two alternatives are shown using the red bombs.

The one on top works for the two red $2$ ’s shown, but then it fails for the black $3$ as there is no way to add two more bombs.

So the second arrangement is the only one possible.

Once that’s decided, every remaining placement of a bomb is forced by the numbers.

The two red $2$ ’s and the red $1$ are satisfied by the first three bombs.

The two green $3$ ’s require the green bombs to be added.

Now the green $2$ ’s are full up.

The blue $1$ is the only one left, and one blue bomb in the corner will take care of that.

Counting the bombs, there are $3$ red, $4$ green, and $1$ blue, that is $\boxed{8}$ total.

Consider the four squares that encompass all sixteen nodes, without overlapping. You do not need to know which specific nodes have bombs because you know the total from each of these four squares (3, 1, 1, and 3) for a total of 8 . Four Squares

it is just like playing minesweeper, the hints that show you how many bombs are there, are actually useful when you figure out the boxes which are not bombed. This is a classic problem of elimination and deduction. Just like Enstein's riddle.

I found the following to be simple and direct.

Start with the center 3. There are 4 ways to fill it.

Place bombs in (2,2) and (3,3) adjacent 3-square corners.

Place a bomb in (3,2) . (2,3) is equivalent and produces a reflection of the solution below.

Placing bombs in both (2,3) and (3,2) will fail so there are only the two possible choices above.

Placing bombs in (2,2), (3,3) and (3,2) eliminates a few nodes.

(2,3) (3-square has 3)

(2,1) and (3,1) (2-square has 2)

(4,2) and (4,3) (2-square has 2)

(4,1) (1 square has 1)

3 filled, 6 eliminated by force.

Bombs must go into (1,1) and 1,2) to satisfy the corner 3-square. The eliminates (1,3) (2-square has 2)

Bombs must go into (3,4) and (4,4) to satisfy the corner 3-square. The eliminates (2,4) (2-square has 2)

7 filled, 8 eliminated by force.

This leaves only a bomb in (1,4) to satisfy the 1-square and account for all 16 nodes.

All forced once the initial state of the center 3 is fixed.

Thus 8 is the unique answer found in two symmetrical solutions depending on the initial choice of node adjoining a 1-coner from the center 3-square.

3+3+1+1 of course

I added the numbers in the 4 corners together..