I discovered an answer for

3×4:

I declare that is optimum.

Proof:

Consider the alternative drawback. Instead of inserting Ts, we’ll place partitions to dam off the cells.

Note that each dot right here has a “path to freedom”. If this wasn’t true, then you possibly can circle the world across the separated “dots” to kind a loop.

So, we should place as many inner blockers as we’ve got dots. In different phrases, given an $m×n$ grid, we should place not less than $(m-1)(n-1)$ inner blockers.

Of course, no sq. can have two blockers round it, or then it would not be a T anymore. We can solely place one inner blocker per two cells within the grid.

If we had a $3×5$ grid, we would want to put 8 (from $2×4$) inner blockers – however we would solely be ready to put 7.5 (from $3×5/2$). So 3×5 is unattainable.

If we had a $4×4$ grid, we would want to put 9 (from $3×3$) inner blockers. But we would solely be ready to put 8 (from $4×4/2$). So 4×4 is unattainable.

All different bigger instances fail in the identical manner: if we attempt to enhance this by including on some extra rows and columns, to get a $(3+m)×(4+n)$ grid, then we discover:
$$textual content{quantity wanted} ≤ textual content{quantity usable}$$
$$(2+m)(3+n) ≤ (3+m)(4+n)/2$$
$$6+3m+2n+mn ≤ frac12(12 + 3n + 4m + mn)$$
$$6+3m+2n+mn ≤ 6 + frac32n + 2m + frac12mn$$
$$ m + frac12n + frac12mn ≤ 0$$
If $m$ and $n$ are constructive, or if one is constructive and the opposite is zero, the left facet will at all times be constructive, and this inequality breaks. Therefore we will not add any extra rows or columns on while not having extra blockers than we’ve got.

What about different preparations of the identical measurement?

On a 3×4, we want to put 6 blockers internally, and we have 6 blockers to work with. So each blocker should be inner. (Or within the T-pipe model, all segments that lead exterior the grid should be used.)

This means we’ve got to divide the 12 squares into 6 adjoining pairs. This means we’re in search of a domino tiling of a 3×4 grid.

There are solely 5 such tilings (as much as rotation and reflection). Of these 5, three do not work – they might give loops. One of the others (the second on the precise column) corresponds to the answer I discovered; the opposite is new.