To reason about this we can look at the 1-d case.

If `a = [2, 4]`

and `b = [1, 2, 3]`

, our output should be `[True, False]`

. Intuitively we want to create some kind of 2 by 3 table some we can compute the pairwise equalities all at once.

To do this we can use `torch.expand`

:

`torch.expand(a, (3, 2))`

gives us `[[2, 4], [2, 4], 2, 4]]`

`torch.expand(b, (2, 3))`

gives us `[[1, 2, 3], [1, 2, 3]]`

Note that here, the shapes of a and b, are (3, 2), and (2, 3), and simply transposing b will allow us to have the same shape.`a1 = [[2, 4], [2, 4], [2, 4]]`

`b1 = [[1, 1], [2, 2], [3, 3]]`

.

As you can see here, if we now do `a1 == b1`

, which computes as `[[False, False], [True, False], [False, False]]`

, we are simply computing the equality across all pairs in the cartesian product. If we simply find max of this tensor along `dim=0`

we have `[True, False]`

as desired.

Now to generalize to cases where dimension > 1. If you have tensors `a`

and `b`

, by definition their shapes must be the same along all dimensions except one. For example if we want to compute `a in b`

along `dim=1`

their shapes could be `(x, y, z)`

and `(x, w, z)`

. Working backwards, we know that our output needs to be one-dimensional and have shape `(y)`

. So the the idea here could be to compute some kind of table of with y rows and w columns where the ith output corresponds max value of the ith row.

Some non-tested code below based on our intuitions above is:

```
a1 = torch.expand(a, (w, x, y, z))
b1 = torch.expand(b, (y, x, w, z)).permute(0, 2)
t = (a1 == b1).min(dim=(1, 3)).values # shape: (w, y)
t.max(dim=0).values # shape: (y, )
```