This is the way.

A beautiful solution and a great explanation! Thanks!

This code wouldn't work correctly if the sampling rate for the bits is high enough to expose the slight variations in timing that a human generating morse code would naturally make.
So this code only works when the sampling rate for the bits is low and the bit string has no slight variation in timings dots, dashes, and the gaps between dot and dashes and spaces.
For example, if the sample rate is high enough that the dots are sometimes 20 1's long but sometimes 21 1's long (which would happen regularly with a human sending morse code) then this code won't
decode the bits correctly.
Same thing if a space was sometimes 140 0's long and sometimes 139 or 141 zeroes zeroes long: this code wouldn't work correctly.

Hi clcraig,

could you answer to me issue above, saying that and giving the list of the languages you updated, please?

yep... And at least 95% of the users who solved that kata did exactly like you... ;)

Oh, I misread a statement again. This condition kills the simple beauty of so many solutions :/

Uh...whoops. Why? Only the center key was pressed

You need to tell whether the given binary grid is symmetric according to the centre point. The grid might not be square.

and that's actually correct.

OMG laughing so hard :D Nice trick!