Posted this in a YouTube comment but I'll also post it here. Considering that the jump is completely diagonal, it's pretty much just jumping the length of the hypotenuse of a 3x3 square. Using c = √(a^2)+(b^2), The jump itself would be √(3^2)+(3^2) or √18 which is about 4.24 blocks, so HHaz is technically correct that the jump is 4 blocks long, even though you only need to place 3 blocks to fill in the gap.

Unfortunately, the actual formula isn't simply Pythagoras' theorem. It turns out that you have to count the extra 0.3 blocks you can stand on by shifting. The formula then comes out as: √[(x-0.6)² + (y-0.6)²] + 0.6 In our case 3x3: √[(3-0.6)² + (3-0.6)²] + 0.6 = √(11.52) + 0.6 = 3.99 blocks As provided by the parkour legend EntityWolf. The formula might not be correct though, and he doesn't remember if the 0.6 has to be 0.3 or 0.6... lol. Someone may have done a better one, but we don't know. If it was -0.3 instead of -0.6 (+0.6 unchanged) the result is 4.41. Though, the jump itself is much harder than a regular 4b jump, so we may take in consideration the actual result with Pythagoras' theorem (4.24). If you happen to find the correct formula post it here!