As Ava’s reply factors out, the ‘variety of main zeros’ is only a simplification, so I assumed I’d give an instance to extra concretely illustrate this:
Block hashes are normally represented in hex format, which makes use of the character set [0,1,2,3,4,5,6,7,8,9,a,b,c,d,e,f]. However for simplicity, we will simply use a base 10 quantity set [0,1,2,3,4,5,6,7,8,9], the very same rules apply.
The block hash is the output of a SHA256 operation, so it’s a 256-bit quantity, which is a really giant quantity. This implies the output of the SHA256 operation will probably be a quantity within the vary of 0 to 2^256. Once more although, for simplicity sake we will take into account a a lot smaller quantity vary in our instance, so we’ll use the vary 0 to 999,999 as an alternative.
Now, lets think about that the hash goal is ≤500, thus any in any other case legitimate block enter that provides an output of 0000500 or much less can be thought-about a legitimate block. Lets additionally think about that the hash goal adjustments to be ≤450 for the subsequent problem interval, making any output of 0000450 or much less a legitimate block.
Discover that these two targets would “require the identical variety of main zeros”, however a block that hashes to 0000475 would solely be legitimate within the earlier problem interval, and never the later. It’s because the hash goal is a selected quantity, the variety of main zeros is just a simplification/tough proxy for this.