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No, that's actually pretty easy to compute, because you only have to compute all possible hashes of possible two strings, which would be $n^2$, where $n$ is the number of possible characters / symbols.

Edit:

As fgrieu correctly explained in the comment: Because of the birthday attack we would expect a natural collision between two SHA-1 hashes to occur at $\approx 2^{80}$ hashes.

And $2^{80}$ is a lot larger than $1,111,998^2$.

No, that's actually pretty easy to compute, because you only have to compute all possible hashes of possible two strings, which would be $n^2$, where $n$ is the number of possible characters / symbols.

Edit:

As fgrieu correctly explained in the comment: Because of the birthday attack we would expect a natural collision between two SHA-1 hashes to occur at $\approx 2^{80}$ hashes.

And $2^{80}$ is a lot larger than $1,111,998^2$.

2nd Edit:

There are a lot of different sources for the exact amount of Unicode characters / symbols there could be (see comments below), but most of them are around $1'100'000$ and are either way far too small to reach $2^{80}$.

No, that's actually pretty easy to compute, because you only have to compute all possible hashes of possible two strings, which would be $n^2$, where $n$ is the number of possible characters / symbols.

Edit:

As fgrieu correctly explained in the comment: Because of the birthday attack we would expect a natural collision between two SHA-1 hashes to occur at $\approx 2^{80}$ hashes.

And $2^{80}$ is a lot larger than $1112064^2$.

No, that's actually pretty easy to compute, because you only have to compute all possible hashes of possible two strings, which would be $n^2$, where $n$ is the number of possible characters / symbols.

Edit:

As fgrieu correctly explained in the comment: Because of the birthday attack we would expect a natural collision between two SHA-1 hashes to occur at $\approx 2^{80}$ hashes.

And $2^{80}$ is a lot larger than $1,111,998^2$.

No, that's actually pretty easy to compute, because you only have to compute all possible hashes of possible two strings, which would be $n^2$, where $n$ is the number of possible characters / symbols.

No, that's actually pretty easy to compute, because you only have to compute all possible hashes of possible two strings, which would be $n^2$, where $n$ is the number of possible characters / symbols.

Edit:

As fgrieu correctly explained in the comment: Because of the birthday attack we would expect a natural collision between two SHA-1 hashes to occur at $\approx 2^{80}$ hashes.

And $2^{80}$ is a lot larger than $1112064^2$.