<p>The smaller the target, the fewer successful values below it, thus making it harder to find. There's a direct mathematical relationship between the target and the expected number of hashes you must attempt before you have a 50% chance of finding one below the target.</p>

CC BY-SA 3.0
2012-12-05T08:35:19.447Z
2012-12-05T08:35:19.447Z
David Schwartz
<p>Hashes are 256-bit integers, meaning they are whole numbers between 0 and 115792089237316195423570985008687907853269984665640564039457584007913129639935. There are no infinitely many different hashes. In particular, the number of hashes smaller than the target is exactly equal to the target.</p>

<p>In any case, Cantor's diagonal has nothing to do with it - if hashes were real numbers, there would be infinitely many of them even without any reference to their particular cardinality.</p>

<p>What makes mining difficult is that calculating a hash takes time, and that each hash only has a small probability to be less than the target. For practical purposes, each hash is a uniform random number between 0 and 2^256-1, so if the target is X it has a probability of (X/2^256) to be smaller than the target. This means that, on average, 2^256/X hash calculations will be needed in order to find a valid block. The lower X is, the more calculations will be needed.</p>

CC BY-SA 3.0
2012-12-05T09:01:42.007Z
2012-12-05T17:51:30.683Z
2012-12-05T17:51:30.683Z
Meni Rosenfeld
<p>Miners attempt to hash block data such that it is below a certain target value? By <a href="https://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument" rel="nofollow">Cantor's diagonal proof</a> aren't there infinitely many successful values below any given target?</p>

<protocol><hash><difficulty>
What makes difficulty more difficult?
Kinnard Hockenhull
# What makes difficulty more difficult?

(Asked by: username_0 on 2012-12-05)

Miners attempt to hash block data such that it is below a certain target value? By [Cantor's diagonal proof](https://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument) aren't there infinitely many successful values below any given target?

**Tags:** protocol, hash, difficulty


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## 2 answers

### Answer by username_2 on 2012-12-05. Score: 9 | Accepted answer

Hashes are 256-bit integers, meaning they are whole numbers between 0 and 115792089237316195423570985008687907853269984665640564039457584007913129639935. There are no infinitely many different hashes. In particular, the number of hashes smaller than the target is exactly equal to the target.

In any case, Cantor's diagonal has nothing to do with it - if hashes were real numbers, there would be infinitely many of them even without any reference to their particular cardinality.

What makes mining difficult is that calculating a hash takes time, and that each hash only has a small probability to be less than the target. For practical purposes, each hash is a uniform random number between 0 and 2^256-1, so if the target is X it has a probability of (X/2^256) to be smaller than the target. This means that, on average, 2^256/X hash calculations will be needed in order to find a valid block. The lower X is, the more calculations will be needed.


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### Answer by username_1 on 2012-12-05. Score: 1

The smaller the target, the fewer successful values below it, thus making it harder to find. There's a direct mathematical relationship between the target and the expected number of hashes you must attempt before you have a 50% chance of finding one below the target.


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