Editing Middle-square method (section)

== The method == 
To generate a sequence of ''n''-digit pseudorandom numbers, an ''n''-digit starting value is created and squared, producing a 2''n''-digit number. If the result has fewer than 2''n'' digits, [[leading zero]]es are added to compensate. The middle ''n'' digits of the result would be the next number in the sequence and returned as the result. This process is then repeated to generate more numbers. 

The value of ''n'' must be even in order for the method to work{{snd}} if the value of ''n'' is odd, then there will not necessarily be a uniquely defined "middle ''n''-digits" to select from. Consider the following: If a 3-digit number is squared, it can yield a 6-digit number (e.g. 540<sup>''2''</sup> = 291600). If there were to be middle 3&nbsp;digits, that would leave 6 − 3 = 3&nbsp;digits to be distributed to the left and right of the middle. It is impossible to evenly distribute these digits equally on both sides of the middle number, and therefore there are no "middle digits". It is acceptable to pad the seeds with zeros to the left in order to create an even valued ''n''-digit number (e.g. 540&nbsp;→&nbsp;0540).

For a generator of ''n''-digit numbers, the period can be no longer than 8<sup>''n''</sup>. If the middle ''n'' digits are all zeroes, the generator then outputs zeroes forever. If the first half of a number in the sequence is zeroes, the subsequent numbers will be decreasing to zero. While these runs of zero are easy to detect, they occur too frequently for this method to be of practical use. The middle-squared method can also get stuck on a number other than zero.  For ''n''&nbsp;=&nbsp;4, this occurs with the values 0100, 2500, 3792, and 7600. Other seed values form very short repeating cycles, e.g., 0540 → 2916 → 5030 → 3009. These phenomena are even more obvious when ''n''&nbsp;=&nbsp;2, as none of the 100 possible seeds generates more than 14 iterations without reverting to 0, 10, 50, 60, or a 24 ↔ 57 loop.

=== Example implementation ===
Here, the algorithm is rendered in [[Python 3|Python 3.12]].
 
<syntaxhighlight lang="python">
seed_number = int(input("Please enter a four-digit number:\n[####] "))
number = seed_number
already_seen = set()
counter = 0

while number not in already_seen:
    counter += 1
    already_seen.add(number)
    number = int(str(number * number).zfill(8)[2:6])  # zfill adds padding of zeroes
    print(f"#{counter}: {number}")

print(f"We began with {seed_number} and"
      f" have repeated ourselves after {counter} steps"
      f" with {number}.")
</syntaxhighlight>