If the data is in binary, the data will be even if its LSB is 0. The mod2 logic does exactly that. For data in the binary format, it is checking the LSB instead of converting the number to decimal or doing other exotic things. You're right about one bit test but then I'm using the mod2 logic to handle all kinds of numbers (where all kinds == {decimal, octal, hexadecimal, binary}).

```my \$n = 534587;
print "even\n" if(\$n % 2 == 0);

This will print nothing on the screen (obviously, it indicates that the number is odd) and no matter how fast the processor is, the condition inside "if" will actually be evaluated i.e. 534587 will actually be divided by 2 and the remainder will be compared with 0. Do any compiler/processor optimization, this fact will not change (afaik, that is to say).

OTOH,

```my \$n = "534587";
print "even\n" if (\$n =~ /[02468]\$/);

there are no mathematical evaluations here, no matter how big the string is, the regex here just checks last character. There will be no need of actually calculating the remainder and comparing.

I'm not saying that processor will take time to evaluate a mathematical expression, it will be derogatory to think that about modern processors. I'm just trying to stress on the fact mod2 machine circumvents any mathematical operation.

This algorithm does fail if the bases are not of the form of 2^n or base 10. Also, as ww said, it will fail for base 3 numbers on a quantum computer

In reply to Re^2: A mod2 Machine. by code-ninja
in thread A mod2 Machine. by code-ninja

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