Modular arithmetic

Richard Haseltine

I would expect any operation mod n to return a value in [ 0 , n - 1 ]. Using the % operator in Daz Script that works for addition but not for subtraction:

for ( var i = 0 ; i < 4 ; i++ ) {	print ( ( i + 1 ) % 4 );	print ( ( i - 1 ) % 4 );}

returns:

1
-1
2
0
3
1
0
2

instead of  the expected (by me):

1
3
2
0
3
1
0
2

Which of us is wrong, Daz Script or me? And if it's the latter, is there a neater way to get my expected behaviour than testing for 0 in subtraction operations and giving it a special result of n - 1 instead of -1?

MarcCCTx

I think the way you stacked the prints DAZ is correct.

MarcCCTx

0+1 mod 4 = 1

0-1 mod 4 = -1

1+1 mod 4 = 2

1- 1 mod 4 = 0

2+1 mod 4 = 3

2-1 mod 4 -= 1

3+1 mod 4 = 0

3-1 mod 4 = 2

Richard Haseltine

OK, thanks.

MarcCCTx

for ( var i = 4 ; i > 0 ; i-- ) {
print ( ( i + 1 ) % 4 );
print ( ( i - 1 ) % 4 );
}

Would give you your expected: 1, 3, 0, 2, 3, 1, 2, 0

murgatroyd314

MarcCCTx said:

I think the way you stacked the prints DAZ is correct.

Speaking as a mathematician, no. It's seriously messing up on handling negative values.

Richard Haseltine

Murgatroyd said:

MarcCCTx said:

I think the way you stacked the prints DAZ is correct.

Speaking as a mathematician, no. It's seriously messing up on handling negative values.

Thanks, that is reassuring - though if it is a bug it's presumably a Qt bug, not soemthign that Daz can fix.

Richard Haseltine

MarcCCTx said:

for ( var i = 4 ; i > 0 ; i-- ) {
print ( ( i + 1 ) % 4 );
print ( ( i - 1 ) % 4 );
}

Would give you your expected: 1, 3, 0, 2, 3, 1, 2, 0

not useful though, unfortunately, as I am dealing in the real case with array indices which start at 0 and run to n-1.

Richard Haseltine

Aha, Rob points out that the operator is a remainder operator, not a true modulo operator https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Arithmetic_Operators#Remainder_() . So the issue is not "bad modular maths" but "thinking it was modular maths when it wasn't".

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Arithmetic_Operators#Remainder_()

murgatroyd314

((x % y) + y) % y should give you a true modulus, if I'm understanding the function properly.

Richard Haseltine

Murgatroyd said:

((x % y) + y) % y should give you a true modulus, if I'm understanding the function properly.

Yes, that looks logical.