Hi all, I would like to ask the following qn:
In a computer, will
x / (a * b * c * d * e)
give a different solution from
x / a / b / c / d / e
where x,a,b,c,d,e are all integer data types? Since values are rounded down, I was wondering if that would cost the 2nd method of division to reach a smaller value than the 1st method if too much rounding is done.
Intuitively, it seems that it should not matter, but I cannot really prove it. Can someone help me out? Thanks!
In a computer, will
x / (a * b * c * d * e)
give a different solution from
x / a / b / c / d / e
where x,a,b,c,d,e are all integer data types? Since values are rounded down, I was wondering if that would cost the 2nd method of division to reach a smaller value than the 1st method if too much rounding is done.
Intuitively, it seems that it should not matter, but I cannot really prove it. Can someone help me out? Thanks!
Then x/a is (b*c*d*e)*q + [r/a], where [x] is rounding down of float value.
x/a/b is (c*d*e)*q + [[r/a]/b]
...
x/a/b/c/d/e is q+[[[[[r/a]/b]/c]/d]/e].
As r is strictly less than a*b*c*d*e, [[[[[r/a]/b]/c]/d]/e]=0.
Thanks for your comments. Actually, what I was asking is whether
floor(x / (a * b * c * d * e))
in normal arithmetic is the same as
x / a / b / c / d / e
in computer arithmetic.
But you have still managed to answer my question. Thanks everyone! So unless I am dealing with negative numbers in Python, this should work fine.