Reputation: 77
I am aware that there are better methods to handle division by zero in integer division. However, for educational purposes, I am curious about why adding a small epsilon value (1e-7) to the divisor in my function causes runtime errors instead of preventing division by zero. Here's the code snippet:
def kangaroo(x1, v1, x2, v2):
if (x2-x1) % (v1-v2) == 0 and (x2-x1) / ((v1-v2) + 1e-7) >= 0: # adding epsilon
return "YES"
else:
return "NO"
In this function, I attempted to add 1e-7 to the divisor (v1-v2) to prevent a division by zero. My expectation was that this would allow the division to proceed without error even when (v1-v2) is zero. I also tried using a larger epsilon value like 1e-1, but this approach still leads to runtime errors in certain scenarios. Can someone explain why this specific implementation fails and discuss the implications of using such a small epsilon in integer arithmetic, and why even larger epsilon values like 1e-1 do not solve the issue?
Upvotes: -2
Views: 178
Reputation: 967
This approach WILL cause runtime errors. When v1-v2
is zero, the first condition (x2-x1) % (v1-v2)
will fail to execute and cause the runtime error. The code you have written will work well when v1-v2
is non zero and divides x2-x1
. In that case the epsilon
would automatically get added to v1-v2
. This error is independent of the magnitude of epsilon
chosen, since the first condition would've failed even before we evaluate the second part.
A better solution would be:
if (x2-x1) % ((v2-v1) + epsilon) == 0:
return "YES"
else:
return "NO"
Or maybe this:
if (v2-v1) == 0:
v2 += epsilon
if (x2-x1) % (v2-v1) == 0:
return "YES"
else:
return "NO"
The second solution adds epsilon only when the difference is zero, while the other when does it irrespective of what the value is, since for a difference of order 1e-4
or higher, an addition of 1e-7
would not create much of a difference. The first approach may some milliseconds on execution depending on how you structure your code.
Upvotes: 1