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Thanks for the comment, I will revise the kata tomorrow.
Yes, but currently you have a fully documented function
calculate_power_factor
that says it returns the power factor rounded by 4 decimal places, and is also not tested. There are two problems with this:If we're supposed to use this function, then when input contains an item which PF calculates to be something like
0.8499
it'll be rounded up to0.85
and hence is incorrectly removed from the results. So using this function with the behaviour it has been documented actually fails the kata. It's sort of supposed to be a helper function, except it's actually sabotaging the main function, so we're supposed and not supposed to use it at the same time?In TDD (Test-Driven Development), if you don't test something, it doesn't exist. So the kata requirement is telling us to implement something but it doesn't check it is actually implemented.
Speaking of this, there should be a test case with input which PF is between
0.8495
and0.95
exclusive.Answer to the first part:
Answer to the second part of the comment:
Okay, so what should the power factor be when
S = 0
then...?But the kata tells us to use a sub-function which rounds by 4 decimal places. This means a power factor of
0.84999
will be considered not below0.85
, even though it shouldn't.But then,
calculate_power_factor
is not tested anyway, so why are we rounding in the first place? Is testingP/S < 0.85
so complicated?If I am not wrong, Unnamed was using a mathematical approach and it levarages the periodicity of the sequence in terms of odd and even terms. While most of us is using iterative approach.The provided solution is more efficient for larger values.
Regarding you question:
First part (n mod 6)
If you know the parity of two numbers, you also know the parity of their sum. So the sequence is defined as: f(n) = f(n-1)+f(n-2)+f(n-3)+f(n-4)+f(n-5)
Initial values: f(0)=0, f(1)=1, f(2)=1, f(3)=2, f(4)=4. If you compute the parity, you will be able to see a pattern (i.e.odd, odd, even, even, even, even, odd, odd, even, even, even, even,...). So you will be able to notice the pattern repeats every 6 terms.
The list [0,1,2,2,2,2],
The above pattern (section 1) gives us the cumulative count of odd numbers.
For n, there are (n/6) complete segments of 6 terms. Therefore, 2x(n/6) gives the number of odd terms in these complete segments.
The remainder, n mod 6, tells us where in the 6-term segment we are. The list [0,1,2,2,2,2] gives the cumulative count of odd numbers up to that point in the cycle.
In summary, the sequence has a repeating pattern of parity due its linear definition and the properties of of parity over addition. Therefore, this repeating pattern allowed him, Unnamed, to use a mathematical approach.