This bc uses the math algorithms below:
This bc uses brute force addition, which is linear (O(n)) in the number of digits.
This bc uses brute force subtraction, which is linear (O(n)) in the number of digits.
This bc uses two algorithms: Karatsuba and brute force.
Karatsuba is used for "large" numbers. ("Large" numbers are defined as any number with BC_NUM_KARATSUBA_LEN digits or larger. BC_NUM_KARATSUBA_LEN has a sane default, but may be configured by the user.) Karatsuba, as implemented in this bc, is superlinear but subpolynomial (bounded by O(n^log_2(3))).
Brute force multiplication is used below BC_NUM_KARATSUBA_LEN digits. It is polynomial (O(n^2)), but since Karatsuba requires both more intermediate values (which translate to memory allocations) and a few more additions, there is a "break even" point in the number of digits where brute force multiplication is faster than Karatsuba. There is a script ($ROOT/karatsuba.py) that will find the break even point on a particular machine.
WARNING: The Karatsuba script requires Python 3.
This bc uses Algorithm D (long division). Long division is polynomial (O(n^2)), but unlike Karatsuba, any division "divide and conquer" algorithm reaches its "break even" point with significantly larger numbers. "Fast" algorithms become less attractive with division as this operation typically reduces the problem size.
While the implementation of long division may appear to use the subtractive chunking method, it only uses subtraction to find a quotient digit. It avoids unnecessary work by aligning digits prior to performing subtraction and finding a starting guess for the quotient.
Subtraction was used instead of multiplication for two reasons:
bc is small code).Using multiplication would make division have the even worse algorithmic complexity of O(n^(2*log_2(3))) (best case) and O(n^3) (worst case).
This bc implements Exponentiation by Squaring, which (via Karatsuba) has a complexity of O((n*log(n))^log_2(3)) which is favorable to the O((n*log(n))^2) without Karatsuba.
This bc implements the fast algorithm Newton's Method (also known as the Newton-Raphson Method, or the Babylonian Method) to perform the square root operation. Its complexity is O(log(n)*n^2) as it requires one division per iteration.
bc Only)This bc uses the series
x - x^3/3! + x^5/5! - x^7/7! + ...
to calculate sin(x) and cos(x). It also uses the relation
cos(x) = sin(x + pi/2)
to calculate cos(x). It has a complexity of O(n^3).
Note: this series has a tendency to occasionally produce an error of 1 ULP. (It is an unfortunate side effect of the algorithm, and there isn't any way around it; this article explains why calculating sine and cosine, and the other transcendental functions below, within less than 1 ULP is nearly impossible and unnecessary.) Therefore, I recommend that users do their calculations with the precision (scale) set to at least 1 greater than is needed.
bc Only)This bc uses the series
1 + x + x^2/2! + x^3/3! + ...
to calculate e^x. Since this only works when x is small, it uses
e^x = (e^(x/2))^2
to reduce x. It has a complexity of O(n^3).
Note: this series can also produce errors of 1 ULP, so I recommend users do their calculations with the precision (scale) set to at least 1 greater than is needed.
bc Only)This bc uses the series
a + a^3/3 + a^5/5 + ...
(where a is equal to (x - 1)/(x + 1)) to calculate ln(x) when x is small and uses the relation
ln(x^2) = 2 * ln(x)
to sufficiently reduce x. It has a complexity of O(n^3).
Note: this series can also produce errors of 1 ULP, so I recommend users do their calculations with the precision (scale) set to at least 1 greater than is needed.
bc Only)This bc uses the series
x - x^3/3 + x^5/5 - x^7/7 + ...
to calculate atan(x) for small x and the relation
atan(x) = atan(c) + atan((x - c)/(1 + x * c))
to reduce x to small enough. It has a complexity of O(n^3).
Note: this series can also produce errors of 1 ULP, so I recommend users do their calculations with the precision (scale) set to at least 1 greater than is needed.
bc Only)This bc uses the series
x^n/(2^n * n!) * (1 - x^2 * 2 * 1! * (n + 1)) + x^4/(2^4 * 2! * (n + 1) * (n + 2)) - ...
to calculate the bessel function (integer order only).
It also uses the relation
j(-n,x) = (-1)^n * j(n,x)
to calculate the bessel when x < 0, It has a complexity of O(n^3).
Note: this series can also produce errors of 1 ULP, so I recommend users do their calculations with the precision (scale) set to at least 1 greater than is needed.
dc Only)This dc uses the Memory-efficient method to compute modular exponentiation. The complexity is O(e*n^2), which may initially seem inefficient, but n is kept small by maintaining small numbers. In practice, it is extremely fast.