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/scripts/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:
- Division and subtraction can share code (one of the less important goals of
this
bcis small code). - It minimizes algorithmic complexity.
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, and
it doubles the amount of correct digits per iteration.
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.
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.
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.
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.
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.
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.
This is implemented in the function p(x,y).
The algorithm used is to use the formula e(y*l(x)).
It has a complexity of O(n^3) because both e() and l() do.
However, there are details to this algorithm, described by the author, TediusTimmy, in GitHub issue #69.
First, check if the exponent is 0. If it is, return 1 at the appropriate
scale.
Next, check if the number is 0. If so, check if the exponent is greater than zero; if it is, return 0. If the exponent is less than 0, error (with a divide by 0) because that is undefined.
Next, check if the exponent is actually an integer, and if it is, use the exponentiation operator.
At the z=0 line is the start of the meat of the new code.
z is set to zero as a flag and as a value. What I mean by that will be clear
later.
Then we check if the number is less than 0. If it is, we negate the exponent
(and the integer version of the exponent, which we calculated earlier to check
if it was an integer). We also save the number in z; being non-zero is a flag
for later and a value to be used. Then we store the reciprocal of the number in
itself.
All of the above paragraph will not make sense unless you remember the
relationship l(x) == -l(1/x); we negated the exponent, which is equivalent to
the negative sign in that relationship, and we took the reciprocal of the
number, which is equivalent to the reciprocal in the relationship.
But what if the number is negative? We ignore that for now because we eventually
call l(x), which will raise an error if x is negative.
Now, we can keep going.
If at this point, the exponent is negative, we need to use the original formula
(e(y * l(x))) and return that result because the result will go to zero
anyway.
But if we did not return, we know the exponent is not negative, so we can get clever.
We then compute the integral portion of the power by computing the number to power of the integral portion of the exponent.
Then we have the most clever trick: we add the length of that integer power (and
a little extra) to the scale. Why? Because this will ensure that the next part
is calculated to at least as many digits as should be in the integer plus any
extra scale that was wanted.
Then we check z, which, if it is not zero, is the original value of the
number. If it is not zero, we need to take the take the reciprocal again
because now we have the correct scale. And we also have to calculate the
integer portion of the power again.
Then we need to calculate the fractional portion of the number. We do this by
using the original formula, but we instead of calculating e(y * l(x)), we
calculate e((y - a) * l(x)), where a is the integer portion of y. It's
easy to see that y - a will be just the fractional portion of y (the
exponent), so this makes sense.
But then we multiply it into the integer portion of the power. Why? Because
remember: we're dealing with an exponent and a power; the relationship is
x^(y+z) == (x^y)*(x^z).
So we multiply it into the integer portion of the power.
Finally, we set the result to the scale.
This is implemented in the function r(x,p).
The algorithm is a simple method to check if rounding away from zero is
necessary, and if so, adds 1e10^p.
It has a complexity of O(n) because of add.
This is implemented in the function ceil(x,p).
The algorithm is a simple add of one less decimal place than p.
It has a complexity of O(n) because of add.
This is implemented in the function f(n).
The algorithm is a simple multiplication loop.
It has a complexity of O(n^3) because of linear amount of O(n^2)
multiplications.
This is implemented in the function perm(n,k).
The algorithm is to use the formula n!/(n-k)!.
It has a complexity of O(n^3) because of the division and factorials.
This is implemented in the function comb(n,r).
The algorithm is to use the formula n!/r!*(n-r)!.
It has a complexity of O(n^3) because of the division and factorials.
This is implemented in the function log(x,b).
The algorithm is to use the formula l(x)/l(b) with double the scale because
there is no good way of knowing how many digits of precision are needed when
switching bases.
It has a complexity of O(n^3) because of the division and l().
This is implemented in the function l2(x).
This is a convenience wrapper around log(x,2).
This is implemented in the function l10(x).
This is a convenience wrapper around log(x,10).
This is implemented in the function root(x,n).
The algorithm is Newton's method. The initial guess is calculated as
10^ceil(length(x)/n).
Like square root, its complexity is O(log(n)*n^2) as it requires one division
per iteration, and it doubles the amount of correct digits per iteration.
This is implemented in the function cbrt(x).
This is a convenience wrapper around root(x,3).
This is implemented in the function gcd(a,b).
The algorithm is an iterative version of the Euclidean Algorithm.
It has a complexity of O(n^4) because it has a linear number of divisions.
This function ensures that a is always bigger than b before starting the
algorithm.
This is implemented in the function lcm(a,b).
The algorithm uses the formula a*b/gcd(a,b).
It has a complexity of O(n^4) because of gcd().
This is implemented in the function pi(s).
The algorithm uses the formula 4*a(1).
It has a complexity of O(n^3) because of arctangent.
This is implemented in the function t(x).
The algorithm uses the formula s(x)/c(x).
It has a complexity of O(n^3) because of sine, cosine, and division.
This is implemented in the function a2(y,x).
The algorithm uses the standard formulas.
It has a complexity of O(n^3) because of arctangent.