Calcium example programs¶
See Examples for general information about example programs. Running:
make examples
will compile the programs and place the binaries in
build/examples. The examples related to the Calcium module are
documented below.
elementary.c¶
This program evaluates several elementary expressions.
For some inputs,
Calcium’s arithmetic should produce
a simplified result automatically.
Some inputs do not yet automatically simplify as much
as one might hope.
Calcium may still able to prove that such a number is zero or nonzero;
the output of ca_check_is_zero() is then T_TRUE or T_FALSE.
Sample output:
> build/examples/elementary
>>> Exp(Pi*I) + 1
0
>>> Log(-1) / (Pi*I)
1
>>> Log(-I) / (Pi*I)
-0.500000 {-1/2}
>>> Log(1 / 10^123) / Log(100)
-61.5000 {-123/2}
>>> Log(1 + Sqrt(2)) / Log(3 + 2*Sqrt(2))
0.500000 {1/2}
>>> Sqrt(2)*Sqrt(3) - Sqrt(6)
0
>>> Exp(1+Sqrt(2)) * Exp(1-Sqrt(2)) / (Exp(1)^2)
1
>>> I^I - Exp(-Pi/2)
0
>>> Exp(Sqrt(3))^2 - Exp(Sqrt(12))
0
>>> 2*Log(Pi*I) - 4*Log(Sqrt(Pi)) - Pi*I
0
>>> -I*Pi/8*Log(2/3-2*I/3)^2 + I*Pi/8*Log(2/3+2*I/3)^2 + Pi^2/12*Log(-1-I) + Pi^2/12*Log(-1+I) + Pi^2/12*Log(1/3-I/3) + Pi^2/12*Log(1/3+I/3) - Pi^2/48*Log(18)
0
>>> Sqrt(5 + 2*Sqrt(6)) - Sqrt(2) - Sqrt(3)
0e-1126 {a-c-d where a = 3.14626 [Sqrt(9.89898 {2*b+5})], b = 2.44949 [b^2-6=0], c = 1.73205 [c^2-3=0], d = 1.41421 [d^2-2=0]}
>>> Is zero?
T_TRUE
>>> Sqrt(I) - (1+I)/Sqrt(2)
0e-1126 + 0e-1126*I {(2*a-b*c-b)/2 where a = 0.707107 + 0.707107*I [Sqrt(1.00000*I {c})], b = 1.41421 [b^2-2=0], c = I [c^2+1=0]}
>>> Is zero?
T_TRUE
>>> Exp(Pi*Sqrt(163)) - (640320^3 + 744)
-7.49927e-13 {a-262537412640768744 where a = 2.62537e+17 [Exp(40.1092 {b*c})], b = 3.14159 [Pi], c = 12.7671 [c^2-163=0]}
>>> Erf(2*Log(Sqrt(1/2-Sqrt(2)/4))+Log(4)) - Erf(Log(2-Sqrt(2)))
0
cpu/wall(s): 0.022 0.022
virt/peak/res/peak(MB): 36.45 36.47 9.37 9.37
binet.c¶
This program computes the n-th Fibonacci number using Binet’s formula \(F_n = (\varphi^n - (1-\varphi)^n)/\sqrt{5}\) where \(\varphi = \tfrac{1}{2} (1+\sqrt{5})\). The program takes n as input.
Sample output:
> build/examples/binet 250
7.89633e+51 {7896325826131730509282738943634332893686268675876375}
cpu/wall(s): 0.002 0.001
virt/peak/res/peak(MB): 36.14 36.14 5.81 5.81
This illustrates exact arithmetic in algebraic number fields. The program also illustrates another aspect of Calcium arithmetic: evaluation limits. For example, trying to compute the index \(n = 10^6\) Fibonacci number hits an evaluation limit, so the value is not expanded to an explicit integer:
> build/examples/binet 1000000
1.95328e+208987 {(a*c-b*c)/5 where a = 4.36767e+208987 [Pow(1.61803 {(c+1)/2}, 1.00000e+6 {1000000})], b = 2.28955e-208988 [Pow(-0.618034 {(-c+1)/2}, 1.00000e+6 {1000000})], c = 2.23607 [c^2-5=0]}
cpu/wall(s): 0.006 0.005
virt/peak/res/peak(MB): 36.14 36.14 9.05 9.05
Calling the program with -limit B n raises the bit evaluation
limit to B. Setting this large enough allows \(F_{10^6}\) to expand
to an integer (the following output has been truncated to avoid
reproducing all 208988 digits):
> build/examples/binet -limit 10000000 1000000
1.95328e+208987 {1953282128...8242546875}
cpu/wall(s): 0.229 0.242
virt/peak/res/peak(MB): 36.79 37.29 7.13 7.13
The exact mechanisms and interfaces for evaluation limits are still a work in progress.
machin.c¶
This program checks several variations of Machin’s formula
expressing \(\pi\) or logarithms of small integers in terms of arctangents or hyperbolic arctangents of rational numbers. The program actually evaluates \(4 \operatorname{atan}\left(\tfrac{1}{5}\right) - \operatorname{atan}\left(\tfrac{1}{239}\right) - \tfrac{\pi}{4}\) (etc.) and prints the result, which should be precisely 0, proving the identity. Inverse trigonometric functions are not yet implemented in Calcium, so the example program evaluates them using logarithms.
Sample output:
> build/examples/machin
[(1)*Atan(1/1) - Pi/4] = 0
[(1)*Atan(1/2) + (1)*Atan(1/3) - Pi/4] = 0
[(2)*Atan(1/2) + (-1)*Atan(1/7) - Pi/4] = 0
[(2)*Atan(1/3) + (1)*Atan(1/7) - Pi/4] = 0
[(4)*Atan(1/5) + (-1)*Atan(1/239) - Pi/4] = 0
[(1)*Atan(1/2) + (1)*Atan(1/5) + (1)*Atan(1/8) - Pi/4] = 0
[(1)*Atan(1/3) + (1)*Atan(1/4) + (1)*Atan(1/7) + (1)*Atan(1/13) - Pi/4] = 0
[(12)*Atan(1/49) + (32)*Atan(1/57) + (-5)*Atan(1/239) + (12)*Atan(1/110443) - Pi/4] = 0
[(14)*Atanh(1/31) + (10)*Atanh(1/49) + (6)*Atanh(1/161) - Log(2)] = 0
[(22)*Atanh(1/31) + (16)*Atanh(1/49) + (10)*Atanh(1/161) - Log(3)] = 0
[(32)*Atanh(1/31) + (24)*Atanh(1/49) + (14)*Atanh(1/161) - Log(5)] = 0
[(144)*Atanh(1/251) + (54)*Atanh(1/449) + (-38)*Atanh(1/4801) + (62)*Atanh(1/8749) - Log(2)] = 0
[(228)*Atanh(1/251) + (86)*Atanh(1/449) + (-60)*Atanh(1/4801) + (98)*Atanh(1/8749) - Log(3)] = 0
[(334)*Atanh(1/251) + (126)*Atanh(1/449) + (-88)*Atanh(1/4801) + (144)*Atanh(1/8749) - Log(5)] = 0
[(404)*Atanh(1/251) + (152)*Atanh(1/449) + (-106)*Atanh(1/4801) + (174)*Atanh(1/8749) - Log(7)] = 0
cpu/wall(s): 0.016 0.016
virt/peak/res/peak(MB): 35.57 35.57 8.80 8.80
swinnerton_dyer_poly.c¶
This program computes the coefficients of the Swinnerton-Dyer polynomial
where \(p_n\) denotes the \(n\)-th prime number and all combinations
of signs are taken. This polynomial has degree \(2^n\).
The polynomial is expanded from its roots
using naive polynomial multiplication over ca_t coefficients.
There are far more efficient ways to construct this polynomial;
this program simply illustrates that arithmetic in
multivariate number fields works smoothly.
The program prints the coefficients of \(S_n\), from the constant term to the coefficient of \(x^{2^n}\).
Sample output:
> build/examples/swinnerton_dyer_poly 3
576
0
-960
0
352
0
-40
0
1
cpu/wall(s): 0.002 0.002
virt/peak/res/peak(MB): 35.07 35.11 5.40 5.40
A big benchmark problem (output truncated):
> build/examples/swinnerton_dyer_poly 10
4.35675e+809 {43567450015...212890625}
0
...
0
1
cpu/wall(s): 9.296 9.307
virt/peak/res/peak(MB): 38.95 38.95 10.01 10.01
huge_expr.c¶
This program proves equality of two complicated algebraic numbers. More precisely, the program verifies that \(N = -(1 - |M|^2)^2\) where N and M are given by huge symbolic expressions involving nested square roots (about 7000 operations in total).
By default, the program runs the computation using qqbar_t arithmetic:
> build/examples/huge_expr
Evaluating N...
cpu/wall(s): 7.205 7.206
Evaluating M...
cpu/wall(s): 0.933 0.934
Evaluating E = -(1-|M|^2)^2...
cpu/wall(s): 0.391 0.391
N ~ -0.16190853053311203695842869991458578203473645660641
E ~ -0.16190853053311203695842869991458578203473645660641
Testing E = N...
cpu/wall(s): 0.001 0
Equal = T_TRUE
Total: cpu/wall(s): 8.53 8.531
virt/peak/res/peak(MB): 54.50 64.56 24.64 34.61
To run the computation using ca_t arithmetic instead,
pass the -ca flag:
> build/examples/huge_expr -ca
Evaluating N...
cpu/wall(s): 0.193 0.193
Evaluating M...
cpu/wall(s): 0.024 0.024
Evaluating E = -(1-|M|^2)^2...
cpu/wall(s): 0.008 0.009
N ~ -0.16190853053311203695842869991458578203473645660641
E ~ -0.16190853053311203695842869991458578203473645660641
Testing E = N...
cpu/wall(s): 8.017 8.019
Equal = T_TRUE
Total: cpu/wall(s): 8.243 8.246
virt/peak/res/peak(MB): 61.67 65.29 33.97 37.54
This simplification problem was posted in a help request for Sage (https://ask.sagemath.org/question/52653). The C code has been generated from the symbolic expressions using a Python script.
hilbert_matrix.c¶
This program constructs the Hilbert matrix \(H_n = (1/(i+j-1))_{i=1,j=1}^n\), computes its eigenvalues \(\lambda_1, \ldots, \lambda_n\), as exact algebraic numbers, and verifies the exact trace and determinant formulas
Sample output:
> build/examples/hilbert_matrix 6
Trace:
1.87821 {6508/3465}
1.87821 {6508/3465}
Equal: T_TRUE
Det:
5.36730e-18 {1/186313420339200000}
5.36730e-18 {1/186313420339200000}
Equal: T_TRUE
cpu/wall(s): 0.07 0.069
virt/peak/res/peak(MB): 36.56 36.66 9.69 9.69
The program accepts the following optional arguments:
With
-vieta, force use of Vieta’s formula internally (by default, Calcium uses Vieta’s formulas when working with algebraic conjugates, but only up to some bound on the degree).With
-novieta, force Calcium not to use Vieta’s formulas internally.With
-qqbar, do a similar computation usingqqbar_tarithmetic.
dft.c¶
This program demonstrates the discrete Fourier transform (DFT) in exact arithmetic. For the input vector \(\textbf{x} = (x_n)_{n=0}^{N-1}\), it verifies the identity
where
The program computes the DFT by naive \(O(N^2)\) summation (not using FFT). It uses repeated multiplication of \(\omega\) to precompute an array of roots of unity \(1,\omega,\omega^2,\ldots,\omega^{2N-1}\) for use in both the DFT and the inverse DFT.
Usage:
build/examples/dft [-verbose] [-input i] [-limit B] [-timing T] N
The required parameter N selects the length of the vector.
The optional flag -verbose chooses whether to print the arrays.
The optional parameter -timing T selects a timing method (default = 0).
0: run the computation once and time it
1: run the computation repeatedly if needed to get an accurate timing, creating a new context object for each iteration so that fields are not cached
2: run the computation once, then run the computation at least one more time (repeatedly if needed to get an accurate timing), recycling the same context object to measure the performance with cached fields
The optional parameter -input i selects an input sequence (default = 0).
0: \(x_n = n+2\)
1: \(x_n = \sqrt{n+2}\)
2: \(x_n = \log(n+2)\)
3: \(x_n = e^{2 \pi i / (n+2)}\)
The optional parameter -limit B sets the internal degree limit for algebraic numbers.
Sample output:
> build/examples/dft 4 -input 1 -verbose
DFT benchmark, length N = 4
[x] =
1.41421 {a where a = 1.41421 [a^2-2=0]}
1.73205 {a where a = 1.73205 [a^2-3=0]}
2
2.23607 {a where a = 2.23607 [a^2-5=0]}
DFT([x]) =
7.38233 {a+b+c+2 where a = 2.23607 [a^2-5=0], b = 1.73205 [b^2-3=0], c = 1.41421 [c^2-2=0]}
-0.585786 + 0.504017*I {a*d-b*d+c-2 where a = 2.23607 [a^2-5=0], b = 1.73205 [b^2-3=0], c = 1.41421 [c^2-2=0], d = I [d^2+1=0]}
-0.553905 {-a-b+c+2 where a = 2.23607 [a^2-5=0], b = 1.73205 [b^2-3=0], c = 1.41421 [c^2-2=0]}
-0.585786 - 0.504017*I {-a*d+b*d+c-2 where a = 2.23607 [a^2-5=0], b = 1.73205 [b^2-3=0], c = 1.41421 [c^2-2=0], d = I [d^2+1=0]}
IDFT(DFT([x])) =
1.41421 {c where a = 2.23607 [a^2-5=0], b = 1.73205 [b^2-3=0], c = 1.41421 [c^2-2=0], d = I [d^2+1=0]}
1.73205 {b where a = 2.23607 [a^2-5=0], b = 1.73205 [b^2-3=0], c = 1.41421 [c^2-2=0], d = I [d^2+1=0]}
2
2.23607 {a where a = 2.23607 [a^2-5=0], b = 1.73205 [b^2-3=0], c = 1.41421 [c^2-2=0], d = I [d^2+1=0]}
[x] - IDFT(DFT([x])) =
0 (= 0 T_TRUE)
0 (= 0 T_TRUE)
0 (= 0 T_TRUE)
0 (= 0 T_TRUE)
cpu/wall(s): 0.009 0.009
virt/peak/res/peak(MB): 36.28 36.28 9.14 9.14