[En] What is CoCoA?
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- CoCoA is a program to compute with numbers and polynomials.
- It is free.
- It works on many operating systems.
- It is used by many researchers, but can be
useful even for "simple" computations.
[En] What can we compute with CoCoA?
[En] Very Big Integers
[En] The biggest "machine integer" you can use on a
32-bit computer is 2^32-1, but CoCoA, thanks to the powerful GMP
library, can even compute numbers as big as 2^300000: try it!
[En] Rational Numbers
[En] CoCoA is very precise with fractions: it never
approximates them! So 1/3 is quite different from 0.3333333333333
(1/3) * 3;
0.3333333333333 * 3;
[En] CoCoA is specialized in polynomial computations:
it can multiply, divide, factorize, ...
(x-y)^2 * (x^4-4*z^4) / (x^2+2*z^2);
x^4 -2*x^3*y +x^2*y^2 -2*x^2*z^2 +4*x*y*z^2 -2*y^2*z^2
Factor(x^4 -2*x^3*y +x^2*y^2 -2*x^2*z^2 +4*x*y*z^2 -2*y^2*z^2);
RemainingFactor := 1,
factors := [x^2 -2*z^2, x -y],
multiplicities := [1, 2]]
[En] Linear Systems
[En] CoCoA can solve linear systems. You just need to
write every equation
f = c
as the polynomial
. CoCoA can also solve polynomial systems, but this is a
bit more difficult and we'll see it later. Now we solve
System := ideal(x-y+z-2, 3*x-z+6, x+y-1);
[x +3/5, y -8/5, z -21/5]
[En] Hence the solution is (z=21/5, x=-3/5, y=8/5)
[En] Non-negative Integer Solutions
[En] Can you find the triples of non-negative integer
solutions of the following system?
|3x - 4y + 7z||=2|
| 2x - 2y + 5z||=10|
M := mat([[3, -4, 7, -2], [2, -2, 5, -10]]);
H := HilbertBasisKer(M);
L := [h In H | h <= 1];
[[0, 10, 6, 1], [6, 11, 4, 1], [12, 12, 2, 1], [18, 13, 0, 1]]
[En] The interpretation is that there are
just four solutions:
(0, 10, 6), (6, 11, 4), (12, 12, 2), (18, 13, 0).
[En] Logic Example
[En] A says: "B lies."
B says: "C lies."
C says: "A and B lie."
Now, just who is lying here?
To answer this question we code TRUE with 1 and FALSE with 0 in
I1 := ideal(a, b-1);
I2 := ideal(a-1, b);
A := intersect(I1, I2);
I3 := ideal(b, c-1);
I4 := ideal(b-1, c);
B := intersect(I3, I4);
I5 := ideal(a, b, c-1);
I6 := ideal(b-1, a, c);
I7 := ideal(b, a-1, c);
I8 := ideal(b-1, a-1, c);
C := IntersectList([I5, I6, I7, I8]);
ReducedGBasis(A + B + C);
[b +1, a, c]
[En] The unique solution is that A and C were
lying, and B was telling the truth.
[En] Geographical Map Colouring
[En] Can the countries on a map be coloured with three
colours in such a way that no two adjacent countries have the same
use P ::= ZZ/(3)[x[1..6]];
define F(X) return X*(X-1)*(X+1); enddefine;
VerticesEq := [ F(x[i]) | i in 1..6 ];
edges := [[1,2],[1,3], [2,3],[2,4],[2,5], [3,4],[3,6],
EdgesEq := [ (F(x[edge])-F(x[edge]))/(x[edge]-x[edge])
| edge in edges ];
I := ideal(VerticesEq) + ideal(EdgesEq) + ideal(x-1, x);
[x, x -1, x +1, x -1, x, x +1]
[En] The interpretation is that there is indeed
a colouring in this case. For instance, if 0 means blue, 1 means red,
and -1 means green, we get [country 1 = red; country 2 = blue; country
3 = green; country 4 = red; country 5 = green; country 6 = blue]
[En] Heron's Formula
[En] Is it possible to express the area of a triangle as
a function of the length of its sides?
A := [x, 0];
B := [x, 0];
C := [ 0, y];
Hp := ideal(a^2 - (x^2+y^2), b^2 - (x^2+y^2),
c - (x-x), 2*s - c*y);
E := elim(x..y, Hp);
f := monic(gens(E));
a^4 -2*a^2*b^2 +b^4 -2*a^2*c^2 -2*b^2*c^2 +c^4 +16*s^2
factor(f - 16*s^2);
RemainingFactor := 1,
factors := [a +b -c, a -b +c, a +b +c, a -b -c],
multiplicities := [1, 1, 1, 1]
[En] The interpretation is that we have
s^2 = -(1/16)(a+b+c)(a+b-c)(a-b+c)(a-b-c).
This means that the square of the area of a triangle with sides a, b,
c is p(p-a)(p-b)(p-c) where p = 1/2(a+b+c) is the semiperimeter.
Hence the answer is YES.
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Last Update: 20 November 2018