CoCoA System
Computations in
Commutative
Algebra
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.
What can we compute with CoCoA?
Very Big Integers
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!
2^32-1;
4294967295
2^64-1;
18446744073709551615
Rational Numbers
CoCoA is very precise with fractions: it never
approximates them! So 1/3 is quite different from 0.3333333333333
(1/3) * 3;
1
0.3333333333333 * 3;
9999999999999/10000000000000
Polynomials
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);
Record[
Exponents := [1, 2],
Factors := [x^2 -2*z^2, x -y],
RemainingFactor := 1
]
Linear Systems
CoCoA can solve linear systems. You just need to
write every equation
f = c
as the polynomial
f -
c
. 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);
ReducedGBasis(System);
[z - 21/5, x + 3/5, y - 8/5]
Hence the solution is (z=21/5, x=-3/5, y=8/5)
Non-negative Integer Solutions
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 := HilbertBasis(M);
L := [A In H | A[4] <= 1];
L;
[[0, 10, 6, 1], [6, 11, 4, 1], [12, 12, 2, 1], [18, 13, 0, 1]]
The interpretation is that there are
just four solutions:
(0, 10, 6), (6, 11, 4), (12, 12, 2), (18, 13, 0).
Logic Example
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
ZZ/(2):
Use ZZ/(2)[a,b,c];
I1 := Ideal(a, b-1);
I2 := Ideal(a-1, b);
A := Intersection(I1, I2);
I3 := Ideal(b, c-1);
I4 := Ideal(b-1, c);
B := Intersection(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 := Intersection(I5, I6, I7, I8);
ReducedGBasis(A + B + C);
[a, b+1, c]
The unique solution is that A and C were
lying, and B was telling the truth.
Geographical Map Colouring
Can the countries on a map be coloured with three
colours in such a way that no two adjacent countries have the same
colour?
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],
[4,5], [4,6], [5,6]];
EdgesEq := [ (F(x[A[1]])-F(x[A[2]]))/(x[A[1]]-x[A[2]])
| A In Edges ];
I := Ideal(VerticesEq) + Ideal(EdgesEq) + Ideal(x[1]-1, x[2]);
ReducedGBasis(I);
[x[2], x[1] - 1, x[3] + 1, x[4] - 1, x[6], x[5] + 1]
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]
Heron's Formula
Is it possible to express the area of a triangle as
a function of the length of its sides?
Use QQ[x[1..2],y,a,b,c,s];
A := [x[1], 0];
B := [x[2], 0];
C := [ 0, y];
Hp := Ideal(a^2 - (x[2]^2+y^2), b^2 - (x[1]^2+y^2),
c - (x[2]-x[1]), 2*s - c*y);
E := Elim(x[1]..y,Hp);
F := Monic(Comp(Gens(E),1));
F;
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);
[[a + b + c, 1], [a + b - c, 1], [a - b + c, 1], [a - b - c, 1]]
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.
Written by Anna Bigatti
Please send comments or suggestions to cocoa(at)dima.unige.it
Last Update: 19 September 2011