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Computations in Commutative Algebra

What is CoCoA?


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What can we compute with CoCoA?

  • Very Big Integers
  • Rational Numbers
  • Polynomials
  • Linear Systems
  •  
  • Non-negative Integer Solutions
  • Logic Example
  • Geographical Map Colouring
  • Heron's Formula

  • 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[
      RemainingFactor := 1,
      factors := [x^2 -2*z^2,  x -y],
      multiplicities := [1,  2]]
    ]

    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
    x-y+z=2
    3x-z=-6
    x+y=1
    System := ideal(x-y+z-2, 3*x-z+6, x+y-1);
    ReducedGBasis(System);
    [x +3/5,  y -8/5,  z -21/5]
    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 := HilbertBasisKer(M);
    L := [h In H | h[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 := 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]
    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[edge[1]])-F(x[edge[2]]))/(x[edge[1]]-x[edge[2]])
                      |  edge 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(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);
    record[
      RemainingFactor := 1,
      factors := [a +b -c,  a -b +c,  a +b +c,  a -b -c],
      multiplicities := [1,  1,  1,  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: 20 November 2018