Question

这是Maple中扩展欧几里得算法的代码，它应返回l的整数pi,ri,si,ti，多项式0<=i<=l+1。 qi的多项式为1<=i<=l，si(f)+ti(g) = ri和sl(f)+tl(g)=rl=GCD(f,g)。

问题是，我一直被零除。此外，它每次都会将pi = lcoeff(ri-1 - qiri)评估为零。即使我删除了它，它仍然表示存在零除法，它必须来自qi:=quo(ri-1,ri, x);但是我不知道为什么考虑循环的要求是r[i]不等于零。我真的可以用一双新眼睛来看看我做错了什么。

＆＃13;

ExtEuclidean := proc (f, g) local p, r, s, t, i, q, l;
p[0] := lcoeff(f); 
p[1] := lcoeff(g); 
r[0] := f/p[0]; 
r[1] := g/p[1]; 
s[0] := 1; 
s[1] := 0; 
t[0] := 0; 
t[1] := 1; 
i := 1; 
while r[i] <> 0 do 
l := i; 
q[i] := quo(r[i-1], r[i], x, 'remainder');
simplify(q[i]); 
p[i+1] := lcoeff(r[i-1]-q[i]*r[i]);
r[i+1] := (r[i-1]-q[i]*r[i])/p[i+1]; 
s[i+1] := (s[i-1]-q[i]*s[i])/p[i+1]; 
t[i+1] := (t[i-1]-q[i]*t[i])/p[i+1]; 
i := i+1; 
simplify(r[i]) 
end do; 
return l, p[i], r[i], s[i], t[i], q[i] 
end proc;

＆＃13;

Answer 1

您的索引看起来有问题。我没想太多，但这似乎适用于基本的例子。我在循环中将p [i + 1]更改为p [i]并更改了输出

ExtEuclidean := proc (f, g) local p, r, s, t, i, q, l;
    p[0] := lcoeff(f); 
    p[1] := lcoeff(g); 
    r[0] := f/p[0]; 
    r[1] := g/p[1]; 
    s[0] := 1; 
    s[1] := 0; 
    t[0] := 0; 
    t[1] := 1; 
    i := 1; 
    while r[i] <> 0 do 
    l := i; 
    q[i] := quo(r[i-1], r[i], x, 'remainder');
    simplify(q[i]); 
    p[i+1] := lcoeff(r[i-1]-q[i]*r[i]);
    r[i+1] := (r[i-1]-q[i]*r[i])/p[i]; 
    s[i+1] := (s[i-1]-q[i]*s[i])/p[i]; 
    t[i+1] := (t[i-1]-q[i]*t[i])/p[i]; 
    i := i+1; 
    simplify(r[i]) 
    end do; 
    return l, p[i-1], r[i-1], s[i-1], t[i-1], q[i-1] 
    end proc;

Maple中的扩展欧氏算法

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