Solution of Diophantine equation ax + by = c, by means of Euclidean algorithm
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function [d,p,q,r,s] = euklidalg(a,b,c);
% Solution of Diophantine equation ax + by = c
% by means of Euclidean algorithm
% The greatest common divisor is the last nonzero remainder
% Back substitution:
% coefficients of the linear combination p, q
% r = -b/d, s = a/d
%
% test of equation correctness:
if isempty(find(a)) | isempty(find(b))
display('Diophantine equation is not given correctly!')
p=0; q=0; d=0; s=0; r=0;
return
end
alfa=a; beta=b; rem=a;
% search for the greatest common divisor:
i=0;
n=abs(length(a)-length(b))+1;
if n == 1
n=n+1;
end
% test of zero remainder:
while norm(rem,inf) > 100*eps
i=i+1;
% elimination the zero leading coefficients:
ind=find(beta);
beta=beta(ind(1):length(beta));
% quotient and remainder:
[quot,rem]=deconv(alfa,beta);
i0=1+n-length(quot);
% storing of quotients:
qq(i,i0:n)=quot;
% shift of polynomials:
alfa=beta; beta=rem;
end
% recurrent computation of the coefficients
d=alfa; p=0; q=1; m=i-1;
for i=m:-1:1
r=p; p=q
% formal rearrangement for polynomial sum executing:
rr=zeros(1,length(qq(i,:))+length(p)-1);
rr(length(rr)-length(r)+1:length(rr))=r;
% computation of further element of the sequence:
q=rr-conv(qq(i,:),p);
end
% normalization of polynomial:
ind=find(q); q=q(ind(1):length(q));
% general solution of the reduced equation:
r=-deconv(b,d); s=deconv(a,d);
return
I want solve this:
so my insert this:
a = [1, -1];
b = [0, 1, -2];
c = 1;
the result:
p = 1
d = 1
p = 1
q = -1
r = 0 -1 2
but the correct is:


Have you any idea? or some Euclidean algorithm to solve diophatine.
댓글 수: 7
답변 (2개)
Hassaan
2024년 1월 26일
편집: Hassaan
2024년 1월 26일
@Marek Hutta With a = [1, -1] and b = [0, 1, -2], it's unclear how these relate as a system of equations because of the differing lengths.
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Andika
2025년 2월 7일
편집: Andika
2025년 2월 7일
Hi, Marek.
If you have Symbolic Math Toolbox installed, you can use the gcd function with 3 output arguments. The syntax [G,C,D] = gcd(A,B,X) finds the greatest common divisor of A and B, and also returns the Bézout coefficients, C and D, such that G = A*C + B*D. Here, X is the free parameter for the polynomials in A and B. Note that the polynomials must be linear combinations of positive integer powers.
For your specific example, you can try the code below.
syms z zp
a = 1 - zp; % zp represents 1/z
b = (1-2*zp)*zp;
c = 1;
[g,u,v] = gcd(a,b,zp);
if mod(c,g) == 0 % check if the gcd g divides c, or c/gcd(a,b) must be integer
cp = c/g;
disp('Particular solution:')
x1 = u*cp % one of the particular solution
y1 = v*cp % one of the particular solution
disp('General solution:')
syms k integer
x = x1 + b/g*k % find the general solution
y = y1 - a/g*k % find the general solution
% Confirm the general solution is correct
tf = isAlways(simplify(a*x + b*y) == c)
% Substitute zp to 1/z
x = subs(x,zp,1/z)
y = subs(y,zp,1/z)
else
disp('This Diophantine equation does not have a solution.')
end
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