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I'm trying to make an exercise which gets a list of numers, an shows a list of elements like this: if A=[a0,a1,a2] then there is U=[u0,u1,u2], knowing that a0*u0 + a1*u1 + a2*u2 = d and d is the gcd of A.

For 2 elements is a pretty simple thing, as Sage has a function to retrieve u0 and u1 out of a0 and a1: 

A=[15,21]
(d,u0,u1)=xgcd(a[0],a[1])

I just don't understand how could I do this with a list of n elements.

Zasito
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2 Answers2

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Note that gcd(a, b, c) = gcd((gcd(a, b), c). This means that you can use the built-in function repeatedly to calculate the coefficients that you want.

Code-Apprentice
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You helped me a lot, came to this:

x1=[1256,5468,5552,1465]
n=-1
for i in x1:
    n=n+1
(d,w,x)=xgcd(x1[n-1],x1[n])
u1=[w,x]
n=n-2
while n>=0:
    div=d
    (d,u,v)=xgcd(x1[n],div)
    position=0
    for j in u1:
        a=j*v
        u1[position]=a
        position=position+1
    u1=[u]+u1
    n=n-1
u1

And it works ;)

Zasito
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