Integer solutions to linear equation?
mwh21 at cam.ac.uk
Tue Apr 18 18:03:51 CEST 2000
grant at nowhere. (Grant Edwards) writes:
> This isn't really a Python question, but my example is in
> Python, and there seem to be plenty of people who know a fair
> bit of math here....
I hope I count ...
> A friend of mine ran across a brain-teaser involving a bunch of
> flowers, some magic bowls and some other camoflage text. What
> you end up with is having to solve the equation
> 64x = 41y + 1
> Where x and y are both integers. After scratching our heads
> for a while, we used the brute force approach:
> for x in range(1,100):
> y = ((64*x)-1)/41
> if 64*x == 41*y+1:
> print (x,y)
> The results:
> (25, 39)
> (66, 103)
> 25 was the expected solution, so we got both the equation and
> the Python snippet correct. Is there a non-iterative way to
> solve a linear equation when the results are contrained to be
> integers? I don't remember anything from any of my math
> classes that seems relevent, but I didn't take any anything
> beyond what is required for all undergrad engineers.
Use Euclid's algorithm; here's some code that does that:
""" return h,q,r st h==q*x0+r*y0 """
if x0<0: qn1=-qn1
if y0<0: rn1=-rn1
""" solve(m,n,a:Integer) -> (Integer,Integer)
find integers (x,y,r,s) such that (m+u*r)*x+(n+u*s)*y==a
for all integers u."""
h,x,y = find_hcf_as_lc(m,n)
if a%h <> 0:
raise "no solution"
(-16, -25, -41, -64)
And your solution is:
(-16 - -41)*64 + (-25 - -64)*-41
I feel I must have neater code knocking about somewhere to do this,
but I can't find it right now.
it's not that perl programmers are idiots, it's that the language
rewards idiotic behavior in a way that no other language or tool
has ever done -- Erik Naggum, comp.lang.lisp
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