I want to cal
https://en.wikipedia.org/wiki/Diophantine_equation#Examples
ax by = c This is a linear Diophantine equation.
i try
I think the way you substituted h is not good. How should I fix it?
x,y integer
(1)(2)5^4x-2^4y=1
(3)5^5x-2^5y=1
(4)11^5x-2^5y=1
from sympy.solvers.inequalities import reduce_rational_inequalities
from sympy import *
import math
var('x y n')
def myEq(f,g,h,myMax):
myN = Symbol('myN', real=True)
myGcdex = gcdex(f, -g)
myX = g * n myGcdex[2]
myY = solve((f * x - g * y - h).subs({x: myX}), y)[0]
myN = math.ceil(reduce_rational_inequalities([[myX.subs({n: myN}) >= myMax]], myN).lhs)
return myX.subs({n: myN}),myY.subs({n: myN})
print("#(1)",myEq ( 5**4,2**4,1, 0) )
print("#(2)",myEq ( 5**4,2**4,1, 10) )
print("#(3)",myEq ( 5**5,2**5,1,100), "????<correct>x=125,y=12207")
print("#(4)",myEq (11**5,2**5,1, 0)," ","????<correct>x= 19,y=95624")
#(1) (1, 39)
#(2) (17, 664)
#(3) (129, 100781/8) ????<correct>x=125,y=12207
#(4) (1, 80525/16) ????<correct>x= 19,y=95624
#
#〇(1)nxy 0 16*n 1 625*n 39 <correct>x = 16 n 1 , y = 625 n 39 , n=0
#〇(2)nxy 1 16*n 1 625*n 39
#×(3)nxy 4 32*n 1 3125*n 781/8 <correct>x = 32 n 29 , y = 3125 n 2832 , n=3
#×(4)nxy 0 32*n 1 161051*n 80525/16 <correct>x = 32 n 19 , y = 161051 n 95624, n=0
ref
I want function convert from xy to cells
https://mathworld.wolfram.com/Congruence.html
(20220407)
MY_diox( 5**4*x-2**4*y-1, 0) # (( 1, 39), (16*n - 15, 625*n - 586))
MY_diox( 5**4*x-2**4*y-1, 10) # (( 17, 664), (16*n 1, 625*n 39))
MY_diox( 5**5*x-2**5*y-1,100) # ((125, 12207), (32*n 93, 3125*n 9082))
MY_diox(11**5*x-2**5*y-1, 0) # (( 19, 95624), (32*n - 13, 161051*n - 65427))
CodePudding user response:
It looks like you are trying to get a specific solution to the equation that has a certain minimum value for x. If you solve the inequality for the parameter you can convert that to an interval on integers and just select the first such integer. A modified equation to give that value at n = 1
can also be given using a shift of origin.
def diox(eq, x):
"""
>>> from sympy.abc import x, y
>>> eq = 5**5*x-2**5*y-1
>>> diox(eq, 30)
((61, 5957), (32*n 29, 3125*n 2832))
"""
s = diophantine(eq)
assert len(s) == 1
a, b = s.pop()
p = a.free_symbols
assert len(p) == 1
p = p.pop()
i = solve(a >= x).as_set().intersection(Integers)
if i.start.is_infinite:
i = i.sup
else:
i = i.start
t = Tuple(a, b).xreplace({p: i})
e = Tuple(a, b).xreplace({p: Symbol('n') i-1})
return t, e
CodePudding user response:
As expected, the following was not possible.
from sympy import *
from sympy.abc import x, y
from sympy.solvers.diophantine import diophantine
var('x a b c')
myFormula= a*x b*y-c
def diox(eq, x):
.......................................
return t, e
print("#",diox(myFormula,0))
# NotImplementedError: No solver has been written for inhomogeneous_general_quadratic.