def pelleq_solution_generator(d, n):
u, v = diop_DN(d, 1)[0]
uv_zero = np.mat([[u], [v]])
uv_mat = np.mat([[u, d * v], [v, u]])
xy_mat = [np.mat([[x, d * y], [y, x]]) for x, y in diop_DN(d, n)]
new_mat = np.identity(2, dtype=int)
while True:
yield [np.abs(x @ new_mat @ uv_zero) for x in xy_mat]
new_mat = new_mat @ uv_mat
def test_DN():
# Most of the test cases were adapted from,
# Solving the generalized Pell equation x**2 - D*y**2 = N, John P. Robertson, July 31, 2004.
# http://www.jpr2718.org/pell.pdf
# others are verified using Wolfram Alpha.
# Covers cases where D <= 0 or D > 0 and D is a square or N = 0
# Solutions are straightforward in these cases.
assert diop_DN(3, 0) == [(0, 0)]
assert diop_DN(-17, -5) == []
assert diop_DN(-19, 23) == [(2, 1)]
assert diop_DN(-13, 17) == [(2, 1)]
assert diop_DN(-15, 13) == []
assert diop_DN(0, 5) == []
assert diop_DN(0, 9) == [(3, t)]
assert diop_DN(9, 0) == [(3 * t, t)]
assert diop_DN(16, 24) == []
assert diop_DN(9, 180) == [(18, 4)]
assert diop_DN(9, -180) == [(12, 6)]
assert diop_DN(7, 0) == [(0, 0)]
# When equation is x**2 + y**2 = N
# Solutions are interchangeable
assert diop_DN(-1, 5) == [(2, 1), (1, 2)]
assert diop_DN(-1, 169) == [(12, 5), (5, 12), (13, 0), (0, 13)]
# D > 0 and D is not a square
# N = 1
assert diop_DN(13, 1) == [(649, 180)]
assert diop_DN(980, 1) == [(51841, 1656)]
assert diop_DN(981, 1) == [(158070671986249, 5046808151700)]
assert diop_DN(986, 1) == [(49299, 1570)]
assert diop_DN(991, 1) == [
(379516400906811930638014896080, 12055735790331359447442538767)
]
assert diop_DN(17, 1) == [(33, 8)]
assert diop_DN(19, 1) == [(170, 39)]
# N = -1
assert diop_DN(13, -1) == [(18, 5)]
assert diop_DN(991, -1) == []
assert diop_DN(41, -1) == [(32, 5)]
assert diop_DN(290, -1) == [(17, 1)]
assert diop_DN(21257, -1) == [(13913102721304, 95427381109)]
assert diop_DN(32, -1) == []
# |N| > 1
# Some tests were created using calculator at
# http://www.numbertheory.org/php/patz.html
assert diop_DN(13, -4) == [(3, 1), (393, 109), (36, 10)]
# Source I referred returned (3, 1), (393, 109) and (-3, 1) as fundamental solutions
# So (-3, 1) and (393, 109) should be in the same equivalent class
assert equivalent(-3, 1, 393, 109, 13, -4) == True
assert diop_DN(13, 27) == [(220, 61), (40, 11), (768, 213), (12, 3)]
assert set(diop_DN(157, 12)) == {
(13, 1),
(10663, 851),
(579160, 46222),
(483790960, 38610722),
(26277068347, 2097138361),
(21950079635497, 1751807067011),
}
assert diop_DN(13, 25) == [(3245, 900)]
assert diop_DN(192, 18) == []
assert diop_DN(23, 13) == [(-6, 1), (6, 1)]
assert diop_DN(167, 2) == [(13, 1)]
assert diop_DN(167, -2) == []
assert diop_DN(123, -2) == [(11, 1)]
# One calculator returned [(11, 1), (-11, 1)] but both of these are in
# the same equivalence class
assert equivalent(11, 1, -11, 1, 123, -2)
assert diop_DN(123, -23) == [(-10, 1), (10, 1)]
assert diop_DN(0, 0, t) == [(0, t)]
assert diop_DN(0, -1, t) == []