Initial commit. TOMS748 1st algorithm coded following strictly the paper.
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Jérôme
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.gitignore
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*.pdf
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__pycache__
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findEnclosingBracket.py
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findEnclosingBracket.py
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# method to find a bracket that encloses a root of f
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from math import *
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def findEnclosingBracket(f, x, factor):
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""" Finds a bracket [a b] that encloses a root of f : f(a)*f(b) < 0.
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A typical value for the factor is 2.
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The initial guess must be on the correct side of the real line : the algorithm will never cross the zero during the search.
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The factor is increased every 10 iterations in order to speed up convergence for when the root is orders of magnitude away from the initial guess.
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If the bracket returned by the function is too wide, try reducing the factor (while always keeping it > 1).
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"""
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a = x
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b = x*factor # try higher
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fa = f(a)
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fb = f(b)
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i = 0
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while i in range(100) and fa*fb > 0:
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if fabs(fa) < fabs(fb): # fa is closer to 0 than fb, extend the interval in the a direction (reducing it)
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a = a / factor
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fa = f(a)
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else: # fb is closer to 0 than fa, extend the interval in the b direction (augmenting it)
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b = b * factor
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fb = f(b)
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i = i+1
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if i % 10 == 0: # every 10 iterations, bump up the factor to speed up convergence
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factor = factor * 5
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return (a, b, fa, fb)
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f = lambda x : exp(sin(x))-2
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a, b, fa, fb = findEnclosingBracket(f, 0.005, 2)
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print((a,b)); print((fa, fb))
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f = lambda x : exp(-x)-1e-200
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a, b, fa, fb = findEnclosingBracket(f, 0.001, 2)
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print((a,b)); print((fa, fb))
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a, b, fa, fb = findEnclosingBracket(f, 0.001, 5)
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print((a,b)); print((fa, fb))
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f = lambda x : x**(1/3)-1000
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a, b, fa, fb = findEnclosingBracket(f, 1e-10, 2)
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print((a,b)); print((fa, fb))
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# from toms748 import TOMS748_solve1
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# def f(x):
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# print(x)
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# return x**(1/3)-1000
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# TOMS748_solve1(f,a,b)
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toms748.py
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toms748.py
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# Implementation of TOMS748
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# https://na.math.kit.edu/alefeld/download/1995_Algorithm_748_Enclosing_Zeros_of_Continuous_Functions.pdf
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from math import sin, fabs, cos
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# finite differences 1st derivative given a, b, fa, fb
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def fbracket1(a, b, fa, fb):
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return (fb-fa)/(b-a)
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# finite differences 2nd derivative given a, b, d, fa, fb, fd
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def fbracket2(a, b, d, fa, fb, fd):
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return (fbracket1(b, d, fb, fd) - fbracket1(a, b, fa, fb))/(d-a)
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# standard bracketing routine
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# returns ahat, bhat, d
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# d is a point outside the new interval
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def bracket(a, b, c, fa, fb, fc):
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# print('bracket')
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if fabs(fc) < 1e-12:
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print("root found : %f ; f(root) = %f" % (c, fc))
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if fa*fc < 0:
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return (a, c, b, fa, fc, fb)
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else:
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return (c, b, a, fc, fb, fa)
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# finds an approximate solution to the quadratic P(x) = fa + f[a,b]*(x-a) + f[a,b,d]*(x-a)(x-b)
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# with f[a,b] = fbracket1(a,b) and f[a,b,d] = fbracket2(a,b,d)
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def NewtonQuadratic(a, b, d, fa, fb, fd, k):
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A = fbracket2(a,b,d,fa,fb,fd)
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B = fbracket1(a,b,fa,fb)
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if A == 0:
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return a - fa/B
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if A*fa > 0:
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r = a
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else:
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r = b
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for i in range(k):
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r = r - B*r/(B + A*(2*r - a - b))
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return r
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# Inverse cubic interpolation evaluated at 0 (modified Aitken-Neville interpolation)
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def ipzero(a, b, c, d, fa, fb, fc, fd):
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Q11 = (c-d)*fc/(fd-fc)
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Q21 = (b-c)*fb/(fc-fb)
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Q31 = (a-b)*fa/(fb-fa)
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D21 = (b-c)*fc/(fc-fb)
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D31 = (a-b)*fb/(fb-fa)
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Q22 = (D21-Q11)*fb/(fd-fb)
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Q32 = (D31-Q21)*fa/(fc-fa)
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D32 = (D31-Q21)*fc/(fc-fa)
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Q33 = (D32-Q22)*fa/(fd-fa)
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return a + Q31 + Q32 + Q33
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def TOMS748_solve1(f, a, b):
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mu = 0.5
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fa, fb = f(a), f(b)
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c = a - fa/fbracket1(a, b, fa, fb) # 4.1.1 secant method
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fc = f(c)
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a, b, d, fa, fb, fd = bracket(a, b, c, fa, fb, fc) # 4.1.2
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e = d
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fe = fd
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# ---
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for n in range(2,10): # 4.1.3
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if n == 2 or (fa == fb or fb == fd or fd == fe):
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c = NewtonQuadratic(a, b, d, fa, fb, fd, 2)
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else:
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c = ipzero(a, b, d, e, fa, fb, fd, fe)
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if (c-a)*(c-b) >= 0:
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c = NewtonQuadratic(a, b, d, fa, fb, fd, 2)
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# ---
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fc = f(c)
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a, b, dbar, fa, fb, fdbar = bracket(a, b, c, fa, fb, fc) # 4.1.4
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# ---
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if fabs(fa) < fabs(fb): # 4.1.5
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u = a
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fu = fa
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else:
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u = b
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fu = fb
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# ---
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c = u - 2*fu/fbracket1(a, b, fa, fb) # 4.1.6
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# ---
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if fabs(c - u) > 0.5*(b - a): # 4.1.7
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c = 0.5*(b + a)
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# ---
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fc = f(c)
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a, b, dhat, fa, fb, fdhat = bracket(a, b, c, fa, fb, fc) # 4.1.8
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# ---
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if b - a < mu*(b - a): # 4.1.9
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d = dhat
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e = dbar
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fd = fdhat
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fe = fdbar
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else:
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e = dhat
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fe = fdhat
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c = 0.5*(a+b)
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fc = f(c)
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a, b, d, fa, fb, fd = bracket(a, b, c, fa, fb, fc)
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# TEST
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f = lambda E : E - 0.5*sin(E) - 0.3
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a, b, c, d = 0, 1, 0.3, 1.2
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fa, fb, fc, fd = f(a), f(b), f(c), f(d)
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print(fbracket1(a, b, fa, fb))
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print(fbracket2(a, b, d, fa, fb, fd))
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print(bracket(a, b, c, fa, fb, fc))
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print(NewtonQuadratic(a, b, d, fa, fb, fd, 2))
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print(ipzero(a, b, c, d, fa, fb, fc, fd))
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# solution using bracket only :
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print('bracket only')
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for i in range(50):
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c = (a+b)/2
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fc = f(c)
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a, b, d, fa, fb, fd = bracket(a, b, c, fa, fb, fc)
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if fabs(b-a) < 1e-12:
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break
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print(i)
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print((a,b), (a+b)/2, b-a)
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print('Newton with derivative')
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df = lambda E : 1 - 0.5*cos(E)
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x = 0.5
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for i in range(50):
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delta = f(x)/df(x)
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x = x - delta
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if fabs(delta) < 1e-12:
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break
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print(i)
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print(x)
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print('TOMS748_solve1 only')
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def fct1(E):
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print("x = %20.15f" % E)
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return E - 0.5*sin(E) - 0.3
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TOMS748_solve1(fct1, 0, 1)
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