# Numerical methods challenge: Day 5

During October (2017) I will write a program per day for some well-known numerical methods in both Python and Julia. It is intended to be an exercise then don't expect the code to be good enough for real use. Also, I should mention that I have almost no experience with Julia, so it probably won't be idiomatic Julia but more Python-like Julia.

## Broyden's method

Today we have Broyden's method. The main idea in this method is to compute the Jacobian matrix just at the first iteration and change it for each iteration doing rank-1 updates.

where we need estimate the Jacobian matrix at step \(n\) by

that correspond to rank-1 updates of the Jacobian matrix.

We will test the method with the function \(\mathbf{F}(x, y) = (x + 2y - 2, x^2 + 4x - 4)\) with solution \(\mathbf{x} = (0, 1)\).

Following are the codes.

### Python

from __future__ import division, print_function from numpy import array, outer, dot from numpy.linalg import solve, norm, det def broyden(fun, jaco, x, niter=50, ftol=1e-12, verbose=False): msg = "Maximum number of iterations reached." J = jaco(x) for cont in range(niter): if det(J) < ftol: x = None msg = "Derivative near to zero." break if verbose: print("n: {}, x: {}".format(cont, x)) f_old = fun(x) dx = -solve(J, f_old) x = x + dx f = fun(x) df = f - f_old J = J + outer(df - dot(J, dx), dx)/dot(dx, dx) if norm(f) < ftol: msg = "Root found with desired accuracy." break return x, msg def fun(x): return array([x[0] + 2*x[1] - 2, x[0]**2 + 4*x[1]**2 - 4]) def jaco(x): return array([ [1, 2], [2*x[0], 8*x[1]]]) print(broyden(fun, jaco, [1.0, 2.0]))

### Julia

function broyden(fun, jaco, x, niter=50, ftol=1e-12, verbose=false) msg = "Maximum number of iterations reached." J = jaco(x) for cont = 1:niter if det(J) < ftol x = nothing msg = "Derivative near to zero." break end if verbose println("n: $(cont), x: $(x)") end f_old = fun(x) dx = -J\f_old x = x + dx f = fun(x) df = f - f_old J = J + (df - J*dx) * dx'/ (dx' * dx) if norm(f) < ftol msg = "Root found with desired accuracy." break end end return x, msg end function fun(x) return [x[1] + 2*x[2] - 2, x[1]^2 + 4*x[2]^2 - 4] end function jaco(x) return [1 2; 2*x[1] 8*x[2]] end println(broyden(fun, jaco, [1.0, 2.0]))

### Comparison Python/Julia

Regarding number of lines we have: 38 in Python and 39 in Julia. The comparison
in execution time is done with `%timeit`

magic command in IPython and
`@benchmark`

in Julia.

For Python:

%timeit broyden(fun, jaco, [1.0, 2.0])

with result

1000 loops, best of 3: 703 µs per loop

For Julia:

@benchmark broyden(fun, jaco, [1.0, 2.0])

with result

BenchmarkTools.Trial: memory estimate: 14.41 KiB allocs estimate: 220 -------------- minimum time: 12.099 μs (0.00% GC) median time: 12.867 μs (0.00% GC) mean time: 15.378 μs (10.78% GC) maximum time: 3.511 ms (97.53% GC) -------------- samples: 10000 evals/sample: 1

In this case, we can say that the Python code is roughly 50 times slower than the Julia one.

### Comparison Newton/Broyden

Following, we are comparing Newton's and Broyden's method. We are using the function \(\mathbf{x}^T \mathbf{x} + \mathbf{x}\) for this test.

#### Python

The code for the function and Jacobian is

from numpy import diag fun = lambda x: x**2 + x jaco = lambda x: diag(2*x + 1)

and the results are:

n |
Newton (μs) |
Broyden (μs) |

2 |
500 |
664 |

10 |
541 |
717 |

100 |
3450 |
4800 |

#### Julia

The code for the function and Jacobian is

fun(x) = x' * x + x jaco(x) = diagm(2*x + 1)

and the results are:

n |
Newton (μs) |
Broyden (μs) |

2 |
1.76 |
1.65 |

10 |
56.42 |
5.12 |

100 |
1782 |
367 |

In this case, we are comparing the mean values of the results from
`@benchmark`

.

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