# Integer step size in scipy optimize minimize

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I have a computer vision algorithm I want to tune up using scipy.optimize.minimize. Right now I only want to tune-up two parameters but the number of parameters might eventually grow so I would like to use a technique that can do high-dimensional gradient searches. The Nelder-Mead implementation in SciPy seemed like a good fit.

I got the code all set up but it seems that the minimize function really wants to use floating-point values with a step size that is less than one. The current set of parameters are both integers and one has a step size of one and the other has a step size of two (i.e. the value must be odd, if it isn't the thing I am trying to optimize will convert it to an odd number). Roughly one parameter is a window size in pixels and the other parameter is a threshold (a value from 0-255).

For what it is worth I am using a fresh build of scipy from the git repo. Does anyone know how to tell scipy to use a specific step size for each parameter? Is there some way I can roll my own gradient function? Is there a scipy flag that could help me out? I am aware that this could be done with a simple parameter sweep, but I would eventually like to apply this code too much larger sets of parameters.

The code itself is dead simple:

import numpy as np

from scipy.optimize import minimize

from ScannerUtil import straightenImg

import bson

def doSingleIteration(parameters):

# do some machine vision magic

# return the difference between my value and the truth value

parameters = np.array([11,10])

res = minimize( doSingleIteration, parameters, method='Nelder-Mead',options={'xtol': 1e-2, 'disp': True,'ftol':1.0,}) #not sure if these params do anything

print "~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~"

print res

This is what my output looks like. As you can see we are repeating a lot of runs and not getting anywhere in the minimization.

[ 11.  10.]  <-- Output from scipy minimize

{'block_size': 11, 'degree': 10} <-- input to my algorithm rounded and made int

120  <-- output of the function I am trying to minimize

[ 11.55  10.  ]

{'block_size': 11, 'degree': 10}

120

[ 11.   10.5]

{'block_size': 11, 'degree': 10}

120

[ 11.55   9.5 ]

{'block_size': 11, 'degree': 9}

120

[ 11.1375  10.25  ]

{'block_size': 11, 'degree': 10}

120

[ 11.275  10.   ]

{'block_size': 11, 'degree': 10}

120

[ 11.    10.25]

{'block_size': 11, 'degree': 10}

120

[ 11.275   9.75 ]

{'block_size': 11, 'degree': 9}

120

SNIP

[ 11.         10.0078125]

{'block_size': 11, 'degree': 10}

120

Optimization terminated successfully.

Current function value: 120.000000

Iterations: 7

Function evaluations: 27

status: 0

nfev: 27

success: True

fun: 120.0

x: array([ 11.,  10.])

message: 'Optimization terminated successfully.'

nit: 7*

## 1 Answer

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by (33.2k points)

Let’s say that the function to minimize is arbitrarily complex (nonlinear), this is a complicated problem in general. You have to try every possible solution to get the most optimal result. If you are not aware of any constrained non-linear optimizer, then Nelder-Mead should work fine if it was a continuous function.

Simply set up a coarse+fine grid search first, if you then feel like trying if your Nelder-Mead works (and converges faster), the points below may help.

But maybe some points that help:

Here, the whole integer constraint is very difficult, maybe it would be an option to do some simple interpolation to improve the optimizer. It should converge to an integer solution. Of course, this requires to calculate extra points, but it might solve various other problems.

Nelder-Mead starts with N+1 points, this is hard-wired in scipy to (1+0.05) * x0[j] (for j in all dimensions, unless x0[j] is 0), which you will see in your first evaluation steps. This can be fulfilled in newer versions, otherwise, you could just change/copy the scipy code and set it to something more reasonable. Or if you feel that it is simpler, scale all input variables down so that (1+0.05)*x0 is of a sensible size.

You should create all function evaluations since if you use Nelder-Mead I would guess you can always run into the duplicate evaluation.

Using Nelder-Mead will just shrink to a single value and give up because it always finds the same result.

This optimization is doomed if the function does not change smoothly over the parameter space, and even then it can easily run into local minima if you should have of those.

For more details, study SciPy Tutorial. One can also go through Python Course to master the domain.

Hope this answer helps.

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