Jason Ansel

Implementing a WeightedIntegerParameter or WeightedEnum/Boolean would be easy to do. You would implement it with a projection from the unweighted space to a weighted space inside the parameter. You could...

You can do this by sub-classing `IntegerParameter` and overriding the `legal_range` method. For example: ``` class DependentIntegerParameter(IntegerParameter): def legal_range(self, config): if self.name == 'a': max_value = 99 - config['b'] else:...

Hrmm... An alternate version would be something like [this](https://gist.github.com/jansel/0e4d3ac54de7f4ff4967403284abd1e1). Note that I changed your example from `a + b < 100` to `a + b + c = 100` with...

@jbosboom is correct. The use case OpenTuner is optimized for is for search spaces far too large for exhaustive search. Adding an exhaustive search technique would be a good (and...

This should be possible already. `search_driver.get_configuration(cfg_data)` will give you a configuration object (if you call it from multiple places with the same data you will get the same object). Then...

Some existing de-duplication code is here: https://github.com/jansel/opentuner/blob/50e5eec9cf934b9117611280ef94a443eaab1f35/opentuner/search/technique.py#L256 On Fri, Jun 3, 2016 at 10:52 PM, Jason Ansel [email protected] wrote: > This should be possible already. > > `search_driver.get_configuration(cfg_data)` will give...

You could either subclass the base class directly or AsyncProceduralSearchTechnique. I would expect it not to matter. Usually training is dominated by testing configurations not the search technique. On Sun,...

Agreed, seems like that crossover code has been special cased for permutations. We should make them more general. On Wed, Jun 1, 2016 at 9:05 PM, Jeffrey Bosboom [email protected] wrote:...

The three versions represent at 5%, 10%, and 20% chance of mutating each parameter. You can select which one you want with `--technique=UniformGreedyMutation20`

It is impossible to try all permutations in a human lifetime except for short lists because the number of permutations grows exponentially. For `[0, 1, 2, 3,...99]` there are `100!...