random cover of a range

[Jason House]

Andrei Alexandrescu Wrote:

> Andrei Alexandrescu wrote:
> > Andrei Alexandrescu wrote:
> >> Given an array of length m, return a range that iterates the array in 
> >> random order such that the entire array is visited without going 
> >> through the same element more than once. Consume minimal amounts of 
> >> memory and time. Baseline solution: allocate an array of indices, 
> >> shuffle it, then follow indices stored in the array.
> > 
> > Ok, here's something that should work.
> > 
> > Start with array a of length a.length, allocate a bitmap with one bit 
> > per element in the map. Also the number of unselected elements left in 
> > the array is kept, call it k.
> > 
> > To implement next():
> > 
> > 1. Walk from the beginning of the array to the first unselected element.
> > 
> > 2. Roll a fair dice with k faces. If the dice chooses a specific face, 
> > choose that first unselected element. Otherwise continue.
> > 
> > 3. Continue walking until the next unselected element.
> > 
> > 4. Roll a fair dice with k-1 faces. If the dice chooses a specific face, 
> > choose that unselected element. Otherwise continue from step 3 using 
> > dices with k-1, k-2, ..., 1 faces.
> > 
> > This has O(n log n) complexity. There is one obvious optimization: 
> > eliminate the first selected elements so we don't need to walk over them 
> > at each iteration. It's unclear to me how this affects complexity.
> > 
> > 
> > Andrei
> 
> This seems to work for forward ranges as well. I'll be darned.
> 
> Andrei

Random ranges and forward ranges should have different implementations. Most notably, random ranges can avoid a lot of unnecessary calls to next.