random cover of a range

[Yigal Chripun]

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

what do you mean by "If the dice chooses a specific face"?
I don't understand this wording..