## digitalmars.D - And here's another interesting algorithm/structure: Randomized Slide

Andrei Alexandrescu <SeeWebsiteForEmail erdani.org> writes:
```Now that we got talking about searching in arrays, allow me to also
share an idea I've had a short while ago.

(Again, we're in the "I'd prefer to use an array if at all possible"
mindset. So let's see how we can help searching an array with as little
work as possible.)

One well-known search strategy is "Bring to front" (described by Knuth
in TAoCP). A BtF-organized linear data structure is searched with the
classic linear algorithm. The difference is what happens after the
search: whenever the search is successful, the found element is brought
to the front of the structure. If we're looking most often for a handful
of elements, in time these will be near the front of the searched structure.

For a linked list, bringing an element to the front is O(1) (just rewire
the pointers). For an array, things are not so pleasant - rotating the
found element to the front of the array is O(n).

So let's see how we can implement a successful BtF for arrays.

The first idea is to just swap the found element with the first element
of the array. That's O(1) but has many disadvantages - if you search
e.g. for two elements, they'll compete for the front of the array and
they'll go back and forth without making progress.

Another idea is to just swap the found element with the one just before
it. The logic is, each successful find will shift the element closer to
the front, in a bubble sort manner. In time, the frequently searched
elements will slowly creep toward the front. The resulting performance
is not appealing - you need O(n) searches to bring a given element to
the front, for a total of O(n * n) steps spent in the n searches. Meh.

So let's improve on that: whenever an element is found in position k,
pick a random number i in the range 0, 1, 2, ..., k inclusive. Then swap
the array elements at indexes i and k. This is the Randomized Slide to
Front strategy.

With RStF, worst case search time remains O(n), as is the unsuccessful
search. However, frequently searched elements migrate quickly to the
front - it only takes O(log n) searches to bring a given value at the
front of the array.

Insertion and removal are both a sweet O(1), owing to the light
structuring: to insert just append the element (and perhaps swap it in a
random position of the array to prime searching for it). Removal by
position simply swaps the last element into the position to be removed
and then reduces the size of the array.

So the RStF is suitable in all cases where BtF would be recommended, but
allows an array layout without considerable penalty.

Related work: Theodoulos Garefalakis' Master's thesis "A Family of
Randomized Algorithms for List Accessing" describes Markov Move to
Front, which brings the searched element to front according to a Markov
chain schedule; and also Randomized Move to Front, which decides whether
a found element is brought to front depending on a random choice. These
approaches are similar in that they both use randomization, but
different because neither has good complexity on array storage.

Andrei
```
Nov 30 2015
Andrei Alexandrescu <SeeWebsiteForEmail erdani.org> writes:
```On 11/30/15 4:33 PM, Andrei Alexandrescu wrote:
[snip]

I just posted to reddit:

Andrei
```
Nov 30 2015
"H. S. Teoh via Digitalmars-d" <digitalmars-d puremagic.com> writes:
```On Mon, Nov 30, 2015 at 04:33:27PM -0500, Andrei Alexandrescu via Digitalmars-d
wrote:
[...]
One well-known search strategy is "Bring to front" (described by Knuth
in TAoCP). A BtF-organized linear data structure is searched with the
classic linear algorithm. The difference is what happens after the
search: whenever the search is successful, the found element is
brought to the front of the structure. If we're looking most often for
a handful of elements, in time these will be near the front of the
searched structure.

[...]
So let's see how we can implement a successful BtF for arrays.

The first idea is to just swap the found element with the first
element of the array. That's O(1) but has many disadvantages - if you
search e.g. for two elements, they'll compete for the front of the
array and they'll go back and forth without making progress.

Another idea is to just swap the found element with the one just
before it.  The logic is, each successful find will shift the element
closer to the front, in a bubble sort manner. In time, the frequently
searched elements will slowly creep toward the front. The resulting
performance is not appealing - you need O(n) searches to bring a given
element to the front, for a total of O(n * n) steps spent in the n
searches. Meh.

So let's improve on that: whenever an element is found in position k,
pick a random number i in the range 0, 1, 2, ..., k inclusive. Then
swap the array elements at indexes i and k. This is the Randomized
Slide to Front strategy.

What about when element i is matched, swap it with the (i/2)'th element?

Then it will take just log(n) searches to bring it to the front of the
array, but it won't (immediately) compete with whatever's currently the
most popular item in the array. Furthermore, when it does compete with
it, it will already have been moved closer to the front of the array, so
the previous most-popular element won't be pushed far back into the deep
recesses of the array, but remain close to the front where it will be
quickly found.

More generally, you could pick the (i/k)'th element for swapping when
you've matched an element, with k being a constant, chosen parameter.
You may be able to optimize for certain use cases (e.g., if you knew
beforehand the average number of "popular" elements) by choosing an
appropriate value of k.

T

--
Nobody is perfect.  I am Nobody. -- pepoluan, GKC forum
```
Nov 30 2015
Andrei Alexandrescu <SeeWebsiteForEmail erdani.org> writes:
```On 11/30/15 4:41 PM, H. S. Teoh via Digitalmars-d wrote:
What about when element i is matched, swap it with the (i/2)'th element?

Randomization is essential - without it you have thrashing if you search
for 2 elements in alternation. -- Andrei
```
Nov 30 2015
Steven Schveighoffer <schveiguy yahoo.com> writes:
```On 11/30/15 4:50 PM, Andrei Alexandrescu wrote:
On 11/30/15 4:41 PM, H. S. Teoh via Digitalmars-d wrote:
What about when element i is matched, swap it with the (i/2)'th element?

Randomization is essential - without it you have thrashing if you search
for 2 elements in alternation. -- Andrei

-Steve
```
Nov 30 2015
"H. S. Teoh via Digitalmars-d" <digitalmars-d puremagic.com> writes:
```On Mon, Nov 30, 2015 at 04:58:16PM -0500, Steven Schveighoffer via
Digitalmars-d wrote:
On 11/30/15 4:50 PM, Andrei Alexandrescu wrote:
On 11/30/15 4:41 PM, H. S. Teoh via Digitalmars-d wrote:
What about when element i is matched, swap it with the (i/2)'th element?

Randomization is essential - without it you have thrashing if you
search for 2 elements in alternation. -- Andrei

[...]

Or selecting the (i/k)'th element for k = uniform(1..i)?

T

--
People walk. Computers run.
```
Nov 30 2015
Andrei Alexandrescu <SeeWebsiteForEmail erdani.org> writes:
```On 11/30/15 4:58 PM, Steven Schveighoffer wrote:
On 11/30/15 4:50 PM, Andrei Alexandrescu wrote:
On 11/30/15 4:41 PM, H. S. Teoh via Digitalmars-d wrote:
What about when element i is matched, swap it with the (i/2)'th element?

Randomization is essential - without it you have thrashing if you search
for 2 elements in alternation. -- Andrei

I think complexity would stay the same. Choosing a tighter range puts a
greater weight on the last search than on recent searches.

One thing I like is that I choose 0..k, not 0..k-1, which means it's
possible that the element stays put (no change in position). That
reduces thrashing for the top (most frequently searched) few elements.

andrei
```
Nov 30 2015
Steven Schveighoffer <schveiguy yahoo.com> writes:
```On 11/30/15 5:07 PM, Andrei Alexandrescu wrote:
On 11/30/15 4:58 PM, Steven Schveighoffer wrote:

I think complexity would stay the same. Choosing a tighter range puts a
greater weight on the last search than on recent searches.

In the case where you search for a very small number of elements, it
puts an upper bound on how soon they make it to the front (log(n)

One thing I like is that I choose 0..k, not 0..k-1, which means it's
possible that the element stays put (no change in position). That
reduces thrashing for the top (most frequently searched) few elements.

I think insignificantly, depending on the number of "frequently
searched" elements. But you could tune it, let's say to 8 elements:

const upperBound = max(k/2, min(8, k));

There are a lot of options for tuning that can be played with, probably
the best way to "prove" what is best is to just try some experiments :)

-Steve
```
Dec 01 2015
Chris Wright <dhasenan gmail.com> writes:
```On Mon, 30 Nov 2015 16:58:16 -0500, Steven Schveighoffer wrote:

On 11/30/15 4:50 PM, Andrei Alexandrescu wrote:
On 11/30/15 4:41 PM, H. S. Teoh via Digitalmars-d wrote:
What about when element i is matched, swap it with the (i/2)'th
element?

Randomization is essential - without it you have thrashing if you
search for 2 elements in alternation. -- Andrei

-Steve

You can use that to put a hard upper bound of O(log n), and maybe you'll
want to use that. However, that also means you have greater odds of a
single rare query making it to the front of the stack.

If you want to prevent that and still get a guarantee of O(log n), you
could choose a range of floor(sqrt(k))..k/2.
```
Nov 30 2015
```On Monday, 30 November 2015 at 21:50:09 UTC, Andrei Alexandrescu
wrote:
On 11/30/15 4:41 PM, H. S. Teoh via Digitalmars-d wrote:
What about when element i is matched, swap it with the
(i/2)'th element?

Randomization is essential - without it you have thrashing if
you search for 2 elements in alternation. -- Andrei

You'd end up swaping the 2 element in front, but keep them both
in front, so that sounds like it would have the same behavior as
the randomized algorithm.

Where it gets hairy, is when you access 2 elements in the array
that would swap each other without getting in the front (because
they are at 2n and 2n + 1 with n big).
```
Nov 30 2015
Denis Koroskin <2korden gmail.com> writes:
```On Monday, 30 November 2015 at 22:11:09 UTC, deadalnix wrote:
On Monday, 30 November 2015 at 21:50:09 UTC, Andrei
Alexandrescu wrote:
On 11/30/15 4:41 PM, H. S. Teoh via Digitalmars-d wrote:
What about when element i is matched, swap it with the
(i/2)'th element?

Randomization is essential - without it you have thrashing if
you search for 2 elements in alternation. -- Andrei

You'd end up swaping the 2 element in front, but keep them both
in front, so that sounds like it would have the same behavior
as the randomized algorithm.

Where it gets hairy, is when you access 2 elements in the array
that would swap each other without getting in the front
(because they are at 2n and 2n + 1 with n big).

Imagine that there are 1000 elements, 500th elements is X and
1000th element is Y.

1) search for Y: Y is last, takes 1000 iterations, swaps X<->Y
2) search for X: X is last, takes 1000 iterations, swaps X<->Y
3) back to 1
```
Nov 30 2015
=?UTF-8?Q?Ali_=c3=87ehreli?= <acehreli yahoo.com> writes:
```On 11/30/2015 03:15 PM, Denis Koroskin wrote:

Forget the algorithms! Denis is back... :)

Ali
```
Nov 30 2015
Iain Buclaw via Digitalmars-d <digitalmars-d puremagic.com> writes:
```On 1 December 2015 at 00:21, Ali =C3=87ehreli <digitalmars-d puremagic.com>
wrote:

On 11/30/2015 03:15 PM, Denis Koroskin wrote:

Forget the algorithms! Denis is back... :)

Hurray!
```
Dec 01 2015
```On Monday, 30 November 2015 at 21:33:31 UTC, Andrei Alexandrescu
wrote:
Now that we got talking about searching in arrays, allow me to
also share an idea I've had a short while ago.

(Again, we're in the "I'd prefer to use an array if at all
possible" mindset. So let's see how we can help searching an
array with as little work as possible.)

One well-known search strategy is "Bring to front" (described
by Knuth in TAoCP). A BtF-organized linear data structure is
searched with the classic linear algorithm. The difference is
what happens after the search: whenever the search is
successful, the found element is brought to the front of the
structure. If we're looking most often for a handful of
elements, in time these will be near the front of the searched
structure.

For a linked list, bringing an element to the front is O(1)
(just rewire the pointers). For an array, things are not so
pleasant - rotating the found element to the front of the array
is O(n).

So let's see how we can implement a successful BtF for arrays.

The first idea is to just swap the found element with the first
element of the array. That's O(1) but has many disadvantages -
if you search e.g. for two elements, they'll compete for the
front of the array and they'll go back and forth without making
progress.

Another idea is to just swap the found element with the one
just before it. The logic is, each successful find will shift
the element closer to the front, in a bubble sort manner. In
time, the frequently searched elements will slowly creep toward
the front. The resulting performance is not appealing - you
need O(n) searches to bring a given element to the front, for a
total of O(n * n) steps spent in the n searches. Meh.

So let's improve on that: whenever an element is found in
position k, pick a random number i in the range 0, 1, 2, ..., k
inclusive. Then swap the array elements at indexes i and k.
This is the Randomized Slide to Front strategy.

With RStF, worst case search time remains O(n), as is the
unsuccessful search. However, frequently searched elements
migrate quickly to the front - it only takes O(log n) searches
to bring a given value at the front of the array.

Insertion and removal are both a sweet O(1), owing to the light
structuring: to insert just append the element (and perhaps
swap it in a random position of the array to prime searching
for it). Removal by position simply swaps the last element into
the position to be removed and then reduces the size of the
array.

So the RStF is suitable in all cases where BtF would be
recommended, but allows an array layout without considerable
penalty.

Related work: Theodoulos Garefalakis' Master's thesis "A Family
of Randomized Algorithms for List Accessing" describes Markov
Move to Front, which brings the searched element to front
according to a Markov chain schedule; and also Randomized Move
to Front, which decides whether a found element is brought to
front depending on a random choice. These approaches are
similar in that they both use randomization, but different
because neither has good complexity on array storage.

Andrei

What is the advantage compared to let's say a ringbuffer ? On
find you can put the element to the front, and swap the old
element with the new element ?

I guess randomizing would avoid hitting pathological cases too
often, but would converge more slowly ?
```
Nov 30 2015
Andrei Alexandrescu <SeeWebsiteForEmail erdani.org> writes:
```On 11/30/15 4:55 PM, deadalnix wrote:
I guess randomizing would avoid hitting pathological cases too often,
but would converge more slowly ?

That's it. Problem is with deterministic approaches pathological cases
are easy to hit and relatively common. -- Andrei
```
Nov 30 2015
"H. S. Teoh via Digitalmars-d" <digitalmars-d puremagic.com> writes:
```On Mon, Nov 30, 2015 at 01:41:12PM -0800, H. S. Teoh via Digitalmars-d wrote:
[...]
What about when element i is matched, swap it with the (i/2)'th
element?

Then it will take just log(n) searches to bring it to the front of the
array, but it won't (immediately) compete with whatever's currently
the most popular item in the array. Furthermore, when it does compete
with it, it will already have been moved closer to the front of the
array, so the previous most-popular element won't be pushed far back
into the deep recesses of the array, but remain close to the front
where it will be quickly found.

In fact, it's probably provable that if there are 2 most popular items
in the array, they will eventually migrate to the 1st two positions of
the array. Not so sure about the case of n most popular items for n>2,
as position 3 is a kind of odd case where it gets displaced only by
elements from indices that aren't a power of 2, but it would seem at a
cursory glance that the 3 most popular items would tend to settle around
the first 4 elements of the array.

Hmm... it seems that in the worst case (the most popular n items all lie
precisely at indices of the form 2^j) the most popular items will end up
within the first 2^n positions of the array. Not sure how to compute the
average case; intuitively at least it seems that it should lie somewhere
between the first n positions and the first 2^n positions.

T

--
Любишь кататься - люби и саночки возить.
```
Nov 30 2015
Andrei Alexandrescu <SeeWebsiteForEmail erdani.org> writes:
```On 11/30/15 4:53 PM, H. S. Teoh via Digitalmars-d wrote:
On Mon, Nov 30, 2015 at 01:41:12PM -0800, H. S. Teoh via Digitalmars-d wrote:
[...]
What about when element i is matched, swap it with the (i/2)'th
element?

Then it will take just log(n) searches to bring it to the front of the
array, but it won't (immediately) compete with whatever's currently
the most popular item in the array. Furthermore, when it does compete
with it, it will already have been moved closer to the front of the
array, so the previous most-popular element won't be pushed far back
into the deep recesses of the array, but remain close to the front
where it will be quickly found.

In fact, it's probably provable that if there are 2 most popular items
in the array, they will eventually migrate to the 1st two positions of
the array. Not so sure about the case of n most popular items for n>2,
as position 3 is a kind of odd case where it gets displaced only by
elements from indices that aren't a power of 2, but it would seem at a
cursory glance that the 3 most popular items would tend to settle around
the first 4 elements of the array.

Hmm... it seems that in the worst case (the most popular n items all lie
precisely at indices of the form 2^j) the most popular items will end up
within the first 2^n positions of the array. Not sure how to compute the
average case; intuitively at least it seems that it should lie somewhere
between the first n positions and the first 2^n positions.

With RStF it's easy to prove (e.g. by reductio ad absurdum) that if you
search only for k items out of n, they will end up in the top k
positions of the array. Then they'll churn there :o). The key to the
proof is that in the stationary state no element migrates in our out of
the top k slots. I think it would be difficult to achieve this property
with a deterministic approach.

The more interesting question would be what the element distribution is
if both elements and searches are Gaussian-distributed (probably a
frequent case in practice).

Andrei
```
Nov 30 2015
Ola Fosheim =?UTF-8?B?R3LDuHN0YWQ=?= writes:
```On Monday, 30 November 2015 at 21:33:31 UTC, Andrei Alexandrescu
wrote:
With RStF, worst case search time remains O(n), as is the
unsuccessful search. However, frequently searched elements

If you just do a linear search then shifting down the array in
another pass won't change the complexity. O(2N) == O(N) But you
could also just shift down the array while searching since if the
elements are less than the cacheline-size then you already have
everything in registers/first level cache.

(The write back cost from cache to memory is contextual and
depends on many factors.)
```
Nov 30 2015
Ola Fosheim =?UTF-8?B?R3LDuHN0YWQ=?= writes:
```On Monday, 30 November 2015 at 21:33:31 UTC, Andrei Alexandrescu
wrote:
So let's improve on that: whenever an element is found in
position k, pick a random number i in the range 0, 1, 2, ..., k
inclusive. Then swap the array elements at indexes i and k.
This is the Randomized Slide to Front strategy.

With RStF, worst case search time remains O(n), as is the
unsuccessful search. However, frequently searched elements
migrate quickly to the front - it only takes O(log n) searches
to bring a given value at the front of the array.

Something is wrong with the math here.  The randomization means
that you must assume that you get element k-1 in the worst case,
so if you repeatedly search for the same element you need O(N)
repeats to move it to the front, so you get O(N^2) complexity for
moving any element to the front.

Right?

You are probably thinking about the average case analysis, which
is a more sensible theoretical concept for randomized algorithms
than worst case, but then you need a model for what is typical.
```
Nov 30 2015
Chris Wright <dhasenan gmail.com> writes:
```On Mon, 30 Nov 2015 23:27:24 +0000, Ola Fosheim Grøstad wrote:

On Monday, 30 November 2015 at 21:33:31 UTC, Andrei Alexandrescu wrote:
So let's improve on that: whenever an element is found in position k,
pick a random number i in the range 0, 1, 2, ..., k inclusive. Then
swap the array elements at indexes i and k. This is the Randomized
Slide to Front strategy.

With RStF, worst case search time remains O(n), as is the unsuccessful
search. However, frequently searched elements migrate quickly to the
front - it only takes O(log n) searches to bring a given value at the
front of the array.

Something is wrong with the math here.  The randomization means that you
must assume that you get element k-1 in the worst case, so if you
repeatedly search for the same element you need O(N) repeats to move it
to the front, so you get O(N^2) complexity for moving any element to the
front.

Right?

You are probably thinking about the average case analysis, which is a
more sensible theoretical concept for randomized algorithms than worst
case, but then you need a model for what is typical.

People typically use lax terminology. Here, when someone doesn't specify
whether they're talking about average or worst case complexity for an
algorithm, they're probably talking about average case. Similarly for
amortized algorithms. Andrei is no doubt aware of the worst case (which
in this case is O(inf), not O(N)) and has responded to people talking
about ways to reduce the worst case.

This doesn't mean Andrei was wrong or misspoke; it means that he could
have been marginally clearer. Most people instantly grok the intent, but
some who are blessed with pedantry don't.

Identifying these cases is a skill that took me years to learn.
```
Nov 30 2015
Ola Fosheim =?UTF-8?B?R3LDuHN0YWQ=?= writes:
```On Tuesday, 1 December 2015 at 01:12:39 UTC, Chris Wright wrote:
People typically use lax terminology. Here, when someone
doesn't specify whether they're talking about average or worst
case complexity for an algorithm, they're probably talking

Don't use lax terminology when doing complexity analysis. Average
case analysis is much much much harder to do than worst
case/amortized.
```
Nov 30 2015
Marcelo Juchem <juchem gmail.com> writes:
```On Monday, 30 November 2015 at 21:33:31 UTC, Andrei Alexandrescu
wrote:
[...]
One well-known search strategy is "Bring to front" (described
by Knuth in TAoCP). A BtF-organized linear data structure is
searched with the classic linear algorithm. The difference is
what happens after the search: whenever the search is
successful, the found element is brought to the front of the
structure. If we're looking most often for a handful of
elements, in time these will be near the front of the searched
structure.

[...]
Another idea is to just swap the found element with the one
just before it. The logic is, each successful find will shift
the element closer to the front, in a bubble sort manner. In
time, the frequently searched elements will slowly creep toward
the front. The resulting performance is not appealing - you
need O(n) searches to bring a given element to the front, for a
total of O(n * n) steps spent in the n searches. Meh.

So let's improve on that: whenever an element is found in
position k, pick a random number i in the range 0, 1, 2, ..., k
inclusive. Then swap the array elements at indexes i and k.
This is the Randomized Slide to Front strategy.

[...]
Insertion and removal are both a sweet O(1), owing to the light
structuring: to insert just append the element (and perhaps
swap it in a random position of the array to prime searching
for it). Removal by position simply swaps the last element into
the position to be removed and then reduces the size of the
array.

[...]
Andrei

It seems to me you're trying to implement the array based
equivalent of Splay Trees (Splay Array rhymes, btw). Would that
be a close enough description?

I'm assuming you're trying to optimize for some distribution
where a minority of the elements account for the majority of
queries (say, Zipfian).

Here are some ideas that come to mind. I haven't thought through
them too much so everyone's welcome to destroy me.

Rather than making index 0 always the front, use some rotating
technique similar to what ring buffers do.

Say we initially have elements ABCDE (front at 0) and we search
for C. We swap the left of front (cycling back to the end of the
array, thus index 4) with the new front. We now have the
following array at hand: ABEDC, front at 4 (logically CABED).

Obviously we shouldn't change front if the queried element is

An immediate problem with this technique is that we'll frequently
pollute the front of the array with infrequent items. Say these
are the number of queries made so far for each element: A:7, B:5,
C:2, all others 0. Also, assume that this is the state of the
array at this point: DEABC, front at 2. Say we now query for B.
This is the new state: DBAEC, front at 1 (logically BAECD).
Having E in front of C is undesirable, so we need a way to avoid
that.

From now on I'll refer to indexes as the logical index. That is,
let i be (front + index) % size. For the sake of brevity, let d
be the distance between the element and the front = i - front.
Let q be the number of successful queries performed so far.

What I have in mind boils down to decide between:
- move a newly queried element at logical position i to the left
of front (technique above). Let's call it move-pre-front for the
lack of a better name;
- bubble the element up to some position between [0, i), not
necessarily max(0, i - 1).

Augmenting the array with the number of queries for each element
would tremendously help the decision making, but I'm assuming
that's undesirable for a few reasons like:
- the array can't be transparently used in algorithms unaware of
the structure;
- alignment;
- data bloating.

My immediate thought is to use some heuristic. For instance, say
we have some threshold k. If d <= k, we bubble up s <= d
positions to the left, where s could be computed using some
deterministic formula taking d, q and/or k into account, or just
randomly (Andrei's RStF). If d > k, we move-pre-front the element.

The threshold k could be computed as a factor of q. Say, sqrt(q),
log q or log^2 q (logarithm base 2).

Thoughts?

Marcelo
```
Nov 30 2015
Richard Hodges <hodges.r gmail.com> writes:
```I had a little think about the pathological case of most searches
being for one of a few elements.

I'm sure my idea is not new, but it seems to me that keeping a
'hit count' for each element solves this. The decision of whether
to swap then becomes:

++ Frequency[I]
if I >= Threshold1 then
choose an index J that is [either I/2 or rand(0...I-1),
depending on preference]
if (Frequency[I] - Frequency[J]) >= Threshold2 then
swap Item[I] and Item[J]
swap Frequency[I] and Frequency[J]
I = J
endif
endif
return I

I wrote a little test in c++ (sorry guys, old habits die hard):

template<class T>
struct mrucache
{
using vector_type  = std::vector<T>;
using iterator = typename vector_type::iterator;
using const_iterator = typename vector_type::const_iterator;

_items.push_back(std::move(item));
_frequency.push_back(0);
}

template<class Pred>
iterator find_if(Pred&& pred)
{
using std::begin;
using std::end;
auto iter = std::find_if(begin(_items),
end(_items),
std::forward<Pred>(pred));
if (iter != end(_items))
{
auto i = std::distance(_items.begin(), iter);
++ _frequency[i];
i = maybe_swap(i);
iter = std::next(begin(_items), i);
}
return iter;
}

std::size_t maybe_swap(std::size_t i)
{
if (i >= closeness_threshold)
{
auto candidate_i = i / 2;
if ((_frequency[i] - _frequency[candidate_i]) >=
difference_threshold) {
swap(_items[i], _items[candidate_i]);
swap(_frequency[i], _frequency[candidate_i]);
i = candidate_i;
}
}
return i;
}

auto begin() const { return _items.begin(); }
auto end() const { return _items.end(); }

static const size_t closeness_threshold = 4;
static const int difference_threshold = 1;

std::vector<T> _items;
std::vector<int> _frequency;
};
```
Dec 01 2015
CraigDillabaugh <craig.dillabaugh gmail.com> writes:
```On Monday, 30 November 2015 at 21:33:31 UTC, Andrei Alexandrescu
wrote:
Now that we got talking about searching in arrays, allow me to
also share an idea I've had a short while ago.

[...]

Perhaps some strategy similar to Working Sets:
https://en.wikipedia.org/wiki/Iacono's_working_set_structure
would work (or inspired by the same idea).  You move the element
from where it is found to T_1, move a random element from T_1 to
T_2, from T_2 to T_3 and so on to T_i. In this case rather than
trees you would have lists.  Maybe that has poor cache locality
properties though.
```
Dec 01 2015
Jakob Ovrum <jakobovrum gmail.com> writes:
```On Monday, 30 November 2015 at 21:33:31 UTC, Andrei Alexandrescu
wrote:
Now that we got talking about searching in arrays, allow me to
also share an idea I've had a short while ago.

I wrote a range-based implementation to see how it would look
like.

https://gist.github.com/JakobOvrum/45a37f55ba5c9a7501d6

The tests are very thin but it seems to do the trick.
```
Dec 08 2015