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

- Andrei Alexandrescu (48/48) Nov 30 2015 Now that we got talking about searching in arrays, allow me to also
- Andrei Alexandrescu (5/5) Nov 30 2015 On 11/30/15 4:33 PM, Andrei Alexandrescu wrote:
- H. S. Teoh via Digitalmars-d (19/45) Nov 30 2015 [...]
- Andrei Alexandrescu (3/4) Nov 30 2015 Randomization is essential - without it you have thrashing if you search...
- Steven Schveighoffer (3/7) Nov 30 2015 What about selecting a random element in 0..k/2 instead of 0..k-1?
- H. S. Teoh via Digitalmars-d (6/15) Nov 30 2015 [...]
- Andrei Alexandrescu (7/15) Nov 30 2015 I think complexity would stay the same. Choosing a tighter range puts a
- Steven Schveighoffer (10/18) Dec 01 2015 In the case where you search for a very small number of elements, it
- Chris Wright (6/18) Nov 30 2015 You can use that to put a hard upper bound of O(log n), and maybe you'll...
- deadalnix (8/13) Nov 30 2015 You'd end up swaping the 2 element in front, but keep them both
- Denis Koroskin (6/20) Nov 30 2015 Imagine that there are 1000 elements, 500th elements is X and
- =?UTF-8?Q?Ali_=c3=87ehreli?= (3/3) Nov 30 2015 On 11/30/2015 03:15 PM, Denis Koroskin wrote:
- Iain Buclaw via Digitalmars-d (3/5) Dec 01 2015 Hurray!
- deadalnix (7/63) Nov 30 2015 What is the advantage compared to let's say a ringbuffer ? On
- Andrei Alexandrescu (3/5) Nov 30 2015 That's it. Problem is with deterministic approaches pathological cases
- H. S. Teoh via Digitalmars-d (17/27) Nov 30 2015 In fact, it's probably provable that if there are 2 most popular items
- Andrei Alexandrescu (11/35) Nov 30 2015 With RStF it's easy to prove (e.g. by reductio ad absurdum) that if you
- Ola Fosheim =?UTF-8?B?R3LDuHN0YWQ=?= (9/11) Nov 30 2015 If you just do a linear search then shifting down the array in
- Ola Fosheim =?UTF-8?B?R3LDuHN0YWQ=?= (11/19) Nov 30 2015 Something is wrong with the math here. The randomization means
- Chris Wright (11/33) Nov 30 2015 People typically use lax terminology. Here, when someone doesn't specify...
- Ola Fosheim =?UTF-8?B?R3LDuHN0YWQ=?= (4/8) Nov 30 2015 Don't use lax terminology when doing complexity analysis. Average
- Marcelo Juchem (56/82) Nov 30 2015 On Monday, 30 November 2015 at 21:33:31 UTC, Andrei Alexandrescu
- Richard Hodges (65/65) Dec 01 2015 I had a little think about the pathological case of most searches
- CraigDillabaugh (9/12) Dec 01 2015 Perhaps some strategy similar to Working Sets:
- Jakob Ovrum (6/8) Dec 08 2015 I wrote a range-based implementation to see how it would look

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

On 11/30/15 4:33 PM, Andrei Alexandrescu wrote: [snip] I just posted to reddit: https://www.reddit.com/r/programming/comments/3uwp42/its_my_birthday_so_heres_some_cake_for_thought/ Andrei

Nov 30 2015

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

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

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 selecting a random element in 0..k/2 instead of 0..k-1? -SteveWhat 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

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:[...] Or selecting the (i/k)'th element for k = uniform(1..i)? T -- People walk. Computers run.On 11/30/15 4:41 PM, H. S. Teoh via Digitalmars-d wrote:What about selecting a random element in 0..k/2 instead of 0..k-1?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

On 11/30/15 4:58 PM, Steven Schveighoffer wrote:On 11/30/15 4:50 PM, Andrei Alexandrescu wrote: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. andreiWhat 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

On 11/30/15 5:07 PM, Andrei Alexandrescu wrote:On 11/30/15 4:58 PM, Steven Schveighoffer wrote: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) instead of n)What about selecting a random element in 0..k/2 instead of 0..k-1?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.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

On Mon, 30 Nov 2015 16:58:16 -0500, Steven Schveighoffer wrote:On 11/30/15 4:50 PM, Andrei Alexandrescu wrote: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.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

On Monday, 30 November 2015 at 21:50:09 UTC, Andrei Alexandrescu 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

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: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 1What 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

On 11/30/2015 03:15 PM, Denis Koroskin wrote: Forget the algorithms! Denis is back... :) Ali

Nov 30 2015

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. AndreiWhat 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

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

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

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: [...]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). AndreiWhat 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.

Nov 30 2015

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 elementsIf 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

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

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: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.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

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 about average case.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

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.[...]AndreiIt 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 already it. 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

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; void add(T item) { _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

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

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