There was a great post about XOR filters a while back: What is an XOR filter?
Can someone provide an explanation about the binary fuse filter? How does it differ in construction and what are the reasonings for such choices? I tried to read the paper but got lost in the binary fuse specific details. How does it compare to XOR? Why is it smaller and faster?
Intuitively, a binary fuse filter follows the same strategy as a regular XOR filter but with a slightly different strategy for placing items, so if you haven’t yet read about how XOR filters work, it’s best to start there.
Like an XOR filter, the binary fuse filter (and several other related structures like the Ribbon filter) work by computing a fingerprint f(x) for each item x, as well as computing some hashes h1(x), h2(x), and h3(x) that give positions into an array. The array is then populated so that the value h1(x) xor h2(x) xor h3(x) = f(x) for all elements in the filter.
The difference between XOR filters and binary fuse filters is how that table is populated. Both data structures use an approach called peeling to fill in the table. That’s the strategy outlined in the other post: find a slot that has only one item hashing to it, remove it, recursively place the other elements, then set the value in that slot so that the removed element’s hashes work out correctly.
In an XOR filter, the array of slots needs to have size roughly 1.23n for this process to have a high chance of succeeding. The reason for this is surprisingly mathematically deep: if the hashes are uniformly distributed across the table, then with fewer than 1.23n slots the probability that the peeling strategy works drops very quickly to 0, and with more than 1.23n slots the probability that the peeling strategy works very rapidly trends to 1. Therefore, you can think of 1.23n as a hard theoretical limit on the size of the table using an XOR filter.
The idea behind a fuse filter is to change how hashes are assigned over the table. Rather than having the hashes chosen so that they’re uniformly random across the table, we use another approach. Pick a window size w. Then, for each element x, choose h1(x), h2(x), and h3(x) in the following way:
Select a random window of size w within the array.
Choose h1(x), h2(x), and h3(x) uniformly at random within that window.
(The actual logic of the binary fuse filter is slightly different than this in that it requires h1(x), h2(x), and h3(x) to be spaced out a bit, but let’s ignore that for now.)
Once we’ve assigned hashes this way, we use the same peeling strategy as before to fill in the array: we find a slot where only one item hashes, remove that item, place all the other items, then put the item back in.
What’s beautiful here is how that peeling proceeds. Intuitively, because of how we assigned hashes, the slots closest to the two ends of the array are the ones most likely to only have one item hash to them. Why? Because the only way you can have collisions near the ends is if you have two items whose windows are both very close to the ends and that happen to pick slots within those windows that are far to the sides of the windows. That’s pretty unlikely, and as a result the leftmost and rightmost items are likely to be peeled.
But once those items are peeled, by the same logic, the items that are now on the far left or far right are likely to be peelable. This is where the name “fuse” comes from - it’s like lighting a fuse at both ends and watching it burn in toward the center.
The fact that there’s some predictability here about where items will be peelable means that we need fewer table slots than if the hashes were assigned randomly across the table. The paper cites a space usage of around 1.13n table slots needed, a big improvement over the 1.23n needed for an XOR filter, and it’s done purely by changing the strategy for assigning hashes. Pretty neat!
Related
I was reading: https://en.wikipedia.org/wiki/Counting_sort and https://www.geeksforgeeks.org/counting-sort/
There is one little detail which I don't get at all, why to complicate things where they can be so much easier? What's the problem of allocating an array of size k where the field of numbers is [1...k] and count how many times each number appeared and lastly walking down the array and printing according to the counter in each cell.
What's the problem of allocating an array of size k where the field of numbers is [1...k] and count how many times each number appeared and lastly walking down the array and printing according to the counter in each cell.
From your phrase "how many times each number appeared", it sounds like you're picturing an array of positive integers, where you want to sort them in increasing order, and where you can use those integers directly as indices in your helper array?
But that's not what the Wikipedia article describes. The algorithm in the Wikipedia article is for an array whose elements can have whatever data-type we choose, provided there's a function key that maps from that data-type to the set of indices in the helper array, with the property that we want to stably sort elements according to the result of key (so, if key(x) < key(y) then we want to sort x before y, and if key(x) = key(y) then we want to keep x and y in the same order they originally had).
In particular, the counting-sort algorithm in the Wikipedia article is useful as a component of radix sort: first you sort by the last digit (using a key function that gives the last digit of a number), then by the second-to-last digit, and so on, until an array of numbers is sorted.
There is one little detail which I don't get at all, why to complicate things where they can be so much easier?
A pro tip: we all usually think that our own code is "easier" and that other people are "complicating things", because code is easier to write than to read, so the code that we understand best is the code that we've come up with ourselves.
As it happens, in this case the Wikipedia code really is more complicated, because it serves a much more general use-case than you were picturing; but in general, it's not a good idea to just assume that everyone will agree that your code is the easy version and that others' is unnecessarily complicated.
I have encountered an algorithmic problem but am not able to figure out anything better than brute force or reduce it to a better know problem. Any hints?
There are N bags of variable sizes and N types of items. Each type of items belongs to one bag. There are lots of items of each type and each item may be of a different size. Initially, these items are distributed across all the bags randomly. We have to place the items in their respective bags. However, we can only operate with a pair of bags at one time by exchanging items (as much as possible) and proceeding to the next pair. The aim is to reduce the total number of pairs. Edit: The aim is to find a sequence of transfers that minimizes the total number of bag pairs involved
Clarification:
The bags are not arbitrarily large (You can assume the bag and item sizes to be integers between 0 to 1000 if it helps). You'll frequently encounter scenarios where the all the items between 2 bags cannot be swapped due to the limited capacity of one of the bags. This is where the algorithm needs to make an optimisation. Perhaps, if another pair of bags were swapped first, the current swap can be done in one go. To illustrate this, let's consider Bags A, B and C and their items 1, 2, 3 respectively. The number in the brackets is the size.
A(10) : 3(8)
B(10): 1(2), 1(3)
C(10): 1(4)
The swap orders can be AB, AC, AB or AC, AB. The latter is optimal as the number of swaps is lesser.
Since I cannot come to an idea for an algorithm that will always find an optimal answer, and approximation of the fitness of the solution (amount of swaps) is also fine, I suggest a stochastic local search algorithm with pruning.
Given a random starting configuration, this algorithm considers all possible swaps, and makes a weighed decision based on chance: the better a swap is, the more likely it is chosen.
The value of a swap would be the sum of the value of the transaction of an item, which is zero if the item does not end up in it's belonging bag, and is positive if it does end up there. The value increases as the item's size increases (the idea behind this is that a larger block is hard to move many times in comparison to smaller blocks). This fitness function can be replaced by any other fitness function, it's efficiency is unknown until empirically shown.
Since any configuration can be the consequence of many preceding swaps, we keep track of which configurations we have seen before, along with a fitness (based on how many items are in their correct bag - this fitness is not related to the value of a swap) and the list of preceded swaps. If the fitness function for a configuration is the sum of the items that are in their correct bags, then the amount of items in the problem is the highest fitness (and therefor marks a configuration to be a solution).
A swap is not possible if:
Either of the affected bags is holding more than it's capacity after the potential swap.
The new swap brings you back to the last configuration you were in before the last swap you did (i.e. reversed swap).
When we identify potential swaps, we look into our list of previously seen configurations (use a hash function for O(1) lookup). Then we either set its preceded swaps to our preceded swaps (if our list is shorter than it's), or we set our preceded swaps to its list (if it's list is shorter than ours). We can do this because it does not matter which swaps we did, as long as the amount of swaps is as small as possible.
If there are no more possible swaps left in a configuration, it means you're stuck. Local search tells you 'reset' which you can do in may ways, for instance:
Reset to a previously seen state (maybe the best one you've seen so far?)
Reset to a new valid random solution
Note
Since the algorithm only allows you to do valid swaps, all constraints will be met for each configuration.
The algorithm does not guarantee to 'stop' out of the box, you can implement a maximum number of iterations (swaps)
The algorithm does not guarantee to find a correct solution, as it does it's best to find a better configuration each iteration. However, since a perfect solution (set of swaps) should look closely to an almost perfect solution, a human might be able to finish what the local search algorithm was not after it results in a invalid configuration (where not every item is in its correct bag).
The used fitness functions and strategies are very likely not the most efficient out there. You could look around to find better ones. A more efficient fitness function / strategy should result in a good solution faster (less iterations).
I'm writing a program for a competition and I need to be faster than all the other competitors. For this I need a little algorithm help; ideally I'd be using the fastest algorithm.
For this problem I am given 2 things. The first is a list of tuples, each of which contains exactly two elements (strings), each of which represents an item. The second is an integer, which indicates how many unique items there are in total. For example:
# of items = 3
[("ball","chair"),("ball","box"),("box","chair"),("chair","box")]
The same tuples can be repeated/ they are not necessarily unique.) My program is supposed to figure out the maximum number of tuples that can "agree" when the items are sorted into two groups. This means that if all the items are broken into two ideal groups, group 1 and group 2, what are the maximum number of tuples that can have their first item in group 1 and their second item in group 2.
For example, the answer to my earlier example would be 2, with "ball" in group 1 and "chair" and "box" in group 2, satisfying the first two tuples. I do not necessarily need know what items go in which group, I just need to know what the maximum number of satisfied tuples could be.
At the moment I'm trying a recursive approach, but its running on (n^2), far too inefficient in my opinion. Does anyone have a method that could produce a faster algorithm?
Thanks!!!!!!!!!!
Speed up approaches for your task:
1. Use integers
Convert the strings to integers (store the strings in an array and use the position for the tupples.
String[] words = {"ball", "chair", "box"};
In tuppls ball now has number 0 (pos 0 in array) , chair 1, box 2.
comparing ints is faster than Strings.
2. Avoid recursion
Recursion is slow, due the recursion overhead.
For example look at binarys search algorithm in a recursive implementatiion, then look how java implements binSearch() (with a while loop and iteration)
Recursion is helpfull if problems are so complex that a non recursive implementation is to complex for a human brain.
An iterataion is faster, but not in the case when you mimick recursive calls by implementing your own stack.
However you can start implementing using a recursiove algorithm, once it works and it is a suited algo, then try to convert to a non recursive implementation
3. if possible avoid objects
if you want the fastest, the now it becomes ugly!
A tuppel array can either be stored in as array of class Point(x,y) or probably faster,
as array of int:
Example:
(1,2), (2,3), (3,4) can be stored as array: (1,2,2,3,3,4)
This needs much less memory because an object needs at least 12 bytes (in java).
Less memory becomes faster, when the array are really big, then your structure will hopefully fits in the processor cache, while the objects array does not.
4. Programming language
In C it will be faster than in Java.
Maximum cut is a special case of your problem, so I doubt you have a quadratic algorithm for it. (Maximum cut is NP-complete and it corresponds to the case where every tuple (A,B) also appears in reverse as (B,A) the same number of times.)
The best strategy for you to try here is "branch and bound." It's a variant of the straightforward recursive search you've probably already coded up. You keep track of the value of the best solution you've found so far. In each recursive call, you check whether it's even possible to beat the best known solution with the choices you've fixed so far.
One thing that may help (or may hurt) is to "probe": for each as-yet-unfixed item, see if putting that item on one of the two sides leads only to suboptimal solutions; if so, you know that item needs to be on the other side.
Another useful trick is to recurse on items that appear frequently both as the first element and as the second element of your tuples.
You should pay particular attention to the "bound" step --- finding an upper bound on the best possible solution given the choices you've fixed.
In my lexical analyzer generator I use McNaughton and Yamada algorithm for NFA construction, and one of its properties that transition form I to J marked with char at J position.
So, each node of NFA can be represented simply as list of next possible states.
Which data structure best suit for storing this type of data? It must provide fast lookup for all possible states and use less space, but insertion time is not so important.
My understanding is that you want to encode a graph, where the nodes are states and the edges are transitions, and where every edge is labelled with a character. Is that correct?
The dull but practical answer is to have a object for each state, and to encode the transitions in some little structure in that object.
The simplest one would be an array, indexed by character code: that's as fast as it gets, but not naturally space-efficient. You can make it more space efficient by using a sort of offset, truncated array: store only the part of the array which contains transitions, along with the start and end indices of that part. When looking up a character in it, check that its code is within the bounds; if it isn't, treat it as a null edge (or an edge back to the start state or whatever), and if it is, fetch the element at index (character code - start). Does that make sense?
A more complex option would be a little hashtable, which would be more compact but slightly slower. I would suggest closed hashing, because collision lists will use too much memory; linear probing should be enough. You could look into using perfect hashing (look it up), which takes a lot of time to generate the table but then gives collision-free lookup. The generation process is quite complex, though.
A clever approach is to use both arrays and hashtables, and to pick one or the other based on the number of edges: if the compacted array would be more than, say, a third full, use it, but if not, use a hashtable.
Now, something a bit more radical you could do would be to use arrays, but to overlap them - if they're sparse, they'll have lots of holes in, and if you're clever, you can arrange them so that the entries in each array lines up with holes in the others. That will give you fast lookups, but also excellent memory efficiency. You will need some scheme for distinguishing when a lookup has found something from when it's found an empty slot with some other state's transition in, but i'm sure you can think of something.
This problem is a little similar to that solved by reservoir sampling, but not the same. I think its also a rather interesting problem.
I have a large dataset (typically hundreds of millions of elements), and I want to estimate the number of unique elements in this dataset. There may be anywhere from a few, to millions of unique elements in a typical dataset.
Of course the obvious solution is to maintain a running hashset of the elements you encounter, and count them at the end, this would yield an exact result, but would require me to carry a potentially large amount of state with me as I scan through the dataset (ie. all unique elements encountered so far).
Unfortunately in my situation this would require more RAM than is available to me (nothing that the dataset may be far larger than available RAM).
I'm wondering if there would be a statistical approach to this that would allow me to do a single pass through the dataset and come up with an estimated unique element count at the end, while maintaining a relatively small amount of state while I scan the dataset.
The input to the algorithm would be the dataset (an Iterator in Java parlance), and it would return an estimated unique object count (probably a floating point number). It is assumed that these objects can be hashed (ie. you can put them in a HashSet if you want to). Typically they will be strings, or numbers.
You could use a Bloom Filter for a reasonable lower bound. You just do a pass over the data, counting and inserting items which were definitely not already in the set.
This problem is well-addressed in the literature; a good review of various approaches is http://www.edbt.org/Proceedings/2008-Nantes/papers/p618-Metwally.pdf. The simplest approach (and most compact for very high accuracy requirements) is called Linear Counting. You hash elements to positions in a bitvector just like you would a Bloom filter (except only one hash function is required), but at the end you estimate the number of distinct elements by the formula D = -total_bits * ln(unset_bits/total_bits). Details are in the paper.
If you have a hash function that you trust, then you could maintain a hashset just like you would for the exact solution, but throw out any item whose hash value is outside of some small range. E.g., use a 32-bit hash, but only keep items where the first two bits of the hash are 0. Then multiply by the appropriate factor at the end to approximate the total number of unique elements.
Nobody has mentioned approximate algorithm designed specifically for this problem, Hyperloglog.