I wanted to generate prime numbers between two given numbers ‘a’ and ‘b’ (b > a). What I did was store Boolean values in an array of size b-1 (that is for numbers 2 to b) and then I applied the sieve method.
Is there a better way, that reduces space complexity, if I don't need all prime numbers from 2 to b?
You need to store all primes which are smaller of equal than the square root of b, then for each number between a and b check whether they are divisible by any of these numbers and they don't equal these numbers. So in our case the magic number is sqrt(b)
You can use segmented sieve of Eratosthenes. The basic idea is pretty simple.
In a typical sieve, we start with a large array of Booleans, all set to the same value. These represent odd numbers, starting from 3. We look at the first and see that it's true, so we add it to the list of prime numbers. Then we mark off every multiple of that number as not prime.
Now, the problem with this is that it's not very cache friendly. As we mark off the multiples of each number, we go through the entire array. Then when we reach the end, we start over from the beginning (which is no longer in the cache) and walk through the entire array again. Each time through the array, we read the entire array from main memory again.
For a segmented sieve, we do things a bit differently. We start by by finding only the primes up to the square root of the limit we care about. Then we use those to mark off primes in the main array. The difference here is the order in which we mark off primes. Instead of marking off all the multiples of three, then all the multiples of 5, and so on, we start by marking off the multiples of three for data that will fit in the cache. Then, instead of continuing on to more data in the array, we go back and mark off the multiples of five for the data that fits in the cache. Then the multiples of 7, and so on.
Then, when we've marked off all the multiples in that cache-sized chunk of data, we move on to the next cache-sized chunk of data. We start over with marking off multiples of 3 in this chunk, then multiples of 5, and so on until we've marked off all the multiples in this chunk. We continue that pattern until we've marked off all the non-prime numbers in all the chunks, and we're done.
So, given N primes below the square root of the limit we care about, a naive sieve will read the entire array of Booleans N times. By contrast, a segmented sieve will only read each chunk of the data once. Once a chunk of data is read from main memory, all the processing on that chunk is done before any more data is read from main memory.
The exact speed-up this gives will depend on the ratio of the speed of cache to the speed of main memory, the size of the array you're using vs. the size of the cache, and so on. Nonetheless, it is generally pretty substantial--for example, on my particular machine, looking for the primes up to 100 million, the segmented sieve has a speed advantage of about 10:1.
One thing you must remember, if you're using C++. A well-known issue with std::vector<bool> is Under C++98/03, vector<bool> was required to be a specialization that stored each Boolean as a single bit with some proxy trickery to get bool-like behavior. That requirement has since been lifted, but many libraries still include it.
With a non-segmented sieve, it's generally a useful trade-off. Although it requires a little extra CPU time to compute masks and such to modify only a single bit at a time, it saves enough bandwidth to main memory to more than compensate.
With a segmented sieve, bandwidth to main memory isn't nearly as large a factor, so using a vector<char> generally seems to give better results (at least with the compilers and processors I have handy).
Getting optimal performance from a segmented sieve does require knowledge of the size of your processor's cache, but getting it precisely correct isn't usually critical--if you assume the size is smaller than it really is, you won't necessarily get optimal use of your cache, but you usually won't lose a lot either.
Related
I'm trying to understand hash tables, and from what I've seen the modulo operator is used to select which bucket a key will be placed in. I know that hash algorithms are supposed to minimize the same result for different inputs, however I don't understand how the same results for different inputs can be minimal after the modulo operation. Let's just say we have a near-perfect hash function that gives a different hashed value between 0 and 100,000, and then we take the result modulo 20 (in our example we have 20 buckets), isn't the resulting number very close to a random number between 0 and 19? Meaning roughly the probability that the final result is any of a number between 0 and 19 is about 1 in 20? If this is the case, then the original hash function doesn't seem to ensure minimal collisions because after the modulo operation we end up with something like a random number? I must be wrong, but I'm thinking that what ensures minimal collisions the most is not the original hash function but how many buckets we have.
I'm sure I'm misunderstanding this. Can someone explain?
Don't you get a random number after doing modulo on a hashed number?
It depends on the hash function.
Say you have an identify hash for numbers - h(n) = n - then if the keys being hashed are generally incrementing numbers (perhaps with an occasional ommision), then after hashing they'll still generally hit successive buckets (wrapping at some point from the last bucket back to the first), with low collision rates overall. Not very random, but works out well enough. If the keys are random, it still works out pretty well - see the discussion of random-but-repeatable hashing below. The problem is when the keys are neither roughly-incrementing nor close-to-random - then an identity hash can provide terrible collision rates. (You might think "this is a crazy bad example hash function, nobody would do this; actually, most C++ Standard Library implementations' hash functions for integers are identity hashes).
On the other hand, if you have a hash function that say takes the address of the object being hashed, and they're all 8 byte aligned, then if you take the mod and the bucket count is also a multiple of 8, you'll only ever hash to every 8th bucket, having 8 times more collisions than you might expect. Not very random, and doesn't work out well. But, if the number of buckets is a prime, then the addresses will tend to scatter much more randomly over the buckets, and things will work out much better. This is the reason the GNU C++ Standard Library tends to use prime numbers of buckets (Visual C++ uses power-of-two sized buckets so it can utilise a bitwise AND for mapping hash values to buckets, as AND takes one CPU cycle and MOD can take e.g. 30-40 cycles - depending on your exact CPU - see here).
When all the inputs are known at compile time, and there's not too many of them, then it's generally possible to create a perfect hash function (GNU gperf software is designed specifically for this), which means it will work out a number of buckets you'll need and a hash function that avoids any collisions, but the hash function may take longer to run than a general purpose function.
People often have a fanciful notion - also seen in the question - that a "perfect hash function" - or at least one that has very few collisions - in some large numerical hashed-to range will provide minimal collisions in actual usage in a hash table, as indeed this stackoverflow question is about coming to grips with the falsehood of this notion. It's just not true if there are still patterns and probabilities in the way the keys map into that large hashed-to range.
The gold standard for a general purpose high-quality hash function for runtime inputs is to have a quality that you might call "random but repeatable", even before the modulo operation, as that quality will apply to the bucket selection as well (even using the dumber and less forgiving AND bit-masking approach to bucket selection).
As you've noticed, this does mean you'll see collisions in the table. If you can exploit patterns in the keys to get less collisions that this random-but-repeatable quality would give you, then by all means make the most of that. If not, the beauty of hashing is that with random-but-repeatable hashing your collisions are statistically related to your load factor (the number of stored elements divided by the number of buckets).
As an example, for separate chaining - when your load factor is 1.0, 1/e (~36.8%) of buckets will tend to be empty, another 1/e (~36.8%) have one element, 1/(2e) or ~18.4% two elements, 1/(3!e) about 6.1% three elements, 1/(4!e) or ~1.5% four elements, 1/(5!e) ~.3% have five etc.. - the average chain length from non-empty buckets is ~1.58 no matter how many elements are in the table (i.e. whether there are 100 elements and 100 buckets, or 100 million elements and 100 million buckets), which is why we say lookup/insert/erase are O(1) constant time operations.
I know that hash algorithms are supposed to minimize the same result for different inputs, however I don't understand how the same results for different inputs can be minimal after the modulo operation.
This is still true post-modulo. Minimising the same result means each post-modulo value has (about) the same number of keys mapping to it. We're particularly concerned about in-use keys stored in the table, if there's a non-uniform statistical distribution to the use of keys. With a hash function that exhibits the random-but-repeatable quality, there will be random variation in post-modulo mapping, but overall they'll be close enough to evenly balanced for most practical purposes.
Just to recap, let me address this directly:
Let's just say we have a near-perfect hash function that gives a different hashed value between 0 and 100,000, and then we take the result modulo 20 (in our example we have 20 buckets), isn't the resulting number very close to a random number between 0 and 19? Meaning roughly the probability that the final result is any of a number between 0 and 19 is about 1 in 20? If this is the case, then the original hash function doesn't seem to ensure minimal collisions because after the modulo operation we end up with something like a random number? I must be wrong, but I'm thinking that what ensures minimal collisions the most is not the original hash function but how many buckets we have.
So:
random is good: if you get something like the random-but-repeatable hash quality, then your average hash collisions will statistically be capped at low levels, and in practice you're unlikely to ever see a particularly horrible collision chain, provided you keep the load factor reasonable (e.g. <= 1.0)
that said, your "near-perfect hash function...between 0 and 100,000" may or may not be high quality, depending on whether the distribution of values has patterns in it that would produce collisions. When in doubt about such patterns, use a hash function with the random-but-repeatable quality.
What would happen if you took a random number instead of using a hash function? Then doing the modulo on it? If you call rand() twice you can get the same number - a proper hash function doesn't do that I guess, or does it? Even hash functions can output the same value for different input.
This comment shows you grappling with the desirability of randomness - hopefully with earlier parts of my answer you're now clear on this, but anyway the point is that randomness is good, but it has to be repeatable: the same key has to produce the same pre-modulo hash so the post-modulo value tells you the bucket it should be in.
As an example of random-but-repeatable, imagine you used rand() to populate a uint32_t a[256][8] array, you could then hash any 8 byte key (e.g. including e.g. a double) by XORing the random numbers:
auto h(double d) {
uint8_t i[8];
memcpy(i, &d, 8);
return a[i[0]] ^ a[i[1]] ^ a[i[2]] ^ ... ^ a[i[7]];
}
This would produce a near-ideal (rand() isn't a great quality pseudo-random number generator) random-but-repeatable hash, but having a hash function that needs to consult largish chunks of memory can easily be slowed down by cache misses.
Following on from what [Mureinik] said, assuming you have a perfect hash function, say your array/buckets are 75% full, then doing modulo on the hashed function will probably result in a 75% collision probability. If that's true, I thought they were much better. Though I'm only learning about how they work now.
The 75%/75% thing is correct for a high quality hash function, assuming:
closed hashing / open addressing, where collisions are handled by finding an alternative bucket, or
separate chaining when 75% of buckets have one or more elements linked therefrom (which is very likely to mean the load factor (which many people may think of when you talk about how "full" the table is) is already significantly more than 75%)
Regarding "I thought they were much better." - that's actually quite ok, as evidenced by the percentages of colliding chain lengths mentioned earlier in my answer.
I think you have the right understanding of the situation.
Both the hash function and the number of buckets affect the chance of collisions. Consider, for example, the worst possible hash function - one that returns a constant value. No matter how many buckets you have, all the entries will be lumped to the same bucket, and you'd have a 100% chance of collision.
On the other hand, if you have a (near) perfect hash function, the number of buckets would be the main factor for the chance of collision. If your hash table has only 20 buckets, the minimal chance of collision will indeed be 1 in 20 (over time). If the hash values weren't uniformly spread, you'd have a much higher chance of collision in at least one of the buckets. The more buckets you have, the less chance of collision. On the other hand, having too many buckets will take up more memory (even if they are empty), and ultimately reduce performance, even if there are less collisions.
Why is it recommended to have a load factor of 1.0 in separate chaining?
I've seen plenty of people saying that it is recommended, but not given a clear explanation of why.
With open addressing, I know the load factor should be between 0.5 and 0.7 because it should be a fast operation to find an unoccupied index when dealing with collisions. But I can't see why a load factor of 1 should be better in separate chaining. I mean, if I have a table of size 100, isn't there still a chance that all 100 elements hashes to the same index and get placed in the same list? So I really can't comprehend why this specific load factor for separate chaining should be 1.
tl;dr: To save memory space by not having slots unoccupied and speed up access by minimizing the number of list traversal operation.
If you understand the load factor as n_used_slots / n_total_slots:
Having a load factor of 1 just describes the ideal situation for a well-implemented hash table using Separate Chaining collision handling: no slots are left empty.
The other classical approach, Open Addressing, requires the table to always have a free slot available when adding a new item. Resizing the table is way too costly to do it for each item, but we are also restricted on memory and wouldn’t want to have too many unused slots lying around. One has to find a balance between speed (few table resizes, quick inserts and lookups) and memory (few empty slots) [as ever so often in programming]. The ideal load factor is based on this idea of balancing and can be estimated based on the actual hash function, the value domain and other factors.
With Separate Chaining, on the other hand, we usually expect from the start to have (way) more items than available hash table slots. If a collision occurs, we need to add the item to the linked list stored in a specific slot. Since searching in a linked list is costly, we would like to minimize list traversal operations. For that, the best case is to have all slots filled with lists of ideally the same length! Having all slots filled corresponds to a load factor of 1.
To put it another way: A load factor < 1 means that there are empty slots and items had to be added to a linked list in another slot, increasing the number of list traversal operations and wasting some memory.
Concerning your example of a table with size 100: yes, there is a chance that all items collide and occupy just one single slot. In that case, the effective load factor would be 0.01 and performance would be heavily impacted.
If you understand the load factor as n_items / n_total_slots:
In that case, the load factor can be larger than 1. A factor < 1 means you have empty slots, while factor > 1 means that there are slots holding more than one item and consequently, list traversals are required. In the first case, you are wasting space and in the second case list traversals lead to a (small) performance hit, depending on the size of the lists.
Example: A load factor of 10 means that on average each slot holds 10 items. Searching for an item therefore implies traversing 5 list nodes on average.
A load factor of 1 means you waste no space and have the fastest lookup, if you use a decent hash function that ensures a regular and evenly balanced usage of slots.
Using Matlab, I am going to generate several data files and store them in H5 format as 20x1500xN, where N is an integer that can vary, but typically around 2300. Each file will have 4 different data sets with equal structure. Thus, I will quickly achieve a storage problem. My two questions:
Is there any reason not the split the 4 different data sets, and just save as 4x20x1500xNinstead? I would prefer having them split, since it is different signal modalities, but if there is any computational/compression advantage to not having them separated, I will join them.
Using Matlab's built-in compression, I set deflate=9 (and DataType=single). However, I have now realized that using deflate multiplies my computational time with 5. I realize this could have something to do with my ChunkSize, which I just put to 20x1500x5 - without any reasoning behind it. Is there a strategic way to optimize computational load w.r.t. deflation and compression time?
Thank you.
1- Splitting or merging? It won't make a difference in the compression procedure, since it is performed in blocks.
2- Your choice of chunkshape seems, indeed, bad. Chunksize determines the shape and size of each block that will be compressed independently. The bad is that each chunk is of 600 kB, that is much larger than the L2 cache, so your CPU is likely twiddling its fingers, waiting for data to come in. Depending on the nature of your data and the usage pattern you will use the most (read the whole array at once, random reads, sequential reads...) you may want to target the L1 or L2 sizes, or something in between. Here are some experiments done with a Python library that may serve you as a guide.
Once you have selected your chunksize (how many bytes will your compression blocks have), you have to choose a chunkshape. I'd recommend the shape that most closely fits your reading pattern, if you are doing partial reads, or filling in in a fastest-axis-first if you want to read the whole array at once. In your case, this will be something like 1x1500x10, I think (second axis being the fastest, last one the second fastest, and fist the slowest, change if I am mistaken).
Lastly, keep in mind that the details are quite dependant on the specific machine you run it: the CPU, the quality and load of the hard drive or SSD, speed of RAM... so the fine tuning will always require some experimentation.
I read this page about the time complexity of Scala collections. As it says, Vector's complexity is eC for all operations.
It made me wonder what Vector is. I read the document and it says:
Because vectors strike a good balance between fast random selections and fast random functional updates, they are currently the
default implementation of immutable indexed sequences. It is backed by
a little endian bit-mapped vector trie with a branching factor of 32.
Locality is very good, but not contiguous, which is good for very
large sequences.
As with everything else about Scala, it's pretty vague. How actually does Vector work?
The keyword here is Trie.
Vector is implemented as a Trie datastructure.
See http://en.wikipedia.org/wiki/Trie.
More precisely, it is a "bit-mapped vector trie". I've just found a consise enough description of the structure (along with an implementation - apparently in Rust) here:
https://bitbucket.org/astrieanna/bitmapped-vector-trie
The most relevant excerpt is:
A Bitmapped Vector Trie is basically a 32-tree. Level 1 is an array of size 32, of whatever data type. Level 2 is an array of 32 Level 1's. and so on, until: Level 7 is an array of 2 Level 6's.
UPDATE: In reply to Lai Yu-Hsuan's comment about complexity:
I will have to assume you meant "depth" here :-D. The legend for "eC" says "The operation takes effectively constant time, but this might depend on some assumptions such as maximum length of a vector or distribution of hash keys.".
If you are willing to consider the worst case, and given that there is an upper bound to the maximum size of the vector, then yes indeed we can say that the complexity is constant.
Say that we consider the maximum size to be 2^32, then this means that the worst case is 7 operations at most, in any case.
Then again, we can always consider the worst case for any type of collection, find an upper bound and say this is constant complexity, but for a list by example, this would mean a constant of 4 billions, which is not quite practical.
But Vector is the opposite, 7 operations being more than practical, and this is how we can afford to consider its complexity constant in practice.
Another way to look at this: we are not talking about log(2,N), but log(32,N). If you try to plot that you'll see it is practically an horizontal line. So pragmatically speaking you'll never be able to see much increase in processing time as the collection grows.
Yes, that's still not really constant (which is why it is marked as "eC" and not just "C"), and you'll be able to see a difference around short vectors (but again, a very small difference because the number of operations grows so much slowly).
The other answers re 'Trie' are good. But as a close approximation, just for quick understanding:
Vector internally uses a tree structure - not a binary tree, but a 32-ary tree
Each '32-way node' uses Array[32] and can store either 0-32 references to child nodes or 0-32 pieces of data
The tree is structured to be balanced in a certain way - it is "n" levels deep, but levels 1 to n-1 are "index-only levels" (100% child references; no data) and level n contains all the data (100% data; no child references). So if the number of elements of data is "d" then n = log-base-32(d) rounded upwards
Why this? Simple: for performance.
Instead of doing thousands/millions/gazillions of memory allocations for each individual data element, memory is allocated in 32 element chunks. Instead of walking miles deep to find your data, the structure is quite shallow - it's a very wide, short tree. E.g. 5 levels deep can contain 32^5 data elements (for 4 byte elements = 132GB i.e. pretty big) and each data access would lookup & walk through 5 nodes from the root (whereas a big array would use a single data access). The vector does not proactively allocat memory for all of Level n (data), - it allocates in 32 element chunks as needed. It gives read performance somewhat similar to a huge array, whilst having functional characteristics (power & flexibility & memory-efficiency) somewhat similar to a binary tree.
:)
These may be interesting for you:
Ideal Hash Trees by Phil Bagwell.
Implementing Persistent Vectors in Scala - Daniel Spiewak
More Persistent Vectors: Performance Analysis - Daniel Spiewak
Persistent data structures in Scala
I am parsing a large amount of network trace data. I want to split the trace into chunks, hash each chunk, and store a sequence of the resulting hashes rather than the original chunks. The purpose of my work is to identify identical chunks of data - I'm hashing the original chunks to reduce the data set size for later analysis. It is acceptable in my work that we trade off the possibility that collisions occasionally occur in order to reduce the hash size (e.g. 40 bit hash with 1% misidentification of identical chunks might beat 60 bit hash with 0.001% misidentification).
My question is, given a) number of chunks to be hashed and b) allowable percentage of misidentification, how can one go about choosing an appropriate hash size?
As an example:
1,000,000 chunks to be hashed, and we're prepared to have 1% misidentification (1% of hashed chunks appear identical when they are not identical in the original data). How do we choose a hash with the minimal number of bits that satisifies this?
I have looked at materials regarding the Birthday Paradox, though this is concerned specifically with the probability of a single collision. I have also looked at materials which discuss choosing a size based on an acceptable probability of a single collision, but have not been able to extrapolate from this how to choose a size based on an acceptable probability of n (or fewer) collisions.
Obviously, the quality of your hash function matters, but some easy probability theory will probably help you here.
The question is what exactly are you willing to accept, is it good enough that you have an expected number of collisions at only 1% of the data? Or, do you demand that the probability of the number of collisions going over some bound be something? If its the first, then back of the envelope style calculation will do:
Expected number of pairs that hash to the same thing out of your set is (1,000,000 C 2)*P(any two are a pair). Lets assume that second number is 1/d where d is the the size of the hashtable. (Note: expectations are linear, so I'm not cheating very much so far). Now, you say you want 1% collisions, so that is 10000 total. Well, you have (1,000,000 C 2)/d = 10,000, so d = (1,000,000 C 2)/10,000 which is according to google about 50,000,000.
So, you need a 50 million ish possible hash values. That is a less than 2^26, so you will get your desired performance with somewhere around 26 bits of hash (depending on quality of hashing algorithm). I probably have a factor of 2 mistake in there somewhere, so you know, its rough.
If this is an offline task, you cant be that space constrained.
Sounds like a fun exercise!
Someone else might have a better answer, but I'd go the brute force route, provided that there's ample time:
Run the hashing calculation using incremental hash size and record the collision percentage for each hash size.
You might want to use binary search to reduce the search space.