How to save a matrix in C++ in a non-linear way - performance

I have to program an optimized multi-thread implementation of the Levenshtein distance problem. It can be computed using dynamic programming with a matrix, the wikipedia page on Levenshtein distance covers that well enough.
Now, I can compute diagonal elements concurrently. That is all alright.
My problem now comes with caches. Matrices in c++ are normaly saved in memory row by row, correct? Well, that is not good for me as I need 2 element of the previous row and 1 element of the current row to compute my result, that is horrible cache-wise. The cache will hold the current row (or part of it), then I ask for the previous one which it will probably not hold anymore.
Then for another one, I need a different part of the diagonal, so yet again, I ask for completely different rows and the cache will not have those ready for me.
Therefore, I would like to save my matrix to memory in blocks or maybe diagoals. That will result in fewer cachce misses and make my implementation faster again.
How do you do that? I tried searching the internet, but I could never find anything that would show me the way. Is it possible to tell c++ how to order that type in memory?
EDIT: As some of you seem confused about the nature of my question. I want to save a matrix (does not matter if I will make it a 2D array or any other way) in a custom way into the MEMORY. Normally, a 2D array will save row after row, I need to work with diagonals therefore caches will miss a lot on the huge matrices I will work at (possibly millions of rows and columns).

I believe you may have a mis-perception of (CPU) cache.
It's true that CPU caching is linear - that is, if you access an address in memory, it will bring into the cache some previous and some successive memory locations - which is like "guessing" that subsequent accesses will involve 1-dimensional-close elements. However, this is true on the micro-level. A CPU's cache is made up of a large number of small "lines" (64 Bytes on all cache levels in recent Intel CPUs). The locality is limited to the line; different cache lines can come from completely different places in memory.
Thus, if you "need two elements of the previous row and one element of the current row" of your matrix, then the cache should work very well for you: Some of the cache will hold elements of the previous row, and some will hold elements of the current row. And when you advance to the next element, the cache overall will usually contain the matrix elements you need to access. Just make sure your order of iteration agrees with the order of progression within the cache line.
Also, in some cases you could be faced with a situation where different threads are thrashing the same cache lines due to the mapping from main memory into the cache. Without getting into details, that is something you need to think about (but again, has nothing to do with 2D vs 1D data).
Edit: As geza notes, if your matrix' lines are long, you will still be reading each memory location twice with the straightforward approach: Once as the current-line, then again as the previous-line, since each value will be evicted from the cache before it's used as a previous-line value. If you want to avoid this, you can iterate over tiles of your matrix, whose size (length x width x sizeof(element)) fits into the L1 cache (along with whatever else needs to be there). You can also consider storing your data in tiles, but I don't think that would be too useful.

Preliminary comment: "Levenshtein distance" is edit distance (under the common definition). This is a very common problem; you probably don't even need to bother writing a solution yourself. Look for existing code.
Now, finally, for a proper answer... You don't actually need have a matrix at all, and you certainly don't need to "save" it: It's enough to keep merely a "front" of your dynamic programming matrix rather than the whole thing.
But what "front" shall you choose, and how do you advance it? I suggest you use anti-diagonals as your front, and given each anti-diagonal, compute concurrently the next anti-diagonal. Thus it'll be {(0,0)}, then {(0,1),(1,0)}, then {(0,2),(1,1),(2,0)} and so on. Each anti-diagonal requires at most two earlier anti-diagonals - and if we keep the values of each anti-diagonal consecutively in memory, then the access pattern going up the next anti-diagonal is a linear progression along the previous anti-diagonals - which is great for the cache (see my other answer).
So, you'll "concurrentize" the computation give each thread a bunch of consecutive anti-diagonal elements to compute; that should do the trick. And at any time you will only keep 3 anti-diagonal in memory: the one you're working on and the two previous ones. You can cycle between three such buffers so you don't re-allocate memory all the time (but then make sure to pre-allocate buffers with the maximum anti-diagonal length).
This whole thing should work basically the same for the non-square case.

I'm not absolutely sure, but i think a matrix is stored as a long array one row after the other and is mapped with pointer arithmetic to a matrix, so you always refer to the same address and calculate the distance in the memory where your value is located
Otherwise you can implement it easily as this type and implement operator[int, int] for your matrix

Related

What is the fastest way to intersect two large set of ids

The Problem
On a server, I host ids in a json file. From clients, I need to mandate the server to intersect and sometimes negate these ids (the ids never travel to the client even though the client instructs the server its operations to perform).
I typically have 1000's of ids, often have 100,000's of ids, and have a maximum of 56,000,000 of them, where each value is unique and between -100,000,000 and +100,000,000.
These ids files are stable and do not change (so it is possible to generate a different representation for it that is better adapted for the calculations if needed).
Sample ids
Largest file sizes
I need an algorithm that will intersect ids in the sub-second range for most cases. What would you suggest? I code in java, but do not limit myself to java for the resolution of this problem (I could use JNI to bridge to native language).
Potential solutions to consider
Although you could not limit yourselves to the following list of broad considerations for solutions, here is a list of what I internally debated to resolve the situation.
Neural-Network pre-qualifier: Train a neural-network for each ids list that accepts another list of ids to score its intersection potential (0 means definitely no intersection, 1 means definitely there is an intersection). Since neural networks are good and efficient at pattern recognition, I am thinking of pre-qualifying a more time-consuming algorithm behind it.
Assembly-language: On a Linux server, code an assembly module that does such algorithm. I know that assembly is a mess to maintain and code, but sometimes one need the speed of an highly optimized algorithm without the overhead of a higher-level compiler. Maybe this use-case is simple enough to benefit from an assembly language routine to be executed directly on the Linux server (and then I'd always pay attention to stick with the same processor to avoid having to re-write this too often)? Or, alternately, maybe C would be close enough to assembly to produce clean and optimized assembly code without the overhead to maintain assembly code.
Images and GPU: GPU and image processing could be used and instead of comparing ids, I could BITAND images. That is, I create a B&W image of each ids list. Since each id have unique values between -100,000,000 and +100,000,000 (where a maximum of 56,000,000 of them are used), the image would be mostly black, but the pixel would become white if the corresponding id is set. Then, instead of keeping the list of ids, I'd keep the images, and do a BITAND operation on both images to intersect them. This may be fast indeed, but then to translate the resulting image back to ids may be the bottleneck. Also, each image could be significantly large (maybe too large for this to be a viable solution). An estimate of a 200,000,000 bits sequence is 23MB each, just loading this in memory is quite demanding.
String-matching algorithms: String comparisons have many adapted algorithms that are typically extremely efficient at their task. Create a binary file for each ids set. Each id would be 4 bytes long. The corresponding binary file would have each and every id sequenced as their 4 bytes equivalent into it. The algorithm could then be to process the smallest file to match each 4 bytes sequence as a string into the other file.
Am I missing anything? Any other potential solution? Could any of these approaches be worth diving into them?
I did not yet try anything as I want to secure a strategy before I invest what I believe will be a significant amount of time into this.
EDIT #1:
Could the solution be a map of hashes for each sector in the list? If the information is structured in such a way that each id resides within its corresponding hash key, then, the smaller of the ids set could be sequentially ran and matching the id into the larger ids set first would require hashing the value to match, and then sequentially matching of the corresponding ids into that key match?
This should make the algorithm an O(n) time based one, and since I'd pick the smallest ids set to be the sequentially ran one, n is small. Does that make sense? Is that the solution?
Something like this (where the H entry is the hash):
{
"H780" : [ 45902780, 46062780, -42912780, -19812780, 25323780, 40572780, -30131780, 60266780, -26203780, 46152780, 67216780, 71666780, -67146780, 46162780, 67226780, 67781780, -47021780, 46122780, 19973780, 22113780, 67876780, 42692780, -18473780, 30993780, 67711780, 67791780, -44036780, -45904780, -42142780, 18703780, 60276780, 46182780, 63600780, 63680780, -70486780, -68290780, -18493780, -68210780, 67731780, 46092780, 63450780, 30074780, 24772780, -26483780, 68371780, -18483780, 18723780, -29834780, 46202780, 67821780, 29594780, 46082780, 44632780, -68406780, -68310780, -44056780, 67751780, 45912780, 40842780, 44642780, 18743780, -68220780, -44066780, 46142780, -26193780, 67681780, 46222780, 67761780 ],
"H782" : [ 27343782, 67456782, 18693782, 43322782, -37832782, 46152782, 19113782, -68411782, 18763782, 67466782, -68400782, -68320782, 34031782, 45056782, -26713782, -61776782, 67791782, 44176782, -44096782, 34041782, -39324782, -21873782, 67961782, 18703782, 44186782, -31143782, 67721782, -68340782, 36103782, 19143782, 19223782, 31711782, 66350782, 43362782, 18733782, -29233782, 67811782, -44076782, -19623782, -68290782, 31721782, 19233782, 65726782, 27313782, 43352782, -68280782, 67346782, -44086782, 67741782, -19203782, -19363782, 29583782, 67911782, 67751782, 26663782, -67910782, 19213782, 45992782, -17201782, 43372782, -19992782, -44066782, 46142782, 29993782 ],
"H540" : [...
You can convert each file (list of ids) into a bit-array of length 200_000_001, where bit at index j is set if the list contains value j-100_000_000. It is possible, because the range of id values is fixed and small.
Then you can simply use bitwise and and not operations to intersect and negate lists of ids. Depending on the language and libraries used, it would require operating element-wise: iterating over arrays and applying corresponding operations to each index.
Finally, you should measure your performance and decide whether you need to do some optimizations, such as parallelizing operations (you can work on different parts of arrays on different processors), preloading some of arrays (or all of them) into memory, using GPU, etc.
First, the bitmap approach will produce the required performance, at a huge overhead in memory. You'll need to benchmark it, but I'd expect times of maybe 0.2 seconds, with that almost entirely dominated by the cost of loading data from disk, and then reading the result.
However there is another approach that is worth considering. It will use less memory most of the time. For most of the files that you state, it will perform well.
First let's use Cap'n Proto for a file format. The type can be something like this:
struct Ids {
is_negated #0 :Bool;
ids #1 :List(Int32);
}
The key is that ids are always kept sorted. So list operations are a question of running through them in parallel. And now:
Applying not is just flipping is_negated.
If neither is negated, it is a question of finding IDs in both lists.
If the first is not negated and the second is, you just want to find IDs in the first that are not in the second.
If the first is negated and the second is not, you just want to find IDs in the second that are not in the first.
If both are negated, you just want to find all ids in either list.
If your list has 100k entries, then the file will be about 400k. A not requires copying 400k of data (very fast). And intersecting with another list of the same size involves 200k comparisons. Integer comparisons complete in a clock cycle, and branch mispredictions take something like 10-20 clock cycles. So you should be able to do this operation in the 0-2 millisecond range.
Your worst case 56,000,000 file will take over 200 MB and intersecting 2 of them can take around 200 million operations. This is in the 0-2 second range.
For the 56 million file and a 10k file, your time is almost all spent on numbers in the 56 million file and not in the 10k one. You can speed that up by adding a "galloping" mode where you do a binary search forward in the larger file looking for the next matching number and picking most of them. Do be warned that this code tends to be tricky and involves lots of mispredictions. You'll have to benchmark it to find out how big a size difference is needed.
In general this approach will lose for your very biggest files. But it will be a huge win for most of the sizes of file that you've talked about.

Optimizing Inserting into the Middle of a List

I have algorithms that works with dynamically growing lists (contiguous memory like a C++ vector, Java ArrayList or C# List). Until recently, these algorithms would insert new values into the middle of the lists. Of course, this was usually a very slow operation. Every time an item was added, all the items after it needed to be shifted to a higher index. Do this a few times for each algorithm and things get really slow.
My realization was that I could add the new items to the end of the list and then rotate them into position later. That's one option!
Another option, when I know how many items I'm adding ahead of time, is to add that many items to the back, shift the existing items and then perform the algorithm in-place in the hole I've made for myself. The negative is that I have to add some default value to the end of the list and then just overwrite them.
I did a quick analysis of these options and concluded that the second option is more efficient. My reasoning was that the rotation with the first option would result in in-place swaps (requiring a temporary). My only concern with the second option is that I am creating a bunch of default values that just get thrown away. Most of the time, these default values will be null or a mem-filled value type.
However, I'd like someone else familiar with algorithms to tell me which approach would be faster. Or, perhaps there's an even more efficient solution I haven't considered.
Arrays aren't efficient for lots of insertions or deletions into anywhere other than the end of the array. Consider whether using a different data structure (such as one suggested in one of the other answers) may be more efficient. Without knowing the problem you're trying to solve, it's near-impossible to suggest a data structure (there's no one solution for all problems). That being said...
The second option is definitely the better option of the two. A somewhat better option (avoiding the default-value issue): simply copy 789 to the end and overwrite the middle 789 with 456. So the only intermediate step would be 0123789789.
Your default-value concern is, however, (generally) not a big issue:
In Java, for one, you cannot (to my knowledge) even assign memory for an array that's not 0- or null-filled. C++ STL containers also enforce this I believe (but not C++ itself).
The size of a pointer compared to any moderate-sized class is minimal (thus assigning it to a default value also takes minimal time) (in Java and C# everything is pointers, in C++ you can use pointers (something like boost::shared_ptr or a pointer-vector is preferred above straight pointers) (N/A to primitives, which are small to start, so generally not really a big issue either).
I'd also suggest forcing a reallocation to a specified size before you start inserting to the end of the array (Java's ArrayList::ensureCapacity or C++'s vector::reserve). In case you didn't know - varying-length-array implementations tend to have an internal array that's bigger than what size() returns or what's accessible (in order to prevent constant reallocation of memory as you insert or delete values).
Also note that there are more efficient methods to copy parts of an array than doing it manually with for loops (e.g. Java's System.arraycopy).
You might want to consider changing your representation of the list from using a dynamic array to using some other structure. Here are two options that allow you to implement these operations efficiently:
An order statistic tree is a modified type of binary tree that supports insertions and selections anywhere in O(log n) time, as well as lookups in O(log n) time. This will increase your memory usage quite a bit because of the overhead for the pointers and extra bookkeeping, but should dramatically speed up insertions. However, it will slow down lookups a bit.
If you always know the insertion point in advance, you could consider switching to a linked list instead of an array, and just keep a pointer to the linked list cell where insertions will occur. However, this slows down random access to O(n), which could possibly be an issue in your setup.
Alternatively, if you always know where insertions will happen, you could consider representing your array as two stacks - one stack holding the contents of the array to the left of the insert point and one holding the (reverse) of the elements to the right of the insertion point. This makes insertions fast, and if you have the right type of stack implementation could keep random access fast.
Hope this helps!
HashMaps and Linked Lists were designed for the problem you are having. Given a indexed data structure with numbered items, the difficulty of inserting items in the middle requires a renumbering of every item in the list.
You need a data structure which is optimized to make inserts a constant O(1) complexity. HashMaps were designed to make insert and delete operations lightning quick regardless of dataset size.
I can't pretend to do the HashMap subject justice by describing it. Here is a good intro: http://en.wikipedia.org/wiki/Hash_table

What makes table lookups so cheap?

A while back, I learned a little bit about big O notation and the efficiency of different algorithms.
For example, looping through each item in an array to do something with it
foreach(item in array)
doSomethingWith(item)
is an O(n) algorithm, because the number of cycles the program performs is directly proportional to the size of the array.
What amazed me, though, was that table lookup is O(1). That is, looking up a key in a hash table or dictionary
value = hashTable[key]
takes the same number of cycles regardless of whether the table has one key, ten keys, a hundred keys, or a gigabrajillion keys.
This is really cool, and I'm very happy that it's true, but it's unintuitive to me and I don't understand why it's true.
I can understand the first O(n) algorithm, because I can compare it to a real-life example: if I have sheets of paper that I want to stamp, I can go through each paper one-by-one and stamp each one. It makes a lot of sense to me that if I have 2,000 sheets of paper, it will take twice as long to stamp using this method than it would if I had 1,000 sheets of paper.
But I can't understand why table lookup is O(1). I'm thinking that if I have a dictionary, and I want to find the definition of polymorphism, it will take me O(logn) time to find it: I'll open some page in the dictionary and see if it's alphabetically before or after polymorphism. If, say, it was after the P section, I can eliminate all the contents of the dictionary after the page I opened and repeat the process with the remainder of the dictionary until I find the word polymorphism.
This is not an O(1) process: it will usually take me longer to find words in a thousand page dictionary than in a two page dictionary. I'm having a hard time imagining a process that takes the same amount of time regardless of the size of the dictionary.
tl;dr: Can you explain to me how it's possible to do a table lookup with O(1) complexity?
(If you show me how to replicate the amazing O(1) lookup algorithm, I'm definitely going to get a big fat dictionary so I can show off to all of my friends my ninja-dictionary-looking-up skills)
EDIT: Most of the answers seem to be contingent on this assumption:
You have the ability to access any page of a dictionary given its page number in constant time
If this is true, it's easy for me to see. But I don't know why this underlying assumption is true: I would use the same process to to look up a page by number as I would by word.
Same thing with memory addresses, what algorithm is used to load a memory address? What makes it so cheap to find a piece of memory from an address? In other words, why is memory access O(1)?
You should read the Wikipedia article.
But the essence is that you first apply a hash function to your key, which converts it to an integer index (this is O(1)). This is then used to index into an array, which is also O(1). If the hash function has been well designed, there should only be one (or a few items) stored at each location in the array, so the lookup is complete.
So in massively-simplified pseudocode:
ValueType array[ARRAY_SIZE];
void insert(KeyType k, ValueType v)
{
int index = hash(k);
array[index] = v;
}
ValueType lookup(KeyType k)
{
int index = hash(k);
return array[index];
}
Obviously, this doesn't handle collisions, but you can read the article to learn how that's handled.
Update
To address the edited question, indexing into an array is O(1) because underneath the hood, the CPU is doing this:
ADD index, array_base_address -> pointer
LOAD pointer -> some_cpu_register
where LOAD loads data stored in memory at the specified address.
Update 2
And the reason a load from memory is O(1) is really just because this is an axiom we usually specify when we talk about computational complexity (see http://en.wikipedia.org/wiki/RAM_model). If we ignore cache hierarchies and data-access patterns, then this is a reasonable assumption. As we scale the size of the machine,, this may not be true (a machine with 100TB of storage may not take the same amount of time as a machine with 100kB). But usually, we assume that the storage capacity of our machine is constant, and much much bigger than any problem size we're likely to look at. So for all intents and purposes, it's a constant-time operation.
I'll address the question from a different perspective from every one else. Hopefully this will give light to why the accessing x[45] and accessing x[5454563] takes the same amount of time.
A RAM is laid out in a grid (i.e. rows and columns) of capacitors. A RAM can address a particular cell of memory by activating a particular column and row on the grid, so let's say if you have a 16-byte capacity RAM, laid out in a 4x4 grid (insanely small for modern computer, but sufficient for illustrative purpose), and you're trying to access the memory address 13 (1101), you first split the address into rows and column, i.e row 3 (11) column 1 (01).
Let's suppose a 0 means taking the left intersection and a 1 means taking a right intersection. So when you want to activate row 3, you send an army of electrons in the row starting gate, the row-army electrons went right, right to reach row 3 activation gate; next you send another army of electrons on the column starting gate, the column-army electrons went left then right to reach the 1st column activation gate. A memory cell can only be read/written if the row and column are both activated, so this would allow the marked cell to be read/written.
The effect of all this gibberish is that the access time of a memory address depends on the address length, and not the particular memory address itself; if an architecture uses a 32-bit address space (i.e. 32 intersections), then addressing memory address 45 and addressing memory address 5454563 both will still have to pass through all 32 intersections (actually 16 intersections for the row electrons and 16 intersections for the columns electrons).
Note that in reality memory addressing takes very little amount of time compared to charging and discharging the capacitors, therefore even if we start having a 512-bit length address space (enough for ~1.4*10^130 yottabyte of RAM, i.e. enough to keep everything under the sun in your RAM), which mean the electrons would have to go through 512 intersections, it wouldn't really add that much time to the actual memory access time.
Note that this is a gross oversimplification of modern RAM. In modern DRAM, if you want to access subsequent memory addresses you only change the columns and not spend time changing the rows, therefore accessing subsequent memory is much faster than accessing totally random addresses. Also, this description is totally ignorant about the effect of CPU cache (although CPU cache also uses a similar grid addressing scheme, however since CPU cache uses the much faster transistor-based capacitor, the negative effect of having large cache address space becomes very critical). However, the point still holds that if you're jumping around the memory, accessing any one of them will take the same amount of time.
You're right, it's surprisingly difficult to find a real-world example of this. The idea of course is that you're looking for something by address and not value.
The dictionary example fails because you don't immediately know the location of page say 278. You still have to look that up the same as you would a word because the page locations are not in your memory.
But say I marked a number on each of your fingers and then I told you to wiggle the one with 15 written on it. You'd have to look at each of them (assuming its unsorted), and if it's not 15 you check the next one. O(n).
If I told you to wiggle your right pinky. You don't have to look anything up. You know where it is because I just told you where it is. The value I just passed to you is its address in your "memory."
It's kind of like that with databases, but on a much larger scale than just 10 fingers.
Because work is done up front -- the value is put in a bucket that is easily accessible given the hashcode of the key. It would be like if you wanted to look up your work in the dictionary but had marked the exact page the word was on.
Imagine you had a dictionary where everything starting with letter A was on page 1, letter B on page 2...etc. So if you wanted to look up "balloon" you would know exactly what page to go to. This is the concept behind O(1) lookups.
Arbitrary data input => maps to a specific memory address
The trade-off of course being you need more memory to allocate for all the potential addresses, many of which may never be used.
If you have an array with 999999999 locations, how long does it take to find a record by social security number?
Assuming you don't have that much memory, then allocate about 30% more array locations that the number of records you intend to store, and then write a hash function to look it up instead.
A very simple (and probably bad) hash function would be social % numElementsInArray.
The problem is collisions--you can't guarantee that every location holds only one element. But thats ok, instead of storing the record at the array location, you can store a linked list of records. Then you scan linearly for the element you want once you hash to get the right array location.
Worst case this is O(n)--everything goes to the same bucket. Average case is O(1) because in general if you allocate enough buckets and your hash function is good, records generally don't collide very often.
Ok, hash-tables in a nutshell:
You take a regular array (O(1) access), and instead of using regular Int values to access it, you use MATH.
What you do, is to take the key value (lets say a string) calculate it into a number (some function on the characters) and then use a well known mathematical formula that gives you a relatively good distribution on the array's range.
So, in that case you are just doing like 4-5 calculations (O(1)) to get an object from that array, using a key which isn't an int.
Now, avoiding collisions, and finding the right mathematical formula for good distribution is the hard part. That's what is explained pretty well in wikipedia: en.wikipedia.org/wiki/Hash_table
Lookup tables know exactly how to access the given item in the table before hand.
Completely the opposite of say, finding an item by it's value in a sorted array, where you have to access items to check that it is what you want.
In theory, a hashtable is a series of buckets (addresses in memory) and a function that maps objects from a domain into those buckets.
Say your domain is 3 letter words, you'd block out 26^3=17,576 addresses for all the possible 3 letter words and create a function that maps all 3 letter words to those addresses, e.g., aaa=0, aab=1, etc. Now when you have a word you'd like to look up, say, "and", you know immediately from your O(1) function that it is address number 367.

Find common words from two files

Given two files containing list of words(around million), We need to find out the words that are in common.
Use Some efficient algorithm, also not enough memory availble(1 million, certainly not).. Some basic C Programming code, if possible, would help.
The files are not sorted.. We can use some sort of algorithm... Please support it with basic code...
Sorting the external file...... with minimum memory available,, how can it be implement with C programming.
Anybody game for external sorting of a file... Please share some code for this.
Yet another approach.
General. first, notice that doing this sequentially takes O(N^2). With N=1,000,000, this is a LOT. Sorting each list would take O(N*log(N)); then you can find the intersection in one pass by merging the files (see below). So the total is O(2N*log(N) + 2N) = O(N*log(N)).
Sorting a file. Now let's address the fact that working with files is much slower than with memory, especially when sorting where you need to move things around. One way to solve this is - decide the size of the chunk that can be loaded into memory. Load the file one chunk at a time, sort it efficiently and save into a separate temporary file. The sorted chunks can be merged (again, see below) into one sorted file in one pass.
Merging. When you have 2 sorted lists (files or not), you can merge them into one sorted list easily in one pass: have 2 "pointers", initially pointing to the first entry in each list. In each step, compare the values the pointers point to. Move the smaller value to the merged list (the one you are constructing) and advance its pointer.
You can modify the merge algorithm easily to make it find the intersection - if pointed values are equal move it to the results (consider how do you want to deal with duplicates).
For merging more than 2 lists (as in sorting the file above) you can generalize the algorithm for using k pointers.
If you had enough memory to read the first file completely into RAM, I would suggest reading it into a dictionary (word -> index of that word ), loop over the words of the second file and test if the word is contained in that dictionary. Memory for a million words is not much today.
If you have not enough memory, split the first file into chunks that fit into memory and do as I said above for each of that chunk. For example, fill the dictionary with the first 100.000 words, find every common word for that, then read the file a second time extracting word 100.001 up to 200.000, find the common words for that part, and so on.
And now the hard part: you need a dictionary structure, and you said "basic C". When you are willing to use "basic C++", there is the hash_map data structure provided as an extension to the standard library by common compiler vendors. In basic C, you should also try to use a ready-made library for that, read this SO post to find a link to a free library which seems to support that.
Your problem is: Given two sets of items, find the intersaction (items common to both), while staying within the constraints of inadequate RAM (less than the size of any set).
Since finding an intersaction requires comparing/searching each item in another set, you must have enough RAM to store at least one of the sets (the smaller one) to have an efficient algorithm.
Assume that you know for a fact that the intersaction is much smaller than both sets and fits completely inside available memory -- otherwise you'll have to do further work in flushing the results to disk.
If you are working under memory constraints, partition the larger set into parts that fit inside 1/3 of the available memory. Then partition the smaller set into parts the fit the second 1/3. The remaining 1/3 memory is used to store the results.
Optimize by finding the max and min of the partition for the larger set. This is the set that you are comparing from. Then when loading the corresponding partition of the smaller set, skip all items outside the min-max range.
First find the intersaction of both partitions through a double-loop, storing common items to the results set and removing them from the original sets to save on comparisons further down the loop.
Then replace the partition in the smaller set with the second partition (skipping items outside the min-max). Repeat. Notice that the partition in the larger set is reduced -- with common items already removed.
After running through the entire smaller set, repeat with the next partition of the larger set.
Now, if you do not need to preserve the two original sets (e.g. you can overwrite both files), then you can further optimize by removing common items from disk as well. This way, those items no longer need to be compared in further partitions. You then partition the sets by skipping over removed ones.
I would give prefix trees (aka tries) a shot.
My initial approach would be to determine a maximum depth for the trie that would fit nicely within my RAM limits. Pick an arbitrary depth (say 3, you can tweak it later) and construct a trie up to that depth, for the smaller file. Each leaf would be a list of "file pointers" to words that start with the prefix encoded by the path you followed to reach the leaf. These "file pointers" would keep an offset into the file and the word length.
Then process the second file by reading each word from it and trying to find it in the first file using the trie you constructed. It would allow you to fail faster on words that don't match. The deeper your trie, the faster you can fail, but the more memory you would consume.
Of course, like Stephen Chung said, you still need RAM to store enough information to describe at least one of the files, if you really need an efficient algorithm. If you don't have enough memory -- and you probably don't, because I estimate my approach would require approximately the same amount of memory you would need to load a file whose words were 14-22 characters long -- then you have to process even the first file by parts. In that case, I would actually recommend using the trie for the larger file, not the smaller. Just partition it in parts that are no bigger than the smaller file (or no bigger than your RAM constraints allow, really) and do the whole process I described for each part.
Despite the length, this is sort of off the top of my head. I might be horribly wrong in some details, but this is how I would initially approach the problem and then see where it would take me.
If you're looking for memory efficiency with this sort of thing you'll be hard pushed to get time efficiency. My example will be written in python, but should be relatively easy to implement in any language.
with open(file1) as file_1:
current_word_1 = read_to_delim(file_1, delim)
while current_word_1:
with open(file2) as file_2:
current_word_2 = read_to_delim(file_2, delim)
while current_word_2:
if current_word_2 == current_word_1:
print current_word_2
current_word_2 = read_to_delim(file_2, delim)
current_word_1 = read_to_delim(file_1, delim)
I leave read_to_delim to you, but this is the extreme case that is memory-optimal but time-least-optimal.
depending on your application of course you could load the two files in a database, perform a left outer join, and discard the rows for which one of the two columns is null

Optimizing Conway's 'Game of Life'

To experiment, I've (long ago) implemented Conway's Game of Life (and I'm aware of this related question!).
My implementation worked by keeping 2 arrays of booleans, representing the 'last state', and the 'state being updated' (the 2 arrays being swapped at each iteration). While this is reasonably fast, I've often wondered about how to optimize this.
One idea, for example, would be to precompute at iteration N the zones that could be modified at iteration (N+1) (so that if a cell does not belong to such a zone, it won't even be considered for modification at iteration (N+1)). I'm aware that this is very vague, and I never took time to go into the details...
Do you have any ideas (or experience!) of how to go about optimizing (for speed) Game of Life iterations?
I am going to quote my answer from the other question, because the chapters I mention have some very interesting and fine-tuned solutions. Some of the implementation details are in c and/or assembly, yes, but for the most part the algorithms can work in any language:
Chapters 17 and 18 of
Michael Abrash's Graphics
Programmer's Black Book are one of
the most interesting reads I have ever
had. It is a lesson in thinking
outside the box. The whole book is
great really, but the final optimized
solutions to the Game of Life are
incredible bits of programming.
There are some super-fast implementations that (from memory) represent cells of 8 or more adjacent squares as bit patterns and use that as an index into a large array of precalculated values to determine in a single machine instruction if a cell is live or dead.
Check out here:
http://dotat.at/prog/life/life.html
Also XLife:
http://linux.maruhn.com/sec/xlife.html
You should look into Hashlife, the ultimate optimization. It uses the quadtree approach that skinp mentioned.
As mentioned in Arbash's Black Book, one of the most simple and straight forward ways to get a huge speedup is to keep a change list.
Instead of iterating through the entire cell grid each time, keep a copy of all the cells that you change.
This will narrow down the work you have to do on each iteration.
The algorithm itself is inherently parallelizable. Using the same double-buffered method in an unoptimized CUDA kernel, I'm getting around 25ms per generation in a 4096x4096 wrapped world.
what is the most efficient algo mainly depends on the initial state.
if the majority of cells is dead, you could save a lot of CPU time by skipping empty parts and not calculating stuff cell by cell.
im my opinion it can make sense to check for completely dead spaces first, when your initial state is something like "random, but with chance for life lower than 5%."
i would just divide the matrix up into halves and start checking the bigger ones first.
so if you have a field of 10,000 * 10,000, you´d first accumulate the states of the upper left quarter of 5,000 * 5,000.
and if the sum of states is zero in the first quarter, you can ignore this first quarter completely now and check the upper right 5,000 * 5,000 for life next.
if its sum of states is >0, you will now divide up the second quarter into 4 pieces again - and repeat this check for life for each of these subspaces.
you could go down to subframes of 8*8 or 10*10 (not sure what makes the most sense here) now.
whenever you find life, you mark these subspaces as "has life".
only spaces which "have life" need to be divided into smaller subspaces - the empty ones can be skipped.
when you are finished assigning the "has life" attribute to all possible subspaces, you end up with a list of subspaces which you now simply extend by +1 to each direction - with empty cells - and perform the regular (or modified) game of life rules to them.
you might think that dividn up a 10,000*10,000 spae into subspaces of 8*8 is a lot os tasks - but accumulating their states values is in fact much, much less computing work than performing the GoL algo to each cell plus their 8 neighbours plus comparing the number and storing the new state for the net iteration somewhere...
but like i said above, for a random init state with 30% population this wont make much sense, as there will be not many completely dead 8*8 subspaces to find (leave alone dead 256*256 subpaces)
and of course, the way of perfect optimisation will last but not least depend on your language.
-110
Two ideas:
(1) Many configurations are mostly empty space. Keep a linked list (not necessarily in order, that would take more time) of the live cells, and during an update, only update around the live cells (this is similar to your vague suggestion, OysterD :)
(2) Keep an extra array which stores the # of live cells in each row of 3 positions (left-center-right). Now when you compute the new dead/live value of a cell, you need only 4 read operations (top/bottom rows and the center-side positions), and 4 write operations (update the 3 affected row summary values, and the dead/live value of the new cell). This is a slight improvement from 8 reads and 1 write, assuming writes are no slower than reads. I'm guessing you might be able to be more clever with such configurations and arrive at an even better improvement along these lines.
If you don't want anything too complex, then you can use a grid to slice it up, and if that part of the grid is empty, don't try to simulate it (please view Tyler's answer). However, you could do a few optimizations:
Set different grid sizes depending on the amount of live cells, so if there's not a lot of live cells, that likely means they are in a tiny place.
When you randomize it, don't use the grid code until the user changes the data: I've personally tested randomizing it, and even after a long amount of time, it still fills most of the board (unless for a sufficiently small grid, at which point it won't help that much anymore)
If you are showing it to the screen, don't use rectangles for pixel size 1 and 2: instead set the pixels of the output. Any higher pixel size and I find it's okay to use the native rectangle-filling code. Also, preset the background so you don't have to fill the rectangles for the dead cells (not live, because live cells disappear pretty quickly)
Don't exactly know how this can be done, but I remember some of my friends had to represent this game's grid with a Quadtree for a assignment. I'm guess it's real good for optimizing the space of the grid since you basically only represent the occupied cells. I don't know about execution speed though.
It's a two dimensional automaton, so you can probably look up optimization techniques. Your notion seems to be about compressing the number of cells you need to check at each step. Since you only ever need to check cells that are occupied or adjacent to an occupied cell, perhaps you could keep a buffer of all such cells, updating it at each step as you process each cell.
If your field is initially empty, this will be much faster. You probably can find some balance point at which maintaining the buffer is more costly than processing all the cells.
There are table-driven solutions for this that resolve multiple cells in each table lookup. A google query should give you some examples.
I implemented this in C#:
All cells have a location, a neighbor count, a state, and access to the rule.
Put all the live cells in array B in array A.
Have all the cells in array A add 1 to the neighbor count of their
neighbors.
Have all the cells in array A put themselves and their neighbors in array B.
All the cells in Array B Update according to the rule and their state.
All the cells in Array B set their neighbors to 0.
Pros:
Ignores cells that don't need to be updated
Cons:
4 arrays: a 2d array for the grid, an array for the live cells, and an array
for the active cells.
Can't process rule B0.
Processes cells one by one.
Cells aren't just booleans
Possible improvements:
Cells also have an "Updated" value, they are updated only if they haven't
updated in the current tick, removing the need of array B as mentioned above
Instead of array B being the ones with live neighbors, array B could be the
cells without, and those check for rule B0.

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