I have 93 arrays. Each array has about 18 values in average
I need to make a product of these arrays.
So I have my two dimension array that store these 93 arrays.
Here is what I try to do
DATASET.first.product(*DATASET[1..-1])
Ruby returns
RangeError: too big to product
Does anyone know some workaround to figure out of it?
Some ways to chunk them?
What you want is impossible.
The product of 93 arrays with ~18 elements each is an array with approximately 549975033204266172374216967425209467080301768557741749051999338598022831065169332830885722071173603516904554174087168 elements, each of which is a 93-element array.
This means you need 549975033204266172374216967425209467080301768557741749051999338598022831065169332830885722071173603516904554174087168 * 93 * 64bit of memory to store it, which is roughly 409181424703974032246417423764355843507744515806959861294687507916928986312485983626178977220953161016576988305520852992 bytes. That is about 40 orders of magnitude more than the number of particles in the universe. In other words, even if you were to convert the entire universe into RAM, you would still need to find a way to store on the order of 827180612553027 yobibyte on each and every particle in the universe; that is about 6000000000000000000000000 times the information content of the World Wide Web and 10000000000000000000000 times the information content of the dark web.
Does anyone know some workaround to figure out of it? Some ways to chunk them?
Even if you process them in chunks, that doesn't change the fact that you still need to process 51147678087996754030802177970544480438468064475869982661835938489616123289060747953272372152619145127072123538190106624 elements. Even if you were able to process one element per CPU instruction (which is unrealistic, you will probably need dozens if not hundreds of instructions), and even if each instruction only takes one clock cycle (which is unrealistic, on current mainstream CPUs, each instruction takes multiple clock cycles), and even if you had a terahertz CPU (which is unrealistic, the fastest current CPUs top out at 5 GHz), and even if your CPU had a million cores (which is unrealistic, even GPUs only have a couple of thousand extremely simple cores), and even if your motherboard had a million sockets (which is unrealistic, mainstream motherboards only have a maximum of 4 sockets, and even the biggest supercomputers only have 10 million cores in total), and even if you had a million of those computers in a cluster, and even if you had a million of those clusters in a supercluster, and even if you had a million friends that also have a supercluster like this, it would still take you about 1621000000000000000000000000000000000000000000000000000000000000000000 years to iterate through them.
Right, so as it is hopefully clear that this should not be attempted I'll take a risk and attempt solving your actual problem.
You've mentioned in the comments that you need this array for property testing - I'll take a massive leap of faith here and assume you want to test that every possible combination satisfies some conditions - and this is the mistake here, as the amount of possible combination is just... large...
Instead, you can test that some of the combinations works. You can easily generate a short, randomized list of combinations using:
Array.new(num) { DATASET.map(&:sample) }
Where num is a number of combinations you want to test. Note that there is a chance that some of the entries will be duplicated - but given your dataset size the chances would be comparable with colliding uuids and can be safely ignored.
Generating such a subset of possible solutions is much easier, faster and, most importantly, possible. Since the output is randomized, it will test slightly different combination on each run, so remember to have some randomization setup in your test suite if you want to be able to recreate failures.
Related
I have following question. I set up an camel -project to parse certain xml files. I have to selecting take out certain nodes from a file.
I have two files 246kb and 347kb in size. I am extracting a parent-child pair of 250 nodes in the above given example.
With the default factory here are the times. For the 246kb file respt 77secs and 106 secs. I wanted to improve the performance so switched to saxon and the times are as follows 47secs and 54secs. I was able to cut the time down by at least half.
Is it possible to cut the time further, any other factory or optimizations I can use will be appreciated.
I am using XpathBuilder to cut the xpaths out. here is an example. Is it possible to not to have to create XpathBuilder repeatedly, it seems like it has to be constructed for every xpath, I would have one instance and keep pumping the xpaths into it, maybe it will improve performance further.
return XPathBuilder.xpath(nodeXpath)
.saxon()
.namespace(Consts.XPATH_PREFIX, nameSpace)
.evaluate(exchange.getContext(), exchange.getIn().getBody(String.class), String.class);
Adding more details based on Michael's comments. So I am kind of joining them, will become clear with my example below. I am combining them into a json.
So here we go, Lets say we have following mappings for first and second path.
pData.tinf.rexd: bm:Document/bm:xxxxx/bm:PmtInf[{0}]/bm:ReqdExctnDt/text()
pData.tinf.pIdentifi.instId://bm:Document/bm:xxxxx/bm:PmtInf[{0}]/bm:CdtTrfTxInf[{1}]/bm:PmtId/bm:InstrId/text()
This would result in a json as below
pData:{
tinf: {
rexd: <value_from_xml>
}
pIdentifi:{
instId: <value_from_xml>
}
}
Hard to say without seeing your actual XPath expression, but given the file sizes and execution time my guess would be that you're doing a join which is being executed naively as a cartesian product, i.e. with O(n*m) performance. There is probably some way of reorganizing it to have logarithmic performance, but the devil is in the detail. Saxon-EE is quite good at optimizing join queries automatically; if not, there are often ways of doing it manually -- though XSLT gives you more options (e.g. using xsl:key or xsl:merge) than XPath does.
Actually I was able to bring the time down to 10 secs. I am using apache-camel. So I added threads there so that multiple files can be read in separate threads. Once the file was being read, it had serial operation to based on the length of the nodes that had to be traversed. I realized that it was not necessary to be serial here so introduced parrallelStream and that now gave it enough power. One thing to guard agains is not to have a proliferation of threads since that can degrade the performance. So I try to restrict the number of threads to twice or thrice the number of cores on the operating machine.
I am trying to build a Sieve of Eratosthenes in Lua and i tried several things but i see myself confronted with the following problem:
The tables of Lua are to small for this scenario. If I just want to create a table with all numbers (see example below), the table is too "small" even with only 1/8 (...) of the number (the number is pretty big I admit)...
max = 600851475143
numbers = {}
for i=1, max do
table.insert(numbers, i)
end
If I execute this script on my Windows machine there is an error message saying: C:\Program Files (x86)\Lua\5.1\lua.exe: not enough memory. With Lua 5.3 running on my Linux machine I tried that too, error was just killed. So it is pretty obvious that lua can´t handle the amount of entries.
I don´t really know whether it is just impossible to store that amount of entries in a lua table or there is a simple solution for this (tried it by using a long string aswell...)? And what exactly is the largest amount of entries in a Lua table?
Update: And would it be possible to manually allocate somehow more memory for the table?
Update 2 (Solution for second question): The second question is an easy one, I just tested it by running every number until the program breaks: 33.554.432 (2^25) entries fit in one one-dimensional table on my 12 GB RAM system. Why 2^25? Because 64 Bit per number * 2^25 = 2147483648 Bits which are exactly 2 GB. This seems to be the standard memory allocation size for the Lua for Windows 32 Bit compiler.
P.S. You may have noticed that this number is from the Euler Project Problem 3. Yes I am trying to accomplish that. Please don´t give specific hints (..). Thank you :)
The Sieve of Eratosthenes only requires one bit per number, representing whether the number has been marked non-prime or not.
One way to reduce memory usage would be to use bitwise math to represent multiple bits in each table entry. Current Lua implementations have intrinsic support for bitwise-or, -and etc. Depending on the underlying implementation, you should be able to represent 32 or 64 bits (number flags) per table entry.
Another option would be to use one or more very long strings instead of a table. You only need a linear array, which is really what a string is. Just have a long string with "t" or "f", or "0" or "1", at every position.
Caveat: String manipulation in Lua always involves duplication, which rapidly turns into n² or worse complexity in terms of performance. You wouldn't want one continuous string for the whole massive sequence, but you could probably break it up into blocks of a thousand, or of some power of 2. That would reduce your memory usage to 1 byte per number while minimizing the overhead.
Edit: After noticing a point made elsewhere, I realized your maximum number is so large that, even with a bit per number, your memory requirements would optimally be about 73 gigabytes, which is extremely impractical. I would recommend following the advice Piglet gave in their answer, to look at Jon Sorenson's version of the sieve, which works on segments of the space instead of the whole thing.
I'll leave my suggestion, as it still might be useful for Sorenson's sieve, but yeah, you have a bigger problem than you realize.
Lua uses double precision floats to represent numbers. That's 64bits per number.
600851475143 numbers result in almost 4.5 Terabytes of memory.
So it's not Lua's or its tables' fault. The error message even says
not enough memory
You just don't have enough RAM to allocate that much.
If you would have read the linked Wikipedia article carefully you would have found the following section:
As Sorenson notes, the problem with the sieve of Eratosthenes is not
the number of operations it performs but rather its memory
requirements.[8] For large n, the range of primes may not fit in
memory; worse, even for moderate n, its cache use is highly
suboptimal. The algorithm walks through the entire array A, exhibiting
almost no locality of reference.
A solution to these problems is offered by segmented sieves, where
only portions of the range are sieved at a time.[9] These have been
known since the 1970s, and work as follows
...
In the past I had to work with big files, somewhere about in the 0.1-3GB range. Not all the 'columns' were needed so it was ok to fit the remaining data in RAM.
Now I have to work with files in 1-20GB range, and they will probably grow as the time will pass. That is totally different because you cannot fit the data in RAM anymore.
My file contains several millions of 'entries' (I have found one with 30 mil entries). On entry consists in about 10 'columns': one string (50-1000 unicode chars) and several numbers. I have to sort the data by 'column' and show it. For the user only the top entries (1-30%) are relevant, the rest is low quality data.
So, I need some suggestions about in which direction to head out. I definitively don't want to put data in a DB because they are hard to install and configure for non computer savvy persons. I like to deliver a monolithic program.
Showing the data is not difficult at all. But sorting... without loading the data in RAM, on regular PCs (2-6GB RAM)... will kill some good hours.
I was looking a bit into MMF (memory mapped files) but this article from Danny Thorpe shows that it may not be suitable: http://dannythorpe.com/2004/03/19/the-hidden-costs-of-memory-mapped-files/
So, I was thinking about loading only the data from the column that has to be sorted in ram AND a pointer to the address (into the disk file) of the 'entry'. I sort the 'column' then I use the pointer to find the entry corresponding to each column cell and restore the entry. The 'restoration' will be written directly to disk so no additional RAM will be required.
PS: I am looking for a solution that will work both on Lazarus and Delphi because Lazarus (actually FPC) has 64 bit support for Mac. 64 bit means more RAM available = faster sorting.
I think a way to go is Mergesort, it's a great algorithm for sorting a
large amount of fixed records with limited memory.
General idea:
read N lines from the input file (a value that allows you to keep the lines in memory)
sort these lines and write the sorted lines to file 1
repeat with the next N lines to obtain file 2
...
you reach the end of the input file and you now have M files (each of which is sorted)
merge these files into a single file (you'll have to do this in steps as well)
You could also consider a solution based on an embedded database, e.g. Firebird embedded: it works well with Delphi/Windows and you only have to add some DLL in your program folder (I'm not sure about Lazarus/OSX).
If you only need a fraction of the whole data, scan the file sequentially and keep only the entries needed for display. F.I. lets say you need only 300 entries from 1 million. Scan the first first 300 entries in the file and sort them in memory. Then for each remaining entry check if it is lower than the lowest in memory and skip it. If it is higher as the lowest entry in memory, insert it into the correct place inside the 300 and throw away the lowest. This will make the second lowest the lowest. Repeat until end of file.
Really, there are no sorting algorithms that can make moving 30gb of randomly sorted data fast.
If you need to sort in multiple ways, the trick is not to move the data itself at all, but instead to create an index for each column that you need to sort.
I do it like that with files that are also tens of gigabytes long, and users can sort, scroll and search the data without noticing that it's a huge dataset they're working with.
Please finde here a class which sorts a file using a slightly optimized merge sort. I wrote that a couple of years ago for fun. It uses a skip list for sorting files in-memory.
Edit: The forum is german and you have to register (for free). It's safe but requires a bit of german knowledge.
If you cannot fit the data into main memory then you are into the realms of external sorting. Typically this involves external merge sort. Sort smaller chunks of the data in memory, one by one, and write back to disk. And then merge these chunks.
We are taught that the abstraction of the RAM memory is a long array of bytes. And that for the CPU it takes the same amount of time to access any part of it. What is the device that has the ability to access any byte out of the 4 gigabytes (on my computer) in the same time? As this does not seem as a trivial task for me.
I have asked colleagues and my professors, but nobody can pinpoint to the how this task can be achieved with simple logic gates, and if it isn't just a tricky combination of logic gates, then what is it?
My personal guess is that you could achieve the access of any memory in O(log(n)) speed, where n would be the size of memory. Because each gate would split the memory in two and send you memory access instruction to the next split the memory in two gate. But that requires ALOT of gates. I can't come up with any other educated guess, and I don't even know the name of the device that I should look up in Google.
Please help my anguished curiosity, and thanks in advance.
edit<
This is what I learned!
quote from yours "the RAM can send the value from cell addressed X to some output pins", here is where everyone skip (again) the thing that is not trivial for me. The way that I see it, In order to build a gate that from 64 pins decides which byte out of 2^64 to get, each pin needs to split the overall possible range of memory into two. If bit at index 0 is 0 -> then the address is at memory 0-2^64/2, else address is at memory 2^64/2-2^64. And so on, However the amout of gates (lets call them) that the memory fetch will go through will be 64, (a constant). However the amount of gates needed is N, where N is the number of memory bytes there are.
Just because there is 64 pins, it doesn't mean that you can still decode it into a single fetch from a range of 2^64. Does 4 gigabytes memory come with a 4 gigabytes gates in the memory control???
now this can be improved, because as I read furiously more and more about how this memory is architectured, if you place the memory into a matrix with sqrt(N) rows and sqrt(N) columns, the amount of gates that a fetch memory will need to go through is O(log(sqrt(N)*2) and the amount of gates that will be required will be 2*sqrt(N), which is much better, and I think that its probably a trade secret.
/edit<
What the heck, I might as well make this an answer.
Yes, in the physical world, memory access cannot be constant time.
But it cannot even be logarithmic time. The O(log n) circuit you have in mind ultimately involves some sort of binary (or whatever) tree, and you cannot make a binary tree with constant-length wires in a 3D universe.
Whatever the "bits per unit volume" capacity of your technology is, storing n bits requires a sphere with radius O(n^(1/3)). Since information can only travel at the speed of light, accessing a bit at the other end of the sphere requires time O(n^(1/3)).
But even this is wrong. If you want to talk about actual limitations of our universe, our physics friends say the absolute maximum number of bits you can store in any sphere is proportional to the sphere's surface area, not its volume. So the actual radius of a minimal sphere containing n bits of information is O(sqrt(n)).
As I mentioned in my comment, all of this is pretty much moot. The models of computation we generally use to analyze algorithms assume constant-access-time RAM, which is close enough to the truth in practice and allows a fair comparison of competing algorithms. (Although good engineers working on high-performance code are very concerned about locality and the memory hierarchy...)
Let's say your RAM has 2^64 cells (places where it is possible to store a single value, let's say 8-bit). Then it needs 64 pins to address every cell with a different number. When at the input pins of your RAM there 'appears' a binary number X the RAM can send the value from cell addressed X to some output pins, and your CPU can get the value from there. In hardware the addressing can be done quite easily, for example by using multiple NAND gates (such 'addressing device' from some logic gates is called a decoder).
So it is all happening at the hardware-level, this is just direct addressing. If the CPU is able to provide 64 bits to 64 pins of your RAM it can address every single memory cell (as 64 bit is enough to represent any number up to 2^64 -1). The only reason why you do not get the value immediately is a kind of 'propagation time', so time it takes for the signal to go through all the logic gates in the circuit.
The component responsible for dealing with memory accesses is the memory controller. It is used by the CPU to read from and write to memory.
The access time is constant because memory words are truly layed out in a matrix form (thus, the "byte array" abstraction is very realistic), where you have rows and columns. To fetch a given memory position, the desired memory address is passed on to the controller, which then activates the right column.
From http://computer.howstuffworks.com/ram1.htm:
Memory cells are etched onto a silicon wafer in an array of columns
(bitlines) and rows (wordlines). The intersection of a bitline and
wordline constitutes the address of the memory cell.
So, basically, the answer to your question is: the memory controller figures it out. Of course that, given a memory address, the mapping to column and row must be easy to calculate in a constant time.
To fully understand this topic, I recommend you to read this guide on how memory works: http://computer.howstuffworks.com/ram.htm
There are so many concepts to master that it is difficult to explain it all in one answer.
I've been reading your comments and questions until I answered. I think you are on the right track, but there is some confusion here. The random access in which you are implying doesn't exist in the same way you think it does.
Reading, writing, and refreshing are done in a continuous cycle. A particular cell in memory is only read or written in a certain interval if a signal is detected to do so in that cycle. There is going to be support circuitry that includes "sense amplifiers to amplify the signal or charge detected on a memory cell."
Unless I am misunderstanding what you are implying, your confusion is in how easy it is to read/write to a cell. It's different dependent on chip design but there IS a minimum number of cycles it takes to read or write data to a cell.
These are my sources:
http://www.doc.ic.ac.uk/~dfg/hardware/HardwareLecture16.pdf
http://www.electronics.dit.ie/staff/tscarff/memory/dram_cycles.htm
http://www.ece.cmu.edu/~ece548/localcpy/dramop.pdf
To avoid a humungous answer, I left most of the detail out but all three of these will describe the process you are looking for.
I'd like to get the amount of "free memory" per NUMA node.
When dealing with a whole machine, one usually parses /proc/meminfo like free does (the number wanted is MemFree + Buffers + Cached).
There also exist /sys/devices/system/node/nodex/meminfo, which seem to display numbers per NUMA node. Does anybody know how these numbers can be correlated to the content of /proc/meminfo? My trivial assumption would be to sum up some numbers for all NUMA nodes in the system, and the result is equal to some number in /proc/meminfo. But so far I failed to figure out the relationships, especially for page caches.
The code for proc is in fs/proc/meminfo.c, for the sysfs files it's in drivers/base/node.c. Comparing them might give you some hints.
Note that you'll probably never get the numbers to add up 100%, because you can't atomically read the content of all the files, so the values will change while you're reading them.
There also seems to be an inconsistency in the total RAM reported via both methods. One explanation for that is that free_init_mem doesn't appear to be NUMA aware, and increments total_ram_pages but does not do any NUMA accounting.