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I want to do a large block of code until at least one of the elements of array 1 is equal to 1 of the elements of array 2.
I'm asking the community to share, if possible, the best(fastest to process) ways possible to do this "while"
Sum up:
while (none of the elements from arr1 is equal to any of arr2)
{
(code)
}
Reason: In my code, depending on some dimensions set by the user, my program may need to make this n^2 complexity comparison a lot of times, so I'm looking for a way to make it as light as I can.
I'm sorry in advance, and please let me know, if this type of questions are not suitable for StackOverflow.
Edit: My bad in not giving information about the arrays. As I said, it's dimension may vary based on what the user choses, but each one's size should be between 3 and 1000. Both arrays of integers.
Their values do change, the bigger the dimensions, the more it can happen.
The comments mention a hash and I agree, it could work, but it's very size-dependent. O(n^2)'s overhead will be negligible a lot of the time.
Otherwise, just add all the elements of arr1 into the hashset, then go through arr2's elements to see if they're in there. You'll get O(n) time. Honestly though, unless you're working with elements in the hundred thousands, even millions, I don't think the pay-out will be that tangible, but it's machine dependent and I haven't tested it myself.
C++ has std::unordered_set in the standard. If you're using pure C, I'm sure there's implementations available online.
I'm looking for a hint towards a solution of the problem:
Suppose there's an array with some numbers in ascending order and some in descending, for example [1,2,5,9,6,3,2,4,7,8] has sequences asc [1,2,5,9], desc [(9),6,3,2], asc [(2),4,7,8].
Now this isn't a problem, I could simply loop through an array and add them to some data structure, and when the direction changes - I store this structure somwhere and start filling next one.
What I've found tricky is if I want to have threshold of some sort. For example: [0,50,100,99,98,97,105,160]
So the sequence in descending order [(100), 99, 98, 97] could be neglected, because overall change is -3, whereas the sequence was increasing much more dramatically (+100) and as a result, the algorithm identifies only one sequence in ascending order.
I have tried the same method as above, simply adding all sequences in a data structure and then comparing the change in values of two consequtive items: (100 vs -3 means -3 can be neglected). But then the problem is if I have say this situation:
(example only in change of values from start to end of sequense)
[+100, -3, +1, -50]
in this situation I cannot neglect descending movement, because the numbers start to descend, then slightly ascend and again go down pretty significantly.
and it gets really confusing with stuff like that:
[+100, -3, +3, -3, +3, -50]
this is quick sketch of representation of what I am trying to achieve:
black lines represent initial data in an array, red thin lines are desired resulting output
Could somebody point me out in right direction? How would I approach this situation? Compare multiple sequences at a time slowly combining sequences together? Maybe I would need to go through sequences multiple times?
I'm not sure If I've come across problem like that and don't know working algorithm. This is a problem I've faced myself trying to analyse some data.
If I understand correctly, you expect your curve to be a succession of alternatively increasing and decreasing sequences, with a bit of added noise.
The usual way to get rid of noise is to filter data. There are millions of ways to do that, most of them requiring frequency analysis, but in your case you could probably get good enough results with something simple.
The main point is that the relevant variable is not the values in the array, but their variations.
Given N values, consider the array of N-1 elements holding the differences between two consecutive values.
[0,50,100,99,98,97,105,160] -> 50,100,-1,-1,-1,6,45
Now eliminate all values whose absolute value is below a given threshold (say 10 for instance)
-> 50,100,0,0,0,0,45
you can then detect a rising sequence by looking at streaks of all positive or null values (and the same for decreasing sequences, considering zero or negative values).
As for all filtering processes, you will have to find a sweet spot for your threshold. Too low and it will fail to eliminate insignificant variations, too high and it will wipe out significant slope inversions.
I don't know if I understand your problem correctly, but I had to do this kind of dimensionality reduction many times before, so I wrote a small javascript library to do so. It uses the Perceptually Important Points algorithm.
In the algorithm you can define a custom metric of the distance between three consecutive points (to measure how much a single point adds in entropy).
Here is a demonstration (in JS). It works kind like a heap, where you remove points that do not contribute so much to the overall entropy:
for(var i=0; i<data.length; i++)
heap.add(data[i]);
while(heap.minValue() < threshold)
heap.removeMin();
And here is the library.
Well,
I am confused really.
I have a simple order of features, i.e. all the letters and a few symbols, counting how many times are contained in a string.
My selection as a result is as follows
numberOf_a
numberOf_b
...
numberOf_Z
numberOf_.
numberOf_,
I have a test sample of 65 values, and the MLP can get 46 correct.
Now If I chance the order of features in random order, train with the same data, evaluate the same values, I get a different number of correct predictions, e.g. 49.
Results are consistent (the same order will yield the same accuracy) but the accuracy changes between random orders.
The question is, is this supposed to happen? I cannot see how this is backed up by the theory. I am missing something large here?
PS. I am using WEKA's implementation of the MLP
I'm not familiar with the WEKA implementation of the MLP but that doesn't seem like something that should be happening with a neural network algorithm.
It almost seems like it's getting stuck in some sort of local minimum. The algorithm may be initializing the weights of the individual neurons the same way every time. Changing the parameter order might then cause the algorithm to arrive at the same answer for a certain parameter order each time, dependent on the initial parameter order. The "local minimum" might be determined by the algorithm only going through a certain number of iterations each time.
Say, i have 10 billions of numbers stored in a file. How would i find the number that has already appeared once previously?
Well i can't just populate billions of number at a stretch in array and then keep a simple nested loop to check if the number has appeared previously.
How would you approach this problem?
Thanks in advance :)
I had this as an interview question once.
Here is an algorithm that is O(N)
Use a hash table. Sequentially store pointers to the numbers, where the hash key is computed from the number value. Once you have a collision, you have found your duplicate.
Author Edit:
Below, #Phimuemue makes the excellent point that 4-byte integers have a fixed bound before a collision is guaranteed; that is 2^32, or approx. 4 GB. When considered in the conversation accompanying this answer, worst-case memory consumption by this algorithm is dramatically reduced.
Furthermore, using the bit array as described below can reduce memory consumption to 1/8th, 512mb. On many machines, this computation is now possible without considering either a persistent hash, or the less-performant sort-first strategy.
Now, longer numbers or double-precision numbers are less-effective scenarios for the bit array strategy.
Phimuemue Edit:
Of course one needs to take a bit "special" hash table:
Take a hashtable consisting of 2^32 bits. Since the question asks about 4-byte-integers, there are at most 2^32 different of them, i.e. one bit for each number. 2^32 bit = 512mb.
So now one has just to determine the location of the corresponding bit in the hashmap and set it. If one encounters a bit which already is set, the number occured in the sequence already.
The important question is whether you want to solve this problem efficiently, or whether you want accurately.
If you truly have 10 billion numbers and just one single duplicate, then you are in a "needle in the haystack" type of situation. Intuitively, short of very grimy and unstable solution, there is no hope of solving this without storing a significant amount of the numbers.
Instead, turn to probabilistic solutions, which have been used in most any practical application of this problem (in network analysis, what you are trying to do is look for mice, i.e., elements which appear very infrequently in a large data set).
A possible solution, which can be made to find exact results: use a sufficiently high-resolution Bloom filter. Either use the filter to determine if an element has already been seen, or, if you want perfect accuracy, use (as kbrimington suggested you use a standard hash table) the filter to, eh, filter out elements which you can't possibly have seen and, on a second pass, determine the elements you actually see twice.
And if your problem is slightly different---for instance, you know that you have at least 0.001% elements which repeat themselves twice, and you would like to find out how many there are approximately, or you would like to get a random sample of such elements---then a whole score of probabilistic streaming algorithms, in the vein of Flajolet & Martin, Alon et al., exist and are very interesting (not to mention highly efficient).
Read the file once, create a hashtable storing the number of times you encounter each item. But wait! Instead of using the item itself as a key, you use a hash of the item iself, for example the least significant digits, let's say 20 digits (1M items).
After the first pass, all items that have counter > 1 may point to a duplicated item, or be a false positive. Rescan the file, consider only items that may lead to a duplicate (looking up each item in table one), build a new hashtable using real values as keys now and storing the count again.
After the second pass, items with count > 1 in the second table are your duplicates.
This is still O(n), just twice as slow as a single pass.
How about:
Sort input by using some algorith which allows only portion of input to be in RAM. Examples are there
Seek duplicates in output of 1st step -- you'll need space for just 2 elements of input in RAM at a time to detect repetitions.
Finding duplicates
Noting that its a 32bit integer means that you're going to have a large number of duplicates, since a 32 bit int can only represent 4.3ish billion different numbers and you have "10 billions".
If you were to use a tightly packed set you could represent whether all the possibilities are in 512 MB, which can easily fit into current RAM values. This as a start pretty easily allows you to recognise the fact if a number is duplicated or not.
Counting Duplicates
If you need to know how many times a number is duplicated you're getting into having a hashmap that contains only duplicates (using the first 500MB of the ram to tell efficiently IF it should be in the map or not). At a worst case scenario with a large spread you're not going to be able fit that into ram.
Another approach if the numbers will have an even amount of duplicates is to use a tightly packed array with 2-8 bits per value, taking about 1-4GB of RAM allowing you to count up to 255 occurrances of each number.
Its going to be a hack, but its doable.
You need to implement some sort of looping construct to read the numbers one at a time since you can't have them in memory all at once.
How? Oh, what language are you using?
You have to read each number and store it into a hashmap, so that if a number occurs again, it will automatically get discarded.
If possible range of numbers in file is not too large then you can use some bit array to indicate if some of the number in range appeared.
If the range of the numbers is small enough, you can use a bit field to store if it is in there - initialize that with a single scan through the file. Takes one bit per possible number.
With large range (like int) you need to read through the file every time. File layout may allow for more efficient lookups (i.e. binary search in case of sorted array).
If time is not an issue and RAM is, you could read each number and then compare it to each subsequent number by reading from the file without storing it in RAM. It will take an incredible amount of time but you will not run out of memory.
I have to agree with kbrimington and his idea of a hash table, but first of all, I would like to know the range of the numbers that you're looking for. Basically, if you're looking for 32-bit numbers, you would need a single array of 4.294.967.296 bits. You start by setting all bits to 0 and every number in the file will set a specific bit. If the bit is already set then you've found a number that has occurred before. Do you also need to know how often they occur?Still, it would need 536.870.912 bytes at least. (512 MB.) It's a lot and would require some crafty programming skills. Depending on your programming language and personal experience, there would be hundreds of solutions to solve it this way.
Had to do this a long time ago.
What i did... i sorted the numbers as much as i could (had a time-constraint limit) and arranged them like this while sorting:
1 to 10, 12, 16, 20 to 50, 52 would become..
[1,10], 12, 16, [20,50], 52, ...
Since in my case i had hundreds of numbers that were very "close" ($a-$b=1), from a few million sets i had a very low memory useage
p.s. another way to store them
1, -9, 12, 16, 20, -30, 52,
when i had no numbers lower than zero
After that i applied various algorithms (described by other posters) here on the reduced data set
#include <stdio.h>
#include <stdlib.h>
/* Macro is overly general but I left it 'cos it's convenient */
#define BITOP(a,b,op) \
((a)[(size_t)(b)/(8*sizeof *(a))] op (size_t)1<<((size_t)(b)%(8*sizeof *(a))))
int main(void)
{
unsigned x=0;
size_t *seen = malloc(1<<8*sizeof(unsigned)-3);
while (scanf("%u", &x)>0 && !BITOP(seen,x,&)) BITOP(seen,x,|=);
if (BITOP(seen,x,&)) printf("duplicate is %u\n", x);
else printf("no duplicate\n");
return 0;
}
This is a simple problem that can be solved very easily (several lines of code) and very fast (several minutes of execution) with the right tools
my personal approach would be in using MapReduce
MapReduce: Simplified Data Processing on Large Clusters
i'm sorry for not going into more details but once getting familiar with the concept of MapReduce it is going to be very clear on how to target the solution
basicly we are going to implement two simple functions
Map(key, value)
Reduce(key, values[])
so all in all:
open file and iterate through the data
for each number -> Map(number, line_index)
in the reduce we will get the number as the key and the total occurrences as the number of values (including their positions in the file)
so in Reduce(key, values[]) if number of values > 1 than its a duplicate number
print the duplicates : number, line_index1, line_index2,...
again this approach can result in a very fast execution depending on how your MapReduce framework is set, highly scalable and very reliable, there are many diffrent implementations for MapReduce in many languages
there are several top companies presenting already built up cloud computing environments like Google, Microsoft azure, Amazon AWS, ...
or you can build your own and set a cluster with any providers offering virtual computing environments paying very low costs by the hour
good luck :)
Another more simple approach could be in using bloom filters
AdamT
Implement a BitArray such that ith index of this array will correspond to the numbers 8*i +1 to 8*(i+1) -1. ie first bit of ith number is 1 if we already had seen 8*i+1. Second bit of ith number is 1 if we already have seen 8*i + 2 and so on.
Initialize this bit array with size Integer.Max/8 and whenever you saw a number k, Set the k%8 bit of k/8 index as 1 if this bit is already 1 means you have seen this number already.
While thinking about this question and conversing with the participants, the idea came up that shuffling a finite set of clearly biased random numbers makes them random because you don't know the order in which they were chosen. Is this true and if so can someone point to some resources?
EDIT: I think I might have been a little unclear. Suppose a bad random numbers generator. Take n values. These are biased(the rng is bad). Is there a way through shuffling to make the output of the rng over multiple trials statistically match the output of a known good rng?
False.
There is an easy test: Assume the bias in the original set creation algorithm is "creates sets whose arithmetic average is significantly lower than expected average". Obviously, shuffling the result of the algorithm will not change the averages and thus not remove the bias.
Also, regarding your clarification: How would you shuffle the set? Using the same bad output from the bad RNG that created the set in the first place? Or using a better RNG? Which raises the question why you don't use that directly.
It's not true. In the other question the problem is to select 30 random numbers in [1..9] with a sum of 200. After choosing about on average 20 of them randomly, you reach a point where you can't select nines anymore because this would make the total sum go over 200. Of the remaining 10 numbers, most will be ones and twos. So in the end, ones and twos are very overrepresented in the selected numbers. Shuffling doesn't change that. But it's not clear how the random distribution really should look like, so one could say this is as good a solution as any.
In general, if your "random" numbers will be biased to, say, low numbers, they will be biased that way no matter the ordering.
Just shuffling a set of numbers of already random numbers won't do anything to the probability distribution of course. That would mean false. Perhaps I misunderstand your question though?
I would say false, with a caveat:
I think there is random, and then there is 'random-enough'. For most applications that I have needed to work on, 'random-enough' was more than enough, i.e. picking a 'random' ad to display on a page from a list of 300 or so that have paid to be placed on that site.
I am sure a mathematician could prove my very basic 'random' selection criteria is not truly random at all, but in fact is predictable - for my clients, and for the users, nobody cares.
On the other hand if I was writing a video game to be used in Las Vegas where large amounts of money was at hand I'd define random differently (and may have a hard time coming up with truly random).
False
The set is finite, suppose consists of n numbers. What happens if you choose n+1 numbers? Let's also consider a basic random function as implemented in many languages which gives you a random number in [0,1). However, this number is limited to three digits after the decimal giving you a set of 1000 possible numbers (0.000 - 0.999). However in most cases you will not need to use all these 1000 numbers so this amount of randomness is more than enough.
However for some uses, you will need a better random generator than this. So it all comes down to exactly how many random numbers you are going to need, and how random you need them to be.
Addition after reading original question: in the case that you have some sort of limitation (such as in the original question in which each set of selected numbers must sum up to a certain N) you are not really selected random numbers per se, but rather choosing numbers in a random order from a given set (specifically, a permutation of numbers summing up to N).
Addition to edit: Suppose your bad number generator generated the sequence (1,1,1,2,2,2). Does the permutation (1,2,2,1,1,2) satisfy your definition of random?
Completely and utterly untrue: Shuffling doesn't remove a bias, it just conceals it from the casual observer. It's like removing your dog's fondly-laid present from your carpet by just pushing under the sofa - you really haven't solved the problem, you've just made it less conspicuous. Anyone with a nose knows that there is still a problem that needs removing.
The randomness must be applied evenly over the whole range, so here's one way (off the top of my head, lots of assumptions, yadda yadda. The point is the approach, not the code - start with everything even, then introduce your randomness in a consistent fashion until you're done. The only bias now is dependent on the values chosen for 'target' and 'numberofnumbers', which is part of the question.)
target = 200
numberofnumbers = 30
numbers = array();
for (i=0; i<numberofnumbers; i++)
numbers[i] = 9
while (sum(numbers)>target)
numbers[random(numberofnumbers)]--
False. Consider a bad random number generator producing only zeros (I said it was BAD :-) No amount of shuffling the zeros would change any property of that sequence.