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.
Related
I have an m x n matrix of real numbers. I want to choose a single value from each column such that the sum of my selected values is as close as possible to a pre-specified total.
I am not an experienced programmer (although I have an experienced friend who will help). I would like to achieve this using Matlab, Mathematica or c++ (MySQL if necessary).
The code only needs to run a few times, once every few days - it does not necessarily need to be optimised. I will have 16 columns and about 12 rows.
Normally I would suggest dynamic programming, but there are a few features of this situation suggesting an alternative approach. First, the performance demands are light; this program will be run only a couple times, and it doesn't sound as though a running time on the order of hours would be a problem. Second, the matrix is fairly small. Third, the matrix contains real numbers, so it would be necessary to round and then do a somewhat sophisticated search to ensure that the optimal possibility was not missed.
Instead, I'm going to suggest the following semi-brute-force approach. 12**16 ~ 1.8e17, the total number of possible choices, is too many, but 12**9 ~ 5.2e9 is doable with brute force, and 12**7 ~ 3.6e7 fits comfortably in memory. Compute all possible choices for the first seven columns. Sort these possibilities by total. For each possible choice for the last nine columns, use an efficient search algorithm to find the best mate among the first seven. (If you have a lot of memory, you could try eight and eight.)
I would attempt a first implementation in C++, using std::sort and std::lower_bound from the <algorithm> standard header. Measure it; if it's too slow, then try an in-memory B+-tree (does Boost have one?).
I spent some more time thinking about how to implement what I wrote above in the simplest way possible. Here's an approach that will work well for a 12 by 16 matrix on a 64-bit machine with roughly 4 GB of memory.
The number of choices for the first eight columns is 12**8. Each choice is represented by a 4-byte integer between 0 and 12**8 - 1. To decode a choice index i, the row for the first column is given by i % 12. Update i /= 12;. The row for the second column now is given by i % 12, et cetera.
A vector holding all choices requires roughly 12**8 * 4 bytes, or about 1.6 GB. Two such vectors require 3.2 GB. Prepare one for the first eight columns and one for the last eight. Sort them by sum of the entries that they indicate. Use saddleback search to find the best combination. (Initialize an iterator into the first vector and a reverse iterator into the second. While neither iterator is at its end, compare the current combination against the current best and update the current best if necessary. If the current combination sums to than the target, increment the first iterator. If the sum is greater than the target, increment the second iterator.)
I would estimate that this requires less than 50 lines of C++.
Without knowing the range of values that might fill the arrays, how about something generic like this:
divide the target by the number of remaining columns.
Pick the number from that column closest to that value.
Repeat from 1. Until each column picked.
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.
I've been learning about different algorithms in my spare time recently, and one that I came across which appears to be very interesting is called the HyperLogLog algorithm - which estimates how many unique items are in a list.
This was particularly interesting to me because it brought me back to my MySQL days when I saw that "Cardinality" value (which I always assumed until recently that it was calculated not estimated).
So I know how to write an algorithm in O(n) that will calculate how many unique items are in an array. I wrote this in JavaScript:
function countUniqueAlgo1(arr) {
var Table = {};
var numUnique = 0;
var numDataPoints = arr.length;
for (var j = 0; j < numDataPoints; j++) {
var val = arr[j];
if (Table[val] != null) {
continue;
}
Table[val] = 1;
numUnique++;
}
return numUnique;
}
But the problem is that my algorithm, while O(n), uses a lot of memory (storing values in Table).
I've been reading this paper about how to count duplicates in a list in O(n) time and using minimal memory.
It explains that by hashing and counting bits or something one can estimate within a certain probability (assuming the list is evenly distributed) the number of unique items in a list.
I've read the paper, but I can't seem to understand it. Can someone give a more layperson's explanation? I know what hashes are, but I don't understand how they are used in this HyperLogLog algorithm.
The main trick behind this algorithm is that if you, observing a stream of random integers, see an integer which binary representation starts with some known prefix, there is a higher chance that the cardinality of the stream is 2^(size of the prefix).
That is, in a random stream of integers, ~50% of the numbers (in binary) starts with "1", 25% starts with "01", 12,5% starts with "001". This means that if you observe a random stream and see a "001", there is a higher chance that this stream has a cardinality of 8.
(The prefix "00..1" has no special meaning. It's there just because it's easy to find the most significant bit in a binary number in most processors)
Of course, if you observe just one integer, the chance this value is wrong is high. That's why the algorithm divides the stream in "m" independent substreams and keep the maximum length of a seen "00...1" prefix of each substream. Then, estimates the final value by taking the mean value of each substream.
That's the main idea of this algorithm. There are some missing details (the correction for low estimate values, for example), but it's all well written in the paper. Sorry for the terrible english.
A HyperLogLog is a probabilistic data structure. It counts the number of distinct elements in a list. But in comparison to a straightforward way of doing it (having a set and adding elements to the set) it does this in an approximate way.
Before looking how the HyperLogLog algorithm does this, one has to understand why you need it. The problem with a straightforward way is that it consumes O(distinct elements) of space. Why there is a big O notation here instead of just distinct elements? This is because elements can be of different sizes. One element can be 1 another element "is this big string". So if you have a huge list (or a huge stream of elements) it will take a lot memory.
Probabilistic Counting
How can one get a reasonable estimate of a number of unique elements? Assume that you have a string of length m which consists of {0, 1} with equal probability. What is the probability that it will start with 0, with 2 zeros, with k zeros? It is 1/2, 1/4 and 1/2^k. This means that if you have encountered a string starting with k zeros, you have approximately looked through 2^k elements. So this is a good starting point. Having a list of elements that are evenly distributed between 0 and 2^k - 1 you can count the maximum number of the biggest prefix of zeros in binary representation and this will give you a reasonable estimate.
The problem is that the assumption of having evenly distributed numbers from 0 t 2^k-1 is too hard to achieve (the data we encountered is mostly not numbers, almost never evenly distributed, and can be between any values. But using a good hashing function you can assume that the output bits would be evenly distributed and most hashing function have outputs between 0 and 2^k - 1 (SHA1 give you values between 0 and 2^160). So what we have achieved so far is that we can estimate the number of unique elements with the maximum cardinality of k bits by storing only one number of size log(k) bits. The downside is that we have a huge variance in our estimate. A cool thing that we almost created 1984's probabilistic counting paper (it is a little bit smarter with the estimate, but still we are close).
LogLog
Before moving further, we have to understand why our first estimate is not that great. The reason behind it is that one random occurrence of high frequency 0-prefix element can spoil everything. One way to improve it is to use many hash functions, count max for each of the hash functions and in the end average them out. This is an excellent idea, which will improve the estimate, but LogLog paper used a slightly different approach (probably because hashing is kind of expensive).
They used one hash but divided it into two parts. One is called a bucket (total number of buckets is 2^x) and another - is basically the same as our hash. It was hard for me to get what was going on, so I will give an example. Assume you have two elements and your hash function which gives values form 0 to 2^10 produced 2 values: 344 and 387. You decided to have 16 buckets. So you have:
0101 011000 bucket 5 will store 1
0110 000011 bucket 6 will store 4
By having more buckets you decrease the variance (you use slightly more space, but it is still tiny). Using math skills they were able to quantify the error (which is 1.3/sqrt(number of buckets)).
HyperLogLog
HyperLogLog does not introduce any new ideas, but mostly uses a lot of math to improve the previous estimate. Researchers have found that if you remove 30% of the biggest numbers from the buckets you significantly improve the estimate. They also used another algorithm for averaging numbers. The paper is math-heavy.
And I want to finish with a recent paper, which shows an improved version of hyperLogLog algorithm (up until now I didn't have time to fully understand it, but maybe later I will improve this answer).
The intuition is if your input is a large set of random number (e.g. hashed values), they should distribute evenly over a range. Let's say the range is up to 10 bit to represent value up to 1024. Then observed the minimum value. Let's say it is 10. Then the cardinality will estimated to be about 100 (10 × 100 ≈ 1024).
Read the paper for the real logic of course.
Another good explanation with sample code can be found here:
Damn Cool Algorithms: Cardinality Estimation - Nick's Blog
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.