What are practical uses for STL's 'partial_sum'? - c++

What/where are the practical uses of the partial_sum algorithm in STL?
What are some other interesting/non-trivial examples or use-cases?

I used it to reduce memory usage of a simple mark-sweep garbage collector in my toy lambda calculus interpreter.
The GC pool is an array of objects of identical size. The goal is to eliminate objects that aren't linked to other objects, and condense the remaining objects into the beginning of the array. Since the objects are moved in memory, each link needs to be updated. This necessitates an object remapping table.
partial_sum allows the table to be stored in compressed format (as little as one bit per object) until the sweep is complete and memory has been freed. Since the objects are small, this significantly reduces memory use.
Recursively mark used objects and populate the Boolean array.
Use remove_if to condense the marked objects to the beginning of the pool.
Use partial_sum over the Boolean values to generate a table of pointers/indexes into the new pool.
This works because the Nth marked object has N preceding 1's in the array and acquires pool index N.
Sweep over the pool again and replace each link using the remap table.
It's especially friendly to the data cache to put the remap table in the just-freed, thus still hot, memory.

One thing to note about partial sum is that it is the operation that undoes adjacent difference much like - undoes +. Or better yet if you remember calculus the way differentiation undoes integration. Better because adjacent difference is essentially differentiation and partial sum is integration.
Let's say you have simulation of a car and at each time step you need to know the position, velocity, and acceleration. You only need to store one of those values as you can compute the other two. Say you store the position at each time step you can take the adjacent difference of the position to give the velocity and the adjacent difference of the velocity to give the acceleration. Alternatively, if you store the acceleration you can take the partial sum to give the velocity and the partial sum of the velocity gives the position.
Partial sum is one of those functions that doesn't come up too often for most people but is enormously useful when you find the right situation. A lot like calculus.

Last time I (would have) used it is when converting a discrete probability distribution (an array of p(X = k)) into a cumulative distribution (an array of p(X <= k)). To select once from the distribution, you can pick a number from [0-1) randomly, then binary search into the cumulative distribution.
That code wasn't in C++, though, so I did the partial sum myself.

You can use it to generate a monotonically increasing sequence of numbers. For example, the following generates a vector containing the numbers 1 through 42:
std::vector<int> v(42, 1);
std::partial_sum(v.begin(), v.end(), v.begin());
Is this an everyday use case? Probably not, though I've found it useful on several occasions.
You can also use std::partial_sum to generate a list of factorials. (This is even less useful, though, since the number of factorials that can be represented by a typical integer data type is quite limited. It is fun, though :-D)
std::vector<int> v(10, 1);
std::partial_sum(v.begin(), v.end(), v.begin());
std::partial_sum(v.begin(), v.end(), v.begin(), std::multiplies<int>());

Personal Use Case: Roulette-Wheel-Selection
I'm using partial_sum in a roulette-wheel-selection algorithm (link text). This algorithm choses randomly elements from a container with a probability which is linear to some value given beforehands.
Because all my elements to choose from bringing a not-necessarily normalized value, I use the partial_sum algorithm for constructing something like a "roulette-wheel", because I sum up all the elements. Then I chose a random variable in this range (the last partial_sum is the sum of all) and use stl::lower_bound for searching "the wheel" where my random search landed. The element returned by the lower_bound algorithm is the chosen one.
Besides the advantage of clear and expressive code with the use of partial_sum, I could also gain some speed when experimenting with the GCC parallel mode which brings parallelized versions for some algorithms and one of them is the partial_sum (link text).
Another use I know of: One of the most important algorithmic primitives in parallel processing (but maybe a little bit away from STL)
If you're interested in heavy optimized algorithms which are using partial_sum (in this case maybe more results under the synonyms "scan" or "prefix_sum"), than go to the parallel algorithms community. They need it all the time. You won't find a parallel sorting algorithm based on quicksort or mergesort without using it. This operation is one of the most important parallel primitives used. I think it is most commonly used for calculating offsets in dynamic algorithms. Think of a partition step in quicksort, which is split and fed to the parallel threads. You don't know the number of elements in each slot of the partition before calculating it. So you need some offsets for all the threads for later access.
Maybe you will find more informatin in the now-hot topic of GPU processing. One short article regarding Nvidia's CUDA and the scan-primitive with a few application examples you will find in Chapter 39. Parallel Prefix Sum (Scan) with CUDA.

Personal Use Case: intermediate step in counting sort from CLRS:
COUNTING_SORT (A, B, k)
for i ← 1 to k do
c[i] ← 0
for j ← 1 to n do
c[A[j]] ← c[A[j]] + 1
//c[i] now contains the number of elements equal to i
// std::partial_sum here
for i ← 2 to k do
c[i] ← c[i] + c[i-1]
// c[i] now contains the number of elements ≤ i
for j ← n downto 1 do
B[c[A[i]]] ← A[j]
c[A[i]] ← c[A[j]] - 1

You could build a "moving sum" (precursor to a moving average):
template <class T>
void moving_sum (const vector<T>& in, int num, vector<T>& out)
{
// cummulative sum
partial_sum (in.begin(), in.end(), out.begin());
// shift and subtract
int j;
for (int i = out.size() - 1; i >= 0; i--) {
j = i - num;
if (j >= 0)
out[i] -= out[j];
}
}
And then call it with:
vector<double> v(10);
// fill in v
vector<double> v2 (v.size());
moving_sum (v, 3, v2);

You know, I actually did use partial_sum() once... It was this interesting little problem that I was asked on a job interview. I enjoyed it so much, I went home and coded it up.
The problem was: Given a sequential sequence of integers, find the shortest sub-sequence with the highest value. E.g. Given:
Value: -1 2 3 -1 4 -2 -4 5
Index: 0 1 2 3 4 5 6 7
We would find the subsequence [1,4]
Now the obvious solution is to run with 3 for loops, iterating over all possible starts & ends, and adding up the value of each possible subsequence in turn. Inefficient, but quick to code up and hard to make mistakes. (Especially when the third for loop is just accumulate(start,end,0).)
The correct solution involves a divide-and-conquer / bottom up approach. E.g. Divide the problem space in half, and for each half compute the largest subsequence contained within that section, the largest subsequence including the starting number, the largest subsequence including the ending number, and the entire section's subsequence. Armed with this data we can then combine the two halves together without any further evaluation of either one. Obviously the data for each half can be computed by further breaking each half into halves (quarters), each quarter into halves (eighths), and so on until we have trivial singleton cases. It's all quite efficient.
But all that aside, there's a third (somewhat less efficient) option that I wanted to explore. It's similar to the 3-for-loop case, only we add the adjacent numbers to avoid so much work. The idea is that there's no need to add a+b, a+b+c, and a+b+c+d when we can add t1=a+b, t2=t1+c, and t3=t2+d. It's a space/computation tradeoff thing. It works by transforming the sequence:
Index: 0 1 2 3 4
FROM: 1 2 3 4 5
TO: 1 3 6 10 15
Thereby giving us all possible substrings starting at index=0 and ending at indexes=0,1,2,3,4.
Then we iterate over this set subtracting the successive possible "start" points...
FROM: 1 3 6 10 15
TO: - 2 5 9 14
TO: - - 3 7 12
TO: - - - 4 9
TO: - - - - 5
Thereby giving us the values (sums) of all possible subsequences.
We can find the maximum value of each iteration via max_element().
The first step is most easily accomplished via partial_sum().
The remaining steps via a for loop and transform(data+i,data+size,data+i,bind2nd(minus<TYPE>(),data[i-1])).
Clearly O(N^2). But still interesting and fun...

Partial sums are often useful in parallel algorithms. Consider the code
for (int i=0; N>i; ++i) {
sum += x[i];
do_something(sum);
}
If you want to parallelise this code, you need to know the partial sums. I am using GNUs parallel version of partial_sum for something very similar.

I often use partial sum not to sum but to calculate the current value in the sequence depending on the previous.
For example, if you integrate a function. Each new step is a previous step, vt += dvdt or vt = integrate_step(dvdt, t_prev, t_prev+dt);.

In nonparametric Bayesian methods there is a Metropolis-Hastings step (per observation) that determines to sample a new or an existing cluster. If an existing cluster has to be sampled this needs to be done with different weights. These weighted likelihoods are simulated in the following example code.
#include <random>
#include <iostream>
#include <algorithm>
int main() {
std::default_random_engine generator(std::random_device{}());
std::uniform_real_distribution<double> distribution(0.0,1.0);
int K = 8;
std::vector<double> weighted_likelihood(K);
for (int i = 0; i < K; ++i) {
weighted_likelihood[i] = i*10;
}
std::cout << "Weighted likelihood: ";
for (auto i: weighted_likelihood) std::cout << i << ' ';
std::cout << std::endl;
std::vector<double> cumsum_likelihood(K);
std::partial_sum(weighted_likelihood.begin(), weighted_likelihood.end(), cumsum_likelihood.begin());
std::cout << "Cumulative sum of weighted likelihood: ";
for (auto i: cumsum_likelihood) std::cout << i << ' ';
std::cout << std::endl;
std::vector<int> frequency(K);
int N = 280000;
for (int i = 0; i < N; ++i) {
double pick = distribution(generator) * cumsum_likelihood.back();
auto lower = std::lower_bound(cumsum_likelihood.begin(), cumsum_likelihood.end(), pick);
int index = std::distance(cumsum_likelihood.begin(), lower);
frequency[index]++;
}
std::cout << "Frequencies: ";
for (auto i: frequency) std::cout << i << ' ';
std::cout << std::endl;
}
Note that this is not different from the answer by https://stackoverflow.com/users/13005/steve-jessop. It's added to give a bit more context about a particular situation (nonparametric Bayesian mehods, e.g. the algorithms by Neal using the Dirichlet process as prior) and the actual code which uses partial_sum in combination with lower_bound.

Related

Pick a matrix cell according to its probability

I have a 2D matrix of positive real values, stored as follow:
vector<vector<double>> matrix;
Each cell can have a value equal or greater to 0, and this value represents the possibility of the cell to be chosen. In particular, for example, a cell with a value equals to 3 has three times the probability to be chosen compared to a cell with value 1.
I need to select N cells of the matrix (0 <= N <= total number of cells) randomly, but according to their probability to be selected.
How can I do that?
The algorithm should be as fast as possible.
I describe two methods, A and B.
A works in time approximately N * number of cells, and uses space O(log number of cells). It is good when N is small.
B works in time approximately (number of cells + N) * O(log number of cells), and uses space O(number of cells). So, it is good when N is large (or even, 'medium') but uses a lot more memory, in practice it might be slower in some regimes for that reason.
Method A:
The first thing you need to do is normalize the entries. (It's not clear to me if you assume they are normalized or not.) That means, sum all the entries and divide by the sum. (This part is potentially slow, so it's better if you assume or require that it already happened.)
Then you sample like this:
Choose a random [i,j] entry of the matrix (by choosing i,j each uniformly randomly from the range of integers 0 to n-1).
Choose a uniformly random real number p in the range [0, 1].
Check if matrix[i][j] > p. If so, return the pair [i][j]. If not, go back to step 1.
Why does this work? The probability that we end at step 3 with any particular output, is equal to, the probability that [i][j] was selected (this is the same for each entry), times the probality that the number p was small enough. This is proportional to the value matrix[i][j], so the sampling is choosing each entry with the correct proportions. It's also possible that at step 3 we go back to the start -- does that bias things? Basically, no. The reason is, suppose we arbitrarily choose a number k and then consider the distribution of the algorithm, conditioned on stopping exactly after k rounds. Conditioned on the assumption that we stop at the k'th round, no matter what value k we choose, the distribution we sample has to be exactly right by the above argument. Since if we eliminate the case that p is too small, the other possibilities all have their proportions correct. Since the distribution is perfect for each value of k that we might condition on, and the overall distribution (not conditioned on k) is an average of the distributions for each value of k, the overall distribution is perfect also.
If you want to analyze the number of rounds that typically needed in a rigorous way, you can do it by analyzing the probability that we actually stop at step 3 for any particular round. Since the rounds are independent, this is the same for every round, and statistically, it means that the running time of the algorithm is poisson distributed. That means it is tightly concentrated around its mean, and we can determine the mean by knowing that probability.
The probability that we stop at step 3 can be determined by considering the conditional probability that we stop at step 3, given that we chose any particular entry [i][j]. By the formulas for conditional expectation, you get that
Pr[ stop at step 3 ] = sum_{i,j} ( 1/(n^2) * Matrix[i,j] )
Since we assumed the matrix is normalized, this sum reduces to just 1/n^2. So, the expected number of rounds is about n^2 (that is, n^2 up to a constant factor) no matter what the entries in the matrix are. You can't hope to do a lot better than that I think -- that's about the same amount of time it takes to just read all the entries of the matrix, and it's hard to sample from a distribution that you cannot even read all of.
Note: What I described is a way to correctly sample a single element -- to get N elements from one matrix, you can just repeat it N times.
Method B:
Basically you just want to compute a histogram and sample inversely from it, so that you know you get exactly the right distribution. Computing the histogram is expensive, but once you have it, getting samples is cheap and easy.
In C++ it might look like this:
// Make histogram
typedef unsigned int uint;
typedef std::pair<uint, uint> upair;
typedef std::map<double, upair> histogram_type;
histogram_type histogram;
double cumulative = 0.0f;
for (uint i = 0; i < Matrix.size(); ++i) {
for (uint j = 0; j < Matrix[i].size(); ++j) {
cumulative += Matrix[i][j];
histogram[cumulative] = std::make_pair(i,j);
}
}
std::vector<upair> result;
for (uint k = 0; k < N; ++k) {
// Do a sample (this should never repeat... if it does not find a lower bound you could also assert false quite reasonably since it means something is wrong with rand() implementation)
while(1) {
double p = cumulative * rand(); // Or, for best results use std::mt19937 or boost::mt19937 and sample a real in the range [0,1] here.
histogram_type::iterator it = histogram::lower_bound(p);
if (it != histogram.end()) {
result.push_back(it->second);
break;
}
}
}
return result;
Here the time to make the histogram is something like number of cells * O(log number of cells) since inserting into the map takes time O(log n). You need an ordered data structure in order to get cheap lookup N * O(log number of cells) later when you do repeated sampling. Possibly you could choose a more specialized data structure to go faster, but I think there's only limited room for improvement.
Edit: As #Bob__ points out in comments, in method (B) a written there is potentially going to be some error due to floating point round-off if the matrices are quite large, even using type double, at this line:
cumulative += Matrix[i][j];
The problem is that, if cumulative is much larger than Matrix[i][j] beyond what the floating point precision can handle then these each time this statement is executed you may observe significant errors which accumulate to introduce significant inaccuracy.
As he suggests, if that happens, the most straightforward way to fix it is to sort the values Matrix[i][j] first. You could even do this in the general implementation to be safe -- sorting these guys isn't going to take more time asymptotically than you already have anyways.

How do I print out vectors in different order every time

I'm trying to make two vectors. Where vector1 (total1) is containing some strings and vector2(total2) is containing some random unique numbers(that are between 0 and total1.size() - 1)
I want to make a program that print out total1s strings, but in different order every turn. I don't want to use iterators or something because I want to improve my problem solving capacity.
Here is the specific function that crash the program.
for (unsigned i = 0; i < total1.size();)
{
v1 = rand() % total1.size();
for (unsigned s = 0; s < total1.size(); ++s)
{
if (v1 == total2[s])
;
else
{
total2.push_back(v1);
++i;
}
}
}
I'm very grateful for any help that I can get!
Can I suggest you change of algorithm?. Because, even if your current one is correctly implemented ("s", in your code, must go from 0 to total2.size not total1.size and if element is found, break and generate a new random), it has the following drawback: assume vectors of 1.000.000 elements and you are trying the last random number. You have one probability in 1.000.000 of find a random number not previously used. That is a very small amount.Last but one number has a probability of 2 in 1.000.000 also small. In conclusion, your program will loop and expend lots of CPU resources.
Your best alternative is follow #NathanOliver suggestion and look for function std::shuffle. The manual page shows the implementation algorithm, that is what you are looking for.
Another simple algorithm, with some pros and cons, is:
init total2 with sequence 0,1,2,...,n where n is the size total1 - 1
choice two random numbers, i1 and i2, in range [0,n-1].
Swap elements i1 and i2 in total2.
repeat from (2) a fixed number of times "R".
This method allows to known a priori the necessary steps and to control the level of "randomness" of the final vector (bigger R is more random). However, it is far to be good in its randomness quality.
Another method, better in the probabilistic distribution:
fill a list L with number 0,1,2,...size total1-1.
choice a random number i between 0 and the size of list L - 1 .
Store in total2 the i-th element in list L.
Remove this element from L.
repeat from (2) until L is empty.
If you just want to shuffle vector<string> total1, you can do this without using helping vector<int> total2. Here is an implementation based on Fisher–Yates shuffle.
for(int i=n-1; i>=1; i--) {
int j=rand()%(i+1);
swap(total1[j], total1[i]); // your prof might not allow use of swap:)
}
If you must use vector<int> total2 then shuffle it using above algorithm. Next you can use it to create a new vector<string> result from total1 where result[i]=total1[total2[i]].

2 player team knowing maximum moves

Given a list of N players who are to play a 2 player game. Each of them are either well versed in making a particular move or they are not. Find out the maximum number of moves a 2-player team can know.
And also find out how many teams can know that maximum number of moves?
Example Let we have 4 players and 5 moves with ith player is versed in jth move if a[i][j] is 1 otherwise it is 0.
10101
11100
11010
00101
Here maximum number of moves a 2-player team can know is 5 and their are two teams that can know that maximum number of moves.
Explanation : (1, 3) and (3, 4) know all the 5 moves. So the maximal moves a 2-player team knows is 5, and only 2 teams can acheive this.
My approach : For each pair of players i check if any of the players is versed in ith move or not and for each player maintain the maximum pairs he can make with other players with his local maximum move combination.
vector<int> pairmemo;
for(int i=0;i<n;i++){
int mymax=INT_MIN;
int countpairs=0;
for(int j=i+1;j<n;j++){
int count=0;
for(int k=0;k<m;k++){
if(arr[i][k]==1 || arr[j][k]==1)
{
count++;
}
}
if(mymax<count){
mymax=count;
countpairs=0;
}
if(mymax==count){
countpairs++;
}
}
pairmemo.push_back(countpairs);
maxmemo.push_back(mymax);
}
Overall maximum of all N players is answer and count is corresponding sum of the pairs being calculated.
for(int i=0;i<n;i++){
if(maxi<maxmemo[i])
maxi=maxmemo[i];
}
int countmaxi=0;
for(int i=0;i<n;i++){
if(maxmemo[i]==maxi){
countmaxi+=pairmemo[i];
}
}
cout<<maxi<<"\n";
cout<<countmaxi<<"\n";
Time complexity : O((N^2)*M)
Code :
How can i improve it?
Constraints : N<= 3000 and M<=1000
If you represent each set of moves by a very large integer, the problem boils down to finding pair of players (I, J) which have maximum number of bits set in MovesI OR MovesJ.
So, you can use bit-packing and compress all the information on moves in Long integer array. It would take 16 unsigned long integers to store according to the constraints. So, for each pair of players you OR the corresponding arrays and count number of ones. This would take O(N^2 * 16) which would run pretty fast given the constraints.
Example:
Lets say given matrix is
11010
00011
and you used 4-bit integer for packing it.
It would look like:
1101-0000
0001-1000
that is,
13,0
1,8
After OR the moves array for 2 player team becomes 13,8, now count the bits which are one. You have to optimize the counting of bits also, for that read the accepted answer here, otherwise the factor M would appear in complexity. Just maintain one count variable and one maxNumberOfBitsSet variable as you process the pairs.
What Ill do is:
1. Do logical OR between all the possible pairs - O(N^2) and store it's SUM in a 2D array with the symmetric diagonal ignored. (thats we save half of the calc - see example)
2. find the max value in the 2D Array (can be done while doing task 1) -> O(1)
3. count how many cells in the 2D array equals to the maximum value in task 2 O(N^2)
sum: 2*O(N^2)+ O(1) => O(N^2)
Example (using the data in the question (with letters indexes):
A[10101] B[11100] C[11010] D[00101]
Task 1:
[A|B] = 11101 = SUM(4)
[A|C] = 11111 = SUM(5)
[A|D] = 10101 = SUM(3)
[B|C] = 11110 = SUM(4)
[B|D] = 11101 = SUM(4)
[C|D] = 11111 = SUM(5)
Task 2 (Done while is done 1):
Max = 5
Task 3:
Count = 2
By the way, O(N^2) is the minimum possible since you HAVE to check all the possible pairs.
Since you have to find all solutions, unless you find a way to find a count without actually finding the solutions themselves, you have to actually look at or eliminate all possible solutions. So the worst case will always be O(N^2*M), which I'll call O(n^3) as long as N and M are both big and similar size.
However, you can hope for much better performance on the average case by pruning.
Don't check every case. Find ways to eliminate combinations without checking them.
I would sum and store the total number of moves known to each player, and sort the array rows by that value. That should provide an easy check for exiting the loop early. Sorting at O(n log n) should be basically free in an O(n^3) algorithm.
Use Priyank's basic idea, except with bitsets, since you obviously can't use a fixed integer type with 3000 bits.
You may benefit from making a second array of bitsets for the columns, and use that as a mask for pruning players.

Generating random integers with a difference constraint

I have the following problem:
Generate M uniformly random integers from the range 0-N, where N >> M, and where no pair has a difference less than K. where M >> K.
At the moment the best method I can think of is to maintain a sorted list, then determine the lower bound of the current generated integer and test it with the lower and upper elements, if it's ok to then insert the element in between. This is of complexity O(nlogn).
Would there happen to be a more efficient algorithm?
An example of the problem:
Generate 1000 uniformly random integers between zero and 100million where the difference between any two integers is no less than 1000
A comprehensive way to solve this would be to:
Determine all the combinations of n-choose-m that satisfy the constraint, lets called it set X
Select a uniformly random integer i in the range [0,|X|).
Select the i'th combination from X as the result.
This solution is problematic when the n-choose-m is large, as enumerating and storing all possible combinations will be extremely costly. Hence an efficient online generating solution is sought.
Note: The following is a C++ implementation of the solution provided by pentadecagon
std::vector<int> generate_random(const int n, const int m, const int k)
{
if ((n < m) || (m < k))
return std::vector<int>();
std::random_device source;
std::mt19937 generator(source());
std::uniform_int_distribution<> distribution(0, n - (m - 1) * k);
std::vector<int> result_list;
result_list.reserve(m);
for (int i = 0; i < m; ++i)
{
result_list.push_back(distribution(generator));
}
std::sort(std::begin(result_list),std::end(result_list));
for (int i = 0; i < m; ++i)
{
result_list[i] += (i * k);
}
return result_list;
}
http://ideone.com/KOeR4R
.
EDIT: I adapted the text for the requirement to create ordered sequences, each with the same probability.
Create random numbers a_i for i=0..M-1 without duplicates. Sort them. Then create numbers
b_i=a_i + i*(K-1)
Given the construction, those numbers b_i have the required gaps, because the a_i already have gaps of at least 1. In order to make sure those b values cover exactly the required range [1..N], you must ensure a_i are picked from a range [1..N-(M-1)*(K-1)]. This way you get truly independent numbers. Well, as independent as possible given the required gap. Because of the sorting you get O(M log M) performance again, but this shouldn't be too bad. Sorting is typically very fast. In Python it looks like this:
import random
def random_list( N, M, K ):
s = set()
while len(s) < M:
s.add( random.randint( 1, N-(M-1)*(K-1) ) )
res = sorted( s )
for i in range(M):
res[i] += i * (K-1)
return res
First off: this will be an attempt to show that there's a bijection between the (M+1)- compositions (with the slight modification that we will allow addends to be 0) of the value N - (M-1)*K and the valid solutions to your problem. After that, we only have to pick one of those compositions uniformly at random and apply the bijection.
Bijection:
Let
Then the xi form an M+1-composition (with 0 addends allowed) of the value on the left (notice that the xi do not have to be monotonically increasing!).
From this we get a valid solution
by setting the values mi as follows:
We see that the distance between mi and mi + 1 is at least K, and mM is at most N (compare the choice of the composition we started out with). This means that every (M+1)-composition that fulfills the conditions above defines exactly one valid solution to your problem. (You'll notice that we only use the xM as a way to make the sum turn out right, we don't use it for the construction of the mi.)
To see that this gives a bijection, we need to see that the construction can be reversed; for this purpose, let
be a given solution fulfilling your conditions. To get the composition this is constructed from, define the xi as follows:
Now first, all xi are at least 0, so that's alright. To see that they form a valid composition (again, every xi is allowed to be 0) of the value given above, consider:
The third equality follows since we have this telescoping sum that cancels out almost all mi.
So we've seen that the described construction gives a bijection between the described compositions of N - (M-1)*K and the valid solutions to your problem. All we have to do now is pick one of those compositions uniformly at random and apply the construction to get a solution.
Picking a composition uniformly at random
Each of the described compositions can be uniquely identified in the following way (compare this for illustration): reserve N - (M-1)*K spaces for the unary notation of that value, and another M spaces for M commas. We get an (M+1)- composition of N - (M-1)*K by choosing M of the N - (M-1)*K + M spaces, putting commas there, and filling the rest with |. Then let x0 be the number of | before the first comma, xM+1 the number of | after the last comma, and all other xi the number of | between commas i and i+1. So all we have to do is pick an M-element subset of the integer interval[1; N - (M-1)*K + M] uniformly at random, which we can do for example with the Fisher-Yates shuffle in O(N + M log M) (we need to sort the M delimiters to build the composition) since M*K needs to be in O(N) for any solutions to exist. So if N is bigger than M by at least a logarithmic factor, then this is linear in N.
Note: #DavidEisenstat suggested that there are more space efficient ways of picking the M-element subset of that interval; I'm not aware of any, I'm afraid.
You can get an error-proof algorithm out of this by doing the simple input validation we get from the construction above that N ≥ (M-1) * K and that all three values are at least 1 (or 0, if you define the empty set as a valid solution for that case).
Why not do this:
for (int i = 0; i < M; ++i) {
pick a random number between K and N/M
add this number to (N/M)* i;
Now you have M random numbers, distributed evenly along N, all of which have a difference of at least K. It's in O(n) time. As an added bonus, it's already sorted. :-)
EDIT:
Actually, the "pick a random number" part shouldn't be between K and N/M, but between min(K, [K - (N/M * i - previous value)]). That would ensure that the differences are still at least K, and not exclude values that should not be missed.
Second EDIT:
Well, the first case shouldn't be between K and N/M - it should be between 0 and N/M. Just like you need special casing for when you get close to the N/M*i border, we need special initial casing.
Aside from that, the issue you brought up in your comments was fair representation, and you're right. As my pseudocode is presented, it currently completely misses the excess between N/M*M and N. It's another edge case; simply change the random values of your last range.
Now, in this case, your distribution will be different for the last range. Since you have more numbers, you have slightly less chance for each number than you do for all the other ranges. My understanding is that because you're using ">>", this shouldn't really impact the distribution, i.e. the difference in size in the sample set should be nominal. But if you want to make it more fair, you divide the excess equally among each range. This makes your initial range calculation more complex - you'll have to augment each range based on how much remainder there is divided by M.
There are lots of special cases to look out for, but they're all able to be handled. I kept the pseudocode very basic just to make sure that the general concept came through clearly. If nothing else, it should be a good starting point.
Third and Final EDIT:
For those worried that the distribution has a forced evenness, I still claim that there's nothing saying it can't. The selection is uniformly distributed in each segment. There is a linear way to keep it uneven, but that also has a trade-off: if one value is selected extremely high (which should be unlikely given a very large N), then all the other values are constrained:
int prevValue = 0;
int maxRange;
for (int i = 0; i < M; ++i) {
maxRange = N - (((M - 1) - i) * K) - prevValue;
int nextValue = random(0, maxRange);
prevValue += nextValue;
store previous value;
prevValue += K;
}
This is still linear and random and allows unevenness, but the bigger prevValue gets, the more constrained the other numbers become. Personally, I prefer my second edit answer, but this is an available option that given a large enough N is likely to satisfy all the posted requirements.
Come to think of it, here's one other idea. It requires a lot more data maintenance, but is still O(M) and is probably the most fair distribution:
What you need to do is maintain a vector of your valid data ranges and a vector of probability scales. A valid data range is just the list of high-low values where K is still valid. The idea is you first use the scaled probability to pick a random data range, then you randomly pick a value within that range. You remove the old valid data range and replace it with 0, 1 or 2 new data ranges in the same position, depending on how many are still valid. All of these actions are constant time other than handling the weighted probability, which is O(M), done in a loop M times, so the total should be O(M^2), which should be much better than O(NlogN) because N >> M.
Rather than pseudocode, let me work an example using OP's original example:
0th iteration: valid data ranges are from [0...100Mill], and the weight for this range is 1.0.
1st iteration: Randomly pick one element in the one element vector, then randomly pick one element in that range.
If the element is, e.g. 12345678, then we remove the [0...100Mill] and replace it with [0...12344678] and [12346678...100Mill]
If the element is, e.g. 500, then we remove the [0...100Mill] and replace it with just [1500...100Mill], since [0...500] is no longer a valid range. The only time we will replace it with 0 ranges is in the unlikely event that you have a range with only one number in it and it gets picked. (In that case, you'll have 3 numbers in a row that are exactly K apart from each other.)
The weight for the ranges are their length over the total length, e.g. 12344678/(12344678 + (100Mill - 12346678)) and (100Mill - 12346678)/(12344678 + (100Mill - 12346678))
In the next iterations, you do the same thing: randomly pick a number between 0 and 1 and determine which of the ranges that scale falls into. Then randomly pick a number in that range, and replace your ranges and scales.
By the time it's done, we're no longer acting in O(M), but we're still only dependent on the time of M instead of N. And this actually is both uniform and fair distribution.
Hope one of these ideas works for you!

Find pair of elements in integer array such that abs(v[i]-v[j]) is minimized

Lets say we have int array with 5 elements: 1, 2, 3, 4, 5
What I need to do is to find minimum abs value of array's elements' subtraction:
We need to check like that
1-2 2-3 3-4 4-5
1-3 2-4 3-5
1-4 2-5
1-5
And find minimum abs value of these subtractions. We can find it with 2 fors. The question is, is there any algorithm for finding value with one and only for?
sort the list and subtract nearest two elements
The provably best performing solution is assymptotically linear O(n) up until constant factors.
This means that the time taken is proportional to the number of the elements in the array (which of course is the best we can do as we at least have to read every element of the array, which already takes O(n) time).
Here is one such O(n) solution (which also uses O(1) space if the list can be modified in-place):
int mindiff(const vector<int>& v)
{
IntRadixSort(v.begin(), v.end());
int best = MAX_INT;
for (int i = 0; i < v.size()-1; i++)
{
int diff = abs(v[i]-v[i+1]);
if (diff < best)
best = diff;
}
return best;
}
IntRadixSort is a linear time fixed-width integer sorting algorithm defined here:
http://en.wikipedia.org/wiki/Radix_sort
The concept is that you leverage the fixed-bitwidth nature of ints by paritioning them in a series of fixed passes on the bit positions. ie partition them on the hi bit (32nd), then on the next highest (31st), then on the next (30th), and so on - which only takes linear time.
The problem is equivalent to sorting. Any sorting algorithm could be used, and at the end, return the difference between the nearest elements. A final pass over the data could be used to find that difference, or it could be maintained during the sort. Before the data is sorted the min difference between adjacent elements will be an upper bound.
So to do it without two loops, use a sorting algorithm that does not have two loops. In a way it feels like semantics, but recursive sorting algorithms will do it with only one loop. If this issue is the n(n+1)/2 subtractions required by the simple two loop case, you can use an O(n log n) algorithm.
No, unless you know the list is sorted, you need two
Its simple Iterate in a for loop
keep 2 variable "minpos and maxpos " and " minneg" and "maxneg"
check for the sign of the value you encounter and store maximum positive in maxpos
and minimum +ve number in "minpos" do the same by checking in if case for number
less than zero. Now take the difference of maxpos-minpos in one variable and
maxneg and minneg in one variable and print the larger of the two . You will get
desired.
I believe you definitely know how to find max and min in one for loop
correction :- The above one is to find max difference in case of minimum you need to
take max and second max instead of max and min :)
This might be help you:
end=4;
subtractmin;
m=0;
for(i=1;i<end;i++){
if(abs(a[m]-a[i+m])<subtractmin)
subtractmin=abs(a[m]-a[i+m];}
if(m<4){
m=m+1
end=end-1;
i=m+2;
}}