I have a situation where I am using openMP for the Xeon Phi Coprocessor, and I have an opportunity to parallelize an "embarrassingly parallel" double for loop.
However, the for loop is looping through the upper triangle (including the diagonal):
for (int i = 0; i < n; i++)
// access some arrays with the value of i
for (int j = i; j < n; j++)
// do some stuff with the values of j and i
So, I've got the total size of the loop,
for (int q = 0; q < ((n*n - n)/2)+n; q++)
// Do stuff
But where I am struggling is:
How do I calculate i and j from q? Is it possible?
In the meantime, I'm just doing the full matrix, calculating i and j from there, and only doing my stuff when j >= i...but that still leaves a hefty amount of thread overhead.
If I restate your problem, to find i and j from q, you need the greatest i such that
q >= i*n - (i-1)*i/2
to define j as
j = i + (q - i*n - (i-1)*i/2)
If you have such a greatest i, then
(i+1)*n - i*(i+1)/2 > q >= i*n - (i-1)*i/2
n-i > (q - i*n - (i-1)*i/2) >= 0
n > j = i + (q - i*n - (i-1)*i/2) >= i
Here is a first iterative method to find i:
for (i = 0; q >= i*n - (i-1)*i/2; ++i);
i = i-1;
If you iterate over q, the computation of i is likely to exploit the iterative process.
A second method could use sqrt since
i*n - i²/2 + i/2 ~ q
i²/2 - i(n+1/2) + q ~ 0
delta = (n+0.5)² - 2q
i ~ (n+0.5) - sqrt(delta)
i could be defined as floor((n+0.5) - sqrt((n+0.5)² - 2q))
OK, this isn't an answer as far as I can tell. But, it is a workaround (for now).
I was thinking about it, and creating an array (or corresponding pointer), a[(n*n + n)/2 + n][2], and reading in the corresponding i and j values in my calling code, and passing this to my function would allow for the speed up.
You can make your loop to iterate as it is iterating on the whole matrix.
And just to keep track on your current line and each time you are entering a new line increment the index with the value of that line.
Then: i == line and j == i%n.
See this code:
int main() {
int n = 10;
int line = 0;
for (int i=0; i<n*n; i++){
if (i%n == 0 && i!=0){
line++;
i += line;
cout << endl;
}
cout << "("<<line<<","<<i%n<<")";
}
return 0;
}
Running Example
Related
This is code -
for (i = 1; i<=1000000 ; i++ ) {
for ( j = 1; j<= 1000000; j++ ) {
for ( k = 1; k<= 1000000; k++ ) {
if (i % j == k && j%k == 0)
count++;
}
}
}
or is it better to reduce any % operation that goes upto million times in any programme ??
edit- i am sorry ,
initialized by 0, let say i = 1 ok!
now, if i reduce the third loop as #darshan's answer then both the first
&& second loop can run upto N times
and also it calculating % , n*n times. ex- 2021 mod 2022 , then 2021 mod 2023..........and so on
so my question is- % modulus is twice (and maybe more) as heavy as compared to +, - so there's any other logic can be implemented here ?? which is alternate for this question. and gives the same answer as this logic will give..
Thank you so much for knowledgeable comments & help-
Question is:
3 integers (A,B,C) is considered to be special if it satisfies the
following properties for a given integer N :
A mod B=C
B mod C=0
1≤A,B,C≤N
I'm so curious if there is any other smartest solution which can greatly reduces time complexity.
A much Efficient code will be the below one , but I think it can be optimized much more.
First of all modulo (%) operator is quite expensive so try to avoid it on a large scale
for(i = 0; i<=1000000 ; i++ )
for( j = 0; j<= 1000000; j++ )
{
a = i%j;
for( k = 0; k <= j; k++ )
if (a == k && j%k == 0)
count++;
}
We placed a = i%j in second loop because there is no need for it to be calculated every time k changes as it is independent of k and for the condition j%k == 0 to be true , k should be <= j hence change looping restrictions
First of all, your code has undefined behavior due to division by zero: when k is zero then j%k is undefined, so I assume that all your loops should start with 1 and not 0.
Usually the % and the / operators are much slower to execute than any other operation. It is possible to get rid of most invocations of the % operators in your code by several simple steps.
First, look at the if line:
if (i % j == k && j%k == 0)
The i % j == k has a very strict constrain over k which plays into your hands. It means that it is pointless to iterate k at all, since there is only one value of k that passes this condition.
for (i = 1; i<=1000000 ; i++ ) {
for ( j = 1; j<= 1000000; j++ ) {
k = i % j;
// Constrain k to the range of the original loop.
if (k <= 1000000 && k > 0 && j%k == 0)
count++;
}
}
To get rid of "i % j" switch the loop. This change is possible since this code is affected only by which combinations of i,j are tested, not in the order in which they are introduced.
for ( j = 1; j<= 1000000; j++ ) {
for (i = 1; i<=1000000 ; i++ ) {
k = i % j;
// Constrain k to the range of the original loop.
if (k <= 1000000 && k > 0 && j%k == 0)
count++;
}
}
Here it is easy to observe how k behaves, and use that in order to iterate on k directly without iterating on i and so getting rid of i%j. k iterates from 1 to j-1 and then does it again and again. So all we have to do is to iterate over k directly in the loop of i. Note that i%j for j == 1 is always 0, and since k==0 does not pass the condition of the if we can safely start with j=2, skipping 1:
for ( j = 2; j<= 1000000; j++ ) {
for (i = 1, k=1; i<=1000000 ; i++, k++ ) {
if (k == j)
k = 0;
// Constrain k to the range of the original loop.
if (k <= 1000000 && k > 0 && j%k == 0)
count++;
}
}
This is still a waste to run j%k repeatedly for the same values of j,k (remember that k repeats several times in the inner loop). For example, for j=3 the values of i and k go {1,1}, {2,2}, {3,0}, {4,1}, {5,2},{6,0},..., {n*3, 0}, {n*3+1, 1}, {n*3+2, 2},... (for any value of n in the range 0 < n <= (1000000-2)/3).
The values beyond n= floor((1000000-2)/3)== 333332 are tricky - let's have a look. For this value of n, i=333332*3=999996 and k=0, so the last iteration of {i,k}: {n*3,0},{n*3+1,1},{n*3+2, 2} becomes {999996, 0}, {999997, 1}, {999998, 2}. You don't really need to iterate over all these values of n since each of them does exactly the same thing. All you have to do is to run it only once and multiply by the number of valid n values (which is 999996+1 in this case - adding 1 to include n=0).
Since that did not cover all elements, you need to continue the remainder of the values: {999999, 0}, {1000000, 1}. Notice that unlike other iterations, there is no third value, since it would set i out-of-range.
for (int j = 2; j<= 1000000; j++ ) {
if (j % 1000 == 0) std::cout << std::setprecision(2) << (double)j*100/1000000 << "% \r" << std::flush;
int innerCount = 0;
for (int k=1; k<j ; k++ ) {
if (j%k == 0)
innerCount++;
}
int innerLoopRepeats = 1000000/j;
count += innerCount * innerLoopRepeats;
// complete the remainder:
for (int k=1, i= j * innerLoopRepeats+1; i <= 1000000 ; k++, i++ ) {
if (j%k == 0)
count++;
}
}
This is still extremely slow, but at least it completes in less than a day.
It is possible to have a further speed up by using an important property of divisibility.
Consider the first inner loop (it's almost the same for the second inner loop),
and notice that it does a lot of redundant work, and does it expensively.
Namely, if j%k==0, it means that k divides j and that there is pairK such that pairK*k==j.
It is trivial to calculate the pair of k: pairK=j/k.
Obviously, for k > sqrt(j) there is pairK < sqrt(j). This implies that any k > sqrt(j) can be extracted simply
by scanning all k < sqrt(j). This feature lets you loop over only a square root of all interesting values of k.
By searching only for sqrt(j) values gives a huge performance boost, and the whole program can finish in seconds.
Here is a view of the second inner loop:
// complete the remainder:
for (int k=1, i= j * innerLoopRepeats+1; i <= 1000000 && k*k <= j; k++, i++ ) {
if (j%k == 0)
{
count++;
int pairI = j * innerLoopRepeats + j / k;
if (pairI != i && pairI <= 1000000) {
count++;
}
}
}
The first inner loop has to go over a similar transformation.
Just reorder indexation and calculate A based on constraints:
void findAllSpecial(int N, void (*f)(int A, int B, int C))
{
// 1 ≤ A,B,C ≤ N
for (int C = 1; C < N; ++C) {
// B mod C = 0
for (int B = C; B < N; B += C) {
// A mod B = C
for (int A = C; A < N; A += B) {
f(A, B, C);
}
}
}
}
No divisions not useless if just for loops and adding operations.
Below is the obvious optimization:
The 3rd loop with 'k' is really not needed as there is already a many to One mapping from (I,j) -> k
What I understand from the code is that you want to calculate the number of (i,j) pairs such that the (i%j) is a factor of j. Is this correct or am I missing something?
I am having trouble deciding between N^2 and NlogN as the Big O? Whats throwing me off is the third nested for loop from k <=j. How do I reconcile this?
int Max_Subsequence_Sum( const int A[], const int N )
{
int This_Sum = 0, Max_Sum = 0;
for (int i=0; i<N; i++)
{
for (int j=i; j<N; j++)
{
This_Sum = 0;
for (int k=i; k<=j; k++)
{
This_Sum += A[k];
}
if (This_Sum > Max_Sum)
{
Max_Sum = This_Sum;
}
}
}
return Max_Sum;
}
This can be done with estimation or analysis. Looking at the inner most loop there are j-i operations inside the second loop. To get the total number of operations one would sum to get :
(1+N)(2 N + N^2) / 6
Making the algorithm O(N^3). To estimate one can see that there are three loops which at some point have O(N) calls thus it's O(N^3).
Let us analyze the most inner loop first:
for (int k=i; k <= j; k++) {
This_Sum += A[k];
}
Here the counter k iterates from i (inclusive) to j (inclusive), this thus means that the body of the for loop is performed j-i+1 times. If we assume that fetching the k-th number from an array is done in constant time, and the arithmetic operations (incrementing k, calculating the sum of This_Sum and A[k], and comparking k with j), then this thus runs in O(j-i).
The initialization of This_Sum and the if statement is not significant:
This_Sum = 0;
// ...
if (This_Sum > Max_Sum) {
Max_Sum = This_Sum;
}
indeed, if we can compare two numbers in constant time, and set one variable to the value hold by another value in constant time, then regardless whether the condition holds or not, the number of operations is fixed.
Now we can take a look at the loop in the middle, and abstract away the most inner loop:
for (int j=i; j < N; j++) {
// constant number of oprations
// j-i+1 operations
// constant number of operations
}
Here j ranges from i to N, so that means that the total number of operations is:
N
---
\
/ j - i + 1
---
j=i
This sum is equivalent to:
N
---
\
(N-j) * (1 - i) + / j
---
j=i
This is an arithmetic sum [wiki] and it is equivalent to:
(N - i + 1) × ((1 - i) + (i+N) / 2) = (N - i + 1) × ((N-i) / 2 + 1)
or when we expand this:
i2/2 + 3×N/2 - 3×i/2 + N2/2 - N×i + 1
So that means that we can now focus on the outer loop:
for (int i=0; i<N; i++) {
// i2/2 + 3×N/2 - 3×i/2 + N2/2 - N×i + 1
}
So now we can again calculate the number of operations with:
N
---
\
/ i2/2 + 3×N/2 - 3×i/2 + N2/2 - N×i + 1
---
i=0
We can use Faulhaber's formula [wiki] here to solve this sum, and obtain:
(N+1)×(N2+5×N+6)/6
or in expanded form:
N3/6 + N2 + 11×N/6 + 1
which is thus an O(n3) algorithm.
Hexagonal grid is represented by a two-dimensional array with R rows and C columns. First row always comes "before" second in hexagonal grid construction (see image below). Let k be the number of turns. Each turn, an element of the grid is 1 if and only if the number of neighbours of that element that were 1 the turn before is an odd number. Write C++ code that outputs the grid after k turns.
Limitations:
1 <= R <= 10, 1 <= C <= 10, 1 <= k <= 2^(63) - 1
An example with input (in the first row are R, C and k, then comes the starting grid):
4 4 3
0 0 0 0
0 0 0 0
0 0 1 0
0 0 0 0
Simulation: image, yellow elements represent '1' and blank represent '0'.
This problem is easy to solve if I simulate and produce a grid each turn, but with big enough k it becomes too slow. What is the faster solution?
EDIT: code (n and m are used instead R and C) :
#include <cstdio>
#include <cstring>
using namespace std;
int old[11][11];
int _new[11][11];
int n, m;
long long int k;
int main() {
scanf ("%d %d %lld", &n, &m, &k);
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) scanf ("%d", &old[i][j]);
}
printf ("\n");
while (k) {
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
int count = 0;
if (i % 2 == 0) {
if (i) {
if (j) count += old[i-1][j-1];
count += old[i-1][j];
}
if (j) count += (old[i][j-1]);
if (j < m-1) count += (old[i][j+1]);
if (i < n-1) {
if (j) count += old[i+1][j-1];
count += old[i+1][j];
}
}
else {
if (i) {
if (j < m-1) count += old[i-1][j+1];
count += old[i-1][j];
}
if (j) count += old[i][j-1];
if (j < m-1) count += old[i][j+1];
if (i < n-1) {
if (j < m-1) count += old[i+1][j+1];
count += old[i+1][j];
}
}
if (count % 2) _new[i][j] = 1;
else _new[i][j] = 0;
}
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) old[i][j] = _new[i][j];
}
k--;
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
printf ("%d", old[i][j]);
}
printf ("\n");
}
return 0;
}
For a given R and C, you have N=R*C cells.
If you represent those cells as a vector of elements in GF(2), i.e, 0s and 1s where arithmetic is performed mod 2 (addition is XOR and multiplication is AND), then the transformation from one turn to the next can be represented by an N*N matrix M, so that:
turn[i+1] = M*turn[i]
You can exponentiate the matrix to determine how the cells transform over k turns:
turn[i+k] = (M^k)*turn[i]
Even if k is very large, like 2^63-1, you can calculate M^k quickly using exponentiation by squaring: https://en.wikipedia.org/wiki/Exponentiation_by_squaring This only takes O(log(k)) matrix multiplications.
Then you can multiply your initial state by the matrix to get the output state.
From the limits on R, C, k, and time given in your question, it's clear that this is the solution you're supposed to come up with.
There are several ways to speed up your algorithm.
You do the neighbour-calculation with the out-of bounds checking in every turn. Do some preprocessing and calculate the neighbours of each cell once at the beginning. (Aziuth has already proposed that.)
Then you don't need to count the neighbours of all cells. Each cell is on if an odd number of neighbouring cells were on in the last turn and it is off otherwise.
You can think of this differently: Start with a clean board. For each active cell of the previous move, toggle the state of all surrounding cells. When an even number of neighbours cause a toggle, the cell is on, otherwise the toggles cancel each other out. Look at the first step of your example. It's like playing Lights Out, really.
This method is faster than counting the neighbours if the board has only few active cells and its worst case is a board whose cells are all on, in which case it is as good as neighbour-counting, because you have to touch each neighbours for each cell.
The next logical step is to represent the board as a sequence of bits, because bits already have a natural way of toggling, the exclusive or or xor oerator, ^. If you keep the list of neigbours for each cell as a bit mask m, you can then toggle the board b via b ^= m.
These are the improvements that can be made to the algorithm. The big improvement is to notice that the patterns will eventually repeat. (The toggling bears resemblance with Conway's Game of Life, where there are also repeating patterns.) Also, the given maximum number of possible iterations, 2⁶³ is suspiciously large.
The playing board is small. The example in your question will repeat at least after 2¹⁶ turns, because the 4×4 board can have at most 2¹⁶ layouts. In practice, turn 127 reaches the ring pattern of the first move after the original and it loops with a period of 126 from then.
The bigger boards may have up to 2¹⁰⁰ layouts, so they may not repeat within 2⁶³ turns. A 10×10 board with a single active cell near the middle has ar period of 2,162,622. This may indeed be a topic for a maths study, as Aziuth suggests, but we'll tacke it with profane means: Keep a hash map of all previous states and the turns where they occurred, then check whether the pattern has occurred before in each turn.
We now have:
a simple algorithm for toggling the cells' state and
a compact bitwise representation of the board, which allows us to create a hash map of the previous states.
Here's my attempt:
#include <iostream>
#include <map>
/*
* Bit representation of a playing board, at most 10 x 10
*/
struct Grid {
unsigned char data[16];
Grid() : data() {
}
void add(size_t i, size_t j) {
size_t k = 10 * i + j;
data[k / 8] |= 1u << (k % 8);
}
void flip(const Grid &mask) {
size_t n = 13;
while (n--) data[n] ^= mask.data[n];
}
bool ison(size_t i, size_t j) const {
size_t k = 10 * i + j;
return ((data[k / 8] & (1u << (k % 8))) != 0);
}
bool operator<(const Grid &other) const {
size_t n = 13;
while (n--) {
if (data[n] > other.data[n]) return true;
if (data[n] < other.data[n]) return false;
}
return false;
}
void dump(size_t n, size_t m) const {
for (size_t i = 0; i < n; i++) {
for (size_t j = 0; j < m; j++) {
std::cout << (ison(i, j) ? 1 : 0);
}
std::cout << '\n';
}
std::cout << '\n';
}
};
int main()
{
size_t n, m, k;
std::cin >> n >> m >> k;
Grid grid;
Grid mask[10][10];
for (size_t i = 0; i < n; i++) {
for (size_t j = 0; j < m; j++) {
int x;
std::cin >> x;
if (x) grid.add(i, j);
}
}
for (size_t i = 0; i < n; i++) {
for (size_t j = 0; j < m; j++) {
Grid &mm = mask[i][j];
if (i % 2 == 0) {
if (i) {
if (j) mm.add(i - 1, j - 1);
mm.add(i - 1, j);
}
if (j) mm.add(i, j - 1);
if (j < m - 1) mm.add(i, j + 1);
if (i < n - 1) {
if (j) mm.add(i + 1, j - 1);
mm.add(i + 1, j);
}
} else {
if (i) {
if (j < m - 1) mm.add(i - 1, j + 1);
mm.add(i - 1, j);
}
if (j) mm.add(i, j - 1);
if (j < m - 1) mm.add(i, j + 1);
if (i < n - 1) {
if (j < m - 1) mm.add(i + 1, j + 1);
mm.add(i + 1, j);
}
}
}
}
std::map<Grid, size_t> prev;
std::map<size_t, Grid> pattern;
for (size_t turn = 0; turn < k; turn++) {
Grid next;
std::map<Grid, size_t>::const_iterator it = prev.find(grid);
if (1 && it != prev.end()) {
size_t start = it->second;
size_t period = turn - start;
size_t index = (k - turn) % period;
grid = pattern[start + index];
break;
}
prev[grid] = turn;
pattern[turn] = grid;
for (size_t i = 0; i < n; i++) {
for (size_t j = 0; j < m; j++) {
if (grid.ison(i, j)) next.flip(mask[i][j]);
}
}
grid = next;
}
for (size_t i = 0; i < n; i++) {
for (size_t j = 0; j < m; j++) {
std::cout << (grid.ison(i, j) ? 1 : 0);
}
std::cout << '\n';
}
return 0;
}
There is probably room for improvement. Especially, I'm not so sure how it fares for big boards. (The code above uses an ordered map. We don't need the order, so using an unordered map will yield faster code. The example above with a single active cell on a 10×10 board took significantly longer than a second with an ordered map.)
Not sure about how you did it - and you should really always post code here - but let's try to optimize things here.
First of all, there is not really a difference between that and a quadratic grid. Different neighbor relationships, but I mean, that is just a small translation function. If you have a problem there, we should treat this separately, maybe on CodeReview.
Now, the naive solution is:
for all fields
count neighbors
if odd: add a marker to update to one, else to zero
for all fields
update all fields by marker of former step
this is obviously in O(N). Iterating twice is somewhat twice the actual run time, but should not be that bad. Try not to allocate space every time that you do that but reuse existing structures.
I'd propose this solution:
at the start:
create a std::vector or std::list "activated" of pointers to all fields that are activated
each iteration:
create a vector "new_activated"
for all items in activated
count neighbors, if odd add to new_activated
for all items in activated
set to inactive
replace activated by new_activated*
for all items in activated
set to active
*this can be done efficiently by putting them in a smart pointer and use move semantics
This code only works on the activated fields. As long as they stay within some smaller area, this is far more efficient. However, I have no idea when this changes - if there are activated fields all over the place, this might be less efficient. In that case, the naive solution might be the best one.
EDIT: after you now posted your code... your code is quite procedural. This is C++, use classes and use representation of things. Probably you do the search for neighbors right, but you can easily make mistakes there and therefore should isolate that part in a function, or better method. Raw arrays are bad and variables like n or k are bad. But before I start tearing your code apart, I instead repeat my recommendation, put the code on CodeReview, having people tear it apart until it is perfect.
This started off as a comment, but I think it could be helpful as an answer in addition to what has already been stated.
You stated the following limitations:
1 <= R <= 10, 1 <= C <= 10
Given these restrictions, I'll take the liberty to can represent the grid/matrix M of R rows and C columns in constant space (i.e. O(1)), and also check its elements in O(1) instead of O(R*C) time, thus removing this part from our time-complexity analysis.
That is, the grid can simply be declared as bool grid[10][10];.
The key input is the large number of turns k, stated to be in the range:
1 <= k <= 2^(63) - 1
The problem is that, AFAIK, you're required to perform k turns. This makes the algorithm be in O(k). Thus, no proposed solution can do better than O(k)[1].
To improve the speed in a meaningful way, this upper-bound must be lowered in some way[1], but it looks like this cannot be done without altering the problem constraints.
Thus, no proposed solution can do better than O(k)[1].
The fact that k can be so large is the main issue. The most anyone can do is improve the rest of the implementation, but this will only improve by a constant factor; you'll have to go through k turns regardless of how you look at it.
Therefore, unless some clever fact and/or detail is found that allows this bound to be lowered, there's no other choice.
[1] For example, it's not like trying to determine if some number n is prime, where you can check all numbers in the range(2, n) to see if they divide n, making it a O(n) process, or notice that some improvements include only looking at odd numbers after checking n is not even (constant factor; still O(n)), and then checking odd numbers only up to √n, i.e., in the range(3, √n, 2), which meaningfully lowers the upper-bound down to O(√n).
int s_dynamic(int n,int k) {
int maxj = n-k;
int *arr = new int[maxj+1];
for (int i = 0; i <= maxj; ++i)
arr[i] = 1;
for (int i = 1; i <= k; ++i)
for(int j = 1; j <= maxj; ++j)
arr[j] += i*arr[j-1];
return arr[maxj];
}
Here's my attempt at determining Stirling numbers using Dynamic Programming.
It is defined as follows:
S(n,k) = S(n-1,k-1) + k S(n-1,k), if 1 < k < n
S(n,k) = 1, if k=1 ou k=n
Seems ok, right? Except when I run my unit test...
partitioningTest ..\src\Test.cpp:44 3025 == s_dynamic(9,3) expected: 3025 but was: 4414
Can anyone see what I'm doing wrong?
Thanks!
BTW, here's the recursive solution:
int s_recursive(int n,int k) {
if (k == 1 || k == n)
return 1;
return s_recursive(n-1,k-1) + k*s_recursive(n-1,k);
}
Found the bug.
You already computed your dynamic array of Stirlings numbers for k=1 (S(n,1)=1 for all n).
You should start computing S(n,2) - that is:
for (int i = 2; i <= k; ++i) //i=2, not 1
for(int j = 1; j <= maxj; ++j)
arr[j] += i*arr[j-1];
Your approach is just fine, except you seem to have made a simple indexing error. If you think about what indexes i and j represent, and what the inner loop transforms arr[j] to, you'll see it easy enough (I lie, it took me a good half hour to figure out what was what :)).
From what I can decode, i represents the value k during calculations, and arr[j] is transformed from S(i+j, i-1) to S(i+1+j, i). The topmost for loop that initializes arr sets it up as S(1+j, 1). According to these loops, the calculations look just fine. Except for one thing: The very first i loop assumes that arr[j] contains S(0+j, 0), and so it is where your problem lies. If you change the starting value of i from 1 to 2, all should be OK (you may need an if or two for edge cases). The initial i=2 will transform arr[j] from S(1+j, 1) to S(2+j, 2), and the rest of the transformations will be just fine.
Alternatively, you could have initialized arr[j] to S(0+j, 0) if it were defined, but unfortunately, Stirling's numbers are undefined at k=0.
EDIT: Apparently I was wrong in my last comment. If you initialize arr to {1, 0, 0, ...}, you can leave starting value of i as 1. For this, you use the initial values S(0, 0)=1, and S(n, 0)=0, n>0 instead.
1) for (i = 1; i < n; i++) { > n
2) SmallPos = i; > n-1
3) Smallest = Array[SmallPos]; > n-1
4) for (j = i+1; j <= n; j++) > n*(n+1 -i-1)??
5) if (Array[j] < Smallest) { > n*(n+1 -i-1 +1) ??
6) SmallPos = j; > n*(n+1 -i-1 +1) ??
7) Smallest = Array[SmallPos] > n*(n+1 -i-1 +1) ??
}
8) Array[SmallPos] = Array[i]; > n-1
9) Array[i] = Smallest; > n-1
}
i know the big O notation is n^2 ( my bad its not n^3)
i am not sure between line 4-7 anyone care to help out?
im not sure how to get the out put for the second loop since j = i +1 as i changes so does j
also for line 4 the ans suppose to be n(n+1)/2 -1 i want to know why as i can never get that
i am not really solving for the big O i am trying to do the steps that gets to big O as constant and variables are excuded in big O notations.
I would say this is O(n^2) (although as Fred points out above, O(n^2) is a subset of O(n^3), so it's not wrong to say that it's O(n^3)).
Note that it's generally not necessary to compute the number of executions of every single line; as Big-O notation discards low-order terms, it's sufficient to focus only on the most-executed section (which will typically be inside the innermost loop).
So in your case, none of the loops are affected by the values in Array, so we can safely ignore all that. The innermost loop runs (n-1) + (n-2) + (n-3) + ... times; this is an arithmetic series, and so has a term in n^2.
Is this an algorithm given to you, or one you wrote?
I think your loop indexes are wrong.
for (i = 1; i < n; i++) {
should be either
for (i = 0; i < n; i++) {
or
for (i = 1; i <= n; i++) {
depending on whether your array indexes start at 0 or 1 (it's 0 in C and Java).
Assuming we correct it to:
for (i = 0; i < n; i++) {
SmallPos = i;
Smallest = Array[SmallPos];
for (j = i+1; j < n; j++)
if (Array[j] < Smallest) {
SmallPos = j;
Smallest = Array[SmallPos];
}
Array[SmallPos] = Array[i];
Array[i] = Smallest;
}
Then I think the complexity is n2-3/2n = O(n2).
Here's how...
The most costly operation in the innermost loop (my lecturer called this the "basic operation") is key comparison at line 5. It is done once per loop.
So now, you create a summation:
Sum(i=0 to n-1) of Sum(j=i+1 to n-1) of 1.
Now expand the innermost (rightmost) Sum to get:
Sum(i=0 to n-1) of (n-1)-(i+1)+1
and then:
Sum(i=0 to n-1) of n-i-1
and then:
[Sum(i=0 to n-1) of n] - [Sum(i=0 to n-1) of i] - [Sum (i=0 to n-1) of 1]
and then:
n[Sum(i=0 to n-1) of 1] - [(n-1)(n)/2] - [(n-1)-0+1]
and then:
n[(n-1)-0+1] - [(n^2-n)/2] - [n]
and then:
n^2 - [(n^2/2) - n/2] - n
equals:
1/2n^2 - 1/2n
is in:
O(n^2)
If you're asking why it's not O(n3)...
Consider the worst case. if (Array[j] < Smallest) will be true the most times if Array is reverse sorted.
Then you have an inner loop that looks like this:
Array[j] < Smallest;
SmallPos = j;
Smallest = Array[SmallPos];
Now we've got a constant three operations for every inner for (j...) loop.
And O(3) = O(1).
So really, it's i and j that determine how much work we do. Nothing in the inner if loop changes anything.
You can think of it as you should only count while and for loops.
As to why for (j = i+1; j <= n; j++) is n(n+1)/2. It's called an arithmetic series.
You're doing n-1 passes of the for (j...) loop when i==0, n-2 passes when i==1, n-3, etc, until 0.
So the summation is
n-1 + n-2 + n-3 + ... 3 + 2 + 1
now, you sum pairs from outside in, re-writing it as:
n-1+1 + n-2+2 + n-3+3 + ...
equals:
n + n + n + ...
and there are n/2 of these pairs, so you have:
n*(n/2)
Two for() loops, the outer loop from 1 to n, the inner loop runs between 1..n, to n. This makes it O(n^2).
If you 'draw this out', it'll be triangular, rather than rectangular, so O(n^2), while true, is hiding the fact that the constant factor term is smaller than if the inner loop also iterated from 1 to n.
It is O(n^2).
For each of the n iterations of the outer loop you have n iterations in the inner loop.