Related
I found this post How to do iterative quicksort without using stack in c?
but the answer suggested does use a inline stack array! (Only constant amount of extra space is permitted)
The code in the page in reference makes a bold claim:
STACK My implementation does not use the stack to store data...
Yet the function definition has many variables with automatic storage, among them 2 arrays with 1000 entries, which will end up using a fixed but substantial amount of stack space:
// quickSort
//
// This public-domain C implementation by Darel Rex Finley.
//
// * Returns YES if sort was successful, or NO if the nested
// pivots went too deep, in which case your array will have
// been re-ordered, but probably not sorted correctly.
//
// * This function assumes it is called with valid parameters.
//
// * Example calls:
// quickSort(&myArray[0],5); // sorts elements 0, 1, 2, 3, and 4
// quickSort(&myArray[3],5); // sorts elements 3, 4, 5, 6, and 7
bool quickSort(int *arr, int elements) {
#define MAX_LEVELS 1000
int piv, beg[MAX_LEVELS], end[MAX_LEVELS], i=0, L, R ;
beg[0]=0; end[0]=elements;
while (i>=0) {
L=beg[i]; R=end[i]-1;
if (L<R) {
piv=arr[L]; if (i==MAX_LEVELS-1) return NO;
while (L<R) {
while (arr[R]>=piv && L<R) R--; if (L<R) arr[L++]=arr[R];
while (arr[L]<=piv && L<R) L++; if (L<R) arr[R--]=arr[L]; }
arr[L]=piv; beg[i+1]=L+1; end[i+1]=end[i]; end[i++]=L; }
else {
i--; }}
return YES; }
The indentation style is very confusing. Here is a reformatted version:
#define MAX_LEVELS 1000
bool quickSort(int *arr, int elements) {
int piv, beg[MAX_LEVELS], end[MAX_LEVELS], i = 0, L, R;
beg[0] = 0;
end[0] = elements;
while (i >= 0) {
L = beg[i];
R = end[i] - 1;
if (L < R) {
piv = arr[L];
if (i == MAX_LEVELS - 1)
return NO;
while (L < R) {
while (arr[R] >= piv && L < R)
R--;
if (L < R)
arr[L++] = arr[R];
while (arr[L] <= piv && L < R)
L++;
if (L < R)
arr[R--] = arr[L];
}
arr[L] = piv;
beg[i + 1] = L + 1;
end[i + 1] = end[i];
end[i++] = L;
} else {
i--;
}
}
return YES;
}
Note that 1000 is large but not sufficient for pathological cases on moderately large arrays that are already sorted. The function returns NO on such arrays with a size of 1000 only, which is unacceptable.
A much lower value would suffice with an improved version of the algorithm where the larger range is pushed into the array and the loop iterates on the smaller range. This ensures that an array of N entries can handle a set of 2N entries. It still has quadratic time complexity on sorted arrays but at least would sort arrays of all possible sizes.
Here is a modified and instrumented version:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define MAX_LEVELS 64
int quickSort(int *arr, size_t elements) {
size_t beg[MAX_LEVELS], end[MAX_LEVELS], L, R;
int i = 0;
beg[0] = 0;
end[0] = elements;
while (i >= 0) {
L = beg[i];
R = end[i];
if (L + 1 < R--) {
int piv = arr[L];
if (i == MAX_LEVELS - 1)
return -1;
while (L < R) {
while (arr[R] >= piv && L < R)
R--;
if (L < R)
arr[L++] = arr[R];
while (arr[L] <= piv && L < R)
L++;
if (L < R)
arr[R--] = arr[L];
}
arr[L] = piv;
if (L - beg[i] > end[i] - R) {
beg[i + 1] = L + 1;
end[i + 1] = end[i];
end[i++] = L;
} else {
beg[i + 1] = beg[i];
end[i + 1] = L;
beg[i++] = L + 1;
}
} else {
i--;
}
}
return 0;
}
int testsort(int *a, size_t size, const char *desc) {
clock_t t = clock();
size_t i;
if (quickSort(a, size)) {
printf("%s: quickSort failure\n", desc);
return 1;
}
for (i = 1; i < size; i++) {
if (a[i - 1] > a[i]) {
printf("%s: sorting error: a[%zu]=%d > a[%zu]=%d\n",
desc, i - 1, a[i - 1], i, a[i]);
return 2;
}
}
t = clock() - t;
printf("%s: %zu elements sorted in %.3fms\n",
desc, size, t * 1000.0 / CLOCKS_PER_SEC);
return 0;
}
int main(int argc, char *argv[]) {
size_t i, size = argc > 1 ? strtoull(argv[1], NULL, 0) : 1000;
int *a = malloc(sizeof(*a) * size);
if (a != NULL) {
for (i = 0; i < size; i++)
a[i] = rand();
testsort(a, size, "random");
for (i = 0; i < size; i++)
a[i] = i;
testsort(a, size, "sorted");
for (i = 0; i < size; i++)
a[i] = size - i;
testsort(a, size, "reverse sorted");
for (i = 0; i < size; i++)
a[i] = 0;
testsort(a, size, "constant");
free(a);
}
return 0;
}
Output:
random: 100000 elements sorted in 7.379ms
sorted: 100000 elements sorted in 2799.752ms
reverse sorted: 100000 elements sorted in 2768.844ms
constant: 100000 elements sorted in 2786.612ms
Here is a slighlty modified version more resistant to pathological cases:
#define MAX_LEVELS 48
int quickSort(int *arr, size_t elements) {
size_t beg[MAX_LEVELS], end[MAX_LEVELS], L, R;
int i = 0;
beg[0] = 0;
end[0] = elements;
while (i >= 0) {
L = beg[i];
R = end[i];
if (R - L > 1) {
size_t M = L + ((R - L) >> 1);
int piv = arr[M];
arr[M] = arr[L];
if (i == MAX_LEVELS - 1)
return -1;
R--;
while (L < R) {
while (arr[R] >= piv && L < R)
R--;
if (L < R)
arr[L++] = arr[R];
while (arr[L] <= piv && L < R)
L++;
if (L < R)
arr[R--] = arr[L];
}
arr[L] = piv;
M = L + 1;
while (L > beg[i] && arr[L - 1] == piv)
L--;
while (M < end[i] && arr[M] == piv)
M++;
if (L - beg[i] > end[i] - M) {
beg[i + 1] = M;
end[i + 1] = end[i];
end[i++] = L;
} else {
beg[i + 1] = beg[i];
end[i + 1] = L;
beg[i++] = M;
}
} else {
i--;
}
}
return 0;
}
Output:
random: 10000000 elements sorted in 963.973ms
sorted: 10000000 elements sorted in 167.621ms
reverse sorted: 10000000 elements sorted in 167.375ms
constant: 10000000 elements sorted in 9.335ms
As a conclusion:
yes quick sort can be implemented without recursion,
no it cannot be implemented without any local automatic storage,
yes only a constant amount of extra space is necessary, but only because we live is a small world where the maximum size of the array is bounded by available memory. A size of 64 for the local objects handles arrays larger than the size of the Internet, much larger than current 64-bit systems could address.
Apparently, it is possible to implement a non-recursive quicksort with only constant amount of extra space as stated here. This builds upon the Sedgewick's work for non-recursive formulation of quicksort. Instead of preserving the boundary values(low and high) it essentially performs a linear scan to determine these bounds.
Can quicksort be implemented in C without stack and recursion?
Quicksort requires two paths be followed forward from each non-trivial partitioning: a new partitioning of each (sub)partition. Information about the previous partitioning (the bounds of one of the resulting partitions) needs to be carried forward to each new partitioning. The question, then, is where does that information live? In particular, where does the information about one partition live while the program is working on the other?
For a serial algorithm, the answer is that the information is stored on a stack or a queue or a functional equivalent of one of those. Always, because those are our names for data structures that serve the needed purpose. In particular, recursion is a special case, not an alternative. In a recursive quicksort, the data are stored on the call stack. For an iterative implementation you can implement a stack in a formal sense, but it's possible to instead use a simple and relatively small array as a makeshift stack.
But stack and queue equivalents can go a lot farther than that. You could append data to a file, for example, for later read-back. You could write it to a pipe. You could transmit it to yourself asynchronously over a communications network.
If you're clever, you can even make the input array itself serve the need for a stack, by encoding the partition bounds using relative element order or some other element property, as described by Ďurian, for example. This involves a space vs speed tradeoff that is probably not a good deal in most cases. However, it has lower space overhead (O(1)) than do typical quicksort implementations (O(log N)), and it does not change the algorithm's O(N log N) asymptotic time complexity.
If you wanted to go crazy, you could even nest iterations in place of recursing. That would impose a hard upper bound on the size of the arrays that could be handled, but not as tight of one as you might think. With some care and a few tricks, you could handle billion-element arrays with a 25-loop nest. Such a deep nest would be ugly and crazy, but nevertheless conceivable. A human could write it by hand. And in that case, the series of nested loop scopes, with their block-scoped variables, serves as a stack equivalent.
So the answer depends on what exactly you mean by "without stack":
yes, you can use a queue instead, though it would need to have about the same capacity as there are elements to sort;
yes, you can use an array or some other kind of sequential data storage, including the input array itself, to emulate a formal stack or queue;
yes, you can encode a suitable stack equivalent directly into the structure of your program;
yes, you can probably come up with other, more esoteric versions of stacks and queues;
but no, you cannot perform a quicksort without something filling the multi-level data-storage role for which a stack or stack-equivalent is conventionally used.
Well, it can, because I implemented a quicksort in fortran IV (it was a long time ago, and before the language supported recursion - and it was for a bet). However you do need somewhere (a large array would do) to remember your state as you do individual bits of work.
It's a lot easier recursively...
Quicksort is by definition a "divide and conquer" searching algorithm, the idea is that you split the given array into smaller partitions. So you are dividing the problem into subproblems, that is easier to solve.
When using Quicksort without recursion you need a struct of some sort to store the partitions you are not using at the time.
That's why the answer of the post uses an array to make quicksort non recursive.
I'm in my first few months of learning to code in C through a high school program. Someone recently mentioned to me that there's often a way to make code more efficient and I think I have a problem that could be made more efficient. I'm not sure how but I have a hunch that it could be made faster.
We're given a 2D square array of integers with row and col size n. We have subsquares within the 2D square array with row and col size s. We can always assume that s will evenly divide n. I've written the following code to iterate over each subsquare
Currently my code looks something like this:
int **grid;
int s, i, j, k, l;
// reading in inputs, other processing
for (i = 0; i < n; i += s) {
for (j = 0; j < n; j += s) {
for (k = 0; k < s; k++) {
for (l = 0; l < s; l++) {
printf("%d \n", grid[i + k][j + l]);
}
}
printf("next subsquare: \n");
}
}
As you can see, I've got 4 nested for loops and I feel like it's a bit messy to have it in this format. Is there a better way to do this? Later on I might be summing each subsquare or performing some other operation with each subsquare.
Could you explain me how the following two algorithms work?
int countSort(int arr[], int n, int exp)
{
int output[n];
int i, count[n] ;
for (int i=0; i < n; i++)
count[i] = 0;
for (i = 0; i < n; i++)
count[ (arr[i]/exp)%n ]++;
for (i = 1; i < n; i++)
count[i] += count[i - 1];
for (i = n - 1; i >= 0; i--)
{
output[count[ (arr[i]/exp)%n] - 1] = arr[i];
count[(arr[i]/exp)%n]--;
}
for (i = 0; i < n; i++)
arr[i] = output[i];
}
void sort(int arr[], int n)
{
countSort(arr, n, 1);
countSort(arr, n, n);
}
I wanted to apply the algorithm at this array:
After calling the function countSort(arr, n, 1) , we get this:
When I call then the function countSort(arr, n, n) , at this for loop:
for (i = n - 1; i >= 0; i--)
{
output[count[ (arr[i]/exp)%n] - 1] = arr[i];
count[(arr[i]/exp)%n]--;
}
I get output[-1]=arr[4].
But the array doesn't have such a position...
Have I done something wrong?
EDIT:Considering the array arr[] = { 10, 6, 8, 2, 3 }, the array count will contain the following elements:
what do these numbers represent? How do we use them?
Counting sort is very easy - let's say you have an array which contains numbers from range 1..3:
[3,1,2,3,1,1,3,1,2]
You can count how many times each number occurs in the array:
count[1] = 4
count[2] = 2
count[3] = 3
Now you know that in a sorted array,
number 1 will occupy positions 0..3 (from 0 to count[1] - 1), followed by
number 2 on positions 4..5 (from count[1] to count[1] + count[2] - 1), followed by
number 3 on positions 6..8 (from count[1] + count[2] to count[1] + count[2] + count[3] - 1).
Now that you know final position of every number, you can just insert every number at its correct position. That's basically what countSort function does.
However, in real life your input array would not contain just numbers from range 1..3, so the solution is to sort numbers on the least significant digit (LSD) first, then LSD-1 ... up to the most significant digit.
This way you can sort bigger numbers by sorting numbers from range 0..9 (single digit range in decimal numeral system).
This code: (arr[i]/exp)%n in countSort is used just to get those digits. n is base of your numeral system, so for decimal you should use n = 10 and exp should start with 1 and be multiplied by base in every iteration to get consecutive digits.
For example, if we want to get third digit from right side, we use n = 10 and exp = 10^2:
x = 1234,
(x/exp)%n = 2.
This algorithm is called Radix sort and is explained in detail on Wikipedia: http://en.wikipedia.org/wiki/Radix_sort
It took a bit of time to pick though your countSort routine and attempt to determine just what it was you were doing compared to a normal radix sort. There are some versions that split the iteration and the actual sort routine which appears to be what you attempted using both countSort and sort functions. However, after going though that exercise, it was clear you had just missed including necessary parts of the sort routine. After fixing various compile/declaration issues in your original code, the following adds the pieces you overlooked.
In your countSort function, the size of your count array was wrong. It must be the size of the base, in this case 10. (you had 5) You confused the use of exp and base throughout the function. The exp variable steps through the powers of 10 allowing you to get the value and position of each element in the array when combined with a modulo base operation. You had modulo n instead. This problem also permeated you loop ranges, where you had a number of your loop indexes iterating over 0 < n where the correct range was 0 < base.
You missed finding the maximum value in the original array which is then used to limit the number of passes through the array to perform the sort. In fact all of your existing loops in countSort must fall within the outer-loop iterating while (m / exp > 0). Lastly, you omitted a increment of exp within the outer-loop necessary to applying the sort to each element within the array. I guess you just got confused, but I commend your effort in attempting to rewrite the sort routine and not just copy/pasting from somewhere else. (you may have copied/pasted, but if that's the case, you have additional problems...)
With each of those issues addressed, the sort works. Look though the changes and understand what it is doing. The radix sort/count sort are distribution sorts relying on where numbers occur and manipulating indexes rather than comparing values against one another which makes this type of sort awkward to understand at first. Let me know if you have any questions. I made attempts to preserve your naming convention throughout the function, with the addition of a couple that were omitted and to prevent hardcoding 10 as the base.
#include <stdio.h>
void prnarray (int *a, int sz);
void countSort (int arr[], int n, int base)
{
int exp = 1;
int m = arr[0];
int output[n];
int count[base];
int i;
for (i = 1; i < n; i++) /* find the maximum value */
m = (arr[i] > m) ? arr[i] : m;
while (m / exp > 0)
{
for (i = 0; i < base; i++)
count[i] = 0; /* zero bucket array (count) */
for (i = 0; i < n; i++)
count[ (arr[i]/exp) % base ]++; /* count keys to go in each bucket */
for (i = 1; i < base; i++) /* indexes after end of each bucket */
count[i] += count[i - 1];
for (i = n - 1; i >= 0; i--) /* map bucket indexes to keys */
{
output[count[ (arr[i]/exp) % base] - 1] = arr[i];
count[(arr[i]/exp)%n]--;
}
for (i = 0; i < n; i++) /* fill array with sorted output */
arr[i] = output[i];
exp *= base; /* inc exp for next group of keys */
}
}
int main (void) {
int arr[] = { 10, 6, 8, 2, 3 };
int n = 5;
int base = 10;
printf ("\n The original array is:\n\n");
prnarray (arr, n);
countSort (arr, n, base);
printf ("\n The sorted array is\n\n");
prnarray (arr, n);
printf ("\n");
return 0;
}
void prnarray (int *a, int sz)
{
register int i;
printf (" [");
for (i = 0; i < sz; i++)
printf (" %d", a[i]);
printf (" ]\n");
}
output:
$ ./bin/sort_count
The original array is:
[ 10 6 8 2 3 ]
The sorted array is
[ 2 3 6 8 10 ]
I saw an interview question which asked to
Interchange arr[i] and i for i=[0,n-1]
EXAMPLE :
input : 1 2 4 5 3 0
answer :5 0 1 4 2 3
explaination : a[1]=2 in input , so a[2]=1 in answer so on
I attempted this but not getting correct answer.
what i am able to do is : for a pair of numbers p and q , a[p]=q and a[q]=p .
any thoughts how to improve it are welcome.
FOR(j,0,n-1)
{
i=j;
do{
temp=a[i];
next=a[temp];
a[temp]=i;
i=next;
}while(i>j);
}
print_array(a,i,n);
It would be easier for me to to understand your answer if it contains a pseudocode with some explaination.
EDIT : I came to knpw it is cyclic permutation so changed the question title.
Below is what I came up with (Java code).
For each value x in a, it sets a[x] to x, and sets x to the overridden value (to be used for a[a[x]]), and repeats until it gets back to the original x.
I use negative values as a flag to indicate that the value's already been processed.
Running time:
Since it only processes each value once, the running time is O(n).
Code:
int[] a = {1,2,4,5,3,0};
for (int i = 0; i < a.length; i++)
{
if (a[i] < 0)
continue;
int j = a[i];
int last = i;
do
{
int temp = a[j];
a[j] = -last-1;
last = j;
j = temp;
}
while (i != j);
a[j] = -last-1;
}
for (int i = 0; i < a.length; i++)
a[i] = -a[i]-1;
System.out.println(Arrays.toString(a));
Here's my suggestion, O(n) time, O(1) space:
void OrderArray(int[] A)
{
int X = A.Max() + 1;
for (int i = 0; i < A.Length; i++)
A[i] *= X;
for (int i = 0; i < A.Length; i++)
A[A[i] / X] += i;
for (int i = 0; i < A.Length; i++)
A[i] = A[i] % X;
}
A short explanation:
We use X as a basic unit for values in the original array (we multiply each value in the original array by X, which is larger than any number in A- basically the length of A + 1). so at any point we can retrieve the number that was in a certain cell of the original array by array by doing A[i] / X, as long as we didn't add more than X to that cell.
This lets us have two layers of values, where A[i] % X represents the value of the cell after the ordering. these two layers don't intersect through the process.
When we finished, we clean A from the original values multiplied by X by performing A[i] = A[i] % X.
Hopes that's clean enough.
Perhaps it is possible by using the images of the input permutation as indices:
void inverse( unsigned int* input, unsigned int* output, unsigned int n )
{
for ( unsigned int i = 0; i < n; i++ )
output[ input[ i ] ] = i;
}
Given a snipplet of code, how will you determine the complexities in general. I find myself getting very confused with Big O questions. For example, a very simple question:
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
System.out.println("*");
}
}
The TA explained this with something like combinations. Like this is n choose 2 = (n(n-1))/2 = n^2 + 0.5, then remove the constant so it becomes n^2. I can put int test values and try but how does this combination thing come in?
What if theres an if statement? How is the complexity determined?
for (int i = 0; i < n; i++) {
if (i % 2 ==0) {
for (int j = i; j < n; j++) { ... }
} else {
for (int j = 0; j < i; j++) { ... }
}
}
Then what about recursion ...
int fib(int a, int b, int n) {
if (n == 3) {
return a + b;
} else {
return fib(b, a+b, n-1);
}
}
In general, there is no way to determine the complexity of a given function
Warning! Wall of text incoming!
1. There are very simple algorithms that no one knows whether they even halt or not.
There is no algorithm that can decide whether a given program halts or not, if given a certain input. Calculating the computational complexity is an even harder problem since not only do we need to prove that the algorithm halts but we need to prove how fast it does so.
//The Collatz conjecture states that the sequence generated by the following
// algorithm always reaches 1, for any initial positive integer. It has been
// an open problem for 70+ years now.
function col(n){
if (n == 1){
return 0;
}else if (n % 2 == 0){ //even
return 1 + col(n/2);
}else{ //odd
return 1 + col(3*n + 1);
}
}
2. Some algorithms have weird and off-beat complexities
A general "complexity determining scheme" would easily get too complicated because of these guys
//The Ackermann function. One of the first examples of a non-primitive-recursive algorithm.
function ack(m, n){
if(m == 0){
return n + 1;
}else if( n == 0 ){
return ack(m-1, 1);
}else{
return ack(m-1, ack(m, n-1));
}
}
function f(n){ return ack(n, n); }
//f(1) = 3
//f(2) = 7
//f(3) = 61
//f(4) takes longer then your wildest dreams to terminate.
3. Some functions are very simple but will confuse lots of kinds of static analysis attempts
//Mc'Carthy's 91 function. Try guessing what it does without
// running it or reading the Wikipedia page ;)
function f91(n){
if(n > 100){
return n - 10;
}else{
return f91(f91(n + 11));
}
}
That said, we still need a way to find the complexity of stuff, right? For loops are a simple and common pattern. Take your initial example:
for(i=0; i<N; i++){
for(j=0; j<i; j++){
print something
}
}
Since each print something is O(1), the time complexity of the algorithm will be determined by how many times we run that line. Well, as your TA mentioned, we do this by looking at the combinations in this case. The inner loop will run (N + (N-1) + ... + 1) times, for a total of (N+1)*N/2.
Since we disregard constants we get O(N2).
Now for the more tricky cases we can get more mathematical. Try to create a function whose value represents how long the algorithm takes to run, given the size N of the input. Often we can construct a recursive version of this function directly from the algorithm itself and so calculating the complexity becomes the problem of putting bounds on that function. We call this function a recurrence
For example:
function fib_like(n){
if(n <= 1){
return 17;
}else{
return 42 + fib_like(n-1) + fib_like(n-2);
}
}
it is easy to see that the running time, in terms of N, will be given by
T(N) = 1 if (N <= 1)
T(N) = T(N-1) + T(N-2) otherwise
Well, T(N) is just the good-old Fibonacci function. We can use induction to put some bounds on that.
For, example, Lets prove, by induction, that T(N) <= 2^n for all N (ie, T(N) is O(2^n))
base case: n = 0 or n = 1
T(0) = 1 <= 1 = 2^0
T(1) = 1 <= 2 = 2^1
inductive case (n > 1):
T(N) = T(n-1) + T(n-2)
aplying the inductive hypothesis in T(n-1) and T(n-2)...
T(N) <= 2^(n-1) + 2^(n-2)
so..
T(N) <= 2^(n-1) + 2^(n-1)
<= 2^n
(we can try doing something similar to prove the lower bound too)
In most cases, having a good guess on the final runtime of the function will allow you to easily solve recurrence problems with an induction proof. Of course, this requires you to be able to guess first - only lots of practice can help you here.
And as f final note, I would like to point out about the Master theorem, the only rule for more difficult recurrence problems I can think of now that is commonly used. Use it when you have to deal with a tricky divide and conquer algorithm.
Also, in your "if case" example, I would solve that by cheating and splitting it into two separate loops that don; t have an if inside.
for (int i = 0; i < n; i++) {
if (i % 2 ==0) {
for (int j = i; j < n; j++) { ... }
} else {
for (int j = 0; j < i; j++) { ... }
}
}
Has the same runtime as
for (int i = 0; i < n; i += 2) {
for (int j = i; j < n; j++) { ... }
}
for (int i = 1; i < n; i+=2) {
for (int j = 0; j < i; j++) { ... }
}
And each of the two parts can be easily seen to be O(N^2) for a total that is also O(N^2).
Note that I used a good trick trick to get rid of the "if" here. There is no general rule for doing so, as shown by the Collatz algorithm example
In general, deciding algorithm complexity is theoretically impossible.
However, one cool and code-centric method for doing it is to actually just think in terms of programs directly. Take your example:
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
System.out.println("*");
}
}
Now we want to analyze its complexity, so let's add a simple counter that counts the number of executions of the inner line:
int counter = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
System.out.println("*");
counter++;
}
}
Because the System.out.println line doesn't really matter, let's remove it:
int counter = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
counter++;
}
}
Now that we have only the counter left, we can obviously simplify the inner loop out:
int counter = 0;
for (int i = 0; i < n; i++) {
counter += n;
}
... because we know that the increment is run exactly n times. And now we see that counter is incremented by n exactly n times, so we simplify this to:
int counter = 0;
counter += n * n;
And we emerged with the (correct) O(n2) complexity :) It's there in the code :)
Let's look how this works for a recursive Fibonacci calculator:
int fib(int n) {
if (n < 2) return 1;
return fib(n - 1) + fib(n - 2);
}
Change the routine so that it returns the number of iterations spent inside it instead of the actual Fibonacci numbers:
int fib_count(int n) {
if (n < 2) return 1;
return fib_count(n - 1) + fib_count(n - 2);
}
It's still Fibonacci! :) So we know now that the recursive Fibonacci calculator is of complexity O(F(n)) where F is the Fibonacci number itself.
Ok, let's look at something more interesting, say simple (and inefficient) mergesort:
void mergesort(Array a, int from, int to) {
if (from >= to - 1) return;
int m = (from + to) / 2;
/* Recursively sort halves */
mergesort(a, from, m);
mergesort(m, m, to);
/* Then merge */
Array b = new Array(to - from);
int i = from;
int j = m;
int ptr = 0;
while (i < m || j < to) {
if (i == m || a[j] < a[i]) {
b[ptr] = a[j++];
} else {
b[ptr] = a[i++];
}
ptr++;
}
for (i = from; i < to; i++)
a[i] = b[i - from];
}
Because we are not interested in the actual result but the complexity, we change the routine so that it actually returns the number of units of work carried out:
int mergesort(Array a, int from, int to) {
if (from >= to - 1) return 1;
int m = (from + to) / 2;
/* Recursively sort halves */
int count = 0;
count += mergesort(a, from, m);
count += mergesort(m, m, to);
/* Then merge */
Array b = new Array(to - from);
int i = from;
int j = m;
int ptr = 0;
while (i < m || j < to) {
if (i == m || a[j] < a[i]) {
b[ptr] = a[j++];
} else {
b[ptr] = a[i++];
}
ptr++;
count++;
}
for (i = from; i < to; i++) {
count++;
a[i] = b[i - from];
}
return count;
}
Then we remove those lines that do not actually impact the counts and simplify:
int mergesort(Array a, int from, int to) {
if (from >= to - 1) return 1;
int m = (from + to) / 2;
/* Recursively sort halves */
int count = 0;
count += mergesort(a, from, m);
count += mergesort(m, m, to);
/* Then merge */
count += to - from;
/* Copy the array */
count += to - from;
return count;
}
Still simplifying a bit:
int mergesort(Array a, int from, int to) {
if (from >= to - 1) return 1;
int m = (from + to) / 2;
int count = 0;
count += mergesort(a, from, m);
count += mergesort(m, m, to);
count += (to - from) * 2;
return count;
}
We can now actually dispense with the array:
int mergesort(int from, int to) {
if (from >= to - 1) return 1;
int m = (from + to) / 2;
int count = 0;
count += mergesort(from, m);
count += mergesort(m, to);
count += (to - from) * 2;
return count;
}
We can now see that actually the absolute values of from and to do not matter any more, but only their distance, so we modify this to:
int mergesort(int d) {
if (d <= 1) return 1;
int count = 0;
count += mergesort(d / 2);
count += mergesort(d / 2);
count += d * 2;
return count;
}
And then we get to:
int mergesort(int d) {
if (d <= 1) return 1;
return 2 * mergesort(d / 2) + d * 2;
}
Here obviously d on the first call is the size of the array to be sorted, so you have the recurrence for the complexity M(x) (this is in plain sight on the second line :)
M(x) = 2(M(x/2) + x)
and this you need to solve in order to get to a closed form solution. This you do easiest by guessing the solution M(x) = x log x, and verify for the right side:
2 (x/2 log x/2 + x)
= x log x/2 + 2x
= x (log x - log 2 + 2)
= x (log x - C)
and verify it is asymptotically equivalent to the left side:
x log x - Cx
------------ = 1 - [Cx / (x log x)] = 1 - [C / log x] --> 1 - 0 = 1.
x log x
Even though this is an over generalization, I like to think of Big-O in terms of lists, where the length of the list is N items.
Thus, if you have a for-loop that iterates over everything in the list, it is O(N). In your code, you have one line that (in isolation all by itself) is 0(N).
for (int i = 0; i < n; i++) {
If you have a for loop nested inside another for loop, and you perform an operation on each item in the list that requires you to look at every item in the list, then you are doing an operation N times for each of N items, thus O(N^2). In your example above you do in fact, have another for loop nested inside your for loop. So you can think about it as if each for loop is 0(N), and then because they are nested, multiply them together for a total value of 0(N^2).
Conversely, if you are just doing a quick operation on a single item then that would be O(1). There is no 'list of length n' to go over, just a single one time operation.To put this in context, in your example above, the operation:
if (i % 2 ==0)
is 0(1). What is important isn't the 'if', but the fact that checking to see if a single item is equal to another item is a quick operation on a single item. Like before, the if statement is nested inside your external for loop. However, because it is 0(1), then you are multiplying everything by '1', and so there is no 'noticeable' affect in your final calculation for the run time of the entire function.
For logs, and dealing with more complex situations (like this business of counting up to j or i, and not just n again), I would point you towards a more elegant explanation here.
I like to use two things for Big-O notation: standard Big-O, which is worst case scenario, and average Big-O, which is what normally ends up happening. It also helps me to remember that Big-O notation is trying to approximate run-time as a function of N, the number of inputs.
The TA explained this with something like combinations. Like this is n choose 2 = (n(n-1))/2 = n^2 + 0.5, then remove the constant so it becomes n^2. I can put int test values and try but how does this combination thing come in?
As I said, normal big-O is worst case scenario. You can try to count the number of times that each line gets executed, but it is simpler to just look at the first example and say that there are two loops over the length of n, one embedded in the other, so it is n * n. If they were one after another, it'd be n + n, equaling 2n. Since its an approximation, you just say n or linear.
What if theres an if statement? How is the complexity determined?
This is where for me having average case and best case helps a lot for organizing my thoughts. In worst case, you ignore the if and say n^2. In average case, for your example, you have a loop over n, with another loop over part of n that happens half of the time. This gives you n * n/x/2 (the x is whatever fraction of n gets looped over in your embedded loops. This gives you n^2/(2x), so you'd get n^2 just the same. This is because its an approximation.
I know this isn't a complete answer to your question, but hopefully it sheds some kind of light on approximating complexities in code.
As has been said in the answers above mine, it is clearly not possible to determine this for all snippets of code; I just wanted to add the idea of using average case Big-O to the discussion.
For the first snippet, it's just n^2 because you perform n operations n times. If j was initialized to i, or went up to i, the explanation you posted would be more appropriate but as it stands it is not.
For the second snippet, you can easily see that half of the time the first one will be executed, and the second will be executed the other half of the time. Depending on what's in there (hopefully it's dependent on n), you can rewrite the equation as a recursive one.
The recursive equations (including the third snippet) can be written as such: the third one would appear as
T(n) = T(n-1) + 1
Which we can easily see is O(n).
Big-O is just an approximation, it doesn't say how long an algorithm takes to execute, it just says something about how much longer it takes when the size of its input grows.
So if the input is size N and the algorithm evaluates an expression of constant complexity: O(1) N times, the complexity of the algorithm is linear: O(N). If the expression has linear complexity, the algorithm has quadratic complexity: O(N*N).
Some expressions have exponential complexity: O(N^N) or logarithmic complexity: O(log N). For an algorithm with loops and recursion, multiply the complexities of each level of loop and/or recursion. In terms of complexity, looping and recursion are equivalent. An algorithm that has different complexities at different stages in the algorithm, choose the highest complexity and ignore the rest. And finally, all constant complexities are considered equivalent: O(5) is the same as O(1), O(5*N) is the same as O(N).