SOLUTION the solution is unique to my code -- I placed srand(time(NULL)); inside the loop when it should've been placed outside
I'm trying to count the number of comparisons in a quick sort algorithm. I had a recursive version working fine, but it kept seg faulting because I was using large array sizes -- running out of stack space.
So now I've resulted to an iterative approach, and it works. That is, except for my counter for the number of comparisons.
It's returning intermittent results such as...
unsorted: [9][8][7][6][5][4][3][2][1][0]
sorted: [0][1][2][3][4][5][6][7][8][9]
Numer of comparisons: 22
unsorted: [9][8][7][6][5][4][3][2][1][0]
sorted: [0][1][2][3][4][5][6][7][8][9]
Numer of comparisons: 19749794
unsorted: [9][8][7][6][5][4][3][2][1][0]
sorted: [0][1][2][3][4][5][6][7][8][9]
Numer of comparisons: 6088231
my code for the iterative quick sort is...
#include <time.h>
#define BUFLEN 6400
extern int buf[BUFLEN];
extern int quick_count; //comparison count
struct stack {
int stk[BUFLEN];
int top;
};
struct stack s;
void push(int x);
int pop();
void iterative_quick_sort (int buf[], int n) {
int left_ptr, right_ptr, pivot_index, pivot, temp, l, r;
if (n < 2) //case the partitioning has reached the atomic element
return;
r = n - 1;
l = 0;
s.top = -1;
loop: do{
srand(time(NULL));
if ((r - l) == 0)
pivot_index = 1;
else {
pivot_index = rand() % (r - l);
pivot_index += l;
}
pivot = buf[pivot_index]; //pivot holds the value of the pivot element
left_ptr = l;
right_ptr = r;
if ((r - l) != 0 || (r - l) != 1){
while (1) {
while (buf[left_ptr] < pivot){ //loop and increment left_ptr until an element on the left side is larger than the pivot
left_ptr++;
} //now left_ptr holds the index of the value that needs to be swapped with an element from the right side
while (pivot < buf[right_ptr]){ //loop and increment right_ptr until an element on the right side is smaller than the pivot
right_ptr--;
} //now right_ptr holds the index of the value that needs to be swapped with an element from the left side
quick_count++;
if (left_ptr >= right_ptr)
break; //once the pivots reach or overlap each other, break the loop
//perform swap with temporary variable temp
temp = buf[left_ptr];
buf[left_ptr] = buf[right_ptr];
buf[right_ptr] = temp;
}
}
if (l == (n - 2))
break;
else if ((r - l) >= 2){
//goto loop with left side values
push(r);
r = pivot_index + 1;
goto loop;
}
else {
//goto loop with right side values
l = r;
r = pop();
goto loop;
}
}while(1);
}
//cite http://www.sanfoundry.com/c-program-stack-implementation/
void push (int x){
s.top = s.top + 1;
s.stk[s.top] = x;
}
int pop(){
int x = s.stk[s.top];
s.top = s.top - 1;
return x;
}
per request, I've added the function that calls quick sort (Note: quick_count is initialized to zero as a global variable -- used as an extern)
int unsorted_quick[] = {9,8,7,6,5,4,3,2,1,0}; //n = 10
//print unsorted_quick
printf("\nSecond, we sort the following array by using the quick sort algorithm\n");
for (i = 0; i < 10; i++){
printf("[%d]", unsorted_quick[i]);
}
printf("\n");
//fill buf with the unsorted quick array
for (i = 0; i < 10; i++){
buf[i] = unsorted_quick[i];
}
iterative_quick_sort(buf, 10); //call quick_sort()
//print sorted
for (i = 0; i < 10; i++){
printf("[%d]", buf[i]);
}
printf("\nNumber of comparisons: %d\n", quick_count); //print count
You are calling srand(time(NULL)) inside the loop that choose the random pivot. This function must be called once to initialise the state of the random number generator.
The generator needs a starting seed which is set by calling srand(). Then, given the seed, each subsequent call to rand() will give you a random number in a reproducible sequence.
Starting from the same seed you will get the same random sequence.
The problem is that you set the seed in the loop and the seed is the same number so you will always get the same "random" value. This happens because time(NULL) is taken from current time in seconds which means that the random number it's the same in the same second.
You must put it before the loop: do {
Here there is a nice explanation of what is happening: Problems when calling srand(time(NULL)) inside rollDice function
And also here: srand() — why call it only once?
Related
I have a recursive function that I wrote in C that looks like this:
void findSolutions(int** B, int n, int i) {
if (i > n) {
printBoard(B, n);
} else {
for (int x = 1; x <= n; x++) {
if (B[i][x] == 0) {
placeQueen(B, n, i, x);
findSolutions(B, n, i + 1);
removeQueen(B, n, i, x);
}
}
}
}
The initial call is (size is an integer given by user and B is a 2D array):
findSolutions(B, size, 1);
I tried to convert it into a iteration function but there is another function called removeQueen after findSolutions. I got stuck on where to put this function call. How to solve this problem? Stack is also fine but I'm also having trouble doing that.
I'm going to assume that placeQueen(B, n, i, x) makes a change to B and that removeQueen(B, n, i, x) undoes that change.
This answer shows how to approach the problem generically. It doesn't modify the algorithm like Aconcagua has.
Let's start by defining a state structure.
typedef struct {
int **B;
int n;
int i;
} State;
The original code is equivalent to the following:
void _findSolutions(State *state) {
if (state->i >= state->n) {
printBoard(state->B, state->n);
} else {
for (int x = 1; x <= state->n; ++x) {
if (state->B[state->i][x] == 0) {
State *state2 = State_clone(state); // Deep clone.
placeQueen(state2);
++state2->i;
findSolutions(state2);
}
}
}
State_free(state); // Frees the board too.
}
void findSolutions(int** B, int n, int i) {
State *state = State_new(B, n, i); // Deep clones B.
_findSolutions(state);
}
Now, we're in position to eliminate the recursion.
void _findSolutions(State *state) {
StateStack *S = StateStack_new();
do {
if (state->i >= state->n) {
printBoard(state->B, state->n);
} else {
for (int x = state->n; x>=1; --x) { // Reversed the loop to maintain order.
if (state->B[state->i][x] == 0) {
State *state2 = State_clone(state); // Deep clone.
placeQueen(state2);
++state2->i;
StateStack_push(S, state2);
}
}
}
State_free(state); // Frees the board too.
} while (StateStack_pop(&state));
StateStack_free(S);
}
void findSolutions(int** B, int n, int i) {
State *state = State_new(B, n, i); // Deep clones B.
_findSolutions(state);
}
We can eliminate the helper we no longer need.
void findSolutions(int** B, int n, int i) {
StateStack *S = StateStack_new();
State *state = State_new(B, n, i); // Deep clones B.
do {
if (state->i >= state->n) {
printBoard(state->B, state->n);
} else {
for (int x = state->n; x>=1; --x) { // Reversed the loop to maintain order.
if (state->B[state->i][x] == 0) {
State *state2 = State_clone(state); // Deep clone.
placeQueen(state2);
++state2->i;
StateStack_push(S, state2);
}
}
}
State_free(state); // Frees the board too.
} while (StateStack_pop(S, &state));
StateStack_free(S);
}
Functions you need to implement:
StateStack *StateStack_new(void)
void StateStack_free(StateStack *S)
void StateStack_push(StateStack *S, State *state)
int StateStack_pop(StateStack *S, State **p)
State *State_new(int **B, int n, int i) (Note: Clones B)
State *State_clone(const State *state) (Note: Clones state->B)
void State_free(State *state) (Note: Frees state->B)
Structures you need to implement:
StateStack
Tip:
It would be best if you replaced
int **B = malloc((n+1)*sizeof(int*));
for (int i=1; i<=n; ++i)
B[i] = calloc(n+1, sizeof(int));
...
for (int x = 1; x <= n; ++x)
...
B[i][x]
with
char *B = calloc(n*n, 1);
...
for (int x = 0; x < n; ++x)
...
B[(i-1)*n+(x-1)]
What you get by the recursive call is that you get stored the location of the queen in current row before you advance to next row. You will have to re-produce this in the non-recursive version of your function.
You might use another array storing these positions:
unsigned int* positions = calloc(n + 1, sizeof(unsigned int));
// need to initialise all positions to 1 yet:
for(unsigned int i = 1; i <= n; ++i)
{
positions[i] = 1;
}
I reserved a dummy element so that we can use the same indices...
You can now count up last position from 1 to n, and when reaching n there, you increment next position, restarting with current from 1 – just the same way as you increment numbers in decimal, hexadecimal or octal system: 1999 + 1 = 2000 (zero based in this case...).
for(;;)
{
for(unsigned int i = 1; i <= n; ++i)
{
placeQueen(B, n, i, positions[i]);
}
printBoard(B, n);
for(unsigned int i = 1; i <= n; ++i)
{
removeQueen(B, n, i, positions[i]);
}
for(unsigned int i = 1; i <= n; ++i)
{
if(++positions[i] <= n)
// break incrementing if we are in between the numbers:
// 1424 will get 1431 (with last position updated already before)
goto CONTINUE;
positions[i] = 1;
}
// we completed the entire positions list, i. e. we reset very
// last position to 1 again (comparable to an overflow: 4444 got 1111)
// so we are done -> exit main loop:
break;
CONTINUE: (void)0;
}
It's untested code, so you might find a bug in, but it should clearly illustrate the idea. It's the naive aproach, always placing the queens and removing them again.
You can do it a bit cleverer, though: place all queens at positions 1 initially and only move the queens if you really need:
for(unsigned int i = 1; i <= n; ++i)
{
positions[i] = 1;
placeQueen(B, n, i, 1);
}
for(;;)
{
printBoard(B, n);
for(unsigned int i = 1; i <= n; ++i)
{
removeQueen(B, n, i, positions[i]);
++positions[i]
if(++positions[i] <= n)
{
placeQueen(B, n, i, positions[i]);
goto CONTINUE;
}
placeQueen(B, n, i, 1);
positions[i] = 1;
}
break;
CONTINUE: (void)0;
}
// cleaning up the board again:
for(unsigned int i = 1; i <= n; ++i)
{
removeQueen(B, n, i, 1);
}
Again, untested...
You might discover that now the queens move within first row first, different to your recursive approach before. If that disturbs you, you can count down from n to 1 while incrementing the positions and you get original order back...
At the very end (after exiting the loop), don't forget to free the array again to avoid memory leak:
free(positions);
If n doesn't get too large (eight for a typical chess board?), you might use a VLA to prevent that problem.
Edit:
Above solutions will print any possible combinations to place eight queens on a chess board. For an 8x8 board, you get 88 possible combinations, which are more than 16 millions of combinations. You pretty sure will want to filter out some of these combinations, as you did in your original solution as well (if(B[i][x] == 0)), e. g.:
unsigned char* checks = malloc(n + 1);
for(;;)
{
memset(checks, 0, (n + 1));
for(unsigned int i = 1; i <= n; ++i)
{
if(checks[positions[i]] != 0)
goto SKIP;
checks[positions[i]] = 1;
}
// place queens and print board
SKIP:
// increment positions
}
(Trivial approach! Including the filter in the more elaborate approach will get more tricky!)
This will even be a bit more strict than your test, which would have allowed
_ Q _
Q _ _
_ Q _
on a 3x3 board, as you only compare against previous column, whereas my filter wouldn't (leaving a bit more than 40 000 boards to be printed for an 8x8 board).
Edit 2: The diagonals
To filter out those boards where the queens attack each other on the diagonals you'll need additional checks. For these, you'll have to find out what the common criterion is for the fields on the same diagonal. At first, we have to distinguish two types of diagonals, those starting at B[1][1], B[1][2], ... as well as B[2][1], B[3][1], ... – all these run from top left to bottom right direction. On the main diagonal, you'll discover that the difference between row and column index does not differ, on next neighbouring diagonals the indices differ by 1 and -1 respectively, and so on. So we'll have differences in the range [-(n-1); n-1].
If we make the checks array twice as large and shift all differences by n, can re-use do exactly the same checks as we did already for the columns:
unsigned char* checks = (unsigned char*)malloc(2*n + 1);
and after we checked the columns:
memset(checks, 0, (2 * n + 1));
for(unsigned int i = 1; i <= n; ++i)
{
if(checks[n + i - positions[i]] != 0)
goto SKIP;
checks[n + i - positions[i]] = 1;
}
Side note: Even if the array is larger, you still can just memset(checks, 0, n + 1); for the columns as we don't use the additional entries...
Now next we are interested in are the diagonals going from bottom left to top right. Similarly to the other direction, you'll discover that the difference between n - i and positions[i] remains constant for fields on the same diagonal. Again we shift by n and end up in:
memset(checks, 0, (2 * n + 1));
for(unsigned int i = 1; i <= n; ++i)
{
if(checks[2 * n - i - positions[i]] != 0)
goto SKIP;
checks[2 * n - i - positions[i]] = 1;
}
Et voilà, only boards on which queens cannot attack each other.
You might discover that some boards are symmetries (rotational or reflection) of others. Filtering these, though, is much more complicated...
I am new to programming and C is the only language I know. Read a few answers for the same question written in other programming languages. I have written some code for the same but I only get a few test cases correct (4 to be precise). How do I edit my code to get accepted?
I have tried comparing one element of the array with the rest and then I remove the element (which is being compared with the initial) if their sum is divisible by k and then this continues until there are two elements in the array where their sum is divisible by k. Here is the link to the question:
https://www.hackerrank.com/challenges/non-divisible-subset/problem
#include<stdio.h>
#include<stdlib.h>
void remove_element(int array[],int position,long int *n){
int i;
for(i=position;i<=(*n)-1;i++){
array[i]=array[i+1];
}
*n=*n-1;
}
int main(){
int k;
long int n;
scanf("%ld",&n);
scanf("%d",&k);
int *array=malloc(n*sizeof(int));
int i,j;
for(i=0;i<n;i++)
scanf("%d",&array[i]);
for(i=n-1;i>=0;i--){
int counter=0;
for(j=n-1;j>=0;j--){
if((i!=j)&&(array[i]+array[j])%k==0)
{
remove_element(array,j,&n);
j--;
continue;
}
else if((i!=j)&&(array[i]+array[j])%k!=0){
counter++;
}
}
if(counter==n-1){
printf("%ld",n);
break;
}
}
return 0;
}
I only get about 4 test cases right from 20 test cases.
What Gerhardh in his comment hinted at is that
for(i=position;i<=(*n)-1;i++){
array[i]=array[i+1];
}
reads from array[*n] when i = *n-1, overrunning the array. Change that to
for (i=position; i<*n-1; i++)
array[i]=array[i+1];
Additionally, you have
remove_element(array,j,&n);
j--;
- but j will be decremented when continuing the for loop, so decrementing it here is one time too many, while adjustment of i is necessary, since remove_element() shifted array[i] one position to the left, so change j-- to i--.
Furthermore, the condition
if(counter==n-1){
printf("%ld",n);
break;
}
makes just no sense; remove that block and place printf("%ld\n", n); before the return 0;.
To solve this efficiently, you have to realize several things:
Two positive integer numbers a and b are divisible by k (also positive integer number) if ((a%k) + (b%k))%k = 0. That means, that either ((a%k) + (b%k)) = 0 (1) or ((a%k) + (b%k)) = k (2).
Case (1) ((a%k) + (b%k)) = 0 is possible only if both a and b are multiples of k or a%k=0 and b%k=0. For case (2) , there are at most k/2 possible pairs. So, our task is to pick elements that don't fall in case 1 or 2.
To do this, map each number in your array to its corresponding remainder by modulo k. For this, create a new array remainders in which an index stands for a remainder, and a value stands for numbers having such remainder.
Go over the new array remainders and handle 3 cases.
4.1 If remainders[0] > 0, then we can still pick only one element from the original (if we pick more, then sum of their remainders 0, so they are divisible by k!!!).
4.2 if k is even and remainders[k/2] > 0, then we can also pick only one element (otherwise their sum is k!!!).
4.3 What about the other numbers? Well, for any remainder rem > 0 make sure to pick max(remainders[rem], remainders[k - rem]). You can't pick both since rem + k - rem = k, so numbers from such groups can be divisible by k.
Now, the code:
int nonDivisibleSubset(int k, int s_count, int* s) {
static int remainders[101];
for (int i = 0; i < s_count; i++) {
int rem = s[i] % k;
remainders[rem]++;
}
int maxSize = 0;
bool isKOdd = k & 1;
int halfK = k / 2;
for (int rem = 0; rem <= halfK; rem++) {
if (rem == 0) {
maxSize += remainders[rem] > 0;
continue;
}
if (!isKOdd && (rem == halfK)) {
maxSize++;
continue;
}
int otherRem = k - rem;
if (remainders[rem] > remainders[otherRem]) {
maxSize += remainders[rem];
} else {
maxSize += remainders[otherRem];
}
}
return maxSize;
}
Going through CS50, Pset3 and desperate for help/patience.
I'm trying to implemented helpers.c so that find.c has the correct functions to call.. However it is not connecting..
I did a separate piece I titled testBinSearch and that did work. With the same code.. can someone tell me why..?
/**
* helpers.c
*
* Computer Science 50
* Problem Set 3
*
* Helper functions for Problem Set 3.
*/
#include <stdio.h>
#include <cs50.h>
#include "helpers.h"
/**
* Returns true if value is in array of n values, else false.
*/
//search(needle, haystack, size)
bool search(int value, int values[], int n)
{
// TODO: implement a Binary searching algorithm (You are welcome to take an iterative approach (as with a loop) or a recursive approach (wherein a function calls itself).)
//define startPoint. numberOfArrayElements(aka size) - (numberOfArrayElements(aka size) - 1) or Element[0]
//define endPoint. numberOfArrayElements(aka size)
int endPoint = n - 1; //element! we -1 because array start from 0th element. last element of array that is 5 elements big will thus be (total number of Elements - 1)th element.
//define midPoint. numberOfArrayElements(aka size)/2
int midPoint = endPoint/2; //element!
//while loop?
while(n > 0)
{
//if midPoint == needle, return 0
if(values[midPoint] == value)
{
return 0;
}
//////////(if midPoint is smaller(to the left) or larger(to the right) than needle)
//ELSE IF midPoint > than needle(look left), keep startPoint, change endPoint element to values[midPoint - 1], define midPoint again.
else if(values[midPoint] > value)
{
endPoint = midPoint - 1;
midPoint = endPoint/2;
n = endPoint;
printf("mid point is more than needle\n");
}
//ELSE midPoint < than needle(look right), keep endPoint, change Startpoint element to values[midPoint + 1], define mindPoint again.
else if(values[midPoint] < value)
{
int startPoint = midPoint + 1;
//define midpoint again
midPoint = (endPoint + startPoint)/2;
n = endPoint - startPoint + 1;
printf("mid point is less than needle\n");
}
}
printf("cued the while loop return 1\n");
return 1;
}
/**
* Sorts array of n values. Done with Insertion sort*
*/
void sort(int values[], int n)
{
//declare variable
int element;
//number of iterations (or passes?). Skip first because first array is already sorted
for (int i = 1; i < n; i++)
{
//value of element moving into sorted portion
element = values[i];
//declare variable
int j = 0;
//index into the unsorted portion
j = i;
//iterate sorted portion from right to left while sorted portion is greater than 'Element' being compared in this iteration of i.
//basically, it stops this loop once the 'Element' is placed to the left of all greater&&sorted numbers.
while(j > 0 && values[j - 1] > element)
{
//shift all sorted positions to the right
values[j] = values[j - 1];
// this enables the loop to move left through the sorted portion
j = j - 1;
}
//insert temp holder value into the position which is now empty because all sorted&&greater number are to the right of 'Element'
values[j] = element;
}
for(int k = 0; k < n; k++)
//print to check
{
printf("{%i}<-- number in %i-th array (sorted)\n", values[k], k);
}
}
Here is the find.c code:
/**
* find.c
*
* Computer Science 50
* Problem Set 3
*
* Prompts user for as many as MAX values until EOF is reached,
* then proceeds to search that "haystack" of values for given needle.
*
* Usage: ./find needle
*
* where needle is the value to find in a haystack of values
*/
#include <cs50.h>
#include <stdio.h>
#include <stdlib.h>
#include "helpers.h"
// maximum amount of hay
const int MAX = 65536;
int main(int argc, string argv[])
{
// ensure proper usage
if (argc != 2)
{
printf("Usage: ./find needle\n");
return -1;
}
// remember needle
int needle = atoi(argv[1]);
// fill haystack
int size;
int haystack[MAX];
for (size = 0; size < MAX; size++)
{
// wait for hay until EOF
printf("\nhaystack[%i] = ", size);
int straw = GetInt();
if (straw == INT_MAX)
{
break;
}
// add hay to stack
haystack[size] = straw;
}
printf("\n");
// sort the haystack
sort(haystack, size);
// try to find needle in haystack
if (search(needle, haystack, size))
{
printf("\nFound needle in haystack!\n\n");
return 0;
}
else
{
printf("\nDidn't find needle in haystack.\n\n");
return 1;
}
}
And finally, here's the code that worked(or at least it seems to work) separately when I keyed them all in one file... titled testBinSearch below
#include <stdio.h>
#include <cs50.h>
void sort(int array[], int NumberOfElements);
bool search(int value, int values[], int n);
int main(void)
{
//decalre variable
int NumberOfElements;
printf("how many Element would you like in this array?\n");
NumberOfElements = GetInt();
//declare variable for array
int array[NumberOfElements];
for(int i = 0; i < NumberOfElements; i++)
{
printf("alright, please key in value of each element\n");
array[i] = GetInt();
}
sort(array, NumberOfElements);
for (int i = 0; i < NumberOfElements; i++)
{
printf("alright, here is your array sorted, element %i is %i\n", i, array[i]);
}
printf("value ot search for?\n");
int value = GetInt();
search(value, array, NumberOfElements);
}
//----------
void sort(int array[], int NumberOfElements)
{
//declare variable
int element;
//number of iterations (or passes?). Skip first because first array is already sorted
for (int i = 1; i < NumberOfElements; i++)
{
//value of element moving into sorted portion
element = array[i];
//declare variable
int j = 0;
//index into the unsorted portion
j = i;
//iterate sorted portion from right to left while sorted portion is greater than 'Element' being compared in this iteration of i.
//basically, it stops this loop once the 'Element' is placed to the left of all greater&&sorted numbers.
while(j > 0 && array[j - 1] > element)
{
//shift all sorted positions to the right
array[j] = array [j - 1];
// this enables the loop to move left through the sorted portion
j = j - 1;
}
//insert temp holder value into the position which is now empty because all sorted&&greater number are to the right of 'Element'
array[j] = element;
}
}
//--------------
bool search(int value, int values[], int n)
{
// TODO: implement a Binary searching algorithm (You are welcome to take an iterative approach (as with a loop) or a recursive approach (wherein a function calls itself).)
//variables declaration
//int startPoint;
//int endPoint;
//int midPoint;
//define startPoint. numberOfArrayElements(aka size) - (numberOfArrayElements(aka size) - 1) or Element[0]
//define endPoint. numberOfArrayElements(aka size)
int endPoint = n - 1; //element!
//define midPoint. numberOfArrayElements(aka size)/2
int midPoint = endPoint/2; //element!
//while loop?
while(n > 0)
{
//if midPoint == needle, return 0
if(values[midPoint] == value)
{
printf("found it!\n");
return 0;
}
//////////(if midPoint is smaller(to the left) or larger(to the right) than needle)
//ELSE IF midPoint > than needle(look left), keep startPoint, change endPoint element to values[midPoint - 1], define midPoint again.
else if(values[midPoint] > value)
{
endPoint = midPoint - 1;
midPoint = endPoint/2;
n = endPoint;
}
//ELSE midPoint < than needle(look right), keep endPoint, change Startpoint element to values[midPoint + 1], define mindPoint again.
else if(values[midPoint] < value)
{
int startPoint = midPoint + 1;
//define midpoint again
midPoint = (endPoint + startPoint)/2;
n = endPoint - startPoint + 1;
}
}
printf("could not find it\n");
return 1;
}
Can someone help me out and tell me where I went wrong? I came up with the code and copied it right over, but one worked(testBinSearch) and one didn't(helpers.c).. ?
I'm not sure if this covers the whole problem but anyway...
This calculation
midPoint = endPoint/2;
is wrong.
Assume you have an array of 100 elements. The code may bring you to a situation where you look at index 75 to 99 with midpoint in between (e.g. 87), i.e. you have taken the smaller than path a couple of times.
Now if you take the greater than part you calculate a midpoint (e.g. 43) being outside the range of interest
Further, the startpoint variable is not to be a variable inside the smaller than case. It must be at the same level as endpoint. In each loop you must change either startpoint or endpoint. Calculation of midpoint shall always depend on both startpoint and endpoint.
I have a problem, then given some input number n, we have to check whether the no is factorial of some other no or not.
INPUT 24, OUTPUT true
INPUT 25, OUTPUT false
I have written the following program for it:-
int factorial(int num1)
{
if(num1 > 1)
{
return num1* factorial(num1-1) ;
}
else
{
return 1 ;
}
}
int is_factorial(int num2)
{
int fact = 0 ;
int i = 0 ;
while(fact < num2)
{
fact = factorial(i) ;
i++ ;
}
if(fact == num2)
{
return 0 ;
}
else
{
return -1;
}
}
Both these functions, seem to work correctly.
When we supply them for large inputs repeatedly, then the is_factorial will be repeatedly calling factorial which will be really a waste of time.
I have also tried maintaining a table for factorials
So, my question, is there some more efficient way to check whether a number is factorial or not?
It is wasteful calculating factorials continuously like that since you're duplicating the work done in x! when you do (x+1)!, (x+2)! and so on.
One approach is to maintain a list of factorials within a given range (such as all 64-bit unsigned factorials) and just compare it with that. Given how fast factorials increase in value, that list won't be very big. In fact, here's a C meta-program that actually generates the function for you:
#include <stdio.h>
int main (void) {
unsigned long long last = 1ULL, current = 2ULL, mult = 2ULL;
size_t szOut;
puts ("int isFactorial (unsigned long long num) {");
puts (" static const unsigned long long arr[] = {");
szOut = printf (" %lluULL,", last);
while (current / mult == last) {
if (szOut > 50)
szOut = printf ("\n ") - 1;
szOut += printf (" %lluULL,", current);
last = current;
current *= ++mult;
}
puts ("\n };");
puts (" static const size_t len = sizeof (arr) / sizeof (*arr);");
puts (" for (size_t idx = 0; idx < len; idx++)");
puts (" if (arr[idx] == num)");
puts (" return 1;");
puts (" return 0;");
puts ("}");
return 0;
}
When you run that, you get the function:
int isFactorial (unsigned long long num) {
static const unsigned long long arr[] = {
1ULL, 2ULL, 6ULL, 24ULL, 120ULL, 720ULL, 5040ULL,
40320ULL, 362880ULL, 3628800ULL, 39916800ULL,
479001600ULL, 6227020800ULL, 87178291200ULL,
1307674368000ULL, 20922789888000ULL, 355687428096000ULL,
6402373705728000ULL, 121645100408832000ULL,
2432902008176640000ULL,
};
static const size_t len = sizeof (arr) / sizeof (*arr);
for (size_t idx = 0; idx < len; idx++)
if (arr[idx] == num)
return 1;
return 0;
}
which is quite short and efficient, even for the 64-bit factorials.
If you're after a purely programmatic method (with no lookup tables), you can use the property that a factorial number is:
1 x 2 x 3 x 4 x ... x (n-1) x n
for some value of n.
Hence you can simply start dividing your test number by 2, then 3 then 4 and so on. One of two things will happen.
First, you may get a non-integral result in which case it wasn't a factorial.
Second, you may end up with 1 from the division, in which case it was a factorial.
Assuming your divisions are integral, the following code would be a good starting point:
int isFactorial (unsigned long long num) {
unsigned long long currDiv = 2ULL;
while (num != 1ULL) {
if ((num % currDiv) != 0)
return 0;
num /= currDiv;
currDiv++;
}
return 1;
}
However, for efficiency, the best option is probably the first one. Move the cost of calculation to the build phase rather than at runtime. This is a standard trick in cases where the cost of calculation is significant compared to a table lookup.
You could even make it even mode efficient by using a binary search of the lookup table but that's possibly not necessary given there are only twenty elements in it.
If the number is a factorial, then its factors are 1..n for some n.
Assuming n is an integer variable, we can do the following :
int findFactNum(int test){
for(int i=1, int sum=1; sum <= test; i++){
sum *= i; //Increment factorial number
if(sum == test)
return i; //Factorial of i
}
return 0; // factorial not found
}
now pass the number 24 to this function block and it should work. This function returns the number whose factorial you just passed.
You can speed up at least half of the cases by making a simple check if the number is odd or even (use %2). No odd number (barring 1) can be the factorial of any other number
#include<stdio.h>
main()
{
float i,a;
scanf("%f",&a);
for(i=2;a>1;i++)
a/=i;
if(a==1)
printf("it is a factorial");
else
printf("not a factorial");
}
You can create an array which contains factorial list:
like in the code below I created an array containing factorials up to 20.
now you just have to input the number and check whether it is there in the array or not..
#include <stdio.h>
int main()
{
int b[19];
int i, j = 0;
int k, l;
/*writing factorials*/
for (i = 0; i <= 19; i++) {
k = i + 1;
b[i] = factorial(k);
}
printf("enter a number\n");
scanf("%d", &l);
for (j = 0; j <= 19; j++) {
if (l == b[j]) {
printf("given number is a factorial of %d\n", j + 1);
}
if (j == 19 && l != b[j]) {
printf("given number is not a factorial number\n");
}
}
}
int factorial(int a)
{
int i;
int facto = 1;
for (i = 1; i <= a; i++) {
facto = facto * i;
}
return facto;
}
public long generateFactorial(int num){
if(num==0 || num==1){
return 1;
} else{
return num*generateFactorial(num-1);
}
}
public int getOriginalNum(long num){
List<Integer> factors=new LinkedList<>(); //This is list of all factors of num
List<Integer> factors2=new LinkedList<>(); //List of all Factorial factors for eg: (1,2,3,4,5) for 120 (=5!)
int origin=1; //number representing the root of Factorial value ( for eg origin=5 if num=120)
for(int i=1;i<=num;i++){
if(num%i==0){
factors.add(i); //it will add all factors of num including 1 and num
}
}
/*
* amoong "factors" we need to find "Factorial factors for eg: (1,2,3,4,5) for 120"
* for that create new list factors2
* */
for (int i=1;i<factors.size();i++) {
if((factors.get(i))-(factors.get(i-1))==1){
/*
* 120 = 5! =5*4*3*2*1*1 (1!=1 and 0!=1 ..hence 2 times 1)
* 720 = 6! =6*5*4*3*2*1*1
* 5040 = 7! = 7*6*5*4*3*2*1*1
* 3628800 = 10! =10*9*8*7*6*5*4*3*2*1*1
* ... and so on
*
* in all cases any 2 succeding factors inf list having diff=1
* for eg: for 5 : (5-4=1)(4-3=1)(3-2=1)(2-1=1)(1-0=1) Hence difference=1 in each case
* */
factors2.add(i); //in such case add factors from 1st list " factors " to " factors2"
} else break;
//else if(this diff>1) it is not factorial number hence break
//Now last element in the list is largest num and ROOT of Factorial
}
for(Integer integer:factors2){
System.out.print(" "+integer);
}
System.out.println();
if(generateFactorial(factors2.get(factors2.size()-1))==num){ //last element is at "factors2.size()-1"
origin=factors2.get(factors2.size()-1);
}
return origin;
/*
* Above logic works only for 5! but not other numbers ??
* */
}
I am participating in Harvard's opencourse ware and attempting the homework questions. I wrote (or tried to) write a program in C to sort an array using bubble sort implementation. After I finished it, I tested it with an array of size 5, then 6 then 3 etc. All worked. then, I tried to test it with an array of size 11, and then that's when it started bugging out. The program was written to stop getting numbers for the array after it hits the array size entered by the user. But, when I tested it with array size 11 it would continuously try to get more values from the user, past the size declared. It did that to me consistently for a couple days, then the third day I tried to initialize the array size variable to 0, then all of a sudden it would continue to have the same issues with an array size of 4 or more. I un-did the initialization and it continues to do the same thing for an array size of over 4. I cant figure out why the program would work for some array sizes and not others. I used main to get the array size and values from the keyboard, then I passed it to a function I wrote called sort. Note that this is not homework or anything I need to get credit, It is solely for learning. Any comments will be very much appreciated. Thanks.
/****************************************************************************
* helpers.c
*
* Computer Science 50
* Problem Set 3
*
* Helper functions for Problem Set 3.
***************************************************************************/
#include <cs50.h>
#include <stdio.h>
#include "helpers.h"
void
sort(int values[], int n);
int main(){
printf("Please enter the size of the array \n");
int num = GetInt();
int mystack[num];
for (int z=0; z < num; z++){
mystack[z] = GetInt();
}
sort(mystack, num);
}
/*
* Sorts array of n values.
*/
void
sort(int values[], int n)
{
// this is a bubble sort implementation
bool swapped = false; // initialize variable to check if swap was made
for (int i=0; i < (n-1);){ // loops through all array values
if (values[i + 1] > values [i]){ // checks the neighbor to see if it's bigger
i++; // if bigger do nothing except to move to the next value in the array
}
else{ // if neighbor is not bigger then out of order and needs sorting
int temp = values[i]; // store current array value in temp variable for swapping purposes
values[i] = values[i+1]; //swap with neighbor
values[i+1] = temp; // swap neighbor to current array value
swapped = true; // keep track that swap was made
i++;
}
// if we are at the end of array and swap was made then go back to beginning
// and start process again.
if((i == (n-1) && (swapped == true))){
i = 0;
swapped = false;
}
// if we are at the end and swap was not made then array must be in order so print it
if((i == (n-1) && (swapped == false))){
for (int y =0; y < n; y++){
printf("%d", values[y]);
}
// exit program
break;
}
} // end for
// return;
}
You can easily use 2 nested for loops :
int i, j, temp ;
for ( i = 0 ; i < n - 1 ; i++ )
{
for ( j = 0 ; j <= n - 2 - i ; j++ )
{
if ( arr[j] > arr[j + 1] )
{
temp = arr[j] ;
arr[j] = arr[j + 1] ;
arr[j + 1] = temp ;
}
}
}
also you should now it's a c++ code not a c, because c doesn't have something like :
int mystack[num];
and you should enter a number when you're creating an array and you can't use a variable (like "int num" in your code). This is in C, but in C++ you're doing right.
The first thing to do when debugging a problem like this is ensure that the computer is seeing the data you think it should be seeing. You do that by printing out the data as it is entered. You're having trouble with the inputs; print out what the computer is seeing:
static void dump_array(FILE *fp, const char *tag, const int *array, int size)
{
fprintf(fp, "Array %s (%d items)\n", tag, size);
for (int i = 0; i < size; i++)
fprintf(fp, " %d: %d\n", i, array[i]);
}
int main(void)
{
printf("Please enter the size of the array \n");
int num = GetInt();
printf("num = %d\n", num);
int mystack[num];
for (int z = 0; z < num; z++)
{
mystack[z] = GetInt();
printf("%d: %d\n", z, mystack[z]);
}
dump_array(stdout, "Before", mystack, num);
sort(mystack, num);
dump_array(stdout, "After", mystack, num);
}
This will give you direct indications of what is being entered as it is entered, which will probably help you recognize what is going wrong. Printing out inputs is a very basic debugging technique.
Also, stylistically, having a function that should be called sort_array_and_print() suggests that you do not have the correct division of labour; the sort code should sort, and a separate function (like the dump_array() function I showed) should be used for printing an array.
As it turns out the reason why it was doing this is because when comparing an array's neighbor to itself as in:
if (values[i + 1] > values [i])
The fact that I was just checking that it is greater than, without checking if it is '=' then it was causing it to behave undesirably. So if the array is for example [1, 1, 5, 2, 6, 8] then by 1 being next to a 1, my program did not account for this behavior and acted the way it did.