I am simulating a random walk in 2d with periodic boundaries. My code is below:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
/**
*creates the environment in the form of a matrix of ones and zeros, random walk
*/
#define ROWS 3
#define COLS 3
#define MOVES 1
#define DIRECTIONS 4
int main(void)
{
srand(time(NULL)); //seed for the random number generator
int array[ROWS][COLS]; //array for the environment
int x, y; //row and column for loop purposes
int min = 0; //min number in the array
int max = 1; // max number in the array
for(x = 0; x < ROWS; x++) //loop to fill array
{
for(y = 0; y < COLS; y++)
{
array[x][y] = min + rand() % (max - min + 1);
printf("%d", array[x][y]);
}
printf("\n");
}
int steps = 0; //step counter
int direction; //direction will be chosen at random
int i = rand() % ROWS; //Assign start position row
int j = rand() % COLS; //Assign start position column
printf("initial position: %d %d \n", i, j);
while (steps < MOVES)
{
direction = rand() % DIRECTIONS; // 0 = N, 1 = S, 2 = E, 3 = W
printf("direction: %d \n", direction);
switch (direction)
{
case 0: // north
i = ((i - 1) % ROWS);
break;
case 1: // south
i = ((i + 1) % ROWS);
break;
case 2: // east
j = ((j + 1) % COLS);
break;
case 3: // west
j = ((j - 1) % COLS);
break;
}
printf("new position: %d %d \n", i, j);
steps++; //update step counter
printf("number of steps: %d \n", steps);
}
return 0;
}
When I get directions S or E, the boundary condition works ok. The outcome for E with initial position 1 2 is new position 1 0. But if I get positions N or W the boundary condition does not work. The outcome for N with initial position 0 1 is new position -1 1 when it should be 2 1. How can I fix my code? Thank you.
Related
Enter N (dimension of square matrix) such that N is odd and in interval [1,100]. For entered matrix, check if it is a target matrix: if yes, print YES; if no, print NO.
Target Matrix is a matrix that is organized in concentric circles starting from the centre. Each concentric circle has -1 the value of the previous one. Examples:
and
I've tried approaching this problem by using a while loop to increase the radius which starts at centre increases. inside, I've used two for loops to go through that part and check if the values are according the the rule given above.
I'm not really sure if this is a good approach. Do you have some suggestions?
#include <stdio.h>
#define DIM 100
int main() {
int matrix[DIM][DIM];
int N;
int targetMatrix = 1;
int matrixCenter;
int radius;
do{
printf("Enter N: ");
scanf("%d", &N);
if (N % 2 == 0 || N < 1 || N > 100){
printf("Invalid value of N.\n");
}
} while (N % 2 == 0 || N < 1 || N > 100);
// Matrix Entry
printf("Enter the matrix: ");
int i, j;
for (i = 0; i < N; i++){
for (j = 0; j < N; j++){
scanf("%d", &matrix[i][j]);
}
}
// Value at Center
matrixCenter = matrix[N/2][N/2];
radius = 1;
// (N - 1) / 2 is the distance from center of matrix to its side
while (radius <= (N - 1) / 2)
{
for(i = N/2 - radius; i <= N/2 + radius; i++){
for(j = N/2 - radius; j <= N/2 + radius; j++){
if (i == N/2 && j == N/2) // Center Value
continue;
if (matrix[i][j] != matrixCenter - radius)
targetMatrix = 0;
}
}
if (targetMatrix == 0){
printf("NO: This is not a target matrix"); // If not a target matrix
return 1;
}
radius++;
}
printf("YES: this is a target matrix"); // If it is a target matrix
return 0;
}
Testing shell of submatrix only :
int first = N/2 - radius;
int last = N/2 + radius;
for (int i = first; i <= last;i++) {
// first on last line: increment by one
// other : increment by radius * 2
for (int j = first; j <= last;j += ((i==first)||(i==last) ? 1 :radius*2)) {
// test i,j here
}
}
so I've been struggling with this example for a good hour now and I can't even begin to process how should I do this.
Write a program that, for given n and m, forms a matrix as described.
The matrix should be m x m, and it's filled "spirally" with it's
beginning in the upper left corner. The first value in the matrix is
the number n. It's repeated until the "edge" of the matrix, at which
point the number increments. After the number 9 goes 0. 0 ≤ n ≤ 9, 0 ≤
m ≤ 9
Some time ago I had made a function to display the numbers 1 to n on an odd-sized grid.
The principle was to start from the center and to shift by ;
x = 1
x box on the right
x box on the bottom
x++
x box on the left
x box at the top
x++
With this simple algorithm, you can easily imagine to maybe start from the center of your problem and decrement your value, it seems easier to start from the center.
Here is the code that illustrates the above solution, to be adapted of course for your problem, it's only a lead.
#define WE 5
void clock(int grid[WE][WE])
{
int count;
int i;
int reach;
int flag;
int tab[2] = {WE / 2, WE / 2}; //x , y
count = 0;
flag = 0;
i = 0;
reach = 1;
grid[tab[1]][tab[0]] = count;
for (int j = 0; j < WE - 1 && grid[0][WE - 1] != pow(WE, 2) - 1; j++)
for (i = 0; i < reach && grid[0][WE - 1] != pow(WE, 2) - 1; i++, reach++)
{
if(flag % 2 == 0)
{
for(int right = 0 ; right < reach ; right++, tab[0]++, count++, flag = 1)
grid[tab[1]][tab[0]] = count;
if(reach < WE - 1)
for(int bottom = 0; bottom < reach; bottom++, count++, tab[1]++)
grid[tab[1]][tab[0]] = count;
}
else
{
for(int left = 0; left < reach; left++, count++, tab[0]--, flag = 0)
grid[tab[1]][tab[0]] = count;
for(int top = 0; top < reach; top++, tab[1]--, count++)
grid[tab[1]][tab[0]] = count;
}
}
}
I finally solved it. If anybody's interested, here's how I did it:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
//Fills the row number "row" with the number n
int fillRow(int m, int n, int arr[m][m], int row)
{
int j;
for(j=0;j<m;j++)
{
if(arr[row][j] == -1 || arr[row][j] == n-1) arr[row][j] = n;
}
}
//Fills the column number "col" with the number n
int fillCol(int m, int n, int arr[m][m], int col)
{
int i;
for(i=0;i<m;i++)
{
if(arr[i][col] == -1 || arr[i][col] == n-1) arr[i][col] = n;
}
}
int main()
{
int n, m, i, j, r=1, c=1, row=-1, col=-1;
scanf("%d %d",&n, &m);
int arr[m][m];
//Fill array with -1 everywhere
for(i=0;i<m;i++)
{
for(j=0;j<m;j++)
{
arr[i][j] = -1;
}
}
//Calculate which row/column to fill (variables row/col)
//Fill row then column then row than column...
for(i=0;i<2*m;i++)
{
if(i%2==0)
{
row = (r%2==0) ? m-r/2 : r/2;
fillRow(m, n, arr, row);
n++;
r++;
}
else if(i%2==1)
{
col = (c%2==0) ? c/2-1 : m-c/2-1;
fillCol(m, n, arr, col);
n++;
c++;
}
}
//If an element is larger than 9, decrease it by 10
//Prints the elements
for(i=0;i<m;i++)
{
for(j=0;j<m;j++)
{
if(arr[i][j]>9) arr[i][j] -=10;
printf("%d ",arr[i][j]);
}
printf("\n");
}
return 0;
}
Goldbach's conjecture states that every even integer over 4 is the sum of two primes, I am writing a program in C to find these pairs. To do this it first finds all the primes less than a user given number. I have a for loop to iterate from 4 to the user given number and find the pairs within the loop body. When that loop gets to about around 40, suddenly jumps back down by about 30 and then continues to iterate up (with user input 50 it jumped from 38 to 9, with input 60 it jumped from 42 to 7). I can't figure out why this is happening. Here is my code:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <sys/types.h>
#include <unistd.h>
struct pair{
int a;
int b;
}pair_t;
int main(){
int N;
int numPrimes = 1;
int *primes = malloc(100*sizeof(int));
int isPrime = 1;
primes[0] = 2;
int timesRealloc = 0;
int availableSlots = 100;
printf("Please enter the largest even number you want to find the Goldbach pair for: \n");
scanf("%d", &N);
struct pair pairs[N/2 + 4];
int j = 0;
int i;
for (i = 3; i <= N; i+=2){
j = 0;
isPrime = 1;
while (primes[j] <= sqrt(i)) {
if (i%primes[j] == 0) {
isPrime = 0;
break;
}
j++;
}
if (isPrime == 1){
primes[numPrimes] = i;
numPrimes++;
}
if (availableSlots == numPrimes){
timesRealloc++;
availableSlots += 100;
primes = realloc(primes, availableSlots*sizeof(int));
}
}
printf("The largest prime I found was %d\n", primes[(numPrimes-1)]);
int k;
for (i=4; i<=N; i+=2){
printf("i is %d, N is %d\n", i, N);
if (i > N){ break; }
for (j=0; j<numPrimes; j++){
for (k=0; k<numPrimes; k++){
int sum = primes[j] + primes[k];
if(sum == i){
pairs[i].a = primes[j];
pairs[i].b = primes[k];
}
}
}
}
for (i=4; i<=N; i+=2){
printf("%d is the sum of %d and %d\n", i, pairs[i].a, pairs[i].b);
}
return 0;
}
You attempt to be space efficient by compressing the pairs array to just hold every other (even) number and start from 4 instead of zero. However, you miscalculate its size and then when you go to use it, you treat it like it hasn't been compressed and that there's a slot for every natural number.
The code suffers from having the prime array calculation in main() along with the other code, this is best separated out. And when it looks for pairs, it doesn't quit when it finds one, nor when it starts getting sums greater than the target. My rework below attempts to address all of these issues:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <stdbool.h>
#define INITIAL_SLOTS (100)
struct pair {
int a;
int b;
} pair_t;
int compute_primes(int limit, unsigned **primes, int size) {
int numPrimes = 0;
(*primes)[numPrimes++] = 2;
for (int i = 3; i <= limit; i += 2) {
bool isPrime = true;
for (int j = 0; (*primes)[j] <= i / (*primes)[j]; j++) {
if (i % (*primes)[j] == 0) {
isPrime = false;
break;
}
}
if (isPrime) {
(*primes)[numPrimes++] = i;
}
if (numPrimes == size) {
size *= 2;
*primes = realloc(*primes, size * sizeof(unsigned));
}
}
return numPrimes;
}
int main() {
int N;
printf("Please enter the largest even number you want to find the Goldbach pair for: \n");
scanf("%d", &N);
unsigned *primes = calloc(INITIAL_SLOTS, sizeof(unsigned));
int numPrimes = compute_primes(N, &primes, INITIAL_SLOTS);
printf("The largest prime I found was %d\n", primes[numPrimes - 1]);
struct pair pairs[(N - 4) / 2 + 1]; // compressed data structure
for (int i = 4; i <= N; i += 2) {
int offset = (i - 4) / 2; // compressed index
bool found = false;
for (int j = 0; ! found && j < numPrimes; j++) {
for (int k = 0; ! found && k < numPrimes; k++) {
int sum = primes[j] + primes[k];
if (sum == i) {
pairs[offset].a = primes[j];
pairs[offset].b = primes[k];
found = true;
} else if (sum > i) {
break;
}
}
}
}
for (int i = 4; i <= N; i += 2) {
int offset = (i - 4) / 2; // compressed index
printf("%d is the sum of %d and %d\n", i, pairs[offset].a, pairs[offset].b);
}
free(primes);
return 0;
}
OUTPUT
> ./a.out
Please enter the largest even number you want to find the Goldbach pair for:
10000
The largest prime I found was 9973
4 is the sum of 2 and 2
6 is the sum of 3 and 3
8 is the sum of 3 and 5
10 is the sum of 3 and 7
12 is the sum of 5 and 7
14 is the sum of 3 and 11
...
9990 is the sum of 17 and 9973
9992 is the sum of 19 and 9973
9994 is the sum of 53 and 9941
9996 is the sum of 23 and 9973
9998 is the sum of 31 and 9967
10000 is the sum of 59 and 9941
>
I have a binary matrix (zeros and ones) D[][] of dimension nxn where n is large (approximately around 1500 - 2000). I want to find the inverse of this matrix in C.
Since I'm new to C, I started with a 3 x 3 matrix and working around to generalize it to N x N. This works for int values, however since I'm working with binary 1's and 0's. In this implementation, I need unsigned int values.
I could find many solutions for int values but I didn't come across any solution for unsigned int. I'd like to find the inverse of a N x N binary matrix without using any external libraries like blas/lapack. It'd be great if anyone could provide a lead on M x N matrix.
Please note that I need inverse of a matrix, not the pseudo-inverse.
/* To find the inverse of a matrix using LU decomposition */
/* standard Headers */
#include<math.h>
#include<stdio.h>
int main() {
/* Variable declarations */
int i,j;
unsigned int n,m;
unsigned int rows,cols;
unsigned int D[3][3], d[3], C[3][3];
unsigned int x, s[3][3];
unsigned int y[3];
void LU();
n = 2;
rows=3;cols=3;
/* the matrix to be inverted */
D[0][0] = 1;
D[0][1] = 1;
D[0][2] = 0;
D[1][0] = 0;
D[1][1] = 1;
D[1][2] = 0;
D[2][0] = 1;
D[2][1] = 1;
D[2][2] = 1;
/* Store the matrix value for camparison later.
this is just to check the results, we don't need this
array for the program to work */
for (m = 0; m <= rows-1; m++) {
for (j = 0; j <= cols-1; j++) {
C[m][j] = D[m][j];
}
}
/* Call a sub-function to calculate the LU decomposed matrix. Note that
we pass the two dimensional array [D] to the function and get it back */
LU(D, n);
printf(" \n");
printf("The matrix LU decomposed \n");
for (m = 0; m <= rows-1; m++) {
for (j = 0; j <= cols-1; j++){
printf(" %d \t", D[m][j]);
}
printf("\n");
}
/* TO FIND THE INVERSE */
/* to find the inverse we solve [D][y]=[d] with only one element in
the [d] array put equal to one at a time */
for (m = 0; m <= rows-1; m++) {
d[0] = 0;
d[1] = 0;
d[2] = 0;
d[m] = 1;
for (i = 0; i <= n; i++) {
x = 0;
for (j = 0; j <= i - 1; j++){
x = x + D[i][j] * y[j];
}
y[i] = (d[i] - x);
}
for (i = n; i >= 0; i--) {
x = 0;
for (j = i + 1; j <= n; j++) {
x = x + D[i][j] * s[j][m];
}
s[i][m] = (y[i] - x) / D[i][i];
}
}
/* Print the inverse matrix */
printf("The Inverse Matrix\n");
for (m = 0; m <= rows-1; m++) {
for (j = 0; j <= cols-1; j++){
printf(" %d \t", s[m][j]);
}
printf("\n");
}
/* check that the product of the matrix with its iverse results
is indeed a unit matrix */
printf("The product\n");
for (m = 0; m <= rows-1; m++) {
for (j = 0; j <= cols-1; j++){
x = 0;
for (i = 0; i <= 2; i++) {
x = x + C[m][i] * s[i][j];
}
//printf(" %d %d %f \n", m, j, x);
printf("%d \t",x);
}
printf("\n");
}
return 0;
}
/* The function that calcualtes the LU deomposed matrix.
Note that it receives the matrix as a two dimensional array
of pointers. Any change made to [D] here will also change its
value in the main function. So there is no need of an explicit
"return" statement and the function is of type "void". */
void LU(int (*D)[3][3], int n) {
int i, j, k;
int x;
printf("The matrix \n");
for (j = 0; j <= 2; j++) {
printf(" %d %d %d \n", (*D)[j][0], (*D)[j][1], (*D)[j][2]);
}
for (k = 0; k <= n - 1; k++) {
for (j = k + 1; j <= n; j++) {
x = (*D)[j][k] / (*D)[k][k];
for (i = k; i <= n; i++) {
(*D)[j][i] = (*D)[j][i] - x * (*D)[k][i];
}
(*D)[j][k] = x;
}
}
}
This is just a sample example that I tried and I have -1 values in the inverse matrix which is my main concern. I have 1000 x 1000 matrix of binary values and the inverse should also be in binary.
The matrix:
1 1 0
0 1 0
1 1 1
The matrix LU decomposed:
1 1 0
0 1 0
1 0 1
The Inverse Matrix:
1 -1 0
0 1 0
-1 0 1
The product:
1 0 0
0 1 0
0 0 1
I need to know how to find the biggest sum of a given scope in a 2D array, preferably in C to improve the efficiency of the code give below and solve the problem.
To understand this better, read the problem I need to solve below.
Problem
The great city X is a grid of N rows and M columns. There are given
number of people living in each cell. You are asked to position the
telecommunication tower so that as many as people are satisfied. The
cellular tower can cover a rectangular area of Y rows and X columns.
Find the maximum number of people you can satisfy.
Constrains
1 <= N, M <= 1000
1 <= Y <= N, 1 <= X <= M
1 <= number of people in a cell <= 1000
Rectangular area covered by the celluar tower should not cover any cell partially.
Input
First line of the input will contain 4 digits N, M, Y and X respectively separated by spaces. Each of next N lines with contains integers of row 1 to N. Each row will M integers giving the number of people living in each cell separated by spaces.
Output
Output should contain only one integer, the maximum number of people you can satisfy.
Sample Input
4 5 2 3
3 1 1 1 2
2 5 6 7 1
1 2 9 9 1
1 1 1 1 1
Sample Output
38
Explanation
Maximum number of people can be satisfied by placing the tower covering 2x3 area that consists of 5, 6, 7, 2, 9 and 9 cells.
5 + 6 + 7 + 2 + 9 + 9 = 38
My code
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
int N, M, Y, X;
scanf("%d %d %d %d", &N, &M, &Y, &X);
int max = 0;
int total = 0;
int data[N][M];
for(int i = 0; i < N; i++)
for(int j = 0; j < M; j++)
scanf("%d",&(data[i][j]));
for(int i = 0; i < N; i++)
{
for(int j = 0; j < M; j++)
{
total = 0;
for(int l = 0; (l < Y) && (i + Y) <= N; l++)
{
for(int k = 0; (k < X) && (j + X <= M); k++)
{
total += data[i+l][j+k];
}
if(total > max)
max = total;
}
}
}
printf("%d",max);
return 0;
}
This code fails because it's too linear and takes a lot of time when a larger input is used.
You can try out the problem yourself, here
I suppose the main problem in your solution of Number Grid problem is nested for loops. The simplest optimization is to minimaze number of recalculations for each move of the scope.
I tryed the following changes in the original code:
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
int N, M, Y, X;
scanf("%d %d %d %d", &N, &M, &Y, &X);
int max = 0;
int total = 0;
int data[N][M];
for(int i = 0; i < N; i++)
for(int j = 0; j < M; j++)
scanf("%d",&(data[i][j]));
////////////////////////////////////////////////////////////
// calculation of the first total and initial max
int startTotal = 0;
int r, c;
for(r = 0; r < Y-1; r++)
{
for(c = 0; c < X-1; c++)
{
startTotal += data[r][c];
}
}
max = startTotal;
for(int i = 0; i+Y <= N; i++)
{
// add next line
for(int c = 0; c < X-1; c++)
{
startTotal += data[i+Y-1][c];
}
total = startTotal;
for(int j = 0; j+X <= M; j++)
{
// add next column
for(int r = i; r < i+Y; r++)
total += data[r][j+X-1];
// compare
if(total > max)
{
max = total;
}
// subtract the first column
for(int r = i; r < i+Y; r++)
total -= data[r][j];
}
// subtract the first line
for(int c = 0; c < X-1; c++)
{
startTotal -= data[i][c];
}
}
////////////////////////////////////////////////////////
printf("%d",max);
return 0;
}
I have tryed to run the program at hackerrank.com, and received