could any one explain me the difference between one dimensional for loop and two dimensional for loop. and how could i change following one dimensional for loop to two dimensional for loop on the second snippet code. thank you
i think following is one dimensional for loop
Capture video;
for(int i = 0; i < video.pixels.length; i++){
// encoded so blue is > 0 if a pixel is within threshold
if(blue(video.pixels[i]) > 0){
count++;
// processing takes 0-1 (float) color values from shader to 0-255 (int) values for color
// to decode, we need to divide the color by 255 to get the original value
avg.add(red(video.pixels[i]) / 255.0, green(video.pixels[i]) / 255.0);
}
}
and following snippet code is 2 dimensional for loop
Capture video;
for (int x = 0; x < video.width && x < 100; x++ ) {
for (int y = 240; y < video.height; y++ ) {
int loc = x + y*video.width;
}
}
A for loop has 3 parts initialization; condition; increment/decrement
Typically used as
for (int x = 0; x < video.width && x < 100; x++ )
Where int x = 0 initializes the variable x.
x < video.width && x < 100 checks whether the loop should continue.
And x++ increases x every time it finishes one iteration of the loop.
A 2D for is just 2 for loops nested. In the example
for (int x = 0; x < video.width && x < 100; x++ ) {
for (int y = 240; y < video.height; y++ ) {
int loc = x + y*video.width;
}
}
For every increase of x, the for loop will loop through every value 240 ≤ y < video.height.
The 2D for loop is assigning a 1D position to each pixel in the video. You can use this position to read each pixel in video.pixels.
for (int x = 0; x < video.width && x < 100; x++ ) {
for (int y = 240; y < video.height; y++ ) {
int loc = x + y*video.width;
if(blue(video.pixels[loc]) > 0){
count++;
avg.add(red(video.pixels[i]) / 255.0, green(video.pixels[i]) / 255.0);
}
}
}
Both the code above and your original code achieve the same thing but this way is using a 2D for loop
[Edit 1]
To have the same limits on a 1D for loop is a bit more complicated
(untested)
for(int i = 0; i < video.pixels.length; i++){
const pX = i % video.width;
const pY = (i - pX) / video.width;
if(pX > minX && pX < maxX && pY > minY && pY < maxY){
... //It is within minX and maxX and within minY and maxY
}
}
I need help with my Game of Life implementation in C. Other posts on Stackoverflow lead me to believe that my problem was to do with dangling pointers, but after modifying my program to use a global 2D array for the game grid instead of passing it to functions which return new 2D arrays, I realized that it was a problem with my update function.
I have tried hard-coding a number of simple patterns, including gliders and oscillators, and the grid doesn't update correctly. The patterns do update the same way every time the program is run, so I don't think it's a problem of uninitialized memory causing problems. I also know that there are no cells which contain values greater than 1. Therefore, the problem must lie in my mechanisms for updating the grid.
Can someone help me find the problem? I can't find anything wrong with my code and I believe I have programmed the rules correctly.
Here are my neighbors and update functions, along with the relevant variable and constant declarations.
#define MAX_Y 10 /* height */
#define MAX_X 30 /* width */
int grid[MAX_Y][MAX_X];
int neighbors(int x, int y) {
int dx, dy, dstx, dsty;
int n = 0;
for (dy = -1; dy <= 1; ++dy) {
for (dx = -1; dx <= 1; ++dx) {
dsty = y + dy;
dstx = x + dx;
if (dsty >= 0 && dsty < MAX_Y && dstx >= 0 && dstx < MAX_X)
n += !!grid[dsty][dstx]; /* use !! so that non-zero values eval to 1 */
}
}
/* (n > 0) ? printf("Point (%d,%d) has %d neighbors!\n", x, y, n) : 0; */
return n;
}
void update(void) {
int new[MAX_Y][MAX_X];
memset(new, 0, sizeof(int) * MAX_Y * MAX_X);
int i, j, n;
for (i = 0; i < MAX_Y; ++i) {
for (j = 0; j < MAX_X; ++j) {
n = neighbors(i, j);
/* alive, 2 or 3 neighbors -> alive!
* dead, 3 neighbors -> alive!
* anything else -> dead :(
*/
if (grid[i][j] && (n == 2 || n == 3))
new[i][j] = 1;
else if (!grid[i][j] && n == 3)
new[i][j] = 1;
else
new[i][j] = 0;
}
}
memcpy(grid, new, sizeof grid);
}
In your neighbors function, you need to think carefully about the loop iteration where dx and dy are both zero. Conway's Game of Life does not consider a cell to be neighbor of itself, so you need to avoid counting it.
You're also confusing yourself by using the letters i and j. You're allowing j to go all the way up to MAX_X, but then you are using j as the y coordinate when you call neighbors, so that will cause overflows and incorrect calculations. (Starting with the easier case of a 10x10 grid would sometimes save you from bugs like this.)
You should adjust the neighbors() function to omit the cell itself.
Here is a modified version:
#define MAX_Y 10 /* height */
#define MAX_X 30 /* width */
unsigned char grid[MAX_Y][MAX_X];
int neighbors(int x, int y) {
int n = -!!grid[y][x];
for (int dy = -1; dy <= 1; ++dy) {
for (int dx = -1; dx <= 1; ++dx) {
int dsty = y + dy;
int dstx = x + dx;
if (dsty >= 0 && dsty < MAX_Y && dstx >= 0 && dstx < MAX_X && grid[dsty][dstx])
n++;
}
}
return n;
}
void update(void) {
int new[MAX_Y][MAX_X] = { 0 };
for (int y = 0; y < MAX_Y; ++y) {
for (int x = 0; x < MAX_X; ++x) {
int n = neighbors(y, x);
/* alive, 2 or 3 neighbors -> alive!
* dead, 3 neighbors -> alive!
* anything else -> dead :(
*/
new[y][x] = (grid[y][x] && n == 2) || n == 3;
}
}
memcpy(grid, new, sizeof grid);
}
The neighbors() function can be simplified with fewer tests:
int neighbors(int x, int y) {
int n = -(grid[y][x] != 0);
int x1 = x - (x > 0);
int x2 = x + (x < MAX_X - 1);
int y1 = y - (y > 0);
int y2 = y + (y < MAX_Y - 1);
for (y = y1; y <= y2; y++) {
for (x = x1; x <= x2; x++) {
n += grid[y][x] != 0;
}
}
return n;
}
I have an exercise in Objective-C making use of for, if-else and printf to print a Zig-Zag like this in the console:
Please see image:
I have tried a code with C programming and print a triangle then try to edit this one, but can't do anything more to get my Zig-Zag.
I have solve my problems already. Thank you guys so much.
for (int i = 0; i < 5; i++) {
for(int j = 1; j<= 21; j++ ){
if(j<=9){
if(j - i == 5 || j+ i == 5){
printf("*") ;
}else{
printf(" ");
}
}else{
if(j+i == 13 || j - i == 13 || j + i == 21){
printf("*") ;
}
else{
printf(" ");
}
}
}
printf("\n");
}
OK, attempt 2 based on your updated question. Let's start with the basic structure. Set some consts for the width and height - we will refer to these in multiple places, so consts make for nice easy changes - and create a loop with basic logic to start to get the display you want.
const int numRows = 5;
const int numCols = 21;
for (int row = 0; row < numRows; ++row)
{
for (int col = 0; col < numCols; ++col)
{
if (col == row)
{
printf("X");
}
else
{
printf(" ");
}
}
printf("\n");
}
This gives us:
X
X
X
X
X
which is the wrong way up and only gives us one "downstroke", but it is a good start. Let's use the modulus function to make the pattern repeat. The repeat happens on the 9th character, which suggests we need to use ((numRows - 1) * 2) for our modulus.
const int modulusVal = ((numRows - 1) * 2);
for (int row = 0; row < numRows; ++row)
{
for (int col = 0; col < numCols; ++col)
{
int modCol = (col % modulusVal);
if (modCol == row)
{
printf("X");
}
else
{
printf(" ");
}
}
printf("\n");
}
So now we get:
X X X
X X X
X X X
X X X
X X X
which is starting to get a lot closer to what we want. So, on to the upstrokes. The downstrokes are displayed when modCol is in the range of row, which is 0-4. When modCol is past this range, we bring it back into range by subtracting the number of rows from it.
Then we "invert" it by subtracting it from the highest value that row can be, which is numRows - 1. This means that when modCol is 0 it will become 4, when it is 1 it will become 3, when it is 2 it will be unchanged.
for (int row = 0; row < numRows; ++row)
{
for (int col = 0; col < numCols; ++col)
{
int modCol = (col % modulusVal);
if (modCol >= numRows)
{
modCol -= numRows;
modCol = ((numRows - 1) - modCol);
}
if (modCol == row)
{
printf("X");
}
else
{
printf(" ");
}
}
printf("\n");
}
Now we get:
X X X
X X X
X X X X X
X X X X X
XX XX X
Close, but no cigar. Column 5 is being treated as column 0, but we need it to be treated as column 1. Make this change to fix it:
modCol = ((numRows - 1) - (modCol + 1));
Output:
X X X
X X X X X
X X X X X
X X X X X
X X X
This has the pattern we want, but starts at the top left when we want the bottom left. Easy fix. Invert modCol after it has been calculated in the same we we invert it when it is greater/equal than/to numRows.
for (int row = 0; row < numRows; ++row)
{
for (int col = 0; col < numCols; ++col)
{
int modCol = (col % modulusVal);
if (modCol >= numRows)
{
modCol -= numRows;
modCol = ((numRows - 1) - (modCol + 1));
}
modCol = ((numRows - 1) - modCol);
if (modCol == row)
{
printf("X");
}
else
{
printf(" ");
}
}
printf("\n");
}
Finally we get the output we want:
X X X
X X X X X
X X X X X
X X X X X
X X X
I hope this helps and is reasonably easy to understand.
I'm not familiar with console programming using Objectionable C and you haven't explained exactly what you've tried, what has worked and what has not.
Assuming your problem is the code to choose the appropriate position to display the '+' character and not displaying text at a particular location on the console, here's what I would do:
int currentY = MIN_Y;
int incrementY = 1;
for (int currentX = MIN_X; currentX <= MAX_X; ++currentX)
{
SetCursorPosition(currentX, currentY);
printf("+");
currentY += incrementY;
if (currentY == MAX_Y)
{
incrementY = -1;
}
else if (currentY == MIN_Y)
{
incrementY = 1;
}
}
This assumes you have a function for SetCursor() and consts/defines for MIN/MAX X/Y.
The 'if' statement could be optimised to:
if (currentY == MAX_Y || currentY == MIN_Y)
{
incrementY -= incrementY;
}
but you required the use of the 'else' statement. Hopefully this code is pretty self explanatory and of some use to you.
I have run this program and get the error expression result unused. I may be doing something simple wrong, but I have spent the day trying to figure it out to no avail. Any help you can provide is greatly appreciated.
#include <stdio.h>
#include <cs50.h>
int main()
{
int x, y = 0;
printf("Enter the amount of change ");
x = GetFloat() * 100;
while (x != 0)
{
if (x >= 25)
{
x - 25;
y = y + 1;
}
if (x >= 10 && x < 25)
{
x - 10;
} y = y + 1;
if (x >= 5 && x < 10)
{
x - 5;
} y = y + 1;
if (x >= 1 && x < 5)
{ x - 1;
y= y + 1;
}
}
printf("The number of coins neccessary is %d", y);
}
if (x >= 25)
{
x - 25; // This accomplishes nothing
y = y + 1;
}
if (x >= 10 && x < 25)
{
x - 10; // This accomplishes nothing
} y = y + 1;
if (x >= 5 && x < 10)
{
x - 5; // This accomplishes nothing
} y = y + 1;
if (x >= 1 && x < 5)
{
x - 1; // This accomplishes nothing
y= y + 1;
}
In each of those lines you're subtracting a number from x, but you're doing nothing with the result. If you're trying to update x with the result, you need to do just like you're doing with y, and put x = in front of the expression.
So if you want x to go down by 25, you should write:
x = x - 25;
Alternatively, you can write the shorthand:
x -= 25; // Note the equal sign
All the 4 statements x - 25, x- 10, x- 5, x - 1 will turn out to be useless unless you assign that value to x;
Because you are trying to subtract the value from x, but you are not assigning the new value to x.
Here is the solution of your problem:
#include <stdio.h>
#include <cs50.h>
int main()
{
int x, y = 0;
printf("Enter the amount of change ");
x = GetFloat() * 100;
while (x != 0)
{
if (x >= 25)
{
x = x - 25; //or x-=25;
y = y + 1;
}
if (x >= 10 && x < 25)
{
x = x - 10; //or x-=10;
y = y + 1;
}
if (x >= 5 && x < 10)
{
x = x - 5; //or x-=5;
y = y + 1;
}
if (x >= 1 && x < 5)
{
x = x - 1; //or x-=1; or x--; or --x; :)
y = y + 1;
}
}
printf("The number of coins neccessary is %d", y);
}
I would remain to be convinced about your loop structure. There's the division operator that can be used to good effect:
int total = 0;
int ncoins;
int amount = GetFloat() * 100;
assert(amount >= 0);
ncoins = amount / 25;
total += ncoins;
amount -= ncoins * 25;
assert(amount < 25);
ncoins = amount / 10;
total += ncoins;
amount -= ncoins * 10;
assert(amount < 10);
ncoins = amount / 5;
total += ncoins;
amount -= ncoins * 5;
assert(amount < 5);
total += amount;
That's written out longhand; you could devise a loop, too:
int values[] = { 25, 10, 5, 1 };
enum { N_VALUES = sizeof(values) / sizeof(values[0]) };
int total = 0;
int ncoins;
int amount = GetFloat() * 100;
assert(amount >= 0);
for (int i = 0; i < N_VALUES && amount > 0; i++)
{
ncoins = amount / values[i];
total += ncoins;
amount -= ncoins * values[i];
}
assert(amount == 0);
For an assignment of a course called High Performance Computing, I required to optimize the following code fragment:
int foobar(int a, int b, int N)
{
int i, j, k, x, y;
x = 0;
y = 0;
k = 256;
for (i = 0; i <= N; i++) {
for (j = i + 1; j <= N; j++) {
x = x + 4*(2*i+j)*(i+2*k);
if (i > j){
y = y + 8*(i-j);
}else{
y = y + 8*(j-i);
}
}
}
return x;
}
Using some recommendations, I managed to optimize the code (or at least I think so), such as:
Constant Propagation
Algebraic Simplification
Copy Propagation
Common Subexpression Elimination
Dead Code Elimination
Loop Invariant Removal
bitwise shifts instead of multiplication as they are less expensive.
Here's my code:
int foobar(int a, int b, int N) {
int i, j, x, y, t;
x = 0;
y = 0;
for (i = 0; i <= N; i++) {
t = i + 512;
for (j = i + 1; j <= N; j++) {
x = x + ((i<<3) + (j<<2))*t;
}
}
return x;
}
According to my instructor, a well optimized code instructions should have fewer or less costly instructions in assembly language level.And therefore must be run, the instructions in less time than the original code, ie calculations are made with::
execution time = instruction count * cycles per instruction
When I generate assembly code using the command: gcc -o code_opt.s -S foobar.c,
the generated code has many more lines than the original despite having made some optimizations, and run-time is lower, but not as much as in the original code. What am I doing wrong?
Do not paste the assembly code as both are very extensive. So I'm calling the function "foobar" in the main and I am measuring the execution time using the time command in linux
int main () {
int a,b,N;
scanf ("%d %d %d",&a,&b,&N);
printf ("%d\n",foobar (a,b,N));
return 0;
}
Initially:
for (i = 0; i <= N; i++) {
for (j = i + 1; j <= N; j++) {
x = x + 4*(2*i+j)*(i+2*k);
if (i > j){
y = y + 8*(i-j);
}else{
y = y + 8*(j-i);
}
}
}
Removing y calculations:
for (i = 0; i <= N; i++) {
for (j = i + 1; j <= N; j++) {
x = x + 4*(2*i+j)*(i+2*k);
}
}
Splitting i, j, k:
for (i = 0; i <= N; i++) {
for (j = i + 1; j <= N; j++) {
x = x + 8*i*i + 16*i*k ; // multiple of 1 (no j)
x = x + (4*i + 8*k)*j ; // multiple of j
}
}
Moving them externally (and removing the loop that runs N-i times):
for (i = 0; i <= N; i++) {
x = x + (8*i*i + 16*i*k) * (N-i) ;
x = x + (4*i + 8*k) * ((N*N+N)/2 - (i*i+i)/2) ;
}
Rewritting:
for (i = 0; i <= N; i++) {
x = x + ( 8*k*(N*N+N)/2 ) ;
x = x + i * ( 16*k*N + 4*(N*N+N)/2 + 8*k*(-1/2) ) ;
x = x + i*i * ( 8*N + 16*k*(-1) + 4*(-1/2) + 8*k*(-1/2) );
x = x + i*i*i * ( 8*(-1) + 4*(-1/2) ) ;
}
Rewritting - recalculating:
for (i = 0; i <= N; i++) {
x = x + 4*k*(N*N+N) ; // multiple of 1
x = x + i * ( 16*k*N + 2*(N*N+N) - 4*k ) ; // multiple of i
x = x + i*i * ( 8*N - 20*k - 2 ) ; // multiple of i^2
x = x + i*i*i * ( -10 ) ; // multiple of i^3
}
Another move to external (and removal of the i loop):
x = x + ( 4*k*(N*N+N) ) * (N+1) ;
x = x + ( 16*k*N + 2*(N*N+N) - 4*k ) * ((N*(N+1))/2) ;
x = x + ( 8*N - 20*k - 2 ) * ((N*(N+1)*(2*N+1))/6);
x = x + (-10) * ((N*N*(N+1)*(N+1))/4) ;
Both the above loop removals use the summation formulas:
Sum(1, i = 0..n) = n+1
Sum(i1, i = 0..n) = n(n + 1)/2
Sum(i2, i = 0..n) = n(n + 1)(2n + 1)/6
Sum(i3, i = 0..n) = n2(n + 1)2/4
y does not affect the final result of the code - removed:
int foobar(int a, int b, int N)
{
int i, j, k, x, y;
x = 0;
//y = 0;
k = 256;
for (i = 0; i <= N; i++) {
for (j = i + 1; j <= N; j++) {
x = x + 4*(2*i+j)*(i+2*k);
//if (i > j){
// y = y + 8*(i-j);
//}else{
// y = y + 8*(j-i);
//}
}
}
return x;
}
k is simply a constant:
int foobar(int a, int b, int N)
{
int i, j, x;
x = 0;
for (i = 0; i <= N; i++) {
for (j = i + 1; j <= N; j++) {
x = x + 4*(2*i+j)*(i+2*256);
}
}
return x;
}
The inner expression can be transformed to: x += 8*i*i + 4096*i + 4*i*j + 2048*j. Use math to push all of them to the outer loop: x += 8*i*i*(N-i) + 4096*i*(N-i) + 2*i*(N-i)*(N+i+1) + 1024*(N-i)*(N+i+1).
You can expand the above expression, and apply sum of squares and sum of cubes formula to obtain a close form expression, which should run faster than the doubly nested loop. I leave it as an exercise to you. As a result, i and j will also be removed.
a and b should also be removed if possible - since a and b are supplied as argument but never used in your code.
Sum of squares and sum of cubes formula:
Sum(x2, x = 1..n) = n(n + 1)(2n + 1)/6
Sum(x3, x = 1..n) = n2(n + 1)2/4
This function is equivalent with the following formula, which contains only 4 integer multiplications, and 1 integer division:
x = N * (N + 1) * (N * (7 * N + 8187) - 2050) / 6;
To get this, I simply typed the sum calculated by your nested loops into Wolfram Alpha:
sum (sum (8*i*i+4096*i+4*i*j+2048*j), j=i+1..N), i=0..N
Here is the direct link to the solution. Think before coding. Sometimes your brain can optimize code better than any compiler.
Briefly scanning the first routine, the first thing you notice is that expressions involving "y" are completely unused and can be eliminated (as you did). This further permits eliminating the if/else (as you did).
What remains is the two for loops and the messy expression. Factoring out the pieces of that expression that do not depend on j is the next step. You removed one such expression, but (i<<3) (ie, i * 8) remains in the inner loop, and can be removed.
Pascal's answer reminded me that you can use a loop stride optimization. First move (i<<3) * t out of the inner loop (call it i1), then calculate, when initializing the loop, a value j1 that equals (i<<2) * t. On each iteration increment j1 by 4 * t (which is a pre-calculated constant). Replace your inner expression with x = x + i1 + j1;.
One suspects that there may be some way to combine the two loops into one, with a stride, but I'm not seeing it offhand.
A few other things I can see. You don't need y, so you can remove its declaration and initialisation.
Also, the values passed in for a and b aren't actually used, so you could use these as local variables instead of x and t.
Also, rather than adding i to 512 each time through you can note that t starts at 512 and increments by 1 each iteration.
int foobar(int a, int b, int N) {
int i, j;
a = 0;
b = 512;
for (i = 0; i <= N; i++, b++) {
for (j = i + 1; j <= N; j++) {
a = a + ((i<<3) + (j<<2))*b;
}
}
return a;
}
Once you get to this point you can also observe that, aside from initialising j, i and j are only used in a single mutiple each - i<<3 and j<<2. We can code this directly in the loop logic, thus:
int foobar(int a, int b, int N) {
int i, j, iLimit, jLimit;
a = 0;
b = 512;
iLimit = N << 3;
jLimit = N << 2;
for (i = 0; i <= iLimit; i+=8) {
for (j = i >> 1 + 4; j <= jLimit; j+=4) {
a = a + (i + j)*b;
}
b++;
}
return a;
}
OK... so here is my solution, along with inline comments to explain what I did and how.
int foobar(int N)
{ // We eliminate unused arguments
int x = 0, i = 0, i2 = 0, j, k, z;
// We only iterate up to N on the outer loop, since the
// last iteration doesn't do anything useful. Also we keep
// track of '2*i' (which is used throughout the code) by a
// second variable 'i2' which we increment by two in every
// iteration, essentially converting multiplication into addition.
while(i < N)
{
// We hoist the calculation '4 * (i+2*k)' out of the loop
// since k is a literal constant and 'i' is a constant during
// the inner loop. We could convert the multiplication by 2
// into a left shift, but hey, let's not go *crazy*!
//
// (4 * (i+2*k)) <=>
// (4 * i) + (4 * 2 * k) <=>
// (2 * i2) + (8 * k) <=>
// (2 * i2) + (8 * 512) <=>
// (2 * i2) + 2048
k = (2 * i2) + 2048;
// We have now converted the expression:
// x = x + 4*(2*i+j)*(i+2*k);
//
// into the expression:
// x = x + (i2 + j) * k;
//
// Counterintuively we now *expand* the formula into:
// x = x + (i2 * k) + (j * k);
//
// Now observe that (i2 * k) is a constant inside the inner
// loop which we can calculate only once here. Also observe
// that is simply added into x a total (N - i) times, so
// we take advantange of the abelian nature of addition
// to hoist it completely out of the loop
x = x + (i2 * k) * (N - i);
// Observe that inside this loop we calculate (j * k) repeatedly,
// and that j is just an increasing counter. So now instead of
// doing numerous multiplications, let's break the operation into
// two parts: a multiplication, which we hoist out of the inner
// loop and additions which we continue performing in the inner
// loop.
z = i * k;
for (j = i + 1; j <= N; j++)
{
z = z + k;
x = x + z;
}
i++;
i2 += 2;
}
return x;
}
The code, without any of the explanations boils down to this:
int foobar(int N)
{
int x = 0, i = 0, i2 = 0, j, k, z;
while(i < N)
{
k = (2 * i2) + 2048;
x = x + (i2 * k) * (N - i);
z = i * k;
for (j = i + 1; j <= N; j++)
{
z = z + k;
x = x + z;
}
i++;
i2 += 2;
}
return x;
}
I hope this helps.
int foobar(int N) //To avoid unuse passing argument
{
int i, j, x=0; //Remove unuseful variable, operation so save stack and Machine cycle
for (i = N; i--; ) //Don't check unnecessary comparison condition
for (j = N+1; --j>i; )
x += (((i<<1)+j)*(i+512)<<2); //Save Machine cycle ,Use shift instead of Multiply
return x;
}