Yes, this is a basic C coding homework problem. No, I am not just looking for someone to do it for me. Considering that this is my first programming class, I'm not surprised that I can't get it to work, and I'm certain there is plenty wrong with it. I just want some help pointing out the problems in my code and the things that are missing so that I can fix them on my own.
Homework Question:
Write a program to read ONLY one integer number (your input must be
one 3 digit number from 100 to 999), and to think of a number as
being ABC (where A, B, and C are the 3 digits of a number). Now,
form the number to become ABC, BCA, and CAB, then find out the
remainder of these three numbers when they are divided by 11.
Assume remainders would respectively be X, Y, and Z and add them
up as X+Y, Y+Z, and Z+X. Now if any of these summations is odd
number, increase it by 11 if the summation plus 11 is less than 20,
otherwise decrease the summation by 11 (this summation operation
must be positive number but less than 20). Finally, divide each
of the sums in half. Now, print out all the resulting digits.
My Code:
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
int main()
{
//Declare all variables
int OrigNumber;
int x, y, z;
int number;
number = x, y, z;
int sum;
//
printf("Input a three digit number");
//
int c;
c = OrigNumber %10;
//
int b;
b=((OrigNumber - c) % 100)/10;
//
int a;
a = (OrigNumber - (b + c))/100;
//
int abc, bca, cab;
abc = (a*100) + (10*b) + c;
bca = (10*b) + c + (a*100);
cab = c + (a*100) + (10*b);
//
if((number % 2) == 1)
{
if(number + 11 < 20)
number += 11;
else if((100 - 11 > 0) && (100 - 11 < 20))
number -= 11;
}
//
x = abc/11;
y = bca/11;
z = cab/11;
//
sum = (x + y),
(y + z),
(z + x);
}
To start with, you need to read the input. Start with a prompt that includes a carriage return:
printf("Input a three digit number: \n");
Since it's a three digit number, you could add the following line to read the input:
scanf("%3d", &OrigNumber);
The next bit of code works quite well until you get to your if (number % 2) which is meaningless since you didn't really define number - well, you did, but the line
number = x, y, z;
does NOT do what you think it does. If you add
printf("So far I have abc=%d, bca=%d, cab=%d\n", abc, bca, cab);
after you first read in the number and computed those three, you will see you are well on your way.
Note that
number = x, y, z;
Uses a thing called the "comma operator". All the things (a,b,c) are "evaluated" but their values are not returned. At any rate, where you have that line, you didn't yet assign a value to x,y and z.
Is that enough to get your started?
update now that you have had a few hours to mull this over, here are a few more pointers.
Your computation of abc, cab, bca makes no sense. I will show you just one of them:
cab = c*100 + a*10 + b;
Next you need to compute each of x, y and z. Again, here is one of the three:
y = bca%11;
Now you have to make the sums - I call them xy, yz, and zx. Just one of them:
zx = z + x;
Next, to deal with the instruction: "Now if any of these summations is odd number, increase it by 11 if the summation plus 11 is less than 20, otherwise decrease the summation by 11:
if(xy % 2 == 1) {
if(xy + 11 < 20) xy += 11; else xy -= 11;
}
use similar code for all three sums. Then "divide by 2":
xy /= 2;
repeat as needed.
Finally, print out the result:
printf("xy: %d, yz: %d, zx: %d\n", xy, yz, zx);
The amazing thing is that if you did this right, you get the original numbers back...
You could make the code more compact by using an array of values and looping through it - rather than repeating the code snippets I wrote above with different variables. But I suspect that is well outside the scope of what you are expected to know at this point.
Can you take it from here?
#include <stdio.h>
int main()
{
//Declare all variables
int OrigNumber;
int a, b, c;
int abc, bca, cab;
int x, y, z;
int xplusy , yplusz, xplusz;
printf(" A program to read ONLY one integer number.\n Input must be one 3 digit number from 100 to 999 : ");
scanf("%d", &OrigNumber); // Get input from console
if(OrigNumber > 999 || OrigNumber < 100) {
printf("Invalid number. Quiting program. This is error handling. Important while learning programming.");
return 0;
}
c = OrigNumber %10; // digit at unit's place
b=((OrigNumber) % 100)/10; //digit at the ten's place
a = (OrigNumber)/100; //digit at the 100's place. Note: 734/100 = 7. NOT 7.34.
printf("\n Three numbers say A,B, C : %d, %d , %d ", a, b, c);
abc = a*100 + 10*b + c;
bca = 100*b + 10*c + a;
cab = c*100 + a*10 + b;
printf("\n Three numbers say ABC, BCA, CAB : %d, %d , %d ", abc, bca, cab);
x = abc % 11; // Reminder when divided by 11.
y = bca % 11;
z = cab % 11;
printf("\n Three numbers say X, Y, Z : %d, %d , %d ", x, y, z);
xplusy = x + y; // Adding reminders two at a time.
yplusz = y + z;
xplusz = x + z;
printf("\n Three numbers X+Y, Y+Z, X+Z : %d, %d , %d ", xplusy, yplusz, xplusz);
if((xplusy % 2) == 1) {
if(xplusy + 11 < 20)
xplusy += 11;
else
xplusy -= 11;
}
if((yplusz % 2) == 1) {
if(yplusz + 11 < 20)
yplusz += 11;
else
yplusz -= 11;
}
if((xplusz % 2) == 1) {
if(xplusz + 11 < 20)
xplusz += 11;
else
xplusz -= 11;
}
xplusy /= 2; // Finally, divide each of the sum in half.
yplusz /= 2;
xplusz /= 2;
printf("\n Now print out all the resulting digits : %d, %d , %d \n", xplusy, yplusz, xplusz);
return 0;
}
int abc, bca, cab;
abc = (a*100) + (10*b) + c;
bca = (10*b) + c + (a*100);
cab = c + (a*100) + (10*b);
I suggest printing out the numbers at this point in the code.
printf( "%d %d %d", abc, bca, cab );
I think you'll see one of the problems you need to solve.
#include <iostream>
#include <cstdio>
using namespace std;
int main() {
int n, a, b, c, abc, bca, cab, x, y, z, p, q, r;
scanf("%d", &n);
c=n%10;
b=(n/10)%10;
a=n/100;
abc=a*100+b*10+c;
bca=b*100+c*10+a;
cab=c*100+a*10+b;
x=abc%11;
y=bca%11;
z=cab%11;
p=x+y;
q=y+z;
r=z+x;
return 0;
}
Now if any of these summations is odd number, increase it by 11 if the
summation plus 11 is less than 20, otherwise decrease the summation by
11 (this summation operation must be positive number but less than
20). Finally, divide each of the sums in half. Now, print out all the
resulting digits.
i didnt get the final part, can you explain it more clearly?
So we're reading a matrix and saving it in an array sequentially. We read the matrix from a starting [x,y] point which is provided. Here's an example of some code I wrote to get the values of [x-1,y] [x+1,y] [x,y-1] [x,y+1], which is a cross.
for(i = 0, n = -1, m = 0, array_pos = 0; i < 4; i++, n++, array_pos++) {
if(x+n < filter_matrix.src.columns && x+n >= 0 )
if(y+m < filter_matrix.src.lines && y+m >= 0){
for(k = 0; k < numpixels; k++) {
arrayToProcess[array_pos].rgb[h] = filter_matrix.src.points[x+n][y+m].rgb[h];
}
}
m = n;
m++;
}
(The if's are meant to avoid reading null positions, since it's an image we're reading the origin pixel can be located in a corner. Not relevant to the issue here.)
Now is there a similar generic algorithm which can read ALL the elements around as a square (not just a cross) based on a single parameter, which is the size of the square's side squared?
If it helps, the only values we're dealing with are 9, 25 and 49 (a 3x3 5x5 and 7x7 square).
Here is a generalized code for reading the square centered at (x,y) of size n
int startx = x-n/2;
int starty = y-n/2;
for(int u=0;u<n;u++) {
for(int v=0;v<n;v++) {
int i = startx + u;
int j = starty + v;
if(i>=0 && j>=0 && i<N && j<M) {
printf(Matrix[i][j]);
}
}
}
Explanation: Start from top left value which is (x - n/2, y-n/2) now consider that you are read a normal square matrix from where i and j are indices of Matrix[i][j]. So we just added startx & starty to shift the matrix at (0,0) to (x-n/2,y-n/2).
Given:
static inline int min(int x, int y) { return (x < y) ? x : y; }
static inline int max(int x, int y) { return (x > y) ? x : y; }
or equivalent macros, and given that:
the x-coordinates range from 0 to x_max (inclusive),
the y-coordinates range from 0 to y_max (inclusive),
the centre of the square (x,y) is within the bounds,
the square you are creating has sides of (2 * size + 1) (so size is 1, 2, or 3 for the 3x3, 5x5, and 7x7 cases; or if you prefer to have sq_side = one of 3, 5, 7, then size = sq_side / 2),
the integer types are all signed (so x - size can produce a negative value; if they're unsigned, you will get the wrong result using the expressions shown),
then you can ensure that you are within bounds by setting:
x_lo = max(x - size, 0);
x_hi = min(x + size, x_max);
y_lo = max(y - size, 0);
y_hi = min(y + size, y_max);
for (x_pos = x_lo; x_pos <= x_hi; x_pos++)
{
for (y_pos = y_lo; y_pos <= y_hi; y_pos++)
{
// Process the data at array[x_pos][y_pos]
}
}
Basically, the initial assignments determine the bounds of the the array from [x-size][y-size] to [x+size][y+size], but bounded by 0 on the low side and the maximum sizes on the high end. Then scan over the relevant rectangular (usually square) sub-section of the matrix. Note that this determines the valid ranges once, outside the loops, rather than repeatedly within the loops.
If the integer types are signed, you have ensure you never try to create a negative number during subtraction. The expressions could be rewritten as:
x_lo = x - min(x, size);
x_hi = min(x + size, x_max);
y_lo = y - min(y, size);
y_hi = min(y + size, y_max);
which isn't as symmetric but only uses the min function.
Given the coordinates (x,y), you first need to find the surrounding elements. You can do that with a double for loop, like this:
for (int i = x-1; i <= x+1; i++) {
for (int j = y-1; j <= y+1; j++) {
int elem = square[i][j];
}
}
Now you just need to do a bit of work to make sure that 0 <= i,j < n, where n is the length of a side;
I don't know whether the (X,Y) in your code is the center of the square. I assume it is.
If the side of the square is odd. generate the coordinates of the points on the square. I assume the center is (0,0). Then the points on the squares are
(-side/2, [-side/2,side/2 - 1]); ([-side/2 + 1,side/2], -side/2); (side/2,[side/2 - 1,-side/2]);([side/2 - 1, side/2],-side/2);
side is the length of the square
make use of this:
while(int i<=0 && int j<=0)
for (i = x-1; i <= x+1; i++) {
for (j = y-1; j <= y+1; j++) {
int elem = square[i][j];
}
}
}
I am using Bresenham's circle algorithm for fast circle drawing. However, I also want to (at the request of the user) draw a filled circle.
Is there a fast and efficient way of doing this? Something along the same lines of Bresenham?
The language I am using is C.
Having read the Wikipedia page on Bresenham's (also 'Midpoint') circle algorithm, it would appear that the easiest thing to do would be to modify its actions, such that instead of
setPixel(x0 + x, y0 + y);
setPixel(x0 - x, y0 + y);
and similar, each time you instead do
lineFrom(x0 - x, y0 + y, x0 + x, y0 + y);
That is, for each pair of points (with the same y) that Bresenham would you have you plot, you instead connect with a line.
Just use brute force. This method iterates over a few too many pixels, but it only uses integer multiplications and additions. You completely avoid the complexity of Bresenham and the possible bottleneck of sqrt.
for(int y=-radius; y<=radius; y++)
for(int x=-radius; x<=radius; x++)
if(x*x+y*y <= radius*radius)
setpixel(origin.x+x, origin.y+y);
Here's a C# rough guide (shouldn't be that hard to get the right idea for C) - this is the "raw" form without using Bresenham to eliminate repeated square-roots.
Bitmap bmp = new Bitmap(200, 200);
int r = 50; // radius
int ox = 100, oy = 100; // origin
for (int x = -r; x < r ; x++)
{
int height = (int)Math.Sqrt(r * r - x * x);
for (int y = -height; y < height; y++)
bmp.SetPixel(x + ox, y + oy, Color.Red);
}
bmp.Save(#"c:\users\dearwicker\Desktop\circle.bmp");
You can use this:
void DrawFilledCircle(int x0, int y0, int radius)
{
int x = radius;
int y = 0;
int xChange = 1 - (radius << 1);
int yChange = 0;
int radiusError = 0;
while (x >= y)
{
for (int i = x0 - x; i <= x0 + x; i++)
{
SetPixel(i, y0 + y);
SetPixel(i, y0 - y);
}
for (int i = x0 - y; i <= x0 + y; i++)
{
SetPixel(i, y0 + x);
SetPixel(i, y0 - x);
}
y++;
radiusError += yChange;
yChange += 2;
if (((radiusError << 1) + xChange) > 0)
{
x--;
radiusError += xChange;
xChange += 2;
}
}
}
Great ideas here!
Since I'm at a project that requires many thousands of circles to be drawn, I have evaluated all suggestions here (and improved a few by precomputing the square of the radius):
http://quick-bench.com/mwTOodNOI81k1ddaTCGH_Cmn_Ag
The Rev variants just have x and y swapped because consecutive access along the y axis are faster with the way my grid/canvas structure works.
The clear winner is Daniel Earwicker's method ( DrawCircleBruteforcePrecalc ) that precomputes the Y value to avoid unnecessary radius checks. Somewhat surprisingly that negates the additional computation caused by the sqrt call.
Some comments suggest that kmillen's variant (DrawCircleSingleLoop) that works with a single loop should be very fast, but it's the slowest here. I assume that is because of all the divisions. But perhaps I have adapted it wrong to the global variables in that code. Would be great if someone takes a look.
EDIT: After looking for the first time since college years at some assembler code, I managed find that the final additions of the circle's origin are a culprit.
Precomputing those, I improved the fastest method by a factor of another 3.7-3.9 according to the bench!
http://quick-bench.com/7ZYitwJIUgF_OkDUgnyMJY4lGlA
Amazing.
This being my code:
for (int x = -radius; x < radius ; x++)
{
int hh = (int)std::sqrt(radius_sqr - x * x);
int rx = center_x + x;
int ph = center_y + hh;
for (int y = center_y-hh; y < ph; y++)
canvas[rx][y] = 1;
}
I like palm3D's answer. For being brute force, this is an amazingly fast solution. There are no square root or trigonometric functions to slow it down. Its one weakness is the nested loop.
Converting this to a single loop makes this function almost twice as fast.
int r2 = r * r;
int area = r2 << 2;
int rr = r << 1;
for (int i = 0; i < area; i++)
{
int tx = (i % rr) - r;
int ty = (i / rr) - r;
if (tx * tx + ty * ty <= r2)
SetPixel(x + tx, y + ty, c);
}
This single loop solution rivals the efficiency of a line drawing solution.
int r2 = r * r;
for (int cy = -r; cy <= r; cy++)
{
int cx = (int)(Math.Sqrt(r2 - cy * cy) + 0.5);
int cyy = cy + y;
lineDDA(x - cx, cyy, x + cx, cyy, c);
}
palm3D's brute-force algorithm I found to be a good starting point. This method uses the same premise, however it includes a couple of ways to skip checking most of the pixels.
First, here's the code:
int largestX = circle.radius;
for (int y = 0; y <= radius; ++y) {
for (int x = largestX; x >= 0; --x) {
if ((x * x) + (y * y) <= (circle.radius * circle.radius)) {
drawLine(circle.center.x - x, circle.center.x + x, circle.center.y + y);
drawLine(circle.center.x - x, circle.center.x + x, circle.center.y - y);
largestX = x;
break; // go to next y coordinate
}
}
}
Next, the explanation.
The first thing to note is that if you find the minimum x coordinate that is within the circle for a given horizontal line, you immediately know the maximum x coordinate.
This is due to the symmetry of the circle. If the minimum x coordinate is 10 pixels ahead of the left of the bounding box of the circle, then the maximum x is 10 pixels behind the right of the bounding box of the circle.
The reason to iterate from high x values to low x values, is that the minimum x value will be found with less iterations. This is because the minimum x value is closer to the left of the bounding box than the centre x coordinate of the circle for most lines, due to the circle being curved outwards, as seen on this image
The next thing to note is that since the circle is also symmetric vertically, each line you find gives you a free second line to draw, each time you find a line in the top half of the circle, you get one on the bottom half at the radius-y y coordinate. Therefore, when any line is found, two can be drawn and only the top half of the y values needs to be iterated over.
The last thing to note is that is that if you start from a y value that is at the centre of the circle and then move towards the top for y, then the minimum x value for each next line must be closer to the centre x coordinate of the circle than the last line. This is also due to the circle curving closer towards the centre x value as you go up the circle. Here is a visual on how that is the case.
In summary:
If you find the minimum x coordinate of a line, you get the maximum x coordinate for free.
Every line you find to draw on the top half of the circle gives you a line on the bottom half of the circle for free.
Every minimum x coordinate has to be closer to the centre of the circle than the previous x coordinate for each line when iterating from the centre y coordinate to the top.
You can also store the value of (radius * radius), and also (y * y) instead of calculating them
multiple times.
Here's how I'm doing it:
I'm using fixed point values with two bits precision (we have to manage half points and square values of half points)
As mentionned in a previous answer, I'm also using square values instead of square roots.
First, I'm detecting border limit of my circle in a 1/8th portion of the circle. I'm using symetric of these points to draw the 4 "borders" of the circle. Then I'm drawing the square inside the circle.
Unlike the midpoint circle algorith, this one will work with even diameters (and with real numbers diameters too, with some little changes).
Please forgive me if my explanations were not clear, I'm french ;)
void DrawFilledCircle(int circleDiameter, int circlePosX, int circlePosY)
{
const int FULL = (1 << 2);
const int HALF = (FULL >> 1);
int size = (circleDiameter << 2);// fixed point value for size
int ray = (size >> 1);
int dY2;
int ray2 = ray * ray;
int posmin,posmax;
int Y,X;
int x = ((circleDiameter&1)==1) ? ray : ray - HALF;
int y = HALF;
circlePosX -= (circleDiameter>>1);
circlePosY -= (circleDiameter>>1);
for (;; y+=FULL)
{
dY2 = (ray - y) * (ray - y);
for (;; x-=FULL)
{
if (dY2 + (ray - x) * (ray - x) <= ray2) continue;
if (x < y)
{
Y = (y >> 2);
posmin = Y;
posmax = circleDiameter - Y;
// Draw inside square and leave
while (Y < posmax)
{
for (X = posmin; X < posmax; X++)
setPixel(circlePosX+X, circlePosY+Y);
Y++;
}
// Just for a better understanding, the while loop does the same thing as:
// DrawSquare(circlePosX+Y, circlePosY+Y, circleDiameter - 2*Y);
return;
}
// Draw the 4 borders
X = (x >> 2) + 1;
Y = y >> 2;
posmax = circleDiameter - X;
int mirrorY = circleDiameter - Y - 1;
while (X < posmax)
{
setPixel(circlePosX+X, circlePosY+Y);
setPixel(circlePosX+X, circlePosY+mirrorY);
setPixel(circlePosX+Y, circlePosY+X);
setPixel(circlePosX+mirrorY, circlePosY+X);
X++;
}
// Just for a better understanding, the while loop does the same thing as:
// int lineSize = circleDiameter - X*2;
// Upper border:
// DrawHorizontalLine(circlePosX+X, circlePosY+Y, lineSize);
// Lower border:
// DrawHorizontalLine(circlePosX+X, circlePosY+mirrorY, lineSize);
// Left border:
// DrawVerticalLine(circlePosX+Y, circlePosY+X, lineSize);
// Right border:
// DrawVerticalLine(circlePosX+mirrorY, circlePosY+X, lineSize);
break;
}
}
}
void DrawSquare(int x, int y, int size)
{
for( int i=0 ; i<size ; i++ )
DrawHorizontalLine(x, y+i, size);
}
void DrawHorizontalLine(int x, int y, int width)
{
for(int i=0 ; i<width ; i++ )
SetPixel(x+i, y);
}
void DrawVerticalLine(int x, int y, int height)
{
for(int i=0 ; i<height ; i++ )
SetPixel(x, y+i);
}
To use non-integer diameter, you can increase precision of fixed point or use double values.
It should even be possible to make a sort of anti-alias depending on the difference between dY2 + (ray - x) * (ray - x) and ray2 (dx² + dy² and r²)
If you want a fast algorithm, consider drawing a polygon with N sides, the higher is N, the more precise will be the circle.
I would just generate a list of points and then use a polygon draw function for the rendering.
It may not be the algorithm yo are looking for and not the most performant one,
but I always do something like this:
void fillCircle(int x, int y, int radius){
// fill a circle
for(int rad = radius; rad >= 0; rad--){
// stroke a circle
for(double i = 0; i <= PI * 2; i+=0.01){
int pX = x + rad * cos(i);
int pY = y + rad * sin(i);
drawPoint(pX, pY);
}
}
}
The following two methods avoid the repeated square root calculation by drawing multiple parts of the circle at once and should therefore be quite fast:
void circleFill(const size_t centerX, const size_t centerY, const size_t radius, color fill) {
if (centerX < radius || centerY < radius || centerX + radius > width || centerY + radius > height)
return;
const size_t signedRadius = radius * radius;
for (size_t y = 0; y < radius; y++) {
const size_t up = (centerY - y) * width;
const size_t down = (centerY + y) * width;
const size_t halfWidth = roundf(sqrtf(signedRadius - y * y));
for (size_t x = 0; x < halfWidth; x++) {
const size_t left = centerX - x;
const size_t right = centerX + x;
pixels[left + up] = fill;
pixels[right + up] = fill;
pixels[left + down] = fill;
pixels[right + down] = fill;
}
}
}
void circleContour(const size_t centerX, const size_t centerY, const size_t radius, color stroke) {
if (centerX < radius || centerY < radius || centerX + radius > width || centerY + radius > height)
return;
const size_t signedRadius = radius * radius;
const size_t maxSlopePoint = ceilf(radius * 0.707106781f); //ceilf(radius * cosf(TWO_PI/8));
for (size_t i = 0; i < maxSlopePoint; i++) {
const size_t depth = roundf(sqrtf(signedRadius - i * i));
size_t left = centerX - depth;
size_t right = centerX + depth;
size_t up = (centerY - i) * width;
size_t down = (centerY + i) * width;
pixels[left + up] = stroke;
pixels[right + up] = stroke;
pixels[left + down] = stroke;
pixels[right + down] = stroke;
left = centerX - i;
right = centerX + i;
up = (centerY - depth) * width;
down = (centerY + depth) * width;
pixels[left + up] = stroke;
pixels[right + up] = stroke;
pixels[left + down] = stroke;
pixels[right + down] = stroke;
}
}
This was used in my new 3D printer Firmware, and it is proven the
fastest way for filled circle of a diameter from 1 to 43 pixel. If
larger is needed, the following memory block(or array) should be
extended following a structure I wont waste my time explaining...
If you have questions, or need larger diameter than 43, contact me, I
will help you drawing the fastest and perfect filled circles... or
Bresenham's circle drawing algorithm can be used above those
diameters, but having to fill the circle after, or incorporating the
fill into Bresenham's circle drawing algorithm, will only result in
slower fill circle than my code. I already benchmarked the different
codes, my solution is 4 to 5 times faster. As a test I have been
able to draw hundreds of filled circles of different size and colors
on a BigTreeTech tft24 1.1 running on a 1-core 72 Mhz cortex-m4
https://www.youtube.com/watch?v=7_Wp5yn3ADI
// this must be declared anywhere, as static or global
// as long as the function can access it !
uint8_t Rset[252]={
0,1,1,2,2,1,2,3,3,1,3,3,4,4,2,3,4,5,5,5,2,4,5,5,
6,6,6,2,4,5,6,6,7,7,7,2,4,5,6,7,7,8,8,8,2,5,6,7,
8,8,8,9,9,9,3,5,6,7,8,9,9,10,10,10,10,3,5,7,8,9,
9,10,10,11,11,11,11,3,5,7,8,9,10,10,11,11,12,12,
12,12,3,6,7,9,10,10,11,12,12,12,13,13,13,13,3,6,
8,9,10,11,12,12,13,13,13,14,14,14,14,3,6,8,9,10,
11,12,13,13,14,14,14,15,15,15,15,3,6,8,10,11,12,
13,13,14,14,15,15,15,16,16,16,16,4,7,8,10,11,12,
13,14,14,15,16,16,16,17,17,17,17,17,4,7,9,10,12,
13,14,14,15,16,16,17,17,17,18,18,18,18,18,4,7,9,
11,12,13,14,15,16,16,17,17,18,18,18,19,19,19,19,
19,7,9,11,12,13,15,15,16,17,18,18,19,19,20,20,20,
20,20,20,20,20,7,9,11,12,14,15,16,17,17,18,19,19
20,20,21,21,21,21,21,21,21,21};
// SOLUTION 1: (the fastest)
void FillCircle_v1(uint16_t x, uint16_t y, uint16_t r)
{
// all needed variables are created and set to their value...
uint16_t radius=(r<1) ? 1 : r ;
if (radius>21 ) {radius=21; }
uint16_t diam=(radius*2)+1;
uint16_t ymir=0, cur_y=0;
radius--; uint16_t target=(radius*radius+3*radius)/2; radius++;
// this part draws directly into the ILI94xx TFT buffer mem.
// using pointers..2 versions where you can draw
// pixels and lines with coordinates will follow
for (uint16_t yy=0; yy<diam; yy++)
{ ymir= (yy<=radius) ? yy+target : target+diam-(yy+1);
cur_y=y-radius+yy;
uint16_t *pixel=buffer_start_addr+x-Rset[ymir]+cur_y*buffer_width;
for (uint16_t xx= 0; xx<=(2*Rset[ymir]); xx++)
{ *pixel++ = CANVAS::draw_color; }}}
// SOLUTION 2: adaptable to any system that can
// add a pixel at a time: (drawpixel or add_pixel,etc_)
void FillCircle_v2(uint16_t x, uint16_t y, uint16_t r)
{
// all needed variables are created and set to their value...
uint16_t radius=(r<1) ? 1 : r ;
if (radius>21 ) {radius=21; }
uint16_t diam=(radius*2)+1;
uint16_t ymir=0, cur_y=0;
radius--; uint16_t target=(radius*radius+3*radius)/2; radius++;
for (uint16_t yy=0; yy<diam; yy++)
{ ymir= (yy<=radius) ? yy+target : target+diam-(yy+1);
cur_y=y-radius+yy;
uint16_t Pixel_x=x-Rset[ymir];
for (uint16_t xx= 0; xx<=(2*Rset[ymir]); xx++)
{ //use your add_pixel or draw_pixel here
// using those coordinates:
// X position will be... (Pixel_x+xx)
// Y position will be... (cur_y)
// and add those 3 brackets at the end
}}}
// SOLUTION 3: adaptable to any system that can draw fast
// horizontal lines
void FillCircle_v3(uint16_t x, uint16_t y, uint16_t r)
{
// all needed variables are created and set to their value...
uint16_t radius=(r<1) ? 1 : r ;
if (radius>21 ) {radius=21; }
uint16_t diam=(radius*2)+1;
uint16_t ymir=0, cur_y=0;
radius--; uint16_t target=(radius*radius+3*radius)/2; radius++;
for (uint16_t yy=0; yy<diam; yy++)
{ ymir= (yy<=radius) ? yy+target : target+diam-(yy+1);
cur_y=y-radius+yy;
uint16_t start_x=x-Rset[ymir];
uint16_t width_x=2*Rset[ymir];
// ... then use your best drawline function using those values:
// start_x: position X of the start of the line
// cur_y: position Y of the current line
// width_x: length of the line
// if you need a 2nd coordinate then :end_x=start_x+width_x
// and add those 2 brackets after !!!
}}
I did pretty much what AlegGeorge did but I changed three lines. I thought that this is faster but these are the results am I doing anything wrong? my function is called DrawBruteforcePrecalcV4. here's the code:
for (int x = 0; x < radius ; x++) // Instead of looping from -radius to radius I loop from 0 to radius
{
int hh = (int)std::sqrt(radius_sqr - x * x);
int rx = center_x + x;
int cmx = center_x - x;
int ph = center_y+hh;
for (int y = center_y-hh; y < ph; y++)
{
canvas[rx][y] = 1;
canvas[cmx][y] = 1;
}
}