problem met when checking collision between circle and polygon using SAT - c

I try to use SAT to check collision between circle and polygon, it works well when the circle is near the edges (of polygon), however when circle is closet to the corners (of polygon), it is push inside, and outside after that.
under this it failed
but under such cases it works well
How to solve this?
the algorithm looks like this:
double mtv = DBL_MAX; // minimum translation vector
Vector2 minAxis;
foreach(Edge e in p.Edges)
{
var axis = e.normal;
var circleInterval = ProjectInterval(c, axis);
var polygonInterval = ProjectInterval(p, axis);
float overlap = OverlapIntervals(circleInterval, polygonInterval);
if(overlap < 0)
{
//the intervals overlap if it returns a positive number
return -1.0f;
}
if(overlap < mtv)
{
mtv = overlap;
minAxis = e.normal;
}
}
and my code is here:
static AS_UNUSED inline AS_Bool AS_SAT_test_polygon_circle(AS_Polygon *poly, AS_Circle *circle, AS_Result *result) {
AS_Vector *axis_normed = AS_vector_create(0, 0);
if (result) {
result->is_overlapping = AS_TRUE;
result->shape_a = poly;
result->shape_b = circle;
result->type_a = AS_POLYGON;
result->type_b = AS_CIRCLE;
}
for (int i = 0; i < array_len(poly->coords); i++) {
AS_Vector *coord1 =poly->coords[i];
AS_Vector *coord2 = poly->coords[i + 1 >= array_len(poly->coords) ? 0 : i + 1];
axis_normed->x = coord1->x - coord2->x;
axis_normed->y = coord1->y - coord2->y;
AS_vector_perp(axis_normed);
AS_vector_norm(axis_normed);
double min1 = DBL_MAX, max1 = -DBL_MAX;
double dot_product;
for (int j = 0; j < array_len(poly->coords); j++) {
dot_product = AS_vector_dot(poly->coords[j], axis_normed);
if (dot_product > max1) max1 = dot_product;
if (dot_product < min1) min1 = dot_product;
}
double b = AS_vector_dot(circle->position, axis_normed);
double a = b - circle->radius, c = b + circle->radius;
double min2, max2;
min2 = a < b ? a : b;
min2 = min2 < c ? min2 : c;
max2 = a > b ? a : b;
max2 = max2 > c ? max2 : c;
if (result) {
double overlap;
if (min1 < min2) {
result->a_in_b = AS_FALSE;
if (max1 < min2) {
result->is_overlapping = AS_FALSE;
return AS_FALSE;
} else if (max1 < max2) {
overlap = max1 - min2;
} else {
double option1 = max1 - min2;
double option2 = max2 - min1;
overlap = option1 < option2 ? option1 : option2;
}
} else {
result->b_in_a = AS_FALSE;
if (max2 < min1) {
result->is_overlapping = AS_FALSE;
return AS_FALSE;
} else if (max2 < max1) {
overlap = max2 - min1;
} else {
double option1 = max2 - min1;
double option2 = max1 - min2;
overlap = option1 < option2 ? option1 : option2;
}
}
if (overlap < result->overlap) {
result->overlap = overlap;
result->overlap_x = axis_normed->x;
result->overlap_y = axis_normed->y;
}
} else {
if (min1 > max2 || min2 > max1) {
return AS_FALSE;
}
}
}
return AS_TRUE;
}

Simply add add a check about closest point of the polygon, that will be okay.

Related

Ray Casting (GONE WRONG)

i'm building a raycasting engine and i've done the 2d projection, with wall collisions, now shifting to 3d i'm not being able to render a proper 3d image, i'll throw my code below:
i've tried all types of things by now, even changing my line function
include "../incs/Cub3d.h"
#define DOWN 0
#define UP 1
#define LEFT 2
#define RIGHT 3
#define RAY_NUM 816
t_point horizontalinter(t_mlx *m, float ang)
{
t_point inter;
inter.hit = 1;
double xinter, yinter;
double xstep, ystep;
double xwall, ywall;
xwall = 0;
ywall = 0;
int hit = 0;
float r_angle;
r_angle = (ang);
int r_dir = upordown(r_angle);
int r_lor = leftorright(r_angle);
yinter = floor(m->p->y / TILES) * TILES;
if (r_dir == DOWN)
yinter += TILES;
xinter = m->p->x + (yinter - m->p->y)/tan(r_angle);
ystep = TILES;
if (r_dir == UP)
ystep *= -1;
xstep = TILES / tan(r_angle);
if ((r_lor == LEFT && xstep > 0) || (r_lor == RIGHT && xstep < 0))
xstep *= -1;
double nextx = xinter;
double nexty = yinter;
while (nextx >= 0 && nextx <= WIDTH && nexty >= 0 && nexty <= HEIGHT)
{
if (has_wall(nextx, nexty - (r_dir == UP ? 1 : 0) , m))
{
hit = 1;
xwall = nextx;
ywall = nexty;
break;
}
else
{
nextx += xstep;
nexty += ystep;
}
}
inter.x = xwall;
inter.y = ywall;
if (hit)
{
inter.dist2pl = distancebetween(m->p->x, m->p->y, xwall, ywall);
return(inter);
}
inter.dist2pl = INT_MAX;
return(inter);
}
t_point verticalinter(t_mlx *m, float ang)
{
t_point inter;
inter.hit = 0;
double xinter, yinter;
double xstep, ystep;
double xwall, ywall;
int hit = 0;
xwall = 0;
ywall = 0;
float r_angle = normalize_ang(ang);
int r_dir = upordown(r_angle);
int r_lor = leftorright(r_angle);
//FIND X-interseption
xinter = floor(m->p->x / TILES) * TILES;
if (r_lor == RIGHT)
xinter += TILES;
//FIND Y-interseption
yinter = m->p->y + (xinter - m->p->x) * tan(r_angle);
//find STEPS
//xstep
xstep = TILES;
if (r_lor == LEFT)
xstep *= -1;
//ystep
ystep = TILES * tan(r_angle);
if ((r_dir == UP && ystep > 0) || (r_dir == DOWN && ystep < 0))
ystep *= -1;
// next intersection
double nextx = xinter;
double nexty = yinter;
//make sure that mf hits a wall
// loop untill it finds a wall or leaves canvas, increments with x and y step
while (nextx >= 0 && nextx <= WIDTH && nexty >= 0 && nexty <= HEIGHT)
{
if (has_wall(nextx - (r_lor == LEFT ? 1 : 0) , nexty, m))
{
hit = 1;
xwall = nextx;
ywall = nexty;
break;
}
else
{
nextx += xstep;
nexty += ystep;
}
}
inter.x = xwall;
inter.y = ywall;
if (hit)
{
inter.dist2pl = distancebetween(m->p->x, m->p->y, xwall, ywall);
return(inter);
}
inter.dist2pl = INT_MAX;
return(inter);
}
void calc_rays(t_mlx *m, float ang)
{
int color ;
ang = normalize_ang(ang);
t_point h_hit = horizontalinter(m, ang);
t_point v_hit = verticalinter(m, ang);
t_point closest = v_hit.dist2pl > h_hit.dist2pl ? h_hit : v_hit;
if (closest.hit)
color = rgb_to_int(0,255,0);
else
color = rgb_to_int(0,0,255);
int lineheigth = ((TILES * 5) / closest.dist2pl) * 277;
int drawStart = -lineheigth / 2 + HEIGHT / 2;
if(drawStart < 0)drawStart = 0;
int drawEnd = lineheigth / 2 + HEIGHT / 2;
if(drawEnd >= HEIGHT)drawEnd = HEIGHT - 1;
mlx_line_to(m, closest.x,drawStart , closest.x, drawEnd, color);
printf("%f\n", closest.dist2pl);
}
void cast_rays(t_mlx *m)
{
float r_angle = m->p->ang - (FOV / 2);
for(int i = 0; i < RAY_NUM; i++)
{
calc_rays(m, r_angle);
r_angle += (FOV / RAY_NUM);
}
}
i've tried calculating line heights differently but with no sucess

cell overlap removal in vlsi placement

Below code is to remove overlap among cells residing inside grids.
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>
#define TRUE 1
#define FALSE 0
//coordinate representation
struct Point
{
double x;
double y;
};
//block representation with 4 coordinates
struct Block
{
unsigned int id;
struct Point bottomLeft, topLeft, bottomRight, topRight;
double width, height;
double whitespace;
};
struct Overlap
{
double overlapLength;
struct Block *b1, *b2;
} OL[1000];
struct Block *b;
struct Block ROW1, ROW2, ROW3;
int comparator(const void *p, const void *q)
{
return (((struct Overlap *)p)->overlapLength -
(((struct Overlap *)q)->overlapLength));
}
//intialise Blocks using bottom left x,y coordinate
void intialiseBlock(struct Block *B)
{
B->topLeft.x = B->bottomLeft.x;
B->topLeft.y = B->height + B->bottomLeft.y;
B->bottomRight.x = B->width + B->bottomLeft.x;
B->bottomRight.y = B->bottomLeft.y;
B->topRight.x = B->width + B->bottomLeft.x;
B->topRight.y = B->height + B->bottomLeft.y;
}
//update coordinate
void move(double xDelta, double yDelta, struct Point *p)
{
// Parameter values are increments to the current coordinates
p->x += xDelta;
p->y += yDelta;
}
//calculate and return distance between 2 points
double distance(const struct Point aPoint, const struct Point bPoint)
{
return sqrt((aPoint.x - bPoint.x) * (aPoint.x - bPoint.x) +
(aPoint.y - bPoint.y) * (aPoint.y - bPoint.y));
}
//generate width of cell using number of cells
double computeCellWidth(int ncell)
{
double width = (0.8 * 400) / ncell;
return width;
}
//create a
unsigned int rand_interval(unsigned int min, unsigned int max)
{
int r;
const unsigned int range = 1 + max - min;
const unsigned int buckets = RAND_MAX / range;
const unsigned int limit = buckets * range;
/* Create equal size buckets all in a row, then fire randomly towards
* the buckets until you land in one of them. All buckets are equally
* likely. If you land off the end of the line of buckets, try again. */
do
{
r = rand();
} while (r >= limit);
return min + (r / buckets);
}
//place blocks in appropriate rows
void placeVertical(struct Block *b, int n, struct Block ROW1, struct Block ROW2,
struct Block ROW3)
{
for (int i = 0; i < n; i++)
{
double distanceFromBottom = 0.0;
double distanceFromTop = 0.0;
// check if Inside Row1 and Row2
if ((b[i].bottomLeft.y < ROW1.topLeft.y) && (b[i].topLeft.y > ROW1.topLeft.y))
{
//printf("ROW1 and ROW2 Executed\n");
distanceFromBottom = ROW1.topLeft.y - b[i].bottomLeft.y;
distanceFromTop = b[i].height - distanceFromBottom;
if (distanceFromBottom > distanceFromTop)
{
move(0, -distanceFromTop, &b[i].bottomLeft);
move(0, -distanceFromTop, &b[i].topLeft);
move(0, -distanceFromTop, &b[i].bottomRight);
move(0, -distanceFromTop, &b[i].topRight);
}
else
{
move(0, distanceFromBottom, &b[i].bottomLeft);
move(0, distanceFromBottom, &b[i].topLeft);
move(0, distanceFromBottom, &b[i].bottomRight);
move(0, distanceFromBottom, &b[i].topRight);
}
}
// check if Inside Row2 and Row3
else if ((b[i].bottomLeft.y < ROW2.topLeft.y) && (b[i].topLeft.y > ROW2.topLeft.y))
{
//System.out.println("ROW2 and ROW3 Executed");
distanceFromBottom = ROW2.topLeft.y - b[i].bottomLeft.y;
distanceFromTop = b[i].height - distanceFromBottom;
if (distanceFromBottom > distanceFromTop)
{
move(0, -distanceFromTop, &b[i].bottomLeft);
move(0, -distanceFromTop, &b[i].topLeft);
move(0, -distanceFromTop, &b[i].bottomRight);
move(0, -distanceFromTop, &b[i].topRight);
}
else
{
move(0, distanceFromBottom, &b[i].bottomLeft);
move(0, distanceFromBottom, &b[i].topLeft);
move(0, distanceFromBottom, &b[i].bottomRight);
move(0, distanceFromBottom, &b[i].topRight);
}
}
else if ((b[i].bottomLeft.y >= ROW1.bottomLeft.y) && (b[i].topLeft.y <= ROW1.topLeft.y))
{
// do nothing in ROW1
}
else if ((b[i].bottomLeft.y >= ROW2.bottomLeft.y) && (b[i].topLeft.y <= ROW2.topLeft.y))
{
// do nothing in ROW2
}
else // ((b[i].bottomLeft.y >=
// ROW2.bottomLeft.y)&&(b[i].topLeft.y <=
// ROW2.topLeft.y))
{
// do nothing in ROW3
}
}
}
//use when floating point operations create run time problems
void linkfloat()
{
float a = 0.0, *x;
x = &a;
a = *x;
}
int findOverlaps(struct Block *b, int n)
{
int k, j, p = 0;
//struct Overlap OL[200];
//short flag = FALSE;
double overlapLength = 0.0;
for (int i = 0; i < n - 1; i++)
{
k = i;
for (j = i + 1; j < n; j++)
{
if (k == j)
continue;
overlapLength = 0.0;
//for ROW1
if (b[k].bottomLeft.y >= 0 && b[k].topLeft.y <= 100 && b[j].bottomLeft.y >= 0 && b[j].topLeft.y <= 100)
{
//if ((b[k].bottomRight.x > b[j].bottomLeft.x) && (b[k].bottomRight.y < b[j].topLeft.y || b[k].topRight.y > b[j].bottomLeft.y))
if ((b[k].bottomLeft.y == b[j].bottomLeft.y) && (b[k].bottomRight.x > b[j].bottomLeft.x))
{
overlapLength = b[k].bottomRight.x - b[j].bottomLeft.x;
OL[p].overlapLength = overlapLength;
OL[p].b1 = &b[k];
OL[p].b2 = &b[j];
p++;
}
else if (((b[k].topLeft.y > b[j].bottomLeft.y) && (b[k].topRight.x > b[j].bottomLeft.x)) || ((b[k].bottomLeft.y < b[j].topLeft.y) && (b[k].bottomRight.x > b[j].topLeft.x)))
{
//flag = TRUE;
overlapLength = b[k].bottomRight.x - b[j].bottomLeft.x;
OL[p].overlapLength = overlapLength;
OL[p].b1 = &b[k];
OL[p].b2 = &b[j];
p++;
}
else
{
}
overlapLength = 0.0;
}
//For Row2
else if (b[k].bottomLeft.y >= 100 && b[k].topLeft.y <= 200 && b[j].bottomLeft.y >= 100 && b[j].topLeft.y <= 200)
{
if ((b[k].bottomLeft.y == b[j].bottomLeft.y) && (b[k].bottomRight.x > b[j].bottomLeft.x))
{
overlapLength = b[k].bottomRight.x - b[j].bottomLeft.x;
OL[p].overlapLength = overlapLength;
OL[p].b1 = &b[k];
OL[p].b2 = &b[j];
p++;
}
else if (((b[k].topLeft.y > b[j].bottomLeft.y) && (b[k].topRight.x > b[j].bottomLeft.x)) || ((b[k].bottomLeft.y < b[j].topLeft.y) && (b[k].bottomRight.x > b[j].topLeft.x)))
{
//flag = TRUE;
overlapLength = b[k].bottomRight.x - b[j].bottomLeft.x;
OL[p].overlapLength = overlapLength;
OL[p].b1 = &b[k];
OL[p].b2 = &b[j];
p++;
}
else
{
}
overlapLength = 0.0;
}
//For Row3
else if (b[k].bottomLeft.y >= 200 && b[k].topLeft.y <= 300 && b[j].bottomLeft.y >= 200 && b[j].topLeft.y <= 300)
{
if ((b[k].bottomLeft.y == b[j].bottomLeft.y) && (b[k].bottomRight.x > b[j].bottomLeft.x))
{
overlapLength = b[k].bottomRight.x - b[j].bottomLeft.x;
OL[p].overlapLength = overlapLength;
OL[p].b1 = &b[k];
OL[p].b2 = &b[j];
p++;
}
else if (((b[k].topLeft.y > b[j].bottomLeft.y) && (b[k].topRight.x > b[j].bottomLeft.x)) || ((b[k].bottomLeft.y < b[j].topLeft.y) && (b[k].bottomRight.x > b[j].topLeft.x)))
{
//flag = TRUE;
overlapLength = b[k].bottomRight.x - b[j].bottomLeft.x;
OL[p].overlapLength = overlapLength;
OL[p].b1 = &b[k];
OL[p].b2 = &b[j];
p++;
}
else
{
}
overlapLength = 0.0;
}
else
{
//do nothing
}
}
}
return p;
}
struct Block *
getBlock(struct Block *b, int n, int id)
{
int i;
for (i = 0; i < n; i++)
{
if (id == b[i].id)
return (b + i);
}
return NULL;
}
void placeHorizontalGreedy(struct Block *b, int n)
{
struct Block *t = NULL;
int p = 0, k = 0;
p = findOverlaps(b, n);
qsort(OL, p, sizeof(struct Overlap), comparator);
while (TRUE)
{
t = getBlock(b, 12, OL[k].b2->id);
if (t != NULL)
{
//t->bottomLeft.x = t->bottomLeft.x+OL[0].overlapLength;
if ((t->bottomLeft.x + OL[0].overlapLength + t->width) <= ROW1.bottomRight.x)
{
move(OL[0].overlapLength, 0, &t->bottomLeft);
intialiseBlock(t);
}
}
p = findOverlaps(b, n);
if (p == 0)
break;
else
qsort(OL, p, sizeof(struct Overlap), comparator);
}
}
void placeHorizontalNonGreedy(struct Block *b, int n, struct Block ROW1,
struct Block ROW2, struct Block ROW3)
{
int k, j, p = 0;
double overlapLength = 0.0;
for (int i = 0; i < n - 1; i++)
{
k = i;
for (j = i + 1; j < n; j++)
{
if (k == j)
continue;
overlapLength = 0.0;
//check if in Row1
if (b[k].bottomLeft.y >= ROW1.bottomLeft.y && b[k].topLeft.y <= ROW1.topLeft.y && b[j].bottomLeft.y >= ROW1.bottomLeft.y && b[j].topLeft.y <= ROW1.topLeft.y)
{
if ((b[k].bottomRight.x > b[j].bottomLeft.x) && (b[k].bottomRight.y < b[j].topLeft.y || b[k].topRight.y > b[j].bottomLeft.y))
{
//flag = TRUE;
overlapLength = b[k].bottomRight.x - b[j].bottomLeft.x;
if (overlapLength != 0)
{
if (b[j].bottomLeft.x + overlapLength + b[j].width <=
ROW1.bottomRight.x)
{
b[j].bottomLeft.x += overlapLength;
intialiseBlock(&b[j]);
}
else
{
b[j].bottomLeft.x =
b[j].bottomLeft.x + (ROW1.bottomRight.x -
b[j].bottomRight.x);
}
}
}
}
//check if in Row2
else if (b[k].bottomLeft.y >= ROW2.bottomLeft.y && b[k].topLeft.y <= ROW2.topLeft.y && b[j].bottomLeft.y >= ROW2.bottomLeft.y && b[j].topLeft.y <= ROW2.topLeft.y)
{
if ((b[k].bottomRight.x > b[j].bottomLeft.x) && (b[k].bottomRight.y < b[j].topLeft.y || b[k].topRight.y > b[j].bottomLeft.y))
{
//flag = TRUE;
overlapLength = b[k].bottomRight.x - b[j].bottomLeft.x;
if (overlapLength != 0)
{
if (b[j].bottomLeft.x + overlapLength + b[j].width <=
400)
{
b[j].bottomLeft.x += overlapLength;
intialiseBlock(&b[j]);
}
}
}
}
//check if in Row3
else if (b[k].bottomLeft.y >= ROW3.bottomLeft.y && b[k].topLeft.y <= ROW3.topLeft.y && b[j].bottomLeft.y >= ROW3.bottomLeft.y && b[j].topLeft.y <= ROW3.topLeft.y)
{
if ((b[k].bottomRight.x > b[j].bottomLeft.x) && (b[k].bottomRight.y < b[j].topLeft.y || b[k].topRight.y > b[j].bottomLeft.y))
{
//flag = TRUE;
overlapLength = b[k].bottomRight.x - b[j].bottomLeft.x;
if (overlapLength != 0)
{
if (b[j].bottomLeft.x + overlapLength + b[j].width <=
400)
{
b[j].bottomLeft.x += overlapLength;
intialiseBlock(&b[j]);
}
}
}
}
else
{
//printf("Invalid input\n");
continue;
}
}
}
}
int main()
{
//struct Block b[16];
//struct Block b[4];
//struct Block *b;
//struct Block ROW1, ROW2, ROW3;
//FILE *fp = fopen("/home/jeetu/experimentprograms/recordcoord.txt", "w+");
//FILE *fp = fopen("/home/jeetu/experimentprograms/blockcoordinates.txt", "r");
int ncell = 0, width = 0, i = 0, totalncells = 0, id = 0;
double distanceFromBottom = 0.0;
double distanceFromTop = 0.0;
int x = 0, y = 0, temp = 0, d = 0;
printf("Enter number of cells: ");
scanf("%d", &ncell);
width = computeCellWidth(ncell);
//width = 40;
b = (struct Block *)malloc(3 * ncell * sizeof(struct Block));
//b = (struct Block *)malloc(12 * sizeof(struct Block));
srand(time(0));
ROW1.bottomLeft.x = 0;
ROW1.bottomLeft.y = 0;
ROW1.height = 100;
ROW1.width = 400;
ROW1.whitespace = 40000;
intialiseBlock(&ROW1);
ROW2.bottomLeft.x = 0;
ROW2.bottomLeft.y = 100;
ROW2.height = 100;
ROW2.width = 400;
ROW2.whitespace = 40000;
intialiseBlock(&ROW2);
ROW3.bottomLeft.x = 0;
ROW3.bottomLeft.y = 200;
ROW3.height = 100;
ROW3.width = 400;
ROW3.whitespace = 40000;
intialiseBlock(&ROW3);
for (i = 0; i < ncell; i++)
{
if (i == 0)
{
// x = rand_interval(0, width / 2);
x = rand() % (width / 2);
//y = rand_interval(0, 70);
y = rand() % 70;
temp = x;
}
//m = rand();
else
{
x = rand_interval((temp + (int)width / 2), temp + width);
//x = (rand()%((temp+width)-(temp+width/2)))+(temp+width/2);
//y = rand_interval(0, 70);
y = rand() % 70;
temp = x;
}
b[totalncells].bottomLeft.x = x;
b[totalncells].bottomLeft.y = y;
b[totalncells].width = width;
b[totalncells].height = 50;
b[totalncells].id = id;
//initial white space calculation for ROW1
if (b[totalncells].bottomLeft.y >= ROW1.bottomLeft.y && b[totalncells].topLeft.y <= ROW1.topLeft.y)
{
ROW1.whitespace =
ROW1.whitespace - b[totalncells].width * b[totalncells].height;
}
id++;
intialiseBlock(&b[totalncells]);
totalncells++;
//printf("x=%d\ty=%d\n", x, y);
}
temp = 0;
//Generate random coordinates For Row2
for (i = 0; i < ncell; i++)
{
//int x = 0, y = 0, m = 0, d = 0;
if (i == 0)
{
/*x = rand_interval(0, width / 2);
y = rand_interval(100, 170); */
//x = rand() % (width / 2);
x = rand() % (width / 2);
y = (rand() % (170 - 100)) + 100;
temp = x;
}
//m = rand();
else
{
/*x = rand_interval((temp + (int)width / 2), 400 - (int)width);
y = rand_interval(100, 170); */
//x = (rand()%((temp+width)-(temp+width/2)))+(temp+width/2);
x = rand_interval((temp + (int)width / 2), temp + width);
y = (rand() % (170 - 100)) + 100;
temp = x;
}
b[totalncells].bottomLeft.x = x;
b[totalncells].bottomLeft.y = y;
b[totalncells].width = width;
b[totalncells].height = 50;
b[totalncells].id = id;
//initial white space calculation for ROW2
if (b[totalncells].bottomLeft.y >= ROW2.bottomLeft.y && b[totalncells].topLeft.y <= ROW2.topLeft.y)
{
ROW2.whitespace =
ROW2.whitespace - b[totalncells].width * b[totalncells].height;
}
id++;
intialiseBlock(&b[totalncells]);
totalncells++;
//printf("x=%d\ty=%d\n", x, y);
}
temp = 0;
//Generate random coordinates For Row3
for (i = 0; i < ncell; i++)
{
//int x = rand_interval(0, 100 - (int)width);
//int y = rand_interval(20, 25); //cell height is 5 so max will reach 30
//int x = 0, y = 0, m = 0, d = 0;
if (i == 0)
{
/*x = rand_interval(0, width / 2);
y = rand_interval(200, 250); */
x = rand() % (width / 2);
y = (rand() % (250 - 200)) + 200;
temp = x;
}
//m = rand();
else
{
/*x = rand_interval((temp + (int)width / 2), 400 - (int)width);
y = rand_interval(200, 250); */
x = rand_interval((temp + (int)width / 2), temp + width);
//x = (rand()%((temp+width)-(temp+width/2)))+(temp+width/2);
y = (rand() % (250 - 200)) + 200;
temp = x;
}
b[totalncells].bottomLeft.x = x;
b[totalncells].bottomLeft.y = y;
b[totalncells].width = width;
b[totalncells].height = 50;
b[totalncells].id = id;
//initial white space calculation for ROW3
if (b[totalncells].bottomLeft.y >= ROW3.bottomLeft.y && b[totalncells].topLeft.y <= ROW3.topLeft.y)
{
ROW3.whitespace =
ROW3.whitespace - b[totalncells].width * b[totalncells].height;
}
id++;
intialiseBlock(&b[totalncells]);
totalncells++;
//printf("x=%d\ty=%d\n", x, y);
}
for (i = 0; i < totalncells; i++)
{
//printf("%f %f %f %f\n", b[i].bottomLeft.x, b[i].bottomLeft.y, b[i].width, b[i].height);
printf("%d\t%d\t%d\t%d\n", (int)b[i].bottomLeft.x,
(int)b[i].bottomLeft.y, (int)b[i].width, (int)b[i].height);
}
placeVertical(b, totalncells, ROW1, ROW2, ROW3);
//placeVerticalWithDensityConstraints(b,totalncells,ROW1,ROW2,ROW3);
printf("\nAfter vertical Updation\n");
for (i = 0; i < totalncells; i++)
{
printf("%d\t%d\t%d\t%d\n", (int)b[i].bottomLeft.x,
(int)b[i].bottomLeft.y, (int)b[i].width, (int)b[i].height);
}
/*placeHorizontalNonGreedy(b, 12, ROW1, ROW2, ROW3);
printf("\nAfter Horizontal Updation(Non greedy)\n");
for (i = 0; i < 12; i++)
{
printf("%d\t%d\t%d\t%d\n", (int)b[i].bottomLeft.x, (int)b[i].bottomLeft.y, (int)b[i].width, (int)b[i].height);
} */
placeHorizontalGreedy(b, totalncells);
printf("\nAfter Horizontal Updation(Greedy)\n");
for (i = 0; i < totalncells; i++)
{
printf("%d\t%d\t%d\t%d\n", (int)b[i].bottomLeft.x,
(int)b[i].bottomLeft.y, (int)b[i].width, (int)b[i].height);
}
return 0;
}
Explanation about the code
in this program I have defined three structures
one to represent a Point,one to represent rectangular block with 4 coordinates,one to represent overlap between pair of cells.
There is function(computeCellWidth) to dynamically compute width of cells according to a formula.
There are three rows ROW1,ROW2,ROW3 in which cells will reside.
This program will first ask for number of cells and this is numbers of cells per row.At first x,y coordinates for each cell is randomly generated.Then placeVertical() function will move cells to different rows which are spanning across two rows.Then there is a placeHorizontalGreedy() function which will remove overlap among cells by first finding overlaplength and then sorting them and removing the overlap between cells which has least overlap and this process will continue till either there will be no overlaps or overlap removal is not possible anymore.Here I would like to bring to the notice of everybody that cells will be moved towards the right of the grid to remove overlaps.In placeHorizontalGreedy() function I have used a variable p which will indicate number of overlaps after each iteration.for small number of cells testing for p to be zero works fine but not as I increase the numbers.Plus how can I make sure that overlaps exists but overlap removal is not possible anymore,so that I can stop the loop. Please help me out.

function in queen's Attack II returns wrong answer

I'm solving Queen's Attack II problem in Hackerrank. My idea is traverse through every obstacles to find the closest obstacle in each directions (left, right, up down and 4 diagons. My function only passed half of testcases. Then I search on the Internet and I found a similar solution but different in code structure, obsiously it passed all testcases.
My function:
int queensAttack(int n, int k, int r_q, int c_q, int obstacles_rows, int obstacles_columns, int **obstacles) {
long a[8];
a[0]=c_q-1;
a[1]=n-c_q;
a[2]=r_q-1;
a[3]=n-r_q;
a[4]=min(a[3], a[1]);
a[5]=min(a[2], a[1]);
a[6]=min(a[2], a[0]);
a[7]=min(a[3], a[0]);
int x=obstacles[i][1], y=obstacles[i][0];
for (int i=0; i<obstacles_rows; i++) {
if(y==r_q && x<c_q) a[0]=min(a[0], c_q-x-1);
else if(y==r_q && x>c_q) a[1]=min(a[1], x-c_q-1);
else if(x==c_q && y<r_q) a[2]=min(a[2], r_q-y-1);
else if(x==c_q && y>r_q) a[3]=min(a[3], y-r_q-1);
else if(y-r_q==x-c_q) a[4]=min(a[4], y-r_q-1);
else if(r_q-y==x-c_q) a[5]=min(a[5], r_q-y-1);
else if(r_q-y==c_q-x) a[6]=min(a[6], r_q-y-1);
else if(y-r_q==c_q-x) a[7]=min(a[7], y-r_q-1);
}
int result=0;
for (int i=0; i<8; i++) {
result+=a[i];
}
return result;
}
Their fucntion:
int queensAttack(int n, int k, int r_q, int c_q, int obstacles_rows, int obstacles_columns, int **obstacles) {
a[0] = c_q -1 ;
a[1] = n -r_q;
a[2] = n -c_q;
a[3] = r_q -1;
a[4] = min(a[0],a[1]);
a[5] = min(a[1],a[2]);
a[6] = min(a[2],a[3]);
a[7] = min(a[3],a[0]);
int x,y;
while (--k >= 0)
{
y = obstacles[k][0] - r_q;
x = obstacles[k][1] - c_q;
if (x == 0)
{
if (y > 0)
{
a[1] = min(a[1], y - 1);
}
else
{
a[3] = min(a[3], -(y + 1));
}
}
else if (y == 0)
{
if (x > 0)
{
a[2] = min(a[2], x - 1);
}
else
{
a[0] = min(a[0], -(x + 1));
}
}
else
{
float m = (float)y / x;
if (m == 1.0)
{
if (x > 0)
{
a[5] = min(a[5], x - 1);
}
else
{
a[7] = min(a[7], -y - 1);
}
}
else if (m == -1.0)
{
if (x > 0)
{
a[6] = min(a[6], x - 1);
}
else
{
a[4] = min(a[4], y - 1);
}
}
}
}
int result=0;
for (int i=0; i<8; i++) {
result+=a[i];
}
return result;
}
Where am I wrong?
Any help would be appreciated.
Some issues:
1) This line should not occur before the loop, but as a first statement inside the loop body:
int x=obstacles[i][1], y=obstacles[i][0];
2) The second half of the loop has conditions which cause a problem. For instance when y-r_q==x-c_q is true, then also r_q-y==c_q-x is true, and vice versa. So this expression tells you whether an obstacle is on a queen's diagonal, but not on which side of the queen. As a consequence, the argument you pass to the min function could be negative, and this should never happen. You need an additional condition, much like you have in the first half of that if-chain. So change the second half to this:
else if(y-r_q==x-c_q && y>r_q) a[4]=min(a[4], y-r_q-1);
else if(r_q-y==x-c_q && y<r_q) a[5]=min(a[5], r_q-y-1);
else if(r_q-y==c_q-x && y<r_q) a[6]=min(a[6], r_q-y-1);
else if(y-r_q==c_q-x && y>r_q) a[7]=min(a[7], y-r_q-1);
It took me some time to figure out this one. My first code couldn't get through the time execution time limits. So, some math comes handy. First I calculate the total number of moves as there are no obstacles, just using math. If there are no obstacles that is the answer.
If there are obstacles, loop through them, using boolean to close each direction that is found and using math calculate the non moves. But I still had error, because the obstacles had to be sorted from the queen to the farthest obstacles position.
Here is my code in Java.
int vertHor = n - 1;
int d1 = Math.min(n - c_q, n - r_q) + Math.min(r_q - 1, c_q - 1);
int d2 = Math.min(c_q - 1, n - r_q) + Math.min(n - c_q, r_q - 1);
int total = vertHor * 2 + d1 + d2;
if (k == 0) {
return total;
}
boolean u = false;
boolean d = false;
boolean l = false;
boolean r = false;
boolean ul = false;
boolean ur = false;
boolean dl = false;
boolean dr = false;
Arrays.sort(obstacles, (int[] x, int[] y) -> Math.min(Math.abs(x[0] - r_q), Math.abs(x[1] - c_q))
- Math.min(Math.abs(y[0] - r_q), Math.abs(y[1] - c_q)));
for (int[] obs : obstacles) {
int oRow = obs[0];
int oCol = obs[1];
if (oRow == r_q) {
if (oCol < c_q && !l) {
total -= oCol;
l = true;
} else if (!r) {
total -= n - oCol + 1;
r = true;
}
} else if (oCol == c_q) {
if (oRow < r_q && !d) {
total -= oRow;
d = true;
} else if (!u) {
total -= n - oRow + 1;
u = true;
}
} else if (Math.abs(c_q - oCol) == Math.abs(r_q - oRow)) {//oCol != c_q && oRow != r_q) {
if (oCol < c_q && oRow > r_q && !ul) {
total -= Math.min(n - oRow + 1, oCol);
ul = true;
} else if (oCol < c_q && oRow < r_q && !dl) {
total -= Math.min(oRow, oCol);
dl = true;
} else if (oCol > c_q && oRow < r_q && !dr) {
total -= Math.min(oRow, n - oCol + 1);
dr = true;
} else if (oCol > c_q && oRow > r_q && !ur) {
total -= Math.min(n - oRow, n - oCol) + 1;
ur = true;
}
}
if (u && d && l && r && ul && ur && dl && dr) {
break;
}
}
return total;

C language bresenhamLine Algorithm

I am confused about what could be the problem in this algorithm:
With these parameters:
ine temp = {(Vector){10, 150}, (Vector){120, 40}, polygon->color};
drawaLine(monitor, &temp);
However, when I change the coordinates to 120, 20 it fails:
Line temp = {(Vector){10, 150}, (Vector){120, 20}, polygon->color};
drawaLine(monitor, &temp);
My code is:
int changed = 0;
int x = line->start.x;
int y = line->start.y;
int dx = (line->end.x - line->start.x);
int dy = (line->end.y - line->start.y);
dx = (dx < 0) ? -dx : dx;
dy = (dy < 0) ? -dy : dy;
int signx;
int signy;
if ((line->end.x - line->start.x) == 0) {
signx = 0;
} else {
int tempsignx = (line->end.x - line->start.x);
signx = (tempsignx < 0) ? -1 : 1;
}
if ((line->end.y - line->start.y) == 0) {
signy = 0;
} else {
int tempsingy = (line->end.y - line->start.y);
signy = (tempsingy < 0) ? -1 : 1;
}
//swap values
if (dy > dx) {
int temp = dy;
dy = dx;
dx = temp;
changed = 1; //changed = true
}
int e = 2 * dy - dx;
for (int i = 1; i <= dx; i++) {
Vector temp = {x, y};
drawPixel(monitor, temp, line->color);
while (e >= 0) {
if (changed) {
x = x + signx;
} else {
y = y + signy;
e = e - 2 * dx;
}
}
if (changed) {
y += signy;
} else {
x += signx;
e = e + 2 * dy;
}
Vector x2y2 = {line->end.x, line->end.y};
drawPixel(monitor, x2y2, line->color);
}
}

Rotate a 2D NxN matrix in Concentric Circles

Given a 2D NxN matrix, visualize it as concentric circles. You have to find the rotated matrix where each element in the circle is rotated by 1 position layer by layer in an alternate clockwise and anticlockwise direction. All rotations should be in-place.
2 3 4 5
1 6 7 8
4 2 1 9
5 3 2 4
should get transformed to
1 2 3 4
4 7 1 5
5 6 2 8
3 2 4 9
I thought about the solution
1> For clockwise circle rotation, read elements in the order
i -> 0 to n-1 and j = 0
j -> 0 to n-1 and i = n-1
i -> n-1 to 0 and j = n-1
j -> n-1 to 0 and i = 0
2> For anti-clockwise circle rotation, read elements in the order
j -> 0 to n-1 and i = 0
i -> 0 to n-1 and j = n-1
j -> n-1 to 0 and i = n-1
i -> n-1 to 0 and j = 0
Code
for(int cnt = 0; cnt < n/2; cnt++)
{
if(cnt%2 == 0) // Clockwise
{
i = cnt; j = cnt;
// while loops for each case
}
else // anti-clockwise
{
i = cnt; j = cnt;
// while loops for each case
}
}
Is there any better approach to solve this problem in O(n2) or better ?
As your array is of size N*N and the desired computation demands each element to be visited atleast once, there cannot be a solution better than O(n^2) which uses 2 dimensional arrays.
I think that your solution will be fine if the operation has to be done single time on the same array.
If you have to do this operation multiple times on the same input array, better create circles from the input array. The data structure of circle should be a CLL (circular linked list). So doing the operation multiple times will be piece of cake as you have to change the root element of the CLL storing the circle info depending on the direction.
I think that this can be solved easily in-place in O(n) time.
(NOTE: O(N) where N is the total number of matrix elements)
The following solution doesn't use four consecutive loops, but uses a small table of [X, Y] deltas that describe the difference between the current coordinate and the next one, when the next coordinate is invalid (i.e outside of the current window) the index to the table itself is advanced and the look-up repeated. The algorithm starts with full window and decreases it by one element from each side every time the nested loop finishes. The whole process repeats until the window defined as [minX, minY, maxX, maxY] is valid (i.e. contains at least 2x2 elements).
This solution doesn't implement swapping cw and ccw, but this is the easiest part to add.
function pad(s, n) {
while(s.length < n)
s = " " + s;
return s;
}
// Create a matrix of [WxH] size.
function Matrix(w, h, data) {
if (Array.isArray(data)) {
if (data.length !== w * h)
throw new Error("Data.length has to match the size " + (w * h) + ".");
}
else {
var n = typeof data === "number" ? data : 0.0;
data = [];
for (var i = 0; i < w*h; i++) data.push(n);
}
this.w = w;
this.h = h;
this.data = data;
}
// Get value at [x, y]
Matrix.prototype.get = function(x, y) {
if (x < 0 || x >= this.w || y < 0 || y >= this.h)
throw new Error("Index [" + x + ", " + y + "] out of bounds");
return this.data[y * this.w + x];
}
// Set value at [x, y] and return the previous value.
Matrix.prototype.set = function(x, y, value) {
if (x < 0 || x >= this.w || y < 0 || y >= this.h)
throw new Error("Index [" + x + ", " + y + "] out of bounds");
var i = y * this.w + x;
var prev = this.data[i];
this.data[i] = value;
return prev;
}
// Log the matrix data.
Matrix.prototype.dump = function() {
var s = "["
var i = 0;
for (var y = 0; y < this.h; y++) {
for (var x = 0; x < this.w; x++, i++) {
s += pad("" + this.data[i], 2);
if (x !== this.w - 1)
s += ","
}
if (y !== this.h - 1)
s += ",\n ";
}
s += "]";
console.log(s);
}
// Shift, `dir` can be "cw" or "ccw".
Matrix.prototype.shift = function(dir) {
var instructions = {
cw : [1, 0, 0, 1,-1, 0, 0,-1],
ccw: [0, 1, 1, 0, 0,-1,-1, 0]
};
var inst = instructions[dir];
// A window, shrink by one from each side after the nested loop is done.
var minX = 0;
var minY = 0;
var maxX = this.w - 1;
var maxY = this.h - 1;
while (minX < maxX && minY < maxY) {
// Always start at the top-left corner and iterate.
var x0 = minX;
var y0 = minY;
var v0 = this.get(x0, y0);
var n = 0;
for (;;) {
var x1 = x0 + inst[n + 0];
var y1 = y0 + inst[n + 1];
if (x1 < minX || x1 > maxX || y1 < minY || y1 > maxY) {
n += 2;
x1 = x0 + inst[n + 0];
y1 = y0 + inst[n + 1];
}
v0 = this.set(x1, y1, v0);
// Last one.
if (x1 === minX && y1 === minY)
break;
x0 = x1;
y0 = y1;
}
minX++;
minY++;
maxX--;
maxY--;
}
}
var a = new Matrix(3, 3, [
1,2,3,
4,5,6,
7,8,9,
]);
a.dump();
a.shift("cw");
a.dump();
var b = new Matrix(4, 4, [
1 ,2 ,3 ,4 ,
5 ,6 ,7 ,8 ,
9 ,10,11,12,
13,14,15,16
]);
b.dump();
b.shift("ccw");
b.dump();
I faced this problem recently in a job interview and I failed to solve it in under one hour.
Then I got home and produced the code below in java. It is recursive and I believe it has O(n^2) complexity. You can also see the code here: https://github.com/lotif/rotateMatrix
You will have first to input the dimension of the (square) matrix and then the numbers of the matrix itself separated by spaces, line by line. For instance:
3
1 2 3
4 5 6
7 8 9
It will return you the matrix rotated clockwise in concentric circles.
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class RotateMatrix {
public static void main(String args[] ) throws Exception {
Scanner s = new Scanner(System.in);
int d = s.nextInt();
int[][] matrix = new int[d][d];
for(int i = 0; i < d; i++) {
for(int j = 0; j < d; j++) {
matrix[i][j] = Integer.parseInt(s.next());
}
}
matrix = rotate(matrix);
for(int i = 0; i < matrix.length; i++) {
for(int j = 0; j < matrix.length; j++) {
System.out.print(matrix[i][j] + " ");
}
System.out.println();
}
s.close();
}
public static int[][] rotate(int[][] matrix) {
if(matrix == null || matrix.length == 0 || matrix.length == 1) {
return matrix;
}
List<Integer> outerCircle = getOuterCircle(matrix);
matrix = removeOuterCircle(matrix);
//rotating outer circle
outerCircle.add(0, outerCircle.remove(outerCircle.size() - 1));
matrix = rotate(matrix);
matrix = addOuterCircle(outerCircle, matrix);
return matrix;
}
private static int[][] addOuterCircle(List<Integer> outerCircle, int[][] matrix) {
int d = matrix.length + 2;
int[][] newMatrix = new int[d][d];
//Adding the outer circle to the matrix
for(int j = 0; j < d; j++) {
newMatrix[0][j] = outerCircle.remove(0);
}
for(int i = 1; i < d; i++) {
newMatrix[i][d-1] = outerCircle.remove(0);
}
for(int j = d-2; j >= 0; j--) {
newMatrix[d-1][j] = outerCircle.remove(0);
}
for(int i = d-2; i >= 1; i--) {
newMatrix[i][0] = outerCircle.remove(0);
}
//Adding the inner matrix
for(int i = 0; i < matrix.length; i++) {
for(int j = 0; j < matrix[i].length; j++) {
newMatrix[i + 1][j + 1] = matrix[i][j];
}
}
return newMatrix;
}
private static List<Integer> getOuterCircle(int[][] matrix) {
int d = matrix.length;
List<Integer> outerCircle = new ArrayList<Integer>();
for(int j = 0; j < d; j++) {
outerCircle.add(matrix[0][j]);
}
for(int i = 1; i < d; i++) {
outerCircle.add(matrix[i][d-1]);
}
for(int j = d-2; j >= 0; j--) {
outerCircle.add(matrix[d-1][j]);
}
for(int i = d-2; i >= 1; i--) {
outerCircle.add(matrix[i][0]);
}
return outerCircle;
}
private static int[][] removeOuterCircle(int[][] matrix) {
int d = matrix.length;
int[][] newMatrix = new int[d-2][d-2];
for(int i = 1; i < d-1; i++) {
for(int j = 1; j < d-1; j++) {
newMatrix[i-1][j-1] = matrix[i][j];
}
}
return newMatrix;
}
}
public class ShiftArray {
static void shiftArray(int[][]a, int index, int n) {
if ((n%2 == 0) && (index >= n/2))
return;
if ((n%2 != 0) && (index > n/2))
return;
int tempRowTopLast = a[index][n-1-index];
int tempColRightLast = a[n-1-index][n-1-index];
int tempRowBottomLast = a[n-1-index][index];
int tempColLeftLast = a[index][index];
int temp, temp2;
temp = tempColLeftLast;
for (int k = index + 1; k < n-index; k++) {
temp2 = a[index][k];
a[index][k] = temp;
temp = temp2;
}
temp = tempRowTopLast;
for (int k = index + 1; k < n-index; k++) {
temp2 = a[k][n-1-index];
a[k][n-1-index] = temp;
temp = temp2;
}
temp = tempColRightLast;
for (int k = n-2-index; k >=index; k--) {
temp2 = a[n-1-index][k];
a[n-1-index][k] = temp;
temp = temp2;
}
temp = tempRowBottomLast;
for (int k = n-2-index; k >=index; k--) {
temp2 = a[k][index];
a[k][index] = temp;
temp = temp2;
}
shiftArray(a, index+1, n);
}
public static void main(String[] args) {
int a[][] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
shiftArray(a, 0, 3);
System.out.println("Rotated array...");
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
System.out.print(a[i][j] + ",");
}
System.out.println();
}
}
}
import java.util.Scanner;
public class RotateMatrix
{
static int rows = 0;
static int cols = 0;
public static void main(String[] args)
{
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
rows = scan.nextInt();
cols = scan.nextInt();
int rots = scan.nextInt();
int[][] matrix = new int[rows][cols];
for (int i = 0; i < rows; i++)
{
for (int j = 0; j < cols; j++)
{
matrix[i][j] = scan.nextInt();
}
}
for (int i = 0; i < rots; i++)
rotate(matrix, 0, rows - 1, 0, cols - 1);
for (int i = 0; i < rows; i++)
{
for (int j = 0; j < cols; j++)
{
System.out.print(matrix[i][j] + " ");
}
System.out.println();
}
scan.close();
}
public static int[][] rotate(int[][] arr, int rowStart, int rowEnd, int colStart, int colEnd)
{
if (rowStart == rowEnd && colStart == colEnd)
{
return arr;
}
if (rowStart > rowEnd || colStart > colEnd)
{
return arr;
}
int temp = arr[rowStart][colStart];
for (int j = colStart; j < colEnd; j++)
{
arr[colStart][j] = arr[colStart][j + 1];
}
for (int i = rowStart; i < rowEnd; i++)
{
arr[i][colEnd] = arr[i + 1][colEnd];
}
for (int i = colEnd; i > colStart; i--)
{
arr[rowEnd][i] = arr[rowEnd][i - 1];
}
for (int i = rowEnd; i > rowStart; i--)
{
arr[i][colStart] = arr[i - 1][colStart];
}
if (rows == 1)
{
arr[colEnd][rowStart] = temp;
}
else
arr[rowStart + 1][colStart] = temp;
System.out.println("-----------------------------------------\n");
for (int i = 0; i < rows; i++)
{
for (int j = 0; j < cols; j++)
{
System.out.print(arr[i][j] + " ");
}
System.out.println();
}
System.out.println("-----------------------------------------\n");
rotate(arr, rowStart + 1, rowEnd - 1, colStart + 1, colEnd - 1);
return arr;
}
}
Clockwise rotation by 90 degrees. O(n^2) time and O(1) memory in Python3:
# #param A : list of list of integers
# #return the same list modified
def rotate(A):
for row in range(len(A) // 2):
for col in range(row, len(A)-1 - row): # First col already takes care of last.
r = row
c = col
tmp1 = A[r][c]
while True:
next_r = c
next_c = len(A) - 1 - r
tmp2 = A[next_r][next_c]
A[next_r][next_c] = tmp1
if next_r == row and next_c == col:
break
tmp1 = tmp2
r = next_r
c = next_c
return A

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