190 - Circle Through Three Points

[Lord Nemrod]

I think you've done it too complicatedly. Try to solve the eqations for thee points - you will need only basic mathemathic operations +-* and/. That'd be more precise.

[Larry]

Also, hypot returns an "int" in ASCII C unless you explicitly declare the function header.

[emotional blind]

what is wrong with my solution
it gets wrong answer
my code is very simple
so anybody can look at a glance
have any idea why wa
or have any input for which my solution produce wrong
here is my code.
thanks

Code: Select all

#include <iostream>
#include <stdio.h>
#include <math.h>

using namespace std;

int main (void)
{
	float x1, x2, x3, y1, y2, y3;
	float c, d, e, h, k, r;
	while(scanf("%f%f%f%f%f%f",&x1,&y1,&x2,&y2,&x3,&y3)==6){
		c=((x2*x2+y2*y2-x3*x3-y3*y3)*(y1-y2) - (x1*x1 + y1*y1 - x2*x2 - y2*y2)*(y2-y3))/((x1-x2)*(y2-y3)-(x2-x3)*(y1-y2));
		d=(x1*x1+y1*y1-x3*x3-y3*y3-c*(x3-x1))/(y3-y1);
		e=-1*(x1*x1+y1*y1+c*x1+d*y1);
		h=c/(-2.0);
		k=d/(-2.0);
		r=sqrt(h*h+k*k-e);
		printf("(x %c %.3f)^2 + (y %c %.3f)^2 = %.3f^2\n",(h<0)?'+':'-', (h<0)?(-h):h, (k<0)?'+':'-',(k<0)?(-k):k,r);
		printf("x^2 + y^2 %c %.3fx %c %.3fy %c %.3f = 0\n\n",(c<0)?'-':'+',(c<0)?(-c):c, (d<0)?'-':'+',(d<0)?(-d):d, (e<=0)?'-':'+', (e<0)?(-e):e);
	}
	return 0;
}

[Ghust_omega]

Hi emotional blind !! try with this input
in:

Code: Select all

5 5 4 7 16 5

out:

Code: Select all

(x - 10.500)^2 + (y - 9.000)^2 = 6.801^2
x^2 + y^2 - 21.000x - 18.000y + 145.000 = 0

Hope it helps
Keep posting !!

[emotional blind]

dear Ghust_omega
thanks a lot ...................................
I got ac
thanks again

[emotional blind]

dear Ghust_omega
thanks a lot ...................................
I got ac
thanks again

[Oyhama Hora]

what is wrong with my solution
it gets wrong answer
my code is very simple
so anybody can look at a glance
have any idea why wa
or have any input for which my solution produce wrong
here is my code.
thanks


#include<stdio.h>
#include<math.h>

int z;
float x[3],y[3];
double ak,bk,ck;

void output(void)
{
double r;
r=sqrt(ak*ak+bk*bk-ck);

printf("(x ");
if (-1*ak>0) printf("- ");
else printf("+ ");
printf("%.3f)^2 + (y ",fabs(ak));
if (-1*bk>0) printf("- ");
else printf("+ ");
printf("%.3f)^2 = %.3f",fabs(bk),r);
if (r*r!=r) printf("^2\n");
else printf("\n");

printf("x^2 + y^2 ");
if (2*ak<0) printf("- ");
else printf("+ ");
printf("%.3fx ",fabs(2*ak));
if (2*bk<0) printf("- ");
else printf("+ ");
printf("%.3fy ",fabs(2*bk));
if (ck<0) printf("- ");
else printf("+ ");
printf("%.3f = 0",fabs(ck));
}

void hitung()
{
int i;
double a[6],b[6],p[2],bil[6];
for (i=0;i<3;i++) {
bil=x*x+y*y;
a=x*2;
b=y*2;
}
for (i=3;i<5;i++) {
bil=bil[i-3]-bil[i-2];
a[i]=a[i-3]-a[i-2];
b[i]=b[i-3]-b[i-2];
}
p[0]=a[3];
p[1]=a[4];
for (i=3;i<5;i++) {
bil[i]=bil[i]*fabs(p[4-i]);
b[i]=b[i]*fabs(p[4-i]);
a[i]=a[i]*fabs(p[4-i]);
}
if((p[0]<0&&p[1]<0)||(p[0]>0&&p[1]>0)) {
bil[5]=bil[3]-bil[4];
b[5]=b[3]-b[4];
} else {
bil[5]=bil[3]+bil[4];
b[5]=b[3]+b[4];
}
if(b[5] == 0)
{
bk=0.0;
} else{
bk=-1*bil[5]/b[5];}

if(a[4]== 0.0){
ak=0.0;} else{
ak=(bil[4]*-1+bk*b[4]*-1)/a[4];
}

ck=bil[0]*-1+ak*a[0]*-1+bk*b[0]*-1;
}

main()
{

while (!feof(stdin)) {
fscanf(stdin,"%f %f %f %f %f %f",&x[0],&y[0],&x[1],&y[1],&x[2],&y[2]);
hitung();
output();
printf("\n\n");

}
return 0;
}

[Sedefcho]

Hi, Oyhama Hora,

First, suppose we have found the equation of the circle
in the following form:

Code: Select all

(X + A)^2 + (Y + B)^2 = R^2

Here we have that R>0 but A and B could have any sign
( they can be zeros as well ).

Then let's suppose that your algorithm is OK so that you
find the correct values for A and B.

Then my question is: what do you print if A==0 ?
And same question for B: if B==0 what do you print ?

Looking at your code ( I haven't compiled and run it )
I think you will print something like:

Code: Select all

(X + 0.000)^2 ... 

which is not correct. Correct output should be

Code: Select all

X^2 ... 

That means the judge wants NO brackets and NO 0.000
in the case A==0 ( B==0 ).

See also the two test cases I have posted here:
http://online-judge.uva.es/board/viewtopic.php?p=35401
I hope this makes the things clear.

Try to change this and see if you still get WA.

[Sedefcho]

I think the two test cases which follow may be helpful
to anyone who still has WA on this problem.

INPUT

Code: Select all

7 0 11 0 9 2
0 9 0 15 3 12

OUTPUT

Code: Select all

(x - 9.000)^2 + y^2 = 2.000^2
x^2 + y^2 - 18.000x + 77.000 = 0

x^2 + (y - 12.000)^2 = 3.000^2
x^2 + y^2 - 24.000y + 135.000 = 0

Good luck to everyone !

[nayeemmeeyan]

why i get wrong answer or compile error
this is my code:
#include<stdio.h>
#include<math.h>
int main()
{
double x1,x2,x3,y1,y2,y3,h,k,r,d1,e;
while(scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3)==6)
{

d1=2*((x2-x1)*(y3-y2)-(x3-x2)*(y2-y1));
h=((y2-y1)*(x3*x3+y3*y3-x2*x2-y2*y2)-(y3-y2)*(x2*x2+y2*y2-x1*x1-y1*y1))/d1;
k=((x3-x2)*(x2*x2+y2*y2-x1*x1-y1*y1)-(x2-x1)*(x3*x3+y3*y3-x2*x2-y2*y2))/d1;
r=sqrt(pow((x1-(-1.0*h)),2)+pow((y1-(-1.0*k)),2));
e=r*r-(h*h+k*k);
if( fabs(h) < 1e-8 && fabs(k)<1e-8)
{
printf("(x + 0.000)^2 + (y + 0.000)^2 = 0.000^2\n",r);
printf("x^2 + y^2 + 0.000x + 0.000y + 0.000 = 0\n",e);
}
else if( h <0 && k <0 && e < 0)
{
h=-1.0*h;
k=-1.0*k;
e=-1.0*e;
printf("(x - %.3lf)^2 + (y - %.3lf)^2 = %.3lf^2\n",h,k,r);
printf("x^2 + y^2 - %.3lfx - %.3lfy + %.3lf = 0\n",2.0*h,2.0*k,e);
}
else if(h < 0 && k< 0 && e>=0)
{
h=-1.0*h;
k=-1.0*k;
printf("(x - %.3lf)^2 + (y - %.3lf)^2 = %.3lf^2\n",h,k,r);
printf("x^2 + y^2 - %.3lfx - %.3lfy - %.3lf = 0\n",2.0*h,2.0*k,e);
}
else if( h >=0 && k <0 && e<0)
{
k=-1.0*k;
e=-1.0*e;
printf("(x + %.3lf)^2 + (y - %.3lf)^2 = %.3lf^2\n",h,k,r);
printf("x^2 + y^2 + %.3lfx - %.3lfy + %.3lf = 0\n",2.0*h,2.0*k,e);
}
else if(h>=0 && k<0 && e>=0)
{
k=-1.0*k;
printf("(x + %.3lf)^2 + (y - %.3lf)^2 = %.3lf^2\n",h,k,r);
printf("x^2 + y^2 + %.3lfx - %.3lfy - %.3lf = 0\n",2.0*h,2.0*k,e);
}
else if(h <0 && k>=0 && e< 0)
{
e=-1.0*e;
h=-1.0*h;
printf("(x - %.3lf)^2 + (y + %.3lf)^2 = %.3lf^2\n",h,k,r);
printf("x^2 + y^2 - %.3lfx + %.3lfy + %.3lf = 0\n",2.0*h,2.0*k,e);
}
else if(h<0 && k>=0 && e>=0)
{
h=-1.0*h;
printf("(x - %.3lf)^2 + (y + %.3lf)^2 = %.3lf^2\n",h,k,r);
printf("x^2 + y^2 - %.3lfx + %.3lfy - %.3lf = 0\n",2.0*h,2.0*k,e);
}
else if(h>=0 && k>=0 && e<0)
{
printf("(x + %.3lf)^2 + (y + %.3lf)^2 = %.3lf^2\n",h,k,r);
printf("x^2 + y^2 + %.3lfx + %.3lfy + %.3lf = 0\n",2.0*h,2.0*k,e);

}
else//(h>=0 && k>=0 && e>=0)
{
printf("(x + %.3lf)^2 + (y + %.3lf)^2 = %.3lf^2\n",h,k,r);
printf("x^2 + y^2 + %.3lfx + %.3lfy - %.3lf = 0\n",2.0*h,2.0*k,e);
}
printf("\n");
}
return 0;
}

[chunyi81]

I have three points to make:

1) Which problem is this code supposed to solve?
2) What is the algorithm used?
3) I believe you are solving problem 190. You can search for past threads on problem 190.

[Isaac]

well.. thank you for your answer
and i have tried your advice
but i still got WA
this is my program and may be you can tell me why i got WA

...

int main()
{
while (!feof(stdin)) {
fscanf(stdin,"%f %f %f %f %f %f",&x[0],&y[0],&x[1],&y[1],&x[2],&y[2]);
hitung();
output();
printf("\n\n");
}
return 0;
}

Dear gracia,

I know it's a little late.
but I think your program has some problem when it reads six numbers from the system.

Because the lack of practice in I/O,
I tried to use your method after many failed.

It's fine when I tested it on my PC,
but when I pressed Ctrl + C to end the program in Dev-C++

It printed something on the screen before close.

then I found another article
http://online-judge.uva.es/board/viewtopic.php?p=35401

I tried the method that 'emotional blind' san use in his program,
and I got an AC.

I think the hint that 'Sedefcho' san post in that article is also useful, but I don't know if it's really matter because i changed my program lots of times before I found the problem about input.

--
Please forgive my poor english...

[nukeu666]

where am i going wrong or gimma an input that doesn work

Code: Select all

#include<math.h>
#include<stdio.h>
int main()
{
  double con,cenx,ceny,rad,ax,ay,bx,by,cx,cy,mab,mac,p1,q1,r1,p2,q2,r2;
  int sx,sy,sc;
while(scanf("%lf %lf %lf %lf %lf %lf",&ax,&ay,&bx,&by,&cx,&cy)==6)
  {
    sx=0,sy=0,sc=0;
    mab=(by-ay)/(bx-ax);
    mac=(cy-ay)/(cx-ax);
    p1=mab;p2=mac;
    q1=1;q2=1;
    r1=(ax+bx)/2+mab*(ay+by)/2;
    r2=(ax+cx)/2+mac*(ay+cy)/2;
    cenx=(r1*p2-r2*p1)/(q1*p2-q2*p1);
    ceny=(q2*r1-q1*r2)/(q2*p1-p2*q1);
    rad=pow(pow((cenx-ax),2)+pow((ceny-ay),2),0.5);
    if(cenx<0)
      sx=1;
    if(ceny<0)
      sy=1;
    if((con=pow(cenx,2)+pow(ceny,2)-pow(rad,2))<0)
      sc=1;
    if(fabs(cenx)<.001)
      printf("x^2 ");
    else
      {
        if(sx==0)
          printf("(x - %.3lf)^2 + ",cenx);
        else
          printf("(x + %.3lf)^2 + ",-cenx);
      }
    if(fabs(ceny)<.001)
      printf("y^2 = %.3lf^2",fabs(rad));
    else
      {
        if(sy==0)
          printf("(y - %.3lf)^2 = %.3lf^2",ceny,fabs(rad));
        else
          printf("(y + %.3lf)^2 = %.3lf^2",-ceny,fabs(rad));
      }

      printf("\nx^2 + y^2 ");
    if(sx==0&&fabs(cenx)>.001)
      printf("- %.3lfx ",2*fabs(cenx));
    if(sx==1&&fabs(cenx)>.001)
      printf("+ %.3lfx ",-(2*cenx));
    if(sy==0&&fabs(ceny)>.001)
      printf("- %.3lfy ",2*fabs(ceny));
    if(sy==1&&fabs(ceny)>.001)
      printf("+ %.3lfy ",-(2*ceny));
    if(sc==1&&fabs(con)>.001)
      printf("- %.3lf ",-con);
    if(sc==0&&fabs(con)>.001)
      printf("+ %.3lf ",fabs(con));
    printf("= 0\n\n");
  }
}

input

Code: Select all

7.0 -5.0 -1.0 1.0 0.0 -6.0
1.0 7.0 8.0 6.0 7.0 -2.0
0 0 5 0 2.5 2.5

output

Code: Select all

(x - 3.000)^2 + (y + 2.000)^2 = 5.000^2
x^2 + y^2 - 6.000x + 4.000y - 12.000 = 0

(x - 3.921)^2 + (y - 2.447)^2 = 5.409^2
x^2 + y^2 - 7.842x - 4.895y - 7.895 = 0

(x - 2.500)^2 + y^2 = 2.500^2
x^2 + y^2 - 5.000x = 0

[Rocky]

i get the following output for those input from accepted code..

Code: Select all

(x - 3.000)^2 + (y + 2.000)^2 = 5.000^2
x^2 + y^2 - 6.000x + 4.000y - 12.000 = 0

(x - 3.921)^2 + (y - 2.447)^2 = 5.409^2
x^2 + y^2 - 7.842x - 4.895y - 7.895 = 0

(x - 2.500)^2 + (y + 0.000)^2 = 2.500^2
x^2 + y^2 - 5.000x - 0.000y - 0.000 = 0

may be it help

GOOD LUCK
Rocky

[xintactox]

Hi everybody!

I'm getting WA in problem 190 - Circle through three points over and over again...

Can someone please check my code, maybe u can find where it's wrong....
I need some critical test cases or something...

For the following input:

Code: Select all

7.0 -5.0 -1.0 1.0 0.0 -6.0 
1.0 7.0 8.0 6.0 7.0 -2.0 
0 0 5 0 2.5 2.5 

my program generates the following output

Code: Select all

(x - 3.000)^2 + (y + 2.000)^2 = 5.000^2 
x^2 + y^2 - 6.000x + 4.000y - 12.000 = 0 

(x - 3.921)^2 + (y - 2.447)^2 = 5.409^2 
x^2 + y^2 - 7.842x - 4.895y - 7.895 = 0 

(x - 2.500)^2 + (y + 0.000)^2 = 2.500^2 
x^2 + y^2 - 5.000x - 0.000y - 0.000 = 0 

I already read all the past posts about this problem, sorry for bothering you all

I've also read about the output with zeros, like the third case of my output... some people say it must me made like i did, other people says that it must be without the brackets or zeros.... What's correct?

My code follows

Code: Select all

#include <iostream>
#include <stdio.h>
#include <math.h>

using namespace std;

class point
{
	public:
		double x, y;
		
};

int main()
{
	point p1, p2, p3;
	double ma, mb, h, k, r, c, d, e;
	while(cin>>p1.x)
	{
		cin>>p1.y>>p2.x>>p2.y>>p3.x>>p3.y;
		ma = (p2.y - p1.y)/(p2.x - p1.x);
		mb = (p3.y - p2.y)/(p3.x - p2.x);
		h = (ma*mb*(p1.y - p3.y) + mb*(p1.x + p2.x) - ma*(p2.x + p3.x))/(2.0*(mb - ma));		
		k = (-1.0/(ma+0.000001))*(h - ((p1.x + p2.x))/2) + (p1.y + p2.y)/2.0;					
		r = sqrt((h - p1.x)*(h - p1.x)+(k - p1.y)*(k - p1.y));
								
		//cout<<"c = "<<c<<endl;
		//cout<<"d = "<<d<<endl;
				
		if(h >= 0.000001)
			printf("(x - %.3f)^2 + (y ", fabs(h));
		else
			printf("(x + %.3f)^2 + (y ", fabs((-1)*h));
		
		if(k >= 0.000001)
			printf("- %.3f)^2 = %.3f^2", fabs(k), fabs(r));
		else
			printf("+ %.3f)^2 = %.3f^2", fabs((-1)*k),fabs(r));		
		
		c = 2.0*h;
		d = 2.0*k;
		e = (h*h) + (k*k) - (r*r);
		
		
		if(c >= 0.001){
		if(c >= 0.00000001)
			printf("\nx^2 + y^2 - %.3fx ", fabs(c));
		else
			printf("\nx^2 + y^2 + %.3fx ", fabs(c));}
		else printf("\nx^2 + y^2 + %.3fx ", fabs(c));	
		
		if(d >= 0.001){
		if(d >= 0.00000001)
			printf("- %.3fy ", fabs(d));
		else
			printf("+ %.3fy ", fabs(d));}
		else if(d == 0.000) printf("- %.3fy ", fabs(d));
		else printf("+ %.3fy ", fabs(d));
		
		if(e > 0.00000001)
			printf("+ %.3f = 0\n", fabs(e));
		else
			printf("- %.3f = 0\n", fabs(e));
		
		printf("\n");		
	}
	return 0;
}

Please help me!

Thank you all and happy new year!