Problem:
Problem I: Most Distant Point from the Sea

The main land of Japan called Honshu is an island surrounded by the sea. In such an island, it
is natural to ask a question: "Where is the most distant point from the sea?" The answer to
this question for Honshu was found in 1996. The most distant point is located in former Usuda
Town, Nagano Prefecture, whose distance from the sea is 114.86 km.

In this problem, you are asked to write a program which, given a map of an island, finds the
most distant point from the sea in the island, and reports its distance from the sea. In order to
simplify the problem, we only consider maps representable by convex polygons.

Input

The input consists of multiple datasets. Each dataset represents a map of an island, which is a convex polygon. The format of a dataset is as follows.

n
x1 y1
.
.
.
xn yn

Every input item in a dataset is a non-negative integer. Two input items in a line are separated by a space.

n in the first line is the number of vertices of the polygon, satisfying 3 ≤ n ≤ 100. Subsequent
n lines are the x- and y-coordinates of the n vertices. Line segments (xi, yi) - (xi+1, yi+1) (1 ≤ i ≤ n - 1) and the line segment (xn, yn) - (x1, y1) form the border of the polygon in counterclockwise order. That is, these line segments see the inside of the polygon in the left of their directions. All coordinate values are between 0 and 10000, inclusive.

You can assume that the polygon is simple, that is, its border never crosses or touches itself. As stated above, the given polygon is always a convex one.

The last dataset is followed by a line containing a single zero.

Output

For each dataset in the input, one line containing the distance of the most distant point from
the sea should be output. An output line should not contain extra characters such as spaces.

The answer should not have an error greater than 0.00001 (10-5 ). You may output any number
of digits after the decimal point, provided that the above accuracy condition is satisfied.

Sample Input

4
0 0
10000 0
10000 10000
0 10000
3
0 0
10000 0
7000 1000
6
0 40
100 20
250 40
250 70
100 90
0 70
3
0 0
10000 10000
5000 5001
0

Output for the Sample Input

5000.000000
494.233641
34.542948
0.353553

Solution in C++ (judge verdict: Accepted)

#include <stdio.h>
#include <cmath>
#include <algorithm>
#include <cfloat>
#include <stack>
#include <queue>
#include <vector>
#include <string>
#include <iostream>
#include <set>
#include <map>
#include <time.h>
typedef long long int ll;
typedef unsigned long long int ull;
#define BIG_NUM 2000000000
#define MOD 1000000007
#define EPS 0.000000001
using namespace std;


struct Point{
	Point(double arg_x,double arg_y){
		x = arg_x;
		y = arg_y;
	}

	Point(){
		x = y = 0.0;
	}

	Point operator + (Point p){ return Point(x+p.x,y+p.y); }
	Point operator - (Point p){ return Point(x-p.x,y-p.y);}
	Point operator * (double a){ return Point(a*x,a*y); }
	Point operator / (double a){ return Point(x/a,y/a); }

	double abs(){ return sqrt(norm()); }
	double norm(){ return x*x + y*y; }

	bool operator<(const Point &p) const{
		return x != p.x? x < p.x: y < p.y;
	}

	bool operator == (const Point &p) const{
		return fabs(x-p.x) < EPS && fabs(y-p.y) < EPS;
	}

	double x,y;
};

typedef Point Vector;
typedef vector<Point> Polygon;

struct Line{
	Line(){
		p[0].x = p[0].y = p[1].x = p[1].y = 0;
	}

	Line(Point a,Point b){
		p[0] = a;
		p[1] = b;
	}
	Point p[2];
};

int N;
static const int COUNTER_CLOCKWISE = 1;
static const int CLOCKWISE = -1;
static const int ONLINE_BACK = 2;
static const int ONLINE_FRONT = -1;
static const int ON_SEGMENT = 0;
static const double PI = acos(-1);

Point rotate(Point point,double theta){
    return Point(point.x*cos(theta)-point.y*sin(theta),point.x*sin(theta)+point.y*cos(theta));
}

double norm(Vector a){
	return a.x*a.x+a.y*a.y;
}

double abs(Vector a){
	return sqrt(norm(a));
}

double cross(Vector a,Vector b){
    return a.x*b.y-a.y*b.x;
}

double dot(Vector a,Vector b){
    return a.x*b.x + a.y*b.y;
}

int ccw(Point p0,Point p1,Point p2){

	Vector a = p1 - p0;
	Vector b = p2 - p0;

	if(cross(a,b) > EPS)return COUNTER_CLOCKWISE;
	if(cross(a,b) < -EPS)return CLOCKWISE;
	if(dot(a,b) < -EPS)return ONLINE_BACK;
	if(a.norm() < b.norm())return ONLINE_FRONT;

	return ON_SEGMENT;
}

int func(double x1,double y1,double x2, double y2, double xp, double yp){
	double naiseki,norm1,norm2,gaiseki;
	norm1 = sqrt((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1));
	norm2 = sqrt((xp-x1)*(xp-x1)+(yp-y1)*(yp-y1));
	naiseki = (xp-x1)*(x2-x1)+(yp-y1)*(y2-y1);
	gaiseki = (x2-x1)*(yp-y1)-(xp-x1)*(y2-y1);
	if(gaiseki > EPS){
		return 1;
	}else if(gaiseki < -EPS){
		return -1;
	}
	if(naiseki < -EPS){
		return 2;
	}

	if(norm1 < norm2){
		return -2;
	}
	return 0;
}

bool is_Cross(Line a,Line b){

	if(func(a.p[0].x,a.p[0].y,a.p[1].x,a.p[1].y,b.p[0].x,b.p[0].y)*
			func(a.p[0].x,a.p[0].y,a.p[1].x,a.p[1].y,b.p[1].x,b.p[1].y) <= 0 &&
			func(b.p[0].x,b.p[0].y,b.p[1].x,b.p[1].y,a.p[0].x,a.p[0].y)*
			func(b.p[0].x,b.p[0].y,b.p[1].x,b.p[1].y,a.p[1].x,a.p[1].y) <= 0){
		return true;
	}
	return false;
}

Point calc_Cross_Point(double x1,double x2,double x3,double x4,double y1,double y2,double y3,double y4){
	Point ret;
	ret.x = ((x2-x1)*(y3*(x4-x3)+x3*(y3-y4))-(x4-x3)*(y1*(x2-x1)+x1*(y1-y2)))/((y2-y1)*(x4-x3)-(y4-y3)*(x2-x1));
	if(x1 != x2){
		ret.y = ((y2-y1)*ret.x+y1*(x2-x1)+x1*(y1-y2))/(x2-x1);
	}else{
		ret.y = ((y4-y3)*ret.x+y3*(x4-x3)+x3*(y3-y4))/(x4-x3);
	}
	return ret;
}

Point calc_Cross_Point(Point a,Point b,Point c,Point d){
	return calc_Cross_Point(a.x,b.x,c.x,d.x,a.y,b.y,c.y,d.y);
}

Polygon ConvexCut(Polygon g,Line line){

	Polygon ret;
	Point a = line.p[0];
	Point b = line.p[1];

	int N = g.size();

	for(int i = 0; i < g.size(); i++){
		Point A = g[i], B = g[(i+1)%N];
		if(ccw(a,b,A) != -1)ret.push_back(A);
		if(ccw(a,b,A)*ccw(a,b,B) == -1)ret.push_back(calc_Cross_Point(a,b,A,B));
	}

	return ret;
}

Polygon G;



void func(){

	G.clear();

	Point point;

	for(int loop = 0; loop < N; loop++){
		scanf("%lf %lf",&point.x,&point.y);
		G.push_back(point);
	}

	double left = 0,right = 999999999.0,m = (left+right)/2;
	double ans = 0;
	Line line;

	for(int i = 0; i < 100; i++){
		Polygon CALC = G;
		for(int k = 0; k < N; k++){
			Point vec = G[(k+1)%N]-G[k];
			vec = vec/abs(vec);
			vec = rotate(vec,PI/2);
			line.p[0] = G[k]+vec*m;
			line.p[1] = G[(k+1)%N]+vec*m;
			CALC = ConvexCut(CALC,line);
		}
		if(CALC.empty()){
			right = m-EPS;
		}else{
			ans = m;
			left = m+EPS;
		}
		m = (left+right)/2;
	}

	printf("%.10lf\n",ans);
}


int main(){

	while(true){
		scanf("%d",&N);
		if(N == 0)break;

		func();
	}

	return 0;
}