Let's learn geometry

Question

Today our lesson is about rectangles and triangles.

you will be given as an input an n x n grid that is based on two characters # and * you have to classify it into : triangle or rectangle. where # represents a point and * represents a blank space.

Example input:

 #***
 ##**
 ###*
 ####

Output:

 triangle

Another example

  ##**
  ##**
  ##**
  ##**

Output:

  rectangle

Rules:

• You should assume that the input is always either a triangle or a rectangle.
• The only two characters you will receive are: * and #.
• Assume that n < 100.
• Be creative, this is a popularity-contest so the most up-voted answer will win.

Answer 1

TSQL

Image recognition? Sounds like a job made for SQL :)

WITH cte AS (
  SELECT LEN(data)-LEN(REPLACE(data,'#','')) c FROM test_t
)
SELECT 
  CASE WHEN COUNT(DISTINCT c)>2                           THEN 'triangle' 
       WHEN SUM(CASE WHEN c=0 THEN 0 ELSE 1 END) = MAX(c) THEN 'square'
                                                          ELSE 'rectangle'
       END result
FROM cte;

An SQLfiddle with sample data and test.

Answer 2

Python

Creativity

Triangle

Forward     Reverse
#***        ####      ####
##**  +     *###  =   ####
###*        **##      ####
####        ***#      ####

Rectangle

Forward  Reverse
##**     ##**     ##**
##**  +  ##**  =  ##**
##**     ##**     ##**
##**     ##**     ##**

Implementation

s=raw_input();print["tri","rect"][all(a!=b for a,b in zip(s,s[::-1])if a!='\n')]+"angle" 

Answer 3

Python + Sympy

A problem in Geometry should be only solved mathematically

Creativity

if s if the number of "#" in the input stream

Triangle: n**2+n-2s=0 -> n should be an integer

Implementation

from sympy import *
s,n=2*raw_input().count("#"),Symbol('n');print["rect","tri"][float(solve(n**2+n-s)[0]).is_integer()]+"angle" 

Answer 4

C

Assumption: the input does not represent a single-line degenerate rectangle, e.g.

****
*##*
****
****

as arguably this is also a degenerate triangle, and violates "the input is always one of the two cases."

Logic: Count the number of points in the first two non-empty lines. If they are the same, it is a rectangle. Otherwise, it is a triangle.

Code:

#include <stdio.h>
#include <string.h>
int main() {
    char s[99];
    int cnt1=0, cnt2=0;
    int i;
    while(!cnt1) {
        gets(s);
        for(i=0;i<strlen(s);i++) cnt1+=(s[i]=='#');
    };
    gets(s);
    for(i=0;i<strlen(s);i++) cnt2+=(s[i]=='#');
    puts(cnt1==cnt2?"rectangle":"triangle");
    return 0;
}

Answer 5

Java

A simple straightforward java implementation. It scans the input looking for some 2x2 square which have # in two opposing corners with something not a # (presumable a *) is some other corner. I.E, it searchs for something that shows that the drawing has a diagonal-lined border.

import java.io.ByteArrayOutputStream;
import java.io.IOException;

public class TrianglesAndRectangles {
    public static void main(String[] args) throws IOException {
        ByteArrayOutputStream baos = new ByteArrayOutputStream();
        int r;
        while ((r = System.in.read()) != -1) {
            baos.write(r);
        }
        String[] lines = baos.toString().split("\n");
        System.out.println(hasTriangle(lines) ? "triangle" : "rectangle");
    }

    public static boolean hasTriangle(String[] lines) {
        for (int i = 0; i < lines.length - 1; i++) {
            for (int j = 0; j < lines[i].length() - 1; j++) {
                boolean a = lines[i].charAt(j) == '#';
                boolean b = lines[i].charAt(j + 1) == '#';
                boolean c = lines[i + 1].charAt(j) == '#';
                boolean d = lines[i + 1].charAt(j + 1) == '#';
                if ((a && d && (!c || !b)) || (b && c && (!a || !d))) return true;
            }
        }
        return false;
    }
}