Answer / sunil

if a, b and c are the three sides of a triangle, then a + b > c
if this is not satisfied, then its not a valid triangle.

To check for right angle, use Pythagoras theorem. Assume
that the longest side is the hypotenuse.

Issosless and Equilateral can be found by simply comparing
the sides.

Answer / giri

One should check for valid sides also. Side values dhould
be greated than ZERO.

Here is the correct routine:
public static String checkTriangle(int[]
triangleSide){
boolean validTriangle = false;
boolean validSides = true;
String result = "NOT VALID TRIANGLE";

for(int side: triangleSide)
if(side <= 0)
validSides = false;

if(validSides){
for(int count= 0; count< 3 ;
count++){
if(((triangleSide[count%3]
+ triangleSide[(count+1)%3]) > triangleSide[(count+2)%3]))
validTriangle =
true;
}

if(validTriangle){
if( triangleSide[0] ==
triangleSide[1] && triangleSide[2] == triangleSide[1])
result
= "EQUILATERAL";
else{
for(int count= 0;
count< 3 ; count++){

if( (
triangleSide[count%3] * triangleSide[count%3] + triangleSide
[(count+1)%3] * triangleSide[(count+1)%3]) == (triangleSide
[(count+2)%3] * triangleSide[(count+2)%3])){

result = "RIGHANGLED";

break;
}else
if
((triangleSide[count%3] == triangleSide[(count+1)%3]))

result = "ISOSCALAUS";
}
}
if("NOT VALID
TRIANGLE".equals(result))
result = "NOT
SPECIAL";

}
}

System.out.println(result);
return result;
}

Answer / ganesh bhat

I believe, this is more efficient one. solves all the cases.. comments please. Written in java

public static Map validateTriange(int a,int b,int c)
{
Map props = new HashMap();
boolean isValidTriangle;

int bigSide = a;
if(bigSide<b){bigSide = b;}
if(bigSide<c){bigSide = c;}

boolean isSpecial = false;

if((a+b+c-bigSide)>bigSide)
{
props.put("VALID","YES");
}
else
{
props.put("VALID","NO");
return props;
}

if(a == b||b==c||c == a)
{
props.put("ISOSCELES","YES");
isSpecial = true;
}

if(a == b && b == c)
{
props.put("EQUALATERAL","YES");
isSpecial = true;
}


if(((a*a+b*b+c*c)-bigSide*bigSide) == bigSide*bigSide)
{
props.put("RIGHT_ANGLED","YES");
isSpecial = true;
}

return props;
}

I am work in it aompenei

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I'm trying to solve this. But I'm not figuring the right solution. Can some one help what the answer is for the question below? You can use as many variables as you need, there are no negative numbers, all numbers are integers. You do not know the size of the integers, they could be infinitely large, so you can't count on truncating at any point. There are NO comparisons allowed, no if statements or anything like that. There are only four operations you can do on a variable. 1) You can set a variable to 0. 2) You can set a variable = another variable. 3) You can increment a variable (only by 1), and it's a post increment. 4) You can loop. So, if you were to say loop(v1) and v1 = 10, your loop would execute 10 times, but the value in v1 wouldn't change so the first line in the loop can change value of v1 without changing the number of times you loop. You need to do 3 things. 1) Write a function that decrements by 1. 2) Write a function that subtracts one variable from another. 3) Write a function that divides one variable by another. 4) See if you can implement all 3 using at most 4 variables. Meaning, you're not making function calls now, you're making macros. And at most you can have 4 variables. The restriction really only applies to divide, the other 2 are easy to do with 4 vars or less. Division on the other hand is dependent on the other 2 functions, so, if subtract requires 3 variables, then divide only has 1 variable left unchanged after a call to subtract. Basically, just make your function calls to decrement and subtract so you pass your vars in by reference, and you can't declare any new variables in a function, what you pass in is all it gets.

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