Pythagoras Triplet

#1

What kind of formula required in calculating pythagoras triplet if any one length is given

#2

@shubhamkumarsingh,

You can use the following approach:

  1. If n<0, n=1, n=2. Print -1 and return because squares aren’t possible for these numbers.
  2. Now if n is even. We will simplify the Pythagorean equation.

a^2 + b^2 = c^2. Hence we manipulate the equation as a^2 = (c+b)(c-b). Now if the given number is even i.e. a is even, we can express it as product of 2 even numbers say 2 and (a^2)/2.

(a^2)/2 * 2 = (c-b)(c+b)
Hence c+b = 2 and c-b = (a^2)/2. Now solve for b and c and do similarly if given n is odd. (NOTE: I have taken a = n)

Also, when the complexity and calculations are huge in any problem. There’s usually a simple equation behind that, you can also try to deduce that in questions of such type.

#3

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#4

This is my code Scanner scn = new Scanner (System.in); int n= scn.nextInt(); int b,c,count=0,counter=0; if(n>2){ //finding the number if(n%2==0){ c= nn/4 + 1; b= nn/4 -1; }else{ c=(nn + 1)/2; b=(nn - 1)/2; } // finding greater if(b>c){ count=1; counter=1; } // finding greatest among all if(count==1){ if(n>c){ count=2; }else{ count=3; } }else{ if(n>c){ count=4; }else{ count=5; } } switch(count){ case 2: if(nn==bb + cc){ if(b>c) System.out.print(c + " " + b); else System.out.print(b + " " + c); } case 3: if(cc==bb + nn){ System.out.print(b + " " + c); } case 4: if(nn==bb + cc){ if(b>c) System.out.print(c + " " + b); else System.out.print(b + " " + c); } case 5: if(cc==bb + nn){ System.out.print(b + " " + c); } //default : // System.out.print(“hello”); } }else{ System.out.print("-1"); } not able to find out the problem why 2 test cases are not satisfied

#5

@shubhamkumarsingh,
I have corrected your code. https://ide.codingblocks.com/s/184205.

You were unnecessarily complicating the code. You don’t need to use switch or check if b>c. (n* n)/4 + 1 is always greater than (n* n)/4 - 1 under the given constraints on inputs.

Also next time for code sharing please use https://ide.codingblocks.com/

#6

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