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  1. #1
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    May 2012
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    Computing Large Numbers and Squares

    I'm just starting out Computer Science again and I'm working on my first assignment however, for some reason I can't get the numbers to work out because I know what should happen and what the output should be but the number is turning negative.

    Squares: In the world of Mathematics, the following identity is valid for all positive integers N:
    12 + 22 + 32 + ... + N2 = N(N + 1)(2N + 1)/6
    However, when a math model is implemented as a computer program, the results often depend on what data types and algorithms are used in the program. Write a program that receives the positive integer N as command-line input. Use the int data type for all integer variables. Your program should calculate the values of two arithmetic expressions:
    (a) LHS = 12 + 22 + 32 + ... + N2
    (b) RHS = N(N + 1)(2N + 1)/6
    Compile and run your program using the following input values:
    N = 1000 N = 1250 N = 1500 N = 1750 N = 2000
    Special questions: For each formula (LHS and RHS), what is the largest value of N that gives the correct output? Why are the two largest values different?
    The issue I think is that since I have to use INT it's too big of a number? But I'm not entirely sure.

    public static int LHS(int num)
        int sum=0;
        for(int i=1; i<=num; i++)
          sum = sum + (i*i);
        System.out.print("LHS = ");
        return sum;
      public static int RHS(int num)
        int sum = num*(num + 1)*(2*num+1);
        sum = sum / 6;
        System.out.print("RHS = ");
        return sum;
    When I use num = 2000, the answer should be 26968667000, but my output that happens is LHS = -1626300296 & RHS = -194644530.

    Does anyone have any idea why this is? Thanks for any help you can offer.

  • #2
    God Emperor Fou-Lu's Avatar
    Join Date
    Sep 2002
    Saskatoon, Saskatchewan
    Thanked 2,659 Times in 2,628 Posts
    Yep, that's correct. Integers are 32-bits in size, and java does not allow you to specify as an unsigned integer, so the maximum size you have available is 1 << 31 - 1 or 2,147,483,647. Any number larger than this uses the 32nd bit, which dictates the signed of the number and it becomes negative (see two's complement notation for information on this) effectively counting up towards 0. This is integer overflow.
    I can't give the answer for the special question though as that would defeat the purpose. You can write an algorithm that effectively converges a known high and a known low though. Choose a divide and conquer approach and it can calculate the max correct in both the algorithms. They have potential for difference since one utilizes a division which may at one point truncate the values when stuffing the float into an int.
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