find the sum of all the numbers between two numbers

[Anton Plotnikov]

How tp solve this task?
Given two integers a and b, which can be positive or negative,
find the sum of all the numbers between including them too
and return it. If the two numbers are equal return a or b.
Note: a and b are not ordered!

  public class Sum
  {
     public int GetSum(int a, int b)
     {
     
      //Good luck!
      return 0;
     }
  }

Thanks.

[Knute Snortum]

What have you tried?  What happened?

[Anton Plotnikov]

i want write this function.
what code do I need to insert?

[Henry Wong]

Anton Plotnikov wrote:i want write this function.
what code do I need to insert?

Knute asked what have you done with this homework assignment so far... basically, we can't tell what kinda help you need, without knowing what you are confused with, and of course, this requires knowing what you did so far.

Henry

[Jim Venolia]

This is an easy question to answer, even for someone fairly new to Java like me.

It's a homework assignment, and a pretty easy one at that.  Read your textbook.

[Campbell Ritchie]

JV, I don't think that post of yours does anything to help AP.

AP: welcome to the Ranch
Do you remember from maths at school how to total progressions of numbers?

[Jim Venolia]

You're right, I apologize.

[Campbell Ritchie]

Apology accepted

[Zach Rode]

Anton, for this problem I think you might want to consider thinking about the steps to go through counting the numbers, write those down. Next, you need to match code to achieve these steps you have written. I would suggest using your textbook and looking up for loops.

See what you can get from this point and then post what you need help on if you have anymore questions.

-Zach

[Quazi Irfan]

Anton Plotnikov wrote:i want write this function.
what code do I need to insert?

Okay, lets work together.

Before we jump start writing the code lets look at the steps we need to take to solve the problem. You can write these steps in many different way: flow chart, pseudo code etc. I like pseudo code because it's more similar to written description. You can also add code in a pseudo code for brevity which I find very handy. For example, I can write a := 0 which means, variable a is set to value zero. Here is the pseudo code for GetSum function,

// start of function GetSum
Function GetSum(Number1, Number2):
   summation := 0
   In a loop
       summation := summation + Number1
       Increment Number1
       Check if Number1 == Number2
             True: break the loop
             False: continue the loop
   return summation
// end of function GetSum

Different language provides different ways to write this function. For example, Java has three different loop constructs: For, while, do-while. Which one do you think would work best for our situation? You also know need to know how to use variables since you need to store the result of the summation somewhere. You would also need to know how to use use arithmetic operator to modify the value stored in a variable.

Once you read those tutorials try writing the function. Post here if you are stuck or have trouble reading the java tutorials. (Note that this function doesn't do exactly what you want, it just gives you a starting point.)

         
       
       

[Campbell Ritchie]

Quazi Irfan wrote:. . . Before we jump start writing the code lets look at the steps we need to take to solve the problem. . . .

That is a good start but then you go straight on to pseudocode for loops. How do you know you will want loops? Isn't that jumping ahead a bit?

[Quazi Irfan]

Campbell Ritchie wrote:How do you know you will want loops? Isn't that jumping ahead a bit?

Most people would repeatedly add numbers to get a sum of a range. I am assuming he knows that much. And I am instructing him to repeat the same tasks in a programming language.

Is that what you are asking?

[Campbell Ritchie]

Quazi Irfan wrote:. . . Most people would repeatedly add numbers to get a sum of a range. . . .

But there is a more efficient way to total those numbers. We don't know how much the OP knows, nor whether he wants efficiency or whether he has been told to write a loop.

[Piet Souris]

In other words: wait for a reply from OP, whether he has gotten some ideas yet.

[Campbell Ritchie]

Piet Souris wrote:. . . wait . . . whether he has gotten some ideas yet.

. . . or whether the deadline for this assignment has passed

[salvin francis]

Hi Anton Plotnikov,
Welcome to Code ranch. I think the problem is simple, but here's a clause that's missed by others above :

Anton Plotnikov wrote:
Note: a and b are not ordered!

This means that you might need to swap the parameters if a is greater than b
With regards to the solution, are you familiar with IntStream ? Hint : There are 2 methods in the IntStream interface that can help you with your assignment.

[Quazi Irfan]

Campbell Ritchie wrote:But there is a more efficient way to total those numbers.

Can you tell me more about it?

[Piet Souris]

The formula for the series 1 + 2 + 3 + ... + n is well known. Try to deduce a similar formula for the similar series from a to b, inclusive. Does it matter here whether a is smaller or larger than b?

[Lakshman Arun]

Does it matter here whether a is smaller or larger than b?

Good catch

[Campbell Ritchie]

Lakshman Arun wrote:Does it matter here whether a is smaller or larger than b? . . .

Yes, but the corrections required for a loop will be more than those required for a simple formula.

[salvin francis]

salvin francis wrote:...but here's a clause that's missed by others above ...

It's rude on my part to simply assume all have missed reading that. I am sorry about that.

Lakshman Arun wrote:Does it matter here whether a is smaller or larger than b?

Let me demonstrate with a program:

public class Main {
    public static void main(String... salvin) {
        int start = 10;
        int end = 20;

        // -- The loop does not execute even once if you uncomment the lines below:

        // start = 20;
        // end = 10;

        for (int n = start; n < end; n++) {
            System.out.println("Reading : " + n);
        }
    }
}

[ras oscar]

salvin francis wrote:Hi Anton Plotnikov,
Welcome to Code ranch. I think the problem is simple, but here's a clause that's missed by others above :

Anton Plotnikov wrote:
Note: a and b are not ordered!

This means that you might need to swap the parameters if a is greater than b
With regards to the solution, are you familiar with IntStream ? Hint : There are 2 methods in the IntStream interface that can help you with your assignment.

And also, one or more of the boundary numbers might be negative. Not sure if all the loop structures will accept negative integers?

[Campbell Ritchie]

ras oscar wrote:. . . . Not sure if all the loop structures will accept negative integers?

They all will if you threaten to hit them if they don't.

[salvin francis]

ras oscar wrote:...Not sure if all the loop structures will accept negative integers?

Does this solve your doubt ?

public class Main {
    public static void main(String... salvin) {
        int start = -10, end = -5;

        // -- looping (You can try while and do-while too):
        System.out.println("Looping with negative values:");
        for (int n = start; n <= end; n++) {
            System.out.println("Reading value: " + n);
        }

        // -- Streams:
        System.out.println("Instream with negative values:");
        IntStream.range(start, end).mapToObj(n -> "Reading value: " + n).forEach(System.out::println);
    }
}

output:

Looping with negative values:
Reading value: -10
Reading value: -9
Reading value: -8
Reading value: -7
Reading value: -6
Reading value: -5
Instream with negative values:
Reading value: -10
Reading value: -9
Reading value: -8
Reading value: -7
Reading value: -6

[Piet Souris]

Or even much easier:

int a = 5;
int b = -12;
for (int i = a; Math.abs(i - a) <= Math.abs(b - a); i += a < b ? 1 : -1) {
    System.out.format("a = %d, b = %d, i = %d%n", a, b, i);
}

[Campbell Ritchie]

Piet Souris wrote:Or even much easier . . .

Easier?

[Piet Souris]

Hmm.. yes.

Since OP STILL hasn't replied, I was afraid we were still writing things too difficult. So I thought about this easy way, and put in a smiley as well, to convince everyone of my sincerity.

[salvin francis]

Piet Souris wrote:Or even much easier: ...

Adding 2 lines to your code more makes it easier to read:

int a = 5;
int b = -12;
int min = Math.min(a, b);
int max = Math.max(a, b);
for (int i = min; i <= max; i++) {
    System.out.format("a = %d, b = %d, i = %d%n", a, b, i);
}

[Stephan van Hulst]

I agree with Salvin. The increment statement of a loop should be as simple as possible, which in this case is achieved by determining the range to iterate over before the loop starts.

[Piet Souris]

Allright, I confess.

Since OP didn't react to all the simple things we wrote, I tried to be funny. But the mrs is right, she always tell me I have no talent at all for being funny. So from now on I'll stick to being the old grumpy man I always was.    

[Campbell Ritchie]

Since OP hasn't come back, I have my own take on that:-

public static final int sumBetween(int i, int j)
{
   return (i + j) * (Math.max(i, j) - Math.min(i, j) + 1) / 2;
}

No need for a loop. Only possible error: overflow. Alternative version:-

public static final int sumBetween(int i, int j)
{
   return (i + j) * (Math.abs(i - j) + 1) / 2;
}

At least I think those methods are correct.

[Knute Snortum]

Piet Souris wrote:Since OP didn't react to all the simple things we wrote, I tried to be funny. But the mrs is right, she always tell me I have no talent at all for being funny. So from now on I'll stick to being the old grumpy man I always was.    

In your defence, you did put a winky-face on the post.

[salvin francis]

Adding to what Campbell has posted, the sum of an arithmetic progression is

n  x  ( (A1 + An) ÷ 2 )

Where
n is the total number of elements,
A1 is the minimum
An is the maximum

Our progression has a common difference of 1, hence its easy to calculate n given the min and max :
n = (An - A1) + 1

[Scott Shipp]

Now we just need the Big-O analysis of space and time complexity of each algorithm.