multiplication of two int value in Dafny

Question

Can anyone help me with this program? I am new to Dafny. I just need to know what will be invariant and decreases values. Thanks in advance

method MultiplyTwoNumber(N: int, M: nat) returns (Result: int)
  ensures Result == M*N;
  requires N>=0 && M>=0;
{
  var a := N;
  var b := M;
  var x := 0;
  var i :=0;
  while (b > 0)    
  invariant ?? // what will be the invariant value
  decreases ?? // what will be the decreases value
  {
    while (b % 15 != 0)   
    invariant ?? // what will be the invariant value
    decreases ?? what will be the decreases value
    {
      x := x + a; 
      b := b - 1; 
      
    }
    a := 15 * a; 
    b := b / 15; 
  
  }
  Result := x; 
}

Can you point out what did you tried and where did you get stuck — Divyanshu Ranjan, Oct 27 '22 at 04:29
I wanted to know in every iteration of the while loop what will be the invariant value and decreases the value — Engr Aminul Islam, Oct 27 '22 at 08:17
You still have not provided what is your thought process regarding coming up with invariant is ? — Divyanshu Ranjan, Oct 27 '22 at 11:26
@DivyanshuRanjan thanks for your comments. Here, I have two while loops for the first while loop I am just checking the b value and at the end, I am assigning a value to my multiplication with the old value of a and also updating the b value. but I am confused how I can implement this logic in invariant. — Engr Aminul Islam, Oct 27 '22 at 11:34
Here is hint: Think of numbers as written in base 15. Inner loop calculates rows in long multiplication. And outer loops sum those row. Can you try what to come up with invariant for base 2. Can you generalize it to base 15 — Divyanshu Ranjan, Oct 27 '22 at 12:01
I am totally new to invariant. could you please help me to write it? @DivyanshuRanjan — Engr Aminul Islam, Oct 27 '22 at 12:23

score 1 · Accepted Answer · answered Oct 27 '22 at 12:27

1

In outer loop b will decrease, since every loop iteration divides b by 15. In inner loop b and b % 15 will decrease.

For invariant observe that in inner loop M * N == a * b + x holds, since every iteration will decrease b by 1 and add a to x as a result a * b + x does not change. Same invariant can be used in outer loop as after loop b must be divisible by 15. And multiplying a by 15 and dividing b by 15 does not change value of a * b + x.

answered Oct 27 '22 at 12:27

Divyanshu Ranjan

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so for the first loop inveriant will be invariant M * N == a * b + x and decreases =b%15 and for the second loop invariant M * N == a * b + x same as first one and decreases will be ?? – Engr Aminul Islam Oct 27 '22 at 13:02

multiplication of two int value in Dafny

1 Answers1