Generate Pattern 2, 33, 222, 5555, 22222, 777777 using constraints

SystemVerilog
,
1

I am trying write a constraint to generate this pattern 2, 33, 222, 5555, 22222, 777777, 2222222, 99999999, 2222222222 and so on
This is my constraint code

    rand longint a;
    int count=0;
    
    constraint value {

        //ODD
        if(count%2 == 1)
            a == odd(count);
        //EVEN
        else
            a == even(count);
    }
    

    function void post_randomize();
        count++;
    endfunction

    //ODD
    function longint odd(int x);
        longint num = 0;
        int value = x+2;
        for(int i=0;i <= x;i++) begin
            num = num*10+value;
        end
        return num;
    endfunction
    
    //EVEN
    function longint even(int x);
        longint num=0;
        for(int i=0;i <= x;i++) begin
            num = num*10+1;
        end
        return 2*num;
    endfunction

This is my output

#2

33

222

5555

22222

777777

2222222

99999999

222222222

12222222221

22222222222

1444444444443

2222222222222

166666666666665

222222222222222

The even pattern of 2s are coming properly.
But odd pattern is coming wrong

2

In reply to bachan21:

What output are you expecting after 99999999?

3

In reply to dave_59:

Yes.
Next output is 222222222 followed by 11111111111111111111 so on

4

In reply to bachan21:

Since this is not really a random pattern, you should be able to debug your function alone by single stepping or printing intermediate results.

5

In reply to bachan21:

i think we can use hexadecimal format and shift operations to print this pattern

  1. take the input patterns like “3”,“5”,“7”,“9”,“11”,“13” …
  2. convert it to hex decimal numbers
    longint value = str.atohex();
  3. using below logic we can get format(num = 0 )
    for (int i = 1; i <= x; i++)
    num = (num << no_of_bits) + value;
    Note: number of bits to shift → no of bits needed to represent a input number in hexadecimal
    if ( ip_value < 16 ) no_of_bits = 4;
    else if ( (ip_value < 256) && ( ip_value >= 16)) no_of_bits = 8;

Hope it helps

6

In reply to Desam:

Hi Desam,
can put this into code

7

In reply to bachan21:

Plz find the code in below link:

8 
// Generate Pattern 2, 33, 222, 5555, 22222, 777777 using constraints
class c_1;
  rand bit [127:0] arr[];
  rand bit [127:0] arr2[];
  int unsigned arr3[10] = '{2, 3, 4, 5, 6, 7, 8, 9, 0, 1};
  rand int n;
  
  function new (string name = "c_1");  
  endfunction : new

  constraint a_1 {
    n == 22;
    arr.size() == n;
    arr2.size() == n;
    
    foreach(arr2[i]){
      if(i == 0){
        arr2[i] == 1;
      } else {
        arr2[i] == arr2[i - 1] * 10 + 1;
      }
    }
      
      foreach(arr[i]){
        if(i % 2 == 0){
          if(i == 0) {
            arr[i] == 2 ;
          } else {
            arr[i] == arr2[i] * 2;
          }
        } else {
          arr[i] == arr2[i] * arr3[(i % 10)];
        }
      }
  }
endclass : c_1        

    module tb;
      initial begin
        c_1 cc;
        cc = new();       
        cc.randomize();
        foreach(cc.arr[i])
          $display("cc.arr[%0d] = \t %d", i, cc.arr[i]);        
      end      
    endmodule : tb

//output
cc.arr[0] = 	                                       2
cc.arr[1] = 	                                      33
cc.arr[2] = 	                                     222
cc.arr[3] = 	                                    5555
cc.arr[4] = 	                                   22222
cc.arr[5] = 	                                  777777
cc.arr[6] = 	                                 2222222
cc.arr[7] = 	                                99999999
cc.arr[8] = 	                               222222222
cc.arr[9] = 	                              1111111111
1 Like