VLSI DV Interview Puzzles · All levels

Bidirectional Solving of x+y==10

For this class, what is P(x==0)? Why is the solver not 'choosing x first, then computing y' procedurally?

Puzzle

Difficulty: Easy · Puzzle 2 of 6 · Topic: Constraint Solving Order Puzzles

For this class, what is P(x==0)? Why is the solver not 'choosing x first, then computing y' procedurally?

Code

systemverilog

class sum_pair;
  rand int unsigned x;
  rand int unsigned y;
  constraint c_dom {
    x inside {[0:10]};
    y inside {[0:10]};
  }
  constraint c_eq { x + y == 10; }
endclass

Hint

Each x value determines exactly one y value in the domain.

Step-by-step solution

diagram

1) For each x in 0..10, y is fixed to 10-x and remains in 0..10.
2) That gives exactly 11 legal tuples.
3) Exactly one tuple has x==0, so P(x==0)=1/11.
4) The equality is bidirectional relation solving, not assignment sequencing.

Answer

Answer: P(x==0)=1/11 (~9.09%), and x/y are solved as a joint relation rather than left-to-right execution.

Why candidates get it wrong

Interviewees often say 'y depends on x only'; in constraints, either variable can be decided first by the solver.

Interviewer follow-up

What is P(x is even), and does adding solve x before y change the legal set?

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