illegal cross tuple does not close coverage

Puzzle

Difficulty: Medium · Puzzle 5 of 6 · Topic: Cross Coverage Puzzles

Calculate legal cross coverage and identify illegal activity.

Code

covergroup cg with function sample(bit [2:0] len, bit [1:0] typ);
  cp_len: coverpoint len {
    bins short = {[1:2]};
    bins long  = {[3:4]};
  }
  cp_typ: coverpoint typ {
    bins rd    = {0};
    bins wr    = {1};
    bins probe = {2};
  }
  x_len_typ: cross cp_len, cp_typ {
    illegal_bins short_probe = binsof(cp_len.short) && binsof(cp_typ.probe);
  }
endgroup

initial begin
  cg c = new();
  c.sample(1,0);
  c.sample(4,1);
  c.sample(2,2);
  c.sample(3,2);
  c.sample(1,1);
end

Hint

Start from 2x3 tuples, then remove the illegal tuple from legal denominator.

Step-by-step solution

1) Raw tuple count is 2 * 3 = 6.
2) short_probe is illegal, so legal scored bins are 5.
3) Legal tuples hit: short-rd, long-wr, long-probe, short-wr -> 4 legal bins.
4) short-probe is hit once but is illegal activity, not legal progress.
5) Legal cross coverage = 4/5 = 80%.

Answer

Answer: Legal cross coverage is 80% with one illegal short-probe hit.

Why candidates get it wrong

An illegal tuple can be frequently exercised while legal closure remains incomplete.

Interviewer follow-up

Would you also assert that short+probe is forbidden, and why is that useful beyond coverage?

Related topics

• Back to Cross Coverage Puzzles

• Functional Coverage Puzzles section