The three prompts

Drill 1. rand bit [15:0] x: x is a power of two.

Drill 2. rand bit [7:0] m: exactly three bits of m are set.

Drill 3. rand bit [7:0] a, b: b differs from a in exactly one bit.

Write the three answers before opening the reveal. Copy the starter into the Code Lab, fill in each empty answer_c, and run it on Verilator. The grader solves each class hundreds of times and checks every result, then pins a few values that must be legal; it prints ok or the first rule your answer broke.

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// Drill starter: write each answer_c, then run it.
// Drills: powers of two, set-bit counts and gray neighbours.
class drill1;                                   // x is a power of two
  rand bit [15:0] x;
  constraint answer_c { /* your answer here */ }
endclass

class drill2;                                   // exactly 3 of the 8 bits of m are set
  rand bit [7:0] m;
  constraint answer_c { /* your answer here */ }
endclass

class drill3;                                   // b differs from a in exactly one bit
  rand bit [7:0] a, b;
  constraint answer_c { /* your answer here */ }
endclass

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// Drills: powers of two, set-bit counts and gray neighbours.
class drill1;                                   // x is a power of two
  rand bit [15:0] x;
  constraint answer_c { x != 0; (x & (x - 1)) == 0; }
endclass

class drill2;                                   // exactly 3 of the 8 bits of m are set
  rand bit [7:0] m;
  constraint answer_c { $countones(m) == 3; }
endclass

class drill3;                                   // b differs from a in exactly one bit
  rand bit [7:0] a, b;
  constraint answer_c { $countones(a ^ b) == 1; }
endclass

drill 1, power of two: ok
drill 2, three bits set: ok
drill 3, gray neighbour: ok
PASS: all 3 drills right

Why these answers

A power of two has exactly one bit set, so $countones(x) == 1 and the bit trick with x != 0 are the same set of 16 values. Exact counts and bands are one system function: $countones(m) == 3, or inside {[3:5]} for a band. Candidates who hand-roll a foreach sum over bits signal that they missed it.

XOR turns differing bits into ones, so a gray neighbour is a one-hot XOR: $countones(a ^ b) == 1. It composes two primitives, which interviewers reward.

A note on the grader: Verilator 5.048 spread drill 3 very unevenly. In 300 calls the flipped bit was bit 6 in 272 of them and bits 0, 2 and 4 never, though all eight are equally legal. The pinned values catch an answer that is too tight; only coverage would catch a solver that is.

Related topics

• Unique Values and Ordering