easy 10 pts Solved

Divide-by-3 Clock Pulse

Generate a pulse that fires once every 3 input clock cycles — dividing by a non-power-of-2 needs a counter, not a toggle flip-flop.

Verilog problems / Sequential Design

What you must build

Generate a pulse that fires once every 3 input clock cycles — dividing by a non-power-of-2 needs a counter, not a toggle flip-flop.

Engineers use “Divide-by-3 Clock Pulse” as a building block in sequential design. Interviewers ask for the same ports and the same corner cases this judge covers. Completing it in the browser is the same skill as writing synthesizable RTL at work, minus the EDA license.

Concept: A 2-bit counter wraps at 2 (0,1,2,0,1,2,...), and the output pulses high exactly when the counter is 0: clk_div3 = (cnt == 0). The wrap condition cnt==2 is what makes this divide-by-3 instead of divide-by-4 — an easy off-by-one to get wrong.

Port contract

The judge instantiates exactly these ports. Extra ports or a different module name fail to elaborate.

NameDirWidthDescription
clkinput1Input clock
rstinput1Sync active-high reset
clk_div3output1High for 1 cycle out of every 3

How to approach this kata

This is a hard kata. Sketch the state bits and the illegal overlaps (full/empty, wrap, simultaneous enable) on paper first. A design that “usually works” in your head will fail a directed corner in the hidden tests.

Hidden tests instantiate top_module, drive the ports, and compare every sample against a golden model. They do not grade coding style. They do grade X/Z, off-by-one counters, and ignoring enables. Sign in only when you want the run saved on the leaderboard — the specification below is public.

Starter shape

Copy this skeleton into the editor (or press Reset starter). Fill the body; do not rename the module.

module top_module(
  input  clk,
  input  rst,
  output clk_div3
);

  // Your code here — a 2-bit counter that wraps at 2 works well.

endmodule

Why this shows up in interviews

Divide-by-3 Clock Pulse sits under Sequential Design (sequential, clock). Concept: A 2-bit counter wraps at 2 (0,1,2,0,1,2,...), and the output pulses high exactly when the counter is 0: clk_div3 = (cnt == 0) . The wrap condition cnt==2 is what makes this divide-by-3 instead of divide-by-4 — an easy off-by-one to get wrong.

A passing solution is synthesizable intent: no delays in the DUT, no initial blocks inside top_module, and no reference to testbench tasks. Use blocking assignments only in combinational always blocks; use non-blocking for registers clocked by clk.

Related problems

  • D Flip-Flop with Asynchronous Reset — Standard D flip-flop with an active-low asynchronous reset. Reset clears q immediately, without waiting for a clock edge.
  • 4-Bit Shift Register (SIPO) — Serial-in, parallel-out shift register. Each clock, shift left and load sin into the LSB. Sync active-high reset clears q.
  • JK Flip-Flop — The flip-flop with no forbidden state: j=k=1 toggles instead of racing. Classic building block for counters.
  • 4-Bit Up/Down Counter — A synchronous counter that increments or decrements each clock edge depending on a direction input, with a synchronous reset.

FAQ

What does this problem require?

What does the Divide-by-3 Clock Pulse problem ask for? Generate a pulse that fires once every 3 input clock cycles — dividing by a non-power-of-2 needs a counter, not a toggle flip-flop. Implement it as Verilog module top_module with the listed ports.

Combinational or sequential?

Is Divide-by-3 Clock Pulse combinational or sequential? Tags: sequential, clock. Follow the clock/reset ports if they appear in the table; if there is no clock, use continuous assignment or combinational always @(*).

How does the auto-grader work?

How is Divide-by-3 Clock Pulse graded? A hidden SystemVerilog/Verilog testbench in the EcrioniX judge simulates your module in the browser. You pass when every directed vector matches, including the waveform contract shown on this page.

Write a module named top_module matching the ports below exactly.
Expected waveform
Your solution
Judge output
// Output appears after you run tests.