easy 10 pts Solved

JK Flip-Flop

The flip-flop with no forbidden state: j=k=1 toggles instead of racing. Classic building block for counters.

Verilog problems / Sequential Design

What you must build

The flip-flop with no forbidden state: j=k=1 toggles instead of racing. Classic building block for counters.

Engineers use “JK Flip-Flop” 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: Four cases on {j,k} at every rising clock edge: 00 holds, 01 clears, 10 sets, 11 toggles (q <= ~q). Unlike an SR latch, JK has a well-defined behavior for every input combination.

Port contract

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

NameDirWidthDescription
clkinput1Clock
jinput1Set control
kinput1Reset control
qoutput1Registered output

How to approach this kata

This is an introductory kata. Prefer a clear continuous assignment or a small combinational always block. Name the module top_module and keep the port list identical to the table — the hidden testbench instantiates that name.

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      j,
  input      k,
  output reg q
);

  // Your code here

endmodule

Why this shows up in interviews

JK Flip-Flop sits under Sequential Design (sequential, flip-flop). Concept: Four cases on {j,k} at every rising clock edge: 00 holds, 01 clears, 10 sets, 11 toggles ( q <= ~q ). Unlike an SR latch, JK has a well-defined behavior for every input combination.

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.
  • 4-Bit Up/Down Counter — A synchronous counter that increments or decrements each clock edge depending on a direction input, with a synchronous reset.
  • Modulo-6 Counter — A counter that wraps back to 0 after reaching 5, instead of overflowing at the natural binary boundary — the pattern behind clock dividers and digit counters.

FAQ

What does this problem require?

What does the JK Flip-Flop problem ask for? The flip-flop with no forbidden state: j=k=1 toggles instead of racing. Classic building block for counters. Implement it as Verilog module top_module with the listed ports.

Combinational or sequential?

Is JK Flip-Flop combinational or sequential? Tags: sequential, flip-flop. 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 JK Flip-Flop 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.