Verilog problems / Sequential Design
What you must build
One register, four behaviors: hold, shift left, shift right, or parallel load, all picked by a 2-bit mode select.
Engineers use “4-Bit Universal Shift Register” 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.
case on mode inside the clocked always block: 00 holds (q<=q), 01 shifts left bringing sin into the LSB, 10 shifts right bringing sin into the MSB, 11 loads din directly.Port contract
The judge instantiates exactly these ports. Extra ports or a different module name fail to elaborate.
| Name | Dir | Width | Description |
|---|---|---|---|
| clk | input | 1 | Clock |
| rst | input | 1 | Sync active-high reset |
| mode | input | 2 | 00=hold, 01=shift-left, 10=shift-right, 11=load |
| sin | input | 1 | Serial input for either shift direction |
| din | input | 4 | Parallel load data |
| q | output | 4 | Register contents |
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, input [1:0] mode, input sin, input [3:0] din, output reg [3:0] q ); // Your code here endmodule
Why this shows up in interviews
4-Bit Universal Shift Register sits under Sequential Design (sequential, shift-register). Concept: A case on mode inside the clocked always block: 00 holds ( q<=q ), 01 shifts left bringing sin into the LSB, 10 shifts right bringing sin into the MSB, 11 loads din directly.
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 4-Bit Universal Shift Register problem ask for? One register, four behaviors: hold, shift left, shift right, or parallel load, all picked by a 2-bit mode select. Implement it as Verilog module top_module with the listed ports.
Combinational or sequential?
Is 4-Bit Universal Shift Register combinational or sequential? Tags: sequential, shift-register. 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 4-Bit Universal Shift Register 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.