Verilog problems / Combinational Design
What you must build
Convert an 8-bit binary number (0-255) into three separate BCD digits — hundreds, tens, ones — using the classic shift-and-add-3 algorithm.
Engineers use “8-Bit Binary to 3-Digit BCD (Double Dabble)” as a building block in combinational 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.
Port contract
The judge instantiates exactly these ports. Extra ports or a different module name fail to elaborate.
| Name | Dir | Width | Description |
|---|---|---|---|
| bin | input | 8 | Binary value, 0-255 |
| hundreds | output | 4 | Hundreds BCD digit |
| tens | output | 4 | Tens BCD digit |
| ones | output | 4 | Ones BCD digit |
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 [7:0] bin, output [3:0] hundreds, output [3:0] tens, output [3:0] ones ); // Your code here — double-dabble: 20-bit shift register, add-3-if->=5 to each BCD nibble, shift left, repeat 8 times. endmodule
Why this shows up in interviews
8-Bit Binary to 3-Digit BCD (Double Dabble) sits under Combinational Design (combinational, algorithmic). Concept: The "double dabble" algorithm: load the binary value into the low bits of a wider shift register, then repeat (once per input bit) — if any BCD nibble holds 5 or more, add 3 to it, then shift the whole register left by 1. After 8 iterations, the upper nibbles hold valid BCD digits. Skipping the add-3 check on any one digit silently corrupts every result above single digits.
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
- 2-to-1 Multiplexer — Select between two 1-bit inputs using sel. When sel=0 output a; when sel=1 output b.
- 1-Bit Full Adder — Compute sum and carry-out for a + b + cin.
- 4-to-1 Multiplexer — Select one of four 1-bit inputs using a 2-bit sel. sel=00→in0, 01→in1, 10→in2, 11→in3.
- 4-to-2 Priority Encoder — Output the index of the highest-priority (MSB-most) set bit in a 4-bit input, plus a valid flag when any bit is set.
FAQ
What does this problem require?
What does the 8-Bit Binary to 3-Digit BCD (Double Dabble) problem ask for? Convert an 8-bit binary number (0-255) into three separate BCD digits — hundreds, tens, ones — using the classic shift-and-add-3 algorithm. Implement it as Verilog module top_module with the listed ports.
Combinational or sequential?
Is 8-Bit Binary to 3-Digit BCD (Double Dabble) combinational or sequential? Tags: combinational, algorithmic. 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 8-Bit Binary to 3-Digit BCD (Double Dabble) 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.