Verilog problems / Combinational Design
What you must build
Add two 4-bit two's-complement numbers and flag whether the result overflowed the representable signed range.
Engineers use “4-Bit Signed Overflow Detector” 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.
overflow = (a[3]==b[3]) && (sum[3]!=a[3]). Adding numbers of opposite sign can never overflow — the result always fits between them.Port contract
The judge instantiates exactly these ports. Extra ports or a different module name fail to elaborate.
| Name | Dir | Width | Description |
|---|---|---|---|
| a | input | 4 | Signed operand (two's complement) |
| b | input | 4 | Signed operand (two's complement) |
| sum | output | 4 | a + b, wrapped |
| overflow | output | 1 | 1 if the true sum doesn't fit in 4-bit signed range |
How to approach this kata
This is a medium kata: you will need sequential logic or a small FSM. Decide what is registered versus combinational before you type. Reset polarity and clock edge must match the spec; the judge will fail you on the first mismatched cycle.
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 [3:0] a, input [3:0] b, output [3:0] sum, output overflow ); // Your code here endmodule
Why this shows up in interviews
4-Bit Signed Overflow Detector sits under Combinational Design (combinational, arithmetic). Concept: Overflow can only happen when both operands share a sign but the result doesn't: overflow = (a[3]==b[3]) && (sum[3]!=a[3]) . Adding numbers of opposite sign can never overflow — the result always fits between them.
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 4-Bit Signed Overflow Detector problem ask for? Add two 4-bit two's-complement numbers and flag whether the result overflowed the representable signed range. Implement it as Verilog module top_module with the listed ports.
Combinational or sequential?
Is 4-Bit Signed Overflow Detector combinational or sequential? Tags: combinational, arithmetic. 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 Signed Overflow Detector 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.