State encoding (binary, one-hot, gray) in Verilog - Step-by-Step Execution

Concept Flow - State encoding (binary, one-hot, gray)

Define States

↓

Choose Encoding Type

↓

Binary

↓

Assign Codes to States

↓

Use Encoded States in FSM

↓

Simulate and Verify

This flow shows how to define states, pick an encoding type (binary, one-hot, or gray), assign codes, and use them in a finite state machine.

Execution Sample

Verilog

typedef enum logic [1:0] {
  IDLE = 2'b00,
  LOAD = 2'b01,
  EXEC = 2'b11,
  DONE = 2'b10
} state_t;

state_t state, next_state;

Defines a 2-bit binary encoded state machine with four states.

Execution Table

Step	State Name	Binary Code	One-Hot Code	Gray Code
1	IDLE	00	0001	00
2	LOAD	01	0010	01
3	EXEC	11	0100	10
4	DONE	10	1000	11
5	End	-	-	-

💡 All states assigned codes in each encoding scheme.

Variable Tracker

State	Binary Code	One-Hot Code	Gray Code
IDLE	00	0001	00
LOAD	01	0010	01
EXEC	11	0100	10
DONE	10	1000	11

Key Moments - 3 Insights

Why does one-hot encoding use more bits than binary encoding?

One-hot encoding assigns one bit per state, so for N states it uses N bits, unlike binary which uses log2(N) bits. See execution_table rows 1-4 for bit widths.

How does Gray code differ from binary in state transitions?

Gray code changes only one bit between consecutive states, reducing glitches. Look at execution_table rows 1-4 to compare Gray and Binary codes.

Can two states have the same code in these encodings?

No, each state must have a unique code in all encoding schemes to avoid confusion. The execution_table shows unique codes for each state.

Visual Quiz - 3 Questions

Look at the execution_table, what is the one-hot code for the EXEC state?

A 0010

B 1000

C 0100

D 0001

💡 Hint

Check the 'One-Hot Code' column at the row where State Name is EXEC.

At which step does the Gray code for the state change from '01' to '11'?

A Step 1 to 2

B Step 3 to 4

C Step 2 to 3

D Step 4 to 5

💡 Hint

Look at the 'Gray Code' column between rows 3 and 4 in execution_table.

If the FSM had 8 states, how many bits would one-hot encoding require?

A 8 bits

B 3 bits

C 4 bits

D 1 bit

💡 Hint

One-hot uses one bit per state; see key_moments about bit width differences.

Concept Snapshot

State encoding assigns binary codes to FSM states.
Binary uses few bits (log2 states).
One-hot uses one bit per state.
Gray changes one bit between states.
Choose encoding by speed, area, or glitch needs.

Full Transcript

State encoding in Verilog means giving each FSM state a unique code. Binary encoding uses the fewest bits by counting in binary. One-hot encoding uses one bit per state, so only one bit is high at a time. Gray code changes only one bit between states to reduce glitches. We define states, pick an encoding, assign codes, and use them in the FSM. The execution table shows codes for four states in each encoding. One-hot uses more bits but can be faster. Gray code helps with smooth transitions. Each state must have a unique code to avoid errors.

Practice

1.

Which state encoding method uses exactly one bit set to 1 for each state, and all other bits are 0?

easy

A. One-hot encoding

B. Binary encoding

C. Gray encoding

D. Thermometer encoding

Solution

1. Step 1: Understand one-hot encoding definition

   One-hot encoding assigns a unique bit to each state, where only one bit is high (1) and the rest are low (0).

2. Step 2: Compare with other encodings

   Binary encoding uses multiple bits to represent states in binary form; Gray encoding changes only one bit between states; Thermometer encoding is different and not the one-hot method.

3. Final Answer:

   One-hot encoding -> Option A

4. Quick Check:

   One-hot = single 1 bit per state [OK]

Hint: One-hot means only one bit is 1 at a time [OK]

Common Mistakes:

• Confusing binary encoding with one-hot
• Thinking Gray encoding sets only one bit
• Mixing thermometer encoding with one-hot

2.

Which of the following is the correct Verilog declaration for a 3-bit binary encoded state register named state?

easy

A. reg state[2:0];

B. reg [3:0] state;

C. reg [2:0] state;

D. wire [2:0] state;

Solution

1. Step 1: Understand Verilog register declaration

   For a 3-bit register, the correct syntax is reg [2:0] name; where bits are indexed from 2 down to 0.

2. Step 2: Check options for correctness

   reg [2:0] state; uses reg [2:0] state; which is correct. reg [3:0] state; declares 4 bits, wire [2:0] state; uses wire (not register), and reg state[2:0]; has incorrect syntax.

3. Final Answer:

   reg [2:0] state; -> Option C

4. Quick Check:

   3-bit reg = reg [2:0] name [OK]

Hint: Use reg [N-1:0] for N-bit registers in Verilog [OK]

Common Mistakes:

• Using wire instead of reg for state storage
• Incorrect bit range syntax
• Confusing array declaration with bit vector

3.

Given the following Verilog snippet using one-hot encoding for 4 states, what is the value of state after the first clock cycle if the initial state is 4'b0001 and the next state logic shifts the 1 bit left by one position?

reg [3:0] state = 4'b0001;
always @(posedge clk) begin
  state <= {state[2:0], state[3]};
end

medium

A. 4'b0001

B. 4'b0100

C. 4'b1000

D. 4'b0010

Solution

1. Step 1: Understand the shift operation in the code

   The assignment state <= {state[2:0], state[3]}; shifts bits left by one, moving bit 0 to bit 1, bit 1 to bit 2, bit 2 to bit 3, and bit 3 to bit 0.

2. Step 2: Apply the shift to initial state 4'b0001

   Initial bits: bit3=0, bit2=0, bit1=0, bit0=1. After shift: bit3=bit2=0, bit2=bit1=0, bit1=bit0=1, bit0=bit3=0 -> 4'b0010.

3. Final Answer:

   4'b0010 -> Option D

4. Quick Check:

   Shift left one bit in one-hot = 4'b0010 [OK]

Hint: Shift bits left, wrap last bit to first [OK]

Common Mistakes:

• Forgetting bit wrap-around
• Confusing bit order in concatenation
• Assuming no change after clock

4.

Identify the error in the following Verilog code snippet that attempts to implement Gray code state encoding for 3 states:

reg [1:0] state;
always @(posedge clk) begin
  case(state)
    2'b00: state <= 2'b01;
    2'b01: state <= 2'b11;
    2'b11: state <= 2'b10;
    2'b10: state <= 2'b01;
    default: state <= 2'b00;
  endcase
end

medium

A. State transitions do not follow Gray code rules

B. State register width is too small for 3 states

C. Missing reset condition for state register

D. Using blocking assignment instead of non-blocking

Solution

1. Step 1: Check state transitions for Gray code property

   Gray code requires only one bit change between consecutive states. Check transitions: 00->01 (1 bit change), 01->11 (1 bit change), 11->10 (1 bit change), 10->01 (2 bits change).

2. Step 2: Identify violation in transitions

   The transition 10->01 changes two bits, violating Gray code rules.

3. Final Answer:

   State transitions do not follow Gray code rules -> Option A

4. Quick Check:

   Gray code = one bit change per transition [OK]

Hint: Gray code changes only one bit between states [OK]

Common Mistakes:

• Ignoring multi-bit changes in transitions
• Assuming register width is incorrect
• Confusing blocking and non-blocking assignments

5.

You need to design a finite state machine (FSM) with 8 states. Which state encoding method is best to minimize the number of flip-flops used, and why?

hard

A. One-hot encoding, because it uses fewer flip-flops

B. Binary encoding, because it uses the minimum number of flip-flops

C. Gray encoding, because it reduces glitches in output

D. Thermometer encoding, because it is simplest to implement

Solution

1. Step 1: Calculate flip-flops needed for each encoding

   For 8 states, binary encoding needs 3 flip-flops (2^3=8). One-hot encoding needs 8 flip-flops (one per state). Gray encoding also uses binary bits, so 3 flip-flops.

2. Step 2: Choose encoding minimizing flip-flops

   Binary encoding uses the fewest flip-flops (3) compared to one-hot (8). Gray encoding is similar to binary but mainly reduces glitches, not flip-flop count.

3. Final Answer:

   Binary encoding, because it uses the minimum number of flip-flops -> Option B

4. Quick Check:

   Binary encoding = minimum flip-flops for many states [OK]

Hint: Binary encoding uses log2(states) flip-flops [OK]

Common Mistakes:

• Choosing one-hot for fewer flip-flops
• Confusing glitch reduction with flip-flop count
• Assuming thermometer encoding is minimal