State encoding (binary, one-hot, gray) in Verilog - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a 2-bit binary state register.

Verilog

reg [[1]:0] state;

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A1

B2

C3

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using 2 instead of 1 as the most significant bit index.

Forgetting that bit indexing starts at 0.

The state register is 2 bits wide, so the index goes from 1 down to 0.

2fill in blank

medium

Complete the code to assign a one-hot encoded state for state 2.

Verilog

assign one_hot_state = 4'b[1];

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A0010

B0001

C1000

D0100

Attempts:

3 left

💡 Hint

Common Mistakes

Setting the wrong bit high for the given state.

Confusing binary index order.

In one-hot encoding, only one bit is high. For state 2, the third bit (index 2) is set: 0100.

3fill in blank

hard

Fix the error in the Gray code assignment for 3-bit states.

Verilog

assign gray_state = state ^ (state >> [1]);

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A3

B2

C1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using shift amounts other than 1.

Forgetting to use XOR operator.

The Gray code is generated by XORing the state with itself shifted right by 1 bit.

4fill in blank

hard

Fill both blanks to complete the binary to one-hot conversion.

Verilog

assign one_hot = 1 << [1]; // shift 1 by state index [2]

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Astate

B1

Cstate + 1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using a fixed number instead of the state variable for shifting.

Incorrect comment that does not match the code.

To convert binary state to one-hot, shift 1 left by the state value.

5fill in blank

hard

Fill all three blanks to complete the Gray code to binary conversion.

Verilog

always @(*) begin
  binary[WIDTH-1] = [1];
  for (int i = WIDTH-2; i >= [2]; i = i - 1) begin
    binary[i] = binary[i+1] ^ [3][i];
  end
end

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Agray[WIDTH-1]

B0

Cgray

Dbinary

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing binary incorrectly.

Using wrong indices in the loop.

Mixing up Gray and binary variables.

The binary MSB is the Gray MSB. Then each binary bit is XOR of previous binary bit and current Gray bit.