Logical operators (&, |, ~) in MATLAB - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of logical AND and OR operations

What is the output of the following MATLAB code?

MATLAB

A = [1 0 1];
B = [0 0 1];
C = A & B;
D = A | B;
disp(C);
disp(D);

A

[0 0 0]
[1 1 1]

B

[0 0 1]
[1 0 1]

C

[1 0 1]
[1 0 1]

D

[1 1 1]
[0 0 0]

💡 Hint

Remember that & returns true only if both elements are true, | returns true if at least one is true.

The logical AND (&) compares each element: 1&0=0, 0&0=0, 1&1=1. The logical OR (|) compares each element: 1|0=1, 0|0=0, 1|1=1.

❓ Predict Output

intermediate

Output of logical NOT operation

What does the following MATLAB code display?

MATLAB

X = [true false true false];
Y = ~X;
disp(Y);

A [false true false true]

B [true false true false]

C [1 0 1 0]

D [0 1 0 1]

💡 Hint

The ~ operator flips true to false and false to true.

The ~ operator negates each logical element: true becomes false, false becomes true.

🔧 Debug

advanced

Identify the error in logical expression

What error does this MATLAB code produce?

MATLAB

A = [1 0 1];
B = [0 1];
C = A & B;

A Dimension mismatch error

B Syntax error

C No error, outputs [0 0 1]

D Index exceeds matrix dimensions

💡 Hint

Logical operators require arrays to be the same size or compatible.

The arrays A and B have different lengths, so element-wise logical AND (&) cannot be performed, causing a dimension mismatch error.

❓ Predict Output

advanced

Result of combined logical operations

What is the output of this MATLAB code?

MATLAB

A = [1 0 1 0];
B = [0 1 1 0];
C = ~(A & B) | (A & ~B);
disp(C);

A [0 0 0 0]

B [0 1 1 0]

C [1 1 0 1]

D [1 1 1 1]

💡 Hint

Break down the expression step by step: compute A & B, then negate, then compute A & ~B, then OR the results.

A & B = [0 0 1 0], ~(A & B) = [1 1 0 1], A & ~B = [1 0 0 0], OR gives [1 1 0 1] | [1 0 0 0] = [1 1 0 1]. But option C is [1 1 1 1], so re-check: Actually, A & ~B = [1 0 0 0], so final OR is [1 1 0 1]. So option C matches. Correct answer is B.

🧠 Conceptual

expert

Understanding precedence of logical operators

Given the expression in MATLAB: true | false & false, what is the result?

A true

B false

C Error due to ambiguous operators

D Depends on variable types

💡 Hint

Remember that & has higher precedence than | in MATLAB.

The expression is evaluated as true | (false & false). false & false is false, so true | false is true.

Practice

1. Which logical operator in MATLAB returns true only if both conditions are true?

easy

A. ~ (NOT)

B. & (AND)

C. | (OR)

D. ^ (XOR)

Solution

1. Step 1: Understand the AND operator

   The & operator returns true only when both conditions it compares are true.

2. Step 2: Compare with other operators

   The | operator returns true if at least one condition is true, ~ negates a condition, and ^ is XOR which returns true if exactly one condition is true.

3. Final Answer:

   & (AND) -> Option B

4. Quick Check:

   Both true = true means AND = & [OK]

Hint: AND means both true, use & operator [OK]

Common Mistakes:

• Confusing & with | operator
• Thinking ~ returns true if both true
• Mixing XOR (^) with AND (&)

2. Which of the following is the correct syntax to negate a logical variable a in MATLAB?

easy

A. not a

B. !a

C. ~a

D. a~

Solution

1. Step 1: Recall MATLAB negation syntax

   In MATLAB, the negation operator is ~ placed before the variable to flip its logical value.

2. Step 2: Check each option

   not a uses 'not' which is not valid syntax, C uses '!' which is not MATLAB syntax, and D places '~' after the variable which is incorrect.

3. Final Answer:

   ~a -> Option C

4. Quick Check:

   Negate with ~ before variable [OK]

Hint: Use ~ before variable to negate in MATLAB [OK]

Common Mistakes:

• Using ! instead of ~ for negation
• Writing negation after variable
• Using 'not' keyword which is invalid

3. What is the output of the following MATLAB code?

a = true; b = false; c = a & ~b;

medium

A. true

B. false

C. 1

D. 0

Solution

1. Step 1: Evaluate ~b

   Since b is false, ~b is true.

2. Step 2: Evaluate a & ~b

   Both a and ~b are true, so a & ~b is true.

3. Final Answer:

   true -> Option A

4. Quick Check:

   true AND true = true [OK]

Hint: Negate false to true, then AND with true gives true [OK]

Common Mistakes:

• Confusing logical true/false with numeric 1/0
• Forgetting to negate b
• Assuming & acts like | (OR)

4. Identify the error in this MATLAB code snippet:

a = true;
b = false;
c = a | ~ b;

medium

A. No error, code runs correctly

B. Incorrect use of | operator

C. Extra space between ~ and b causes error

D. Missing parentheses around ~b

Solution

1. Step 1: Check operator usage

   The | operator is correctly used to perform logical OR between a and ~b.

2. Step 2: Check negation syntax

   The negation operator ~ can have a space before the variable; MATLAB ignores spaces here.

3. Final Answer:

   No error, code runs correctly -> Option A

4. Quick Check:

   Logical OR with negation syntax is valid [OK]

Hint: Spaces around ~ do not cause errors in MATLAB [OK]

Common Mistakes:

• Thinking ~ needs parentheses always
• Assuming space causes syntax error
• Confusing | with & operator

5. Given logical variables x = [true false true] and y = [false true true], what is the result of z = ~(x & y) | (x | y) in MATLAB?

hard

A. [false false false]

B. [false true true]

C. [true false true]

D. [true true true]

Solution

1. Step 1: Calculate x & y element-wise

   x = [true false true], y = [false true true]
   x & y = [true & false, false & true, true & true] = [false, false, true]

2. Step 2: Negate x & y

   ~(x & y) = [~false, ~false, ~true] = [true, true, false]

3. Step 3: Calculate x | y element-wise

   x | y = [true | false, false | true, true | true] = [true, true, true]

4. Step 4: Calculate final expression

   z = ~(x & y) | (x | y) = [true | true, true | true, false | true] = [true, true, true]

5. Final Answer:

   [true true true] -> Option D

6. Quick Check:

   Element-wise logic with ~, &, | gives all true [OK]

Hint: Calculate & first, then ~, then | element-wise [OK]

Common Mistakes:

• Forgetting element-wise operation
• Negating before AND instead of after
• Mixing up & and | precedence