Colon operator for ranges in MATLAB - Step-by-Step Execution

Concept Flow - Colon operator for ranges

Start with start_value

↓

Check if next_value <= end_value

Yes↓

Add next_value to range

↓

Increment next_value by step

Back to Check↓

End range

↩Back to Check

The colon operator creates a range starting from a start value, increasing by a step, until it reaches or passes the end value.

Execution Sample

MATLAB

range = 1:2:7;
disp(range);

Creates a range starting at 1, stepping by 2, up to 7, then displays it.

Execution Table

Step	Current Value	Condition (<=7)	Action	Range So Far
1	1	1 <= 7 (True)	Add 1	[1]
2	3	3 <= 7 (True)	Add 3	[1, 3]
3	5	5 <= 7 (True)	Add 5	[1, 3, 5]
4	7	7 <= 7 (True)	Add 7	[1, 3, 5, 7]
5	9	9 <= 7 (False)	Stop	[1, 3, 5, 7]

💡 Next value 9 is greater than end value 7, so range creation stops.

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	Final
next_value	1	3	5	7	9	9
range	[]	[1]	[1, 3]	[1, 3, 5]	[1, 3, 5, 7]	[1, 3, 5, 7]

Key Moments - 3 Insights

Why does the range include 7 even though the step is 2?

Because the condition checks if the current value is less than or equal to the end value (7). Since 7 <= 7 is true, 7 is included (see execution_table step 4).

What happens if the next value after increment is greater than the end value?

The range stops growing and the loop ends (see execution_table step 5 where 9 > 7 stops the process).

Can the colon operator create ranges with negative steps?

Yes, but the condition changes to check if the current value is greater than or equal to the end value to stop correctly.

Visual Quiz - 3 Questions

Look at the execution_table, what is the range after step 3?

A [1, 3, 5, 7]

B [1, 3, 5]

C [1, 3]

D [1]

💡 Hint

Check the 'Range So Far' column at step 3 in the execution_table.

At which step does the condition become false and stop the range creation?

A Step 3

B Step 4

C Step 5

D Step 2

💡 Hint

Look at the 'Condition' column where it shows false in the execution_table.

If the step was changed to 3, what would be the last value included in the range?

A 7

B 3

C 4

D 6

💡 Hint

Think about adding 3 each time starting from 1 until exceeding 7.

Concept Snapshot

Colon operator syntax: start:step:end
Creates a vector from start to end, adding step each time.
Includes end if it matches exactly.
Stops when next value passes end.
Step can be positive or negative.

Full Transcript

The colon operator in MATLAB creates a range of numbers starting from a start value, increasing by a step, until it reaches or passes the end value. For example, 1:2:7 creates the numbers 1, 3, 5, and 7. Each step adds 2 to the previous number. The process stops when the next number would be greater than 7. This is shown step-by-step in the execution table. The variable tracker shows how the current value and the range change after each step. Key points include that the end value is included if the current value equals it, and the range stops when the next value exceeds the end. The colon operator can also work with negative steps to create decreasing ranges.

Practice

1. What does the MATLAB expression 1:5 produce?

easy

A. [1 3 5]

B. [5 4 3 2 1]

C. [1 2 3 4 5]

D. [1 2 3 4]

Solution

1. Step 1: Understand the colon operator syntax

   The expression start:end creates a sequence from start to end with step 1 by default.

2. Step 2: Apply to given expression

   Here, 1:5 means numbers from 1 to 5 increasing by 1: 1, 2, 3, 4, 5.

3. Final Answer:

   [1 2 3 4 5] -> Option C

4. Quick Check:

   1 to 5 by 1 = [1 2 3 4 5] [OK]

Hint: Colon with two numbers counts up by 1 [OK]

Common Mistakes:

• Thinking it skips numbers
• Confusing start and end order
• Assuming it counts down automatically

2. Which of the following is the correct syntax to create a vector from 10 down to 2 with step -2 in MATLAB?

easy

A. 10:-2:2

B. 10:2:-2

C. 10--2:2

D. 10:-2:-2

Solution

1. Step 1: Recall colon operator with step syntax

   The syntax is start:step:end. For counting down, step must be negative.

2. Step 2: Check each option

   10:-2:2 uses 10:-2:2, which counts down from 10 to 2 by -2 correctly.

3. Final Answer:

   10:-2:2 -> Option A

4. Quick Check:

   Descending with negative step = 10:-2:2 [OK]

Hint: Use negative step for descending ranges [OK]

Common Mistakes:

• Putting step after end value
• Using positive step for descending
• Syntax errors with double minus signs

3. What is the output of the MATLAB code:

vec = 3:3:12;

disp(vec);

medium

A. [3 4 5 6 7 8 9 10 11 12]

B. [3 6 9 12]

C. [3 6 9]

D. [12 9 6 3]

Solution

1. Step 1: Understand the colon operator with step

   The syntax start:step:end creates numbers from start to end increasing by step.

2. Step 2: Calculate the sequence

   Starting at 3, add 3 each time: 3, 6, 9, 12. Since 12 is included, the sequence ends there.

3. Final Answer:

   [3 6 9 12] -> Option B

4. Quick Check:

   3 to 12 by 3 = [3 6 9 12] [OK]

Hint: Multiply step by count until reaching or passing end [OK]

Common Mistakes:

• Stopping before including end if exact
• Confusing step with end value
• Assuming step is always 1

4. Identify the error in this MATLAB code:

range = 5:0.5:3;

medium

A. Empty vector because start > end with positive step

B. Syntax error due to decimal step

C. Vector counts down correctly

D. Step must be integer

Solution

1. Step 1: Analyze start, step, and end values

   Start is 5, end is 3, step is 0.5 (positive).

2. Step 2: Understand colon operator behavior

   With positive step, MATLAB counts up. Since 5 > 3, no values satisfy the condition, so the vector is empty.

3. Final Answer:

   Empty vector because start > end with positive step -> Option A

4. Quick Check:

   Positive step but start > end = empty vector [OK]

Hint: Step sign must match direction from start to end [OK]

Common Mistakes:

• Assuming decimal steps cause syntax errors
• Expecting vector to count down with positive step
• Thinking step must be integer

5. You want to create a vector in MATLAB that starts at 0, goes up to 1, but only includes values less than 1 with step 0.2. Which expression correctly does this?

hard

A. 0:0.2:1

B. 0:0.2:<1

C. 0:0.2:1.1

D. 0:0.2:0.9

Solution

1. Step 1: Understand the requirement

   The vector should start at 0, increase by 0.2, and include values less than 1 (not including 1).

2. Step 2: Check which end value excludes 1

   Using 0:0.2:1 includes 1 since it matches the sequence (0, 0.2, 0.4, 0.6, 0.8, 1). Since 1 is included, it violates the condition.

3. Step 3: Choose end value less than 1

   Using 0:0.2:0.9 stops at 0.8 (0.8 + 0.2 = 1.0 > 0.9), which is less than 1, so it meets the condition.

4. Final Answer:

   0:0.2:0.9 -> Option D

5. Quick Check:

   End less than 1 excludes 1 from vector [OK]

Hint: Set end value just below the limit to exclude it [OK]

Common Mistakes:

• Using end value equal to limit includes it
• Trying invalid syntax like <1
• Assuming floating point precision excludes end automatically