Check if Number is Even or Odd Using Bits in DSA Python - Execution Trace

Concept Flow - Check if Number is Even or Odd Using Bits

Start with number n

↓

Extract least significant bit (n & 1)

↓

Is bit 0?

Yes→Number is Even

No↓

Number is Odd

We check the last bit of the number using bitwise AND with 1. If it is 0, the number is even; if 1, it is odd.

Execution Sample

DSA Python

n = 5
if (n & 1) == 0:
    print("Even")
else:
    print("Odd")

This code checks the last bit of 5 to decide if it is even or odd.

Execution Table

Step	Operation	Value of n	Bitwise n & 1	Condition (n & 1 == 0)	Result	Output
1	Start	5
2	Calculate n & 1	5	1
3	Check if n & 1 == 0	5	1	False	Odd	Odd
4	End	5	1	False	Odd	Odd

💡 Condition n & 1 == 0 is False, so number is Odd and output is printed.

Variable Tracker

Variable	Start	After Step 2	After Step 3	Final
n	5	5	5	5
n & 1		1	1	1
Condition (n & 1 == 0)			False	False
Output			Odd	Odd

Key Moments - 3 Insights

Why do we use 'n & 1' to check even or odd?

Because the least significant bit (rightmost bit) of a binary number tells if it is even (0) or odd (1). See execution_table step 2 where n & 1 extracts this bit.

Why does 'n & 1 == 0' mean the number is even?

If the last bit is 0, the number is divisible by 2, so it is even. This is shown in execution_table step 3 where the condition is checked.

What happens if the number is 0?

0 & 1 equals 0, so the condition is True and the number is even. This is consistent with the logic in execution_table.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of 'n & 1' at step 2 for n=5?

A 1

B 0

C 5

D Undefined

💡 Hint

Check the 'Bitwise n & 1' column at step 2 in execution_table.

At which step does the condition 'n & 1 == 0' evaluate to False?

A Step 1

B Step 3

C Step 2

D Step 4

💡 Hint

Look at the 'Condition (n & 1 == 0)' column in execution_table.

If n was 4 instead of 5, what would be the output according to the execution_table logic?

A Odd

B Error

C Even

D No output

💡 Hint

Recall that even numbers have last bit 0, so n & 1 == 0 would be True.

Concept Snapshot

Check even or odd using bits:
Use 'n & 1' to get last bit.
If last bit is 0, number is even.
If last bit is 1, number is odd.
Fast and efficient bitwise check.

Full Transcript

This concept shows how to check if a number is even or odd using bitwise operations. We use the bitwise AND operator '&' with 1 to extract the least significant bit of the number. If this bit is 0, the number is even; if it is 1, the number is odd. The execution table traces the steps for n=5, showing the calculation of n & 1, the condition check, and the final output. The variable tracker shows how variables change during execution. Key moments clarify why the last bit determines even or odd and what happens with zero. The visual quiz tests understanding of these steps. This method is a quick and efficient way to check parity without division or modulus.