I2C scanner sketch in Arduino - Time & Space Complexity

Time Complexity: I2C scanner sketch

O(n)

Understanding Time Complexity

We want to understand how the time to run an I2C scanner changes as it checks more addresses.

How does the number of addresses affect the total work done?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

#include <Wire.h>

void setup() {
  Wire.begin();
  Serial.begin(9600);
  for (byte address = 1; address <= 127; address++) {
    Wire.beginTransmission(address);
    byte error = Wire.endTransmission();
    if (error == 0) {
      Serial.print("Device found at address 0x");
      Serial.println(address, HEX);
    }
  }
}

void loop() {}
    

This code scans all possible I2C addresses to find connected devices by trying to start communication with each address.

Identify Repeating Operations

• Primary operation: The for-loop that tries to communicate with each I2C address.
• How many times: It runs once for each address from 1 to 127, so about 127 times.

How Execution Grows With Input

As the number of addresses to check grows, the total time grows roughly the same way.

Input Size (n)	Approx. Operations
10	About 10 address checks
100	About 100 address checks
1000	About 1000 address checks

Pattern observation: If you double the number of addresses, the work doubles too.

Final Time Complexity

Time Complexity: O(n)

This means the time to scan grows directly in proportion to how many addresses you check.

Common Mistake

[X] Wrong: "The scanner checks all addresses instantly, so time does not depend on number of addresses."

[OK] Correct: Each address requires a communication attempt, which takes time, so more addresses mean more total time.

Interview Connect

Understanding how loops affect time helps you explain how your code scales, a key skill in programming and problem solving.

Self-Check

"What if we only scanned every other address instead of all? How would the time complexity change?"

Practice

1. What is the main purpose of an I2C scanner sketch in Arduino programming?

easy

A. To reset the Arduino board automatically

B. To detect and list all devices connected on the I2C bus

C. To send data to all I2C devices simultaneously

D. To measure voltage levels on the I2C pins

Solution

1. Step 1: Understand the role of I2C scanner

   An I2C scanner checks each possible address on the I2C bus to find connected devices.

2. Step 2: Identify the main function

   The main function is to detect and list all devices so you know which addresses are in use.

3. Final Answer:

   To detect and list all devices connected on the I2C bus -> Option B

4. Quick Check:

   I2C scanner = device finder [OK]

Hint: I2C scanner finds devices by checking addresses [OK]

Common Mistakes:

• Thinking it sends data to devices
• Confusing it with board reset functions
• Assuming it measures electrical signals

2. Which of the following is the correct way to start the I2C communication in an Arduino I2C scanner sketch?

easy

A. Wire.begin();

B. Wire.start();

C. Wire.init();

D. Wire.connect();

Solution

1. Step 1: Recall the Wire library initialization

   The Wire library uses Wire.begin(); to start I2C communication.

2. Step 2: Check other options

   Functions like Wire.start();, Wire.init();, and Wire.connect(); do not exist in the Wire library.

3. Final Answer:

   Wire.begin(); -> Option A

4. Quick Check:

   Start I2C with Wire.begin() [OK]

Hint: Use Wire.begin() to start I2C communication [OK]

Common Mistakes:

• Using non-existent Wire functions
• Confusing with other library methods
• Omitting Wire.begin() causing no communication

3. Consider this snippet from an I2C scanner sketch:

for (byte address = 1; address < 127; address++) {
  Wire.beginTransmission(address);
  byte error = Wire.endTransmission();
  if (error == 0) {
    Serial.print("Device found at 0x");
    Serial.println(address, HEX);
  }
}

What will this code print if devices are found at addresses 8 and 42?

medium

A. Device found at 0x8\nDevice found at 0x2A

B. Device found at 0x08\nDevice found at 0x2A

C. Device found at 0x08\nDevice found at 0x42

D. Device found at 0x8\nDevice found at 0x42

Solution

1. Step 1: Understand the address printing format

   The code prints the address in hexadecimal using Serial.println(address, HEX);. Hex 8 is '8', hex 42 decimal is '2A'.

2. Step 2: Check leading zeros

   Since Serial.println() with HEX does not pad zeros, 8 prints as '8', 42 prints as '2A'.

3. Final Answer:

   Device found at 0x8\nDevice found at 0x2A -> Option A

4. Quick Check:

   Hex 8 = 8, Hex 42 = 2A [OK]

Hint: Hex print does not add leading zeros by default [OK]

Common Mistakes:

• Assuming leading zeros appear automatically
• Confusing decimal 42 with hex 42
• Expecting uppercase letters always

4. You wrote this I2C scanner code but it never finds any devices:

for (byte address = 0; address <= 127; address++) {
  Wire.beginTransmission(address);
  byte error = Wire.endTransmission();
  if (error == 0) {
    Serial.print("Found device at ");
    Serial.println(address, HEX);
  }
}

What is the likely problem?

medium

A. Wire.endTransmission() must be called twice

B. Address 0 is reserved and should not be scanned

C. The loop should start from 1, not 0

D. Serial.print() cannot print HEX values

Solution

1. Step 1: Check the address range in the loop

   The I2C valid addresses range from 1 to 127. Starting at 0 is invalid and may cause issues.

2. Step 2: Understand scanning addresses

   Starting from 1 avoids reserved addresses and ensures proper scanning.

3. Final Answer:

   The loop should start from 1, not 0 -> Option C

4. Quick Check:

   Scan addresses 1 to 127 [OK]

Hint: Start scanning from address 1, not 0 [OK]

Common Mistakes:

• Including address 0 which is reserved
• Calling Wire.endTransmission() multiple times unnecessarily
• Thinking Serial.print() can't print HEX

5. You want to modify an I2C scanner sketch to skip addresses where Wire.endTransmission() returns 4 (other error) and print a summary count of devices found at the end. Which code snippet correctly implements this?

hard

A. int count = 0;
for (byte addr = 1; addr < 128; addr++) {
Wire.beginTransmission(addr);
byte error = Wire.endTransmission();
if (error == 0) {
Serial.print("Device at 0x"); Serial.println(addr, HEX);
count++;
} else if (error == 4) {
count--;
}
}
Serial.print("Total devices: "); Serial.println(count);

B. int count = 0;
for (byte addr = 0; addr <= 127; addr++) {
Wire.beginTransmission(addr);
byte error = Wire.endTransmission();
if (error == 0) {
Serial.print("Device at 0x"); Serial.println(addr, HEX);
count++;
} else if (error == 4) {
break;
}
}
Serial.print("Total devices: "); Serial.println(count);

C. int count = 0;
for (byte addr = 1; addr < 127; addr++) {
Wire.beginTransmission(addr);
byte error = Wire.endTransmission();
if (error == 0) {
Serial.print("Device at 0x"); Serial.println(addr, HEX);
count++;
} else if (error == 4) {
Serial.println("Error 4 at address " + addr);
}
}
Serial.print("Total devices: "); Serial.println(count);

D. int count = 0;
for (byte addr = 1; addr < 127; addr++) {
Wire.beginTransmission(addr);
byte error = Wire.endTransmission();
if (error == 0) {
Serial.print("Device at 0x"); Serial.println(addr, HEX);
count++;
} else if (error == 4) {
continue;
}
}
Serial.print("Total devices: "); Serial.println(count);

Solution

1. Step 1: Set correct address range and initialize count

   The loop must scan addresses 1 to 126 (less than 127). Initialize count to zero before loop.

2. Step 2: Handle error 4 by skipping and count devices found

   If error is 0, print address and increment count. If error is 4, skip with continue to avoid stopping the scan.

3. Step 3: Print total devices after loop

   After scanning, print the total count of devices found.

4. Final Answer:

   Code snippet A correctly implements the requirements -> Option D

5. Quick Check:

   Skip error 4 with continue, count devices [OK]

Hint: Use continue to skip error 4 and count devices found [OK]

Common Mistakes:

• Using break instead of continue on error 4
• Starting loop at 0 instead of 1
• Decreasing count on error instead of skipping