Fast Exponentiation Power in Log N in DSA Python - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind Fast Exponentiation (Power) algorithm?

It reduces the number of multiplications by repeatedly squaring the base and using the binary representation of the exponent, achieving a time complexity of O(log n).

intermediate

How does Fast Exponentiation handle odd exponents differently than even exponents?

For odd exponents, it multiplies the result by the base once and then reduces the exponent by 1 to make it even, allowing repeated squaring to continue.

beginner

What is the time complexity of Fast Exponentiation compared to the naive method?

Fast Exponentiation runs in O(log n) time, which is much faster than the naive O(n) method that multiplies the base n times.

intermediate

Show the binary representation of exponent 13 and explain how it helps in Fast Exponentiation.

13 in binary is 1101. Fast Exponentiation uses this to multiply powers of base corresponding to bits set to 1 (2^3, 2^2, 2^0), skipping unnecessary multiplications.

beginner

Write the base case for the recursive Fast Exponentiation algorithm.

When the exponent is 0, return 1 because any number raised to the power 0 is 1.

What is the time complexity of Fast Exponentiation?

A O(log n)

B O(n)

C O(n log n)

D O(1)

Fast Exponentiation reduces the number of multiplications by using repeated squaring, resulting in O(log n) time.

In Fast Exponentiation, what do you do when the exponent is odd?

A Multiply result by base and subtract 1 from exponent

B Divide exponent by 2

C Return 0

D Multiply base by itself twice

For odd exponents, multiply the result by base once and reduce exponent by 1 to make it even.

Which of these is NOT a benefit of Fast Exponentiation?

A Faster computation for large exponents

B Reduces time complexity from O(n) to O(log n)

C Uses binary representation of exponent

D Always uses more multiplications than naive method

Fast Exponentiation uses fewer multiplications than the naive method, not more.

What is the result of 2^0 using Fast Exponentiation?

A 0

B 1

C 2

D Undefined

Any number raised to the power 0 is 1.

Which binary bit positions are used in Fast Exponentiation for exponent 13 (binary 1101)?

A Positions 0 and 1

B Positions 1 and 2

C Positions 0, 2, and 3

D Positions 1, 2, and 3

Bits set to 1 are at positions 0, 2, and 3 in binary 1101 (right to left, zero-based).

Explain how Fast Exponentiation reduces the number of multiplications compared to the naive method.

Think about how powers of two help skip many multiplications.

Describe the steps to compute 3^13 using Fast Exponentiation.

Use the binary bits to decide when to multiply.

Practice

1. What is the main advantage of using Fast Exponentiation to calculate a^b compared to the simple repeated multiplication?

easy

A. It works only for positive integers

B. It uses less memory than repeated multiplication

C. It reduces the time complexity from O(b) to O(log b)

D. It avoids using multiplication altogether

Solution

1. Step 1: Understand repeated multiplication complexity

   Calculating a^b by multiplying a by itself b times takes O(b) steps.

2. Step 2: Analyze fast exponentiation approach

   Fast exponentiation uses the idea of squaring and halving the exponent, reducing steps to O(log b).

3. Final Answer:

   It reduces the time complexity from O(b) to O(log b) -> Option C

4. Quick Check:

   Fast exponentiation = O(log b) [OK]

Hint: Remember: halve exponent each step for log speed [OK]

Common Mistakes:

• Thinking it avoids multiplication
• Confusing memory use with speed
• Assuming it only works for positive exponents

2. Which of the following is the correct base case for a recursive fast exponentiation function power(a, b) that calculates a^b?

easy

A. if b == 0: return 1

B. if a == 0: return 0

C. if b == 1: return 0

D. if a == 1: return b

Solution

1. Step 1: Recall mathematical rule for exponent zero

   Any number to the power 0 equals 1, so power(a, 0) should return 1.

2. Step 2: Check options for correct base case

   if b == 0: return 1 matches this rule: if b == 0: return 1.

3. Final Answer:

   if b == 0: return 1 -> Option A

4. Quick Check:

   Base case b=0 returns 1 [OK]

Hint: Base case: exponent zero returns 1 [OK]

Common Mistakes:

• Returning 0 when exponent is 1
• Confusing base and exponent in conditions
• Returning b instead of 1 for base case

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

def fast_power(a, b):
    if b == 0:
        return 1
    half = fast_power(a, b // 2)
    if b % 2 == 0:
        return half * half
    else:
        return half * half * a

print(fast_power(2, 5))

medium

A. 32

B. 10

C. 25

D. 16

Solution

1. Step 1: Trace recursive calls for fast_power(2, 5)

   fast_power(2,5) calls fast_power(2,2), which calls fast_power(2,1), then fast_power(2,0) returns 1.

2. Step 2: Calculate values bottom-up

   fast_power(2,1) = 1*1*2 = 2
   fast_power(2,2) = 2*2 = 4
   fast_power(2,5) = 4*4*2 = 32

3. Final Answer:

   32 -> Option A

4. Quick Check:

   2^5 = 32 [OK]

Hint: Remember: odd exponent multiplies extra base [OK]

Common Mistakes:

• Confusing multiplication order
• Forgetting to multiply by base when exponent is odd
• Incorrect base case handling

4. The following code attempts to implement fast exponentiation but has a bug. What is the bug?

def fast_power(a, b):
    if b == 0:
        return 1
    half = fast_power(a, b // 2)
    if b % 2 == 1:
        return half * half
    else:
        return half * half * a

print(fast_power(3, 3))

medium

A. Function does not handle negative exponents

B. Base case should return 0 instead of 1

C. Multiplication order is incorrect

D. The condition for odd exponent is reversed

Solution

1. Step 1: Analyze the condition for odd exponent

   The code checks if b % 2 == 1 but returns half * half which is for even exponent.

2. Step 2: Identify correct logic for odd exponent

   For odd exponent, it should return half * half * a. The conditions are swapped.

3. Final Answer:

   The condition for odd exponent is reversed -> Option D

4. Quick Check:

   Odd exponent must multiply extra base [OK]

Hint: Odd exponent means multiply extra base factor [OK]

Common Mistakes:

• Swapping odd/even conditions
• Wrong base case return value
• Ignoring negative exponent cases

5. You want to compute 3^1000 mod 1000000007 efficiently using fast exponentiation. Which of the following modifications is necessary to the standard fast exponentiation algorithm to handle this?

hard

A. Convert exponent to binary string before calculation

B. Add modulo operation after each multiplication to keep numbers small

C. Increase recursion depth limit to handle large exponent

D. Use floating point arithmetic to handle large numbers

Solution

1. Step 1: Understand the problem with large powers

   Calculating 3^1000 directly results in a huge number that can cause overflow or slow computation.

2. Step 2: Apply modulo at each multiplication step

   To keep numbers manageable and get correct modulo result, apply modulo after every multiplication.

3. Final Answer:

   Add modulo operation after each multiplication to keep numbers small -> Option B

4. Quick Check:

   Modulo after multiplication keeps numbers small [OK]

Hint: Apply modulo after every multiplication step [OK]

Common Mistakes:

• Using floating point causing precision errors
• Not applying modulo inside recursion
• Trying to convert exponent to string unnecessarily