Mastering the Josephus Problem: A Simple and Effective Approach

Imagine you’re trapped in a cave with 50 others, and every second person is eliminated until only one survives. Where should you stand to ensure survival?

Sounds like a scene from a thriller, right? This isn’t just a historical curiosity—it’s an important problem in recursion, mathematical induction, and coding interviews. It’s actually a classic algorithmic puzzle called the Josephus Problem—a problem that has puzzled mathematicians and computer scientists for centuries. Today, I’ll break it down step by step, making it super easy to understand, even if you're a beginner.

By the end of this article, you’ll not only know how to solve the Josephus Problem, but also why it works and how to optimize it efficiently—and maybe even impress your next interviewer!

What is the Josephus Problem?

The problem is based on a real (or at least legendary) story from 67 A.D. Josephus, a Jewish historian, and 40 other rebels were trapped in a cave during the Jewish-Roman War. Instead of surrendering, they decided to form a circle and eliminate every k-th person until only one was left.

Josephus, being a clever man, figured out exactly where he needed to stand to survive.

Now, the computational version of this problem is:

Given n people standing in a circle, and eliminating every k-th person, determine which position will be the last survivor.

Step 1: Understanding the Problem with an Example

Let’s take n = 5 and k = 2 (i.e., every 2nd person is eliminated).

We number the people 0 to 4 (zero-based index for easy computation).

Initial order: 0 1 2 3 4

Now, let’s start eliminating every 2nd person:

1. First kill: 1 (Now we have: 0 2 3 4)

2. Second kill: 3 (Now we have: 0 2 4)

3. Third kill: 0 (Now we have: 2 4)

4. Fourth kill: 4 (Now we have: 2 – Survivor!)

So, for n = 5, k = 2 → Position 2 survives.

Step 2: Identifying the Pattern

Instead of manually going through eliminations, can we find a formula to predict the survivor’s position?

Let’s consider smaller cases to observe the pattern:

Do you notice something? When n is a power of 2 (like 2, 4, 8), the survivor is always at position 0.

For non-power-of-2 numbers, the survivor follows a specific pattern based on the remaining extra people after removing the largest power of 2.

Power of 2 Case:

If n is a power of 2 (like 2, 4, 8, 16...), the survivor is always at position 0.

🔍 Why does this happen?

When n is a power of 2, the elimination follows a perfect doubling pattern, always leaving the first person (index 0) alive.

Non-Power-of-2 Case:

If n is not a power of 2, we can break it into two parts:

1. The largest power of 2 ≤ n (let’s call it 2^t)
2. The extra remaining people, which is l = n - 2^t

The survivor follows this formula:

J(n,2) = 2l

💡 Why does this work?

• The first l people (those beyond 2^t) will be eliminated first.
• The remaining 2^t people now form a perfect power of 2 situation, which we already know leaves the first person alive (position 0).
• Since l extra people were removed first, the numbering shifts forward by 2l.

Example: Josephus(10,2)

• Largest power of 2 ≤ 10 is 8
• Extra people: l = 10 - 8 = 2
• Survivor’s position = 2 × l = 2 × 2 = 4

✅ So, for n = 10, the survivor is at position 4.

Let’s break this down mathematically.

Step 3: Deriving the Recurrence Relation

Now, let’s define the problem in terms of n and k.

1. Suppose J(n, k) is the position of the survivor.
2. If we remove the first eliminated person (at position k-1), the circle shrinks.
3. The problem reduces to Josephus(n-1, k), but the numbering shifts.

Thus, the new numbering follows:

J(n,k) = (J(n−1,k)+ k) mod n

This formula simply adjusts the survivor’s position after each step, making sure it wraps around the circle using the modulo operation.

Breaking it Down Step by Step

Let’s see an example for n = 5, k = 2 using the recurrence relation:

1. Base case: If n = 1, survivor is 0.

J(1, 2) = 0

2. Compute for n = 2:

J(2,2) = (J(1,2)+2) mod2 = (0+2) mod2 = 0

Survivor: 0

3. Compute for n = 3:

J(3,2) = (J(2,2)+2) mod3 = (0+2) mod3 = 2

Survivor: 2

4. Compute for n = 4:

J(4,2) = (J(3,2)+2) mod4 = (2+2) mod4 = 0

Survivor: 0

5. Compute for n = 5:

J(5,2) = (J(4,2)+2) mod5 = (0+2) mod5 = 2

Survivor: 2

The formula works!

Step 4: Implementing in Python (Recursive and Iterative)

Now that we have our formula, let’s implement it in Python.

Recursive Approach

def josephus_problem(n, k):
    if n == 1:
        return 0
    return (josephus_problem(n - 1, k) + k) % n

# Example
print(josephus_problem(5, 2))  # Output: 2
        

• Time Complexity: O(n)
• Space Complexity: O(n) (due to recursion stack)

Iterative Approach (More Efficient)

def josephus_iterative(n, k):
    survivor = 0
    for i in range(2, n + 1):
        survivor = (survivor + k) % i
    return survivor

# Example
print(josephus_iterative(5, 2))  # Output: 2
        

• Time Complexity: O(n)
• Space Complexity: O(1)

This version avoids recursion, making it faster and memory-efficient.

Step 5: Special Case – The Binary Josephus Problem (k=2, O(1) Solution)

As seen in Step 2, k=2 is a special case where every second person is eliminated.

For k=2, a special property helps us solve it in constant time:

1. Find the largest power of 2 ≤ n (call it 2^t).
2. Compute l = n - 2^t.
3. Use the formula:

J(n,2) = 2×l

Why Does This Happen?

For k=2, every alternate person is removed in a predictable way, which allows for a binary pattern (hence the neat power-of-2 trick).

For k>2, eliminations jump by more than 1, leading to irregular shifts. That’s why we must use the recurrence relation for these cases. No shortcut!

Optimized Python Code for k=2

def josephus_binary(n):
    l = n - (1 << (n.bit_length() - 1))  # Largest power of 2 <= n
    return 2 * l

# Example
print(josephus_binary(41))  # Output: 19
        

• Time Complexity: O(1)
• Space Complexity: O(1)

Conclusion

The Josephus Problem is not just a historical puzzle—it’s a test of your problem-solving and logical thinking. This problem tests recursion, modular arithmetic, bit manipulation, and mathematical reasoning—all valuable skills in coding interviews!

Final Thought

Next time you see a circular elimination problem, you’ll know exactly what to do. Whether it’s a coding test or an actual life-or-death survival game (hopefully not!), you’ve got this 😉

Would love to hear your thoughts! Have you encountered the Josephus Problem in an interview? How did you solve it? Let’s discuss!

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