Solution

Answer: 64564225042

Let each codeword correspond to a leaf in a binary tree:

taking a 0 edge costs 1
taking a 1 edge costs 4

So the cost of a codeword is the weighted path length from the root.

A prefix-free code is exactly a set of leaves in such a tree.

We seek the minimum possible total leaf cost for $n$ leaves.

1. Mathematical analysis

Suppose we currently have a leaf of cost $x$.

If we expand it into two children, then:

left child cost = $x+1$
right child cost = $x+4$

The number of leaves increases by $1$, and the total cost changes by

$$(x+1)+(x+4)-x = x+5.$$

Therefore:

To increase the number of leaves as cheaply as possible, we should always expand the currently cheapest leaf.

This greedy strategy is optimal because every expansion adds exactly one new leaf, and the increment $x+5$ is minimized by choosing the smallest available $x$.

The expansion process

Start with one empty codeword of cost $0$.

Expanding a leaf of cost $c$:

$$c \to (c+1,\ c+4)$$

Let $t_c$ be the number of times a leaf of cost $c$ is expanded.

A node of cost $c$ can only arise from:

a cost-$(c-1)$ node via a 0 edge
a cost-$(c-4)$ node via a 1 edge

Hence:

$$t_c = t_{c-1} + t_{c-4},$$

with

$$t_0=1,\qquad t_c=0 \text{ for } c<0.$$

The sequence begins:

$$1,1,1,1,2,3,4,5,7,10,14,19,\dots$$

Relating this to the number of leaves

Every expansion increases the number of leaves by $1$.

After fully processing all costs $<k$, the number of leaves is

$$1+\sum_{i=0}^{k-1} t_i.$$

The total cost accumulated is

$$\sum_{i=0}^{k-1} t_i(i+5),$$

because each expansion at cost $i$ increases total cost by $i+5$.

If we still need $r$ additional expansions at cost $k$, then

$$\text{Cost}(n) = \sum_{i=0}^{k-1} t_i(i+5) + r(k+5),$$

where

$$r = n-\left(1+\sum_{i=0}^{k-1} t_i\right).$$

2. Python implementation

def cost(n):
    # t[c] = number of leaves of cost c that get expanded
    t = [1]   # t_0 = 1

    leaves = 1   # current number of leaves
    total = 0
    c = 0

    while True:
        # extend recurrence if needed
        if c >= len(t):
            val = t[c - 1]
            if c >= 4:
                val += t[c - 4]
            t.append(val)

        # If processing all nodes of cost c would reach/exceed n,
        # finish partially here.
        if leaves + t[c] >= n:
            remaining = n - leaves
            return total + remaining * (c + 5)

        # fully process all nodes of cost c
        leaves += t[c]
        total += t[c] * (c + 5)

        c += 1

print(cost(6))        # 35
print(cost(10**9))    # final answer

3. Code walkthrough

Initialization

t = [1]

This stores the recurrence values $t_c$.

leaves = 1
total = 0

Initially we have one empty codeword of cost $0$.

Generating the recurrence

val = t[c - 1]
if c >= 4:
    val += t[c - 4]

Implements

$$t_c=t_{c-1}+t_{c-4}.$$

Determining when we stop

if leaves + t[c] >= n:

If processing all cost-$c$ leaves would overshoot the required number of codewords, we only process enough of them.

Updating the total cost

total += t[c] * (c + 5)

Each expansion at cost $c$ increases total cost by $c+5$.

4. Verification on the example

For $n=6$:

The algorithm gives

$$\text{Cost}(6)=35,$$

matching the problem statement.

For $n=10^9$, the computation yields:

$$64564225042$$

This magnitude is reasonable:

average codeword cost is around $64$
$10^9 \times 64 \approx 6.4\times10^{10}$

which matches the computed value.

Answer: 64564225042

Project Euler Problem 219