Detailed mathematical approach

Problem Summary

We want a binary prefix code for \(10^9\) symbols. One branch adds cost \(1\) and the other adds cost \(4\), so the cost of a codeword is the sum of the edge costs along its root-to-leaf path. If \(\text{Cost}(n)\) denotes the minimum possible total cost of all codewords for \(n\) symbols, then Problem 219 asks for \(\text{Cost}(10^9)\).

A direct greedy simulation with an explicit priority queue works for tiny values of \(n\), but not for \(n=10^9\). The implementations therefore replace the evolving leaf set by a counting argument on how many nodes exist at each possible cost.

Mathematical Approach

The right model is an infinite binary tree rooted at cost \(0\). From any node of cost \(c\), its two children have costs \(c+1\) and \(c+4\). A finite prefix-free code is obtained by choosing a finite full subtree and taking its leaves as the codewords.

Full Binary Trees and Internal Costs

In an optimal code tree, every internal node has exactly two children. A node with only one child would add positive cost without helping the prefix condition, so it can be contracted away. Therefore a code with \(n\) leaves has exactly \(n-1\) internal nodes.

Now start from a single leaf of cost \(0\) and expand leaves until \(n\) leaves exist. Expanding a leaf of cost \(c\) removes that leaf and creates two new leaves of costs \(c+1\) and \(c+4\). The total leaf-cost sum therefore changes by

$\Delta=(c+1)+(c+4)-c=c+5.$

If the internal-node costs, listed in the order they are expanded, are \(c_1,c_2,\dots,c_{n-1}\), then

$\text{Cost}(n)=\sum_{k=1}^{n-1}(c_k+5)=5(n-1)+\sum_{k=1}^{n-1}c_k.$

So the whole problem is reduced to finding the smallest possible sum of the \(n-1\) internal-node costs.

Why the Greedy Order Is Correct

Every descendant of a node is strictly more expensive than the node itself, because each edge adds a positive amount. That means any feasible set of internal nodes is automatically prefix-closed: if a node is internal, then every ancestor on the path from the root must also be internal.

Consequently, the minimum possible value of \(\sum c_k\) is obtained by taking the \(n-1\) cheapest nodes in the infinite skew-cost tree. This is exactly the same rule as repeatedly expanding the currently cheapest available leaf. The greedy process is therefore not just a heuristic; it is the canonical order in which internal nodes appear in the optimum.

Counting How Many Nodes Have a Given Cost

Let \(f(c)\) be the number of nodes in the infinite tree whose path cost is exactly \(c\). We have \(f(0)=1\) for the root and \(f(c)=0\) for negative \(c\). For \(c\ge 1\), any node of cost \(c\) must end either with a \(1\)-cost edge from a node of cost \(c-1\) or with a \(4\)-cost edge from a node of cost \(c-4\). These two cases are disjoint, so

$f(c)=f(c-1)+f(c-4),\qquad f(0)=1,\qquad f(c)=0\text{ for }c<0.$

This recurrence is the core object used by all three implementations. It tells us the multiplicity of each cost level without ever constructing the code tree explicitly.

There is also a combinatorial interpretation used as an independent check. If a node has cost \(c\), and exactly \(a\) of its edges are the \(4\)-cost kind, then the remaining number of \(1\)-cost edges is \(c-4a\). The total path length is therefore \(c-3a\), and the number of such words is \(\binom{c-3a}{a}\). Summing over all admissible \(a\) gives

$f(c)=\sum_{a=0}^{\lfloor c/4\rfloor}\binom{c-3a}{a}.$

The C++ implementation checks that this binomial-count formula matches the recurrence on small cost levels, which confirms that the state space has been modeled correctly.

From Multiplicities to the Optimal Total

Set \(m=n-1\), because that is the number of internal nodes we must choose. If we scan cost levels in increasing order, then at level \(c\) there are exactly \(f(c)\) nodes of that cost available to become internal nodes. Let

$A(C)=\sum_{c=0}^{C}f(c),\qquad B(C)=\sum_{c=0}^{C}c\,f(c).$

Let \(C_\star\) be the first cost level with \(A(C_\star)\ge m\). Then all nodes with cost below \(C_\star\) are taken completely, while the last layer is only partially used. Therefore

$\sum_{k=1}^{m}c_k=B(C_\star-1)+C_\star\bigl(m-A(C_\star-1)\bigr),$

and the final answer is

$\boxed{\text{Cost}(n)=5(n-1)+B(C_\star-1)+C_\star\bigl((n-1)-A(C_\star-1)\bigr).}$

The implementations compute this same quantity incrementally: they keep a running count of how many internal nodes have already been consumed and a running sum of their costs, and they stop as soon as the quota \(n-1\) is filled.

Worked Example: \(\text{Cost}(6)=35\)

The checkpoint \(\text{Cost}(6)=35\) is built into the main implementation. Here \(m=n-1=5\), so we need the five cheapest internal-node costs. The recurrence begins

$f(0)=1,\quad f(1)=1,\quad f(2)=1,\quad f(3)=1,\quad f(4)=2.$

Thus the first five internal nodes have costs \(0,1,2,3,4\), whose sum is \(10\). Plugging that into the master formula gives

$\text{Cost}(6)=5\cdot 5+(0+1+2+3+4)=25+10=35.$

The same example can be seen directly through leaf expansions: \(0\to(1,4)\to(2,4,5)\to(3,4,5,6)\to(4,4,5,6,7)\to(4,5,5,6,7,8)\), and the final leaf-cost sum is indeed \(35\).

How the Code Works

The C++, Python, and Java implementations all follow the same production method. They treat \(n-1\) as the number of internal nodes that must be selected, generate the multiplicities \(f(c)\) level by level from the recurrence \(f(c)=f(c-1)+f(c-4)\), and accumulate the sum of the cheapest internal-node costs.

At each cost level \(c\), the implementation takes

$\min\bigl(f(c),\text{remaining internal nodes}\bigr)$

nodes from that layer, adds \(c\) times that amount to the running total of internal-node costs, and advances to the next level. Once the required \(n-1\) nodes have been consumed, it adds the constant term \(5(n-1)\) and prints the result.

The arithmetic is intentionally wide. Python uses built-in arbitrary-precision integers, Java uses arbitrary-precision integer objects, and the C++ implementation uses a wider integer type because both multiplicities and totals quickly outgrow ordinary 32-bit arithmetic. The C++ implementation also includes validation logic: it checks the known value \(\text{Cost}(6)=35\), compares the recurrence against the binomial formula for \(f(c)\), and matches the formula-based solver against a direct greedy priority-queue simulation on small inputs.

Complexity Analysis

Let \(C_\star\) be the largest cost level that must be visited before the first \(n-1\) internal nodes have been collected. The main solver performs a single forward pass through costs \(0,1,\dots,C_\star\), so its running time is \(O(C_\star)\).

The table of multiplicities also has size \(C_\star+1\), so the memory usage is \(O(C_\star)\). This is far smaller than an explicit greedy heap simulation up to \(n=10^9\), which would require maintaining an enormous frontier and would scale like \(O(n\log n)\). The priority-queue approach appears only in the small checkpoint code, not in the real computation.

Footnotes and References

1. Problem page: https://projecteuler.net/problem=219
2. Prefix code: Wikipedia - Prefix code
3. Huffman coding: Wikipedia - Huffman coding
4. Recurrence relation: Wikipedia - Recurrence relation
5. Binomial coefficient: Wikipedia - Binomial coefficient
6. Full binary tree: Wikipedia - Types of binary trees