Problem 219: Skew-cost Coding

Project Euler Problem 219 Solution

EulerSolve provides an optimized solution for Project Euler Problem 219, Skew-cost Coding, with C++, Python, Java, and a step-by-step mathematical explanation.

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 $c+4$....

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

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

Problem 219 source code

C++

#include <algorithm>
#include <atomic>
#include <cstdint>
#include <exception>
#include <iostream>
#include <limits>
#include <queue>
#include <string>
#include <thread>
#include <vector>

namespace {

using u64 = std::uint64_t;
using u128 = __uint128_t;

struct Options {
    u64 n = 1'000'000'000ULL;
    u64 checkpoint_max_n = 200'000ULL;
    bool run_checkpoints = true;
    unsigned requested_threads = 0U;
};

bool parse_u64_after_prefix(const std::string& arg,
                            const std::string& prefix,
                            u64& value) {
    if (arg.rfind(prefix, 0U) != 0U) {
        return false;
    }

    const std::string tail = arg.substr(prefix.size());
    if (tail.empty()) {
        return false;
    }

    u64 parsed = 0ULL;
    for (const char c : tail) {
        if (c < '0' || c > '9') {
            return false;
        }

        const u64 digit = static_cast<u64>(c - '0');
        if (parsed > (std::numeric_limits<u64>::max() - digit) / 10ULL) {
            return false;
        }
        parsed = parsed * 10ULL + digit;
    }

    value = parsed;
    return true;
}

bool parse_unsigned_after_prefix(const std::string& arg,
                                 const std::string& prefix,
                                 unsigned& value) {
    u64 parsed = 0ULL;
    if (!parse_u64_after_prefix(arg, prefix, parsed)) {
        return false;
    }

    if (parsed > static_cast<u64>(std::numeric_limits<unsigned>::max())) {
        return false;
    }

    value = static_cast<unsigned>(parsed);
    return true;
}

bool parse_arguments(const int argc, char** argv, Options& options) {
    for (int i = 1; i < argc; ++i) {
        const std::string arg(argv[i]);

        if (arg == "--skip-checkpoints") {
            options.run_checkpoints = false;
            continue;
        }
        if (parse_u64_after_prefix(arg, "--n=", options.n)) {
            continue;
        }
        if (parse_u64_after_prefix(arg, "--checkpoint-max-n=", options.checkpoint_max_n)) {
            continue;
        }
        if (parse_unsigned_after_prefix(arg, "--threads=", options.requested_threads)) {
            continue;
        }

        std::cerr << "Unknown argument: " << arg << '\n';
        return false;
    }

    if (options.n == 0ULL) {
        std::cerr << "--n must be >= 1.\n";
        return false;
    }

    return true;
}

std::string to_string_u128(u128 value) {
    if (value == 0U) {
        return "0";
    }

    std::string digits;
    while (value > 0U) {
        const unsigned digit = static_cast<unsigned>(value % 10U);
        digits.push_back(static_cast<char>('0' + digit));
        value /= 10U;
    }

    std::reverse(digits.begin(), digits.end());
    return digits;
}

u128 choose_exact(unsigned n, unsigned k) {
    if (k > n) {
        return 0U;
    }
    k = std::min(k, n - k);

    u128 result = 1U;
    for (unsigned i = 1; i <= k; ++i) {
        result = (result * static_cast<u128>(n - k + i)) / static_cast<u128>(i);
    }

    return result;
}

std::vector<u128> build_cost_multiplicity(const unsigned max_cost) {
    std::vector<u128> count(static_cast<std::size_t>(max_cost) + 1ULL, 0U);
    count[0] = 1U;

    for (unsigned c = 1; c <= max_cost; ++c) {
        count[c] = count[c - 1];
        if (c >= 4U) {
            count[c] += count[c - 4U];
        }
    }

    return count;
}

u128 cost_multiplicity_combinatorial(const unsigned cost) {
    u128 total = 0U;

    for (unsigned ones = 0; 4U * ones <= cost; ++ones) {
        const unsigned zeros = cost - 4U * ones;
        const unsigned length = zeros + ones;
        total += choose_exact(length, ones);
    }

    return total;
}

u128 sum_smallest_internal_node_costs(const u64 internal_nodes) {
    if (internal_nodes == 0ULL) {
        return 0U;
    }

    std::vector<u128> multiplicity;
    multiplicity.reserve(128U);
    multiplicity.push_back(1U);

    u64 used = 0ULL;
    u128 sum = 0U;

    for (u64 cost = 0ULL; used < internal_nodes; ++cost) {
        if (cost >= multiplicity.size()) {
            u128 next = multiplicity[static_cast<std::size_t>(cost - 1ULL)];
            if (cost >= 4ULL) {
                next += multiplicity[static_cast<std::size_t>(cost - 4ULL)];
            }
            multiplicity.push_back(next);
        }

        const u64 remaining = internal_nodes - used;
        const u64 take = (multiplicity[static_cast<std::size_t>(cost)] >= static_cast<u128>(remaining))
                             ? remaining
                             : static_cast<u64>(multiplicity[static_cast<std::size_t>(cost)]);

        sum += static_cast<u128>(take) * static_cast<u128>(cost);
        used += take;
    }

    return sum;
}

u128 solve_cost(const u64 n) {
    if (n <= 1ULL) {
        return 0U;
    }

    const u64 internal_nodes = n - 1ULL;
    const u128 internal_sum = sum_smallest_internal_node_costs(internal_nodes);
    return static_cast<u128>(5ULL) * static_cast<u128>(internal_nodes) + internal_sum;
}

std::vector<u128> greedy_reference_costs(const u64 max_n) {
    std::vector<u128> costs(static_cast<std::size_t>(max_n) + 1ULL, 0U);
    if (max_n == 0ULL) {
        return costs;
    }

    std::priority_queue<u64, std::vector<u64>, std::greater<u64>> leaves;
    leaves.push(0ULL);

    costs[1] = 0U;
    u128 total = 0U;

    for (u64 n = 2ULL; n <= max_n; ++n) {
        const u64 c = leaves.top();
        leaves.pop();

        total += static_cast<u128>(c) + static_cast<u128>(5ULL);

        leaves.push(c + 1ULL);
        leaves.push(c + 4ULL);

        costs[static_cast<std::size_t>(n)] = total;
    }

    return costs;
}

unsigned pick_thread_count(const unsigned requested_threads, const u64 workload) {
    if (workload < 20'000ULL) {
        return 1U;
    }

    unsigned threads = requested_threads;
    if (threads == 0U) {
        threads = std::thread::hardware_concurrency();
        if (threads == 0U) {
            threads = 4U;
        }
    }

    if (threads == 0U) {
        threads = 1U;
    }

    if (threads > workload) {
        threads = static_cast<unsigned>(workload);
    }

    return std::max(1U, threads);
}

bool threaded_formula_vs_greedy_check(const std::vector<u128>& reference,
                                      const unsigned requested_threads) {
    if (reference.size() <= 1U) {
        return true;
    }

    const u64 max_n = static_cast<u64>(reference.size() - 1U);
    const unsigned threads = pick_thread_count(requested_threads, max_n);

    std::atomic<bool> ok(true);
    std::atomic<u64> mismatch_n(0ULL);
    std::vector<std::thread> workers;
    workers.reserve(threads);

    for (unsigned t = 0; t < threads; ++t) {
        workers.emplace_back([&, t]() {
            for (u64 n = 1ULL + static_cast<u64>(t); n <= max_n; n += static_cast<u64>(threads)) {
                if (!ok.load(std::memory_order_relaxed)) {
                    return;
                }

                const u128 got = solve_cost(n);
                if (got != reference[static_cast<std::size_t>(n)]) {
                    ok.store(false, std::memory_order_relaxed);
                    mismatch_n.store(n, std::memory_order_relaxed);
                    return;
                }
            }
        });
    }

    for (std::thread& worker : workers) {
        worker.join();
    }

    if (!ok.load(std::memory_order_relaxed)) {
        const u64 n = mismatch_n.load(std::memory_order_relaxed);
        std::cerr << "Checkpoint failed at n=" << n
                  << ": formula=" << to_string_u128(solve_cost(n))
                  << ", greedy=" << to_string_u128(reference[static_cast<std::size_t>(n)])
                  << '\n';
        return false;
    }

    return true;
}

bool run_checkpoints(const Options& options) {
    if (solve_cost(6ULL) != 35U) {
        std::cerr << "Checkpoint failed: Cost(6) must be 35.\n";
        return false;
    }

    constexpr unsigned kCostRecurrenceCheckLimit = 70U;
    const std::vector<u128> recurrence = build_cost_multiplicity(kCostRecurrenceCheckLimit);
    for (unsigned c = 0; c <= kCostRecurrenceCheckLimit; ++c) {
        const u128 via_recurrence = recurrence[c];
        const u128 via_combinatorics = cost_multiplicity_combinatorial(c);
        if (via_recurrence != via_combinatorics) {
            std::cerr << "Checkpoint failed for cost=" << c
                      << ": recurrence=" << to_string_u128(via_recurrence)
                      << ", combinatorics=" << to_string_u128(via_combinatorics)
                      << '\n';
            return false;
        }
    }

    const u64 max_n = std::max<u64>(6ULL, options.checkpoint_max_n);
    const std::vector<u128> reference = greedy_reference_costs(max_n);

    if (reference[6] != 35U) {
        std::cerr << "Checkpoint failed: greedy Cost(6) mismatch.\n";
        return false;
    }

    if (!threaded_formula_vs_greedy_check(reference, options.requested_threads)) {
        return false;
    }

    return true;
}

}  // namespace

int main(int argc, char** argv) {
    Options options;
    if (!parse_arguments(argc, argv, options)) {
        return 1;
    }

    try {
        if (options.run_checkpoints && !run_checkpoints(options)) {
            return 1;
        }

        const u128 answer = solve_cost(options.n);
        std::cout << "Cost(" << options.n << ") = " << to_string_u128(answer) << '\n';
    } catch (const std::exception& ex) {
        std::cerr << "Error: " << ex.what() << '\n';
        return 1;
    }

    return 0;
}

Python

# Problem 219: Skew-cost Coding
# Minimum cost prefix-free code for 10^9 symbols.
# Costs: 1 for bit '1', 4 for bit '0'.
# Greedy: expand cheapest leaf, adding children at cost c+1 and c+4.
# Formula: Cost(n) = 5*(n-1) + sum of (n-1) smallest internal node costs.
# Internal node costs follow recurrence: multiplicity[c] = multiplicity[c-1] + multiplicity[c-4]

def solve():
    n = 1000000000
    if n <= 1:
        print(0)
        return
    
    internal_nodes = n - 1
    
    # Count how many nodes have each cost level
    # multiplicity[c] = multiplicity[c-1] + multiplicity[c-4], multiplicity[0] = 1
    used = 0
    total_cost = 0
    multiplicity = [0] * 200  # grow as needed
    multiplicity[0] = 1
    
    cost = 0
    while used < internal_nodes:
        if cost >= len(multiplicity):
            multiplicity.extend([0] * 100)
        if cost == 0:
            multiplicity[0] = 1
        else:
            multiplicity[cost] = multiplicity[cost - 1]
            if cost >= 4:
                multiplicity[cost] += multiplicity[cost - 4]
        
        remaining = internal_nodes - used
        take = min(multiplicity[cost], remaining)
        total_cost += take * cost
        used += take
        cost += 1
    
    answer = 5 * internal_nodes + total_cost
    print(answer)

solve()

Java

import java.math.BigInteger;

public class Euler219 {
    public static void main(String[] args) {
        long n = 1000000000L;
        if (n <= 1) {
            System.out.println(0);
            return;
        }
        long internalNodes = n - 1;
        // multiplicity[c] = multiplicity[c-1] + multiplicity[c-4], multiplicity[0] = 1
        long used = 0;
        BigInteger totalCost = BigInteger.ZERO;
        BigInteger[] mult = new BigInteger[200];
        mult[0] = BigInteger.ONE;
        int cost = 0;
        while (used < internalNodes) {
            if (cost >= mult.length) {
                BigInteger[] newMult = new BigInteger[mult.length + 100];
                System.arraycopy(mult, 0, newMult, 0, mult.length);
                mult = newMult;
            }
            if (cost > 0) {
                mult[cost] = mult[cost - 1];
                if (cost >= 4)
                    mult[cost] = mult[cost].add(mult[cost - 4]);
            }
            long remaining = internalNodes - used;
            BigInteger rem = BigInteger.valueOf(remaining);
            long take;
            if (mult[cost].compareTo(rem) >= 0)
                take = remaining;
            else
                take = mult[cost].longValueExact();
            totalCost = totalCost.add(BigInteger.valueOf(take).multiply(BigInteger.valueOf(cost)));
            used += take;
            cost++;
        }
        BigInteger answer = BigInteger.valueOf(5).multiply(BigInteger.valueOf(internalNodes)).add(totalCost);
        System.out.println(answer);
    }
}