mgrettondann/partridge-cpp
1
0
Fork 0
Code Issues Activity

Code tidy as reported by clang-tidy

Browse Source
This commit is contained in:
Matthew Gretton-Dann
2025-09-04 11:35:37 +02:00
parent ff43ba5287
commit 5a01260281
1 changed files with 22 additions and 35 deletions
Download Patch File Download Diff File Expand all files Collapse all files

57
main.cc
View File

@@ -1,6 +1,6 @@
/** \file main.cc
* \author Matthew Gretton-Dann
* \brief Solves the partirige problem for user specified size.
* \brief Solves the Partridge problem for user specified size.
*
* Copyright 2025, Matthew-Gretton-Dann
* SPDX: Apache-2.0
@@ -11,13 +11,12 @@
#include <string_view>
#include <vector>
#include <iostream>
#include <set>
using size_t = std::size_t;
namespace {
using size_t = std::uint64_t;
/** (x, y) pair storing a position. */
using Pos = std::size_t;
using Pos = size_t;
/** A square - consisting of position of closest corner to origin, and side-length.
*/
@@ -40,10 +39,10 @@ namespace {
~Square() noexcept = default;
/** Get x co-ordinate of closest corner to origin. */
auto pos() const noexcept -> Pos { return pos_; }
[[nodiscard]] auto pos() const noexcept -> Pos { return pos_; }
/** Get side length. */
auto length() const noexcept -> size_t { return length_; }
[[nodiscard]] auto length() const noexcept -> size_t { return length_; }
private:
Pos pos_; ///< Position of corner closest to origin
@@ -66,7 +65,7 @@ namespace {
~Results() noexcept = default;
auto length() const noexcept -> size_t { return length_; }
[[nodiscard]] auto length() const noexcept -> size_t { return length_; }
/** Output the grid. */
auto output() const -> void {
@@ -87,8 +86,8 @@ namespace {
s[x + y * length_] = c;
}
auto sq_x(Square const &sq) const noexcept -> size_t { return sq.pos() % length_; }
auto sq_y(Square const &sq) const noexcept -> size_t { return sq.pos() / length_; }
[[nodiscard]] auto sq_x(Square const &sq) const noexcept -> size_t { return sq.pos() % length_; }
[[nodiscard]] auto sq_y(Square const &sq) const noexcept -> size_t { return sq.pos() / length_; }
auto prettify_sq(std::string &s, Square const &sq) const noexcept -> void {
switch (sq.length()) {
@@ -117,7 +116,7 @@ namespace {
size_t i = sq_x(sq) + n - 1;
while (n != 0) {
set(s, --i, sq_y(sq) + 1, '0' + (n % 10));
set(s, --i, sq_y(sq) + 1, static_cast<char>('0' + static_cast<char>(n % 10)));
n /= 10;
}
}
@@ -134,7 +133,7 @@ namespace {
using T = std::int_fast64_t;
/** Construct a grid of given side-length. */
Grid(size_t length) : grid_(length * length, empty), length_(length) {
explicit Grid(size_t length) : grid_(length * length, empty), length_(length) {
}
Grid(Grid const &other) = delete;
@@ -148,12 +147,12 @@ namespace {
~Grid() noexcept = default;
/** Get grid length */
auto end() const noexcept -> size_t { return grid_.size(); }
[[nodiscard]] auto end() const noexcept -> size_t { return static_cast<size_t>(grid_.size()); }
/** Add a square to the grid. */
auto add(Square const &sq) noexcept -> void {
/* One would expect the fastest way to do this would be to have x be the
* fastest increasing index so we stores [pos, pos + 1,..., pos+length, ...]
* fastest increasing index so we store [pos, pos + 1,..., pos+length, ...]
* But experimentation tells us this isn't so, and storing
* [pos, pos + length, ..., pos + 1, ...] is faster!
*/
@@ -175,7 +174,7 @@ namespace {
/** \brief Get length of the largest square that fits at \a pos in the grid.
*/
auto largest_square(Pos pos, size_t n) const noexcept -> size_t {
[[nodiscard]] auto largest_square(Pos pos, size_t n) const noexcept -> size_t {
assert(pos < end());
/* Because of how we walk through the grid (starting at 0,0 then increasing
@@ -203,27 +202,15 @@ namespace {
/** Get the next position to check starting at pos.
*
* Returns grid_.length() if no more positions avaialble.
* Returns grid_.length() if no more positions available.
*/
auto next_pos(Pos pos) const noexcept -> Pos {
auto const b = grid_.begin() + pos;
[[nodiscard]] auto next_pos(Pos pos) const noexcept -> Pos {
auto const b = grid_.begin() + static_cast<std::ptrdiff_t>(pos);
auto const p = std::find(b, grid_.end(), empty);
return p - grid_.begin();
}
private:
/** Set the grid position (x, y) to the character c.
*
* It is an error if (x, y) is already set to c - as that means we have overlapping
* squares.
*/
auto set(size_t x, size_t y, T c) noexcept -> void {
assert(x < length_);
assert(y < length_);
assert(grid_[x + y * length_] != c);
grid_[x + y * length_] = c;
}
std::vector<T> grid_; ///< The grid
size_t length_; ///< Side length
@@ -231,7 +218,7 @@ namespace {
static constexpr char filled = 1; ///< Character used for a filled cell,
};
/** Get the n'th triangular number. */
/** Get the n-th triangular number. */
auto triangle_num(size_t n) noexcept -> size_t { return (n * (n + 1)) / 2; }
/** Vector used to identify the available squares. */
@@ -253,7 +240,7 @@ namespace {
* available squares until we find one that fits.
*/
// grid is our inprogress grid of square positions.
// grid is our in-progress grid of square positions.
auto const length = triangle_num(n);
Grid grid(length);
@@ -276,7 +263,7 @@ namespace {
while (true) {
/* If the idx is 0 we've looked at all possible square lengths for this
* position, and they've failed. Pop the last square of the stack, remove
* it from the grid and try the next smaller one in the same position.
* it from the grid and try the next smaller size in the same position.
*/
if (idx == 0) {
// No squares on the stack -> failed to find a solution.
@@ -312,13 +299,13 @@ namespace {
if (pos == grid.end()) { break; }
}
return Results(length, sqs);
return {length, sqs};
}
} // anon namespace
int main(int argc, char **argv) {
auto n = (argc == 1) ? 8 : std::atol(argv[1]);
auto grid = find_solution(n);
auto const grid = find_solution(n);
std::cout << "Partridge problem " << n << " side length " << grid.length() << '\n';
grid.output();
return 0;