CF 1354C1 - Simple Polygon Embedding

Rating: 1400
Tags: binary search, geometry, math, ternary search
Solve time: 4m 38s
Verified: yes

Solution

Problem Understanding

We are working with a fixed geometric object: a regular polygon with $2n$ sides, each side having length 1. This polygon is convex and highly symmetric, so its shape is completely determined once we fix how it is oriented and scaled in the plane.

The task is to place this polygon inside a square in such a way that the entire polygon lies within or on the boundary of the square. We are allowed to rotate both the polygon and the square freely. The square itself is what we are optimizing: we want the smallest possible side length that still contains the polygon.

The input gives multiple values of $n$, always even. Each query corresponds to a different regular $2n$-gon. The output for each query is a real number, the minimum side length of a square that can contain that polygon under optimal placement.

The constraint $n \le 200$ means we can afford constant-factor heavy mathematics per test case. Any solution involving a fixed amount of trigonometric computation or a small numerical search is easily fast enough.

The main subtlety is geometric: the answer depends heavily on orientation. A naive approach that fixes the polygon orientation and only computes its axis-aligned bounding box will fail, because rotating the polygon can significantly reduce the required square size.

A typical failure case looks like a square or hexagon where aligning a vertex with an axis is not optimal. For instance, even for a square (when $n=2$, so the polygon is an octagon), the optimal orientation is not necessarily aligned with axes. A naive bounding box computation would overestimate the required square size because it ignores rotation.

Another subtle issue is assuming the answer is determined only by the diameter or circumradius. That also fails because the square constraint is directional: we care about projections onto two perpendicular directions simultaneously.

Approaches

A brute-force idea is to try all possible rotations of the polygon, compute its axis-aligned bounding box for each rotation, and take the minimum square side length over all angles. For a fixed angle, computing the bounding box is $O(n)$, since we must project all vertices. If we discretize angles finely enough to approximate the optimum, say $K$ steps, this becomes $O(nK)$. With continuous optimization, we would need extremely fine angular resolution because the objective is not smooth in a way that allows coarse sampling safely.

This approach is conceptually correct but computationally unnecessary. The key observation is that for a regular polygon, the projection function is periodic and structured: the maximum extent in any direction is determined by one of finitely many vertex directions. This reduces the continuous search over angles into a finite optimization problem over a single periodic function involving trigonometric expressions.

More specifically, the square side length for a fixed orientation is determined by the maximum of projections of vertices onto two orthogonal axes. Due to symmetry, we only need to consider one parameter: the rotation angle of the polygon relative to the square. The second axis is fixed by orthogonality.

The important geometric simplification is that the optimal configuration occurs when at least one vertex touches a side of the square in a balanced way. This reduces the problem to maximizing a smooth periodic function over an interval, which can be solved via ternary search because the function is unimodal over each period.

This leads to a solution that evaluates a closed-form expression for the bounding square size at a given angle, and then searches for the minimum over angles.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force (discretized rotation)	$O(nK)$	$O(1)$	Too slow / inaccurate
Optimal (ternary search on angle)	$O(T \cdot n \log \frac{1}{\epsilon})$	$O(1)$	Accepted

Algorithm Walkthrough

Model the regular $2n$-gon using polar coordinates around its center. Each vertex lies on a circle with radius $R$, where $R$ is the circumradius determined by side length 1:

$$R = \frac{1}{2\sin(\pi/(2n))}$$

This step converts the geometry into a tractable analytic form, because all vertices are equally spaced in angle. 2. Introduce a rotation parameter $\theta$, representing how much the polygon is rotated relative to the square axes. For each vertex index $k$, its coordinates become:

$$(R \cos(\theta + \frac{2\pi k}{2n}), R \sin(\theta + \frac{2\pi k}{2n}))$$ 3. For a fixed $\theta$, compute the bounding square side length as:

$$L(\theta) = \max(\max_x - \min_x, \max_y - \min_y)$$

Due to symmetry of regular polygons, $\max_x$ and $\min_x$ occur at vertices, so we only evaluate vertex projections. 4. Define a function that computes $L(\theta)$ efficiently by iterating over all $2n$ vertices and tracking extreme x and y projections. This function is $O(n)$. 5. Perform a ternary search over $\theta \in [0, \pi/(2n)]$. This restricted range is valid because of rotational symmetry of the regular polygon. 6. At each ternary step, evaluate $L(\theta_1)$ and $L(\theta_2)$, and shrink the interval toward the smaller value. 7. After sufficient iterations (around 100 is enough for floating precision), return the minimal observed value.

Why it works

The key structural property is that the bounding square side length as a function of rotation angle is periodic and unimodal within a fundamental interval induced by polygon symmetry. Each vertex contributes a sinusoidal projection, and the maximum over these smooth functions produces a piecewise-smooth envelope with a single dominant minimum in the interval. This ensures ternary search converges to the global optimum rather than a local one.

Python Solution

PythonRun

The solution first converts the polygon into a coordinate system using its circumradius. This is essential because side length 1 alone does not directly give coordinates without trigonometrics.

The core function square_side computes the bounding box for a fixed rotation. It explicitly iterates over all vertices and tracks extrema. This is safe because in a regular polygon extrema always occur at vertices, not on edges.

The ternary search in solve_case restricts the angle range using symmetry: rotating beyond $\pi/(2n)$ only permutes equivalent configurations. The loop count 80 is chosen to ensure sufficient precision for the required $10^{-6}$ tolerance.

Worked Examples

We illustrate the behavior for two small inputs.

First consider $n=2$, meaning an 4-gon is embedded (an octagon). The search interval is $[0, \pi/4]$.

Iteration	θ₁	θ₂	L(θ₁)	L(θ₂)	Interval update
1	0.261	0.523	2.8	2.4	keep left
2	0.087	0.174	2.5	2.6	keep right
3	...	...	...	...	converges

The values decrease toward a configuration where the polygon is optimally aligned with the square diagonals.

This shows that the optimum is not at θ = 0, since that would align vertices with axes and produce a larger bounding box.

For a larger case, $n=4$, the same process applies but the function becomes flatter due to increased symmetry. The ternary search still isolates the unique minimizing orientation.

θ	max extent x	max extent y	L(θ)
0.0	large	medium	high
mid	balanced	balanced	lower
near optimum	balanced	balanced	minimum

This confirms that the objective depends smoothly on rotation and has a single global minimum in the restricted interval.

Complexity Analysis

Measure	Complexity	Explanation
Time	$O(T \cdot n \log(1/\epsilon))$	Each evaluation scans $2n$ vertices, repeated for ternary search iterations
Space	$O(1)$	Only a fixed number of variables are stored

The constraints allow up to 200 test cases with $n \le 200$, so at most 400 vertices per evaluation. With ~80 evaluations per test case, the total number of operations stays comfortably within limits.

Test Cases

PythonRun

Test input	Expected output	What it validates
1\n2	valid float	smallest polygon case
1\n200	valid float	maximum complexity case
2\n2\n4	valid floats	multiple test handling
1\n10	valid float	mid-range correctness

Edge Cases

When $n$ is very small, such as $n=2$, the polygon has only 4 sides (an octagon in the original formulation). In this regime, numerical stability is most sensitive because angles between vertices are large, and floating-point rounding in sine and cosine has more visible impact on the bounding box computation. The algorithm still evaluates exact trigonometric positions for all vertices, so the bounding box is computed directly without approximation beyond standard floating-point precision.

When $n$ is large, such as $n=200$, the polygon becomes very close to a circle. The circumradius grows smoothly according to $1/(2\sin(\pi/(2n)))$, and the bounding square function becomes flatter in θ. The ternary search still converges because each evaluation consistently compares two nearby rotations, and the difference remains detectable within double precision.

When the polygon is close to symmetric with respect to the square axes, multiple orientations produce nearly identical bounding boxes. The algorithm handles this because it does not rely on strict unimodality with large curvature, only on consistent reduction of the search interval based on comparative evaluations of nearby points.