0737MediumMathGeometry

Four Pegs Marking a Square Plot

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A land surveyor has hammered four pegs into a field and recorded each one as a coordinate pair [x, y] on the site grid. The pegs are handed to you as p1, p2, p3 and p4 in no particular order: the recording sequence says nothing about how the pegs sit relative to one another, so the corners of the plot may appear in any of the four positions.

Decide whether the four pegs are exactly the four corners of a square whose side length is greater than zero. All four sides must have the same length and all four corners must be right angles. Pegs that coincide, or that all lie on one straight line, cannot mark such a square.

Return true when the pegs mark a square plot and false otherwise.

Examples

Example 1

Input: p1 = [6, 1], p2 = [1, 6], p3 = [1, 1], p4 = [6, 6]
Output: true

Walking the pegs as `[1, 1]`, `[6, 1]`, `[6, 6]`, `[1, 6]` gives four sides of length 5 that meet at four right angles, so the pegs mark a square of side 5.

Example 2

Input: p1 = [2, 3], p2 = [2, 8], p3 = [2, 13], p4 = [9, 8]
Output: false

Three of the pegs sit on the vertical line `x = 2`, so no arrangement of the four gives a shape with four right angles.

Example 3

Input: p1 = [3, 0], p2 = [0, 3], p3 = [-3, 0], p4 = [0, -3]
Output: true

Walking the pegs as `[3, 0]`, `[0, 3]`, `[-3, 0]`, `[0, -3]` gives four sides that each span 3 across and 3 up or down, and each corner turns through a right angle. A square does not have to be lined up with the grid axes.

Constraints

p1.length == 2
p2.length == 2
p3.length == 2
p4.length == 2
-10^4 <= p1[i], p2[i], p3[i], p4[i] <= 10^4

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def is_square_plot(p1: list[int], p2: list[int], p3: list[int], p4: list[int]) -> bool:

Java

public boolean isSquarePlot(int[] p1, int[] p2, int[] p3, int[] p4)

The intended approach

Time O(1): six distances and a fixed-size sort.Space O(1).

Because the pegs arrive unordered, work with the six pairwise squared distances, a quantity that does not depend on the recording sequence at all. Sort those six numbers: for a square the four smallest are the equal sides and the two largest are the equal diagonals, and by the right-angle relation a diagonal squared is exactly twice a side squared. Reject a smallest distance of zero, which is what coincident pegs and any zero-area arrangement produce. Squared distances keep every comparison in exact integer arithmetic.

Pythonentry point: is_square_plot

def is_square_plot(p1, p2, p3, p4):
    # the six pairwise squared distances of a square are four equal sides and
    # two equal diagonals, each diagonal twice a side. O(1) time and space.
    pts = [p1, p2, p3, p4]
    d = []
    for i in range(4):
        for j in range(i + 1, 4):
            dx = pts[i][0] - pts[j][0]
            dy = pts[i][1] - pts[j][1]
            d.append(dx * dx + dy * dy)
    d.sort()
    if d[0] == 0:
        return False
    if not (d[0] == d[1] == d[2] == d[3]):
        return False
    if d[4] != d[5]:
        return False
    return d[4] == 2 * d[0]

The obvious version

Enumerate all 24 orders of the four pegs. For each order, walk the closed quadrilateral and require four equal non-zero edge lengths plus a zero dot product at three consecutive corners. Accept if any order qualifies.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_is_square_plot(p1, p2, p3, p4):
    """Try every order the four pegs could be walked in and check the walk is a
    closed quadrilateral with four equal non-zero sides and right corners."""
    from itertools import permutations

    def sub(a, b):
        return (a[0] - b[0], a[1] - b[1])

    def dot(u, v):
        return u[0] * v[0] + u[1] * v[1]

    def norm2(u):
        return u[0] * u[0] + u[1] * u[1]

    for order in permutations([p1, p2, p3, p4]):
        a, b, c, e = order
        edges = [sub(b, a), sub(c, b), sub(e, c), sub(a, e)]
        lens = [norm2(u) for u in edges]
        if lens[0] == 0:
            continue
        if len(set(lens)) != 1:
            continue
        if dot(edges[0], edges[1]) != 0:
            continue
        if dot(edges[1], edges[2]) != 0:
            continue
        if dot(edges[2], edges[3]) != 0:
            continue
        return True
    return False

How this problem was checked

Reference solution executed against 12 test cases — 12 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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