Similar boomerang

Mary-Kate Olsen
Mary-Kate OlsenOriginal
2024-11-04 05:03:02693browse

Similar boomerang

Weekly Challenge 293

Each week Mohammad S. Anwar sends out The Weekly Challenge, a chance for all of us to come up with solutions to two weekly tasks. My solutions are written in Python first, and then converted to Perl. It's a great way for us all to practice some coding.

Challenge, My solutions

Task 1: Similar Dominos

Task

You are given a list of dominos, @dominos.

Write a script to return the number of dominoes that are similar to any other domino.

$dominos[i] = [a, b] and $dominos[j] = [c, d] are same if either (a = c and b = d) or (a = d and b = c).

My solution

I'm not sure if it's a British/US English thing or something else, but I'm using Dominoes as the plural of Domino. Dominos what what you get when you are hungry.

For this task, I take integers from the command line and convert them to a list of lists (arrays of arrays in Perl). If this was an actual real-world project, I'd probably use a Dataclass, and have an equality operator.

I have a double loop. The outer loop - called i - is from 0 to one less than the number of dominoes. The inner loop - called j - is also the same. I skip the case when i and j are the same. If the dominoes at position i and j are the same (either same number or opposite numbers), I add one to count and exit the inner loop.

def similar_dominoes(dominoes: list) -> int:
    count = 0

    for i in range(len(dominoes)):
        for j in range(len(dominoes)):
            if i == j:
                continue
            if (dominoes[i][0] == dominoes[j][0] and dominoes[i][1] == dominoes[j][1]) \
                    or (dominoes[i][0] == dominoes[j][1] and dominoes[i][1] == dominoes[j][0]):
                count += 1
                break

    return count

Examples

$ ./ch-1.py 1 3 3 1 2 4 6 8
2

$ ./ch-1.py 1 2 2 1 1 1 1 2 2 2
3

Task 2: Boomerang

Task

You are given an array of points, (x, y).

Write a script to find out if the given points are a boomerang.

A boomerang is a set of three points that are all distinct and not in a straight line.

My solution

Like with the last task, I take integers from the command line and convert them to a list of lists (arrays of arrays in Perl).

I remember enough from high school math to know we can get the slope (gradient) of two points with the formula (x2 - x1) ÷ (y2 - y1). However if y1 and y2 are the same, we get a division by zero error.

Therefore I use the following checks:

  1. If all y values (second item in each list) are the same, return False as the points make a flat line.
  2. If any y values are the same as the first y value, return True (as we know at least one y value is different).
  3. Calculate the absolute slop between the first point and the other points using the above formula. Store this as a set. Sets don't store duplicate values.
  4. If the set only has one value, it's a straight line and return False. If there is more than one value, then it's a boomerang. I'm not here to determine if it is a good boomerang :)
def is_boomerang(points: list) -> bool:
    if all(points[0][1] == points[i][1] for i in range(1, len(points))):
           return False

    if any(points[0][1] == points[i][1] for i in range(1, len(points))):
           return True

    degrees = set(abs((points[0][0] - points[i][0]) / (points[0][1] - points[i][1])) for i in range(1, len(points)))
    return False if len(degrees) == 1 else True

Examples

$ ./ch-2.py 1 1 2 3 3 2
true

$ ./ch-2.py 1 1 2 2 3 3
false

$ ./ch-2.py 1 1 1 2 2 3
true

$ ./ch-2.py 1 1 1 2 1 3
false

$ ./ch-2.py 1 1 2 1 3 1
false

$ ./ch-2.py 0 0 2 3 4 5
true

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