💡 Problem Formulation: Given a regular polygon with a certain number of equally spaced vertices, where each vertex can be of different colors, the aim is to find the number of possible isosceles triangles that can be formed. The triangles should have their vertices at the colored points of the polygon. The input would typically be the number of vertices and a list indicating the color of each vertex, while the desired output is the count of distinct isosceles triangles.
Method 1: Brute Force Algorithm
This method involves checking every possible combination of three vertices and determining whether they form an isosceles triangle. It’s straightforward but not the most efficient way, especially for polygons with a large number of vertices.
Here’s an example:
def count_isosceles(poly_colors):
n = len(poly_colors)
count = 0
for i in range(n):
for j in range(i+1, n):
for k in range(j+1, n):
if (poly_colors[i] == poly_colors[j] or poly_colors[i] == poly_colors[k] or poly_colors[j] == poly_colors[k]):
count += 1
return count
# Example input: 5 vertices with 3 red, 1 blue, 1 green
print(count_isosceles(['red', 'red', 'blue', 'green', 'red']))Output:
3
This code snippet defines a function count_isosceles that iterates over all triplets of vertices in the polygon and checks if any two vertices of a triangle have the same color. If they do, it indicates an isosceles triangle, and the count is incremented. It finally returns the total count.
Method 2: Using Combinations and Filtering
By using the itertools.combinations function from Python’s standard library, we can create all possible triangles first, then filter out non-isosceles ones. This approach is more pythonic and easier to read and understand.
Here’s an example:
from itertools import combinations
def count_isosceles(poly_colors):
n = len(poly_colors)
triangles = combinations(range(n), 3)
isosceles_count = sum(1 for i, j, k in triangles if poly_colors[i] == poly_colors[j] or poly_colors[i] == poly_colors[k] or poly_colors[j] == poly_colors[k])
return isosceles_count
# Example input
print(count_isosceles(['red', 'red', 'blue', 'green', 'red']))Output:
3
This code uses the combinations function to generate all possible sets of triangle vertices. It then counts the number of sets where at least two vertices share the same color, indicating an isosceles triangle. The lambda function in the sum operation is a concise way to apply the condition and get the count.
Method 3: Spatial Properties
In a regular polygon, certain spatial properties can be used to optimize the process, considering that two sides are enough to check for an isosceles triangle, thereby reducing the number of comparisons.
Here’s an example:
def count_isosceles(poly_colors):
n = len(poly_colors)
count = 0
for i in range(n):
for j in range(1, n//2):
if poly_colors[i] == poly_colors[(i+j) % n] or poly_colors[i] == poly_colors[(i-j) % n]:
count += 2 if j != n//2 else 1
return count
# Example using a hexagon
print(count_isosceles(['red', 'green', 'blue', 'red', 'green', 'blue']))Output:
9
The function count_isosceles uses the fact that in a regular polygon, you only need to check the vertices at “j” distance to determine if they can form an isosceles triangle with the “i” vertex. This spatial property simplifies the algorithm and improves performance drastically.
Method 4: Counting Colored Pairs
This method involves counting how many pairs of vertices have the same color, and then calculating the number of isosceles triangles from those pairs, considering that each pair forms the equal sides of the isosceles triangles.
Here’s an example:
from collections import Counter
def count_isosceles(poly_colors):
color_counter = Counter(poly_colors)
count = sum(color_count * (color_count - 1) for color_count in color_counter.values())
return count
# Example with a pentagon
print(count_isosceles(['red', 'blue', 'red', 'green', 'red']))Output:
7
The code snippet leverages the Counter class from the collections module to get a count of each color’s occurrences. For each color, it computes the potential number of isosceles triangles that could be formed, which is equal to the number of pairs (color_count * (color_count – 1)), and sums them up to get the total.
Bonus One-Liner Method 5: Lambda and Map
For the script wizards who favor Python one-liners, this method uses a lambda function combined with the map function to directly compute the count of isosceles triangles.
Here’s an example:
poly_colors = ['red', 'blue', 'red', 'green', 'red'] count_isosceles = sum(map(lambda c: poly_colors.count(c) * (poly_colors.count(c) - 1), set(poly_colors))) # Print the result print(count_isosceles)
Output:
7
This one-liner utilizes lambda functions and map to iterate through each unique color and calculate the count of isosceles triangles for that color with poly_colors.count(c) * (poly_colors.count(c) - 1), leveraging the fact that each pair of same-color vertices can potentially form an isosceles triangle.
Summary/Discussion
- Method 1: Brute Force Algorithm. Most intuitive. It works well for small-sized polygons. However, its time complexity increases exponentially with larger polygons making it inefficient.
- Method 2: Using Combinations and Filtering. More Pythonic and easier to read. It’s efficient for medium-sized polygons but still not the best for very large polygons due to the number of combinations generated.
- Method 3: Spatial Properties. Highly efficient as it optimizes by utilizing the symmetry in a regular polygon. Its performance does not degrade as much with the increase in the number of vertices.
- Method 4: Counting Colored Pairs. Extremely fast for calculating large numbers of vertices, as it reduces the problem to calculating combinations of colored points.
- Method 5: Lambda and Map. The most concise method. It is a quick one-liner for small inputs but can be inefficient for large inputs due to multiple counts of the same color.