Source: Day 12: Garden Groups
Full solution for today (spoilers!).
Part 1
Given a
Gridof regions, calculate the sum of the productperimeter * areafor each region.
Full solution for today (spoilers!).
Given a
Gridof regions, calculate the sum of the productperimeter * areafor each region.
Full solution for today (spoilers!).
Given a grid with either open tiles (
.) or towers (anything else), for each pair of towers, there is an antinode at each of the points that is 2x as far from one tower as the other. How many antinodes are there still within the bounds of the map?
Part 1: Given a list of sides, determine how many form valid triangles. (Hint: triangle inequality)
Yesterday’s post at /r/dailyprogrammer managed to pique my interest1:
A triangle on a flat plane is described by its angles and side lengths, and you don’t need all of the angles and side lengths to work out everything about the triangle. (This is the same as last time.) However, this time, the triangle will not necessarily have a right angle. This is where more trigonometry comes in. Break out your trig again, people.
Here’s a quick little programming task that I came to via a post on L2Program (who in turn seems to have found it on Reddit). The basic idea is to take a given list of circles and to determine the area enclosed (while correctly accounting for overlap).
Four points, a square?) and comes originally from a Google Code Jam problem. The problem is stated simply enough
Accept three points as input, determine if they form a triangle, and, if they do, classify it at equilateral (all three sides the same), isoceles (two sides the same, the other different), or scalene (all three sides different), and also classify it as acute (all three angles less than 90 degrees), obtuse (one angle greater than 90 degrees) or right (one angle equal 90 degrees).
But once you start implementing it, that’s when things get more interesting. 😄
Another post from Programming Praxis. This one was originally intended for Friday but they posted it early, so I figured I would go ahead and do the same. The problem is actually deceptively straight forward:
Given four points, do they form a square?