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Check if point is on line two variants
(version: 0)
Comparing performance of:
new vs old
Created:
6 years ago
by:
Guest
Jump to the latest result
Script Preparation code:
var sq = function (a) { return a * a; }; function isOnLine1(p, v, w) { if (v[0] == w[0] && p[0] == v[0] || v[1] == w[1] && p[1] == w[1] || v[0] == w[0] && v[1] == w[1]) return true; var yDiff = Math.abs(p[1] - v[1] - (w[1] - v[1]) * ((p[0] - v[0]) / (w[0] - v[0]))); var xDiff = Math.abs(p[0] - v[0] - (w[0] - v[0]) * ((p[1] - v[1]) / (w[1] - v[1]))); var distance = yDiff < xDiff ? yDiff : xDiff; // off by a factor of at most ~0.707 = 1 / sqrt(2) return distance < 0.0017; // fp errors } function isOnLine2(p, v, w) { var distance = ((Math.abs((w[1] - v[1]) * p[0] - (w[0] - v[0]) * p[1] + w[0] * v[1] - w[1] * v[0])) / (Math.sqrt(sq(w[1] - v[1]) + sq(w[0] - v[0])))); return distance < 0.001; // fp errors } var a = false;
Tests:
new
a = isOnLine1([54.34523], [543.23], [123.3]);
old
a = isOnLine2([54.34523], [543.23], [123.3]);
Rendered benchmark preparation results:
Suite status:
<idle, ready to run>
Run tests (2)
Previous results
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Test case name
Result
new
old
Fastest:
N/A
Slowest:
N/A
Latest run results:
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Autogenerated LLM Summary
(model
llama3.2:3b
, generated one year ago):
Let's dive into the world of JavaScript microbenchmarks! The provided JSON represents two benchmark test cases: `isOnLine1` and `isOnLine2`. Both functions are designed to determine whether a given point (`p`) lies on a line defined by another pair of points (`v` and `w`). The main difference between the two approaches lies in their implementation. **Option 1: isOnLine1** This function uses a more traditional, brute-force approach. It calculates the distance from the point to each edge of the line (defined by `v` and `w`) using the Euclidean distance formula. If the distance is less than a certain threshold (`0.0017`), it's considered "on the line". **Pros:** * Easy to understand and implement * Works well for small to medium-sized inputs **Cons:** * May be slower due to the need to calculate distances from multiple points * Not optimized for performance, as it uses a simple threshold value instead of a more sophisticated algorithm **Option 2: isOnLine2** This function takes a different approach, using linear algebra to compute the distance from the point to the line. It's based on the formula: `distance = |(w[1] - v[1]) * p[0] - (w[0] - v[0]) * p[1] + w[0] * v[1] - w[1] * v[0)| / sqrt((w[1] - v[1])^2 + (w[0] - v[0])^2)` This formula is derived from the concept of the shortest distance between a point and a line in 2D space. **Pros:** * Generally faster than `isOnLine1`, as it avoids calculating multiple distances * Optimized for performance using linear algebra **Cons:** * May be harder to understand and implement, especially for those without a strong math background * Can be sensitive to numerical precision issues (due to the use of floating-point arithmetic) The library used in both functions is not explicitly mentioned. However, based on the code structure and naming conventions, it's likely that the `Math` object is being used extensively, as well as basic JavaScript data structures like arrays. There are alternative approaches to solving this problem, such as using geometric transformations (e.g., finding the closest point on the line) or employing more advanced linear algebra techniques (e.g., using vectors and matrix operations). However, these approaches may be overkill for simple benchmarking scenarios like this one. For developers looking to optimize their JavaScript code, measuring the performance differences between `isOnLine1` and `isOnLine2` can provide valuable insights. The latest benchmark results show that `isOnLine2` is significantly faster than `isOnLine1`, which suggests that optimizing this particular implementation could lead to noticeable performance improvements in certain applications.
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