Skip to main content
Code Review

Return to Answer

replace rotating calipers link with web archive version
Source Link
Quentin Pradet
Quentin Pradet
  • 7.1k
  • 1
  • 25
  • 44

Algorithm

This is a classical problem in computational geometry and there are a lot of efficient approaches if you're interested. The longest distance problem (aka. diameter of a polygon) is more interesting though, and will introduce you to a really useful tool in computational geometry: the rotating calipersthe rotating calipers.

Code

It's easy to read, but three improvements can be made:

  1. for (int m=i; m<numOfPoints-1;m++) is indeed an interesting optimization (x2 improvement). It would be more readable to make m go from i+1 to numOfPoints.
  2. You can avoid computing Math.sqrt(dx*dx + dy*dy);. dx*dx + dy*dy is enough to compare the distances. When printing the distance, you can use Math.sqrt.
  3. First set shortestDistance to Double.MAX_VALUE, this will avoid you the if (m == 0 && i == 0) case and the resulting duplication: you'll always find a distance shorter than infinity.

Algorithm

This is a classical problem in computational geometry and there are a lot of efficient approaches if you're interested. The longest distance problem (aka. diameter of a polygon) is more interesting though, and will introduce you to a really useful tool in computational geometry: the rotating calipers.

Code

It's easy to read, but three improvements can be made:

  1. for (int m=i; m<numOfPoints-1;m++) is indeed an interesting optimization (x2 improvement). It would be more readable to make m go from i+1 to numOfPoints.
  2. You can avoid computing Math.sqrt(dx*dx + dy*dy);. dx*dx + dy*dy is enough to compare the distances. When printing the distance, you can use Math.sqrt.
  3. First set shortestDistance to Double.MAX_VALUE, this will avoid you the if (m == 0 && i == 0) case and the resulting duplication: you'll always find a distance shorter than infinity.

Algorithm

This is a classical problem in computational geometry and there are a lot of efficient approaches if you're interested. The longest distance problem (aka. diameter of a polygon) is more interesting though, and will introduce you to a really useful tool in computational geometry: the rotating calipers.

Code

It's easy to read, but three improvements can be made:

  1. for (int m=i; m<numOfPoints-1;m++) is indeed an interesting optimization (x2 improvement). It would be more readable to make m go from i+1 to numOfPoints.
  2. You can avoid computing Math.sqrt(dx*dx + dy*dy);. dx*dx + dy*dy is enough to compare the distances. When printing the distance, you can use Math.sqrt.
  3. First set shortestDistance to Double.MAX_VALUE, this will avoid you the if (m == 0 && i == 0) case and the resulting duplication: you'll always find a distance shorter than infinity.
added 63 characters in body
Source Link
Quentin Pradet
Quentin Pradet
  • 7.1k
  • 1
  • 25
  • 44

Algorithm

This is a classical problem in computational geometry and there are a lot of efficient approaches if you're interested. The longest distance problem (aka. diameter of a polygon) is more interesting though, and will introduce you to a really useful tool in computational geometry: the rotating calipers.

Code

It's easy to read, but three improvements can be made:

  1. for (int m=i; m<numOfPoints-1;m++) is indeed an interesting optimization (x2 improvement). It would be more readable to make m go from i+1 to numOfPoints.
  2. You can avoid computing Math.sqrt(dx*dx + dy*dy);. dx*dx + dy*dy is enough to compare the distances. When printing the distance, you can use Math.sqrt.
  3. First set shortestDistance to Double.MAX_VALUE, this will avoid you the if (m == 0 && i == 0) case and the resulting duplication: you'll always find a distance shorter than infinity.

Algorithm

This is a classical problem in computational geometry and there are a lot of efficient approaches if you're interested. The longest distance problem (aka. diameter of a polygon) is more interesting though, and will introduce you to a really useful tool in computational geometry: the rotating calipers.

Code

It's easy to read, but three improvements can be made:

  1. for (int m=i; m<numOfPoints-1;m++) is indeed an interesting optimization (x2 improvement). It would be more readable to make m go from i+1 to numOfPoints.
  2. You can avoid computing Math.sqrt(dx*dx + dy*dy);. dx*dx + dy*dy is enough to compare the distances. When printing the distance, you can use Math.sqrt.
  3. First set shortestDistance to Double.MAX_VALUE, this will avoid you the if (m == 0 && i == 0) case and the duplication.

Algorithm

This is a classical problem in computational geometry and there are a lot of efficient approaches if you're interested. The longest distance problem (aka. diameter of a polygon) is more interesting though, and will introduce you to a really useful tool in computational geometry: the rotating calipers.

Code

It's easy to read, but three improvements can be made:

  1. for (int m=i; m<numOfPoints-1;m++) is indeed an interesting optimization (x2 improvement). It would be more readable to make m go from i+1 to numOfPoints.
  2. You can avoid computing Math.sqrt(dx*dx + dy*dy);. dx*dx + dy*dy is enough to compare the distances. When printing the distance, you can use Math.sqrt.
  3. First set shortestDistance to Double.MAX_VALUE, this will avoid you the if (m == 0 && i == 0) case and the resulting duplication: you'll always find a distance shorter than infinity.
added 4 characters in body
Source Link
Quentin Pradet
Quentin Pradet
  • 7.1k
  • 1
  • 25
  • 44

Algorithm

This is a classical problem in computationcomputational geometry and there are a lot of efficient approaches if you're interested. The longest distance problem (aka. diameter of a polygon) is more interesting though, and will introduce you to a really useful tool in computational geometry: the rotating calipers.

Code

It's easy to read, but three improvements can be made:

  1. for (int m=i; m<numOfPoints-1;m++) is indeed an interesting optimization (x2 improvement). It would be more readable to make m go from i+1 to numOfPoints.
  2. You can avoid computing Math.sqrt(dx*dx + dy*dy);. dx*dx + dy*dy is enough to compare the distances. When printing the distance, you can use Math.sqrt.
  3. First set shortestDistance to Double.MAX_VALUE, this will avoid you the if (m == 0 && i == 0) case and the duplication.

Algorithm

This is a classical problem in computation geometry and there are a lot of efficient approaches if you're interested. The longest distance problem (aka. diameter of polygon) is more interesting though, and will introduce you to a really useful tool in computational geometry: the rotating calipers.

Code

It's easy to read, but three improvements can be made:

  1. for (int m=i; m<numOfPoints-1;m++) is indeed an interesting optimization (x2 improvement). It would be more readable to make m go from i+1 to numOfPoints.
  2. You can avoid computing Math.sqrt(dx*dx + dy*dy);. dx*dx + dy*dy is enough to compare the distances. When printing the distance, you can use Math.sqrt.
  3. First set shortestDistance to Double.MAX_VALUE, this will avoid you the if (m == 0 && i == 0) case and the duplication.

Algorithm

This is a classical problem in computational geometry and there are a lot of efficient approaches if you're interested. The longest distance problem (aka. diameter of a polygon) is more interesting though, and will introduce you to a really useful tool in computational geometry: the rotating calipers.

Code

It's easy to read, but three improvements can be made:

  1. for (int m=i; m<numOfPoints-1;m++) is indeed an interesting optimization (x2 improvement). It would be more readable to make m go from i+1 to numOfPoints.
  2. You can avoid computing Math.sqrt(dx*dx + dy*dy);. dx*dx + dy*dy is enough to compare the distances. When printing the distance, you can use Math.sqrt.
  3. First set shortestDistance to Double.MAX_VALUE, this will avoid you the if (m == 0 && i == 0) case and the duplication.
Source Link
Quentin Pradet
Quentin Pradet
  • 7.1k
  • 1
  • 25
  • 44