查找三角形中的所有非素数路径

Question

I'm trying to find routes that do not contain prime numbers in a triangle of integers leading from top to bottom.

What is wrong with my code?

For example ( int[][] input = {{1}, {8, 4}, {2, 6, 9}, {8, 5, 9, 3}} ),

   1    
  8 4
 2 6 9
8 5 9 3

This triangle has three paths that meet these criteria:

1 - 8 - 6 - 9  
1 - 4 - 6 - 9  
1 - 4 - 9 - 9

How can I achieve this?

My code:

private static void findPaths(int x, int y, int input[][])
{
   if(x < input.length - 1)
   {
     if(!isPrime(input[x + 1][y]) & !isPrime(input[x + 1][y + 1]))
     {
         System.out.println(input[x + 1][y]);
         findPaths(x + 1, y, input);
            
         System.out.println(input[x + 1][y + 1]);
         findPaths(x + 1, y + 1, input);
     }
     else if(!isPrime(input[x + 1][y]))
     {
         System.out.println(input[x + 1][y]);
         findPaths(x + 1, y, input);
     }
     else if(!isPrime(input[x + 1][y + 1]))
     {
         System.out.println(input[x + 1][y + 1]);
         findPaths(x + 1, y + 1, input);
     }
   }
}

public static boolean isPrime(int num)
{
  if (num <= 1) {return false;}

  for (int i = 2; i * i <= num; i++)
  {
     if ((num % i) == 0) {return false;}
  }
  return true;
}

Results for findPaths(0, 0, input) (array input initialized as shown above):

8
6
9
4
6
9
9
9

Answer 1

This problem can be solved by modeling the relationships between the numbers in the given array as a Graph .

Note that although the elements form a structure which resembles a Binary Tree . It's certainly not a tree because each node of the tree by definition should have only one parent-node , but here it's not the case (remainder: Tree - is a particularization of a Graph, which has no circles, intersections between brunches and disjointed components).

      1 <- root element
     / \   
    8   4 <- level
   / \ / \
  2   6   9
 / \ / \ / \
8   5   9   3

Although it's not a tree, for simplicity I would use terminology which is usually used to describe a tree, like root - which the top vertex ( node ) of this tree-like structure and level - which is a group of vertices ( nodes ) having the same depth (distance from the root to a vertex ).

And like in a Binary Tree , each element can be represented as a vertex having left and right children.

To find all non-prime paths in a graph starting from the root , we need to traverse over the graph . Each path would be represented as a separate List , ie every non-prime child of a non-prime vertex should spawn a new path . And every encountered prime vertex would cause the path that leads to it to be discarded.

To traverse the implement the graph , I've implemented Depth first search algorithm using recursion (because it's relatively easier to digest).

We would have two base cases (conditions when recursion should terminate):

• a prime number has been encountered, path gets invalidated;
• we've reached the last level and current vertex has a non-prime value, current path should be added to the collection of paths .

And in case if current vertex has a non-prime value, and it doesn't belong to the last level (ie it has children), we need to fire the depth first search recursively for both left and right child-vertices . That would be the recursive case of the algorithm.

Note: to avoid shifting the focus from the main goal of finding non-prime paths in the Graph to checking whether a particular value is a prime number, I've used built-in method BigInteger.isProbablePrime() which expects an int argument certainty , denoting the probability of getting correct result according to the formula . You're free to replace this check with any other logic.

Implementation might look like this:

public class NonPrimeTriangle {
    private final Vertex root;
    
    private NonPrimeTriangle(Vertex root) {
        this.root = root;
    }
    
    public List<List<Integer>> getNonPrimePaths() {
        List<List<Integer>> paths = new ArrayList<>();
        dfs(paths, new ArrayList<>(), root);
        
        return paths;
    }
    
    private void dfs(List<List<Integer>> paths, List<Integer> path, Vertex root) {
        if (BigInteger.valueOf(root.getValue()).isProbablePrime(100)) return; // base case - a prime number has been encountered, hence this path is invalidated
        
        path.add(root.getValue());  // vertex is proved to be non-prime and should be added to the path
        
        if (root.hasNoChildren()) { // base case - a non-prime path has been encountered found
            paths.add(path);
            return;
        }
        // recursive case
        dfs(paths, new ArrayList<>(path), root.getLeft());
        dfs(paths, new ArrayList<>(path), root.getRight());
    }
    
    // Not a part of the Algorithm - convenience-method for parsing the given Array into a Graph
    public static NonPrimeTriangle parse(int[][] arr) {
        if (arr.length == 0 || arr[0].length == 0) throw new IllegalArgumentException();
        
        Vertex root = new Vertex(arr[0][0]);
        NonPrimeTriangle triangle = new NonPrimeTriangle(root);
        List<Vertex> prevLevel = List.of(root);
        
        for (int row = 1; row < arr.length; row++) {
            List<Vertex> nextLevel = new ArrayList<>(prevLevel.size() + 1); // next level of Vertices
            for (int col = 0; col < arr[row].length; col++) {
                Vertex newVertex = new Vertex(arr[row][col]);
                // every vertex has two parents, except for the leftmost and the rightmost which have only one and the root which has none
                if (col != 0) prevLevel.get(col - 1).setRight(newVertex);
                if (col != prevLevel.size()) prevLevel.get(col).setLeft(newVertex);
                
                nextLevel.add(newVertex); // updating the next level of Vertices
            }
            prevLevel = nextLevel;
        }
        return triangle;
    }

    public static class Vertex {
        private final int value;
        private Vertex left;
        private Vertex right;
    
        public Vertex(int value) {
            this.value = value;
        }
    
        public boolean hasNoChildren() {
            return left == null && right == null;
        }
        
        // getters & setters
    }
}

main()

public static void main(String[] args) {
    int[][] input = {{1}, {8, 4}, {2, 6, 9}, {8, 5, 9, 3}};
    NonPrimeTriangle triangle = NonPrimeTriangle.parse(input);
    
    triangle.getNonPrimePaths().forEach(System.out::println);
}

Output:

[1, 8, 6, 9]
[1, 4, 6, 9]
[1, 4, 9, 9]

A link to Online Demo

Answer 2

Alternative approach (don't miss comments in code):

private static void traverseNonPrimalAndPrintPaths(int row, int col, int[][] arr, String curPath){
    int curNum = arr[row][col];
    if (!isPrime(curNum)) {//we only want to handle non-prime values

        //lets add value we are in to current path
        if (row == 0){//first value shouldn't have "-" before it
            curPath = String.valueOf(curNum);
        } else {//later values need to also add "-" before them in path
            curPath = curPath + "-" + curNum;
        }

        //If there are rows below check left and right "node" below current node
        if (row < arr.length-1){//we are NOT in last row
            traverseNonPrimalAndPrintPaths(row+1, col, arr, curPath);//check left node
            traverseNonPrimalAndPrintPaths(row+1, col+1, arr, curPath);//check right node
        } else {//here we know there are no more rows.
            //So we all we need to do is print path we took to get here.
            System.out.println(curPath);
        }
    }
}

Demo:

int[][] input1 = {{1}, {8, 4}, {2, 6, 9}, {8, 5, 9, 3}};
traverseNonPrimalAndPrintPaths(0, 0, input1,"");

Output:

1-8-6-9
1-4-6-9
1-4-9-9