Print N terms of Withoff Sequence

Wythoff array is an infinite matrix of integers derived from the Fibonacci sequence. Every positive integer in the matrix occurs only once.
Wythoff array:

  1    2    3    5    8    13    ...
  
  4    7    11   18   29   47    ...
                        
  6    10   16   26   42   68    ...
   
  9    15   24   39   63   102   ...
  
  12   20   32   52   84   136   ...
  
  14   23   37   60   97   157   ...

  .    .    .    .    .    .
  .    .    .    .    .    .

If A_{m, n} denotes the element in the mth row and nth column then

A_{m, 1} = [[m?]?]
A_{m, 2} = [[m?]?²]
A_{m, n} = A_{m, n-2} + A_{m, n-1} for n > 2
? = (1 + ?5) / 2

If we traverse matrix in an anti-diagonal way starting from top-left element then
Wythoff sequence:

1, 2, 4, 3, 7, 6, 5, 11, 10, 9….

For a given N, the task to print first N numbers of the sequence.
Examples:

Input : N = 10
Output : 1, 2, 4, 3, 7, 6, 5, 11, 10, 9
Input : N = 15
Output : 1, 2, 4, 3, 7, 6, 5, 11, 10, 9, 8, 18, 16, 15, 12

Approach:
The above recursions can be modified as

T(n, -1) = n-1, if k = -1
T(n, 0) = [n*?], if k = 0
T(n, k) = T(n, k-1) + T(n, k-2), if k > 0
? = (1 + ?5) / 2

So we can recursively find the value of T(n, k) with two base cases for t = 0 and for t = –1. we will store the values in a map and use it when needed to reduce computation. After we get the array we have to traverse it in an anti- diagonal way, so we set i=0 and j=0 and decrease the j and increase i when the j < 0 we initialise j = i and i = 0.
we also keep a count which is increased when a number is displayed. We break the array when the count reaches the required value.
Below is the implementation of the above approach :

CPP

// C++ program to find Wythoff array
#include <bits/stdc++.h>

using namespace std;
 
// Function to find the n, k term of Wythoff array

int Wythoff(map<int, map<int, int> >& mp, int n, int k)
{

    // tau = (sqrt(5)+1)/2

    double tau = (sqrt(5) + 1) / 2.0, t_n_k;
 
    // Already_stored

    if (mp[n][k] != 0)

        return mp[n][k];
 
    // T(n, -1) = n-1.

    if (k == -1) {

        return n - 1;

    }

    // T(n, 0) = floor(n*tau).

    else if (k == 0) {

        t_n_k = floor(n * tau);

    }

    // T(n, k) = T(n, k-1) + T(n, k-2) for k>=1.

    else

    {

        t_n_k = Wythoff(mp, n, k - 1) +

                             Wythoff(mp, n, k - 2);

    }
 
    // Store

    mp[n][k] = t_n_k;
 
    // Return the ans

    return (int)t_n_k;
}
 
// Function to find  first n terms of Wythoff 
// array by traversing in anti-diagonal

void Wythoff_Array(int n)
{

    int i = 0, j = 0, count = 0;
 
    // Map to store the Wythoff array

    map<int, map<int, int> > mp;
 
    while (count < n) {
 
        cout << Wythoff(mp, i + 1, j + 1);

        count++;

        if(count != n)

            cout << ", ";
 
        // Anti diagonal

        i++;

        j--;
 
        if (j < 0) {

            j = i;

            i = 0;

        }

    }
}
 
// Driver code

int main()
{

    int n = 15;

    // Function call

    Wythoff_Array(n);

    return 0;
}

Java

// Java program to find Wythoff array

import java.util.*; 

public class GFG
{
 
  // Function to find the n, k term of Wythoff array

  static int Wythoff(HashMap<Integer, 

                     HashMap<Integer, Integer>> mp, 

                     int n, int k)

  {
 
    // tau = (sqrt(5)+1)/2

    double tau = (Math.sqrt(5) + 1) / 2.0, t_n_k;
 
    // Already_stored

    if (mp.containsKey(n) && 

        mp.get(n).containsKey(k) &&

        mp.get(n).get(k) != 0)

      return mp.get(n).get(k);
 
    // T(n, -1) = n-1.

    if (k == -1) 

    {

      return n - 1;

    }
 
    // T(n, 0) = floor(n*tau).

    else if (k == 0) 

    {

      t_n_k = Math.floor(n * tau);

    }
 
    // T(n, k) = T(n, k-1) + T(n, k-2) for k>=1.

    else

    {

      t_n_k = Wythoff(mp, n, k - 1) +

        Wythoff(mp, n, k - 2);

    }
 
    // Store

    mp.put(n, new HashMap<Integer, Integer>(k,(int)t_n_k));
 
    // Return the ans

    return (int)t_n_k;

  }
 
  // Function to find  first n terms of Wythoff 

  // array by traversing in anti-diagonal

  static void Wythoff_Array(int n)

  {

    int i = 0, j = 0, count = 0;
 
    // Map to store the Wythoff array

    HashMap<Integer, HashMap<Integer,Integer>> mp = 

      new HashMap<Integer, HashMap<Integer,Integer>>();

    while (count < n) 

    {

      System.out.print(Wythoff(mp, i + 1, j + 1));

      count++;

      if(count != n)

        System.out.print(", ");
 
      // Anti diagonal

      i++;

      j--;
 
      if (j < 0)

      {

        j = i;

        i = 0;

      }

    }

  }
 
  // Driver code 

  public static void main(String[] args)

  {

    int n = 15;
 
    // Function call

    Wythoff_Array(n);

  }
}
 
// This code is contributed by divyeshrabadiya07.

Python3

# Python3 program to find Wythoff array

import math
 
# Function to find the n, k term of Wythoff array

def Wythoff(mp, n, k):
 
    # tau = (sqrt(5)+1)/2

    tau = (math.sqrt(5) + 1) / 2

    t_n_k = 0
 
    # Already_stored

    if ((n in mp) and (k in mp[n])):

        return mp[n][k];
 
    # T(n, -1) = n-1.

    if (k == -1):

        return n - 1;

    # T(n, 0) = floor(n*tau).

    elif (k == 0):

        t_n_k = math.floor(n * tau);

    # T(n, k) = T(n, k-1) + T(n, k-2) for k>=1.

    else:

        t_n_k = Wythoff(mp, n, k - 1) + Wythoff(mp, n, k - 2)

    # Store

    if n not in mp:

        mp[n] = dict()

    mp[n][k] = t_n_k;
 
    # Return the ans

    return int(t_n_k)
 
# Function to find  first n terms of Wythoff 
# array by traversing in anti-diagonal

def Wythoff_Array(n):
 
    i = 0

    j = 0

    count = 0;
 
    # Map to store the Wythoff array

    mp = dict()
 
    while (count < n):
 
        print(Wythoff(mp, i + 1, j + 1), end = '')

        count += 1

        if(count != n):

            print(", ", end = '')
 
        # Anti diagonal

        i += 1

        j -= 1
 
        if (j < 0):

            j = i;

            i = 0;

# Driver code

if __name__=='__main__':
 
    n = 15;

    # Function call

    Wythoff_Array(n);
 
    # This code is contributed by rutvik_56

// C# program to find Wythoff array

using System;

using System.Collections.Generic; 

class GFG 
{
 
  // Function to find the n, k term of Wythoff array

  static int Wythoff(Dictionary<int, Dictionary<int, int>> mp, int n, int k)

  {
 
    // tau = (sqrt(5)+1)/2

    double tau = (Math.Sqrt(5) + 1) / 2.0, t_n_k;
 
    // Already_stored

    if (mp.ContainsKey(n) && mp[n].ContainsKey(k) && mp[n][k] != 0)

      return mp[n][k];
 
    // T(n, -1) = n-1.

    if (k == -1) {

      return n - 1;

    }
 
    // T(n, 0) = floor(n*tau).

    else if (k == 0) 

    {

      t_n_k = Math.Floor(n * tau);

    }
 
    // T(n, k) = T(n, k-1) + T(n, k-2) for k>=1.

    else

    {

      t_n_k = Wythoff(mp, n, k - 1) + Wythoff(mp, n, k - 2);

    }
 
    // Store

    if(!mp.ContainsKey(n))

    {

      mp[n] = new Dictionary<int,int>();

    }

    mp[n][k] = (int)t_n_k;
 
    // Return the ans

    return (int)t_n_k;

  }
 
  // Function to find  first n terms of Wythoff 

  // array by traversing in anti-diagonal

  static void Wythoff_Array(int n)

  {

    int i = 0, j = 0, count = 0;
 
    // Map to store the Wythoff array

    Dictionary<int, Dictionary<int,int>> mp = new Dictionary<int, Dictionary<int,int>>();

    while (count < n) 

    {
 
      Console.Write(Wythoff(mp, i + 1, j + 1));

      count++;

      if(count != n)

        Console.Write(", ");
 
      // Anti diagonal

      i++;

      j--;
 
      if (j < 0) {

        j = i;

        i = 0;

      }

    }

  }
 
  // Driver code

  static void Main()

  {

    int n = 15;
 
    // Function call

    Wythoff_Array(n);

  }
}
 
// This code is contributed by divyesh072019.

Javascript

// JS program to find Wythoff array
 
// Function to find the n, k term of Wythoff array

function Wythoff(mp, n, k)
{
 
    // tau = (sqrt(5)+1)/2

    let tau = (Math.sqrt(5) + 1) / 2.0, t_n_k;
 
    // Already_stored

    if (mp.hasOwnProperty(n) && mp[n].hasOwnProperty(k)

        && mp[n][k] != 0)

        return mp[n][k];
 
    // T(n, -1) = n-1.

    if (k == -1) {

        return n - 1;

    }
 
    // T(n, 0) = floor(n*tau).

    else if (k == 0) {

        t_n_k = Math.floor(n * tau);

    }
 
    // T(n, k) = T(n, k-1) + T(n, k-2) for k>=1.

    else {

        t_n_k

            = Wythoff(mp, n, k - 1) + Wythoff(mp, n, k - 2);

    }
 
    // Store

    if (!mp.hasOwnProperty(n)) {

        mp[n] = {};

    }

    mp[n][k] = Math.floor(t_n_k);
 
    // Return the ans

    return Math.floor(t_n_k);
}
 
// Function to find  first n terms of Wythoff
// array by traversing in anti-diagonal

function Wythoff_Array(n)
{

    let i = 0, j = 0, count = 0;
 
    // Map to store the Wythoff array

    let mp = {};

    while (count < n) {
 
        process.stdout.write(""

                             + Wythoff(mp, i + 1, j + 1));

        count++;

        if (count != n)

            process.stdout.write(", ");
 
        // Anti diagonal

        i++;

        j--;
 
        if (j < 0) {

            j = i;

            i = 0;

        }

    }
}
 
// Driver code
let n = 15;
 
// Function call
Wythoff_Array(n);
 
// This code is contributed by phasing17.

Output:

1, 2, 4, 3, 7, 6, 5, 11, 10, 9, 8, 18, 16, 15, 12,

Reference : https://oeis.org/A035513

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