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# Hosoya’s Triangle

• Difficulty Level : Medium
• Last Updated : 10 Aug, 2022

The Fibonacci triangle or Hosoya’s triangle is a triangular arrangement of numbers based on Fibonacci numbers. Each number is the sum of two numbers above in either the left diagonal or the right diagonal. The first few rows are:

The numbers in this triangle follow the recurrence relations

Relation to Fibonacci numbers
The entries in the triangle satisfy the identity

Thus, the two outermost diagonals are the Fibonacci numbers, while the numbers on the middle vertical lines are the squares of the Fibonacci numbers. All the other numbers in the triangle are the product of two distinct Fibonacci numbers greater than 1. The row sums are the first convolved Fibonacci numbers.
Sources: Stackoverflow, Wikipedia
Given a positive integers n. The task is print Hosoya’s triangle of size n.
Examples:

Input : n = 4
Output :
1
1 1
2 1 2
3 2 2 3

Input : n = 5
Output :
1
1 1
2 1 2
3 2 2 3
5 3 4 3 5

Below is the implementation of printing Hosoya’s triangle of height n:

## C++

 // CPP Program to print Hosoya's // triangle of height n. #include using namespace std;   int Hosoya(int n, int m) {     // Base case     if ((n == 0 && m == 0) ||         (n == 1 && m == 0) ||         (n == 1 && m == 1) ||         (n == 2 && m == 1))         return 1;       // Recursive step     if (n > m)         return Hosoya(n - 1, m)                + Hosoya(n - 2, m);       else if (m == n)         return Hosoya(n - 1, m - 1)                + Hosoya(n - 2, m - 2);       else         return 0; }   // Print the Hosoya triangle of height n. void printHosoya(int n) {     for (int i = 0; i < n; i++) {         for (int j = 0; j <= i; j++)             cout << Hosoya(i, j) << " ";               cout << endl;     } }   // Driven Program int main() {     int n = 5;     printHosoya(n);     return 0; }

## Java

 // Java Program to print Hosoya's // triangle of height n. import java.util.*;   class GFG {           static int Hosoya(int n, int m)     {         // Base case         if ((n == 0 && m == 0) ||             (n == 1 && m == 0) ||             (n == 1 && m == 1) ||             (n == 2 && m == 1))             return 1;                // Recursive step         if (n > m)             return Hosoya(n - 1, m)                    + Hosoya(n - 2, m);                       else if (m == n)             return Hosoya(n - 1, m - 1)                     + Hosoya(n - 2, m - 2);                       else             return 0;     }            // Print the Hosoya triangle of height n.     static void printHosoya(int n)     {         for (int i = 0; i < n; i++)         {             for (int j = 0; j <= i; j++)                 System.out.print(Hosoya(i, j)                                         + " ");                          System.out.println("");         }     }             /* Driver program to test above function */     public static void main(String[] args)     {         int n = 5;         printHosoya(n);                } }   // This code is contributed by  Arnav Kr. Mandal.

## Python3

 # Python3 code to print Hosoya's # triangle of height n.   def Hosoya( n , m ):       # Base case     if ((n == 0 and m == 0) or         (n == 1 and m == 0) or         (n == 1 and m == 1) or         (n == 2 and m == 1)):                 return 1           # Recursive step     if n > m:         return Hosoya(n - 1, m)                     + Hosoya(n - 2, m)       elif m == n:         return Hosoya(n - 1, m - 1)                         + Hosoya(n - 2,    m - 2)       else:         return 0           # Print the Hosoya triangle of height n. def printHosoya( n ):     for i in range(n):         for j in range(i + 1):             print(Hosoya(i, j) , end = " ")         print("\n", end = "")           # Driven Code n = 5 printHosoya(n)   # This code is contributed by Sharad_Bhardwaj

## C#

 // C# Program to print Hosoya's // triangle of height n. using System;   class GFG {           static int Hosoya(int n, int m)     {         // Base case         if ((n == 0 && m == 0) ||             (n == 1 && m == 0) ||             (n == 1 && m == 1) ||             (n == 2 && m == 1))             return 1;               // Recursive step         if (n > m)             return Hosoya(n - 1, m)                  + Hosoya(n - 2, m);                       else if (m == n)             return Hosoya(n - 1, m - 1)                  + Hosoya(n - 2, m - 2);                       else             return 0;     }           // Print the Hosoya triangle of height n.     static void printHosoya(int n)     {         for (int i = 0; i < n; i++)         {             for (int j = 0; j <= i; j++)                 Console.Write(Hosoya(i, j)                                         + " ");                   Console.WriteLine("");         }     }             /* Driver program to test above function */     public static void Main()     {         int n = 5;                   printHosoya(n);               } }   // This code is contributed by vt_m.

## PHP

 \$m)         return Hosoya(\$n - 1,\$m) +                Hosoya(\$n - 2, \$m);       else if (\$m == \$n)         return Hosoya(\$n - 1, \$m - 1) +                Hosoya(\$n - 2, \$m - 2);       else         return 0; }   // Print the Hosoya // triangle of height n. function printHosoya( \$n) {     for ( \$i = 0; \$i < \$n; \$i++)     {         for ( \$j = 0; \$j <= \$i; \$j++)             echo Hosoya(\$i, \$j) , " ";             echo "\n";     } }   // Driven Code \$n = 5; printHosoya(\$n);   // This code is contributed by anuj_67. ?>

## Javascript



Output:

1
1 1
2 1 2
3 2 2 3
5 3 4 3 5

Time Complexity: O(n2)

Auxiliary Space: O(n2)
Below is the implementation of printing Hosoya’s triangle of height n using Dynamic Programming:

## C++

 // CPP Program to print Hosoya's triangle of height n. #include #define N 5 using namespace std;   // Print the Hosoya triangle of height n. void printHosoya(int n) {     int dp[N][N];     memset(dp, 0, sizeof(dp));       // base case.     dp[0][0] = dp[1][0] = dp[1][1] = 1;       // For each row.     for (int i = 2; i < n; i++) {           // for each column;         for (int j = 0; j < n; j++) {               // recursive steps.             if (i > j)                 dp[i][j] = dp[i - 1][j] + dp[i - 2][j];               else                 dp[i][j] = dp[i - 1][j - 1] + dp[i - 2][j - 2];         }     }       // printing the solution     for (int i = 0; i < n; i++) {         for (int j = 0; j <= i; j++)             cout << dp[i][j] << " ";                  cout << endl;     } }   // Driven Program int main() {     int n = 5;     printHosoya(n);     return 0; }

## Java

 // JAVA Code for Hosoya Triangle import java.util.*;   class GFG {           static int N = 5;           // Print the Hosoya triangle     // of height n.     static void printHosoya(int n)     {         int dp[][] = new int[N][N];                   // base case.         dp[0][0] = dp[1][0] = 1;         dp[1][1] = 1;                // For each row.         for (int i = 2; i < n; i++)         {             // for each column;             for (int j = 0; j < n; j++)             {                  // recursive steps.                 if (i > j)                     dp[i][j] = dp[i - 1][j] +                                         dp[i - 2][j];                        else                     dp[i][j] = dp[i - 1][j - 1] +                                     dp[i - 2][j - 2];             }         }                // printing the solution         for (int i = 0; i < n; i++)         {             for (int j = 0; j <= i; j++)                 System.out.print(dp[i][j] + " ");                           System.out.println("");         }     }           /* Driver program*/     public static void main(String[] args)     {         int n = 5;         printHosoya(n);     } }   // This code is contributed by Arnav Kr. Mandal.

## Python3

 # Python3 Program to print # Hosoya's triangle of height n. N = 5   # Print the Hosoya triangle # of height n. def printHosoya(n):     dp = [[0 for i in range(N)]              for i in range(N)]                    # base case.     dp[0][0] = dp[1][0] = dp[1][1] = 1           # For each row.     for i in range(2, n):                   # for each column         for j in range(n):                           # recursive steps.             if (i > j):                 dp[i][j] = (dp[i - 1][j] +                             dp[i - 2][j])             else:                 dp[i][j] = (dp[i - 1][j - 1] +                             dp[i - 2][j - 2])                                   # printing the solution     for i in range(n):         for j in range(i + 1):             print(dp[i][j], end = ' ')         print()   # Driver Code n = 5 printHosoya(n)   # This code is contributed # by sahilshelangia

## C#

 // C# Code for Hosoya Triangle using System;   class GFG {           static int N = 5;           // Print the Hosoya triangle     // of height n.     static void printHosoya(int n)     {         int [,]dp = new int[N,N];                   // base case.         dp[0,0] = dp[1,0] = 1;         dp[1,1] = 1;               // For each row.         for (int i = 2; i < n; i++)         {             // for each column;             for (int j = 0; j < n; j++)             {                 // recursive steps.                 if (i > j)                     dp[i,j] = dp[i - 1,j] +                               dp[i - 2,j];                       else                     dp[i,j] = dp[i - 1,j - 1]                            + dp[i - 2,j - 2];             }         }               // printing the solution         for (int i = 0; i < n; i++)         {             for (int j = 0; j <= i; j++)                 Console.Write(dp[i,j] + " ");                   Console.WriteLine("");         }     }           /* Driver program*/     public static void Main()     {         int n = 5;                   printHosoya(n);     } }   // This code is contributed by Vt_m.

## PHP

 \$j)                 \$dp[\$i][\$j] = \$dp[\$i - 1][\$j]                             + \$dp[\$i - 2][\$j];               else                 \$dp[\$i][\$j] = \$dp[\$i - 1][\$j - 1]                             + \$dp[\$i - 2][\$j - 2];         }     }       // printing the solution     for (\$i = 0; \$i < \$n; \$i++) {         for (\$j = 0; \$j <= \$i; \$j++)             echo \$dp[\$i][\$j]." ";           echo "\n";     } }   // Driven Program       \$n = 5;     printHosoya(\$n);   // This code is contributed by mits ?>

## Javascript



Output:

1
1 1
2 1 2
3 2 2 3
5 3 4 3 5

Time complexity: O(n*n)

space complexity: O(n*n)

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