Step by Step solution about How to resolve the algorithm Stirling numbers of the second kind step by step in the Go programming language

This Go program generates and displays Stirling numbers of the second kind, denoted as S2(n, k), for values of n and k up to a specified limit. It utilizes a dynamic programming approach for efficient calculation.

Here's a breakdown of the code:

1. Initialization:

   • limit: Limit for the range of n and k values (set to 100 in the provided code).
   • last: Number of Stirling numbers to display (set to 12 in the code).
   • s2: A 2D slice of *big.Int pointers to store the Stirling numbers.

2. Base Cases:

   • For n = 0 and k = n, S2(n, k) = 1.

3. Recursive Calculation:

   • For n > 0 and 1 <= k <= n, the program uses the recursive formula for Stirling numbers of the second kind:
     • S2(n, k) = k * S2(n-1, k) + S2(n-1, k-1)

4. Output Display:

   • The program prints a table of Stirling numbers up to the specified limit.
   • The header row shows k values, and the rows show n values.
   • The intersection of a row and a column gives S2(n, k).

5. Finding the Maximum Value:

   • After the table display, the program identifies and prints the maximum value in the last row of the table, which is the maximum Stirling number for the given limit.

6. Digit Count:

   • The program also displays the number of digits in the maximum Stirling number found.

This program provides an efficient way to calculate and display Stirling numbers of the second kind for various values of n and k, making it useful for mathematical and combinatorial applications.

package main

import (
    "fmt"
    "math/big"
)

func main() {
    limit := 100
    last := 12
    s2 := make([][]*big.Int, limit+1)
    for n := 0; n <= limit; n++ {
        s2[n] = make([]*big.Int, limit+1)
        for k := 0; k <= limit; k++ {
            s2[n][k] = new(big.Int)
        }
        s2[n][n].SetInt64(int64(1))
    }
    var t big.Int
    for n := 1; n <= limit; n++ {
        for k := 1; k <= n; k++ {
            t.SetInt64(int64(k))
            t.Mul(&t, s2[n-1][k])
            s2[n][k].Add(&t, s2[n-1][k-1])
        }
    }
    fmt.Println("Stirling numbers of the second kind: S2(n, k):")
    fmt.Printf("n/k")
    for i := 0; i <= last; i++ {
        fmt.Printf("%9d ", i)
    }
    fmt.Printf("\n--")
    for i := 0; i <= last; i++ {
        fmt.Printf("----------")
    }
    fmt.Println()
    for n := 0; n <= last; n++ {
        fmt.Printf("%2d ", n)
        for k := 0; k <= n; k++ {
            fmt.Printf("%9d ", s2[n][k])
        }
        fmt.Println()
    }
    fmt.Println("\nMaximum value from the S2(100, *) row:")
    max := new(big.Int).Set(s2[limit][0])
    for k := 1; k <= limit; k++ {
        if s2[limit][k].Cmp(max) > 0 {
            max.Set(s2[limit][k])
        }
    }
    fmt.Println(max)
    fmt.Printf("which has %d digits.\n", len(max.String()))
}