recursion - Recursive function to create number combinations in golang -
i'm trying figure out recursive function thing. have non-recursive demo works uses static methods not recursive. functions prints out combinations of "number sets" "pool_size". if could, please me make function recursive great.
package main import ( "fmt" ) func combos_of1(pool_size int) { := 1; < pool_size+1; i++ { fmt.println(i) } fmt.println("\n") } func combos_of2(pool_size int) { := 1; < pool_size+1; i++ { j := + 1; j < pool_size+1; j++ { fmt.println(i, j) } } fmt.println("\n") } func combos_of3(pool_size int) { := 1; < pool_size+1; i++ { j := + 1; j < pool_size+1; j++ { k := j + 1; k < pool_size+1; k++ { fmt.println(i, j, k) } } } fmt.println("\n") } func main() { combos_of1(10) combos_of2(10) combos_of3(10) }
for example,
package main import "fmt" func rcombinations(p int, n []int, c []int, ccc [][][]int) [][][]int { if len(n) == 0 || p <= 0 { return ccc } if len(ccc) == 0 { ccc = make([][][]int, p) } p-- := range n { cc := make([]int, len(c)+1) copy(cc, c) cc[len(cc)-1] = n[i] ccc[len(cc)-1] = append(ccc[len(cc)-1], cc) ccc = rcombinations(p, n[i+1:], cc, ccc) } return ccc } func combinations(p int, n []int) [][][]int { return rcombinations(p, n, nil, nil) } func main() { pools := 3 numbers := []int{1, 2, 3, 4, 5, 6, 7, 8, 9, 10} fmt.println(pools, "pools", "for", "numbers", numbers) fmt.println() nc := 0 c := combinations(pools, numbers) fmt.println("pools:") d := " digit : " := range c { fmt.println(i+1, d) d = " digits: " j := range c[i] { nc++ fmt.println(c[i][j], " ") } } fmt.println() fmt.println(nc, "combinations") } output:
3 pools numbers [1 2 3 4 5 6 7 8 9 10] pools: 1 digit : [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] 2 digits: [1 2] [1 3] [1 4] [1 5] [1 6] [1 7] [1 8] [1 9] [1 10] [2 3] [2 4] [2 5] [2 6] [2 7] [2 8] [2 9] [2 10] [3 4] [3 5] [3 6] [3 7] [3 8] [3 9] [3 10] [4 5] [4 6] [4 7] [4 8] [4 9] [4 10] [5 6] [5 7] [5 8] [5 9] [5 10] [6 7] [6 8] [6 9] [6 10] [7 8] [7 9] [7 10] [8 9] [8 10] [9 10] 3 digits: [1 2 3] [1 2 4] [1 2 5] [1 2 6] [1 2 7] [1 2 8] [1 2 9] [1 2 10] [1 3 4] [1 3 5] [1 3 6] [1 3 7] [1 3 8] [1 3 9] [1 3 10] [1 4 5] [1 4 6] [1 4 7] [1 4 8] [1 4 9] [1 4 10] [1 5 6] [1 5 7] [1 5 8] [1 5 9] [1 5 10] [1 6 7] [1 6 8] [1 6 9] [1 6 10] [1 7 8] [1 7 9] [1 7 10] [1 8 9] [1 8 10] [1 9 10] [2 3 4] [2 3 5] [2 3 6] [2 3 7] [2 3 8] [2 3 9] [2 3 10] [2 4 5] [2 4 6] [2 4 7] [2 4 8] [2 4 9] [2 4 10] [2 5 6] [2 5 7] [2 5 8] [2 5 9] [2 5 10] [2 6 7] [2 6 8] [2 6 9] [2 6 10] [2 7 8] [2 7 9] [2 7 10] [2 8 9] [2 8 10] [2 9 10] [3 4 5] [3 4 6] [3 4 7] [3 4 8] [3 4 9] [3 4 10] [3 5 6] [3 5 7] [3 5 8] [3 5 9] [3 5 10] [3 6 7] [3 6 8] [3 6 9] [3 6 10] [3 7 8] [3 7 9] [3 7 10] [3 8 9] [3 8 10] [3 9 10] [4 5 6] [4 5 7] [4 5 8] [4 5 9] [4 5 10] [4 6 7] [4 6 8] [4 6 9] [4 6 10] [4 7 8] [4 7 9] [4 7 10] [4 8 9] [4 8 10] [4 9 10] [5 6 7] [5 6 8] [5 6 9] [5 6 10] [5 7 8] [5 7 9] [5 7 10] [5 8 9] [5 8 10] [5 9 10] [6 7 8] [6 7 9] [6 7 10] [6 8 9] [6 8 10] [6 9 10] [7 8 9] [7 8 10] [7 9 10] [8 9 10] 175 combinations the variation single pool is:
package main import "fmt" func rpool(p int, n []int, c []int, cc [][]int) [][]int { if len(n) == 0 || p <= 0 { return cc } p-- := range n { r := make([]int, len(c)+1) copy(r, c) r[len(r)-1] = n[i] if p == 0 { cc = append(cc, r) } cc = rpool(p, n[i+1:], r, cc) } return cc } func pool(p int, n []int) [][]int { return rpool(p, n, nil, nil) } func main() { pool := 9 numbers := []int{1, 2, 3, 4, 5, 6, 7, 8, 9, 10} p := pool(pool, numbers) fmt.println(pool, "digit pool", "for", "numbers", numbers) := range p { fmt.println(p[i]) } } output:
9 digit pool numbers [1 2 3 4 5 6 7 8 9 10] [1 2 3 4 5 6 7 8 9] [1 2 3 4 5 6 7 8 10] [1 2 3 4 5 6 7 9 10] [1 2 3 4 5 6 8 9 10] [1 2 3 4 5 7 8 9 10] [1 2 3 4 6 7 8 9 10] [1 2 3 5 6 7 8 9 10] [1 2 4 5 6 7 8 9 10] [1 3 4 5 6 7 8 9 10] [2 3 4 5 6 7 8 9 10] 
Comments
Post a Comment