Merge pull request #6389 from pokgopun/pwc172

pwc172 solution in go

4 files changed, 252 insertions, 0 deletions

diff --git a/challenge-172/pokgopun/go/ch-1.go b/challenge-172/pokgopun/go/ch-1.go
new file mode 100644
index 0000000000..dace6e8075
--- /dev/null
+++ b/challenge-172/pokgopun/go/ch-1.go
@@ -0,0 +1,82 @@
+/* https://theweeklychallenge.org/blog/perl-weekly-challenge-172/
+
+Task 1: Prime Partition
+
+Submitted by: [50]Mohammad S Anwar
+     __________________________________________________________________
+
+   You are given two positive integers, $m and $n.
+
+   Write a script to find out the Prime Partition of the given number. No
+   duplicates allowed.
+
+   For example,
+Input: $m = 18, $n = 2
+Output: 5, 13 or 7, 11
+
+Input: $m = 19, $n = 3
+Output: 3, 5, 11
+
+*/
+package main
+
+import (
+	"fmt"
+	"log"
+	"os"
+	"strings"
+
+	"github.com/jbarham/primegen"
+)
+
+func main() {
+	var m, n uint64
+	if len(os.Args) > 2 {
+		_, err := fmt.Sscanf(strings.Join(os.Args[1:], " "), "%d %d", &m, &n)
+		if err != nil {
+			log.Fatal(err)
+		}
+	} else {
+		log.Fatal("need two positive integers m and n")
+	}
+	pg := primegen.New()
+	var (
+		s []uint64
+		p uint64
+	)
+	for {
+		p = pg.Next()
+		if p > m {
+			break
+		}
+		s = append(s, p)
+	}
+	for _, v := range getCombo(int(n), s) {
+		p = 0
+		for _, i := range v {
+			p += i
+		}
+		if p == m {
+			fmt.Println(v)
+		}
+	}
+}
+func getCombo(n int, e []uint64) (r [][]uint64) {
+	var c []uint64
+	var sc []uint64
+	cTree(n, e, c, func(s []uint64) {
+		sc = make([]uint64, len(s))
+		copy(sc, s)
+		r = append(r, sc)
+	})
+	return r
+}
+func cTree(n int, e []uint64, c []uint64, f func(s []uint64)) {
+	if len(c) == n || len(c)+len(e) == n {
+		f(append(c, e...)[:n])
+	} else {
+		for i := 0; len(c)+len(e)-i >= n; i++ {
+			cTree(n, e[i+1:], append(c, e[i]), f)
+		}
+	}
+}
diff --git a/challenge-172/pokgopun/go/ch-2.go b/challenge-172/pokgopun/go/ch-2.go
new file mode 100644
index 0000000000..79635247d5
--- /dev/null
+++ b/challenge-172/pokgopun/go/ch-2.go
@@ -0,0 +1,86 @@
+/* https://theweeklychallenge.org/blog/perl-weekly-challenge-172/
+
+Task 2: Five-number Summary
+
+Submitted by: [51]Mohammad S Anwar
+     __________________________________________________________________
+
+   You are given an array of integers.
+
+   Write a script to compute the five-number summary of the given set of
+   integers.
+
+   You can find the definition and example in the [52]wikipedia page.
+*/
+/*
+lower and upper quartile using Method 1 mentioned in https://en.wikipedia.org/wiki/Quartile#Discrete_distributions
+
+Computing methods
+Discrete distributions
+For discrete distributions, there is no universal agreement on selecting the quartile values.[3]
+
+Method 1
+
+1) Use the median to divide the ordered data set into two-halves.
+ If there is an odd number of data points in the original ordered data set, do not include the median (the central value in the ordered list) in either half.
+ If there is an even number of data points in the original ordered data set, split this data set exactly in half.
+
+2) The lower quartile value is the median of the lower half of the data. The upper quartile value is the median of the upper half of the data.
+
+This rule is employed by the TI-83 calculator boxplot and "1-Var Stats" functions.
+*/
+
+package main
+
+import (
+	"bufio"
+	"fmt"
+	"log"
+	"os"
+	"sort"
+	"strconv"
+)
+
+func main() {
+	var s []int
+	if len(os.Args) > 1 {
+		for _, v := range os.Args[1:] {
+			n, err := strconv.Atoi(v)
+			if err != nil {
+				log.Fatal(err)
+			}
+			s = append(s, n)
+		}
+	} else {
+		s = []int{2, 3, 5, 7, 11, 13}
+	}
+	sort.SliceStable(s, func(i, j int) bool {
+		return s[i] < s[j]
+	})
+	l := len(s)
+	w := bufio.NewWriter(os.Stdout)
+	fmt.Fprintln(w, "Input:", s)
+	name := []string{"sample minimum", "lower quartile", "median", "upper quartile", "sample maximum"}
+	var value []float64
+	if len(s) == 1 {
+		for i := 0; i < 5; i++ {
+			value = append(value, float64(s[0]))
+		}
+	} else {
+		value = []float64{float64(s[0]), median(s[:l/2]), median(s), median(s[l/2+l%2:]), float64(s[l-1])}
+	}
+	for i, v := range name {
+		fmt.Fprintln(w, v, ":", value[i])
+	}
+	w.Flush()
+}
+func median(s []int) float64 {
+	l := len(s)
+	if l == 1 {
+		return float64(s[0])
+	}
+	if l%2 == 0 {
+		return (float64(s[l/2]) + float64(s[l/2-1])) / 2
+	}
+	return float64(s[l/2])
+}
diff --git a/challenge-172/pokgopun/perl/ch-1.pl b/challenge-172/pokgopun/perl/ch-1.pl
new file mode 100644
index 0000000000..b4ad616a31
--- /dev/null
+++ b/challenge-172/pokgopun/perl/ch-1.pl
@@ -0,0 +1,37 @@
+### /* https://theweeklychallenge.org/blog/perl-weekly-challenge-172/
+### 
+### Task 1: Prime Partition
+### 
+### Submitted by: [50]Mohammad S Anwar
+###      __________________________________________________________________
+### 
+###    You are given two positive integers, $m and $n.
+### 
+###    Write a script to find out the Prime Partition of the given number. No
+###    duplicates allowed.
+### 
+###    For example,
+### Input: $m = 18, $n = 2
+### Output: 5, 13 or 7, 11
+### 
+### Input: $m = 19, $n = 3
+### Output: 3, 5, 11
+### 
+### */
+use strict;
+use warnings;
+use Math::Prime::Util qw/forprimes/;
+use Math::Combinatorics;
+
+die "need two positive integers m and n\n" unless @ARGV==2 && join(" ",@ARGV[0,1]) =~ /^\d+ \d+$/;
+my ($m, $n) = @ARGV;
+
+my @p;
+forprimes { push @p, $_ } $m;
+
+my $c = Math::Combinatorics->new( count => $n, data => \@p );
+{
+	last unless my @combo = $c->next_combination();
+	printf("%s\n", join(", ",@combo)) if eval(join(" + ",@combo))==$m;
+	redo;
+}
diff --git a/challenge-172/pokgopun/perl/ch-2.pl b/challenge-172/pokgopun/perl/ch-2.pl
new file mode 100644
index 0000000000..fe31c8bea7
--- /dev/null
+++ b/challenge-172/pokgopun/perl/ch-2.pl
@@ -0,0 +1,47 @@
+### /* https://theweeklychallenge.org/blog/perl-weekly-challenge-172/
+### 
+### Task 2: Five-number Summary
+### 
+### Submitted by: [51]Mohammad S Anwar
+###      __________________________________________________________________
+### 
+###    You are given an array of integers.
+### 
+###    Write a script to compute the five-number summary of the given set of
+###    integers.
+### 
+###    You can find the definition and example in the [52]wikipedia page.
+### */
+### /*
+### lower and upper quartile using Method 1 mentioned in https://en.wikipedia.org/wiki/Quartile#Discrete_distributions
+### 
+### Computing methods
+### Discrete distributions
+### For discrete distributions, there is no universal agreement on selecting the quartile values.[3]
+### 
+### Method 1
+### 
+### 1) Use the median to divide the ordered data set into two-halves.
+###  If there is an odd number of data points in the original ordered data set, do not include the median (the central value in the ordered list) in either half.
+###  If there is an even number of data points in the original ordered data set, split this data set exactly in half.
+### 
+### 2) The lower quartile value is the median of the lower half of the data. The upper quartile value is the median of the upper half of the data.
+### 
+### This rule is employed by the TI-83 calculator boxplot and "1-Var Stats" functions.
+### */
+use strict;
+use warnings;
+
+die "please provide integers to calculate five-number summary" unless @ARGV && join(" ",@ARGV) =~ /^\d+(\s\d+)*$/;
+my @a = sort{$a <=> $b} @ARGV;
+
+printf "Input: (%s)\n", join(", ",@a);
+my @fn = ($a[-1], @a==1 ? $a[0] : median(@a[int(@a/2)+@a%2..$#a]), median(@a), @a==1 ? $a[0] : median(@a[0..int(@a/2)-1]), $a[0]);
+my $i = 5;
+foreach (qw/sample_minimum lower_quartile median upper_quartile sample_maximum/){
+	printf "%s: %s\n", $_, $fn[--$i];
+}
+
+sub median{
+	return @_ % 2 ? $_[int(@_/2)] : ($_[@_/2-1] + $_[@_/2])/2;
+}