Solutions in 6 languages for week 125, part 1

7 files changed, 302 insertions, 23 deletions

diff --git a/challenge-125/abigail/README.md b/challenge-125/abigail/README.md
index 155bfc8302..f8c6f2595b 100644
--- a/challenge-125/abigail/README.md
+++ b/challenge-125/abigail/README.md
@@ -29,35 +29,12 @@ Output: -1
 ~~~~
 
 ### Solutions
-* [AWK](awk/ch-1.awk)
-* [Bash](bash/ch-1.sh)
-* [Basic](basic/ch-1.bas)
-* [bc](bc/ch-1.bc)
-* [Befunge-93](befunge-93/ch-1.bf93)
 * [C](c/ch-1.c)
-* [Cobol](cobol/ch-1.cb)
-* [Csh](csh/ch-1.csh)
-* [Erlang](erlang/ch-1.erl)
-* [Forth](forth/ch-1.fs)
-* [Fortran](fortran/ch-1.f90)
-* [Go](go/ch-1.go)
-* [Java](java/ch-1.java)
 * [Lua](lua/ch-1.lua)
-* [m4](m4/ch-1.m4)
-* [MMIX](mmix/ch-1.mms)
 * [Node.js](node/ch-1.js)
-* [Ocaml](ocaml/ch-1.ml)
-* [Pascal](pascal/ch-1.p)
 * [Perl](perl/ch-1.pl)
-* [Php](php/ch-1.php)
-* [Postscript](postscript/ch-1.ps)
 * [Python](python/ch-1.py)
-* [R](r/ch-1.r)
-* [Rexx](rexx/ch-1.rexx)
 * [Ruby](ruby/ch-1.rb)
-* [Scheme](scheme/ch-1.scm)
-* [SQL](sql/ch-1.sql)
-* [Tcl](tcl/ch-1.tcl)
 
 ### Blog
 [Perl Weekly Challenge 125: Pythagorean Triples][blog1]
diff --git a/challenge-125/abigail/c/ch-1.c b/challenge-125/abigail/c/ch-1.c
new file mode 100644
index 0000000000..63b3a67db1
--- /dev/null
+++ b/challenge-125/abigail/c/ch-1.c
@@ -0,0 +1,52 @@
+# include <stdlib.h>
+# include <stdio.h>
+# include <string.h>
+# include <math.h>
+
+/*
+ * See ../README.md
+ */
+
+/*
+ * Run as: cc -o ch-1.o ch-1.c; ./ch-1.o < input-file
+ */
+
+typedef long long number;
+
+number introot (number square) {
+    return ((number) floorl (.4 + sqrt (square)));
+}
+
+int main (void) {
+    number n;
+
+    while (scanf ("%lld", &n) == 1) {
+        if (n <= 2) {
+            printf ("-1\n");
+            continue;
+        }
+
+        number n_sq = n * n;
+        number c    = n + 1;
+        number c_sq = n_sq + 2 * n + 1;
+        while (2 * c - 1 <= n_sq) {
+            number b_sq = c_sq - n_sq;
+            number b    = introot (b_sq);
+            if (b_sq == b * b) {
+                printf ("%lld %lld %lld\n", n, b, c);
+            }
+            c_sq += 2 * c ++ + 1;
+        }
+
+        number max_a = (number) floorl (n / sqrt (2));
+        for (number a = 3; a <= max_a; a ++) {
+            number b_sq = n_sq - a * a;
+            number b    = introot (b_sq);
+            if (b_sq == b * b) {
+                printf ("%lld %lld %lld\n", a, b, n);
+            }
+        }
+    }
+
+    return (0);
+}
diff --git a/challenge-125/abigail/lua/ch-1.lua b/challenge-125/abigail/lua/ch-1.lua
new file mode 100644
index 0000000000..1a0efb40c8
--- /dev/null
+++ b/challenge-125/abigail/lua/ch-1.lua
@@ -0,0 +1,48 @@
+#!/opt/local/bin/lua
+
+--
+-- See ../README.md
+--
+
+--
+-- Run as: lua ch-1.lua < input-file
+--
+
+function introot (square)
+    return (math . floor (.4 + math . sqrt (square)))
+end
+
+
+for line in io . lines () do
+    local n = tonumber (line)
+    if n <= 2 then
+        print (-1)
+        goto end_loop
+    end
+
+    local n_sq = n * n
+    local c    = n + 1
+    local c_sq = n_sq + 2 * n + 1
+    while 2 * c - 1 <= n_sq do
+        local b_sq = c_sq - n_sq
+        local b    = introot (b_sq)
+
+        if b_sq == b * b then
+            print (n .. " " .. b .. " " .. c)
+        end
+
+        c_sq = c_sq + 2 * c + 1
+        c    = c + 1
+    end
+
+    local max_a = math . floor (n / math . sqrt (2))
+    for a = 3, max_a do
+        local b_sq = n_sq - a * a
+        local b    = introot (b_sq)
+        if b_sq == b * b then
+            print (a .. " " .. b .. " " .. n)
+        end
+    end
+
+    ::end_loop::
+end
diff --git a/challenge-125/abigail/node/ch-1.js b/challenge-125/abigail/node/ch-1.js
new file mode 100644
index 0000000000..a8c8c9c2c4
--- /dev/null
+++ b/challenge-125/abigail/node/ch-1.js
@@ -0,0 +1,47 @@
+#!/usr/local/bin/node
+
+//
+// See ../README.md
+//
+
+//
+// Run as: node ch-1.js < input-file
+//
+
+  require ('readline')
+. createInterface ({input: process . stdin})   
+. on              ('line', line => main (+ line))
+
+function introot (square) {
+    return (Math . floor (.4 + Math . sqrt (square)))
+}
+
+function main (n) {
+    if (n <= 2) {
+        console . log (-1)
+        return
+    }
+
+    let n_sq = n * n
+    let c    = n + 1
+    let c_sq = n_sq + 2 * n + 1
+    while (2 * c - 1 <= n_sq) {
+        let b_sq = c_sq - n_sq
+        let b    = introot (b_sq)
+
+        if (b_sq == b * b) {
+            console . log (n + " " + b + " " + c)
+        }
+
+        c_sq += 2 * c ++ + 1
+    }
+
+    let max_a = Math . floor (n / Math . sqrt (2))
+    for (let a = 3; a <= max_a; a ++) {
+        let b_sq = n_sq - a * a
+        let b    = introot (b_sq)
+        if (b_sq == b * b) {
+            console . log (a + " " + b + " " + n)
+        }
+    }
+}
diff --git a/challenge-125/abigail/perl/ch-1.pl b/challenge-125/abigail/perl/ch-1.pl
new file mode 100644
index 0000000000..c62b9b92d3
--- /dev/null
+++ b/challenge-125/abigail/perl/ch-1.pl
@@ -0,0 +1,69 @@
+#!/opt/perl/bin/perl
+
+use 5.032;
+
+use strict;
+use warnings;
+no  warnings 'syntax';
+
+use experimental 'signatures';
+use experimental 'lexical_subs';
+
+#
+# See ../README.md
+#
+
+#
+# Run as: perl ch-1.pl < input-file
+#
+
+sub introot ($square) {
+    #
+    # Return the largest integer less than, or equal to the
+    # square root of $square.
+    #
+    int (.4 + sqrt ($square));
+}
+
+
+
+while (my $n = <>) {
+    chomp ($n);
+    say (-1), next if $n <= 2;
+    #
+    # First case, $n is not the hypothenuse; wlog, assume n = a.
+    #
+    #   Then, we start searching from c = n + 1 until 
+    #   c^2 - (c - 1)^2 > n^2. Note that c^2 - (c - 1)^2 = 2c - 1
+    #   In each iteration, we calculate b^2 = c^2 - n^2. If b^2 is
+    #   a proper square, we have a Pythagorian triple.
+    #
+    my $n_sq = $n * $n;
+
+    my $c    = $n + 1;
+    my $c_sq = $n_sq + 2 * $n + 1;
+    while (2 * $c - 1 <= $n_sq) {
+        #
+        # We now have a^2 (n_sq) and c^2. We can calculate b^2 (b_sq)
+        # and check whether this is a proper square.
+        #
+        my $b_sq = $c_sq - $n_sq;
+        my $b    = introot ($b_sq);
+
+        say "$n $b $c" if $b_sq == $b * $b;
+        $c_sq += 2 * $c ++ + 1;  # (c + 1)^2 == c^2 + 2 * c + 1
+    }
+
+    #
+    # Handle the case $n is the hypothenuse, so $n == c.
+    #
+    # We now need to search for a, b such that a^2 + b^2 = c^2 ($n_sq).
+    # Wlog, assume a < b (a == b cannot happen). Then a < c / sqrt (2).
+    #
+    my $max_a = int ($n / sqrt (2));
+    for (my $a = 3; $a <= $max_a; $a ++) {
+        my $b_sq = $n_sq - $a * $a;
+        my $b    = introot ($b_sq);
+        say "$a $b $n" if $b_sq == $b * $b;
+    }
+}
diff --git a/challenge-125/abigail/python/ch-1.py b/challenge-125/abigail/python/ch-1.py
new file mode 100644
index 0000000000..f8c07799b3
--- /dev/null
+++ b/challenge-125/abigail/python/ch-1.py
@@ -0,0 +1,41 @@
+#!/opt/local/bin/python
+
+#
+# See ../README.md
+#
+
+#
+# Run as: python ch-1.py < input-file
+#
+
+import fileinput
+import math
+
+def introot (square):
+    return math . floor (.4 + math . sqrt (square))
+
+for line in fileinput . input ():
+    n = int (line)
+    if n <= 2:
+        print (-1)
+        continue
+
+    n_sq = n * n
+    c    = n + 1
+    c_sq = n_sq + 2 * n + 1
+    while 2 * c - 1 <= n_sq:
+        b_sq = c_sq - n_sq
+        b    = introot (b_sq)
+
+        if b_sq == b * b:
+            print (str (n) + " " + str (b) + " " + str (c))
+
+        c_sq = c_sq + 2 * c + 1
+        c    = c + 1
+
+    max_a = math . floor (n / math . sqrt (2))
+    for a in range (3, max_a + 1):
+        b_sq = n_sq - a * a
+        b    = introot (b_sq)
+        if b_sq == b * b:
+            print (str (a) + " " + str (b) + " " + str (n))
diff --git a/challenge-125/abigail/ruby/ch-1.rb b/challenge-125/abigail/ruby/ch-1.rb
new file mode 100644
index 0000000000..91518fc95e
--- /dev/null
+++ b/challenge-125/abigail/ruby/ch-1.rb
@@ -0,0 +1,45 @@
+#!/usr/bin/ruby
+
+#
+# See ../README.md
+#
+ 
+#
+# Run as: ruby ch-1.rb < input-file
+#
+
+def introot (square)
+    return Math . sqrt(square) . floor()
+end
+
+ARGF . each_line do
+    |n|
+    n = n . to_i
+    if (n <= 2) then
+        puts (-1)
+    end
+
+    n_sq = n * n
+    c    = n + 1
+    c_sq = n_sq + 2 * n + 1
+    while (2 * c - 1 <= n_sq) do
+        b_sq = c_sq - n_sq
+        b    = introot (b_sq)
+
+        if (b_sq == b * b) then
+            puts (n . to_s + " " + b . to_s + " " + c . to_s)
+        end
+
+        c_sq += 2 * c + 1
+        c    += 1
+    end
+
+    max_a = (n / Math . sqrt(2)) . floor()
+    for a in 3 .. max_a do
+        b_sq = n_sq - a * a
+        b    = introot (b_sq)
+        if (b_sq == b * b) then
+            puts (a . to_s + " " + b . to_s + " " + n . to_s)
+        end
+    end
+end