Add Python solution to challenge 89

9 files changed, 113 insertions, 52 deletions

diff --git a/challenge-089/paulo-custodio/Makefile b/challenge-089/paulo-custodio/Makefile
new file mode 100644
index 0000000000..c3c762d746
--- /dev/null
+++ b/challenge-089/paulo-custodio/Makefile
@@ -0,0 +1,2 @@
+all:
+	perl ../../challenge-001/paulo-custodio/test.pl
diff --git a/challenge-089/paulo-custodio/forth/ch-1.fs b/challenge-089/paulo-custodio/forth/ch-1.fs
index 4814b6d1e9..b695281395 100644
--- a/challenge-089/paulo-custodio/forth/ch-1.fs
+++ b/challenge-089/paulo-custodio/forth/ch-1.fs
@@ -7,9 +7,6 @@
 \ Write a script to sum GCD of all possible unique pairs between 1 and $N.
 \
 \ This solution uses a recursive algorithm to compute the GCD.
-\
-\ Start the script with N in the stack, i.e.
-\   gforth -e 12345 ch-1.pl
 
 : gcd       { a b -- gcd }
     a 0= IF
@@ -28,4 +25,5 @@
     LOOP
 ;
 
+NEXT-ARG S>NUMBER? 0= THROW DROP
 sum_gcd . CR BYE
diff --git a/challenge-089/paulo-custodio/perl/ch-1.pl b/challenge-089/paulo-custodio/perl/ch-1.pl
index 7a7a1d490d..0e5fe68e05 100644
--- a/challenge-089/paulo-custodio/perl/ch-1.pl
+++ b/challenge-089/paulo-custodio/perl/ch-1.pl
@@ -1,4 +1,4 @@
-#!/usr/bin/perl
+#!/usr/bin/env perl
 
 # Challenge 089
 
diff --git a/challenge-089/paulo-custodio/perl/ch-2.pl b/challenge-089/paulo-custodio/perl/ch-2.pl
index 13f0fb5c8f..03adbe1366 100644
--- a/challenge-089/paulo-custodio/perl/ch-2.pl
+++ b/challenge-089/paulo-custodio/perl/ch-2.pl
@@ -1,4 +1,4 @@
-#!/usr/bin/perl
+#!/usr/bin/env perl
 
 # Challenge 089
 
@@ -46,9 +46,9 @@ use warnings;
 sub square {
     my($a, $b, $c) = @_;
     my @sq;
-    push @sq, [ $c - $b,            $c + ($a + $b), $c - $a         ];
-    push @sq, [ $c - ($a - $b),     $c,             $c + ($a - $b)  ];
-    push @sq, [ $c + $a,            $c - ($a + $b), $c + $b         ];
+    push @sq, [ $c - $b,        $c - ($a - $b), $c + $a        ];
+    push @sq, [ $c + ($a + $b), $c,             $c - ($a + $b) ];
+    push @sq, [ $c - $a,        $c + ($a - $b), $c + $b        ];
     return @sq;
 }
 
diff --git a/challenge-089/paulo-custodio/python/ch-1.py b/challenge-089/paulo-custodio/python/ch-1.py
new file mode 100644
index 0000000000..7b5cc6d4f9
--- /dev/null
+++ b/challenge-089/paulo-custodio/python/ch-1.py
@@ -0,0 +1,28 @@
+#!/usr/bin/env python3
+
+# Challenge 089
+
+# TASK #1 > GCD Sum
+# Submitted by: Mohammad S Anwar
+# You are given a positive integer $N.
+#
+# Write a script to sum GCD of all possible unique pairs between 1 and $N.
+
+# This solution uses a recursive algorithm to compute the GCD.
+
+import sys
+
+def gcd(a, b):
+    if a==0:
+        return b
+    else:
+        return gcd(b%a, a)
+
+def sum_gcd(n):
+    sum = 0
+    for a in range(1, n):
+        for b in range(a+1, n+1):
+            sum += gcd(a, b)
+    return sum
+
+print(sum_gcd(int(sys.argv[1])))
diff --git a/challenge-089/paulo-custodio/python/ch-2.py b/challenge-089/paulo-custodio/python/ch-2.py
new file mode 100644
index 0000000000..95dadc9d10
--- /dev/null
+++ b/challenge-089/paulo-custodio/python/ch-2.py
@@ -0,0 +1,59 @@
+#!/usr/bin/env python3
+
+# Challenge 089
+
+# TASK #2 > Magical Matrix
+# Submitted by: Mohammad S Anwar
+# Write a script to display matrix as below with numbers 1 - 9.
+# Please make sure numbers are used once.
+#
+# [ a b c ]
+# [ d e f ]
+# [ g h i ]
+# So that it satisfies the following:
+#
+# a + b + c = 15
+# d + e + f = 15
+# g + h + i = 15
+# a + d + g = 15
+# b + e + h = 15
+# c + f + i = 15
+# a + e + i = 15
+# c + e + g = 15
+
+# Solution based on Edouard Lucas (https://en.wikipedia.org/wiki/Magic_square#A_method_for_constructing_a_magic_square_of_order_3)
+#
+# Solution is:
+#
+# [   c - b     | c + (a + b) |   c - a     ]
+# [ c - (a - b) |      c      | c + (a - b) ]
+# [   c + a     | c - (a + b) |   c + b     ]
+#
+# The magic constant is 3c, where  0 < a < b < (c - a)  and  b != 2a.
+# Moreover, every 3x3 magic square of distinct positive integers is of this form.
+#
+# In the problem statement it is said that the sum of each row, column and
+# diagonal is 15, this is the square magic constant, therefore we know that:
+#
+#       c = 5
+#
+# We need to find the pairs (a,b) that satisfy the above conditions.
+
+# find a and b for a given c
+def find_ab(c):
+    for a in range(1, 3*c-1+1):
+        for b in range(a+1, 3*c+1):
+            if b < c-a and b != 2*a:
+                return a, b
+
+def square(a, b, c):
+    return [[ c - b,       c - (a - b), c + a       ],
+            [ c + (a + b), c,           c - (a + b) ],
+            [ c - a,       c + (a - b), c + b       ]]
+
+magic_constant = 15
+c = int(magic_constant/3)
+a, b = find_ab(c)
+sq = square(a,b,c)
+for row in (sq):
+    print("[ "+ " ".join([str(x) for x in row]) +" ]")
diff --git a/challenge-089/paulo-custodio/t/test-1.yaml b/challenge-089/paulo-custodio/t/test-1.yaml
new file mode 100644
index 0000000000..ba9c8cbcb3
--- /dev/null
+++ b/challenge-089/paulo-custodio/t/test-1.yaml
@@ -0,0 +1,10 @@
+- setup:
+  cleanup:
+  args:     3
+  input:
+  output:   3
+- setup:
+  cleanup:
+  args:     4
+  input:
+  output:   7
diff --git a/challenge-089/paulo-custodio/t/test-2.yaml b/challenge-089/paulo-custodio/t/test-2.yaml
new file mode 100644
index 0000000000..3c1154ac88
--- /dev/null
+++ b/challenge-089/paulo-custodio/t/test-2.yaml
@@ -0,0 +1,8 @@
+- setup:
+  cleanup:
+  args:
+  input:
+  output:   |
+            [ 2 7 6 ]
+            [ 9 5 1 ]
+            [ 4 3 8 ]
diff --git a/challenge-089/paulo-custodio/test.pl b/challenge-089/paulo-custodio/test.pl
deleted file mode 100644
index 043bbc070d..0000000000
--- a/challenge-089/paulo-custodio/test.pl
+++ /dev/null
@@ -1,44 +0,0 @@
-#!/usr/bin/perl
-
-use strict;
-use warnings;
-use Test::More;
-
-# task 1
-
-ok system("fbc basic/ch-1.bas") == 0;
-for ([3,3], [4,7]) {
-    my($in, $out) = @$_;
-    is capture("basic/ch-1 $in"), " $out\n";
-    is capture("perl perl/ch-1.pl $in"), "$out\n";
-    is capture("gforth -e $in forth/ch-1.fs"), "$out\n";
-}
-
-# task 2
-
-ok system("fbc basic/ch-2.bas") == 0;
-
-is capture("perl perl/ch-2.pl"), <<END;
-[ 2 9 4 ]
-[ 7 5 3 ]
-[ 6 1 8 ]
-END
-
-for ("basic/ch-2",
-     "gforth forth/ch-2.fs") {
-    is capture($_), <<END;
-[ 2 7 6 ]
-[ 9 5 1 ]
-[ 4 3 8 ]
-END
-}
-
-done_testing;
-
-# capture output, normalize eol
-sub capture {
-    my($cmd) = @_;
-    local $_ = `$cmd`;
-    s/[ \t\r]*\n/\n/g;
-    return $_;
-}