Merge remote-tracking branch 'upstream/master'

23 files changed, 638 insertions, 1 deletions

diff --git a/challenge-077/abigail/Part1/input1 b/challenge-077/abigail/Part1/input1
new file mode 100644
index 0000000000..1e8b314962
--- /dev/null
+++ b/challenge-077/abigail/Part1/input1
@@ -0,0 +1 @@
+6
diff --git a/challenge-077/abigail/Part1/input2 b/challenge-077/abigail/Part1/input2
new file mode 100644
index 0000000000..ec635144f6
--- /dev/null
+++ b/challenge-077/abigail/Part1/input2
@@ -0,0 +1 @@
+9
diff --git a/challenge-077/abigail/Part1/input3 b/challenge-077/abigail/Part1/input3
new file mode 100644
index 0000000000..83b33d238d
--- /dev/null
+++ b/challenge-077/abigail/Part1/input3
@@ -0,0 +1 @@
+1000
diff --git a/challenge-077/abigail/Part1/input4 b/challenge-077/abigail/Part1/input4
new file mode 100644
index 0000000000..66a899ac41
--- /dev/null
+++ b/challenge-077/abigail/Part1/input4
@@ -0,0 +1 @@
+377
diff --git a/challenge-077/abigail/Part1/output1.exp b/challenge-077/abigail/Part1/output1.exp
new file mode 100644
index 0000000000..170feacebb
--- /dev/null
+++ b/challenge-077/abigail/Part1/output1.exp
@@ -0,0 +1,2 @@
+1 + 2 + 3 = 6
+1 + 5 = 6
diff --git a/challenge-077/abigail/Part1/output2.exp b/challenge-077/abigail/Part1/output2.exp
new file mode 100644
index 0000000000..d029569c34
--- /dev/null
+++ b/challenge-077/abigail/Part1/output2.exp
@@ -0,0 +1,2 @@
+1 + 3 + 5 = 9
+1 + 8 = 9
diff --git a/challenge-077/abigail/Part1/output3.exp b/challenge-077/abigail/Part1/output3.exp
new file mode 100644
index 0000000000..88b8b236eb
--- /dev/null
+++ b/challenge-077/abigail/Part1/output3.exp
@@ -0,0 +1,15 @@
+2 + 3 + 8 + 21 + 34 + 89 + 233 + 610 = 1000
+2 + 3 + 8 + 55 + 89 + 233 + 610 = 1000
+2 + 3 + 8 + 144 + 233 + 610 = 1000
+2 + 3 + 8 + 377 + 610 = 1000
+2 + 3 + 8 + 987 = 1000
+5 + 8 + 21 + 34 + 89 + 233 + 610 = 1000
+5 + 8 + 55 + 89 + 233 + 610 = 1000
+5 + 8 + 144 + 233 + 610 = 1000
+5 + 8 + 377 + 610 = 1000
+5 + 8 + 987 = 1000
+13 + 21 + 34 + 89 + 233 + 610 = 1000
+13 + 55 + 89 + 233 + 610 = 1000
+13 + 144 + 233 + 610 = 1000
+13 + 377 + 610 = 1000
+13 + 987 = 1000
diff --git a/challenge-077/abigail/Part1/output4.exp b/challenge-077/abigail/Part1/output4.exp
new file mode 100644
index 0000000000..4fc109d510
--- /dev/null
+++ b/challenge-077/abigail/Part1/output4.exp
@@ -0,0 +1,7 @@
+1 + 2 + 5 + 13 + 34 + 89 + 233 = 377
+3 + 5 + 13 + 34 + 89 + 233 = 377
+8 + 13 + 34 + 89 + 233 = 377
+21 + 34 + 89 + 233 = 377
+55 + 89 + 233 = 377
+144 + 233 = 377
+377 = 377
diff --git a/challenge-077/abigail/Part1/solution.js b/challenge-077/abigail/Part1/solution.js
new file mode 100644
index 0000000000..f2d7173403
--- /dev/null
+++ b/challenge-077/abigail/Part1/solution.js
@@ -0,0 +1,97 @@
+//
+// Exercise:
+//     You are given a positive integer $N.
+//     Write a script to find out all possible combination of Fibonacci
+//     Numbers required to get $N on addition.
+//
+//     You are NOT allowed to repeat a number. Print 0 if none found.
+//
+
+//
+// Note:
+//    The "Print 0 if none found." is irrelevant. There is always at
+//    least one way to write any positive integer as a sum of distinct
+//    Fibonacci Numbers. (Zeckendorf's theorem states: "very positive
+//    integer can be represented uniquely as the sum of one or more
+//    distinct Fibonacci numbers in such a way that the sum does not
+//    include any two consecutive Fibonacci numbers")
+//
+
+//
+// Read the input number from STDIN
+//
+let fs = require ("fs");
+let N  = +fs . readFileSync (0) . toString () . trim ();
+
+//
+// Generate a list of Fibonacci numbers, starting with (1, 2),
+// up to the target number. Store this in FIB.
+//
+let FIB = [1, 2];
+while (FIB [FIB . length - 1] + FIB [FIB . length - 2] <= N) {
+    FIB . push (FIB [FIB . length - 1] + FIB [FIB . length - 2]);
+}
+
+//
+// Recursive function to find the sums. First argument is the target
+// number, second argument is the index of smallest number which can
+// be used. 
+//
+function solutions (target, index = 0) {
+    let output = [];
+    //
+    // Iterate over the list of Fibonacci numbers, looking for
+    // candidates. We're starting with the given index.
+    //
+    for (let i = index; i < FIB . length; i ++) {
+        let fib = FIB [i];
+        if (fib > target) {
+            //
+            // If the candidate is larger than the target number,
+            // we can stop looking, as each subsequent number will
+            // be larger.
+            //
+            break;
+        }
+        if (fib == target) {
+            //
+            // If the candidate is equal to the target number, 
+            // then it's a trivial solution (a sum of 1 number).
+            // Add it to the list of possibilities, and stop
+            // searching.
+            //
+            output . push ([fib]);
+            break;
+        }
+        else {
+            //
+            // Find solutions for the difference between the
+            // candidate and the target number, with the restriction
+            // that each number in that sum is larger than the numbers
+            // used so far.
+            //
+            let rec_solutions = solutions (target - fib, i + 1);
+
+            //
+            // For each solution found in recursion, we have a solution
+            // for this call, by adding the candidate to it.
+            //
+            for (let j = 0; j < rec_solutions . length; j ++) {
+                output . push ([fib] . concat (rec_solutions [j]));
+            }
+        }
+    }
+    return output;
+}
+
+//
+// Find the solutions
+//
+let sols = solutions (N);
+
+//
+// And print the results
+//
+for (let i = 0; i < sols . length; i ++) {
+    console . log (sols [i] . join (" + ") + " = " + N);
+}
diff --git a/challenge-077/abigail/Part1/solution.pl b/challenge-077/abigail/Part1/solution.pl
new file mode 100644
index 0000000000..b39028c5b9
--- /dev/null
+++ b/challenge-077/abigail/Part1/solution.pl
@@ -0,0 +1,63 @@
+#!/opt/perl/bin/perl
+
+#
+# Exercise:
+#     You are given a positive integer $N.
+#     Write a script to find out all possible combination of Fibonacci
+#     Numbers required to get $N on addition.
+#
+#     You are NOT allowed to repeat a number. Print 0 if none found.
+#
+
+#
+# Note:
+#    The "Print 0 if none found." is irrelevant. There is always at
+#    least one way to write any positive integer as a sum of distinct
+#    Fibonacci Numbers. (Zeckendorf's theorem states: "very positive
+#    integer can be represented uniquely as the sum of one or more
+#    distinct Fibonacci numbers in such a way that the sum does not
+#    include any two consecutive Fibonacci numbers")
+#
+
+use 5.032;
+
+use strict;
+use warnings;
+no  warnings 'syntax';
+
+use experimental 'signatures';
+use experimental 'lexical_subs';
+
+chomp (my $N = <>);
+
+#
+# Create a list of Fibonacci Numbers up to $N. We will start with (1, 2),
+# skipping the first two numbers 0 and 1. 0 doesn't add anything, and
+# we cannot duplicate numbers, so we just need one 1.
+#
+my @FIB = (1, 2);
+while ($FIB [-1] + $FIB [-2] <= $N) {
+    push @FIB => $FIB [-1] + $FIB [-2];
+}
+
+sub solutions;
+
+#
+# Create all solutions for number $target, with all Fibonnaci numbers
+# having index $index or above.
+#
+sub solutions ($target, $index = 0) {
+    map {
+        my $fib = $FIB [$_];
+        $target  < $fib ? ()
+      : $target == $fib ? [$target]
+      : map {[$fib, @$_]} solutions ($target - $fib, $_ + 1)
+    } $index .. @FIB - 1;
+}
+
+my @all = solutions $N;
+
+local $" = " + ";
+say "@$_ = $N" for @all;
+
+__END__
diff --git a/challenge-077/abigail/Part2/input1 b/challenge-077/abigail/Part2/input1
new file mode 100644
index 0000000000..f8beeb613f
--- /dev/null
+++ b/challenge-077/abigail/Part2/input1
@@ -0,0 +1,3 @@
+O O X
+X O O
+X O O
diff --git a/challenge-077/abigail/Part2/input2 b/challenge-077/abigail/Part2/input2
new file mode 100644
index 0000000000..d81104f1b2
--- /dev/null
+++ b/challenge-077/abigail/Part2/input2
@@ -0,0 +1,4 @@
+O O X O
+X O O O
+X O O X
+O X O O
diff --git a/challenge-077/abigail/Part2/output1.exp b/challenge-077/abigail/Part2/output1.exp
new file mode 100644
index 0000000000..d00491fd7e
--- /dev/null
+++ b/challenge-077/abigail/Part2/output1.exp
@@ -0,0 +1 @@
+1
diff --git a/challenge-077/abigail/Part2/output2.exp b/challenge-077/abigail/Part2/output2.exp
new file mode 100644
index 0000000000..0cfbf08886
--- /dev/null
+++ b/challenge-077/abigail/Part2/output2.exp
@@ -0,0 +1 @@
+2
diff --git a/challenge-077/abigail/Part2/solution.js b/challenge-077/abigail/Part2/solution.js
new file mode 100644
index 0000000000..ee04d5c605
--- /dev/null
+++ b/challenge-077/abigail/Part2/solution.js
@@ -0,0 +1,60 @@
+//
+// Exercise:
+//    You are given m x n character matrix consists of O and X only.
+//    Write a script to count the total number of X surrounded by O only.
+//    Print 0 if none found.
+//
+
+//
+// Read in the board:
+//    - Read STDIN
+//    - Split by newlines
+//    - Split each line on spaces
+//    - Map an 'X' to a 1, 'O' to a 0.
+//    - Add a 0 to the beginning and end of each line.
+//
+let fs    = require ("fs");
+let board = fs . readFileSync (0) . toString () . trim () . split ("\n") .
+            map (line => (line . split (" ")) . map (c => c == "X" ? 1 : 0)) .
+            map (line => ([0] . concat (line) . concat ([0])));
+
+//
+// Add top and bottom with 0s
+//
+board . push    (board [0] . map (x => 0));
+board . unshift (board [0] . map (x => 0));
+
+let count = 0;
+
+//
+// Iterate over the cells of the board board, skipping cells on the edge
+// (as we added them). For each 1, check the 8 cells surrounding the cell
+// (this will never be outside of the board). If one of the neighbouring
+// cells is a 1, move on the next cell. If no neighbouring cell is 1,
+// we add 1 to the count.
+//
+for (let x = 1; x < board . length - 1; x ++) {
+  ELEMENT:
+    for (let y = 1; y < board [x] . length - 1; y ++) {
+        if (!board [x] [y]) {
+            continue;
+        }
+        for (let dx = -1; dx <= 1; dx ++) {
+            for (let dy = -1; dy <= 1; dy ++) {
+                if (dx == 0 && dy == 0) {
+                    continue;
+                }
+                if (board [x + dx] [y + dy]) {
+                    continue ELEMENT;
+                }
+            }
+        }
+        count ++;
+    }
+}
+
+
+//
+// Print the results.
+//
+console . log (count);
diff --git a/challenge-077/abigail/Part2/solution.pl b/challenge-077/abigail/Part2/solution.pl
new file mode 100644
index 0000000000..784f4cd662
--- /dev/null
+++ b/challenge-077/abigail/Part2/solution.pl
@@ -0,0 +1,56 @@
+#!/opt/perl/bin/perl
+
+#
+# Exercise:
+#    You are given m x n character matrix consists of O and X only.
+#    Write a script to count the total number of X surrounded by O only.
+#    Print 0 if none found.
+#
+
+use 5.032;
+
+use strict;
+use warnings;
+no  warnings 'syntax';
+
+use experimental 'signatures';
+use experimental 'lexical_subs';
+
+#
+# Read in the board. Use 1 for an X, and 0 for an 0. Add a band
+# of 0's around the board.
+#
+
+my      @board = map {[0, map ({/O/ ? 0 : 1} /[OX]/g), 0]} <>;
+push    @board => [(0) x @{$board [0]}];
+unshift @board => [(0) x @{$board [0]}];
+
+my $count = 0;
+
+#
+# Iterate over the cells of the board board, skipping cells on the edge
+# (as we added them). For each 1, check the 8 cells surrounding the cell
+# (this will never be outside of the board). If one of the neighbouring
+# cells is a 1, move on the next cell. If no neighbouring cell is 1,
+# we add 1 to the count.
+#
+for (my $x = 1; $x < @board - 1; $x ++) {
+  ELEMENT:
+    for (my $y = 1; $y < @{$board [$x]} - 1; $y ++) {
+        next unless $board [$x] [$y];
+        foreach my $dx (-1 .. 1) {
+            foreach my $dy (-1 .. 1) {
+                next unless $dx || $dy;
+                next ELEMENT if $board [$x + $dx] [$y + $dy];
+            }
+        }
+        $count ++;
+    }
+}
+
+#
+# Print solution.
+#
+say $count;
+
+__END__
diff --git a/challenge-077/abigail/node/ch-1.js b/challenge-077/abigail/node/ch-1.js
new file mode 100644
index 0000000000..f2d7173403
--- /dev/null
+++ b/challenge-077/abigail/node/ch-1.js
@@ -0,0 +1,97 @@
+//
+// Exercise:
+//     You are given a positive integer $N.
+//     Write a script to find out all possible combination of Fibonacci
+//     Numbers required to get $N on addition.
+//
+//     You are NOT allowed to repeat a number. Print 0 if none found.
+//
+
+//
+// Note:
+//    The "Print 0 if none found." is irrelevant. There is always at
+//    least one way to write any positive integer as a sum of distinct
+//    Fibonacci Numbers. (Zeckendorf's theorem states: "very positive
+//    integer can be represented uniquely as the sum of one or more
+//    distinct Fibonacci numbers in such a way that the sum does not
+//    include any two consecutive Fibonacci numbers")
+//
+
+//
+// Read the input number from STDIN
+//
+let fs = require ("fs");
+let N  = +fs . readFileSync (0) . toString () . trim ();
+
+//
+// Generate a list of Fibonacci numbers, starting with (1, 2),
+// up to the target number. Store this in FIB.
+//
+let FIB = [1, 2];
+while (FIB [FIB . length - 1] + FIB [FIB . length - 2] <= N) {
+    FIB . push (FIB [FIB . length - 1] + FIB [FIB . length - 2]);
+}
+
+//
+// Recursive function to find the sums. First argument is the target
+// number, second argument is the index of smallest number which can
+// be used. 
+//
+function solutions (target, index = 0) {
+    let output = [];
+    //
+    // Iterate over the list of Fibonacci numbers, looking for
+    // candidates. We're starting with the given index.
+    //
+    for (let i = index; i < FIB . length; i ++) {
+        let fib = FIB [i];
+        if (fib > target) {
+            //
+            // If the candidate is larger than the target number,
+            // we can stop looking, as each subsequent number will
+            // be larger.
+            //
+            break;
+        }
+        if (fib == target) {
+            //
+            // If the candidate is equal to the target number, 
+            // then it's a trivial solution (a sum of 1 number).
+            // Add it to the list of possibilities, and stop
+            // searching.
+            //
+            output . push ([fib]);
+            break;
+        }
+        else {
+            //
+            // Find solutions for the difference between the
+            // candidate and the target number, with the restriction
+            // that each number in that sum is larger than the numbers
+            // used so far.
+            //
+            let rec_solutions = solutions (target - fib, i + 1);
+
+            //
+            // For each solution found in recursion, we have a solution
+            // for this call, by adding the candidate to it.
+            //
+            for (let j = 0; j < rec_solutions . length; j ++) {
+                output . push ([fib] . concat (rec_solutions [j]));
+            }
+        }
+    }
+    return output;
+}
+
+//
+// Find the solutions
+//
+let sols = solutions (N);
+
+//
+// And print the results
+//
+for (let i = 0; i < sols . length; i ++) {
+    console . log (sols [i] . join (" + ") + " = " + N);
+}
diff --git a/challenge-077/abigail/node/ch-2.js b/challenge-077/abigail/node/ch-2.js
new file mode 100644
index 0000000000..ee04d5c605
--- /dev/null
+++ b/challenge-077/abigail/node/ch-2.js
@@ -0,0 +1,60 @@
+//
+// Exercise:
+//    You are given m x n character matrix consists of O and X only.
+//    Write a script to count the total number of X surrounded by O only.
+//    Print 0 if none found.
+//
+
+//
+// Read in the board:
+//    - Read STDIN
+//    - Split by newlines
+//    - Split each line on spaces
+//    - Map an 'X' to a 1, 'O' to a 0.
+//    - Add a 0 to the beginning and end of each line.
+//
+let fs    = require ("fs");
+let board = fs . readFileSync (0) . toString () . trim () . split ("\n") .
+            map (line => (line . split (" ")) . map (c => c == "X" ? 1 : 0)) .
+            map (line => ([0] . concat (line) . concat ([0])));
+
+//
+// Add top and bottom with 0s
+//
+board . push    (board [0] . map (x => 0));
+board . unshift (board [0] . map (x => 0));
+
+let count = 0;
+
+//
+// Iterate over the cells of the board board, skipping cells on the edge
+// (as we added them). For each 1, check the 8 cells surrounding the cell
+// (this will never be outside of the board). If one of the neighbouring
+// cells is a 1, move on the next cell. If no neighbouring cell is 1,
+// we add 1 to the count.
+//
+for (let x = 1; x < board . length - 1; x ++) {
+  ELEMENT:
+    for (let y = 1; y < board [x] . length - 1; y ++) {
+        if (!board [x] [y]) {
+            continue;
+        }
+        for (let dx = -1; dx <= 1; dx ++) {
+            for (let dy = -1; dy <= 1; dy ++) {
+                if (dx == 0 && dy == 0) {
+                    continue;
+                }
+                if (board [x + dx] [y + dy]) {
+                    continue ELEMENT;
+                }
+            }
+        }
+        count ++;
+    }
+}
+
+
+//
+// Print the results.
+//
+console . log (count);
diff --git a/challenge-077/abigail/perl/ch-1.pl b/challenge-077/abigail/perl/ch-1.pl
new file mode 100644
index 0000000000..b39028c5b9
--- /dev/null
+++ b/challenge-077/abigail/perl/ch-1.pl
@@ -0,0 +1,63 @@
+#!/opt/perl/bin/perl
+
+#
+# Exercise:
+#     You are given a positive integer $N.
+#     Write a script to find out all possible combination of Fibonacci
+#     Numbers required to get $N on addition.
+#
+#     You are NOT allowed to repeat a number. Print 0 if none found.
+#
+
+#
+# Note:
+#    The "Print 0 if none found." is irrelevant. There is always at
+#    least one way to write any positive integer as a sum of distinct
+#    Fibonacci Numbers. (Zeckendorf's theorem states: "very positive
+#    integer can be represented uniquely as the sum of one or more
+#    distinct Fibonacci numbers in such a way that the sum does not
+#    include any two consecutive Fibonacci numbers")
+#
+
+use 5.032;
+
+use strict;
+use warnings;
+no  warnings 'syntax';
+
+use experimental 'signatures';
+use experimental 'lexical_subs';
+
+chomp (my $N = <>);
+
+#
+# Create a list of Fibonacci Numbers up to $N. We will start with (1, 2),
+# skipping the first two numbers 0 and 1. 0 doesn't add anything, and
+# we cannot duplicate numbers, so we just need one 1.
+#
+my @FIB = (1, 2);
+while ($FIB [-1] + $FIB [-2] <= $N) {
+    push @FIB => $FIB [-1] + $FIB [-2];
+}
+
+sub solutions;
+
+#
+# Create all solutions for number $target, with all Fibonnaci numbers
+# having index $index or above.
+#
+sub solutions ($target, $index = 0) {
+    map {
+        my $fib = $FIB [$_];
+        $target  < $fib ? ()
+      : $target == $fib ? [$target]
+      : map {[$fib, @$_]} solutions ($target - $fib, $_ + 1)
+    } $index .. @FIB - 1;
+}
+
+my @all = solutions $N;
+
+local $" = " + ";
+say "@$_ = $N" for @all;
+
+__END__
diff --git a/challenge-077/abigail/perl/ch-2.pl b/challenge-077/abigail/perl/ch-2.pl
new file mode 100644
index 0000000000..784f4cd662
--- /dev/null
+++ b/challenge-077/abigail/perl/ch-2.pl
@@ -0,0 +1,56 @@
+#!/opt/perl/bin/perl
+
+#
+# Exercise:
+#    You are given m x n character matrix consists of O and X only.
+#    Write a script to count the total number of X surrounded by O only.
+#    Print 0 if none found.
+#
+
+use 5.032;
+
+use strict;
+use warnings;
+no  warnings 'syntax';
+
+use experimental 'signatures';
+use experimental 'lexical_subs';
+
+#
+# Read in the board. Use 1 for an X, and 0 for an 0. Add a band
+# of 0's around the board.
+#
+
+my      @board = map {[0, map ({/O/ ? 0 : 1} /[OX]/g), 0]} <>;
+push    @board => [(0) x @{$board [0]}];
+unshift @board => [(0) x @{$board [0]}];
+
+my $count = 0;
+
+#
+# Iterate over the cells of the board board, skipping cells on the edge
+# (as we added them). For each 1, check the 8 cells surrounding the cell
+# (this will never be outside of the board). If one of the neighbouring
+# cells is a 1, move on the next cell. If no neighbouring cell is 1,
+# we add 1 to the count.
+#
+for (my $x = 1; $x < @board - 1; $x ++) {
+  ELEMENT:
+    for (my $y = 1; $y < @{$board [$x]} - 1; $y ++) {
+        next unless $board [$x] [$y];
+        foreach my $dx (-1 .. 1) {
+            foreach my $dy (-1 .. 1) {
+                next unless $dx || $dy;
+                next ELEMENT if $board [$x + $dx] [$y + $dy];
+            }
+        }
+        $count ++;
+    }
+}
+
+#
+# Print solution.
+#
+say $count;
+
+__END__
diff --git a/challenge-077/abigail/test.pl b/challenge-077/abigail/test.pl
new file mode 100755
index 0000000000..115e570c62
--- /dev/null
+++ b/challenge-077/abigail/test.pl
@@ -0,0 +1,44 @@
+#!/opt/perl/bin/perl
+
+#
+# Test the solutions. Either call it with the directory name you
+# want to test in, or call it as "../test.pl" from within the directory.
+#
+
+use 5.032;
+
+use strict;
+use warnings;
+no  warnings 'syntax';
+
+chdir shift if @ARGV;
+
+use experimental 'signatures';
+
+use Test::More;
+
+my %languages = (
+    Perl        =>   ["/opt/perl/bin/perl"  => 'pl'],
+    JavaScript  =>   ["/usr/local/bin/node" => 'js'],
+);
+
+
+my @inputs = <input*>;
+
+foreach my $language (sort keys %languages) {
+    my ($exe, $ext) = @{$languages {$language}};
+    next unless -r "solution.$ext";
+    subtest $language => sub {
+        foreach my $input (@inputs) {
+            my $output_exp = ($input =~ s/input/output/r) . ".exp";
+            my $exp        = `cat $output_exp`;
+            my $got        = `$exe ./solution.$ext < $input`;
+            is $got, $exp, $input;
+        }
+    }
+}
+
+done_testing;
+
+
+__END__
diff --git a/challenge-077/cheok-yin-fung/perl/ch-2.pl b/challenge-077/cheok-yin-fung/perl/ch-2.pl
index e5b4026bed..69b9ee0cf3 100644
--- a/challenge-077/cheok-yin-fung/perl/ch-2.pl
+++ b/challenge-077/cheok-yin-fung/perl/ch-2.pl
@@ -50,9 +50,10 @@ print_matrix(\@matrix);
 
 sub detect {
     my $segment = join "", @_;
+    $segment =~ s/XX/II/g;
     $segment =~ s/XI/II/g;
     $segment =~ s/IX/II/g;
-    $segment =~ s/XX/II/g;
+#    $segment =~ s/XX/II/g;   modify after deadline
     return split //, $segment;
 }
 
diff --git a/challenge-077/mohammad-anwar/blog1.txt b/challenge-077/mohammad-anwar/blog1.txt
new file mode 100644
index 0000000000..dcb1ac636c
--- /dev/null
+++ b/challenge-077/mohammad-anwar/blog1.txt
@@ -0,0 +1 @@
+https://www.youtube.com/watch?v=dgz4T4GwKxQ