Merge remote-tracking branch 'upstream/master'

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diff --git a/challenge-109/abigail/node/ch-2.js b/challenge-109/abigail/node/ch-2.js
new file mode 100644
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+++ b/challenge-109/abigail/node/ch-2.js
@@ -0,0 +1,114 @@
+#!/usr/local/bin/node
+
+//
+// See ../README.md
+//
+
+//
+// Run as: node ch-2.js < input-file
+//
+
+let SIZE = 7
+let printf = require ('printf')
+
+require ('readline')
+. createInterface ({input: process . stdin})   
+. on ('line', _ => main (_ . split (/\s+/) . map (_ => +_)))
+;
+
+
+function main (numbers) {
+    //
+    // First, we create a datastructure to collect all the
+    // differences between two numbers, with the restriction
+    // that the difference is present in numbers
+    //
+    let differences = {}
+    numbers . forEach (n => differences [n] = [])
+
+    numbers . forEach ((x, x_i) => {
+        numbers . forEach ((y, y_i) => {
+            if (x_i != y_i) {
+                let diff = x - y
+                if (differences [diff]) {
+                    differences [diff] . push ([x_i, y_i])
+                }
+            }
+        })
+    })
+
+    //
+    // For each number d in numbers, with index d_i, find all pairs
+    // of distinct numbers, all with index other than d_i, where the
+    // differences of each pair is d.
+    //
+    numbers . forEach ((d, d_i) => {
+        let diffs = differences [d]
+        for (let x = 0; x < diffs . length; x ++) {
+            //
+            // Ignore any difference involving d_i
+            //
+            if (diffs [x] [0] == d_i || diffs [x] [1] == d_i) {
+                continue
+            }
+            for (let y = x + 1; y < diffs . length; y ++) {
+                //
+                // Second diff should not involve d_i, and all its
+                // elemenst should be different from the first diff
+                //
+                if (diffs [y] [0] == d_i || diffs [y] [1] == d_i ||
+                    diffs [x] [0] == diffs [y] [0] ||
+                    diffs [x] [0] == diffs [y] [1] ||
+                    diffs [x] [1] == diffs [y] [0] ||
+                    diffs [x] [1] == diffs [y] [1]) {
+                    continue
+                }
+                //
+                // W.l.o.g we can now assume diff [x] gives us a_i and c_i,
+                // and diff [y] gives use g_i and e_i
+                //
+                let [a_i, c_i] = diffs [x]
+                let [g_i, e_i] = diffs [y]
+
+                //
+                // Find the two possibilities for b_i and f_i
+                //
+                for (let b_i = 0; b_i < numbers . length; b_i ++) {
+                    if (b_i == a_i || b_i == c_i || b_i == d_i ||
+                        b_i == e_i || b_i == g_i) {
+                        continue
+                    }
+                    for (let f_i = 0; f_i < numbers . length; f_i ++) {
+                        if (f_i == a_i || f_i == b_i || f_i == c_i ||
+                            f_i == d_i || f_i == e_i || f_i == g_i) {
+                            continue
+                        }
+                        //
+                        // Do we have a winner?
+                        //
+                        let target  = numbers [a_i] + numbers [b_i]
+                        if (target == numbers [b_i] + numbers [c_i] +
+                                      numbers [d_i] &&
+                            target == numbers [d_i] + numbers [e_i] +
+                                      numbers [f_i] &&
+                            target == numbers [f_i] + numbers [g_i]) {
+                            //
+                            // Print the results, and the reverse
+                            //
+                            printf (process . stdout, "%d %d %d %d %d %d %d\n",
+                                    numbers [a_i], numbers [b_i],
+                                    numbers [c_i], numbers [d_i],
+                                    numbers [e_i], numbers [f_i],
+                                    numbers [g_i])
+                            printf (process . stdout, "%d %d %d %d %d %d %d\n",
+                                    numbers [g_i], numbers [f_i],
+                                    numbers [e_i], numbers [d_i],
+                                    numbers [c_i], numbers [b_i],
+                                    numbers [a_i])
+                        }
+                    }
+                }
+            }
+        }
+    })
+}