Week 151: Simplified algorithm for part 2.

By adding two dummy 0's to the "best" array, and not bothering the
calculate the best values for the first two houses, we don't have
a case study inside the main loop.

6 files changed, 43 insertions, 83 deletions

diff --git a/challenge-151/abigail/awk/ch-2.awk b/challenge-151/abigail/awk/ch-2.awk
index 73d9b19812..65428f81e8 100644
--- a/challenge-151/abigail/awk/ch-2.awk
+++ b/challenge-151/abigail/awk/ch-2.awk
@@ -14,13 +14,11 @@ function max (a, b) {
 
 {
     delete best
-    for (i = NF; i > 0; i --) {
-        best [i] = NF <= 2     ? $i                                     \
-                 : i == NF     ? $i                                     \
-                 : i == 1      ? $i     + best [i + 2]                  \
-                 : i == NF - 1 ? max($i,  best [i + 1])                 \
-                 :               max($i + best [i + 2], best [i + 1])
+    best [NF + 1] = 0
+    best [NF + 2] = 0
+    for (i = NF; i > 2; i --) {
+        best [i] = max($i + best [i + 2], best [i + 1])
     }
-    print best [1]
+    print $1 + best [3]
 }
 
diff --git a/challenge-151/abigail/bash/ch-2.sh b/challenge-151/abigail/bash/ch-2.sh
index 0b8f213fca..6a9eceec29 100644
--- a/challenge-151/abigail/bash/ch-2.sh
+++ b/challenge-151/abigail/bash/ch-2.sh
@@ -13,34 +13,14 @@ set -f
 declare -a best
 
 while read -a houses
-do    if   [[ ${#houses[@]} -lt 2 ]]
-      then echo ${houses[0]}
-           continue
-      fi
-
-      size=${#houses[@]}
-      for  ((i = size - 1; i >= 0; i --))
-      do   ((i1 = i + 1))
-           ((i2 = i + 2))
-           if ((i == size - 1))
-           then best[$i]=${houses[$i]}
-                continue
-           fi
-
-           if ((i == 0))
-           then best[$i]=$((${houses[$i]} + ${best[$i2]}))
-                continue
-           fi
-
-           if   ((i == size - 2))
-           then val1=${houses[$i]}
-                val2=${best[$i1]}
-           else val1=$((${houses[$i]} + ${best[$i2]}))
-                val2=${best[$i1]}
-           fi
-
+do    size=${#houses[@]}
+      best[$((size + 0))]=0
+      best[$((size + 1))]=0
+      for  ((i = size - 1; i >= 2; i --))
+      do   ((val1 = ${houses[$i]} + ${best[$((i + 2))]}))
+           ((val2 = ${best[$((i + 1))]}))
            best[$i]=$((val1 < val2 ? val2 : val1))
       done
 
-      echo ${best[0]}
+      echo $((${houses[0]} + ${best[2]}))
 done
diff --git a/challenge-151/abigail/c/ch-2.c b/challenge-151/abigail/c/ch-2.c
index c2a62f6c92..34e7e24ac1 100644
--- a/challenge-151/abigail/c/ch-2.c
+++ b/challenge-151/abigail/c/ch-2.c
@@ -37,21 +37,20 @@ int main (void) {
             line_ptr += offset;
         }
 
-        if ((best = (int *) malloc (nr_of_houses * sizeof (int))) == NULL) {
+        if ((best = (int *)
+             malloc ((nr_of_houses + 2) * sizeof (int))) == NULL) {
             perror ("Malloc failed");
             exit (1);
         }
 
-        for (int i = nr_of_houses - 1; i >= 0; i --) {
-            best [i] = nr_of_houses < 2      ? houses [i]
-                     : i == nr_of_houses - 1 ? houses [i]
-                     : i == 0                ? houses [i]      + best [i + 2]
-                     : i == nr_of_houses - 2 ? max (houses [i],  best [i + 1])
-                     :                         max (houses [i] + best [i + 2],
-                                                                 best [i + 1]);
+        best [nr_of_houses + 0] = 0;
+        best [nr_of_houses + 1] = 0;
+
+        for (int i = nr_of_houses - 1; i >= 2; i --) {
+            best [i] = max (houses [i] + best [i + 2], best [i + 1]);
         }
 
-        printf ("%d\n", best [0]);
+        printf ("%d\n", houses [0] + best [2]);
 
         free (houses);
         free (best);
diff --git a/challenge-151/abigail/lua/ch-2.lua b/challenge-151/abigail/lua/ch-2.lua
index 98df0aae1d..0e644d0a6c 100644
--- a/challenge-151/abigail/lua/ch-2.lua
+++ b/challenge-151/abigail/lua/ch-2.lua
@@ -15,19 +15,12 @@ for line in io . lines () do
         houses [#houses + 1] = val
     end
 
-    for i = #houses, 1, -1 do
-        if     2 >= #houses then
-            best [i] =             houses [i]
-        elseif i == #houses then
-            best [i] =             houses [i]
-        elseif i == 0       then
-            best [i] =             houses [i] + best [i + 2]
-        elseif i == #houses - 1 then
-            best [i] = math . max (houses [i],  best [i + 1])
-        else
-            best [i] = math . max (houses [i] + best [i + 2],
-                                                best [i + 1])
-        end
+    best [#houses + 1] = 0
+    best [#houses + 2] = 0
+
+    for i = #houses, 3, -1 do
+        best [i] = math . max (houses [i] + best [i + 2], best [i + 1])
     end
-    print (best [1])
+
+    print (houses [1] + best [3])
 end
diff --git a/challenge-151/abigail/node/ch-2.js b/challenge-151/abigail/node/ch-2.js
index f4d698ec49..079c61e0ce 100644
--- a/challenge-151/abigail/node/ch-2.js
+++ b/challenge-151/abigail/node/ch-2.js
@@ -13,14 +13,11 @@
 . on              ('line', line => {
     let houses = line . trim () . split (/ +/) . map (n => +n)
     let best   = []
-    for (let i = houses . length - 1; i >= 0; i --) {
-        best [i] =
-            2 >  houses . length     ?             houses [i]
-          : i == houses . length - 1 ?             houses [i]
-          : i == 0                   ?             houses [i] + best [i + 2]
-          : i == houses . length - 2 ? Math . max (houses [i],  best [i + 1])
-          :                            Math . max (houses [i] + best [i + 2],
-                                                                best [i + 1])
+    let size   = houses . length
+    best [size + 0] = 0
+    best [size + 1] = 0
+    for (let i = size - 1; i >= 2; i --) {
+        best [i] = Math . max (houses [i] + best [i + 2], best [i + 1])
     }
-    console . log (best [0])
+    console . log (houses [0] + best [2])
 })
diff --git a/challenge-151/abigail/perl/ch-2.pl b/challenge-151/abigail/perl/ch-2.pl
index 5f5519381f..9c1e1d3b60 100644
--- a/challenge-151/abigail/perl/ch-2.pl
+++ b/challenge-151/abigail/perl/ch-2.pl
@@ -26,29 +26,22 @@ use List::Util qw [max];
 # the value of the current house, plus the maximum we can get
 # if we started two houses down. The maximum value we can get by
 # not robbing this house is the best we can get by starting from
-# the next house.
+# the next house. We'll add two empty houses to make everything work
+# out nicely.
 #
-# We need some fiddling around the edges:
-#    - Maximizing the value starting from the last house is by robbing it.
-#    - Maximizing the value starting from the penultimate house is
-#      robbing the most valuable of the last two houses.
-#    - If there are only two houses, rob the current house.
-#    - The first house should always be robbed (a requirement); the
-#      maximum value which can be achieved we get by starting from the
-#      third house, adding the value of the first.
+# Note that we don't have to calculate the best option starting
+# from the second house, as we will always skip the second house.
+# And we must always pick the first house.
 #
 
 sub best ($houses) {
     my @best;
-    foreach my $i (reverse keys @$houses) {
-        $best [$i] = @$houses < 2        ?     $$houses [$i]
-                   : $i == $#$houses     ?     $$houses [$i]
-                   : $i == 0             ?     $$houses [$i] + $best [$i + 2]
-                   : $i == $#$houses - 1 ? max $$houses [$i],  $best [$i + 1]
-                   :                       max $$houses [$i] + $best [$i + 2],
-                                                               $best [$i + 1];
+    $best [$#$houses + 1] = 0;
+    $best [$#$houses + 2] = 0;
+    for (my $i = $#$houses; $i >= 2; $i --) {
+        $best [$i] = max $$houses [$i] + $best [$i + 2], $best [$i + 1];
     }
-    $best [0];
+    $$houses [0] + $best [2];
 }