Just another update

Task/Pascals-triangle-Puzzle/00DESCRIPTION

@@ -7,6 +7,7 @@ This puzzle involves a [http://xunor.free.fr/en/riddles/auto/pyramidnb.php Pasca
     [ X][11][ Y][ 4][ Z]
 </pre>
 Each brick of the pyramid is the sum of the two bricks situated below it.<br>
-Of the three missing numbers at the base of the pyramid, the middle one is the sum of the other two (that is, Y = X + Z).
+Of the three missing numbers at the base of the pyramid,
+the middle one is the sum of the other two (that is, Y = X + Z).
 
 Write a program to find a solution to this puzzle.

Task/Pascals-triangle-Puzzle/Curry/pascals-triangle-puzzle.curry

@@ -0,0 +1,25 @@
+import CLPFD
+import Constraint (allC, andC)
+import Findall (findall)
+import List (init, last)
+
+
+solve :: [[Int]] -> Success
+solve body@([n]:rest) =
+    domain (concat body) 1 n
+  & andC (zipWith atop body rest)
+  & labeling [] (concat body)
+  where
+    xs `atop` ys = andC $ zipWith3 tri xs (init ys) (tail ys)
+
+tri :: Int -> Int -> Int -> Success
+tri x y z = x =# y +# z
+
+test (x,y,z) | tri y x z =
+    [ [151]
+    , [ _,  _]
+    , [40,  _, _]
+    , [ _,  _, _, _]
+    , [ x, 11, y, 4, z]
+    ]
+main = findall $ solve . test

Task/Pascals-triangle-Puzzle/D/pascals-triangle-puzzle.d

@@ -1,45 +1,38 @@
 import std.stdio, std.algorithm;
 
-void iterate(bool doPrint=true)(double[] v, double[] diff)
-/*pure nothrow*/ {
+void iterate(bool doPrint=true)(double[] v, double[] diff) @safe {
     static ref T E(T)(T[] x, in size_t row, in size_t col)
-    pure nothrow {
+    pure nothrow @safe @nogc {
         return x[row * (row + 1) / 2 + col];
     }
 
     double tot = 0.0;
     do {
-        // enforce boundary conditions
+        // Enforce boundary conditions.
         E(v, 0, 0) = 151;
         E(v, 2, 0) = 40;
         E(v, 4, 1) = 11;
         E(v, 4, 3) = 4;
 
-        // calculate difference from equilibrium
+        // Calculate difference from equilibrium.
         foreach (immutable i; 1 .. 5) {
             foreach (immutable j; 0 .. i + 1) {
                 E(diff, i, j) = 0;
                 if (j < i)
-                    E(diff, i, j) += E(v, i - 1, j) -
-                                     E(v, i, j + 1) -
-                                     E(v, i, j);
+                    E(diff, i, j) += E(v, i - 1, j) - E(v, i, j + 1) - E(v, i, j);
                 if (j)
-                    E(diff, i, j) += E(v, i - 1, j - 1) -
-                                     E(v, i, j - 1) -
-                                     E(v, i, j);
+                    E(diff, i, j) += E(v, i - 1, j - 1) - E(v, i, j - 1) - E(v, i, j);
             }
         }
 
         foreach (immutable i; 1 .. 4)
             foreach (immutable j; 0 .. i)
-                E(diff, i, j) += E(v, i + 1, j) +
-                                 E(v, i + 1, j + 1) -
-                                 E(v, i, j);
+                E(diff, i, j) += E(v, i + 1, j) + E(v, i + 1, j + 1) - E(v, i, j);
 
         E(diff, 4, 2) += E(v, 4, 0) + E(v, 4, 4) - E(v, 4, 2);
 
-        // do feedback, check if we are close enough
-        // 4: scale down the feedback to avoid oscillations
+        // Do feedback, check if we are close enough.
+        // 4: scale down the feedback to avoid oscillations.
         v[] += diff[] / 4;
         tot = diff.map!q{ a ^^ 2 }.sum;
 
@@ -53,18 +46,16 @@ void iterate(bool doPrint=true)(double[] v, double[] diff)
 }
 
 void main() {
-    static void show(in double[] x) nothrow {
+    static void show(in double[] x) nothrow @nogc {
         int idx;
         foreach (immutable i; 0 .. 5)
             foreach (immutable j; 0 .. i+1) {
-                printf("%4d%c", cast(int)(0.5 + x[idx]),
-                       j < i ? ' ' : '\n');
+                printf("%4d%c", cast(int)(0.5 + x[idx]), j < i ? ' ' : '\n');
                 idx++;
             }
     }
 
-    double[15] v = 0.0;
-    double[15] diff = 0.0;
+    double[15] v = 0.0, diff = 0.0;
     iterate(v, diff);
     show(v);
 }

Task/Pascals-triangle-Puzzle/Ruby/pascals-triangle-puzzle.rb

@@ -7,7 +7,8 @@ pyramid = [
       [nil,nil,nil,nil],
     ["x", 11,"y", 4,"z"]
 ]
-p pyramid
+pyramid.each{|row| p row}
+
 equations = [[1,-1,1,0]]   # y = x + z
 
 def parse_equation(str)
@@ -16,10 +17,10 @@ def parse_equation(str)
   eqn[3] = rhs.to_i
   for term in lhs.split("+")
     case term
-    when "x": eqn[0] += 1
-    when "y": eqn[1] += 1
-    when "z": eqn[2] += 1
-    else      eqn[3] -= term.to_i
+    when "x" then eqn[0] += 1
+    when "y" then eqn[1] += 1
+    when "z" then eqn[2] += 1
+    else          eqn[3] -= term.to_i
     end
   end
   eqn
@@ -28,7 +29,7 @@ end
 -2.downto(-5) do |row|
   pyramid[row].each_index do |col|
     val = pyramid[row][col]
-    sum = "%s+%s" % [pyramid[row+1][col].to_s, pyramid[row+1][col+1].to_s]
+    sum = "%s+%s" % [pyramid[row+1][col], pyramid[row+1][col+1]]
     if val.nil?
       pyramid[row][col] = sum
     else
@@ -42,12 +43,13 @@ reduced = convert_to(reduced_row_echelon_form(equations), :to_i)
 for eqn in reduced
   if eqn[0] + eqn[1] + eqn[2] != 1
     fail "no unique solution! #{equations.inspect} ==> #{reduced.inspect}"
-  elsif eqn[0] == 1: x = eqn[3]
-  elsif eqn[1] == 1: y = eqn[3]
-  elsif eqn[2] == 1: z = eqn[3]
+  elsif eqn[0] == 1 then x = eqn[3]
+  elsif eqn[1] == 1 then y = eqn[3]
+  elsif eqn[2] == 1 then z = eqn[3]
   end
 end
 
+puts
 puts "x == #{x}"
 puts "y == #{y}"
 puts "z == #{z}"
@@ -56,4 +58,5 @@ answer = []
 for row in pyramid
   answer << row.collect {|cell| eval cell.to_s}
 end
-p answer
+puts
+answer.each{|row| p row}