Add preview of 1984-2019

The 2020 winners are hinted it, but the source code is not there,
nor are the hints. Someone could look into the HTML and
see what was commented out, but that is not a problem because
WHO won is known and the name of their AWARD is known already.
Moreover a number of the winners have talked about what their
entirs do, so the commented out spoiler text is not a bit problem.

One will find HTML commented out with:

    <!-- XXX remove when 2020 is ready
    2020 links here
    XXX remove when 2020 is ready -->

The "XXX remove when 2020 is ready" string is found at the
start and end comment blocks.  Again, soneone could read
what was commented out, but that is OK since most of that
information is known or can be known.  The following files
have such HTML comment blocks in them.

    ./winners.html
    ./years-spoiler.html
    ./years.html

In addition, ./all/summary.txt was modified to not
include the 2020 entries for the momemt.

The 2020 winner code AND write-ups are NOT part of this set.

p.s. The all/all.tar.bz2 does not contain the 2020 source code (yet?).

2012/endoh2/find-font-table.rb

@@ -0,0 +1,95 @@
+# This script finds table for 3x5 decimal fonts.
+
+fonts = <<END
+###  .#.  ###  ###  #.#  ###  ###  ###  ###  ###
+#.#  ##.  ..#  ..#  #.#  #..  #..  #.#  #.#  #.#
+#.#  .#.  ###  ###  ###  ###  ###  ..#  ###  ###
+#.#  .#.  #..  ..#  ..#  ..#  #.#  .#.  #.#  ..#
+###  ###  ###  ###  ..#  ###  ###  .#.  ###  ###
+END
+fonts = fonts.lines.map {|l| l.scan(/\S{3}/) }.transpose
+fonts = fonts.join.scan(/.{15}/).map {|l| l.tr(".#", "01").reverse.to_i(2) }
+
+
+def printable?(x)
+  35 <= x && x != 92 && x <= 126
+end
+
+# Given integers d, e, and f, finds (x, y) s.t. x*d - y*e = f, where x and y
+# are both printable.
+def solve_identity(d, e, f)
+  # First we solve Bezout's identity by using the extended gcd algorithm.
+  # See the wikipedia entry in detail.
+  g = d.gcd(e)
+  return if f % g != 0  # No solution
+  d, e, f = d / g, e / g, f / g
+  a, b = d, -e
+  x0, x1 = 1, 0
+  y0, y1 = 0, 1
+  while b != 0
+    q = a / b
+    a, b = b, a % b
+    x0, x1 = x1, x0 - x1 * q
+    y0, y1 = y1, y0 - y1 * q
+  end
+
+  # Now we have solution candidates.
+  #   x = -f * x0 - e * t
+  #   y = -f * y0 - d * t
+  # where t is a free parameter.
+  # Then, we consider another constraint: x and y is both printable.
+
+  # We must satisfy: 35 <= x <= 126
+  # ->  ceil((35+f*x0) / e)            <= -t <= floor((126+f*x0) / e)
+  # ->  ceil((35+f*x0) / e) * d - f*y0 <=  y <= floor((126+f*x0) / e) * d - f*y0
+  ymin = ( 35+f*x0+e+1) / e * d - f*y0
+  ymax = (126+f*x0    ) / e * d - f*y0
+
+  return if ymax < ymin || ymax < 33 || ymin > 126  # No solution.
+
+  ((ymax + f*y0) / -d).upto((ymin + f*y0) / -d) do |t|
+    x = -f * x0 - e * t
+    y = -f * y0 - d * t
+    return [x, y] if printable?(x) && printable?(y) # Solution found.
+  end
+
+  return # No solution.
+end
+
+# Given integers f0, f1, ..., f9, by brute force, finds (u, v, d) s.t. there is
+# some xi and yi:
+#   x0 * u - y0 * v = f0 + d
+#   x1 * u - y1 * v = f1 + d
+#   ...
+#   x9 * u - y9 * v = f9 + d
+# where both xi and yi are printable.
+def find_params(fonts)
+  1.upto(9) do |d|
+    2.times do
+      1.upto(999) do |u|
+        1.upto(999) do |v|
+          if fonts.all? {|f| solve_identity(u, v, f + d) }
+            table = fonts.map do |f|
+              solve_identity(u, v, f + d).reverse
+            end.transpose.flatten.map {|c| c.chr }.join
+            return u, v, d, table
+          end
+        end
+      end
+      d = -d
+    end
+  end
+  puts "Not found."
+  exit
+end
+
+
+puts "Searching..."
+
+u, v, d, s = find_params(fonts)
+
+puts "Found."
+puts
+puts "Let s = %p." % s
+puts "You can get the font data of dicimal character 'n' (0--9):"
+puts "  %d * s[n+10] - %d * s[n] - %d" % [u, v, d]