#! /usr/bin/lua

function popcount(x0) -- adapted from https://gist.github.com/davidm/2065267
   local x = x0
   local c = 0
   while x ~= 0 do
      x = x & (x - 1)
      c = c + 1
   end
   return c
end

function sortbyonebits(a)
   local b = a
   local c = {}
   for _, v in ipairs(b) do
      if c[v] == nil then
         c[v] = popcount(v)
      end
   end
   table.sort(b, function(a, b)
                 if c[a] == c[b] then
                    return a < b
                 else
                    return c[a] < c[b]
                 end
   end)
   return b
end

-- by Michael Anderson at
-- https://stackoverflow.com/questions/8722620/comparing-two-index-tables-by-index-value-in-lua
-- modified by Roger
function recursive_compare(t1,t2)
  -- Use usual comparison first.
  if t1==t2 then return true end
  -- We only support non-default behavior for tables
  if (type(t1)~="table") then return false end
  -- They better have the same metatables
  local mt1 = getmetatable(t1)
  local mt2 = getmetatable(t2)
  if( not recursive_compare(mt1,mt2) ) then return false end
  -- Build list of all keys
  local kk = {}
  for k1, _ in pairs(t1) do
     kk[k1] = true
  end
  for k2, _ in pairs(t2) do
     kk[k2] = true
  end
  -- Check each key that exists in at least one table
  for _, k in ipairs(kk) do
     if (not recursive_compare(t1[k], t2[k])) then
        return false
     end
  end
  return true
end

if recursive_compare(sortbyonebits({0, 1, 2, 3, 4, 5, 6, 7, 8}), {0, 1, 2, 4, 8, 3, 5, 6, 7}) then
  io.write("Pass")
else
  io.write("FAIL")
end
io.write(" ")

if recursive_compare(sortbyonebits({1024, 512, 256, 128, 64}), {64, 128, 256, 512, 1024}) then
  io.write("Pass")
else
  io.write("FAIL")
end
print("")