最佳化 Str Rep

function log2(n)
    local _n = 2
    local x = 1
    if (_n < n) then
        repeat
            x = x + 1
            _n = _n + _n
        until (_n >= n)
    elseif (_n > n) then
        if (n == 1) then
            return 0
        else
            return nil
        end
    end 
    if (_n > n) then
        return x-1
    else
        return x
    end 
end 
    
function get_bits(n)
    local bits = {}
    local rest = n
    repeat
        local major_bit = log2(rest)
        rest = rest - 2^major_bit
        bits[major_bit] = 1
        if (bits.count == nil) then
            bits.count = major_bit
        end
    until (rest == 0)
    return bits
end



function fast_strrep(str, times)
    local bits = get_bits(times)
    local strs = {[0] = str}

    local count = bits.count

    for i = 1, count do
        strs[i] = strs[i-1] .. strs[i-1]
    end

    local result = ''
    for i = 0, count do
        if (bits[i]) then
            result = result .. strs[i]
        end
    end

    return result

end

for numreps = 1024, 30*1024*1024, 1024*64 do

    a = nil
    b = nil
    collectgarbage()

    start = clock()
    a = fast_strrep("a", numreps)
    print("L:"..numreps.." "..(clock() - start))
    start = clock()
    b = strrep("a", numreps)
    print("C:"..numreps.." "..(clock() - start))

    if (a~=b) then
        print("the algorithm is wrong!")
    else
        print("ok")
    end

    flush(_STDOUT)

end

        

  -- Suppose that x = b[n]*2^n + b[n-1]*2^(n-1) + ... + b[0]*2^0
  --   (where every b[i] is either 0 or 1)
  -- This is exactly equivalent to:
  --    b[0] + 2 * (b[1] + 2 * (b[2] + (... + b[n])))
  -- So we've effectively eliminated all the multiplications, replacing them with doubling.

  -- Now, x * y (for any y) can be calculated by distributing multiplication over the
  -- above, which effectively replaces every b[i] with b[i] * y. However, every b[i]
  -- is either 0 or 1, so the product is either 0 or y.

  -- Now, if k is an integer and str1 and str2 are strings, and we write:
  --   str1 + str2       for the concatenation of str1 and str2
  --   k * str1          for "k copies of str1 concatenated"
  -- we can see that we have + and * are "just like" integer arithmetic in the sense that
  -- + and * are commutative and associative, and * distributes over +. So the equivalence
  -- continues to work, except that every term must be either "" (for 0) or y (the string).

  -- All that is left is to compute the expression from the inside out: each step is
  -- either 2 * r or y + 2 * r, where r is the cumulated value and y is the original string.
  -- In string terms, we can write these as result .. result (2 * r) and
  -- result .. result .. str (2 * r + y)

  -- We could use the same idea to compute integer exponents in the minimum number of
  -- multiplications, using * and ^ instead of + and * (which is where this algorithm
  -- comes from.)

  -- This makes use of the fact that Lua optimises a .. b .. c into a single concatenation.
  -- With a bit more work, we could use any base we wanted to, not just base 2. But it would
  -- require more options in the if statement.

function another_strrep(str, times)
  local result = ""
  local high_bit = 1
  while high_bit < times do high_bit = high_bit * 2 end

  -- at this point, high_bit is the largest (integral) power of 2 smaller than times
  -- (unless times < 1 in which case high_bit is 1)
  -- The computation of highbit could be:
  --   local high_bit = 2 ^ floor(log(times) / log(2))
  -- which is probably faster but requires the math library

  -- we are now going to work through times, bit by bit, making use of the above formula:

  while high_bit >= 1 do
    if high_bit <= times then           -- the bit is 1 if times is >= high_bit
      times = times - high_bit          -- we "turn it off" for the next iteration
      result = result .. result .. str  -- and the next step is 2 * r + y
    else                                -- the bit is 0
      result = result .. result         -- so the next step is 2 * r
    end
    high_bit = high_bit / 2             -- Now go for the next bit
  end
  return result
end

luiz/rici = 1.41  (rici is  1.41  times faster than luiz)
c/luiz    = 2.98  (luiz is  2.98  times faster than c)
c/rici    = 4.19  (rici is  4.19  times faster than c)
        

do
  local concats = {
    ["0"] = function(a, b) return a .. a .. a .. a .. a .. a .. a .. a .. a .. a end,
    ["1"] = function(a, b) return a .. a .. a .. a .. a .. a .. a .. a .. a .. a
                                    .. b end,
    ["2"] = function(a, b) return a .. a .. a .. a .. a .. a .. a .. a .. a .. a
                                    .. b .. b end,
    ["3"] = function(a, b) return a .. a .. a .. a .. a .. a .. a .. a .. a .. a
                                    .. b .. b .. b end,
    ["4"] = function(a, b) return a .. a .. a .. a .. a .. a .. a .. a .. a .. a
                                    .. b .. b .. b .. b end,
    ["5"] = function(a, b) return a .. a .. a .. a .. a .. a .. a .. a .. a .. a
                                    .. b .. b .. b .. b .. b end,
    ["6"] = function(a, b) return a .. a .. a .. a .. a .. a .. a .. a .. a .. a
                                    .. b .. b .. b .. b .. b .. b end,
    ["7"] = function(a, b) return a .. a .. a .. a .. a .. a .. a .. a .. a .. a
                                    .. b .. b .. b .. b .. b .. b .. b end,
    ["8"] = function(a, b) return a .. a .. a .. a .. a .. a .. a .. a .. a .. a
                                    .. b .. b .. b .. b .. b .. b .. b .. b end,
    ["9"] = function(a, b) return a .. a .. a .. a .. a .. a .. a .. a .. a .. a
                                    .. b .. b .. b .. b .. b .. b .. b .. b .. b end,
  }

  function decimal_strrep(str, times)
    local state = {r = "" }
    local concats = %concats
    times = tostring(times)
    if strfind(times, "^[0-9]+$") then
      gsub(times, "(.)",
           function(digit)
             %state.r = %concats[digit](%state.r, %str)
           end)
    end
    return state.r
  end
end