How to implement combinations tail-recursively? - lua

Question

I'm teaching myself Lua by reading Ierusalimschy's Programming in Lua (4th edition), and doing the exercises. Exercise 6.5 is
Write a function that takes an array and prints all combinations of the elements in the array.
After this succinct statement the book gives a hint that makes it clear that what one is expected to do is to write a function that prints all the C(n, m) combinations of m elements from an array of n elements.
I implemented the combinations function shown below:
function combinations (array, m)
local append = function (array, item)
local copy = {table.unpack(array)}
copy[#copy + 1] = item
return copy
end
local _combinations
_combinations = function (array, m, prefix)
local n = #array
if n < m then
return
elseif m == 0 then
print(table.unpack(prefix))
return
else
local deleted = {table.unpack(array, 2, #array)}
_combinations(deleted, m - 1, append(prefix, array[1]))
_combinations(deleted, m, prefix)
end
end
_combinations(array, m, {})
end
It works OK, but it is not tail-recursive.
Can someone show me a tail-recursive function that does the same thing as combinations above does?
(For what it's worth, I am using Lua 5.3.)
NB: I realize that the exercise does not require that the function be tail-recursive. This is a requirement I have added myself, out of curiosity.
EDIT: I simplified the function slightly, but removing a couple of nested functions that were not adding much.

Answer 1

There is a third option, one that doesn't have a snake eating it's tail. Although recursion with tail-calls don't lead to stack overflow, I avoid doing so out of personal preference. I use a while loop and a stack that holds the information for each iteration. Within the loop you pop the next task from the stack, do the work, then push next task onto the stack. I feel it looks cleaner and it's easier to visualize the nesting.
Here is how I would translate your code into the way I would write it:
function combinations(sequence, item)
local function append(array, item)
local copy = {table.unpack(array)}
copy[#copy + 1] = item
return copy
end
local stack = {}
local node = { sequence, item, {} }
while true do
local seq = node[ 1 ]
local itm = node[ 2 ]
local pre = node[ 3 ]
local n = #seq
if itm == 0 then
print(table.unpack(pre))
elseif n < itm then
-- do nothing
else
local reserve = {table.unpack(seq, 2, #seq)}
table.insert(stack, { reserve, itm, pre })
table.insert(stack, { reserve, itm-1, append(pre, seq[ 1 ]) })
end
if #stack > 0 then
node = stack[ #stack ] -- LIFO
stack[ #stack ] = nil
else
break
end
end
end
You can use this while-loop stack/node technique for just about any recursive method. Here is an example where it's applied to printing deeply nested tables: https://stackoverflow.com/a/42062321/5113346
My version, using your input example gives the same output:
1 2 3
1 2 4
1 2 5
1 3 4
1 3 5
1 4 5
2 3 4
2 3 5
2 4 5
3 4 5.
Forgive me if it doesn't work with other passed params because I didn't try to solve the answer to the exercise but rather just rewrite the code in your original post.

Answer 2

OK, I think I found one way to do this:
function combinations (array, m)
local dropfirst = function (array)
return {table.unpack(array, 2, #array)}
end
local append = function (array, item)
local copy = {table.unpack(array)}
copy[#copy + 1] = item
return copy
end
local _combinations
_combinations = function (sequence, m, prefix, queue)
local n = #sequence
local newqueue
if n >= m then
if m == 0 then
print(table.unpack(prefix))
else
local deleted = dropfirst(sequence)
if n > m then
newqueue = append(queue, {deleted, m, prefix})
else
newqueue = queue
end
return _combinations(deleted, m - 1,
append(prefix, sequence[1]),
newqueue)
end
end
if #queue > 0 then
newqueue = dropfirst(queue)
local newargs = append(queue[1], newqueue)
return _combinations(table.unpack(newargs))
end
end
_combinations(sequence, m, {}, {})
end
This version is, I think, tail-recursive. Unfortunately, it does not print out the results in as nice an order as did my original non-tail-recursive version (not to mention the added complexity of the code), but one can't have everything!
EDIT: Well, no, one can have everything! The version below is tail-recursive, and prints its results in the same order as does the original non-tail-recursive version:
function combinations (sequence, m, prefix, stack)
prefix, stack = prefix or {}, stack or {}
local n = #sequence
if n < m then return end
local newargs, newstack
if m == 0 then
print(table.unpack(prefix))
if #stack == 0 then return end
newstack = droplast(stack)
newargs = append(stack[#stack], newstack)
else
local deleted = dropfirst(sequence)
if n > m then
newstack = append(stack, {deleted, m, prefix})
else
newstack = stack
end
local newprefix = append(prefix, sequence[1])
newargs = {deleted, m - 1, newprefix, newstack}
end
return combinations(table.unpack(newargs)) -- tail call
end
It uses the following auxiliary functions:
function append (sequence, item)
local copy = {table.unpack(sequence)}
copy[#copy + 1] = item
return copy
end
function dropfirst (sequence)
return {table.unpack(sequence, 2, #sequence)}
end
function droplast (sequence)
return {table.unpack(sequence, 1, #sequence - 1)}
end
Example:
> combinations({1, 2, 3, 4, 5}, 3)
1 2 3
1 2 4
1 2 5
1 3 4
1 3 5
1 4 5
2 3 4
2 3 5
2 4 5
3 4 5
Ironically, this version achieves tail-recursion by implementing its own stack, so I am not sure it is ultimately any better than the non-tail-recursive version... Then again, I guess the function's stack actually lives in the heap (right?), because Lua's tables are passed around by reference (right?), so maybe this is an improvement, after all. (Please correct me if I'm wrong!)