Converting any recursion to stack

I’ve seen a couple of blogs which talk about this, but most of them have very simple examples. I will go with non trivial ones and my solutions will mimic the function call stack.

Lets start with a simple example, the Tower of Hanoi problem.

Tower of Hanoi

The Tower of Hanoi (also called the Tower of Brahma or Lucas’ Tower) is a mathematical game or puzzle. It consists of three rods, and a number of disks of different sizes which can slide onto any rod. The puzzle starts with the disks in a neat stack in ascending order of size on one rod, the smallest at the top, thus making a conical shape.

The objective of the puzzle is to move the entire stack to another rod, obeying the following simple rules: Only one disk can be moved at a time.

Each move consists of taking the upper disk from one of the stacks and placing it on top of another stack i.e. a disk can only be moved if it is the uppermost disk on a stack.

No disk may be placed on top of a smaller disk. Tower of Hanoi wikipedia

The code below implements a recursive solution. The tricky part in converting this to stack is there are 2 recursive calls. If there is only one it would be straight forward using a simple while loop.

def hanoi ( n , fr , to , buf ): if ( n == 0 ): return #print "deb", 'l', n, fr, to, buf hanoi ( n - 1 , fr , buf , to ) #print "deb", 'r', n, fr, to, buf print "{} {}->{}" . format ( n , fr , to ) hanoi ( n - 1 , buf , to , fr )

Now lets think of how the compiler deals with stack. Before the first call to hanoi, it saves the state of the function’s execution, in the call stack’s present entry, pushes a new entry to the stack for recursive call and starts executing the operations in recursive call. The following code implements that.

#Details explained after the code block ends def hanoi_iter ( n , fr , to , buf ): stack = list () #python's list can be used as stack stack . append (( 'full' , n , fr , to , buf )) while len ( stack ) != 0 : op , n , fr , to , buf = stack . pop () if n == 0 : continue #print "deb", op, n, fr, to, buf if op == 'step1' : print "{} {}->{}" . format ( n , fr , to ) stack . append (( 'full' , n - 1 , buf , to , fr )) else : #saving the state stack . append (( 'step1' , n , fr , to , buf )) stack . append (( 'full' , n - 1 , fr , buf , to ))

I’m differentiating between saved state( step1 ) and full function call( full ) by the first value (lets call it operation) in stack entries. Until the stack becomes empty we pop an entry from stack, handle the base cases, then work based on the operation.

For the saved state, print the move and do a full call on a smaller problem. For the full case, save the state and mark the state as 'step1' , and do a full call on the sub problem. That does the job.

This formulation works as long as the functions have no return value. The N Choose K problem will tackle that.

N Choose K

Given a list of elements, return all possible ways of picking K elements as a list of lists. There are 6 ways picking 2 elements from [0,1,2,3] [0,1] [0,2] [0,3] [1,2] [1,3] [2,3]

One recursive solution for this is to

Pick the list’s first element and pick k-1 elements from sublist from 1 to end

Do not pick the first element of list and pick k elements from sublist from 1 to end

Return the union of both

The code below implements that

# Returns all ways of picking 'k' elements in the 'inp' list from index 'start' # (ie k elements in inp[start:]). For all the solutions returned, # elements in to_add will be prepended def selectK ( inp , k , start = 0 , to_add = []): if ( len ( inp ) - start ) < k or k < 0 : #invalid return [] if k == 0 : #one way to pick no elements; pick nothing; but add to_add return [ list ( to_add )] if ( len ( inp ) - start ) == k : #only k present in the array, pick them all; add to_add ret = list ( to_add ) ret . extend ( inp [ start :]) return [ ret ] #Base cases done #pick k elements in inp[start+1:] and not include inp[start] without_pres = selectK ( inp , k , start + 1 , to_add ) #step1(of stack solution) starts here #pick k-1 elements in inp[start+1:] and include inp[start] to_add . append ( inp [ start ]) with_pres = selectK ( inp , k - 1 , start + 1 , to_add ) #step2(of stack solution) starts here to_add . pop () ret = with_pres ret . extend ( without_pres ) return ret

Here we have 2 recursive calls and we need results of both of them to compute the final result. (In the previous problem we did not need the return value at all) So we split the function to 3 (as opposed to 2 for Hanoi). I marked them as 'full' , 'step1' , 'step2'

The main problem here is passing the return value. Lets go back to how compiler does call stacks. In the call stack once an entry is popped then its output is determined and passed to the caller’s stack. So we need exactly one variable which can pass the return variable between caller and callee. The code below does exactly that.

#Details explained after the code block ends def selectK_rec ( inp , k ): stack = list () # 5 elements in the tuple # (operation, k, start_of_list, to_add, extra_info) stack . append (( 'full' , k , 0 , [], None )) ret = None while len ( stack ) != 0 : opn , k , start , to_add , extra = stack . pop () if opn == 'full' : #base cases; replace returns in recursive version #with assignments to tmp if ( len ( inp ) - start ) < k or k < 0 : #invalid ret = [] continue if k == 0 : #one way to pick no elements; #pick nothing; but add to_add ret = [ list ( to_add )] continue if ( len ( inp ) - start ) == k : #only k present in the array, #pick them all; add to_add ret = list ( to_add ) ret . extend ( inp [ start :]) ret = [ ret ] continue #base cases Done #save the state stack . append (( 'step1' , k , start , to_add , None )) stack . append (( 'full' , k , start + 1 , to_add , None )) elif opn == 'step1' : to_add . append ( inp [ start ]) stack . append (( 'step2' , k , start , to_add , ret )) stack . append (( 'full' , k - 1 , start + 1 , to_add , None )) else : to_add . pop () with_pres = ret without_pres = extra with_pres . extend ( without_pres ) ret = with_pres return ret

I initialized the ret var to None. Wherever I return a value or use a return value in recursive code, I assign/use ret in the stack version. Apart from that the logic in the loop is exactly same as the function body of recursive solution. When the stack becomes empty ret contains the solution.

With this approach any recursive code can be converted to stack based iteration. The full code with examples is on gist.