How to recursively increment a binary number (in list form) without converting to integer?

Ah, ha! Okay, you have some idea of what you're doing. The basic outline is

Base case: how do I know when I'm done?

You're done when you run out of digits. number should be a list of individual digits; check its length to figure out when not to recur.

Recursion case: what next?

The general concept of recursion is "do something simple, reduce the problem by a small amount, and recur with the smaller problem." Your job in this part is to do the addition for one digit. If you need to keep going (is there a carry from that digit?), then recur. Otherwise, you have all the info you need to finish.

Specific application

Your recursion step will involve calling increment_helper with one digit less: not number - 1, but number[:-1].

After you return from each recursion, you'lll then want to append the digit you just finished. For instance, if you're incrementing 1101, your first call will see that the right-hand one, incremented, has a carry. The new digit is 0, and you have to recur. Hold onto the 0 for a moment, call yourself with 110, and get the result of that call. Append your saved 0 to that, and return to your main program.

Does that get you moving?

This one is "tricky" because you have two things going at each step:

1. what is the current number I'm looking at? (binary question 0/1)
2. should it be incremented? (do we have carry?) (binary question yes/no)

This leads to 4 cases.

There is the extra "case" of did we even get a list but it isn't that interesting.

So I would describe it as follows:

if not carry:
    # technically meaningless so just return the entire list immediately
    return list

# we have "carry", things will change
if not list:
    # assumes [] == [0]
    return [1]

if list[0]:
    # step on `1` (make it 0 and carry)
    return [0] + increment(rest_of_list)

# step on `0` (make it 1 and no more carry)
return [1] + rest_of_list

I strongly advise to change lists to tuples and work with these.

Also note that the recursion is on the reversed list, so apply as follows:

def increment_helper(lst, carry):
    if not carry:
        return lst

    if not lst:
        return [1]

    if lst[0]:
        return [0] + increment_helper(lst[1:], True)

    return [1] + lst[1:]


def increment(lst):
    # long but really just reverse input and output
    return increment_helper(lst[::-1], True)[::-1]

I used some shortcuts by returning lists immediately (short-circuiting), but you can make it more "pure" by carrying on the recursion even without carry.

Another recursive approach.

• Is the current input an empty list?

  • If yes return [1]
  • If no, continue

• Is the sum (value) of the last element in the list and 1 greater than 1?

  • If so recursively call your function on the list without the last element (number_list[:-1]) and append [0] to the result.
  • If no, set the last element of the list to the sum.

• Return the number_list

Code:

def increment_helper(number_list):
    if not number_list:
        return [1]

    value = number_list[-1] + 1

    if value > 1:
        number_list = increment_helper(number_list[:-1]) + [0]
    else:
        number_list[-1] = value

    return number_list

Example output:

numbers = [[1, 0, 1], [1,1,1], [1,0,0,1]]
for n in numbers:
    print("%r ---> %r" % (n, increment_helper(n)))
#[1, 0, 1] ---> [1, 1, 0]
#[1, 1, 1] ---> [1, 0, 0, 0]
#[1, 0, 0, 1] ---> [1, 0, 1, 0]

Try this:

def add1(listNum):

    if listNum.count(0):
        oneArr = [[0] * (len(listNum) - 1)] + [1]
        sumArr = []
        for i in range(len(listNum)):
            sumArr.append(sum(listNum[i], oneArr[i]))
        newArr = []
        for j in range(len(sumArr) - 1):
            if sumArr[len(sumArr) - 1 - j] < 2:
                newArr.insert(0, sumArr[len(sumArr) - 1 - j])
            else:
                newArr.insert(0, 1)
                sumArr[len(sumArr) - 1 - j] += 1
        return sumArr

    else:
        return [1] + [[0] * len(listNum)]

There aren't many reasons for using recursion for a program as simple as this, which is why I have chosen to provide a non-recursive solution.

In case it interests you, I've calculated the time complexity of this function and it is O(n).

It may be best to use two functions: one to check if a simple increment of the last position would suffice and another to perform the recursion should the previous attempt fail:

vals = [[1, 0, 1], [1,1,1], [1,0,0,1]]
def update_full(d):
   if all(i in [1, 0] for i in d):
      return d
   start = [i for i, a in enumerate(d) if a > 1]
   return update_full([1]+[0 if i > 1 else i for i in d] if not start[0] else [a+1 if i == start[0] -1 else 0 if i == start[0] else a for i, a in enumerate(d)])

def increment(d):
    if not d[-1]:
       return d[:-1]+[1]
    return update_full(d[:-1]+[2])

print(list(map(increment, vals)))

Output:

[[1, 1, 0], [1, 0, 0, 0], [1, 0, 1, 0]]

You can treat the (binary) digits recursively by traversing the number from tail to head (just like addition works).

Before performing the recursion you have to check for two special cases:

• No increment at the current digit is to be performed. Then just return the unmodified digits.
• A single digit remains (you're at the head of the number). Then you possibly need to append the overflow to the head.

For the remaining cases you can increment the current digit and treat all digits before the current one in a recursive manner.

def bin_incr(n, incr=True):
    if not incr:
        return n
    if len(n) == 1:
        return [1, 0] if n[0] == 1 else [1]
    return (
        # `n[-1] == 1` denotes an overflow to the next digit.
        bin_incr(n[:-1], n[-1] == 1)
        + [(n[-1] + 1) % 2]
    )

I understand you don't want to use decimal addition, but if you must use big endian bit order, converting back to decimal first is probably the most practical. Otherwise, you end up with unnecessary reverse calls or awkward negative array indices

def binary (dec): # big endian bit order
  if dec < 2:
    return [ dec ]
  else:
    return binary (dec >> 1) + [ dec & 1 ]

def decimal (bin, acc = 0):
  if not bin:
    return acc
  else:
    return decimal (bin[1:], (acc << 1) + bin[0])

def increment (bin):
  # sneaky cheat
  return binary (decimal (bin) + 1)

for x in range (10):
  print (x, binary (x), increment (binary (x)))

# 0 [0] [1]
# 1 [1] [1, 0]
# 2 [1, 0] [1, 1]
# 3 [1, 1] [1, 0, 0]
# 4 [1, 0, 0] [1, 0, 1]
# 5 [1, 0, 1] [1, 1, 0]
# 6 [1, 1, 0] [1, 1, 1]
# 7 [1, 1, 1] [1, 0, 0, 0]
# 8 [1, 0, 0, 0] [1, 0, 0, 1]
# 9 [1, 0, 0, 1] [1, 0, 1, 0]

If however you can represent your binary numbers in little endian bit order, things change. Instead of converting back to decimal, increment can be defined directly as a beautiful recursive function

def binary (dec): # little endian bit order
  if dec < 2:
    return [ dec ]
  else:
    return [ dec & 1 ] + binary (dec >> 1) 

def increment (bin):
  if not bin:
    return [1]
  elif bin[0] == 0:
    return [1] + bin[1:]
  else:
    return [0] + increment(bin[1:])

for x in range (10):
  print (x, binary (x), increment (binary (x)))

# 0 [0] [1]
# 1 [1] [0, 1]
# 2 [0, 1] [1, 1]
# 3 [1, 1] [0, 0, 1]
# 4 [0, 0, 1] [1, 0, 1]
# 5 [1, 0, 1] [0, 1, 1]
# 6 [0, 1, 1] [1, 1, 1]
# 7 [1, 1, 1] [0, 0, 0, 1]
# 8 [0, 0, 0, 1] [1, 0, 0, 1]
# 9 [1, 0, 0, 1] [0, 1, 0, 1]

Aside: converting the little endian representation back to decimal is a little different. I provide this to show that use cases for recursion exist everywhere

def decimal (bin, power = 0):
  if not bin:
    return 0
  else:
    return (bin[0] << power) + decimal (bin[1:], power + 1)

This part of the answer gives you cake and allows you to eat it too. You get big endian bit order and a recursive increment that steps through the bits in left-to-right order – You should use either implementation above for a number of reasons, but this aims to show you that even though your problem is complex, it's still possible to think about it recursively. No reverse or arr[::-1] was misused in the making of this function.

def binary (dec): # big endian bit order
  if dec < 2:
    return [ dec ]
  else:
    return binary (dec >> 1) + [ dec & 1 ]

def increment (bin, cont = lambda b, carry: [1] + b if carry else b):
  if bin == [0]:
    return cont ([1], 0)
  elif bin == [1]:
    return cont ([0], 1)
  else:
    n, *rest = bin
    return increment (rest, lambda b, carry:
                              cont ([n ^ carry] + b, n & carry))

for x in range (10):
  print (x, binary (x), increment (binary (x)))

# 0 [0] [1]
# 1 [1] [1, 0]
# 2 [1, 0] [1, 1]
# 3 [1, 1] [1, 0, 0]
# 4 [1, 0, 0] [1, 0, 1]
# 5 [1, 0, 1] [1, 1, 0]
# 6 [1, 1, 0] [1, 1, 1]
# 7 [1, 1, 1] [1, 0, 0, 0]
# 8 [1, 0, 0, 0] [1, 0, 0, 1]
# 9 [1, 0, 0, 1] [1, 0, 1, 0]

We start by breaking the problem up into smaller parts; n is the first problem, and rest is the rest of the problems. But the key to thinking with continuations (like cont above) is to think big.

In this particular problem, n gets updated based on whether rest gets updated. So we immediately recur on rest and pass a continuation that will receive the result of the subproblem. Our continuation receives the answer to the subproblem b, and whether or not that subproblem results in a carry.

...
  else:
    n, *rest = bin
    return increment (rest, lambda b, carry:
                              cont ([n ^ carry] + b, n & carry))

The n ^ carry and n & carry expressions determine what the answer to this subproblem is and what the next carry will be. The following truth table shows that ^ and & encodes our answer and next_carry respectively. For example, if n is 0 and carry is 1, the carry can be consumed. The answer will be [1] + the answer to the subproblem and the next carry will be 0.

n   carry   (answer, next_carry)   n ^ carry   n & carry
0   0       ([0] + b, 0)           0           0
0   1       ([1] + b, 0)           1           0
1   0       ([1] + b, 0)           1           0
1   1       ([0] + b, 1)           0           1

The base cases are simple. If the subproblem is [0], the answer is [1] and no carry of 0. If the subproblem is [1], then the answer is [0]with a carry of 1

...
  if bin == [0]:
    return cont ([1], 0)
  elif bin == [1]:
    return cont ([0], 1)

Lastly, design the default continuation – if the answer to the problem b results in a carry, simply prepend [1] to the answer, otherwise just return the answer.

cont = lambda b, carry: [1] + b if carry else b

You are asking for the increment/successor/next function that will generate a sequence of sequences. Since other have given code, I will give a general method for developing such functions.

First, develop a multiple recursion (2 or more recursive calls) for calculating, say, all sequences of the type of length N. For binary sequences (bs) in big-endian order, from N 0s to N 1s, the base case bs(0) expression is [[]], the sequence that contains the one sequence with no binary digits. The double recursion for bs(n) in terms of bs(n-1) is ([0] concatenated to all members of bs(n-1) (in order)) plus ([1] contanenated to all members of bs(n-1)).

Next, focus on the transition between the subsequences returned by adjacent recursive calls. Here there is just one: 0, 1, ..., 1 to 1, 0, ..., 0. To increment across this boundary, we need to replace 0 followed by 0 or more 1s by 1 followed by the same number of 0s. We find such breaks by scanning from the right for the first 0, as others have shown.

It turns out the every increment crosses the boundary between adjacent bs(k) calls for some value of k, which is to say, at some level of the tree of calls resulting from double recursion.

So far, that I know of, the same idea works for designing the increment function for sequences of grey codes, sequences of conbinations (n things taken k at a time), and sequences of permutations.

Note 1: the 1->0 transitions can be done 1 at a time or all at once.

Note 2: the binary bit testing and flipping is the turing machine algorithm for count + 1. Stephen Wolfram, A New Kind of Science, presents, as I remember, 3 different implementations in the TM chapter.

Note 3 (added in edit): If one switches from prepending 0 first to prepending 1 first, or from prepending to appending, or both, one gets 3 other sequence of sequences, with 3 other increment functions.

You do not need recursion in this case. Let us start with the simplest implementation:

1. implement a full adder which will operate on bits.
2. implement a ripple adder using the full adder which will operate on 2 lists of bits.

The full adder implementation is straight forrward.

def full_adder(a,b,cin):
    sum_ = a^(b^cin)
    cout = (a&b) | (a|b)&cin
    return sum_, cout

Tests to make sure the full adder conforms to the specs:

>>> zero_one = (0,1)
>>> [full_adder(*x) for x in [(a,b,c) for a in zero_one for b in zero_one for c in zero_one]]
[(0, 0), (1, 0), (1, 0), (0, 1), (1, 0), (0, 1), (0, 1), (1, 1)]

Since the parameters of the ripple adder are lists, we need to ensure the list lengths match before the addition. This is done by padding the shorter list with leading zeros.

def ripple_adder(xs,ys):
    x, y = map(len, (xs, ys))
    alen = max(x, y)
    ax, by = map(lambda f: f if len(f) == alen else [0]*(alen-len(f)) + f, (xs, ys))
    cout = 0
    res = [0]*(alen)
    for i in range(alen-1, -1, -1):
        a, b, cin = ax[i], by[i], cout
        s, cout = full_adder(a, b, cin)
        res[i] = s
    if cout:
        res = [1] + res
    return res

Finally, we define bin_inc the binary increment function in terms of the ripple adder

def bin_inc(bin_lst):
    return ripple_adder(bin_lst, [1])

>>> bin_inc([1,1,1])
[1, 0, 0, 0]
>>> bin_inc([1,0,0,1])
[1, 0, 1, 0]

Now for a solution that is simpler but requires a little insight. Consider the following obervations for an input xs with length L and output ys

1. if xs is all ones, ys will have length L+1, the first element of ys will be 1 and the rest will be zeros.

2. if xs has a zero element, let p be the index of the last zero element in xs. Then ys will be the list consisting of the first p elements of xs followed by a 1 followed by zeros of length L - p. That is ys = xs[:p] + [1] + [0]*(L-p).

We can translate this to python code easily. We use python's list.index method to find the last occurence of zero by working on the reverse of the list and adjust the algorithm appropriately:

def bin_inc_2(xs):
    if all(xs):
        return [1] + [0]*len(xs)
    p = xs[::-1].index(0)
    return xs[:-p-1] + [1] + [0]*p

This works and is simpler than the ripple_adder based implementation. One minor drawback you might notice is when xs has a zero element, we traverse it to check if it is all ones, then traverse it again to find the first occurence of zero.

We can simplify the implementation and make it more pythonic atthe same time by using a try except block:

def bin_inc_3(bin_lst):
    try:
        p = bin_lst[::-1].index(0)
        return bin_lst[:-p-1] + [1] + [0]*p
    except ValueError: 
        return [1] + [0]*len(bin_lst)

This implementation is simple in terms of source text, and idiomatic python. Now we test it against the ripple_adder based adder to make sure it works well.

>>> z_n = (0,1)
>>> xs = [[a,b,c,d,e,f,g,h] for a in z_n for b in z_n for c in z_n for d in z_n
                            for e in z_n for f in z_n for g in z_n for h in z_n ]

>>> print(all(ripple_adder(x, [1]) == bin_inc_3(x) for x in xs))
True

Fantastic, it works as intended and correctly handles leading zeros as evidenced by the tests (which increments every number from 0 to 255).

There's only a need to recurse when there is a carry:

def f(n):
  # Base cases
  if not n:
    return [1]
  if n == [1]:
    return [1, 0]
  if n[-1] == 0:
    return n[:-1] + [1]
  if n[-2:] == [0, 1]:
    return n[:-2] + [1, 0]

  # Recurse
  return f(n[:-2]) + [0, 0]

numbers = [[1, 0, 1], [1,1,1], [1,0,0,1], [1, 0, 1, 1]]
for n in numbers:
  print n, f(n)