# Integer math algorithms without integer datatypes or math-specific hardware

## March 31, 2021

Categories: Math Programming

I previously posted about simple, programmable, Turing-complete machines. That post discussed a programmable machine that optimized ease-of-programming and ease-of-implementation while still being able to run any program. With those goals in mind I put together a Python package L2 which emulates such a simple machine in software. For the L2 emulator, I used math operations and datatypes that were simply provided by Python to create the mathematical functionality. This is a bit of a cop out, because mathematical operations are a very important part of computing, and a machine that simply borrows them from Python, which uses the CPUs math operations, cannot really said to be simple. With that in mind, and in light of a recent post about implementing exp and other functions without using some standard library, I thought it would be neat to try to bootstrap mathematical operations from the cells, symbols, and primitive operations of L2.

This post will focus on integer math, while floating point math will be a topic for a future post. Because L2 is very simple, not as easy to prototype in, and less comprehensible to people not familiar with LISP, I’ll be writing Python code that uses only features easily portable to L2.

## Representation

Math, especially integer math, is all about counting, and anything that can be counted can be used to represent numbers. For instance, a list (made of Cell objects) could represent a positive integer with its length. To add two lists, one could be concatenated to the other. To subtract, the tail of the longer list would represent the difference. Multiplication and division could be bootstrapped with repeated addition and subtraction. Negative numbers could have a special first element to identify them.

This list representation is not particularly efficient, since the memory used is linear with the size of the integer. The Roman solution to this problem was to introduce several symbols of greater value, which meant that 1 (I), 5 (V), 10 (X), etc., could be represented by one numeral. This pattern continued up to 1000 (M), with some more complicated rules to compress bigger numbers. Manipulating Roman numerals can be quite difficult. Addition and subtraction are manageable with some rules (I plus III is IV, I more than IV is V, and I more than IV is VI), but this is complex. Multiplication (division) by repeated addition (subtraction) would work, taking into account the same (exhaustive) rules.

The “modern” solution to notation is to represent some small quantity of numbers (called the base, $b$) uniquely, and then use the position of those number symbols in a list to represent how many powers of the base are represented. So the quantity $A$ which has a number symbol in the $i$th position, $A_i$, would be the the count of $b^i$ values in the quantity, usually counted from the right. $$A = \{A_n,A_{n-1},…,A_{1},A_{0}\} = 1234$$ $$A = \sum^n_{i=0} A_i b^i = 4\times10^0 + 3\times10^1 + 2\times10^2 + 1\times10^3$$ This allows a number to be represented in approximately $\log_{10} N$ characters - much better than linear! The mathematical operations you learned in elementary school that operate on each position at a time can also be applied to these numbers, as you well know.

To perform those algorithms, one must only know the result for each pair of digits (for operations with two arguments, like addition, etc). This means an operation has roughly $b^2 = 10^2 = 100$ rules that need to be defined. A simple language like L2 would have no problem enumerating these rules, or even keeping track of ten possible symbols in lists. But there are simpler representations than $b=10$. First, note that $b=1$ is essentially the same as the list-length encoding above: each position adds a factor of $1$ to the quantity. The case of $b=2$, known as binary notation, has the same logarithmic growth as all values larger than $b=1$, but keeps the number of possible symbols to a minimum. The smallness of the number of symbols means that the rules for implementing operations are correspondingly fewer. This, in addition to the fact that the two symbols can correspond to “on” and “off” makes binary the most popular way to represent numbers in ordinary computers. Such systems still require a mechanism for converting human-readable base-10 (and often base-8 or base-16) such that the math algorithms implemented in base-2 are useful.

### Pre-bootstrapping

I’ll start with a Python function that converts Python integers into a representations that would work for L2: a list of symbols in position notation. This list will correspond to the $A_i$ values in the summation above, where the base is understood from the number of possible symbols. The value of the symbols will be represented by their position in the list. To be extra clear that no numbers are being used in this representation of numbers, I’ll use 'i' to correspond to one and 'o' to correspond to zero.

bin_rep = ['o','i']

def int_to_rep(i,rep=bin_rep):
'''Turn a python integer into a base rep integer representation'''
if not i:
return []
else:
res = [] if i > 0 else ['-']
if i < 0:
i = -i
while i > 0:
res.append(rep[i%len(rep)])
i = i // len(rep)
return res


This method takes a Python integer and performs Python integer division // and remainder % operations to arrive at the representation. The algorithm here is to calculate the lowest place value as the remainder of integer division by the base. The result of the division then becomes the value used to calculate the next lowest place value. $$A_i = N_{i-1}\mod b$$ $$N_i = \lfloor N_{i-1} / b \rfloor$$ where $N_{-1} = A$, the number to be encoded in this base. These mathematical operations will be re-implemented later; for now, this is just to generate some representations to work with.

int_to_rep(128)

['o', 'o', 'o', 'o', 'o', 'o', 'o', 'i']


Note that only the 7th index ($2^7 = 128$) has the one symbol. This method can also convert a Python integer to any base.

rep_int(128,rep=['o','i','j'])

['j', 'o', 'j', 'i', 'i']


Where a two symbol in the ones place, zero symbol in threes place, twp symbol in nines place, and one symbols in the twenty-sevens and eighty-ones place. If you’re following, that’s base three for $2\times1+0\times3+2\times9+1\times27+1\times81=128$. And since base-10 is included in “any base”:

int_to_rep(128,rep='0123456789')

['8', '2', '1']


Which clearly drives the point home that this representation has the least significant place first: annoying perhaps for a modern person, but convenient for calculation purposes.

I’ll make an array of small numbers in binary encoding for testing purposes as the algorithms for mathematical operations are developed.

numbers = [int_to_rep(i) for i in range(10)]

[[],
['i'],
['o', 'i'],
['i', 'i'],
['o', 'o', 'i'],
['i', 'o', 'i'],
['o', 'i', 'i'],
['i', 'i', 'i'],
['o', 'o', 'o', 'i'],
['i', 'o', 'o', 'i']]


Which clearly drives the point home that this representation has the least significant place first: annoying perhaps for a modern person, but convenient for calculation purposes.

Even though these numbers are least significant bit first, I’ll choose to represent negative numbers as a signed magnitude by appending a special minus symbol '-' to them.

int_to_rep(-128,rep='0123456789')

['-', '8', '2', '1']


With this representation, any negative or positive integer can be stored with an arbitrary number of bits as a list of said bits, with zero is represented as an empty list.

## Manipulation

With a representation of integers as lists of symbols, what’s left is to define algorithms that manipulate the representations like mathematical operators. This will be done in python, but using a subset of the language that is equivalent to L2 code. The datatypes provided by the L2 machine are unique Symbolss and that contain two values, either another Cell or a Symbol. Cells are used to form linked lists, which I’ll use Python lists to replicate. L2’s primitive Cell functions are implemented as follows:

def getl(list):
return list[0]

def getr(list):
return list[1:] if list else None

def cell(item,list):
return [item] + list # 1 of 2 plus signs in this code

def append(item,list): # not a primative
return list + [item] # 2 of 2 plus signs in this code

def reverse(list): # not a primative
if not list:
return []
return append(getl(list),reverse(getr(list)))


Just like L2, an empty list tests as False and a list with elements tests as True.

Methods can be constructed to test for negativity and negate an integer representation.

def is_neg_bin_rep(rep):
return rep and getl(rep) == '-'

def neg_bin_rep(rep):
if not rep:
return rep
elif is_neg_bin_rep(rep):
return getr(rep) #getr pops the negative sign
else:
return cell('-',rep)


To clean up any representations with leading zero symbols, or any other symbols, they can be removed from the right side.

def rep_strip(res,remove='o'):
if not res:
return []
keep = rep_strip(getr(res))
if keep or getl(res) != remove:
return cell(getl(res),keep)
else:
return []


The addition algorithms, like the corresponding algorithm to add base-10 numbers, will add numbers in columns of place value, carrying to the next place value if there is overflow. Summing two bits, $A$ and $B$, there can be a maximum result $R$ and carry $C$ of one.

$A$ $B$ $R$ $C$
0 0 $\rightarrow$ 0 0
0 1 $\rightarrow$ 1 0
1 0 $\rightarrow$ 1 0
1 1 $\rightarrow$ 0 1

This is easy to implement in code with cascading if statements:

def add_bin_bit_nocarry(a,b):
if a == 'o':
if a == b:
return 'o','o'
else:
return 'i','o'
else:
if a == b:
return 'o','i'
else:
return 'i','o'


In practice, adding the bits at a place value also requires including the carry from a less significant place value. This means adding three bits, $A$, $B$, and $C’$ is necessary. This can be accomplished by chaining three copies of the two-bit adder. Since the result of adding three bits is at most three, a result and carry bit is still all that is required as the result.

def add_bin_bit(a,b,carry):
carry,_ = add_bin_bit_nocarry(carry_a,carry_b) # only one operation could carry, so no carry here
return res,carry


Then some recursion can be used to iterate over and add the bits in two representations, with some logic to handle numbers with different bit lengths. The carry is passed to each place value, and initialized to zero for the ones place.

def add_bin_rep(a,b,carry='o'):
if a and b:
elif a: # a was longer
elif b: # b was longer
elif carry != 'o': # sum resulted in more bits than a or b
return cell(carry,[])
else:
return []


So far this only handles unsigned numbers.

Subtraction is fortunately very similar to addition, just with different single bit rules, and borrowing $B$ from higher place values instead of carrying.

$A$ $B$ $R$ $B$
0 0 $\rightarrow$ 0 0
0 1 $\rightarrow$ 1 1
1 0 $\rightarrow$ 1 0
1 1 $\rightarrow$ 0 0
def sub_bin_bit_noborrow(a,b):
if a == 'o':
if a == b:
return 'o','o'
else:
return 'i','i'
else:
if a == b:
return 'o','o'
else:
return 'i','o'

def sub_bin_bit(a,b,borrow):
res,borrow_a = sub_bin_bit_noborrow(a,b)
res,borrow_b = sub_bin_bit_noborrow(res,borrow)
return res,borrow


Unlike addition of positive numbers, which always results in positive numbers, subtraction can result in a negative number. The ring nature of two’s complement makes this a non-issue for hardware implementations with fixed-length representations, since the borrows from higher place value bits eventually truncate. For example subtracting one from an 8-bit zero looks like this in twos complement: $$\begin{array}{r} & 0 0 0 0 0 0 0 0 \\ -& 0 0 0 0 0 0 0 1 \\ \hline & 1 1 1 1 1 1 1 1 \end{array}$$ The series of one bits technically continues to the left in an infinite series as a 2-adic number, but because an 8-bit number is fixed to eight bits, only a finite number of bits is kept, and a borrow flag might be set to indicate the sign changed. Note that with all bits set, the value of this collection of bits as an unsigned number is $255$. Since the signed value of a twos complement number $n$ of length $d$ with the highest bit set is given by $n - 2^d = 255 - 2^8 = -1$, this truncation eight bits works fine. Without a fixed number of bits (infinite bits), this breaks down, though in some conceptual sense, the number that borrowed one from infinity would be one less than infinity.

To get around infinite lengths, I implement an intermediate subtraction function sub_bin_rep_ which will return a truncated 2-adic number with one more bit than either argument if the result is negative.

def sub_bin_rep_(a,b,borrow='o'):
if a and b:
res,borrow = sub_bin_bit(getl(a),getl(b),borrow)
return cell(res,sub_bin_rep_(getr(a),getr(b),borrow))
elif a:
res,borrow = sub_bin_bit_noborrow(getl(a),borrow)
return cell(res,sub_bin_rep_(getr(a),[],borrow))
elif b:
res,borrow = sub_bin_bit('o',getl(b),borrow)
return cell(res,sub_bin_rep_([],getr(b),borrow))
elif borrow != 'o':
return cell(borrow,[]) # this borrow would repeat forever
else:
return []


This result can be tested to see if it has more bits than the arguments, and if so, use the two’s complement formula above to find the magnitude of the 2-adic number.

def sign_correct_bin_rep(a,b,res):
'''Subtraction can return a truncated 2-adic or twos-complement number, convert to a negative'''
res = rep_strip(res)
def longer(a,b,res):
if res and ((not a) and (not b)):
return True
elif not res:
return False
else:
return longer(getr(a) if a else [], getr(b) if b else [], getr(res))
if longer(a,b,res):
# twos is the value 2**d where d is the number of bits in res
twos =  append('i',['o' for b in res]) # L2 can map functions
return neg_bin_rep(rep_strip(sub_bin_rep_(twos,res)))
else:
return res


The true subtraction function sub_bin_rep can then be implemented as:

def sub_bin_rep(a,b,borrow='o'):
return sign_correct(a,b,sub_bin_rep_(a,b,borrow=borrow))


So far these functions have only dealt with positive magnitudes as inputs, and only subtraction can return a negative magnitude. Because I’ve chosen to use signed magnitude with arbitrary bit depth instead of something like two’s complement, additional logic will be necessary to handle adding numbers with different signs. An advantage of the fixed-length representations using twos complement ring is that this distinction is unnecessary (addition is just addition, regardless of sign), but I wanted arbitrary-length numbers here, so the extra logic is required. To write functions that handle signed numbers, each operation should check the sign of the inputs, and perform the correct operation.

def add_bin_rep_signed(a,b):
if is_neg_bin_rep(a) and is_neg_bin_rep(b): #same sign
elif not is_neg_bin_rep(a) and not is_neg_bin_rep(b): #same sign
elif is_neg_bin_rep(a): #opposite sign
return sub_bin_rep(b,getr(a))
elif is_neg_bin_rep(b): #opposite sign
return sub_bin_rep(a,getr(b))

def sub_bin_rep_signed(a,b):
if is_neg_bin_rep(a) and is_neg_bin_rep(b): #same sign
return neg_bin_rep(sub_bin_rep(getr(a),getr(b)))
elif not is_neg_bin_rep(a) and not is_neg_bin_rep(b): #same sign
return sub_bin_rep(a,b)
elif is_neg_bin_rep(a): #opposite sign
elif is_neg_bin_rep(b): #opposite sign


Using the numbers generated earlier, these functions can be tested right away.

print('two',add_bin_rep_signed(numbers[1],numbers[1]))
print('two',sub_bin_rep_signed(numbers[3],numbers[1]))
print('negative one',sub_bin_rep(numbers[1],numbers[2]))

two ['o', 'i']
negative two ['-', 'o', 'i']
two ['o', 'i']
negative one ['-', 'i']


### Multiplication

Multiplication of position notation numbers is also fairly straightforward, and does not have to resort to repeated addition, which would scale linearly with the magnitude of one of the numbers. Instead, one argument $B = \{B_0,…,B_n\}$ is broken up by place value, and each place value is separately multiplies the other argument $A = \{A_0,…,A_m\}$. These intermediate values are shifted left and zero padded such that the ones place lines up with the position of the place value used in multiplication, and then added. With this done in base $b$, and a place value position $i$, this shifting is the same as a multiplication by a power of $b^i$. This can be seen by expanding $A$ and $B$ into their base $b$ representations. $$AB = \sum_{i=0}^n A B_i b^i = \sum_{i=0}^n \sum_{j=0}^m (A_j b^j) B_i b^i$$

In binary this operation is even easier because there are only two possible values to multiply by: one, which is the identity, and zero, which is zero. So, depending on whether the bit in one argument is one or zero, the other argument is either added or not, and the result is shifted to the correct position. This can easily be done with recursion, and a little logic to handle signs.

def mul_bin_rep(a,b):
if not a or not b:
return []
if getl(b) == 'i':
low = a
else:
low = []
high = cell('o',mul_bin_rep(a,getr(b)))

def mul_bin_rep_signed(a,b):
if is_neg_bin_rep(a) and is_neg_bin_rep(b): #same sign
return mul_bin_rep(getr(a),getr(b))
elif not is_neg_bin_rep(a) and not is_neg_bin_rep(b): #same sign
return mul_bin_rep(a,b)
elif is_neg_bin_rep(a): #opposite sign
return neg_bin_rep(mul_bin_rep(getr(a),b))
elif is_neg_bin_rep(b): #opposite sign
return neg_bin_rep(mul_bin_rep(a,getr(b)))


Testing with the numbers generated earlier demonstrates that this works.

print('six',mul_bin_rep_signed(numbers[3],numbers[2]))
print('negative eighty-one',mul_bin_rep_signed(numbers[9],numbers[9]))
print('negative eighty-one',mul_bin_rep_signed(numbers[9],neg_bin_rep(numbers[9])))

six ['o', 'i', 'i']
negative eighty-one ['i', 'o', 'o', 'o', 'i', 'o', 'i']
negative eighty-one ['-', 'i', 'o', 'o', 'o', 'i', 'o', 'i']


### Division

As with everything before, division algorithms are very similar to the long division technique taught in elementary school, with the exception that instead of computing a fractional part if the divisor is not an integer factor of the dividend, the integer remainder will be returned along with the integer result. This is a more efficient position notation algorithm than repeated subtraction, but is more complicated than the multiplication algorithm.

1. The divisor is subtracted from the highest place value in the dividend as many times as possible without going negative.
2. If the number of times the divisor can be subtracted without going negative gives the highest place value of the result.
3. The remaining value after subtraction is multiplied by the base and added to the next place value of the dividend.
4. The subtraction without going negative is performed again, giving the next-highest place value.
5. Repeat at step 3 until there are no more place values in the dividend.
6. The result has been calculated, along with some possibly nonzero remainder

This algorithm simplifies a bit in binary because the subtraction step will require either zero or one subtraction, so an attempted subtraction is either negative or not. Again, recursion can be used to implement this algorithm to iterate over place values.

def div_bin_rep_(a,b):
if not a:
return [],[]
div,rem = div_bin_rep(getr(a),b)
rem = cell(getl(a),rem)
cmp = sub_bin_rep_signed(rem,b)
if is_neg_bin_rep(cmp):
if div: #suppress zeros in high bits
div = cell('o',div)
else:
div = cell('i',div)
rem = cmp
return div,rem

def div_bin_rep(a,b):
div,rem = div_bin_rep_(a,b)
return div,rep_strip(rem) #to prevent accumulating zero high bits

def div_bin_rep_signed(a,b):
if is_neg_bin_rep(a) and is_neg_bin_rep(b): #same sign
return div_bin_rep(getr(a),getr(b))
elif not is_neg_bin_rep(a) and not is_neg_bin_rep(b): #same sign
return div_bin_rep(a,b)
elif getl(a) == '-': #opposite sign
return [neg_bin_rep(x) for x in div_bin_rep(getr(a),b)]
elif getl(b) == '-': #opposite sign
return [neg_bin_rep(x) for x in div_bin_rep(a,getr(b))]


And you can see that the division result and remainder are calculated correctly.

print('one, remainder zero',div_bin_rep(numbers[1],numbers[1]))
print('two, remainder one',div_bin_rep(numbers[5],numbers[2]))

one, remainder zero (['i'], [])
two, remainder one (['o', 'i'], ['i'])


## Converting Representations

With addition, subtraction, multiplication, and division, all the integer math one needs can be bootstrapped. To demonstrate this, I’ll implement some functions to convert representations in one base to another base, which will let L2 handle converting to and from human-readable base-10 numbers. But first, some utilities for representing list lengths, element positions, and list indexing.

b_zero = []
b_one = ['i']
b_two = ['o','i']

def length(list):
if not list:
return b_zero

def position(list,x):
if not list:
return b_zero
elif getl(list) == x:
return b_zero

def element(list,i):
if not list:
return None
elif not i:
return getl(list)
else:
return element(getr(list),sub_bin_rep(i,b_one))


length will return the total length of a list in a binary representation.

length(['0','1','2','3'])

['o', 'o', 'i']


position will return a binary representation of the index of an item in a list.

position(['0','1','2','3'],'2')

['o', 'i']


While take will return the item in a list at binary representation of the index.

arr = ['0','1','2','3']
take(arr,position(arr,'2'))

'2'


With these parts, a rep_to_bin_rep function can be implemented to convert any base into a binary representation, using the same algorithm in the earlier test method int_to_rep. A str_to_bin_rep method wraps this to convert a more standard most-to-least-significant ordering of digits into binary representations.

def rep_to_bin_rep_(i,rep,base):
if not i:
return []
res = position(rep,getl(i))
higher = rep_to_bin_rep_(getr(i),rep,base)

def rep_to_bin_rep(i,rep=['0','1','2','3','4','5','6','7','8','9']):
if not i:
return []
else:
base = length(rep)
if is_neg_bin_rep(i):
return neg_bin_rep(rep_to_bin_rep_(i,rep,base))
else:
return rep_to_bin_rep_(i,rep,base)

def str_to_bin_rep(digits,rep=['0','1','2','3','4','5','6','7','8','9']):
digits = list(digits)
if is_neg_bin_rep(digits):
return neg_bin_rep(rep_to_bin_rep(reverse(neg_bin_rep(digits)),rep=rep))
else:
return rep_to_bin_rep(reverse(digits),rep=rep)


Now, a string in any base (which is really just a list of digit symbols) can be converted to the binary number representations the mathematical algorithms can work with.

str_to_bin_rep('37')

['i', 'o', 'i', 'o', 'o', 'i']


Finally, to convert any base representation to any other base representation, a function bin_rep_to_rep and matching bin_rep_to_str can be written.

def bin_rep_to_rep_(i,rep,base):
if not i:
return []
i,rem = div_bin_rep(i,base)
higher = bin_rep_to_rep_(i,rep,base)
return cell(element(rep,rem),higher)

def bin_rep_to_rep(i,rep=['0','1','2','3','4','5','6','7','8','9']):
if not i:
return []
else:
base = length(rep)
if is_neg_bin_rep(i):
i = neg_bin_rep(i)
return neg_bin_rep(bin_rep_to_rep_(i,rep,base))
else:
return bin_rep_to_rep_(i,rep,base)

def bin_rep_to_str(i,rep=['0','1','2','3','4','5','6','7','8','9']):
i_rep = bin_rep_to_rep(i,rep=rep)
if not i_rep:
return getl(rep)
if is_neg_bin_rep(i_rep):
return ''.join(neg_bin_rep(reverse(neg_bin_rep(i_rep))))
else:
return ''.join(reverse(i_rep))


This can be tested end-to-end to ensure the input is the same as the output.

i = str_to_bin_rep('37')
print(i)
s = bin_rep_to_str(i)
print(s)

['i', 'o', 'i', 'o', 'o', 'i']
37


## Factorial

After putting all this effort into implementing arbitrary length integer representations and the algorithms to manipulate them, one can try out something that is difficult to do even with the 64-bit integer representations available on modern computers: compute large factorials. The factorial of $A$ is written as $A!$ and is the product of all positive integers less than or equal to A. $$A! = \prod_{i=1}^A i$$ Factorials get large fast, and indeed is one of the fastest-growing classes of functions. Something like the factorial of 100 is so large that it will overflow a 64-bit integer, approximately $9.3\times10^{157}$, but should be calculable with the arbitrary-length integer representations developed here.

Code to compute factorials is easy to write with recursion.

    if not i:
return b_one
return mul_bin_rep(i,factorial_bin_rep(sub_bin_rep(i,b_one)))


And with the ability to convert to and from base-10 representations, the result of $100!$ is easy to obtain:

bin_rep_to_str(factorial_bin_rep(str_to_bin_rep('100')))

'93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000'


Which, according to WolframAlpha is correct.

## Next Steps

Ultimately I want to convert all this Python code into L2 code, and have made some progress on that front…

 (defun is-negative (int) (if int (eq (getl int) '-)))
(defun negate (int) (if int (if (eq (getl int) '-) (getr int) (cell '- int))))
(defun bit-strip (int) (if int (let ((keep (bit-strip (getr int))))
(if (or keep (not (eq (getl int) 'o))) (cell (getl int) keep) ) )))

(defun add-bin-bit-nocarry (a b) (if (eq a 'o)
(if (eq a b) (cell 'o 'o) (cell 'i 'o) )
(if (eq a b) (cell 'o 'i) (cell 'i 'o) ) ))

(let ((sum_b (add-bin-bit-nocarry (getl sum_a) carry) ))
(let ((carry_sum (add-bin-bit-nocarry (getr sum_a) (getr sum_b)) ))
(cell (getl sum_b) (getl carry_sum)) ) ) ))

(defun sub-bin-bit-noborrow (a b) (if (eq a 'o)
(if (eq a b) (cell 'o 'o) (cell 'i 'i) )
(if (eq a b) (cell 'o 'o) (cell 'i 'o) ) ))

(defun sub-bin-bit (a b borrow) (let ((diff_a (sub-bin-bit-noborrow a b) ))
(let ((diff_b (sub-bin-bit-noborrow (getl diff_a) borrow) ))
(let ((borrow_sum (add-bin-bit-nocarry (getr diff_a) (getr diff_b)) ))
(cell (getl diff_b) (getl borrow_sum)) ) ) ))

(defun add-bin-unsigned (a b &optional carry) (let ((carry (if carry carry 'o))) (cond
((and a b) (let ((s (add-bin-bit (getl a) (getl b) carry)))
(cell (getl s) (add-bin-unsigned (getr a) (getr b) (getr s))) ) )
(a nil) ;WIP
(b nil) ;WIP
((not (eq carry 'o)) (,carry))
('t '())
)))
`

…but have become distracted by a desire to implement either floating point numbers or some other fractional representation in my mock-up Python code, in order to do real math (pun intended). Because I’m not tied to a particular hardware architecture, IEEE 754 encoding is not very attractive to re-implement. That said, an exponential representation with a maximum precision does have some attractive features, especially since most fractions have non-terminating representations in different bases. Thinking about the choice of representation for fractional numbers in L2 does give some insight into why LISP dialects have a ratio datatype to exactly store any rational number, which could also arbitrarily approximate any real but not rational number. Perhaps ratios in a future post, and floating points later.

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