Right to Left Computability

Written 2020-11-21


A Theorem

In the book Hacker's Delight, Henry S. Warren, Jr states A function mapping words to words can be implemented with word-parallel add, subtract, and, or, and not instructions if and only if each bit of the result depends only on bits at and to the right of each input operand. Of note, add, subtract, and, or, and not also have this same property, that each output bit depends only on input bits in the same position or to the right, never to the left. This theorem has two major topics to unpack. First, any and all functions that are right-to-left computable can be built from a sequence of these more basic instructions. Second, any function whose output bits depend on a bit to the left of them cannot be implemented with only these basic functions.

Part 1: Building right-to-left computable functions

Warren notes that the basic operations can be combined to form left-shift-by-a-constant(repeated additions) and multiply. We can also trivially construct NOR,NAND,XOR,XNOR from the basic binary operations. Now that we have the full set of binary operations, and a mechanism to shift inputs to the left by a constant number of bits, we can build any right-to-left computable function. For each output bit it's possible to implement the boolean logic of a function by taking copies of the input words, shifting them left until the needed input bit lines up with the output bit, then use binary operations to compute the output. Once each output bit is computed, they are combined with addition or word-parallel or. As noted by Warren, this approach is often inefficient, but it does prove that any right-to-left computable function can be implemented with these basic operations.

Part 2: Inability to build left-to-right computable functions from right-to-left computable functions

Many processors contain an instruction to shift a register right by a given amount, dropping bits off that shift right of zero, and inserting most significant bits of either zero(logical right shift) or the previous most significant bit(arithmentic right shift). This is a very useful and simple instruction, but as each output bit depends on bits at or to the left of it, it is not right-to-left computable. As such, it cannot be implemented only by right-to-left computable operations. Since right shift is one of the simplest left-to-right computable functions, it should be obvious we cannot implement it with right-to-left operations.

Warren gives the example of a function that clears the left-most set bit in a word. For each output bit, we need to look at OR(bits to the left) to decide to clear this bit or pass it through unchanged. For fixed-width register machines, this can sometimes be handled with a count-leading-zeros(CLZ) or loop through through each bit searching for the most significant bit, but these approaches require operations that examine bits to the left of the output bit(CLZ) or conditional branching and looping, and a fixed number of bits per register.

Another example is division - the simplest division algorithm is repeated subtraction, whose only arithmetic operation is subtraction, but knowing how many rounds of subtraction requires comparison and looping, which require reading all available bits. Similarly, we can look to long-division (which produces one binary output per loop), and can be done in as many steps as there are bits in the result. But when we work an example, it's plain to see that the rightmost output bit depends on all the bits to the left of it, so we cannot represent division as a function of add, subtract, and, or, not.

A hypothetical machine

Real computers often have instructions like divide, or at least right-shift, which can be used to construct functions outside of right-to-left computable functions. Also, Turing-complete computers have some form of conditional branching. From here, we explore a hypothetical infinite-bit machine, and the problems we can solve with it as we build it, adding instructions at each step to this RISC-style machine.


There are registers, denoted by r followed by a whole number, each capable of storing an infinite number of bits.


In this section, I discuss some discrete instructions that can be added to the machine. The first many instructions will all be right-to-left calculable, but we'll also discuss a limited set of left-to-right instructions that may be supportable. Of note, each instruction is bit-calculable, that is to say that only the bits needed by future instructions can be calculated, and if needed later, the instruction may restart and calculate more bits.


To make things easier, we can create new instructions freely, as long as they can be defined in terms of add, subtract, and, or, and not. As mentioned above, xor, xnor, nand, and nor are trivially implemented from and, or, and not. To perform a left-shift by one we add a register to itself, and can repeat to perform left-shift by a constant amount. Unsigned multiplication by a constant can be represented the same way, but unsigned multiplication by a variable is also right-to-left calculable, so we start with: As Warren mentions, we can start doing some work with just these basic operations. To clear the least-significant-set-bit, the well-known formula x&(x-1) works here as well as it does on realizable machines. (x-1) sets all bits to the right of the least-significant set bit, and clears the least-significant set bit. Computing the binary and with that bitmask leaves only the more significant bits.

The Curious Case of Underflow

You might've wondered what happens when we try to clear the least-significant-set-bit in zero. For that matter, what happens when we compute 0-1? We could trap this and stop the program, but there's really no need to. Decrement effectively searches a bitstring, right to left, toggling/borrowing bits until a set bit is found, which is then set to zero and the operation is complete. When we decrement zero, we product an infinite bitstring of set bits. At first this may seem a little alarming, but consider that each register is an infinite bitstring already, though usually of zeros. Similarly, if we compute (0-1)+1, the result is zero again, as expected. This is very similar to a 2's complement machine but subtly different - knowing the sign of a number would require peeking at the most-significant bit which cannot be done by these operations, but in 2's complement style, we can add:


Interestingly, in some of the examples Warren gives later in Hacker's Delight, not only are the basic operations available, but operations like increment(add 1) and decrement(subtract one) are also available. This is perfectly reasonable, because the output bits of a function that returns a constant do not depend on any input bits. Similarly, it's possible to design a binary adder with an input tied to one. From this, we can load any constant into a register, and denote operations LOAD and LOADREP to do so. Interestingly, it's possible to LOAD both finite constants and LOADREP repeating binary sequences. It should be possible to load other things like recursive sequences as well.

Now that we can load constants and patterns, we can start to implement some vectorized logic. Juha Jarvi suggests this C function, which produces a pattern of 0x80 if a byte-lane is greater than n and less than 128, and 0x00 if a byte-lane is less than n and greater than 0, and may produce nonsense if x is greater than 128. Since the indicator bit is at the top of the byte-lane, it depends only on input bits at and to the right of itself:

#define hasless8(x,n) (((x)-~0UL/255*(n))&~(x)&~0UL/255*128)
When n is a constant, and x is passed in R0, we can compute R6 like so:
	LOADREP R1 = n(repeating every 8 bits) #Represents ~0UL/255 * (n)
	SUB R3 = R0 - R1 #Represents ((x)-~0UL/255*(n))
	INV R4 = ~R0
	LOADREP R2, 0x80(repeating every 8 bits) #Represents ~0UL/255*128
	AND R5 = R3 & R4
	AND R6 = R4 & R2
Similarly, LOADREC(load recursive) could construct arbitrary bit-patterns where each group of bits depends on a group of bits to the right of it in the same register. For example, a repeating counter like 0xFFFEFD...03020100 could be constructed easily by repeating 8-bit increment over each byte. We could also represent all possible states of a pseudo-random number generator in the same way.

8-bit SIMD

As we saw above it is possible to create some simple vectorized expressions. Can we extend our basic operations as needed to create byte-lane operations? Certainly, because each output bit still only depends on input bits at the same location and to the right of it, but for parallel 8-bit arithmetic each output bit now depends on 1-8 input bits instead of all the input bits to the right of it. Similarly we can emulate a SIMD vector of any width.

8-bit SIMD Saturating arithmetic

It can be useful in certain algorithms to clip or saturate an operation instead of overflowing. For example, with 8-bit saturation we would like to produce 0xFF as the result of 0xFF plus 0x01, yet return the normal addition result otherwise. This cannot be represented in right-to-left operations, as the rightmost bit now depends on the bits in a byte-lane to the left of it. But unlike division, the results only depend on a fixed number of left-to-right inputs, so if we were willing to support left-to-right operations with limited number of bits, this might be an interesting extension to play with later.


As a convenience to get data in and out of the machine, we add INPUT and OUTPUT instructions, which are likely connected to some human-facing console or UNIX pipe.

Comparison and branching

So far, the machine is not Turing-complete, being incapable of control-flow. I'm not certain there's a clean way to model this.

Initially I considered some instructions like branch-if-greater-than, but this may require evaluating the entire infinite bit-string, and only works for a handful of cases. For example, comparing two finite numbers is simple enough, but for negative numbers to compare correctly we must know the most-significant bit. I considered a hack allowing the maching to look-ahead to the MSB for a register, but this also breaks down when comparing repeating patterns. Though it's simple enough to compare patterns of the same length, attempting to compare patterns of different lengths quickly leads to a situation of needing to evaluate both inputs entirely to compute the branch condition. Consider computing the signedness of the repeating bitpattern 0b01 - it cannot be done.

Any comparison approach on this machine must not require complete left-to-right evaluation of an entire register for the reasons above, but must support iterative evaluation of said register.

One approach would be a mask-and-branch-on-compare-with-finite-constant. This would be simple enough to construct loops and if-statements, but also seems like a pretty big hack. So, more consideration needed here for sure.

Right shift by a constant

While not right-to-left computable, if we allow ourselves this small treat, we can greatly extend the machine. Right shift by a constant only implies an implementation overhead of evaluating a few more bits earlier.

Division by an arbitrary constant

Many compilers know a trick for division by a constant by multiplication with a reciprocal calculated at compile time. For example, to divide a 32-bit register by 3, we can multiply by 0x55555556(approximately fixed point one-third) into a 64-bit register, then shift right by 32.

Sadly, this approach cannot be scaled up for this machine, as it breaks down once the numerator is larger than a certain value, due to the difference between fixed-point approximately-one-third and one-third. Normally, the result of this difference is in the lower 32-bits, and eliminated by shifting right by 32. This means that to support integer division on arbitrary scales, we would need an infinitely precise reciprocal and infinitely long right shift. We can construct the reciprocal easily enough, but the right-shift cannot be represented by a constant, as the scale of the required denominator changes according to the size of the numerator.

If the scale of possible numerators is known, the fixed-point multiplication-by-reciprocal trick does work well. If the divisor can be represented by a binary terminating decimal(a non-repeating decimal, in binary), multiply then right-shift works well.