Reciprocal Multiplication, a tutorial

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Reciprocal Multiplication, a tutorial

Part of the Arithmetic Tutorial Collection
by Douglas W. Jones
THE UNIVERSITY OF IOWA Department of Computer Science

Copyright © 1999, Douglas. W. Jones, with major revisions made in 2002. This work may be transmitted or stored in electronic form on any computer attached to the Internet or World Wide Web so long as this notice is included in the copy. Individuals may make single copies for their own use. All other rights are reserved.

Index

Introduction

There are many reasons to omit integer division from the instruction set of a computer. In the early days of computing, the cost of this hardware was a dominant factor, and it still dominates in the world of programmable microcontrollers such as the PIC family of chips made by Microchip. Today, the difficulties cleanly integrating divide hardware into pipelined systems has led to the omission of integer divide hardware from a surprising number of high performance microprocessors and DSP chips.

Such an omission can only be justified if division can be done efficiently in software, and fortunately, this is the case! Most integer division operations found in real programs involves division by a constant, and it turns out that division by a constant can always be replaced by multiplication by the reciprocal of that constant. Even when divide hardware is available, it is quite common to find that divide instructions take more than twice as long as multiply instructions, and as a result, reciprocal multiplication is frequently the preferred way to handle integer division!

Elimination of multiply instructions is itself desirable. For example, according to the Pentium instruction timings available from Quantasm Corporation , The base time for a 32-bit integer add or subtract on that machine (discounting operand fetch and other overhead) is 1 clock cycle, and both can sometimes be paired with other instructions for concurrent execution. In contrast, a 32-bit multiply takes 10 cycles and a divide takes 41 cycles, and neither can be paired. Clearly, these instructions should be avoided wherever possible.

The goal of the remainder of this presentation is to investigate how division by constants, particularly by 10, can be efficiently coded for unsigned integers on a machine without using multiply or divide instructions; on the way, we will briefly visit efficient multiplication by constants. Most of this presentation will use C syntax, avoiding the details of actual machine instructions, and after each technique is explored, it will be applied to other divisors.

Our emphasis is on methods that produce exact results for integer division, but these methods are closely related and in many cases based on methods that produce approximate results for division of fixed-point fractions. These latter methods focus on minimizing the expected error, while our focus is on methods that give a remainder that lies between zero and one less than the divisor.

Fixed Point Fractions

Fixed Point Number Representations

To divide by 10, we will multiply by 1/10, but while this is easily said, we need to talk about the representation of 1/10 on a computer before we go into details. The fraction 1/10 can be represented as 0.1₁₀ ; on converting this to binary, we get 0.000110011 ₂ , which is an infinite repeating fraction, not unlike the situation with 1/3 in decimal, which is 0.3333 ₁₀ ; the repeat string in such fractions is traditionally marked by an overbar; here, this is rendered like this . The following table shows the binary reciprocals of some common small integers.

n	1/n base 2
2	.1
3	.010101
4	.01
5	.001100110011
6	.0010101
7	.001001001
8	.001
9	.000111000111
10	.0001100110011
11	.000101110100010111
12	.00010101
13	.0001001110110001001
14	.0001001001
15	.000100010001

Many programmers assume, on seeing a fraction like 0.1₁₀ , that this must be represented as a floating point number, but this is not so! Fixed point representations, in which the point is implicitly placed between any bits of the binary representation of a number, have been used since the dawn of the computer age. When fixed-point numbers are used, all arithmetic operations use integer arithmetic, and this leads to a considerable performance advantage.

For our purpose, the fraction 0.000110011...₂ is represented as follows on a 16 bit machine:

    .0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 

    |_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|

    |       |       |       |       |

If we round instead of truncating, we get:

    .0 0 0 1 1 0 0 1 1 0 0 1 1 0 1 0 

    |_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|

    |       |       |       |       |

If we ignore the question of what to do with the point, these can be expressed in a typical C environment as:

	unsigned int tenth_trunc = 0x1999;

	unsigned int tenth_round = 0x199A;

The extension of these representations to a 32 bit environment should be obvious. For most of this discussion, we will confine ourselves to a 16 bit environment in order to slightly simplify the presentation.

Now, what about the point? When using fixed-point arithmetic, the position of the point is implicit and is never explicitly represented anywhere outside the comments in a program! (Don Gillies quoted John Von Neumann once, saying that a competent programmer never needs floating point hardware because he should always be able to keep the point in his head.) The following brief discussion of fixed-point multiplication illustrates this.

General Rules for Fixed Point Arithmetic

The general rules for fixed-point binary arithmetic are the same rules students learn in elementary school for doing arithmetic on decimal numbers with a fractional part. That is, when adding, first align the points and then add, and when multiplying, the number of places right of the point in the product is the sum of the numbers of places right of the point in the multiplicands. This is illustrated in the following examples.

The examples that follow are based on fixed-point numbers A and B, defined as follows:

	unsigned int A;  /* 4 bits right of the point */

	unsigned int B;  /* 8 bits right of the point */

Given these definitions, if A has the value 0x15, this is interpreted as 1 and 5/16, while if B has the value 0x15, this is is interpreted as 21/256. The sum A+B may be computed in two different ways, depending on the precision desired:

	unsigned int S;

	S = B + (A << 4);        /* 8 bits right of the point */

	S = ((B + 8) >> 4) + A); /* 4 bits right of the point */

In the second case, when discarding fractional bits B prior to the sum, we first added 1 in the most significant of the bits that were discarded in order to round the result instead of truncating. Multiplication is done as follows:

	unsigned int P; /* 12 places right of the point */

	P = A * B; /* places after point: 12 = 4 + 8 */

Here, we have simply taken the integer product of the multiplicands and then, in the commentary, noted that the number of places to the right of the point in the product is the sum of the number of places to the right of the points in the multiplicands.

Basic Reciprocal Multiplication

When we multiply a 16 bit integer by a 16 bit fixed-point fraction approximating 0.1, the result is a 32 bit quantity with 16 places left of the point and 16 places right of the point. If what we want is the integer part of the quotient, with no places to the right of the point, we must therefore shift the product 16 places right.

	unsigned int A;

	unsigned int tenth_trunc = 0x1999; /* 16 bits right of the point */

	unsigned int Q; /* the quotient */



	Q = (A * tenth_trunc) >> 16;

	/* either Q = A/10 or Q+1 = A/10 for all A < 109,230 */

In those cases where the intended divisor has a reciprocal that can be exactly represented as a fixed-point fraction, the result of reciprocal multiplication will be exact. Unfortunately, our multiplier was not exactly represented! The value of P computed by multiplying by an approximation of .1 is only approximately what we want! The quotient may be correct, or it may be one less than the actual value. The same is true if a rounded approximation of 0.1 is used unless we use a bit of extra precision!

It is worth noting that there are two common ways of providing multiply support on an n-bit machine: One uses a single multiply instruction that takes 2 n-bit operands and delivers a product of 2n bits in a pair of registers. The other uses a pair of multiply instructions, one delivering only the most significant n bits of the product and the other delivering only the least significant n bits of the product. In either case, the computation outlined here uses a single multiply instruction and does not require a shift instruction. In the first case, we simply use the register holding the high half of the product and ignore the value in the other register. In the second case, we use the multiply instruction that delivers only the high half of the product.

Correcting the Errors

Using Extra Precision

One way to correct the errors introduced in reciprocal multiplication is simply to use extra precision. Given that our goal is to divide a 16 bit number by 10, it turns out that 4 bits of extra precision, with rounding in the least significant bit, will suffice, as illustrated here:

	unsigned int A;

	unsigned int tenth_round = 0x1999A; /* 20 bits right of the point */

	unsigned int Q; /* the quotient */



	Q = (A * tenth_round) >> 20;

	/* Q = A/10 for all A < 81,920 */

The problem with this approach is that the rounded approximation given here requires 20 bits and the product requires 36 bits prior to shifting. On a 16 bit machine, products over 32 bits cannot be computed directly using the hardware typically available, and we must use a double-word to hold the approximate multiplier. Nonetheless, as we will see, this approach is still quite useful!

First, quick inspection of our approximation of one tenth reveals that it actually has only 16 significant bits! 0.1999A₁₆ or 0.0001100110011001101 ₂ is equal to 0.CCCD₁₆ or 0.1100110011001101 ₂ divided by 8 or shifted three places right. The latter is a fixed point binary approximation of 0.8₁₀ rounded to 16 bits, so we can divide by 10 using the following code using 16x16=32 multiplier:

        unsigned int A;

        unsigned int scaled_reciprocal = 0xCCCD;

		/* 0.8, 16 bits right of the point  */

        unsigned int shift_count = 19;

        unsigned int Q; /* the quotient */



        Q = (A * scaled_reciprocal) >> shift_count;

        /* Q = (A*0.8)/8 = A/10 for all A < 81,920 */

For a general algorithm for computing the approximation to use for an arbitrary divisor, see Section 18.7 of Anger Fog's How to optimize for the Pentium family of microprocessors , available in PDF format from http://www.agner.org/assem/pentopt.pdf , which is indexed on the web at http://www.agner.org/assem/#optimize . This contains a general algorithm for computing the approximation to use for an arbitrary divisor. The following table gives the scaled reciprocals and shift counts for several common divisors:

n	scaled reciprocal	shift count
3	`1010101010101011 AAAB`	17
5	`1100110011001101 CCCD`	18
6	`1010101010101011 AAAB`	18
7	`10010010010010011`	19
9	`1110001110001111 E38F`	19
10	`1100110011001101 CCCD`	19
11	`1011101000101111 BA2F`	19
12	`1010101010101011 AAAB`	19
13	`1001110110001010 9D8A`	19
14	`10010010010010011`	20
15	`1000100010001001 8889`	19

All of these reciprocal approximations have been verified to give the exact quotient for all possible 16 bit dividends. All but two of these use a 16 bit approximation of the reciprocal and so can be substituted into the code given above on a machine that gives a 32-bit product when multiplying two 16-bit operands.

Unfortunately, it is not always possible to use a 16-bit approximation of the reciprocal to get the correct quotient for a 16 bit dividend. In the case of 1/7 and 1/14, for example, it takes a 17-bit approximation, giving a 33-bit product prior to shifting. To handle this case in a machine that only delivers a 32 bit product, we must first compute the product using the least significant 16 bits of the scaled reciprocal and then use shift-and add methods to add in the final partial product.

        unsigned int A;

        unsigned int Q; /* the quotient */



        Q = (((A * 0x2493) >> 16) + A) >> 3

	

        /* Q = A/7 for all A < 104,859 */

The general rule for finding the scaled reciprocal that gives the exact quotient when dividing by a constant is as follows:

Given d, the constant divisor, compute the exact reciprocal:
r = 1/d
1/d will have z leading zeros between the point and its leftmost one:
assert 0.5 < (1/d × 2^z ) < 1
For an n-bit dividend, take the next n+1 bits and add 1 to round:
r = int(1/d × 2^z+n+1 )+ 1
The shift count is z+n+1
int(a/d) = int( [a × (int(1/d × 2^z+n+1 )+ 1)] / 2^z+n+1 )

These rules always give an n+1 bit approximate reciprocal, but if the result of the rounding step is even, we can use only the most significant n bits of the approximate reciprocal and reduce the shift count by one. For 16-bit operands, this works for 3, 5, 6 and 11, but not for 7 and 9. In the case of 9, however, rounding up by two instead of by one worked. This does not work for 7!

The following tables of successively more precise rounded approximations illustrates how adding precision to the approximation extends the range of dividends for which it gives the exact integer part of the quotient:

approximation of A/3	exact so long as
A×0.011	A < 8
A×0.01011	A < 32
A×0.0101011	A < 128
A×0.010101011	A < 512
A×0.01010101011	A < 2,048
A×0.0101010101011	A < 8,192

approximation of A/5	exact so long as
A×0.00111	A < 14
A×0.001101	A < 64
A×0.001100111	A < 174
A×0.0011001101	A < 1,024
A×0.0011001100111	A < 2,734
A×0.00110011001101	A < 16,384

approximation of A/7	exact so long as
A×0.0010011	A < 27
A×0.00100101	A < 90
A×0.0010010011	A < 209
A×0.00100100101	A < 685
A×0.0010010010011	A < 1,644
A×0.00100100100101	A < 5,466

It would be nice if all optimal multipliers worked precisely up to some power of two, but they do not, as is clearly demonstrated in the table for 1/7. Nonetheless, the relationship between the number of significant places in the multiplier and the precision of the result is clear. If there are n places, inclusive, between the most significant one bit and the least significant position that one could have been added in order to round the exact reciprocal to produce this approximation, then multiplying by this approximation will deliver the correct quotient for all numbers up to 2^n-1 and sometimes up to but not including 2ⁿ .

The following table gives the optimal reciprocals for division of 32-bit unsigned numbers by small integers. Although this table is organized in the same way as the table for 16-bit numbers, high-level-language programmers on 32-bit machines will have difficulty using it because most programming languages on 32-bit machines provide no way to access the high 32 bits of the 64-bit product of two 32-bit unsigned numbers.

n	scaled reciprocal	shift count
3	`10101010101010101010101010101011 AAAAAAAB`	33
5	`11001100110011001100110011001101 CCCCCCCD`	34
6	`10101010101010101010101010101011 AAAAAAAB`	34
7	`100100100100100100100100100100101`	35
9	`11100011100011100011100011100100 E38E38E4`	35
10	`11001100110011001100110011001101 CCCCCCCD`	35
11	`10111010001011101000101110100011 BA2E8BA3`	35
12	`10101010101010101010101010101011 AAAAAAAB`	35

Fixing the Approximation

A second approach to fixing the approximate result is to compute the remainder after multiplication and then correct the product if the remainder is too large. This brute force approach is justified in some settings, primarily those where the remainder is needed anyway and where there is no hardware multiply instruction.

	unsigned int A;

	unsigned int tenth_trunc = 0x1999;

	unsigned int Q; /* the quotient */

	unsigned int R; /* the remainder */



	Q = (A * tenth_trunc) >> 16;

	/* either Q = A/10 or Q+1 = A/10 for all A < 109,230 */



	R = A - 10*Q;

	if (R >= 10) {

		Q = Q + 1;

		R = R - 10; /* optional */

	}

	/* Q = A/10 for all A < 109,230 */

This may seem expensive, involving as it does es a second multiply instruction, where the typical hardware divide instruction is only twice as time-consuming as the typical hardware multiply. In fact, as we will see, some modern RISC processors can multiply by 10 with a single add-and-shift instruction, and many can do it in 2 simple instructions, so this cost may not be as great as it seems.

This approach always works! If we use the n bits immediately to the right of the point in the reciprocal as the multiplier, the result will be either the exact quotient or one less than the exact quotient.

Eliminating use of Multiply Hardware

Architectural Features

Several modern RISC processors have register-to-register add instructions equivalent to the following C statements:

	DST = ((SRC1 << S1) + SRC2) << S2;

	DST = ((SRC1 >> S1) + SRC2) >> S2;

Here, SRC1, SRC2 and DST are registers, and S1 and S2 are constant shift-count fields that are part of the instruction; typically, S1 and S2 are severely constrained and only permit small shifts or are limited to powers of 2. There are other RISC machines that have less general add-and-shift instructions, for example, instructions capable of performing one or the other of the following, but not both:

        DST = (SRC1 << S1) + SRC2;

        DST = (SRC1 + SRC2) << S2;

A single shift-and-add instruction of the most general form illustrated above can compute any product involving a constant multiplier with only 2 one bits set, for example:

	DST = ((SRC     ) + SRC)     ; /* times 2 */

	DST = ((SRC << 1) + SRC)     ; /* times 3 */

	DST = ((SRC     ) + SRC) << 1; /* times 4 */

	DST = ((SRC << 2) + SRC)     ; /* times 5 */

	DST = ((SRC << 1) + SRC) << 1; /* times 6 */

	DST = ((SRC     ) + SRC) << 2; /* times 8 */

	DST = ((SRC << 3) + SRC)     ; /* times 9 */

	DST = ((SRC << 2) + SRC) << 1; /* times 10 */

	DST = ((SRC << 1) + SRC) << 2; /* times 12 */

Multipliers such as 7, 11 and 13 require sequences of two instructions. If less powerful shift-and-add instructions are used, we must use multiple instructions for some of the above. For example, we might multiply by 10 using the following two-instruction sequence:

	DST = (SRC << 2) + SRC; /* times 5 */

	DST = (DST     ) + DST; /* 5 + 5 = 10 */

If we are interested in the integer part of a fixed-point product, we must use shift-and-add with right-shift operation instead of left-shifting as above. The following examples illustrate multiplications by fixed-point fractions with no more than 2 ones in the multiplier:

	DST = ((SRC     )      ) >> 3; /* times 1/8 */

	DST = ((SRC     )      ) >> 2; /* times 1/4 */

	DST = ((SRC >> 1) + SRC) >> 2; /* times 3/8 */

	DST = ((SRC     )      ) >> 1; /* times 1/2 */

	DST = ((SRC >> 2) + SRC) >> 1; /* times 5/8 */

	DST = ((SRC >> 1) + SRC) >> 1; /* times 3/4 */

An important property of the shift-and-add instructions that is essential to the correctness of the above code is that, for n-bit arguments, the sum is n+1 bits, and the extra bit of the sum is shifted into the high bit of the result when the result is right shifted.

Multiplication by 7/8 is not included in the above list because this fraction, 0.111₂ has 3 ones and therefore cannot be done in a single instruction. The missing product can be computed using the following 2-instruction sequence:

        DST = ((SRC >> 1) + SRC) >> 1; /* times 3/4 */

        DST = ((DST     ) + SRC) >> 1; /* (3/4 + 1)/2 = 7/8 */

Readers interested in additional ideas in this area should consult Daniel J. Magenheimer, Liz Peters, Karl Pettis, and Dan Zuras. "Integer multiplication and division on the HP precision architecture," Proc. Second International Conference on Architectural Support for Programming Language and Operating Systems (ASPLOS-II), pages 90-99, October 5-8, 1987.

Approximate Multiplication by 1/10

The above general tools allow us to construct several sequences of shift-and-add instructions for multiplying by the 16 bit fixed point binary approximation of 1/10, 0.0001100110011001. The most straightforward of these instruction sequences simply handles each bit of the multiplier in sequence, from right to left, discarding the least significant bits successively with each step in the computation:

	unsigned int A;

	unsigned int Q; /* the quotient */



	Q = ((A >> 3) + A) >> 1; /* Q = A*0.1001 */

	Q = ((Q     ) + A) >> 3; /* Q = A*0.0011001 */

	Q = ((Q     ) + A) >> 1; /* Q = A*0.10011001 */

	Q = ((Q     ) + A) >> 3; /* Q = A*0.00110011001 */

	Q = ((Q     ) + A) >> 1; /* Q = A*0.100110011001 */

	Q = ((Q     ) + A) >> 4; /* Q = A*0.0001100110011001 */

	/* either Q = A/10 or Q+1 = A/10 for all A < 109,230 */

The result in Q from the above code is exact in 16 bits, in the sense that it gives the same result as multiplication by 1999₁₆ ! Several other approaches to this computation are not exact. Because this code discards the fractional bits of the quotient as soon as possible, it never needs more than 16 bits of accumulator for the quotient. If A and Q are 32 bit quantities, we must simply double the number of statements.

This straightforward code, fails to exploit repeating patterns in the binary representation of the multiplier! One way to exploit these patterns is to note that 1/10 can be represented as a sum:

       2/30  = 0.0001000100010001...    2/(10*3)

     + 1/30  = 0.0000100010001000...	1/(10*3)

    ---------------------------------

     = 1/10  = 0.0001100110011001...

Furthermore:

      16/255 = 0.0001000000010000...

     + 1/255 = 0.0000000100000001...

    --------------------------------

     = 1/15  = 0.0001000100010001...

We can tentatively exploit this as follows:

	Q = ((A >> 8) + A) >> 4; /* Q = A*0.000100000001 */

	Q = ((Q >> 4) + Q)     ; /* Q = A*0.0001000100010001 */

	Q = ((Q >> 1) + Q)     ; /* Q = A*0.00011001100110011 */

	/* either Q = A/10 or Q+1 = A/10 for all A < 30 */

The result is awful, but the problem is easy to locate! We discarded too many of the less significant digits of the sum too quickly. If we defer the right shifting that represents simple division by a power of 2 to the very end, the result is far better:

	Q = ((A >> 8) + A) >> 1; /* Q = A*0.100000001 */

	Q = ((Q >> 4) + Q)     ; /* Q = A*0.1000100010001 */

	Q = ((Q >> 1) + Q) >> 3; /* Q = A*0.00011001100110011 */

	/* either Q = A/10 or Q+1 = A/10 for all A < 451,070 */

But even this code performs its largest right-shift first, so we can improve on it by changing the order in which the terms are added, adding the largest terms first:

	unsigned int A;

	unsigned int Q; /* the quotient */



	Q = ((A >> 1) + A) >> 1; /* Q = A*0.11 */

	Q = ((Q >> 4) + Q)     ; /* Q = A*0.110011 */

	Q = ((Q >> 8) + Q) >> 3; /* Q = A*0.00011001100110011 */

        /* either Q = A/10 or Q+1 = A/10 for all A < 534,890 */

This does not compute the exact same quotient as the original code because it actually computes the product of A times a fixed point approximation of 1/10 that is accurate to 19 places (including trailing zeros) instead of the original 16 bit approximation, and this extra precision allows our approximation to remain useful for dividends slightly exceeding 2¹⁹ .

The final code given above also works in 32 bits with only one additional shift and add step:

	unsigned long int A;

	unsigned long int Q; /* the quotient */



	Q = ((A >>  1) + A) >> 1; /* Q = A*0.11 */

	Q = ((Q >>  4) + Q)     ; /* Q = A*0.110011 */

	Q = ((Q >>  8) + Q)     ; /* Q = A*0.11001100110011 */

	Q = ((Q >> 16) + Q) >> 3; /* Q = A*0.000110011001100110011001100110011 */

        /* either Q = A/10 or Q+1 = A/10 for all 32-bit unsigned A */

But note: It is crucial that the carry out of the very first add in the above code be shifted into the most significant bit position of the shift applied to the result. Guaranteeing this in a high level language like C is very cumbersome, but most machine languages trivially provide for this. The final three add operations in the 32 bit version never produce a carry out.

The general ideas in this and the following section are closely related to the material presented in S.-Y. Li, "Fast constant division routines," IEEE Trans. Computers, C-34, 866-869 (Sept. 1985). The primary difference is that Li does not guarantee that the successive approximations of the quotient are all less than the dividend. As a result, for arbitrary n-bit dividends, Li's schemes frequently require n+1 bits in the accumulator.

Approximate Multiplication by Other Small Reciprocals

The approach outlined above works for many other common reciprocals; each of the approximations given below works up to the dividend value cited in the comments; in each case, the example given is the shortest sequence of shift-add-shift steps that has been found adequate for that dividend on a 16-bit machine:

	/* approximate A/5 */

	Q = ((A >>  1) + A) >> 1; /* Q = A*0.11 */

	Q = ((Q >>  4) + Q)     ; /* Q = A*0.110011 */

	Q = ((Q >>  8) + Q) >> 2; /* Q = A*0.0011001100110011 */

        /* either Q = A/5 or Q+1 = A/5 for all A < 185,365 */



	/* approximate A/6 */

	Q = ((A >>  2) + A) >> 1; /* Q = A*0.101 */

	Q = ((Q >>  4) + Q)     ; /* Q = A*0.1010101 */

	Q = ((Q >>  8) + Q) >> 2; /* Q = A*0.00101010101010101 */

        /* either Q = A/6 or Q+1 = A/6 for all A < 222,438 */



	/* approximate A/7 */

	Q = ((A >>  3) + A) >> 1; /* Q = A*0.1001 */

	Q = ((Q >>  6) + Q)     ; /* Q = A*0.1001001001 */

	Q = ((Q >> 12) + Q) >> 2; /* Q = A*0.001001001001001001001001 */

        /* either Q = A/7 or Q+1 = A/7 for all A < 60,577,223 */



	/* approximate A/9 */

	Q = ((A >>  1) + A) >> 1; /* Q = A*0.11 */

	Q = ((Q      ) + A) >> 1; /* Q = A*0.111 */

	Q = ((Q >>  6) + Q)     ; /* Q = A*0.111000111 */

	Q = ((Q >> 12) + Q) >> 3; /* Q = A*0.000111000111000111000111 */

        /* either Q = A/9 or Q+1 = A/9 for all A < 115,638,345 */



	/* approximate A/11 */

	Q = ((A >>  2) + A) >> 1; /* Q = A*0.101 */

	Q = ((Q      ) + A) >> 1; /* Q = A*0.1101 */

	Q = ((Q      ) + A) >> 2; /* Q = A*0.011101 */

	Q = ((Q      ) + A) >> 1; /* Q = A*0.1011101 */

	Q = ((Q >> 10) + Q) >> 3; /* Q = A*0.00010111010001011101 */

        /* either Q = A/11 or Q+1 = A/11 for all A < 10,103,819 */



	/* approximate A/13 */

        Q = ((A >>  1) + A) >> 2; /* Q = A*0.011 */

        Q = ((Q      ) + A) >> 1; /* Q = A*0.1011 */

        Q = ((Q      ) + A) >> 1; /* Q = A*0.11011 */

        Q = ((Q      ) + A) >> 3; /* Q = A*0.00111011 */

        Q = ((Q      ) + A) >> 1; /* Q = A*0.100111011 */

        Q = ((Q >> 12) + Q) >> 3; /* Q = A*0.0001001110110001 ... */

        /* either Q = A/13 or Q+1 = A/13 for all A < 190,894,093 */



	/* approximate A/15 */

        Q = ((A >>  4) + A) >> 1; /* Q = A*0.10001 */

        Q = ((Q >>  8) + Q) >> 3; /* Q = A*0.0001000100010001 */

        /* either Q = A/15 or Q+1 = A/15 for all A < 864,015 */

Code for approximate division by 10, 12, 14, 18, 22, 26 and 30 can be derived from the code for dividing by 5, 6, 7, 9, 11, 13 and 15 by incrementing the final shift count by one. Comparison of the code for division by 5 with the code given previously for division by 10 illustrates this. Similarly, adding two to the final shift count gives code for division by 20, 24, 28, 36, 44, 52 and 60.

It would have been reasonable to expect to be able to divide by 3 using a similar pattern, as follows:

	/* approximate A/3 */

	Q = ((A >>  2) + A) >> 1; /* Q = A*0.101 */

	Q = ((Q >>  4) + Q)     ; /* Q = A*0.1010101 */

	Q = ((Q >>  8) + Q) >> 1; /* Q = A*0.0101010101010101 */

        /* either Q = A/3 or Q+1 = A/3 for all A < 12,723 */

Unfortunately, this does not work correctly for all 16-bit values! The reason is that shifting away the least significant bit in each step discarded too much. If we keep a bit of extra precision, we can solve this problem:

	/* approximate A/3 */

	Q = ((A >>  2) + A)     ; /* Q = A*1.01 */

	Q = ((Q >>  4) + Q)     ; /* Q = A*1.010101 */

	Q = ((Q >>  8) + Q) >> 2; /* Q = A*0.0101010101010101 */

        /* either Q = A/3 or Q+1 = A/3 for all A < 111,219 */

The problem with this is that the intermediate approximations of the quotient will be 4/3 the size of the dividend, so for arbitrary 16-bit dividends, we need a 17-bit accumulator. Lacking this, we can use an extra shift-add-shift step at the beginning to construct a larger initial approximation before the sequence of shift-add-shift steps that double the precision of this approximation.

	/* approximate A/3 */

	Q = ((A >>  2) + A) >> 2; /* Q = A*0.0101 */

	Q = ((Q      ) + A) >> 1; /* Q = A*0.10101 */

	Q = ((Q >>  6) + Q)     ; /* Q = A*0.10101010101 */

	Q = ((Q >> 12) + Q) >> 1; /* Q = A*0.01010101010101010... */

        /* either Q = A/3 or Q+1 = A/3 for all A < 792,771 */

This is a good 16-bit solution, but adding a shift-add-shift step to try to double the precision does not change the outcome! What we must do is significantly increase the precision of the initial approximation. For a 16-bit dividend, we needed a 5-bit initial approximation. For a 32-bit dividend, we need to add an additional 3 shift-add-shift steps to produce an 11-bit initial approximation.

        /* approximate A/3 */

        Q = ((A >>  2) + A) >> 2; /* Q = A*0.0101 */

        Q = ((Q      ) + A) >> 2; /* Q = A*0.010101 */

        Q = ((Q      ) + A) >> 2; /* Q = A*0.01010101 */

        Q = ((Q      ) + A) >> 2; /* Q = A*0.0101010101 */

        Q = ((Q      ) + A) >> 1; /* Q = A*0.10101010101 */

        Q = ((Q >> 12) + Q)     ; /* Q = A*0.10101010101010101010101 */

        Q = ((Q >> 24) + Q) >> 1; /* Q = A*0.010101010101001010101010101 ... */

        /* either Q = A/3 or Q+1 = A/3 for all 32-bit unsigned A */

As mentioned previously, this code only works if the carry out of the add in each shift-add-shift step is shifted into the most significant bit of the final shift. Assuring this in C is difficult!

In contrast to the case for division by 3, the code given above for approximate division by larger divisors extends trivially to 32 bits:

	/* approximate A/5 */

	Q = ((A >>  1) + A) >> 1; /* Q = A*0.11 */

	Q = ((Q >>  4) + Q)     ; /* Q = A*0.110011 */

	Q = ((Q >>  8) + Q)     ; /* Q = A*0.11001100110011 */

	Q = ((Q >> 16) + Q) >> 2; /* Q = A*0.001100110011001100110011001... */

        /* either Q = A/5 or Q+1 = A/5 for all 32-bit unsigned A */



	/* approximate A/6 */

	Q = ((A >>  2) + A) >> 1; /* Q = A*0.101 */

	Q = ((Q >>  4) + Q)     ; /* Q = A*0.1010101 */

	Q = ((Q >>  8) + Q)     ; /* Q = A*0.101010101010101 */

	Q = ((Q >> 16) + Q) >> 2; /* Q = A*0.001010101010101010101010101... */

        /* either Q = A/6 or Q+1 = A/6 for all 32-bit unsigned A */



	/* approximate A/7 */

	Q = ((A >>  3) + A) >> 1; /* Q = A*0.1001 */

	Q = ((Q >>  6) + Q)     ; /* Q = A*0.1001001001 */

	Q = ((Q >> 12) + Q)     ; /* Q = A*0.1001001001001001001001 */

	Q = ((Q >> 24) + Q) >> 2; /* Q = A*0.001001001001001001001001001... */

        /* either Q = A/7 or Q+1 = A/7 for all 32-bit unsigned A */



	/* approximate A/9 */

	Q = ((A >>  1) + A) >> 1; /* Q = A*0.11 */

	Q = ((Q      ) + A) >> 1; /* Q = A*0.111 */

	Q = ((Q >>  6) + Q)     ; /* Q = A*0.111000111 */

	Q = ((Q >> 12) + Q)     ; /* Q = A*0.111000111000111000111 */

	Q = ((Q >> 12) + Q) >> 3; /* Q = A*0.000111000111000111000111000... */

        /* either Q = A/9 or Q+1 = A/9 for all 32-bit unsigned A */



	/* approximate A/11 */

	Q = ((A >>  2) + A) >> 1; /* Q = A*0.101 */

	Q = ((Q      ) + A) >> 1; /* Q = A*0.1101 */

	Q = ((Q      ) + A) >> 2; /* Q = A*0.011101 */

	Q = ((Q      ) + A) >> 1; /* Q = A*0.1011101 */

	Q = ((Q >> 10) + Q)     ; /* Q = A*0.10111010001011101 */

	Q = ((Q >> 20) + Q) >> 3; /* Q = A*0.000101110100010111010001011... */

        /* either Q = A/11 or Q+1 = A/11 for all 32-bit unsigned A */

Fixing the Approximation

We can, of course, fix the results of this code by computing the remainder and then forcing it in bounds. The following code illustrates this for division by 10 and is easily generalized to other divisors:

        unsigned int A;

        unsigned int Q; /* the quotient */

	unsigned int R; /* the remainder */



	Q = ((A >> 1) + A) >> 1; /* Q = A*0.11 */

	Q = ((Q >> 4) + Q)     ; /* Q = A*0.110011 */

	Q = ((Q >> 8) + Q) >> 3; /* Q = A*0.00011001100110011 */

        /* either Q = A/10 or Q+1 = A/10 for all A < 534,890 */



	R = ((Q << 2) + Q) << 1;

	R = A - R; /* R = A - 10*Q */

	if (R >= 10) {

		R = R - 10;

		Q = Q + 1;

	}

        /* Q = A/10 for all A < 534,890 */

It may be reasonable to code directly from this skeleton on a small microcontroller, but on a modern pipelined RISC machine, it is better to avoid the conditional branches implicit in the if statement given above.

Several modern RISC machines and also many microcontrollers have either conditional skip or conditional store instructions; these test the results of an arithmetic instructions and either force the next instruction to be skipped if some condition is met or only store the result if some condition is met. Such instructions have the property that the code runs at exactly the same speed whether or not the condition is met, and there are no branch penalties. Coding with such instructions in mind, we get the following code:

        unsigned int A;

        unsigned int Q; /* the quotient */

        unsigned int R; /* the remainder */

        unsigned int T; /* temporary */



	Q = ((A >> 1) + A) >> 1; /* Q = A*0.11 */

	Q = ((Q >> 4) + Q)     ; /* Q = A*0.110011 */

	Q = ((Q >> 8) + Q) >> 3; /* Q = A*0.00011001100110011 */

        /* either Q = A/10 or Q+1 = A/10 for all A < 534,890 */



	R = ((Q << 2) + Q) << 1;

	R = A - R;

	T = R - 10;

	if (T > 0) Q = Q + 1;

	if (T > 0) R = T;

        /* Q = A/10 for all A < 534,890 */

In the above, we have introduced a temporary variable, T, to eliminate the redundant computations implied by the comparison and subtraction code in the original if statement. A good compiler may be able to perform this optimization automatically.

If we translated the above code to machine code on a RISC machine with general add-shift-and-skip instructions, we can compute our approximate quotient after the 3rd instruction, the approximate remainder after the 5th, and the corrected quotient after the 7th. On such a machine, no more than 2 extra instructions are needed to compute the corrected remainder, so we can get A/10 and A mod 10 in 9 instructions of straight-line code.

Using Extra Precision

As noted, it takes an extra bit of precision to produce an exact quotient using reciprocal multiplication. For a 16-bit product, we need a 17 bit rounded fixed-point multiplier. For division by 10, for example, we multiply by 0x1999A or 0.00011001100110011010₂ . Rounding has destroyed the clean repeating pattern of our truncated multiplier, but if we ignore what we learned above about efficient ways to compute the repeating fraction, we can use brute force to arrive at an exact quotient as follows:

	unsigned int A;

	unsigned int Q; /* the quotient */



	Q = ((A >> 2) + A) >> 1; /* Q = A*0.101 */

	Q = ((Q     ) + A) >> 3; /* Q = A*0.001101 */

	Q = ((Q     ) + A) >> 1; /* Q = A*0.1001101 */

	Q = ((Q     ) + A) >> 3; /* Q = A*0.0011001101 */

	Q = ((Q     ) + A) >> 1; /* Q = A*0.10011001101 */

	Q = ((Q     ) + A) >> 3; /* Q = A*0.00110011001101 */

	Q = ((Q     ) + A) >> 1; /* Q = A*0.100110011001101 */

	Q = ((Q     ) + A) >> 4; /* Q = A*0.0001100110011001101 */

        /* Q = A/10 for all A < 81,920 */

Here, we arrived at an exact 16 bit quotient using 8 generalized shift-and-add instructions, only one instruction more than our estimated instruction count for the more clever approximate-and-correct code given above! On a 16-bit machine, for most small divisors, this exact method is generally comparable in cost to approximate division followed by a fixup step.

For larger word sizes, the the approximate-and-correct methods generally outperform this exact code. This is because the total number of shift-add-shift steps required for the exact method is generally proportional to the word size. In contrast, with the approximate methods given here, each additional instruction typically doubles the precision. The following code, for example, gives A/10 exact to 32 bits:

	Q = ((A >> 2) + A) >> 1; /* Q = A*0.101 */

	Q = ((Q     ) + A) >> 3; /* Q = A*0.001101 */

	Q = ((Q     ) + A) >> 1; /* Q = A*0.1001101 */

	Q = ((Q     ) + A) >> 3; /* Q = A*0.0011001101 */

	Q = ((Q     ) + A) >> 1; /* Q = A*0.10011001101 */

	Q = ((Q     ) + A) >> 3; /* Q = A*0.00110011001101 */

	Q = ((Q     ) + A) >> 1; /* Q = A*0.100110011001101 */

	Q = ((Q     ) + A) >> 3; /* Q = A*0.001100110011001101 */

	Q = ((Q     ) + A) >> 1; /* Q = A*0.1001100110011001101 */

	Q = ((Q     ) + A) >> 3; /* Q = A*0.0011001100110011001101 */

	Q = ((Q     ) + A) >> 1; /* Q = A*0.10011001100110011001101 */

	Q = ((Q     ) + A) >> 3; /* Q = A*0.00110011001100110011001101 */

	Q = ((Q     ) + A) >> 1; /* Q = A*0.100110011001100110011001101 */

	Q = ((Q     ) + A) >> 3; /* Q = A*0.001100110011001100110011001101 */

	Q = ((Q     ) + A) >> 1; /* Q = A*0.1001100110011001100110011001101 */

	Q = ((Q     ) + A) >> 4; /* Q = A*0.00011001100110011001100110011001101 */

	/* Q = A/10 for all 32-bit unsigned A */

If the approximate reciprocal has sufficiently few one bits, however, the exact method will outperform the inexact method. For our small divisors, the case of 15 provides a clear example of this:

	Q = ((A >> 3) + A) >> 4; /* Q = A*0.0001001 */

	Q = ((Q     ) + A) >> 4; /* Q = A*0.00010001001 */

	Q = ((Q     ) + A) >> 4; /* Q = A*0.000100010001001 */

	Q = ((Q     ) + A) >> 4; /* Q = A*0.0001000100010001001 */

	/* Q = A/15 for all A < 74,909 */

These four generalized shift-add-shift instructions for exact 16-bit division by 15 clearly win over our approximate-and-correct method, and extension to a 32-bit dividend remains competitive. Only when operand sizes approach 64 bits does approximation followed by correction become a clear winner.

Another case where the exact methods generally win is when the repeating pattern used in the approximation is long and irregular, as it is for division by 11 and 13. For these cases, the code to compute the irregular pattern dominated the approximation for the 16-bit case. The approximate-and-correct method still wins for these divisors in 32 bits.

Signed Dividends

In all of the above discussion, the focus has been on unsigned integers. Division by negative divisors is relatively trivial. First divide by the positive number, then negate both the remainder and quotient. Division of signed dividends, however, poses more interesting problems.

The classic approach to division of a potentially negative number is to record the sign, take the absolute value, and then restore the sign, as follows:

	int A;

	int Q; /* the quotient */



	{

		int S = A < 0;

		unsigned int AA = A;

		if (S) AA = -AA;	/* AA = abs(A) */



		Q = AA / D;		/* divide */



		if (S) Q = -Q;		/* restore sign */

	}

This logic satisfies the vast majority of programmers, and in fact, this is how division of a negative number by a positive dividend is defined in FORTRAN. Unfortunately, it may be wrong!

The problem with the above algorithm is clear when we examine the following incompatible requirements for integer division:

(-A) / D = -(A / D)
D(A / D) + (A mod D) = A
0 < (A mod D) < D

At a quick glance, these appear to be obvious and true, but for cases where A is negative and the remainder (A mod D) is nonzero, it is not possible for all three to hold. The code framework given above is based on rule 1, and if we define the mod operator as in rule 2, the consequence is that nonzero remainders will have their sign determined by the dividend and therefore, rule 3 will be violated.

Another way of describing the effect is to say that the code outlined above truncates non-integer results towards zero. The effect is to round the absolute value of the quotient down, and then restore the sign.

When we divide by a power of two using a sign-preserving right-shift operator, the result is different. In that case, non-integer results are truncated to the next more negative integer, so that truncation actually increases the value of negative quotients. This preserves rules 2 and 3 above, while allowing rule 1 to be violated.

Which approach is right? This is not the right question. For some applications, one approach may be preferred, while for others, the other approach is likely to be better.

If the shift and add approach to reciprocal multiplication given in the final section above is applied to signed dividends, the result is the correct quotient, adhering to rules 2 and 3 above. Thus, for a 16-bit divide by 10, the following code works:

        int A;

        int Q; /* the quotient */

	int R; /* the remainder */



	Q = ((A >> 1) + A) >> 1; /* Q = A*0.11 */

	Q = ((Q >> 4) + Q)     ; /* Q = A*0.110011 */

	Q = ((Q >> 8) + Q) >> 3; /* Q = A*0.00011001100110011 */

        /* either Q = A/10 or Q+1 = A/10 for -87,381 < A < 534,890 */



	R = ((Q << 2) + Q) << 1;

	R = A - R; /* R = A - 10*Q */

	if (R >= 10) {

		R = R - 10;

		Q = Q + 1;

	}

        /* Q = A/10 for -87,381 < A < 534,890 */

This code assumes that right-shift operators applied to signed operands sign-extend the shifted operand, and it assumes that the very first shift-add-shift sequence correctly accounts for the carry and overflow conditions from the add step as it performs the second shift, so that the result after the shift is as if no data was lost to an overflow in the add step. With properly designed hardware, this is straightforward, but not all computers offer good support in this area.

Notice that the range of values for which this works is not symmetrical! Nonetheless, it is sufficient for all 16-bit signed divisors. This asymmetry also shows up in the case of simple reciprocal multiplication. Thus, where multiplication by 0.1999₁₆ produced an adequate approximation of the quotient for 16-bit unsigned dividends, we must multiply by 0.19998₁₆ to correctly approximate the quotient when dealing with 16-bit signed dividends.

	int tenth_trunc = 0x3333; /* 17 bits right of the point */



	Q = (A * tenth_trunc) >> 17;

        /* either Q = A/10 or Q+1 = A/10 for -65,541 < A < 163,843 */

As noted previously, the above code gives the correct result under rules 2 and 3 given above for defining integer division. If the FORTRAN version, adhering to rule 1 is desired, a fixup step can be added:

	if ((Q < 0) && (R != 0)) {

		R = R - 10;

		Q = Q + 1;

	}

Extension of these methods to 32 bits for signed dividends is straightforward, but only the approximate division methods given for unsigned dividends can be applied. The divisors that gave exact results for unsigned dividends do not give exact results for negative dividends!

Exhaustive Testing

It is easy to forget how fast modern machines are! Exhaustive testing of algorithms for 16 bit arithmetic is not only practical but fast, and given the difficulty of mathematical proof, it is worthwhile. All of the unsigned shift and add code given above has been tested using the following skeleton, written in C:

#include <stdio.h>

main()

{

        unsigned long int A = 0;

        for (;;) {

                unsigned long int Q;

                unsigned long int R;



                /* substitute appropriate code here */



                if ((A/10) != Q) break;

		A++;

		if (A == 0) break;

        }

        printf( "terminates with A = %lu/n", A );

        exit();

}

This code runs to completion in under a second for 16-bit divide routines on an antique IBM RT workstation that, coincidentally, has no integer multiply or divide hardware. On a modern fast machine such as the Power PC G4, exhaustive testing on 32 bit code, or exhaustive tests of 2-argument 16 arithmetic operations is entirely within reason; exhaustive testing of the 32-bit code given above took well under an hour, even with the awkward code required to preserve the carry out in the initial shift-add-shift operations.

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