Reconfigurable Computing - Multipliers: Options in Circuit Design John Morris Chung-Ang University The University of Auckland ‘Iolanthe’ at 13 knots on Cockburn Sound, Western Australia

Multipliers  ‘Long’ multiplication x x x x x x x x x multiplier multiplicand partial products product  In binary, the partial products are trivial – if multiplier bit = 1, copy the multiplicand else 0 Use an ‘and’ gate!

Multipliers  ‘Long’ multiplication a 3 a 2 a 1 a 0 b 3 b 2 b 1 b 0 x x x x x x x x x In binary, the partial products are trivial – if multiplier bit = 1, copy the multiplicand else 0 Use an ‘and’ gate! b0b0 b1b1 b2b2 b3b3 a0a0 b0b0 a1a1 a2a2 a3a3 first row of partial products

Multipliers – Simple binary multiplier  We can add the partial products with FA blocks b0b0 b1b1 a0a0 a1a1 a2a2 a3a3 FA 0 p0p0 p1p1 b2b2 product bits

Parallel Array Adder - VHDL  We can build this adder in VHDL with two GENERATE loops FOR j IN 0 TO n-1 GENERATE -- For each row FOR j IN 0 TO n-1 GENERATE –- Generate a row pjk : full_adder PORT MAP( … ); END GENERATE; This part is straight-forward! SIGNAL pa, pb, cout : ARRAY( 0 TO n-1 ) OF ARRAY( 0 TO n-1 ) OF std_logic; … but you need to fill in the PORT MAP using internal signals!

Multipliers – Adding partial products  We can add the partial products with FA blocks b0b0 b1b1 a0a0 a1a1 a2a2 a3a3 FA 0 p0p0 p1p1 b2b2 product bits Optimization 1: Replace this row of FAs Time? What’s the worst case propagation delay?

Multipliers – Using carry save adders  We can add the partial products with FA blocks b0b0 b1b1 a0a0 a1a1 a2a2 a3a3 FA 0 p0p0 p1p1 b2b2 product bits Try to use a more efficient adder in each row? A simpler scheme uses a ‘carry save’ adder – which pushes the carry out’s down to the next row! Note that an extra adder is needed below the last row to add the last partial products and the carries from the row above! Carry select adder

Multipliers - Tree  Chris Wallace discovered a way to build fast multipliers by reducing the number of carry propagations – and thus the delay  All the partial product bits can be generated directly from the operand bits  A full adder adds 3 input bits to produce a 2 bit result  Use it to add the bits in columns  Produce pairs of ‘first level’ sums  Combine bits in these sums vertically again ······ ······ ······ ······ ······ ······ ······ ······ Combine pp bits vertically! 3 at a time ······ ·· ···· ···· ·· ·· ·· ·· · · · First level results Pairs of bits from FA cells

Multipliers - Tree  Summing the partial products ······ ······ ······ ······ ······ ······ ······ ······ So combine them vertically! ······ ·· ···· ···· ·· ·· ·· ·· · · · First level results

Signed digit arithmetic – Avoiding the carries!  Terminology  First, we need to distinguish carefully between  digits of a number and  bits used in representing the number  In the standard binary representations, one bit is used to represent each binary digit (0 or 1) of a number  However, we can use other representation schemes …  If we use more than one bit to represent each digit of an operand, then we have a redundant system  We’re using more bits than the minimum  log 2 n  needed to represent a number of magnitude, n.  These redundant number systems generally have the ability to avoid carry propagation  This may be exploited in the addition of sequences of numbers  Carries are transferred to the following addition  Concept similar to that used in carry-save multiplier where carries are transferred to the following partial product addition

Booth Recoding  A binary number can be re-coded according to Booth’s scheme to reduce the number of partial products in a multiplier  Original idea  Early computers: shift much faster than add  Observe than when there is a 0 in the multiplier, you can skip the addition and just shift the multiplicand  In a synchronous computer, this doesn’t help – in the worst case, you still have to perform an add for each digit of the multiplier (all or most of them are 1’s) ‬ but  in an asynchronous computer, the ability to skip some additions reduces the average completion time  Booth observed that when there is a long sequence of 1s, eg digits j through (down to) k are 1s, then 2 j + 2 j+1 + … +2 k k = 2 j+1 – 2 k

Booth Recoding  A binary number can be re-coded according to Booth’s scheme to reduce the number of partial products in a multiplier  Booth recoding  Booth observed that when there is a long sequence of 1s, eg digits j through (down to) k are 1s, then 2 j + 2 j+1 + … +2 k k = 2 j+1 – 2 k  Thus the sequence of additions can be replaced by  An addition of the multiplicand shifted by j+1 positions and  A subtraction of the multiplicand shifted by k positions  This is equivalent to recoding the multiplier  from a representation using {0,1}  to one using {-1,0,1} – corresponding to subtract, skip, add  The recoding can be done in O(1) time by inspecting neighbouring digits

Booth Recoding  Booth’s scheme  Radix-2 Booth recoding  For each position, j, inspect x j and x j-1 to determine the bits (2 needed!) of y j  Example x: (0) y:  In practice, this scheme is no use in a synchronous machine,  Worst case: sequence of alternating 0 1  More additions than necessary! ‬ but if we use a higher radix Booth recoding  Noteyjyj x j-1 xjxj No 1’s000 End of a string of 1’s - add110 Start of a string of 1’s - subtract01 Middle of a string of 1’s - skip011

Higher Radix Multiplication  Radix-2 multiplier  Use 1 bit of the multiplier at a time  Form partial product with and gates  Radix-4 multiplier  Use 2 bits of the multiplier at a time  If A is the multiplicand..  Radix-4 Booth recoding …  OperationMultiplier bits none00 +A01 +2A (shift A)10 +3A (precompute A+2A?)11

Radix-4 Booth Recoding  Recode multiplier into a signed digit form  Use 3 bits of the original multiplier at a time  Recoded multiplier has half the number of digits, but each digit is in [-2,2]  Operands to the adders are now formed by shifts alone  Recode  Constant time  Partial products  Shift, and, select  n/2 partial products generated  Potentially 2× speed! Operationyjyj x 2j-1 x 2j x 2j+1 No 1’s0000 +A End of 1’s string A Isolated A End of 1’s string A Beginning of 1’s A End one string, start new one 101 -A Start of 1’s string 011 Middle of 1’s0111

No carries at all?  Residue Number Systems 

Residue Arithmetic  Residue Number Systems  A verse by the Chinese scholar, Sun Tsu, over 1500 years ago posed this problem  What number has remainders 2, 3 and 2 when divided by the numbers 7, 5 and 3, respectively?  This is probably the first documented use of number representations using multiple residues  In a residue number system, a number, x, is represented by the list of its residues (remainders) with respect to k relatively prime moduli, m k-1, m k-2, …, m 0  Thus x is represented by (x k-1, x k-2, …, x 0 )  where  x i = x mod m i  So the puzzle may be re-written What is the decimal representation of (2,3,2) in RNS(7,5,3)?

Residue Number Systems  The dynamic range of a RNS, M = m k-1  m k-2  … m 0  For example, in the system RNS(8,7,5,3) M = 8  7  5  3 = 840  Thus we have  Any RNS can be viewed as a weighted representation  In RNS(8,7,5,3), the weights are:  Thus (1,2,4,0) represents (105     0) 840 = (1689) 840 = 9 DecimalRNS(8,7,5,3) 0 or 840 or -840 or …(0,0,0,0) 1 or 841 or -839 or …(1,1,1,1) 2 or 842 or …(2,2,2,2) 8 or 848 or …(0,1,3,2)

Residue Number Systems - Operations  Complement  To find –x, complement each of the digits with respect to the modulus for that digit 21 = (5,0,1,0)  so -21 = (8-5,0,5-1,0) = (3,0,4,0)  Addition or subtraction is performed on each digit ( 5, 5, 0, 2 ) RNS = 5 10 ( 7, 6, 4, 2 ) RNS = ( (5+7)=4 8, (5+6)=4 7, 4, (2+2)=1 3 ) RNS = 4 10 ( 4, 4, 4, 1 ) RNS = 4 10  Multiplication is also achieved by operations on each digit ( 5, 5, 0, 2 ) RNS = 5 10 ( 7, 6, 4, 2 ) RNS = ( (5x7)=3 8, (5x6)=2 7, 0, (2x2)=1 3 ) RNS = ( 3, 2, 0, 1 ) RNS = -5 10

Residue Arithmetic - Advantages  Parallel independent operations on small numbers of digits  Significant speed ups  Especially for multiplication!  4 bit x 4 bit multiplier (moduli up to 15) much simpler than 16 bit x 16 bit one  Carries are strictly confined to small numbers of bits  Each modulus is only a small number of bits  Can be implemented in Look Up Tables (LUTs)  6 bit residues (moduli up to 64)  64 x 64 x 6 bits required (<4Kbytes)

Residue Arithmetic – Choosing the moduli  Largest modulus determines the overall speed –  Try to make it as small as possible  Simple strategy  Choose sequence of prime numbers until the dynamic range, M, becomes large enough eg Application requires a range of at least 10 5, ie M  10 5  For RNS(13,11,7,5,3,2), M = 30,300  Range is too low, so add one more modulus:  RNS(17,13,11,7,5,3,2), M = 510,510  Now each modulus requires a separate circuit and our range is now ~5 times as large as needed, so remove 5 :  RNS(17,13,11,7,3,2), M = 102,102  Six residues, requiring = 19 bits  The largest modulus (17 requiring 5 bits) determines the speed, so …

Residue Arithmetic – Choosing the moduli Application requires a range of at least 10 5, ie M  10 5  …  RNS(17,13,11,7,3,2), M = 102,102  Six residues, requiring = 19 bits  The largest modulus ( 17 requiring 5 bits) determines the speed, so combine some of the smaller moduli (Remember the requirement is that they be relatively prime!)  Try to produce the largest modulus using only 5 bits – Pair 2 and 13, 3 and 7  RNS(26,21,17, 11), M = 102,102  Four residues, requiring = 19 bits (no improvement in total bit count, but 2 fewer ALUs!)  Better …?

Residue Arithmetic – Choosing the moduli Application requires a range of at least 10 5, ie M  10 5  …  RNS(26,21,17, 11), M = 102,102  Four residues, requiring = 19 bits (no improvement in total bit count, but 2 fewer ALUs!)  Include powers of smaller primes before primes, starting with  RNS(3,2), M = 6  Note that 2 2 is smaller than the next prime, 5, so move to  RNS(2 2,3), M = 12  (trying to minimize the size of the largest modulus)  After including 5 and 7, note that 2 3 and 3 2 are smaller than 11:  RNS(3 2,2 3,7,5), M = 2,520  Add 11  RNS(11,3 2,2 3,7,5), M = 27,720  Add 13  RNS(13,11,3 2,2 3,7,5), M = 360,360

Residue Arithmetic – Choosing the moduli Application requires a range of at least 10 5, ie M  10 5  …  Add 13  RNS(13,11,3 2,2 3,7,5), M = 360,360  M is now 3  larger than needed, so replace 9 with 3, then combine 5 and 3  RNS(15,13,11,2 3,7), M = 360,360  5 moduli,  = 18 bits,  largest modulus has 4 bits  You can actually do somewhat better than this!  Reference: B. Parhami, Computer Arithmetic: Algorithms and Hardware Designs, Oxford University Press, 2000

Residue Numbers - Conversion  Inputs and outputs will invariably be in standard binary or decimal representations,  conversion to and from them is required  Conversion from binary | decimal to RNS  Problem: Given a number, y, find its residues wrt moduli, m i  Divisions would be too time-consuming!  Use this equality:  (y k-1 y k-2 …y 1 y 0 ) 2  mi =   2 k-1 y k-1  mi + … +  2y 1  mi +  y 0  mi  mi  So we only need to precompute the residues  2 j  mi for each of the moduli, m i, used by the RNS

Residue Numbers - Conversion  2 j  3  2 j  5  2 j  7 2 j j For RNS(8,7,5,3) : 8 is trivially calculated (3 LSB bits) For 7, 5 and 3, we need the powers of 2 modulus 7, 5 and 3

Residue Numbers - Conversion  2 j  3  2 j  5  2 j  7 2 j j Find = = in RNS(8,7,5,3) : 8 is = 4 10 Note that the additions are done in a modular adder! Worst case: k additions for each residue for a k -bit number 7 = 7 = 7 = 3

Residue Numbers - Conversion Conversion from RNS to binary  Digits of an RNS representation can be shown to have position weightings, eg for RNS(8,7,5,3) the weightings are  The weightings may be calculated using the Chinese Remainder Theorem x = (x k-1 x k-2 … x 1 x 0 ) RNS =   M i  i x i  m  M where M i = M / m i and  i = m is the multiplicative inverse of M i wrt m i  This means that (x 3, x 2, x 1, x 0 ) RNS = x 3 × x 2 × x 1 × x 0 × 280 i i

Residue Numbers - Conversion Conversion from RNS to binary  Digits of an RNS representation can be shown to have position weightings, eg for RNS(8,7,5,3) the weightings are  Calculate position weights with CRT …  This means that (x 3, x 2, x 1, x 0 ) RNS = x 3 × x 2 × x 1 × x 0 × 280  This is most efficiently done through a LUT  Note that the table for RNS(8,7,5,3) requires only = 23 entries  In general, this requires only ‬  k-1 i=0 m i ‬words – a reasonable number!

Residue Arithmetic - Disadvantages  Range is limited  Division is hard!  Comparison, sign (<0?) are hard  Still suitable for some DSP applications  Only use +, x  Range is limited  Result range is known  Examples: digital filters, Fourier transforms