30000BC Palaeolithic peoples in central Europe and France record numbers on bones. 5000BC A decimal number system is in use in Egypt.

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1 3BC Palaeolithic peoples in central Europe and France record numbers on bones. 5BC A decimal number system is in use in Egypt. 4BC Babylonian and Egyptian calendars in use. 34BC The first symbols for numbers, simple straight lines, are used in Egypt. 3BC The abacus is developed in the Middle East and in areas around the Mediterranean. A somewhat different type of abacus is used in China. 3BC Hieroglyphic numerals in use in Egypt. 3BC Babylonians begin to use a sexagesimal number system for recording financial transactions. It is a place-value system without a zero place value. 2BC Harappans adopt a uniform decimal system of weights and measures. 95BC Babylonians solve quadratic equations.

2 9BC The Moscow papyrus is written. It gives details of Egyptian geometry. 85BC Babylonians know Pythagoras's Theorem. 8BC Babylonians use multiplication tables. 75BC The Babylonians solve linear and quadratic algebraic equations, compile tables of square and cube roots. They use Pythagoras's theorem and use mathematics to extend knowledge of astronomy. 7BC The Rhind papyrus (sometimes called the Ahmes papyrus) is written. It shows that Egyptian mathematics has developed many techniques to solve problems. Multiplication is based on repeated doubling, and division uses successive halving. 36BC A decimal number system with no zero starts to be used in China. BC Chinese use counting boards for calculation. 54BC Counting rods used in China. 5BC The Babylonian sexagesimal number system is used to record and predict the positions of the Sun, Moon and planets.

3 Egyptian Numerals Egyptian number system is additive.

4

5 Mesopotamia Civilization Above: Babylonian sexagesimal (base 6) number. It is the first positional number system. Left: Oldest cuneiform writing by Sumerian.

6 Babylonian numerals

7 4359 Chinese numerals

8 Indian numerals

9 Greek number systems

10 Roman Numerals I II 2 III 3 IV 4 V 5 VI 6 VII 7 VIII 8 IX 9 X L 5 C D 5 M MMMDCCCLXXVIII 3878

11 Mayan mathematics 25 AD to 9 AD, this period was built on top of a civilization which had lived in the region from about 2 BC. [8;4;3;;2] represents 2 + x x 8 x x 8 x x 8 x 2 3 =

12 The numerals from al-sizji's treatise of 969

13 Abaci Chinese Abacus Boethius (Hindu-Arabic) vs Pythagoras (counting board)

14 Logarithm and Slide Rule John Napier of Scotland developed the concept of logarithm around AD 6. If a y =x, then y = log a x Slide rule based on the property of logarithm was invented in the late 7s. log (u v) = log (u) + log(v)

15 Decimal Numbers: Base Digits:,, 2, 3, 4, 5, 6, 7, 8, 9 Example: 327 = (3x 3 ) + (2x 2 ) + (7x ) + (x )

16 Numbers: positional notation Number Base B B symbols per digit: Base (Decimal):,, 2, 3, 4, 5, 6, 7, 8, 9 Base 2 (Binary):, Number representation: d 3 d 3... d d is a 32 digit number value = d 3 B 3 + d 3 B d B + d B Binary:, (In binary digits called bits ) b = = = 26 Here 5 digit binary # turns into a 2 digit decimal # Can we find a base that converts to binary easily? #s often written b

17 Hexadecimal Numbers: Base 6 Hexadecimal:,, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F Normal digits + 6 more from the alphabet In C, written as x (e.g., xfab5) Conversion: Binary Hex hex digit represents 6 decimal values 4 binary digits represent 6 decimal values hex digit replaces 4 binary digits One hex digit is a nibble. Two is a byte Example: (binary) = x?

18 Decimal vs. Hexadecimal vs. Binary Examples: (binary) = xac3 (binary) = (binary) = x7 x3f9 = (binary) How do we convert between hex and Decimal? A B 2 C 3 D 4 E 5 F

19 What to do with representations of numbers? Just what we do with numbers! Add them Subtract them Multiply them Divide them Compare them Example: + 7 = 7 so simple to add in binary that we can build circuits to do it! subtraction just as you would in decimal Comparison: How do you tell if X > Y?

20 Which base do we use? Decimal: great for humans, especially when doing arithmetic Hex: if human looking at long strings of binary numbers, its much easier to convert to hex and look 4 bits/symbol Terrible for arithmetic on paper Binary: what computers use; you will learn how computers do +, -, *, / To a computer, numbers always binary Regardless of how number is written: 32 ten == 32 == x2 == 2 == b Use subscripts ten, hex, two in book, slides when might be confusing

21 BIG IDEA: Bits can represent anything!! Characters? 26 letters 5 bits (2 5 = 32) upper/lower case + punctuation 7 bits (in 8) ( ASCII ) standard code to cover all the world s languages 8,6,32 bits ( Unicode ) Logical values? False, True colors? Ex: locations / addresses? commands? MEMORIZE: N bits at most 2 N things Red () Green () Blue ()

22 How to Represent Negative Numbers? So far, unsigned numbers Obvious solution: define leftmost bit to be sign! +, - Rest of bits can be numerical value of number Representation called sign and magnitude MIPS uses 32-bit integers. + ten would be: And ten in sign and magnitude would be:

23 Shortcomings of sign and magnitude? Arithmetic circuit complicated Special steps depending whether signs are the same or not Also, two zeros x = + ten x8 = - ten What would two s mean for programming? Therefore sign and magnitude abandoned

24 Another try: complement the bits Example: 7 = 2-7 = 2 Called One s Complement Note: positive numbers have leading s, negative numbers have leadings s What is -? Answer: How many positive numbers in N bits? How many negative ones?

25 Shortcomings of One s complement? Arithmetic still a somewhat complicated. Still two zeros x = + ten xffffffff = - ten Although used for awhile on some computer products, one s complement was eventually abandoned because another solution was better.

26 Standard Negative Number Representation What is result for unsigned numbers if tried to subtract large number from a small one? Would try to borrow from string of leading s, so result would have a string of leading s» = With no obvious better alternative, pick representation that made the hardware simple As with sign and magnitude, leading s positive, leading s negative»...xxx is,...xxx is <» except is -, not - (as in sign & mag.) This representation is Two s Complement

27 Sign and Magnitude = + 4 Example: N = = High order bit is sign: = positive (or zero), = negative Remaining low order bits is the magnitude: () thru 7 () Number range for n bits = +/- 2 n- - Representations for? Operations: =, <, >, +, -???

28 Ones Complement (algebraically) N is positive number, then N is its negative 's complement N = (2 n - ) - N 4 2 = - = Example: 's complement of 7-7 = -7 in 's comp. Bit manipulation: simply complement each of the bits ->

29 Ones Complement on the number wheel = = Subtraction implemented by addition & 's complement Sign is easy to determine Closure under negation. If A can be represented, so can -A Still two representations of! If A = B then is A B ==? Addition is almost clockwise advance, like unsigned

30 Twos Complement number wheel like 's comp except shifted one position clockwise = + 4 = Easy to determine sign (?) Only one representation for Addition and subtraction just as in unsigned case Simple comparison: A < B iff A B < One more negative number than positive number - one number has no additive inverse

31 Twos Complement (algebraically) N* = 2 n - N Example: Twos complement of 7 sub 4 2 = 7 = = repr. of -7 Example: Twos complement of -7 Bit manipulation: sub 4 2 = -7 = = repr. of 7 Twos complement: take bitwise complement and add one -> + -> (representation of -7) -> + -> (representation of 7)

32 How is addition performed in each number system? Operands may be positive or negative

33 Sign Magnitude Addition Operand have same sign: unsigned addition of magnitudes result sign bit is the same as the operands' sign (-3) -7 Operands have different signs: subtract smaller from larger and keep sign of the larger

34 Ones complement addition Perform unsigned addition, then add in the end-around carry (-3) 7-7 End around carry End around carry

35 When carry occurs = + 4 = M N where M > N -M - N

36 Why does end-around carry work? Recall: N = (2 n - ) - N End-around carry work is equivalent to subtracting 2 n and adding n n M - N = M + N = M + (2 - - N) = (M - N) (when M > N) n n -M + (-N) = M + N = (2 - M - ) + (2 - N - ) n n = 2 + [2 - - (M + N)] - M + N < 2 n- after end around carry: n = (M + N) this is the correct form for representing -(M + N) in 's comp!

37 Twos Complement Addition Perform unsigned addition and 4-4 Discard the carry out (-3) 7-7 Overflow? Simpler addition scheme makes twos complement the most common choice for integer number systems within digital systems

38 Twos Complement number wheel = + 4 = M + -N where N + M 2n- -M + N when N > M

39 2s Comp: ignore the carry out -M + N when N > M: n n M* + N = (2 - M) + N = 2 + (N - M) Ignoring carry-out is just like subtracting 2 n -M + -N where N + M 2 n- n n -M + (-N) = M* + N* = (2 - M) + (2 - N) n n = 2 - (M + N) + 2 After ignoring the carry, this is just the right twos compl. representation for -(M + N)!

40 2s Complement Overflow How can you tell an overflow occurred? Add two positive numbers to get a negative number or two negative numbers to get a positive number = -8! -7-2 = +7!

41 2s comp. Overflow Detection Overflow Overflow No overflow No overflow Overflow occurs when carry in to sign does not equa l carry out

42 Two s Complement for N=32... two = ten... two = ten... two = 2 ten two = 2,47,483,645 ten... two = 2,47,483,646 ten... two = 2,47,483,647 ten... two = 2,47,483,648 ten... two = 2,47,483,647 ten... two = 2,47,483,646 ten two = 3 ten... two = 2 ten... two = ten One zero; st bit called sign bit extra negative:no positive 2,47,483,648 ten

43 Two s Complement Formula Can represent positive and negative numbers in terms of the bit value times a power of 2: d 3 x -(2 3 ) + d 3 x d 2 x d x 2 + d x 2 Example: two = x-(2 3 ) + x2 2 + x2 + x2 = = = = -3 ten

44 Two s Complement shortcut: Negation Change every to and to (invert or complement), then add to the result Proof: Sum of number and its (one s) complement must be... two However,... two = - ten Let x one s complement representation of x Then x + x = - x + x + = x + = -x Example: -3 to +3 to -3 x : two x : two +: two () : two +: two You should be able to do this in your head

45 Two s comp. shortcut: Sign extension Convert 2 s complement number rep. using n bits to more than n bits Simply replicate the most significant bit (sign bit) of smaller to fill new bits 2 s comp. positive number has infinite s 2 s comp. negative number has infinite s Binary representation hides leading bits; sign extension restores some of them 6-bit -4 ten to 32-bit: two two

46 What if too big? Binary bit patterns above are simply representatives of numbers. Strictly speaking they are called numerals. Numbers really have an number of digits with almost all being same ( or ) except for a few of the rightmost digits Just don t normally show leading digits If result of add (or -, *, / ) cannot be represented by these rightmost HW bits, overflow is said to have occurred. unsigned

47 Kilo, Mega, Giga, Tera, Peta, Exa, Zetta, Yotta physics.nist.gov/cuu/units/binary.html Name Abbr Factor SI size Kilo K 2 =,24 3 =, Mega M 2 2 =,48,576 6 =,, Giga G 2 3 =,73,74,824 9 =,,, Tera T 2 4 =,99,5,627,776 2 =,,,, Peta P 2 5 =,25,899,96,842,624 5 =,,,,, Exa E 2 6 =,52,92,54,66,846,976 8 =,,,,,, Zetta Z 2 7 =,8,59,62,77,4,33,424 2 =,,,,,,, Yotta Y 2 8 =,28,925,89,64,629,74,76,76 24 =,,,,,,,, Confusing! Common usage of kilobyte means 24 bytes, but the correct SI value is bytes Hard Disk manufacturers & Telecommunications are the only computing groups that use SI factors, so what is advertised as a 3 GB drive will actually only hold about 28 x 2 3 bytes, and a Mbit/s connection transfers 6 bps.

48 kibi, mebi, gibi, tebi, pebi, exbi, zebi, yobi en.wikipedia.org/wiki/binary_prefix Name kibi mebi gibi tebi pebi exbi zebi yobi Abbr Ki Mi Gi Ti Pi Ei Zi Yi Factor 2 =, =,48, =,73,74, =,99,5,627, =,25,899,96,842, =,52,92,54,66,846, =,8,59,62,77,4,33, =,28,925,89,64,629,74,76,76 International Electrotechnical Commission (IEC) in 999 introduced these to specify binary quantities. Names come from shortened versions of the original SI prefixes (same pronunciation) and bi is short for binary, but pronounced bee :-( Now SI prefixes only have their base- meaning and never have a base-2 meaning.

49 The way to remember #s What is 2 34? How many bits addresses (I.e., what s ceil log 2 = lg of) 2.5 TiB? Answer! 2 XY means X= --- X= kibi ~ 3 X=2 mebi ~ 6 X=3 gibi ~ 9 X=4 tebi ~ 2 X=5 tebi ~ 5 X=6 exbi ~ 8 X=7 zebi ~ 2 X=8 yobi ~ 24 Y= Y= 2 Y=2 4 Y=3 8 Y=4 6 Y=5 32 Y=6 64 Y=7 28 Y=8 256 Y=9 52 MEMORIZE!

50 Comparing the signed number systems Here are all the 4-bit numbers in the different systems. Positive numbers are the same in all three representations. Signed magnitude and one s complement have two ways of representing. This makes things more complicated. Two s complement has asymmetric ranges; there is one more negative number than positive number. Here, you can represent -8 but not +8. However, two s complement is preferred because it has only one, and its addition algorithm is the simplest. Decimal S.M. s comp. 2 s comp

51 And in Conclusion... We represent things in computers as particular bit patterns: N bits 2 N Decimal for human calculations, binary for computers, hex to write binary more easily s complement - mostly abandoned 2 s complement universal in computing: cannot avoid, so learn Overflow: numbers ; computers finite, errors!

52 Numbers represented in memory Memory is a place to store bits A word is a fixed number of bits (eg, 32) at an address Addresses are naturally represented as unsigned numbers in C = 2 k -

53 Signed vs. Unsigned Variables Java just declares integers int Uses two s complement C has declaration int also Declares variable as a signed integer Uses two s complement Also, C declaration unsigned int Declares a unsigned integer Treats 32-bit number as unsigned integer, so most significant bit is part of the number, not a sign bit

54 Binary Codes for Decimal Digits There are over 8, ways that you can chose elements from the 6 binary numbers of 4 bits. A few are useful: Decimal 8,4,2, Excess3 8,4,-2,

55 Binary Coded Decimal (BCD) Binary Coded Decimal or 8,4,2, Code. This code is the simplest, most intuitive binary code for decimal digits and uses the same weights as a binary number, but only encodes the first ten values from to 9. Examples: is 8 + = 9 is 2 + = 3 is 4 is an illegal code.

56 Other Decimal Codes The Excess-3 Code adds binary to the BCD code. The BCD (8,4, 2, ) Code, and the (8,4,-2,-) Code are examples of weighted codes. Each bit has a "weight" associated with it and you can compute the decimal value by adding the weights where a exists in the code-word. Example: in (8,4,-2,-) is (-2) + (-) = 9

57 Warning: Conversion or Coding? DO NOT mix up CONVERSION of a decimal number to a binary number with CODING a decimal number with a BINARY CODE. 3 = 2 (This is CONVERSION) 3 (This is CODING)

58 Binary Addition: Half Adder Ai Bi Sum Carry Ai Bi Ai Bi Sum = Ai Bi + Ai Bi Carry = Ai Bi A i Sum = Ai + Bi B i Half-adder Schematic Carry But each bit position may have a carry in

59 Full-Adder + Co Cin B A S A B CI S CO CI S CI A B A B S = CI xor A xor B CO CO = B CI + A CI + A B = CI (A + B) + A B Now we can connect them up to do multiple bits

60 Ripple Carry A3 B3 A2 B2 A B A B S3 C3 S2 C2 S C S

61 Full Adder from Half Adders (little aside) Standard Approach: 6 Gates A B S CI A B CI A B CO Alternative Implementation: 5 Gates A B Half Adder S CO A B A + B Half Adder S CO A + B + CI CI (A + B) S CI CO A B + CI (A xor B) = A B + B CI + A CI

62 Delay in the Ripple Carry Adder Critical delay: the propagation of carry from low to high order stages late @ A B CI @ two gate delays to compute CO C 4 stage adder A B A B C A 2 B 2 2 S C A 3 B 3 3 S C final sum and carry

63 Ripple Carry Timing Critical delay: the propagation of carry from low to high order stages S, C Valid S, C2 Valid S2, C3 Valid S3, C4 Valid + worst case addition T T2 T4 T6 T8 T: Inputs to the adder are valid T2: Stage carry out (C) T4: Stage carry out (C2) 2 delays to compute sum but last carry not ready until 6 delays later T6: Stage 2 carry out (C3) T8: Stage 3 carry out (C4)

64 Adders (cont.) Ripple Adder c7 c6 c5 c4 c3 c2 b a c s7 s6 FA Ripple adder is inherently slow because, in general s7 must wait for c7 which must wait for c6 T α n, Cost α n How do we make it faster, perhaps with more cost? Classic approach: Carry Look-Ahead c s Or use a MUX!!!

65 Carry Select Adder b7a7 b6a6 b5a5 b4 a4 b3a3 b2a2 ba ba c FA c8 b7a7 b6a6 b5a5 b4a4 s3 s2 s s FA s7 s6 s5 s4 T = T ripple_adder / 2 + T MUX COST =.5 * COST ripple_adder + (n+) * COST MUX

66 Extended Carry Select Adder b5-b2 a5-a2 b-b8 a-a8 b7-b4 a7-a4 b3-b a3-a 4-bit Adder 4-bit Adder 4-bit Adder cout 4-bit Adder 4-bit Adder 4-bit Adder 4-bit Adder cin What is the optimal # of blocks and # of bits/block? If # blocks too large delay dominated by total mux delay If # blocks too small delay dominated by adder delay per block N stages of N bits T α sqrt(n), Cost 2*ripple + muxes

67 Carry Select Adder Performance b5-b2 a5-a2 b-b8 a-a8 b7-b4 a7-a4 b3-b a3-a 4-bit Adder 4-bit Adder 4-bit Adder cout 4-bit Adder 4-bit Adder 4-bit Adder 4-bit Adder cin Compare to ripple adder delay: T total = 2 sqrt(n) T FA T FA, assuming T FA = T MUX For ripple adder T total = N T FA cross-over at N=3, Carry select faster for any value of N>3. Is sqrt(n) really the optimum? From right to left increase size of each block to better match delays Ex: 64-bit adder, use block sizes [ ] How about recursively defined carry select?

68 What really happens with the carries c7 c6 c5 c4 c3 c2 b a c FA s7 s6 c s A B Cout S Cin Cin ~Cin Carry action kill Propagate Ai Bi Gi Cin ~Cin Cin propagate generate Ai Bi Pi Carry Generate Gi = Ai Bi must generate carry when A = B = Carry Propagate Pi = Ai xor Bi carry in will equal carry out here All generates and propagates in parallel at first stage. No ripple.

69 Carry Look Ahead Logic Carry Generate Gi = Ai Bi must generate carry when A = B = Carry Propagate Pi = Ai xor Bi carry in will equal carry out here Sum and Carry can be reexpressed in terms of generate/propagate: Si = Ai xor Bi xor Ci = Pi xor Ci Ci+ = Ai Bi + Ai Ci + Bi Ci Ci Pi Si = Ai Bi + Ci (Ai + Bi) = Ai Bi + Ci (Ai xor Bi) = Gi + Ci Pi Gi Ci Pi Ci+

70 All Carries in Parallel Reexpress the carry logic for each of the bits: C = G + P C C2 = G + P C = G + P G + P P C C3 = G2 + P2 C2 = G2 + P2 G + P2 P G + P2 P P C C4 = G3 + P3 C3 = G3 + P3 G2 + P3 P2 G + P3 P2 P G + P3 P2 P P C Each of the carry equations can be implemented in a two-level logic network Variables are the adder inputs and carry in to stage!

71 CLA Implementation Ai Bi Ci gate delay 2 gate delays Adder with Propagate and Generate Outputs gate delay Increasingly complex logic C P G C P P G P G C C2 C P P P2 G P P2 G P2 G2 C3 C P P P2 P3 G P P2 P3 G P2 P3 G2 P3 C4 G3

72 How do we extend this to larger adders? A 5-2 B 5-2 A -8 B -8 A 7-4 B 7-4 A 3- B S 5-2 S -8 S 7-4 S 3- Faster carry propagation 4 bits at a time But still linear Can we get to log? Compute propagate and generate for each adder BLOCK

73 Cascaded Carry Lookahead C 6 A [5-2] B [5-2] C A [-8] B [-8] 4-bit Adder 2 C A [7-4] B [7-4] 4-bit Adder 8 C A [3-] B [3-] 4 C 4-bit Adder 4-bit Adder P G P G P G P S [5-2] S [-8] S [7-4] @3 C C 4 P 3 G 3 C 3 P 2 G 2 C 2 P G C P G Lookahead Carry Unit C P 3- G @5 4 bit adders with internal carry lookahead second level carry lookahead unit, extends lookahead to 6 bits One more level to 64 bits

74 Trade-offs in combinational logic design Time vs. Space Trade-offs Doing things fast requires more logic and thus more space Example: carry lookahead logic Simple with lots of gates vs complex with fewer Arithmetic Logic Units Critical component of processor datapath Inner-most "loop" of most computer instructions

75 2s comp. Overflow Detection Overflow Overflow No overflow No overflow Overflow occurs when carry in to sign does not equa l carry out

76 2s Complement Adder/Subtractor A 3 B 3 B 3 A 2 B 2 B 2 A B B A B B Sel Sel Sel Sel A B A B A B A B CO + CI CO + CI CO + CI CO + CI Add/Subtract S S S S S 3 S 2 S S Overflow A - B = A + (-B) = A + B +

77 Summary Circuit design for unsigned addition Full adder per bit slice Delay limited by Carry Propagation» Ripple is algorithmically slow, but wires are short Carry select Simple, resource-intensive Excellent layout Carry look-ahead Excellent asymptotic behavior Great at the board level, but wire length effects are significant on chip Digital number systems How to represent negative numbers Simple operations Clean algorithmic properties 2s complement is most widely used Circuit for unsigned arithmetic Subtract by complement and carry in Overflow when cin xor cout of sign-bit is

78 Basic Arithmetic and the ALU Now Integer multiplication» Booth s algorithm Integer division» Restoring, non-restoring Floating point representation Floating point addition, multiplication

79 Multiplication Flashback to 3 rd grade Multiplier Multiplicand x Partial products Final sum Base : 8 x 9 = 72 PP: = 72 How wide is the result? log(n x m) = log(n) + log(m) 32b x 32b = 64b result

80 Combinational Multiplier Generating partial products 2: mux based on multiplier[i] selects multiplicand or x 32 partial products (!) Summing partial products Build Wallace tree of CSA

81 Carry Save Adder A + B => S Save carries A + B => S, C out Use C in A + B + C => S, S2 (3# to 2# in parallel) Used in combinational multipliers by building a Wallace Tree c b a CSA c s

82 Wallace Tree f e d c b a CSA CSA CSA CSA

83 Multicycle Multipliers Combinational multipliers Very hardware-intensive Integer multiply relatively rare Not the right place to spend resources Multicycle multipliers Iterate through bits of multiplier Conditionally add shifted multiplicand

84 Multiplier (F4.25) x

85 Multiplier (F4.26) Start M ultiplier =. Test M ultip lier M ultiplier = a. Add m ultiplicand to product and place the result in Product register x 2. Shift the M ultiplicand register left bit 3. Shift the M ultiplier register right bit 32nd repetition? N o: < 32 repetitions Yes: 32 repetitions Done

86 Multiplier Improvements Do we really need a 64-bit adder? No, since low-order bits are not involved Hence, just use a 32-bit adder» Shift product register right on every step Do we really need a separate multiplier register? No, since low-order bits of 64-bit product are initially unused Hence, just store multiplier there initially

87 Multiplier (F4.3) x Multiplicand 32 bits 32-bit ALU Product 64 bits Shift right Write Control test

88 Multiplier (F4.32) Start Product =. Test P roduct Product = a. Add m ultiplicand to the left half of the product and place the result in the left half of the Product register x 2. Shift the Product register right bit 32nd repetition? No: < 32 repetitions Yes: 32 repetitions D one

89 Signed Multiplication Recall For p = a x b, if a< or b<, then p < If a< and b<, then p > Hence sign(p) = sign(a) xor sign(b) Hence Convert multiplier, multiplicand to positive number with (n-) bits Multiply positive numbers Compute sign, convert product accordingly Or, Perform sign-extension on shifts for F4.3 design Right answer falls out

90 Booth s Encoding Recall grade school trick When multiplying by 9: E.g.» Multiply by (easy, just shift digits left)» Subtract once» x 9 = x ( ) = » Converts addition of six partial products to one shift and one subtraction Booth s algorithm applies same principle Except no 9 in binary, just and So, it s actually easier!

91 Booth s Encoding Search for a run of bits in the multiplier E.g. has a run of 2 bits in the middle Multiplying by (6 in decimal) is equivalent to multiplying by 8 and subtracting twice, since 6 x m = (8 2) x m = 8m 2m Hence, iterate right to left and: Subtract multiplicand from product at first Add multiplicand to product after first Don t do either for bits in the middle

92 Booth s Algorithm Current bit Bit to right Explanation Example Operation Begins run of Subtract Middle of run of Nothing End of a run of Add Middle of a run of Nothing

93 Integer Division Again, back to 3 rd grade Quotient Divisor Dividend - - Remainder

94 Integer Division How does hardware know if division fits? Condition: if remainder divisor Use subtraction: (remainder divisor) OK, so if it fits, what do we do? Remainder n+ = Remainder n divisor What if it doesn t fit? Have to restore original remainder Called restoring division

95 Start Integer Division (F4.4). Subtract the Divisor register from the Remainder register and place the result in the Remainder register Remainder > Test Remainder Remainder < 2a. Shift the Quotient register to the left, setting the new rightmost bit to Quotient 2b. Restore the original value by adding the Divisor register to the Remainder register and place the sum in the Remainder register. Also shift the Quotient register to the left, setting the new least significant bit to Divisor Dividend - 3. Shift the Divisor register right bit - 33rd repetition? No: < 33 repetitions Remainder Yes: 33 repetitions Done

96 Integer Division Divisor - Quotient Dividend - Remainder Divisor Shift right 64 bits 64-bit ALU Quotient Shift left 32 bits Remainder 64 bits Write Control test

97 Division Improvements Skip first subtract Can t shift into quotient anyway Hence shift first, then subtract» Undo extra shift at end Hardware similar to multiplier Can store quotient in remainder register Only need 32b ALU» Shift remainder left vs. divisor right

98 S ta rt Improved Divider (F4.4). S h ift th e R e m a in d e r re g is t e r le ft b it 2. S u b tra c t t h e D iv is o r re g is te r fro m th e le f t h a lf o f th e R e m a in d e r re g is te r a n d p la c e th e re s u lt in th e le ft h a lf o f th e R e m a in d e r re g is te r R e m a in d e r > T e s t R e m a in d e r R e m a in d e r < 3 a. S h ift th e R e m a in d e r re g is te r to th e le ft, s e ttin g th e n e w rig h tm o s t b it to 3 b. R e s to re th e o rig in a l v a lu e b y a d d in g th e D iv is o r re g is te r to th e le ft h a lf o f th e R e m a in d e r re g is te r a n d p la c e th e s u m in th e le ft h a lf o f th e R e m a in d e r re g is te r. A ls o s h ift th e R e m a in d e r re g is te r to th e le ft, s e t tin g th e n e w rig h tm o s t b it to 3 2 n d re p e titio n? N o : < 3 2 re p e titio n s Y e s : 3 2 r e p e titio n s D o n e. S h ift le ft h a lf o f R e m a in d e r rig h t b it

99 Improved Divider (F4.4) Divisor 32 bits 32-bit ALU Remainder 64 bits Shift right Shift left Write Control test

100 Further Improvements Division still takes: 2 ALU cycles per bit position» to check for divisibility (subtract)» One to restore (if needed) Can reduce to cycle per bit Called non-restoring division Avoids restore of remainder when test fails

101 Non-restoring Division Consider remainder to be restored: R i = R i- d < Since R i is negative, we must restore it, right? Well, maybe not. Consider next step i+: R i+ = 2 x (R i ) d = 2 x (R i d) + d Hence, we can compute R i+ by not restoring R i, and adding d instead of subtracting d Same value for R i+ results Throughput of bit per cycle

102 NR Division Example Iteration Step Divisor Remainder Initial values Shift rem left 2: Rem = Rem - Div 3b: Rem < (add next), sll 2 2: Rem = Rem + Div 3b: Rem < (add next), sll 3 2: Rem = Rem + Div 3a: Rem > (sub next), sll 4 Rem = Rem Div Rem > (sub next), sll Shift Rem right by

103 2 s Complement Number line : N = N- non-negatives 2 N- negatives one zero how many positives?......

104 So what about subtraction? Develop subtraction circuit using the same process Truth table for each bit slice Borrow in from slice of lesser significance Borrow out to slice of greater significance Very much like carry chain Homework exercise

105 Finite representation? What happens when A + B > 2 N -? Overflow Detect?» Carry out What happens when A - B <? Negative numbers? Borrow out?

106 Number Systems Desirable properties: Efficient encoding (2 n bit patterns. How many numbers?) Positive and negative» Closure (almost) under addition and subtraction Except when overflow» Representation of positive numbers same in most systems» Major differences are in how negative numbers are represented Efficient operations» Comparison: =, <, >» Addition, Subtraction» Detection of overflow Algebraic properties?» Closure under negation?» A == B iff A B == Three Major schemes: sign and magnitude ones complement twos complement (excess notation)

107 Booth s Encoding Really just a new way to encode numbers Normally positionally weighted as 2 n With Booth, each position has a sign bit Can be extended to multiple bits Binary + - -bit Booth bit Booth

108 2-bits/cycle Booth Multiplier For every pair of multiplier bits If Booth s encoding is -2» Shift multiplicand left by, then subtract If Booth s encoding is -» Subtract If Booth s encoding is» Do nothing If Booth s encoding is» Add If Booth s encoding is 2» Shift multiplicand left by, then add

109 Booth s Example Negative multiplicand: -6 x 6 = -36 x, in Booth s encoding is +- Hence: x x x x + Final Sum: (-36)

110 Booth s Example Negative multiplier: -6 x -2 = 2 x, in Booth s encoding is - Hence: x x x x Final Sum: (-2)