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- 1. CS/EE 5830/6830 Arithmetic Processing VLSI ARCHITECTURE AP = (operands, operation, results, conditions, singularities) Operands are: Set of numerical values Range Precision (number of bits) Number Representation System (NRS) Operand: +, -, *, , etc. Conditions: Values of results (zero, neg, etc.) Singularities: Illegal results (overflow, NAN, etc.) Chapter 1 – Basic Number Representations and Arithmetic Algorithms Number Representation Basic Fixed Point NRS Need to map numbers to bits Number represented by ordered n-tuple of symbols (or some other representation, but we’ll use bits) Symbols are digits Representation you choose matters! n-tuple is a digit-vector Complexity of arithmetic operations depends heavily Number of digits in digit-vector is precision on representation! But, be careful of conversions Arithmetic that’s easy in one representation may lose its advantage when you convert back and forth… Digit Values Rule of Interpretation A set of numerical values for the digits Mapping of set of digit-vectors to numbers is the set of possible values for is the cardinality of Binary (cardinality 2) is Decimal (cardinality 10) is {0,1,2,3,4,5,6,7,8,9} (1,3) “Thirteen” Balanced ternary (cardinality 3) is Set of integers represented by a digit-vector with n digits is a finite set with max elements of € Digit-Vectors Numbers (N, Z, R, …) 1
- 2. Mappings… Positional Weighted Systems Integer x represented by digit vector Rule of interpretation Not Useful! Digit-vectors N,R,Z,… Nonredundant Where weight vector is Digit-vectors N,R,Z,… Redundant Digit-vectors N,R,Z,… Ambiguous Radix Number Systems Fixed-Radix Systems Weights are not arbitrary In fixed-radix system all elements of the radix they are related to a radix vector vector have the same value r (the radix) Weight vector is So that So Or Radix 2: Radix 4: Radix 10: Mixed-Radix Systems Canonical Systems Time is the most common… Canonical if Hours, Minutes, Seconds Binary= {0,1} Octal= {0,1,2,3,4,5,6,7) X=(5,37,43) = 20,263 seconds Decimal = {0,1,2,3,4,5,6,7,8,9} 5 x 3600 = 18,000 37 x 60 = 2,220 Range of values with n radix-r digits is 43 x 1 = 43 Total = 20,263 seconds 2
- 3. Non-Canonical Systems Conventional Number Systems Digit set that is non canonical… A system with fixed positive radix r and canonical Non-canonical decimal set of digit values Non-canonical binary Radix-rconventional number system After all this fuss, these are what we’ll mostly worry Redundant if non-canonical about… I.e. binary system with Specifically binary (radix 2) (1,1,0,1) and (1,1,1,-1) both represent “thirteen” We’ll also see some signed-digit redundant binary (carry-save, signed-digit) Aside – Residue Numbers Aside – Residue Numbers Example of a non-radix number system P = (17, 13, 11, 7, 5, 3, 2) Weights are not defined recursively Digit-vector (13 4 8 2 0 0 0) Residue Number System (RNS) uses a set of pairwise Number = “thirty” relatively prime numbers 30 mod 17 = 13 A positive integer x is represented by a vector 30 mod 13 = 4 30 mod 11 = 8 Can allow fast add and multiply Etc… No notion of digits on left being more significant than digits on the right (i.e. no weighting) Lots of Choices… Back to Binary Non-negative integers – digits = {0,1} Range with n bits is Higher power of 2 radix – group bits in to groups with bits X = (1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 1) = ((1,1), (0,0), (0,1), (0,1), (1,1), (0,1)) = ( 3, 0, 1, 1, 3, 1) base 4 (quaternary) = ((1,1,0), (0,0,1), (0,1,1), (1,0,1)) = (6,1,3,5) base 8 (octal) = ((1,1,0,0)(0,1,0,1)(1,1,0,1)) = (C, 5, D) base 16 (hexadecimal) 3
- 4. Signed Integers Transformation Three choices for representing signed ints Transform signed numbers into unsigned, then use Directly in the number system conventional systems Signed digit NRS, i.e. {-1, 0, 1} Use extra symbol to represent the sign Sign and Magnitude “minus two” “six” Additional mapping on positive integers (1,1,0) x xR True and Complement system Signed integer X Positive Integer XR Digit-Vector X Z (signed) N (unsigned) Digit vectors True and Complement System Converse Mapping Signed integers in the range Convert back and forth Negative represented by xR Such that C is the Complementation constant Unambiguous if Mapping is True forms Complement forms Boundary conditions Two standard forms If x R = C /2 can be represented, you can assign it to either Range complement system Representationis no longer symmetric Also called Radix Complement Not closed under sign change operation Two’s complement in radix 2 € If can be represented, then there are two Digit Complement System representations of 0 Also called Diminished Radix Complement One’s complement in radix 2 4
- 5. Two’s Compliment One’s Compliment For n bits in the digit vector, For n bits in the digit vector Example three bits: C = 8 Example three bits: C = 7 is outside the range is representable in 3 bits With 3 bits, you can’t represent 8 Two representations of 0 So, only one representation of 0 can be represented, so you have a cannot be represented choice Symmetric range… Usually choose second for sign detection Range is then Range is then (asymmetric) Examples (n=3 bits) Range Comparison (3 bits) Two’s compliment Decimal Binary Sign & Two’s One’s (unsigned) Magnitude compliment Compliment 7 111 -3 represented as 101 6 110 5 101 4 100 111 represents -1 3 011 011 011 011 One’s compliment 2 010 010 010 010 1 001 001 001 001 0 000 000/100 000 000/111 -3 represented as 100 -1 101 111 110 -2 110 110 101 -3 111 101 100 -4 100 Example: 2’s comp, n=4 Example: 1’s comp, n=4 -1 +0 -0 +0 -2 1111 0000 -1 1111 0000 +1 +1 1110 0001 1110 0001 -3 1101 -2 0010 +2 1101 0010 +2 -4 1100 0011 +3 -3 1100 0011 +3 -5 1011 0100 +4 -4 1011 0100 +4 -6 1010 0101 +5 -5 1010 0101 +5 0110 +6 0110 +6 -7 1001 -6 1001 1000 0111 1000 0111 -8 +7 +7 -7 5
- 6. Converse Mapping (2’s comp) Two’s Comp. Example If n −2 Most significant bit has negative x = −X n −1 2 n −1 + ∑ X i 2 i weight, remaining have positive i=0 If n=5: X = 11011 = -16 + 8 + 0 + 2 + 1 = -5 X = 01011 = 0 + 8 + 0 + 2 + 1 = 11 € n −2 Most significant bit has negative x = −X n −1 2 n −1 + ∑ X i 2 i i=0 weight, remaining have positive € Converse Mapping (1’s comp) One’s Comp. Example… Similar in one’s complement (case for Xn-1=1) n −2 Most significant bit has negative x = −X n −1 (2 n −1 −1) + ∑ X i 2 i weight, remaining have positive. i=0 Weight of MSB is different because C=2n-1. Intuition is that You have to add 1 to jump over The extra representation of 0. n −2 € x = −X n −1 (2 n −1 −1) + ∑ X i 2 i n=5: X = 11011 = -(16-1) + 8 + 0 + 2 + 1 = -4 i=0 X = 01011 = 0 + 8 + 0 + 2 + 1 = 11 Remember this! We’ll use it later when we need to adjust things in arrays of signed addition. Think of partial product arrays in multiplication…. € Sign Bits Addition (unsigned) Conveniently, sign is determined by Adding two n-bit operands results in n+1 result bits high-order bit Usually call the n+1 bit Cout Because (Assuming xR = C/2 is In terms of digit vectors assigned to represent Cout = overflow! x = -C/2) 6
- 7. Addition (Signed) Addition: two’s comp. Assume no overflow for a moment… C=2n, and mod2n means ignore Xn (the carry out)! Makes addition simple – add the numbers and ignore Use the property the carry out xR yR cout + cin zR Addition: one’s comp. Addition: one’s comp. C=rn-1, so mod operation is not as easy If the cout is 1, subtract 2n (ignore cout), and add 1 zR=wRmod(2n-1) (end-around carry) xR yR cout + cin So: Ignore cout zR Change of Sign Change of Sign Start with bitwise negation One’s Complement: Flipevery bit in the digit vector Boolean style: {−A = C − A = r n −1 − A = A if A ≥ 0 Fundamental property of n-digit radix-r Two’s Complement: € + + {−A = C − A = r n − A = A +1 if A ≥ 0 € 7
- 8. Another two’s comp. check Two’s comp subtract Verify the property Use two’s complement definition… yR xR cout + 1 cin zR Two’s comp add/subtract Two’s comp add/subtract yR yR Sub xR xR Sub ab c Overflow? 00 0 cout + cin cout + cin 01 1 10 1 11 0 zR zR Overflow (unsigned) Overflow (signed) Overflow condition means that the result can’t be Still the same definition – the result can’t be represented in n bits represented in n bits For unsigned addition, this simply means that the cout But, now not as easy as looking at cout was 1 For 4 bits, and two’s comp, answer was smaller than –8 For n=4, this means the result was bigger than 15 or larger than 7 10102 (1010) + 11002 (1210) = 101102 (2210) Overflow if (pos) + (pos) = (neg) 5+6=11 or (neg) + neg) = (pos) -5+-6=-11 Can you ever have overflow with (pos) + (neg)? 8
- 9. Example: 2’s comp, n=4 Overflow (signed) -1 +0 Overflow only possible if args are same sign -2 1111 0000 Overflow if result is different sign +1 1110 0001 -3 1101 0010 +2 -4 1100 0011 +3 -5 1011 0100 +4 -6 1010 0101 +5 0110 +6 -7 1001 1000 0111 -8 +7 Overflow (signed) Implied Digits (unsigned) Or, consider all possible cases around MSB… Unsigned numbers have an infinite number of leading 0’s Xn-1 Yn-1 Cn-1 Cn Zn-1 OVF 5,243 = …0,000,000,000,005,234 0 0 0 0 0 No 0 0 1 0 1 Yes 1 1010 = …0 0000 0000 0001 1010 0 1 0 0 1 No Changing from n bits to m bits (m>n) is a simple 0 1 1 1 0 No matter of padding with 0’s to the left 1 0 0 0 1 No 1 0 1 1 0 No 1 1 0 1 0 Yes 1 1 1 1 1 No Changing number of bits (signed) Shifting Signed numbers can be thought of as having infinite Shifting corresponds to multiply and divide by replicas of the sign bit to the left powers of 2 Four bits: Left arithmetic shift Shift in 0’s in LSB, OVF if Right arithmetic shift Eight bits: Divide by 2 (integer result)(1-bit shift) Remember to copy the sign bit in empty MSB! 9
- 10. Multiplication (unsigned) Multiplication (unsigned) Pencil and paper method Compute n terms of and then sum them The ith term requires an i-position shift, and a multiplication of x by the single digit Yi Requires n-1 adders Multiplication (unsigned) Multiplication (unsigned) B0 B1 B2 B3 Multiplication (unsigned) Serial Multiplicaton (unsigned) Instead of using n-1 adders, can iterate with 1 takes n steps for n bits 10
- 11. Multiplication (signed!) Division (unsigned) Remember that MSB has negative weight x=qd+w (quotient, divisor, remainder) Add partial products as normal Consider 0<d, x<rnd (precludes /0 and OVF) Subtract multiplicand in last step… Basic division is n iterations of the recurrence w[0] = x w[ j +1] = rw[ j] − d ∗qn −1− j j = 0,...,n −1 n −1 where q = ∑ qi r i and d ∗ = dr n i=0 i.e. divisor is aligned with most-significant half of € residual Division (unsigned) Long Division In each step of the iteration Get one digit of quotient Value of digit is bounded such that This means you find the right digit such that the current remainder is less than (shifted) divisor In binary you only have to guess 1 or 0 Guess 1 and fix if you’re wrong (restoring) Restoring Division Restoring Division 1. Shift current result one bit left 2. Subtract divisor from this result 3. If the result of step 2 is neg, q=0, else q=1 4. If the result of step 2 is neg, restore old value of result by adding divisor back 5. Repeat n times… This is what the recurrence in the book says… 11
- 12. Restoring Division Restoring Division Example Shift Subtract divisor (add negative) use tentative partial residual to decide on quotient bit partial residual was negative, so restore by adding divisor back in “real” partial residual is starting point for next iteration Shift Subtract divisor (add negative) use tentative partial residual to decide on quotient bit partial residual was positive, so residual is correct – no restoration needed Shift Subtract divisor (add negative) use tentative partial residual to decide on quotient bit partial residual was negative, so restore by adding divisor back in “real” partial residual is starting point for next iteration Shift Subtract divisor (add negative) use tentative partial residual to decide on quotient bit partial residual was positive, so residual is correct – no restoration needed Non-performing Division Non-restoring Division Consider what happens Consider again Result at each step is 2r-d (r is current result) At each step 2residual-d If the result is negative, we restore by adding d back in Ifit’s negative, restore to 2r by adding d back in But, if you store the result in a separate place and don’t Then shift to get 4r, then subtract getting 4r-d update the result until you know if it’s negative, then you Suppose you don’t restore, but continued with the shift can save some restoring steps resulting in 4r-2d Now add d instead of subtract resulting in 4r-d That’s what you wanted! Non-restoring Division Non-restoring Division Example For positive partial residual – subtract divisor For negative partial residual – add divisor back in This corrects for the mistake you made on the last iteration… If the last residual is negative – do one final restoration 12
- 13. Whew! Basic number representation systems Unsigned, signed Conversions Basic addition, subtraction of signed numbers Multiplication of unsigned and signed Division of signed Now let’s speed up the operations! 13