june 2007 computer arithmetic, function evaluationslide 1 vi function evaluation topics in this part...
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June 2007 Computer Arithmetic, Function Evaluation Slide 1
VI Function Evaluation
Topics in This PartChapter 21 Square-Rooting Methods
Chapter 22 The CORDIC Algorithms
Chapter 23 Variation in Function Evaluation
Chapter 24 Arithmetic by Table Lookup
Learn hardware algorithms for evaluating useful functions• Divisionlike square-rooting algorithms• Evaluating sin x, tanh x, ln x, . . . by series expansion• Function evaluation via convergence computation• Use of tables: the ultimate in simplicity and flexibility
June 2007 Computer Arithmetic, Function Evaluation Slide 2
23 Variations in Function Evaluation
Chapter Goals
Learning alternate computation methods (convergence and otherwise) for somefunctions computable through CORDIC
Chapter Highlights
Reasons for needing alternate methods: Achieve higher performance or precision Allow speed/cost tradeoffsOptimizations, fit to diverse technologies
June 2007 Computer Arithmetic, Function Evaluation Slide 3
Variations in Function Evaluation: Topics
Topics in This Chapter
23.1. Additive / Multiplicative Normalization
23.2. Computing Logarithms
23.3. Exponentiation
23.4. Division and Square-Rooting, Again
23.5. Use of Approximating Functions
23.6. Merged Arithmetic
June 2007 Computer Arithmetic, Function Evaluation Slide 4
23.1 Additive / Multiplicative Normalization
u (i+1) = f(u
(i), v (i), w
(i))
v (i+1) = g(u
(i), v (i), w
(i))
w (i+1) = h(u
(i), v (i), w
(i))
u (i+1) = f(u
(i), v (i))
v (i+1) = g(u
(i), v (i))
Additive normalization: Normalize u via addition of terms to it
Constant
Desiredfunction
Guide the iteration such that one of the values converges to a constant (usually 0 or 1); this is known as normalization
The other value then converges to the desired function
Multiplicative normalization: Normalize u via multiplication of terms
Additive normalization is more desirable, unless the multiplicative terms are of the form 1 ± 2a (shift-add) or multiplication leads to much faster convergence compared with addition
June 2007 Computer Arithmetic, Function Evaluation Slide 5
Convergence Methods You Already Know
CORDICExample of additive normalization x(i+1) = x(i) – di y(i)
2–i
y(i+1) = y(i) + di x(i) 2–i
z(i+1) = z(i) – di e(i)
Division by repeated multiplicationsExample of multiplicative normalization
d (i+1) = d
(i) (2 d
(i)) Set d (0) = d; iterate until d
(m) 1
z (i+1) = z
(i) (2 d
(i)) Set z (0) = z; obtain z/d = q z
(m)
Force y or z to 0 byadding terms to it
Force d to 1 bymultiplying terms with it
June 2007 Computer Arithmetic, Function Evaluation Slide 6
23.2 Computing Logarithms
x (i+1) = x
(i) c
(i) = x (i)
(1 + di 2–i)
y (i+1) = y
(i) – ln c (i) = y
(i) – ln(1 + di 2–i)
Read out from table
Force x (m) to 1
y (m) converges to y + ln x
di {1, 0, 1}
0
Why does this multiplicative normalization method work?
x (m) = x c
(i) 1 c
(i) 1/x
y (m) = y – ln c
(i) = y – ln (c (i)) = y – ln(1/x) y + ln x
If starting with y = 0, y(m) = lnx
x (0) = x, y
(0) =y
June 2007 Computer Arithmetic, Function Evaluation Slide 7
23.2 Computing Logarithms
Convergence domain: 1/(1 + 2–i) x 1/(1 – 2–i) or 0.21 x 3.45
Number of iterations: k, for k bits of precision; for large i, ln(1 2–i) 2–i
Use directly for x [1, 2).
For x = 2q s, we have:
ln x = q ln 2 + ln s = 0.693 147 180 q + ln s
Radix-4 version can be devised
For log2x = log2(2qs) = q+log2s
= q +log2e x ln s
=q + 1.442 695 041 x ln s
June 2007 Computer Arithmetic, Function Evaluation Slide 8
Computing Binary Logarithms via Squaring 1
For x [1, 2), log2 x is a fractional number y = (. y–1y–2y–3 . . . y–l)two
x = 2y = 2
x 2 = 22y = 2 y–1 = 1 iff x
2 2
(. y–1y–2y–3 . . . y–l)two
(y–1. y–2y–3 . . . y–l)two
Once y–1 has been determined,
if y–1 = 0, we are back at the original situation;
otherwise, divide both sides of the equation above by 2 to get:
x 2/2 = 2 /2 = 2
(1 . y–2y–3 . . . y–l)two (. y–2y–3 . . . y–l)two
June 2007 Computer Arithmetic, Function Evaluation Slide 9
Computing Binary Logarithms via Squaring 2
The complete procedure for computing log2x for 1≦x≦2 is :
For i from 1 to l do x←x2
if x ≧ 2 then y-i = 1 ; x←x/2else y-i = 0
endifendfor
June 2007 Computer Arithmetic, Function Evaluation Slide 10
Computing Binary Logarithms via Squaring 3
log x
Squarer
Initialized to x
value 2 iff this bit is 1
2
Radix Shift 0 1
Point
Fig. 23.1 Hardware elements needed for computing log2 x.
Generalization to base b:
x = b , y–1 = 1 iff x 2 b(. y–1y–2y–3 . . . y–l)two
June 2007 Computer Arithmetic, Function Evaluation Slide 11
23.3 Exponentiation 1
x (i+1) = x
(i) – ln c
(i) = x (i)
– ln(1 + di 2–i)
y (i+1) = y
(i) c
(i) = y (i)
(1 + di 2–i)
Read out from table
Force x (m) to 0
y (m) converges to y ex
di {1, 0, 1}1
Why does this additive normalization method work?
x (m) = x – ln c
(i) 0 ln c (i)
x
y (m) = y c
(i) = y exp(ln c (i)) = y exp( ln c
(i)) y ex
If starting from y = 1y
(m) = ex
Computing ex
June 2007 Computer Arithmetic, Function Evaluation Slide 12
23.3 Exponentiation 2
Convergence domain: ln (1 – 2–i) x ln (1 + 2–i) or –1.24 x 1.56
Number of iterations: k, for k bits of precision; for large i, ln(1 2–i) 2–i
Can eliminate half the iterations becauseln(1 + ) = – 2/2 + 3/3 – . . . for 2 < ulpand we may write y
(k) = y
(k/2) (1 + x
(k/2))
Radix-4 version can be devised
June 2007 Computer Arithmetic, Function Evaluation Slide 13
General Exponentiation, or Computing xy 1
x y = (e
ln x) y
= e y ln x So, compute natural log, multiply, exponentiate
When y is an integer, we can exponentiate by repeated multiplication (need to consider only positive y; for negative y, compute reciprocal)
In particular, when y is a constant, the methods used are reminiscent of multiplication by constants (Section 9.5)
June 2007 Computer Arithmetic, Function Evaluation Slide 14
General Exponentiation, or Computing xy 2
Example: x 25 = ((((x)2x)2)2)2x [4 squarings and 2 multiplications]
Noting that 25 = (1 1 0 0 1)two, leads to a general procedure
Computing x y, when y is an unsigned integer
Initialize the partial result to 1 Scan the binary representation of y, starting at its MSB, and repeat If the current bit is 1, multiply the partial result by x If the current bit is 0, do not change the partial result Square the partial result before the next step (if any)
25 = (1 1 0 0 1)two
x 25 = ((((1x)2x)2 )2 )2x1 1 0 0 1
June 2007 Computer Arithmetic, Function Evaluation Slide 15
Faster Exponentiation via Recoding 1
Example: x 31 = ((((x)2x)2x)2x)2x [4 squarings and 4 multiplications]
Note that 31 = (1 1 1 1 1)two = (1 0 0 0 01)two
x 31 = (((((1x)2)2)2)2)2 / x [5 squarings and 1 division]
Computing x y, when y is an integer encoded in BSD format
Initialize the partial result to 1 Scan the binary representation of y, starting at its MSB, and repeat If the current digit is 1, multiply the partial result by x If the current digit is 0, do not change the partial result If the current digit is 1, divide the partial result by x Square the partial result before the next step (if any)
1 0 0 0 01
June 2007 Computer Arithmetic, Function Evaluation Slide 16
Faster Exponentiation via Recoding 2
Radix-4 example: 31 = (1 1 1 1 1)two = (1 0 0 0 01)two = (2 0 1)four
x 31 = ((x2)4)4 / x
(2 0 1)four
x 31 = ((1x2)4 )4 / x
June 2007 Computer Arithmetic, Function Evaluation Slide 17
23.4 Division and Square-Rooting, Again
s (i+1) = s
(i) –
(i) d
q (i+1) = q
(i) + (i)
Computing q = z / d In digit-recurrence division, (i) is the next
quotient digit and the addition for q turns into concatenation; more generally,
(i) can be any estimate for the difference between the partial quotient q
(i) and the final quotient q
Because s (i)
becomes successively smaller as it converges to 0, scaled versions of the recurrences above are usually preferred. In the following, s
(i) stands for s
(i) r
i and q (i) for q
(i) r
i :
s (i+1) = rs
(i) –
(i) d Set s
(0) = z and keep s (i) bounded
q (i+1) = rq
(i) + (i)
Set q (0) = 0 and find q * = q
(m) r
–m
In the scaled version, (i)
is an estimate for r (r i–m
q – q (i)) = r (r
i q * - q
(i)), where q * = r
–m q represents the true quotient
June 2007 Computer Arithmetic, Function Evaluation Slide 18
Square-Rooting via Multiplicative Normalization
Idea: If z is multiplied by a sequence of values (c (i))2, chosen so that the
product z (c (i))2 converges to 1, then z c
(i) converges to z
x (i+1) = x
(i) (1 + di 2–i)2 = x
(i) (1 + 2di 2–i + di
2 2–2i) x
(0) = z, x
(m) 1
y (i+1) = y
(i) (1 + di 2–i) y
(0) = z, y
(m) z
What remains is to devise a scheme for choosing di values in {–1, 0, 1}
di = 1 for x (i)
< 1 – = 1 – 2–i di = –1 for x (i)
> 1 + = 1 + 2–i
To avoid the need for comparison with a different constant in each step, a scaled version of the first recurrence is used in which u
(i) = 2i (x (i) – 1):
u (i+1) = 2(u
(i) + 2di) + 2–i+1(2di u (i) + di
2) + 2–2i+1di2
u (i) u
(0) = z – 1, u
(m) 0
y (i+1) = y
(i) (1 + di 2–i) y
(0) = z, y
(m) z
Radix-4 version can be devised: Digit set [–2, 2] or {–1, –½, 0, ½, 1}
June 2007 Computer Arithmetic, Function Evaluation Slide 19
Square-Rooting via Additive Normalization
Idea: If a sequence of values c (i) can be obtained such that z – (c
(i))2 converges to 0, then c
(i) converges to z
x (i+1)
= z – (y (i+1))2 = z – (y
(i) + c
(i))2 = x (i)
+ 2di y (i)
2–i – di2
2–2i x
(0) = z, x
(m) 0
y (i+1) = y
(i) + c
(i) = y
(i) – di 2–i y
(0) = 0, y
(m) z
What remains is to devise a scheme for choosing di values in {–1, 0, 1}
di = 1 for x (i)
< – = – 2–i di = –1 for x (i)
> + = + 2–i
To avoid the need for comparison with a different constant in each step, a scaled version of the first recurrence may be used in which u
(i) = 2i x (i):
u (i+1) = 2(u
(i) + 2di y (i) – di
2 2–i
) u (0)
= z , u (i) bounded
y (i+1) = y
(i) – di 2–i y
(0) = 0, y
(m) z
Radix-4 version can be devised: Digit set [–2, 2] or {–1, –½, 0, ½, 1}
June 2007 Computer Arithmetic, Function Evaluation Slide 20
23.5 Use of Approximating Functions
Convert the problem of evaluating the function f to that of function g approximating f, perhaps with a few pre- and postprocessing operations
Approximating polynomials need only additions and multiplications
Polynomial approximations can be derived from various schemes
The Taylor-series expansion of f(x) about x = a is
f(x) = j=0 to f (j)
(a) (x – a) j / j!
The error due to omitting terms of degree > m is:
f (m+1)
(a + (x – a)) (x – a)m+1 / (m + 1)! 0 < < 1
Setting a = 0 yields the Maclaurin-series expansion
f(x) = j=0 to f (j)
(0) x j / j!
and its corresponding error bound:
f (m+1)
(x) xm+1 / (m + 1)! 0 < < 1
Efficiency in computation can be gained via Horner’s method and incremental evaluation
June 2007 Computer Arithmetic, Function Evaluation Slide 21
Some Polynomial Approximations (Table 23.1)–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––Func Polynomial approximation Conditions–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
1/x 1 + y + y 2 + y
3 + . . . + y i + . . . 0 < x < 2, y = 1 – x
ex 1 + x /1! + x 2/2! + x
3/3! + . . . + x i /i ! + . . .
ln x –y – y 2/2 – y
3/3 – y 4/4 – . . . – y
i /i – . . . 0 < x 2, y = 1 – x
ln x 2 [z + z 3/3 + z
5/5 + . . . + z 2i+1/(2i + 1) + . . . ] x > 0, z =
x–1x+1
sin x x – x 3/3! + x
5/5! – x 7/7! +
. . . + (–1)i
x2i+1/(2i + 1)! + . . .
cos x 1 – x 2/2! + x
4/4! – x 6/6! +
. . . + (–1)i
x2i /(2i )! +
. . .
tan–1 x x – x
3/3 + x 5/5 – x
7/7 + . . .
+ (–1)i x2i+1/(2i + 1) +
. . . –1 < x < 1
sinh x x + x 3/3! + x
5/5! + x 7/7! +
. . . + x2i+1/(2i + 1)! +
. . .
cosh x 1 + x 2/2! + x
4/4! + x 6/6! +
. . . + x2i
/(2i )! + . . .
tanh–1x x + x 3/3 + x
5/5 + x 7/7 +
. . . + x2i+1/(2i + 1) +
. . . –1 < x < 1–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
June 2007 Computer Arithmetic, Function Evaluation Slide 22
Function Evaluation via Divide-and-Conquer
Let x in [0, 4) be the (l + 2)-bit significand of a floating-point number or its shifted version. Divide x into two chunks x H and x L:
x = x H + 2–t x L
0 x H < 4 t + 2 bits
0 x L < 1 l – t bits
t bits
x H in [0, 4) x L in [0, 1)
The Taylor-series expansion of f(x) about x = x H is
f(x) = j=0 to f (j)
(x H) (2–t x L)
j / j!
A linear approximation is obtained by taking only the first two terms
f(x) f (x H) + 2–t x L f (x H)
If t is not too large, f and/or f (and other derivatives of f, if needed) can be evaluated via table lookup
June 2007 Computer Arithmetic, Function Evaluation Slide 23
Approximation by the Ratio of Two Polynomials
Example, yielding good results for many elementary functions
f(x)
a(5)x5 + a(4)x4 + a(3)x3 + a(2)x2 + a(1)x + a(0)
b(5)x5 + b(4)x4 + b(3)x3 + b(2)x2 + b(1)x + b(0)
Using Horner’s method, such a “rational approximation” needs 10 multiplications, 10 additions, and 1 division
June 2007 Computer Arithmetic, Function Evaluation Slide 24
23.6 Merged ArithmeticOur methods thus far rely on word-level building-block operations such as addition, multiplication, shifting, . . .
Sometimes, we can compute a function of interest directly without breaking it down into conventional operations
Example: merged arithmetic for inner product computation
z = z (0) + x
(1) y
(1) + x (2)
y (2) + x
(3) y
(3)
x(1) y(1)
x(3) y(3)
x(2) y(2)
z(0)
Fig. 23.2 Merged-arithmetic computation of an inner product followed by accumulation.
June 2007 Computer Arithmetic, Function Evaluation Slide 25
Example of Merged Arithmetic Implementation
Example: Inner product computation
z = z (0) + x
(1) y
(1) + x (2)
y (2) + x
(3) y
(3)
x(1) y(1)
x(3) y(3)
x(2) y(2)
z(0)
Fig. 23.3 Tabular representation of the dot matrix for inner-product computation and its reduction.
1 4 7 10 13 10 7 4 16 FAs 2 4 6 8 8 6 4 2 10 FAs + 1 HA 3 4 4 6 6 3 3 1 9 FAs1 2 3 4 4 3 2 1 1 4 FAs + 1 HA1 3 2 3 3 2 1 1 1 3 FAs + 2 HAs2 2 2 2 2 1 1 1 1 5-bit CPA
Fig. 23.2
June 2007 Computer Arithmetic, Function Evaluation Slide 26
Another Merged Arithmetic Example
Approximation of reciprocal (1/x) and reciprocal square root (1/x) functions with 29-30 bits of precision, so that a long floating-point result can be obtained with just one iteration at the end [Pine02]
u v w 1.
c Table
b Table
a Table
Squarer Radix-4 Booth
Radix-4 Booth
Partial products gen Partial products gen
9 bits 24 bits 19 bits
30 bits 20 bits 12 bits
16 bits
Multioperand adder
30 bits, carry-save
Double-precision significand f(x) = c + bv + av
2
1 square
Comparable to a multiplier
2 mult’s
2 adds