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Algorithms for Best L1 and L_infinity Linear Approximations on a...
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一个简单对于有限集合进行线性规划的算法,优化的目标函数是L1和L无穷
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Numerische Mathematik 8, 295--306 (t966)
Algorithms for Best L x and Loo
Linear Approximations on a Discrete Set
IAN BARRODALE and ANDREW YOUNG
Received December 7, t965
Abstract.
This paper supplies algorithms for the best approximation to a real-
valued function, defined as a table of values, by a linear approximating function
in both the L I and Loo norms. The algorithms are modified simplex algorithms which
due to the particular structures of the tableaux have been condensed and require
minimal storage space. Both algorithms are given as ALGOL procedures and sample
times are noted for several examples.
I. Statement of the Problems
Let ](x)
be a given real-vaiued function defined on a finite subset X-----
{xl, x2 ..... xN}
of an interval I on the real line. Given a set of n real-valued
continuous functions ~v/(x) defined on I we form a linear approximating function
F(A, x) = ~ aicp/(x )
for any real set a = {a 1, a, ..... a.}.
i=x
The L 1 approximation problem is to determine A* such that
,,Xxl F(A*, x)- [(x) I
~ZxlF(A, x)-
I( )1
for all A E E..
The Loo (or Chebyshev) approximation problem is to determine A* such that
maxX IFCA *, x) -- I(~)1
-- < max,,~x
l~(a, x) -1(~)1
for all A ~ E~.
We now restate each problem as a linear programming problem of a type
that can be solved by methods that appear to be more efficient than any published
to date. GOLDSTEIN and CHENEY [1, Th. Q., p. 425] and STIEFEL [3, p. 10]
have reduced the Loo approximation problem, while RICE
[2,
p. 182] has converted
the L x approximation problem, each into linear programming problems. How-
ever STIEFEL'S formulation doubles the number of constraints and he remarks
on the
"'tedious trans/ormations"
necessary to reduce the constraints to the
original number; the number of constraints introduced by RICE is 2 t¢ and, as
he says,
"'this approach does not appear to lead to a practical solution o[ the L 1
approximation problem".
We proceed as follows. To ensure the non-negativity of the unknown vari-
ables we put
~+l=max(0,--nfi'naj) and then
og=ai+o~+ x
for
t~j~_n.
1
296 IAN BARRODALE and ANDREW YOUNG:
Then, for
~i~N,
we define
e i = e (xi) = F(A, xl) -- / (xi)
j=l /=1
= ~Xl 91, i -~''" ~- 0in 9•, i + ~.+1 9.+1, i -- /i
n
where 9n+1(xi)=--~,
9i(xi)
for
i<=i<=N,
and we have written 9i, i and /i re-
/=I
spectively for 9i (x~) and ] (xi). Finally putting ei-----u s- vi where ui~ 0 and vi>= 0
we have N constraints in non-negative variables
/i = 0tl 91, i "~''" -[- 0~.+1
9n+l, i -- Ui 27 Vi
for t --< i ~ N. (A)
N
The L: approximation problem is to find {aj} such that ~.[e,[ is minimized
i=1
and this can be achieved by solving the linear programming problem of mini-
N
mizing ~ (ui+ vi) subject to (A).
i=1
The Loo approximation problem is to find {a 3 such that max [e,I is mini-
l<i~_N
mized. For any A EE. we put w = max [ei[ and obtain the 2N constraints
l~_i<N
Oil
91, i ~-*'" -~- O~ft+l ~9"+1, i -~- W ~_~ /i
for
t<_i<_N
(B)
-
~: 9:,i
..... ~.+19.+1,i + w > -/i
This gives the linear programming problem of minimizing w subject to (B).
However we solve, in practice, the dual problem ~ which is to find non-negative
values of s i and ti for t
<i<N
which maximize
N
- t,)
i=l
subject to the n+ 2 constraints
N
9i.i (si -- ti) =< 0 for
and
t<=j<=n+t
N
Y, (st + t,) =< t.
/=I
These constraints are then expressed as equalities using 0t i and
w,
the original
variables of (B), as the slack variables.
II. The L~ Algorithm
N
The objective function ~
(u~+v~)
and the N constraints
i~l
/i=O~l gl, i Af - "'* ~-O~n+ l gn+l,i-- Ui-~ Vi
are arranged in a simplex tableau as shown in Table i.
I Since this paper was prepared, we have noticed that S.
VAJDA [d,
p. 172] has
also given a method of solving the Leo approximation problem as a dual problem,
but his computational procedure differs from ours.
Algorithms for Best L 1 and Loo 297
The names R, oh ..... ~n+l, ul
..... ug, vl ..... v2v are
obvious choices for the
column vectors and we shall not define them formally. Table t is condensed by
storing only the columns R and ~i for t
<j<=n+
t. These n+ 2 vectors are suf-
ficient to enable us to perform the simplex algorithm on Table t. If every ]i>0
an initial basis is
{vl, v ~
..... vz~} and as at any stage
ui=--v ~
for
l~i<--N
it
is clear that if vi is in the basis then ui must be out and need not even be
stored. If any ]i< 0 we divide the row by -- t and replace the corresponding v i
by u i in the initial basis. Furthermore the sum of the marginal costs of u i and
v i is --2 after each iteration and hence one may be deduced from the other.
Table 1,
Full Initial Simplex Tableau/or the L 1 Algorithm
Costs -+ 0 0 0 t 1 t 1 1 t
$ Basis
R o h % ... ,xn+ x u 1 % ... u N v x vs... v N
1 Vl /1 ~91,1 ~%,1 ~%+1,1 --1 0 0 t 0 0
1 v~
/~ 91, ~ 9s,
~ 9n+1,~ 0 -- 1 0 0 1 0
1 VN /N q)l.N q)2,N
~%+1, N 0 0 --1 0 0 1
Marginal N N N ~"
costs i=xF~/i i=Y1'lq~l'i ~=aYl q~2,i i=~ q',,+~,i --2 --2 --2 0 0 0
Table
2. A Worked Example ( Using the Condensed Tableau) Illustrating the L x Algorithm
4.5 t 0 2.000 -- 3.000 1,000
1.520 1.000 0.000 -- 1.000
t.025 1.000 1.000 --2.000
0.475 1.000 2.000 --3.000
o.o~o
1.000
3.000 --4.000
0.475 --1.o00 --4.000 5.000
1.0o5 --1.00o --5.000 6.000
o 7 8 9
0.305 7.000 0,000 --9.000
0.115 2.000 0.000 --3.o00
0.085
1.000
0.000 --2.000
0.465 -- 1.000 -- 1.000 1.000
1.870 --3.000 --1.000 4.000
0.020
2.000
0.000 -- 1.000
0.085 3.000 0.000 --2.000
0 4 8 3
0.165 --7.000 0.000 --7.000
0.055
--3.000
0.000 --4.000
0.045 --2.000 0.000 --3.000
0.485 t.000 -- 1.000 1.000
t.950
4.000 -- t.000 5.000
0.020 t .000 0.000 2.000
0.045 --2.000 0.000 -- t.000
o --5 8 4
o 4.490
1 1.510
2 1.015
3 0.465
4 0.o10
--5 0.485
--6 1.ot5
0
2 0.235
1 0.095
2 0.075
9 0.475
7 1.900
--5 O.OLO
--6 0.055
0
4 0.O73
1 o.o18
2 0.008
9 0.503
7 2.023
--3 0.038
--6 0.008
0
-- 2.000 -- 9.000 9.000 t
-- 1.000 -- 3.000 3.000 1
-- 1.000 --2.000 2.000 2
-- 1.000 -- 1,000 1.000 3
! .000 3.000 --4.000 7
1.000 --l.O00 1.000 --5
1.000 --2.000 2.000 --6
4 8 9
--3.500 0.000 --5.500 3
-- 1.000 0.000 -- 2.000 t
--0.500 0.000 -- 1.500 2
0.500 -- 1.000 0.500 9
1.500 -- t.000 2.500 7
0.500 0.000
--0.500 4
--1.500 0.000 --0.500 --6
--5 8 3
--1.667 0.000 --0.333 5
0.333 0.000 --1.333 5
--0.667 0.000 --0.333 2
0.333 --1.000 --0.333 9
1.333 --1.000 --0.333 7
0.333 0.000 0.667 --3
--0.667 0.000 1.667 --6
1 8 4
Note : After 3 iterations, no positive marginal costs appear in this condensed tableau,
but some o/ the suppressed marginal costs are still positive,
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