
26  Chapter  1.  Complexity and Approximation 
and  v  is  the  constant  for  ROB3SAT.  We  have  to  define  f,  g,  and  a  so  that 
(f, g, a)  is an AP-reduction from A  to MAX3SAT;  we write  B  for MAX3SAT in 
the following. We have to show that  OPTB (x)  OPTa (x) 
v~(~)  ~< r  implies  -a(~)  ~< l+c~(r--1) for 
a  constant  a  that  we will specify later.  We distinguish  two cases. First,  assume 
that  1 +  c~(r  -  1)  />  c  holds;  in  this  case  the  approximation  given  by 
MA 
is 
already good  enough.  So,  we see it  is  sufficient  to use the  solution  ~  that 
MA 
yields for the  input  x  to fulfill  the  property of an  AP-reduction  and  we define 
g(x, ~1, r) 
:= ~  in this case. 
Let now d  := 1 +  a(r -  1)  <  c hold.  Since we have the approximation algorithm 
MA 
with performance ratio c that yields a solution & for an input x  we know that 
the  value of an optimal solution 
OPTA(X) 
is in the interval  [VA(~), CVA(~)].  We 
divide this interval in k  :=  [log d c]  subintervals with lower bounds 
divA(~)  and 
upper  bounds 
di+lvA(5C) 
for 0  x<  i  <  k.  Since 
A  E  APX  C_ AfPO 
the  problem 
of deciding  whether  the  value  of an  optimal  solution  is  at  least  q  belongs  to 
AlP  for  all  constants  q.  We  now  have  k  AlP-decision  problems  such  that  the 
"membership  proof"  we  mentioned  in  the  discussion  of  Cook's  theorem  is  a 
solution for our optimization problem A  with value at least 
divA(~). 
Using the 
second  tool  mentioned  above we can  compute  in  polynomial time  an  instance 
for  c-ROB3SAT  qo  with  m  clauses  that  is  equivalent  to  the  i-th  A/'T'-decision 
problem  for  all  0  ~<  i  <  k.  Given  an  assignment  Ti  that  satisfies  more  than 
(1 -  e)m  clauses  we  can  compute  in  polynomial time  a  satisfying  assignment 
for  qoi  and  use  this  to  compute  a  solution 
s  E  SA(X) 
to  the  i-th AlP-decision 
problem with 
VA(S)  >1 divA(SC). 
We assume without loss of generality that each instance qoi has the same number 
m  of clauses;  using  an  appropriate  number  of copies  of ~i  suffices to  achieve 
this.  With  io  we  denote  the  maximum  index  i  such  that  qoi  is  satisfiable;  we 
k--1 
have 
diOvA(~)  <<, 
OPTA(X)  • 
d/~  then.  We  define  9  :=  Ai=0  ~i  as 
the  instance  for  MAX3SAT.  Assume  that  T  is  an  assignment  for  9  such  that 
OPTB(~IJ )  ~  rVB(T ) 
holds.  We  can  use  T  as  an  assignment  of qoi  for  all  i  by 
ignoring all assignments for variables that  do not occur in ~i. If T satisfies more 
than  (1  -  e)m clauses  in  qoi  for some i  then  qo,  is satisfiable.  We know that  we 
can  compute  in  polynomial  time  a  satisfying  assignment  for  qoi; from  Cook's 
theorem we know that this assignment can be used to compute deterministically 
in polynomial time a membership proof for the input to II, in this case this proof 
is a  solution for the problem of finding a  solution to our optimization problem A 
with value at least 
divA 
(~). Given the assignment T we are able to compute such 
a  solution  deterministically  in  polynomial time.  It follows, that  if we can show 
that T satisfies at least (1--e)m clauses in ~i0 we can compute in polynomial time 
a  solution  with value at least 
di~ 
that has ratio at most d  =  1 +  a(r -  1). 
So, the defined reduction would indeed be an AP-reduction and the proof would 
be complete. 
We are still free to define the constant a  and will use this freedom to show that 
~- satisfies more than  (1 -e)m  clauses in qo~ o. We know that 
OPTB(~I ~) ~  rUB(T) 
holds 
so  OPTB(~I t)  --VB(7") 
<~  ~ 
OPTB(~I t) 
~ 
r~rlkm 
follows.  We  call  the