 
  
  
  
  
 
The problem of deciding whether a finite ordered set has a fixed point free automorphism was found by Williamson to be NP-complete (cf. [132], Fact 5.1 or Theorem 2.14 here), i.e.,
  
 
We will present two proofs of this theorem here. One by Goddard that reduces the problem to a known NP-complete problem, namely the determination if a given graph has a fixed point free automorphism (cf. [79]). The other is Williamson's original adaptation of Lubiw's proof in [79], which reduces the problem to 3SAT. This proof will be sketched only. For the first proof we need:
  
 
Proof 1 of Theorem 2.14:
(T. Goddard, private communication, also cf. [29])
By [79] it is NP-complete to decide if a given graph has a
fixed point free automorphism. We will show that any
algorithm that decides if an
arbitrary
ordered set P has a fixed point free automorphism in
q(|P|) steps
also decides 
if an arbitrary graph G=(V,E)
has a fixed point free automorphism
in q'(|V|) steps,
where q,q' are polynomials. 
 
Let G=(V,E) be a graph. Order the set
  via the order
  via the order   defined as follows:
  defined as follows:
 is reflexive,
  is reflexive, are related,
  are related, and
  and   , then
v<(e,x)  iff
 , then
v<(e,x)  iff   .
 .
 is a fixed point free automorphism, then
  is a fixed point free automorphism, then
  is a fixed point free automorphism of G.
Conversely, if
  is a fixed point free automorphism of G.
Conversely, if   is a fixed point free automorphism
of G, then
let
  is a fixed point free automorphism
of G, then
let
  
 
and
  
 
f is a fixed point free automorphism of P. \
Thus we even know that the problem to decide if an ordered set of height 1 has a fixed point free automorphism is NP-complete (this is also noted as Fact 5.2 in [132]). This is interesting in the light of Theorem 3.21 and Corollary 3.24, which show that there is an efficient algorithm to detect if an ordered set of height 1 has a fixed point free order-preserving self map. Our second proof is a reduction to 3SAT itself.
    
 
Figure 2: The ordered set used in the second proof of Theorem
2.14:
Consider an instance of 3SAT with m distinct clauses
  , each of which contains
exactly 3 distinct literals
 , each of which contains
exactly 3 distinct literals   out of
 
out of   .
The satisfiabiliy component is an antichain of points
 .
The satisfiabiliy component is an antichain of points
  , where abc counts from zero to 7 in binary.
The truth setting component has comparabilities as indicated.
Finally (not indicated in the figure)
the upper covers of
 , where abc counts from zero to 7 in binary.
The truth setting component has comparabilities as indicated.
Finally (not indicated in the figure)
the upper covers of   are
  are
  and
  and   .
 .
Proof 2 of Theorem 2.14:
(cf. [132], Fact 5.1; we present a sketch)
For a given instance C of 3SAT which we can without loss of generality
choose to have m distinct clauses
  each of which has 3 distinct literals
 
each of which has 3 distinct literals
  out of
  out of
  we construct the ordered set P as indicated in Figure 2.
Note that each
 
we construct the ordered set P as indicated in Figure 2.
Note that each   has a unique set of 6 upper covers.
For
  has a unique set of 6 upper covers.
For   , let
 , let
  .
Then it is easy to see that every automorphism
of P must fix the satisfiability component and every
 .
Then it is easy to see that every automorphism
of P must fix the satisfiability component and every   .
Each
 .
Each   has exactly four fixed point free automorphisms, of which only
  has exactly four fixed point free automorphisms, of which only
  and
 
and
  (we remain consistent with the notation in [132] here)
can possibly extend to an automorphism of P
(the other two map sets that have a lower bound in the satisfiability
component to sets that do not have lower bounds).
 
(we remain consistent with the notation in [132] here)
can possibly extend to an automorphism of P
(the other two map sets that have a lower bound in the satisfiability
component to sets that do not have lower bounds).
 
To show that P has a fixed point free automorphism iff
there is a satisfying assignment for C first assume that
  is a fixed point free automorphism.
If
  is a fixed point free automorphism.
If   , set
 , set   TRUE. Otherwise
we have
  TRUE. Otherwise
we have
  , and we set
 , and we set   FALSE.
Now if
  FALSE.
Now if   and
  and   were all false for some j, then
the definitions of
  were all false for some j, then
the definitions of   and
  and   indicate
that
  indicate
that   must interchange
  must interchange   and
  and   ,
 ,
  and
  and   , and
 , and
  and
  and   .
Since these are the upper covers of
 .
Since these are the upper covers of   we conclude that
  we conclude that
  is a fixed point of
  is a fixed point of   , a contradiction.
 , a contradiction.
 
On the other hand if   is a
satisfying assignment for C (with 0 standing for FALSE and
1 standing for TRUE as usual), we define
  is a
satisfying assignment for C (with 0 standing for FALSE and
1 standing for TRUE as usual), we define   as follows:
Set
  as follows:
Set   if
  if   is true (1),
set
  is true (1),
set   if
  if   is false (0) and
set
  is false (0) and
set
  ,
where
 ,
where   denotes addition modulo 2.
Since one of
  denotes addition modulo 2.
Since one of   is 1 (true),
  is 1 (true),
  has no fixed points.
  has no fixed points.
  trivially is an automorphism on the truth setting component.
Since addition of a 1 or a 0 modulo 2 is injective,
  trivially is an automorphism on the truth setting component.
Since addition of a 1 or a 0 modulo 2 is injective,   is
injective and hence bijective on the satisfiability component.
Finally one checks that
  is
injective and hence bijective on the satisfiability component.
Finally one checks that   is order-preserving. \
  is order-preserving. \
Using the characterization in Remark 2.11 we can also record some known results as easy consequences:
  
 
  
 
Proof:
If there was an algorithm that decides
if a given r-partite
graph G=(V,E) with partition   and
 
and   contains an r-clique
in q(|V|) steps, then by
Remark 2.11
there would be
an
algorithm that determines if the
fixed point free automorphism graph of a
given ordered set P has a
|P|-clique in q'(|P|) steps.
Thus we could determine if
a given ordered set P has a fixed point free automorphism in
q''(|P|) steps (q,q',q'' all polynomials).\
 
contains an r-clique
in q(|V|) steps, then by
Remark 2.11
there would be
an
algorithm that determines if the
fixed point free automorphism graph of a
given ordered set P has a
|P|-clique in q'(|P|) steps.
Thus we could determine if
a given ordered set P has a fixed point free automorphism in
q''(|P|) steps (q,q',q'' all polynomials).\
  
 
Proof:
If there was an algorithm that determines
if a given graph
G=(V,E) with   has an r-clique
in q(|V|) steps, then
this algorithm would
decide if a given r-partite
graph G=(V,E) with partition
  has an r-clique
in q(|V|) steps, then
this algorithm would
decide if a given r-partite
graph G=(V,E) with partition   and
 
and   has an r-clique
in q(|V|) steps. \
 
has an r-clique
in q(|V|) steps. \
 
  
  
  
 