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Combinatorics

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  • Combinatorics
  • Stirling permutation

Symbols

  • CanonicalMultiset
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PeterBurbery`Combinatorics`
PartialOrderGraphQ
​
PartialOrderGraphQ
[g]
yields
True
if the binary relation defined by edges of the graph
g
is a partial order and
False
otherwise.
​
Details and Options

Examples  
(2)
Basic Examples  
(2)
The graph representing the divisibility relation between integers is partial order since the relation is reflexive (each integer divides itself), antisymmetric (
n
cannot divide
m
if
n>m
), and transitive (as
n\m
implies
m=kn
for some integer
k
, so
m\l
implies
n\l)
:
In[1]:=
RelationGraph[(Mod[#1,#2]0)&,Range[8],VertexLabels"Name"]
Out[1]=
In[2]:=
PartialOrderGraphQ
[%]
Out[2]=
True
​
A graph formed by selecting ordered tuples is a partial order:
In[1]:=
tuples=Tuples[CharacterRange["a","e"],2]
Out[1]=
{{a,a},{a,b},{a,c},{a,d},{a,e},{b,a},{b,b},{b,c},{b,d},{b,e},{c,a},{c,b},{c,c},{c,d},{c,e},{d,a},{d,b},{d,c},{d,d},{d,e},{e,a},{e,b},{e,c},{e,d},{e,e}}
In[2]:=
Graph[DirectedEdge@@@Select[tuples,OrderedQ]]
Out[2]=
In[3]:=
PartialOrderGraphQ
[%]
Out[3]=
True
On the other hand, a similar graph constructed from the symmetric set of the tuples is not:
In[4]:=
Graph[DirectedEdge@@@tuples]
Out[4]=
In[5]:=
PartialOrderGraphQ
[%]
Out[5]=
False
SeeAlso
Graph
 
▪
TransitiveGraphQ
RelatedGuides
▪
Combinatorics
RelatedLinks
PartialOrderGraphQ
Resource Function contributed by Wolfram Staff (original content by Sriram V. Pemmaraju and Steven S. Skiena)
""

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