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{{expert-subject|Mathematics|date=June 2009}}
In [[mathematics]], '''Light's associativity test''' is a [[Algorithm|procedure]] invented by F W Light for testing whether a [[binary operation]] defined in a [[finite set]] by a [[Cayley table|Cayley multiplication table]] is [[associative]]. Direct verification  of the associativity of a binary operation specified by a Cayley table is cumbersome and tedious. Light's associativity  test greatly simplifies the task.


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==Description of the procedure==
 
Let a binary operation ' · ' be defined in a finite set ''A'' by a Cayley table. Choosing some element ''a'' in ''A'',  two new binary operations are defined in ''A'' as follows:
:''x'' <math>\star</math> ''y'' = ''x'' &middot; ( ''a'' &middot; ''y'' )
:''x'' <math>\circ</math> ''y'' = ( ''x'' &middot;  ''a'' ) &middot; ''y''
The Cayley tables of these operations are  constructed and compared. If the tables coincide then ''x'' · ( ''a'' · ''y'' ) = ( ''x'' ·  ''a'' ) · ''y'' for all ''x'' and ''y''. This is repeated for every element of the set ''A''.
 
The example below illustrates a further simplification in the procedure for the construction and comparison of the Cayley tables of the operations ' <math>\star</math> ' and ' <math>\circ</math> '.
 
It is not even necessary to construct the Cayley tables of ' <math>\star</math> ' and ' <math>\circ</math> ' for ''all'' elements of ''A''. It is enough to compare Cayley tables of ' <math>\star</math> ' and ' <math>\circ</math> ' corresponding to the elements in a proper generating subset of ''A''.
 
==Example==
 
Consider the binary operation ' · ' in the set ''A'' = { ''a'', ''b'', ''c'', ''d'', ''e'' } defined by the following Cayley table (Table 1):
 
<center>
{| class="wikitable" border="1" width="20%"
|+Table 1
|-
! ·
! ''a''
! ''b''
! ''c''
! ''d''
! ''e''
|-
| &nbsp; ''' ''a'' '''
| &nbsp; ''a''
| &nbsp; ''a''
| &nbsp; ''a''
| &nbsp; ''d''
| &nbsp; ''d''
|-
| &nbsp; ''' ''b'' '''
| &nbsp; ''a''
| &nbsp; ''b''
| &nbsp; ''c''
| &nbsp; ''d''
| &nbsp; ''d''
|-
| &nbsp; ''' ''c'' '''
| &nbsp; ''a''
| &nbsp; ''c''
| &nbsp; ''b''
| &nbsp; ''d''
| &nbsp; ''d''
|-
| &nbsp; ''' ''d'' '''
| &nbsp; ''d''
| &nbsp; ''d''
| &nbsp; ''d''
| &nbsp; ''a''
| &nbsp; ''a''
|-
| &nbsp; ''' ''e'' '''
| &nbsp; ''d''
| &nbsp; ''e''
| &nbsp; ''e''
| &nbsp; ''a''
| &nbsp; ''a''
|}
</center>
 
The set { ''c'', ''e'' } is a generating set for the set ''A'' under the binary operation defined by the above table, for,  ''a'' = ''e'' · ''e'', ''b'' = ''c'' · ''c'', ''d'' = ''c'' · ''e''. Thus it is enough to verify that the binary operations ' <math>\star</math> ' and ' <math>\circ</math> ' corresponding to ''c'' coincide and also that  the binary operations ' <math>\star</math> ' and ' <math>\circ</math> ' corresponding to ''e'' coincide.
 
To verify that the binary operations ' <math>\star</math> ' and ' <math>\circ</math> ' corresponding to ''c'' coincide, choose the row in Table 1 corresponding to the element ''' ''c'' ''':
 
<center>
{| class="wikitable" border="1" width="20%"
|+Table 2
|-
! ·
! ''a''
! ''b''
! ''c''
! ''d''
! ''e''
|-
| &nbsp; ''' ''a'' '''
| &nbsp; ''a''
| &nbsp; ''a''
| &nbsp; ''a''
| &nbsp; ''d''
| &nbsp; ''d''
|-
| &nbsp; ''' ''b'' '''
| &nbsp; ''a''
| &nbsp; ''b''
| &nbsp; ''c''
| &nbsp; ''d''
| &nbsp; ''d''
|-
| &nbsp; ''' ''c'' '''
|style="background:#ADFF2F"| &nbsp; ''a''
|style="background:#ADFF2F"| &nbsp; ''c''
|style="background:#ADFF2F"| &nbsp; ''b''
|style="background:#ADFF2F"| &nbsp; ''d''
|style="background:#ADFF2F"| &nbsp; ''d''
|-
| &nbsp; ''' ''d'' '''
| &nbsp; ''d''
| &nbsp; ''d''
| &nbsp; ''d''
| &nbsp; ''a''
| &nbsp; ''a''
|-
| &nbsp; ''' ''e'' '''
| &nbsp; ''d''
| &nbsp; ''e''
| &nbsp; ''e''
| &nbsp; ''a''
| &nbsp; ''a''
|}
</center>
 
This row is copied as the header row of a new table (Table 3):
 
<center>
{| class="wikitable" border="1" width="20%"
|+Table 3
|-
! &nbsp;&nbsp;&nbsp;
! style="background:#ADFF2F"| &nbsp; ''a''
! style="background:#ADFF2F"| &nbsp; ''c''
! style="background:#ADFF2F"| &nbsp; ''b''
! style="background:#ADFF2F"| &nbsp; ''d''
! style="background:#ADFF2F"| &nbsp; ''d''
|-
| &nbsp;&nbsp;&nbsp;
|
|
|
|
|
|-
| &nbsp;&nbsp;&nbsp;
|
|
|
|
|
|-
| &nbsp;&nbsp;&nbsp;
|
|
|
|
|
|-
| &nbsp;&nbsp;&nbsp;
|
|
|
|
|
|-
| &nbsp;&nbsp;&nbsp;
|
|
|
|
|
|}
</center>
 
Under the header ''' ''a'' ''' copy the corresponding column in Table 1, under the header ''' ''b'' ''' copy the corresponding column in Table 1, etc., and construct Table 4.
 
<center>
{| class="wikitable" border="1" width="20%"
|+Table 4
|-
! &nbsp;&nbsp;&nbsp;
! style="background:#ADFF2F"| &nbsp; ''a''
! style="background:#ADFF2F"| &nbsp; ''c''
! style="background:#ADFF2F"| &nbsp; ''b''
! style="background:#ADFF2F"| &nbsp; ''d''
! style="background:#ADFF2F"| &nbsp; ''d''
|-
|
| &nbsp; ''a''
| &nbsp; ''a''
| &nbsp; ''a''
| &nbsp; ''d''
| &nbsp; ''d''
|-
|
| &nbsp; ''a''
| &nbsp; ''c''
| &nbsp; ''b''
| &nbsp; ''d''
| &nbsp; ''d''
|-
|
| &nbsp; ''a''
| &nbsp; ''b''
| &nbsp; ''c''
| &nbsp; ''d''
| &nbsp; ''d''
|-
|
| &nbsp; ''d''
| &nbsp; ''d''
| &nbsp; ''d''
| &nbsp; ''a''
| &nbsp; ''a''
|-
|
| &nbsp; ''d''
| &nbsp; ''e''
| &nbsp; ''e''
| &nbsp; ''a''
| &nbsp; ''a''
|}
</center>
 
The column headers of Table 4 are now deleted to get Table 5:
 
<center>
{| class="wikitable" border="1" width="20%"
|+Table 5
|-
! &nbsp;&nbsp;&nbsp;
! &nbsp;&nbsp;&nbsp;
! &nbsp;&nbsp;&nbsp;
! &nbsp;&nbsp;&nbsp;
! &nbsp;&nbsp;&nbsp;
! &nbsp;&nbsp;&nbsp;
|-
|
| &nbsp; ''a''
| &nbsp; ''a''
| &nbsp; ''a''
| &nbsp; ''d''
| &nbsp; ''d''
|-
|
| &nbsp; ''a''
| &nbsp; ''c''
| &nbsp; ''b''
| &nbsp; ''d''
| &nbsp; ''d''
|-
|
| &nbsp; ''a''
| &nbsp; ''b''
| &nbsp; ''c''
| &nbsp; ''d''
| &nbsp; ''d''
|-
|
| &nbsp; ''d''
| &nbsp; ''d''
| &nbsp; ''d''
| &nbsp; ''a''
| &nbsp; ''a''
|-
|
| &nbsp; ''d''
| &nbsp; ''e''
| &nbsp; ''e''
| &nbsp; ''a''
| &nbsp; ''a''
|}
</center>
 
The Cayley table of the binary operation  ' <math>\star</math> ' corresponding to the element ''c'' is given by Table 6.
 
<center>
{| class="wikitable" border="1" width="20%"
|+Table 6
|-
! &nbsp;<math>\star</math> (c)
! &nbsp; ''a''
! &nbsp; ''b''
! &nbsp; ''c''
! &nbsp; ''d''
! &nbsp; ''e''
|-
| &nbsp; ''' ''a'' '''
| &nbsp; ''a''
| &nbsp; ''a''
| &nbsp; ''a''
| &nbsp; ''d''
| &nbsp; ''d''
|-
| &nbsp; ''' ''b'' '''
| &nbsp; ''a''
| &nbsp; ''c''
| &nbsp; ''b''
| &nbsp; ''d''
| &nbsp; ''d''
|-
| &nbsp; ''' ''c'' '''
| &nbsp; ''a''
| &nbsp; ''b''
| &nbsp; ''c''
| &nbsp; ''d''
| &nbsp; ''d''
|-
| &nbsp; ''' ''d'' '''
| &nbsp; ''d''
| &nbsp; ''d''
| &nbsp; ''d''
| &nbsp; ''a''
| &nbsp; ''a''
|-
| &nbsp; ''' ''e'' '''
| &nbsp; ''d''
| &nbsp; ''e''
| &nbsp; ''e''
| &nbsp; ''a''
| &nbsp; ''a''
|}
</center>
 
Next choose the ''' ''c'' ''' column of Table 1:
 
<center>
{| class="wikitable" border="1" width="20%"
|+Table 7
|-
! ·
! ''a''
! ''b''
! ''c''
! ''d''
! ''e''
|-
| &nbsp; ''' ''a'' '''
| &nbsp; ''a''
| &nbsp; ''a''
|style="background:#87CEEB"| &nbsp; ''a''
| &nbsp; ''d''
| &nbsp; ''d''
|-
| &nbsp; ''' ''b'' '''
| &nbsp; ''a''
| &nbsp; ''b''
|style="background:#87CEEB"| &nbsp; ''c''
| &nbsp; ''d''
| &nbsp; ''d''
|-
| &nbsp; ''' ''c'' '''
| &nbsp; ''a''
| &nbsp; ''c''
|style="background:#87CEEB"| &nbsp; ''b''
| &nbsp; ''d''
| &nbsp; ''d''
|-
| &nbsp; ''' ''d'' '''
| &nbsp; ''d''
| &nbsp; ''d''
|style="background:#87CEEB"| &nbsp; ''d''
| &nbsp; ''a''
| &nbsp; ''a''
|-
| &nbsp; ''' ''e'' '''
| &nbsp; ''d''
| &nbsp; ''e''
|style="background:#87CEEB"| &nbsp; ''e''
| &nbsp; ''a''
| &nbsp; ''a''
|}
</center>
 
Copy this column to the index column to get Table 8:
 
<center>
{| class="wikitable" border="1" width="20%"
|+Table 8
|-
! &nbsp;&nbsp;&nbsp;
! &nbsp;&nbsp;&nbsp;
! &nbsp;&nbsp;&nbsp;
! &nbsp;&nbsp;&nbsp;
! &nbsp;&nbsp;&nbsp;
! &nbsp;&nbsp;&nbsp;
|-
|style="background:#87CEEB"| &nbsp;''' ''a'' '''
|
|
|
|
|
|-
|style="background:#87CEEB"| &nbsp;''' ''c'' '''
|
|
|
|
|
|-
|style="background:#87CEEB"| &nbsp;''' ''b'' '''
|
|
|
|
|
|-
|style="background:#87CEEB"| &nbsp;''' ''d'' '''
|
|
|
|
|
|-
|style="background:#87CEEB"| &nbsp;''' ''e'' '''
|
|
|
|
|
|}
</center>
 
Against the index entry ''' ''a'' ''' in Table 8 copy the corresponding row in Table 1, against  the index entry ''' ''b'' ''' copy the corresponding row in Table 1, etc., and construct Table 9.
 
<center>
{| class="wikitable" border="1" width="20%"
|+Table 9
|-
! &nbsp;&nbsp;&nbsp;
! &nbsp;&nbsp;&nbsp;
! &nbsp;&nbsp;&nbsp;
! &nbsp;&nbsp;&nbsp;
! &nbsp;&nbsp;&nbsp;
! &nbsp;&nbsp;&nbsp;
|-
| style="background:#87CEEB"| &nbsp; ''' ''a'' '''
| &nbsp; ''a''
| &nbsp; ''a''
| &nbsp; ''a''
| &nbsp; ''d''
| &nbsp; ''d''
|-
| style="background:#87CEEB"|&nbsp; ''' ''c'' '''
| &nbsp; ''a''
| &nbsp; ''c''
| &nbsp; ''b''
| &nbsp; ''d''
| &nbsp; ''d''
|-
| style="background:#87CEEB"|&nbsp; ''' ''b'' '''
| &nbsp; ''a''
| &nbsp; ''b''
| &nbsp; ''c''
| &nbsp; ''d''
| &nbsp; ''d''
|-
| style="background:#87CEEB"|&nbsp; ''' ''d'' '''
| &nbsp; ''d''
| &nbsp; ''d''
| &nbsp; ''d''
| &nbsp; ''a''
| &nbsp; ''a''
|-
| style="background:#87CEEB"|&nbsp; ''' ''e'' '''
| &nbsp; ''d''
| &nbsp; ''e''
| &nbsp; ''e''
| &nbsp; ''a''
| &nbsp; ''a''
|}
</center>
 
The index entries in the first column of Table 9 are now deleted to get Table 10:
 
<center>
{| class="wikitable" border="1" width="20%"
|+Table 10
|-
! &nbsp;&nbsp;&nbsp;
! &nbsp;&nbsp;&nbsp;
! &nbsp;&nbsp;&nbsp;
! &nbsp;&nbsp;&nbsp;
! &nbsp;&nbsp;&nbsp;
! &nbsp;&nbsp;&nbsp;
|-
|&nbsp;&nbsp;&nbsp;
| &nbsp; ''a''
| &nbsp; ''a''
| &nbsp; ''a''
| &nbsp; ''d''
| &nbsp; ''d''
|-
| &nbsp;&nbsp;&nbsp;
| &nbsp; ''a''
| &nbsp; ''c''
| &nbsp; ''b''
| &nbsp; ''d''
| &nbsp; ''d''
|-
| &nbsp;&nbsp;&nbsp;
| &nbsp; ''a''
| &nbsp; ''b''
| &nbsp; ''c''
| &nbsp; ''d''
| &nbsp; ''d''
|-
| &nbsp;&nbsp;&nbsp;
| &nbsp; ''d''
| &nbsp; ''d''
| &nbsp; ''d''
| &nbsp; ''a''
| &nbsp; ''a''
|-
| &nbsp;&nbsp;&nbsp;
| &nbsp; ''d''
| &nbsp; ''e''
| &nbsp; ''e''
| &nbsp; ''a''
| &nbsp; ''a''
|}
</center>
 
The Cayley table of the binary operation  ' <math>\circ</math> ' corresponding to the element ''c'' is given by Table 11.
 
<center>
{| class="wikitable" border="1" width="20%"
|+Table 11
|-
! <math>\circ</math>(c)
! &nbsp; ''a''
! &nbsp; ''b''
! &nbsp; ''c''
! &nbsp; ''d''
! &nbsp; ''e''
|-
| &nbsp; ''' ''a'' '''
| &nbsp; ''a''
| &nbsp; ''a''
| &nbsp; ''a''
| &nbsp; ''d''
| &nbsp; ''d''
|-
| &nbsp; ''' ''b'' '''
| &nbsp; ''a''
| &nbsp; ''c''
| &nbsp; ''b''
| &nbsp; ''d''
| &nbsp; ''d''
|-
| &nbsp; ''' ''c'' '''
| &nbsp; ''a''
| &nbsp; ''b''
| &nbsp; ''c''
| &nbsp; ''d''
| &nbsp; ''d''
|-
| &nbsp; ''' ''d'' '''
| &nbsp; ''d''
| &nbsp; ''d''
| &nbsp; ''d''
| &nbsp; ''a''
| &nbsp; ''a''
|-
| &nbsp; ''' ''e'' '''
| &nbsp; ''d''
| &nbsp; ''e''
| &nbsp; ''e''
| &nbsp; ''a''
| &nbsp; ''a''
|}
</center>
 
One can verify that the entries in the various cells in Table 6 agrees with the entries in the corresponding cells of Table 11. This shows that ''x'' · ( ''c'' · ''y'' ) = ( ''x'' · ''c'' ) · ''y'' for all ''x'' and ''y'' in ''A''. If there were some discrepancy then it would not be true that ''x'' · ( ''c'' · ''y'' ) = ( ''x'' · ''c'' ) · ''y'' for all ''x'' and ''y'' in ''A''.
 
That ''x'' · ( ''e'' · ''y'' ) = ( ''x'' · ''e'' ) · ''y'' for all ''x'' and ''y'' in ''A'' can be verified in a similar way by constructing the following tables (Table 12 and Table 13):
 
<center>
{| class="wikitable" border="1" width="20%"
|+Table 12
|-
! &nbsp;<math>\star</math>(''e'')
! style="background:#ADFF2F"| &nbsp; ''a''
! style="background:#ADFF2F"| &nbsp; ''b''
! style="background:#ADFF2F"| &nbsp; ''c''
! style="background:#ADFF2F"| &nbsp; ''d''
! style="background:#ADFF2F"| &nbsp; ''e''
|-
|style="background:#87CEEB"| &nbsp; ''' ''a'' '''
|&nbsp; ''d''
|&nbsp; ''d''
|&nbsp; ''d''
|&nbsp; ''a''
|&nbsp; ''a''
|-
|style="background:#87CEEB"| &nbsp; ''' ''b'' '''
|&nbsp; ''d''
|&nbsp; ''d''
|&nbsp; ''d''
|&nbsp; ''a''
|&nbsp; ''a''
|-
|style="background:#87CEEB"| &nbsp; ''' ''c'' '''
|&nbsp; ''d''
|&nbsp; ''d''
|&nbsp; ''d''
|&nbsp; ''a''
|&nbsp; ''a''
|-
|style="background:#87CEEB"| &nbsp; ''' ''d'' '''
|&nbsp; ''a''
|&nbsp; ''a''
|&nbsp; ''a''
|&nbsp; ''d''
|&nbsp; ''d''
|-
|style="background:#87CEEB"| &nbsp; ''' ''e'' '''
|&nbsp; ''a''
|&nbsp; ''a''
|&nbsp; ''a''
|&nbsp; ''d''
|&nbsp; ''d''
|}
</center>
 
<center>
{| class="wikitable" border="1" width="20%"
|+Table 13
|-
! &nbsp;<math>\circ</math>(''e'')
! style="background:#ADFF2F"| &nbsp; ''a''
! style="background:#ADFF2F"| &nbsp; ''b''
! style="background:#ADFF2F"| &nbsp; ''c''
! style="background:#ADFF2F"| &nbsp; ''d''
! style="background:#ADFF2F"| &nbsp; ''e''
|-
|style="background:#87CEEB"| &nbsp; ''' ''a'' '''
|&nbsp; ''d''
|&nbsp; ''d''
|&nbsp; ''d''
|&nbsp; ''a''
|&nbsp; ''a''
|-
|style="background:#87CEEB"| &nbsp; ''' ''b'' '''
|&nbsp; ''d''
|&nbsp; ''d''
|&nbsp; ''d''
|&nbsp; ''a''
|&nbsp; ''a''
|-
|style="background:#87CEEB"| &nbsp; ''' ''c'' '''
|&nbsp; ''d''
|&nbsp; ''d''
|&nbsp; ''d''
|&nbsp; ''a''
|&nbsp; ''a''
|-
|style="background:#87CEEB"| &nbsp; ''' ''d'' '''
|&nbsp; ''a''
|&nbsp; ''a''
|&nbsp; ''a''
|&nbsp; ''d''
|&nbsp; ''d''
|-
|style="background:#87CEEB"| &nbsp; ''' ''e'' '''
|&nbsp; ''a''
|&nbsp; ''a''
|&nbsp; ''a''
|&nbsp; ''d''
|&nbsp; ''d''
|}
</center>
 
===A further simplification ===
 
It is not necessary to construct the Cayley tables (Table 6 and table 11) of the binary operations ' <math>\star</math> ' and ' <math>\circ</math> '.  It is enough to copy the column corresponding to the header ''' ''c'' '''  in Table 1 to the index column in Table 5 and form the following table (Table 14) and verify that the ''' ''a'' '''-row of Table 14 is identical with the ''' ''a'' '''-row of Table 1, the ''' ''b'' '''-row of Table 14 is identical with the ''' ''b'' '''-row of Table 1, etc. This is to be repeated mutatis mutandis for all the elements of the generating set of ''A''.
 
<center>
{| class="wikitable" border="1" width="20%"
|+Table 14
|-
! &nbsp;&nbsp;&nbsp;
! style="background:#ADFF2F"| &nbsp; ''a''
! style="background:#ADFF2F"| &nbsp; ''c''
! style="background:#ADFF2F"| &nbsp; ''b''
! style="background:#ADFF2F"| &nbsp; ''d''
! style="background:#ADFF2F"| &nbsp; ''d''
|-
| style="background:#87CEEB"| &nbsp; ''' ''a'' '''
| &nbsp; ''a''
| &nbsp; ''a''
| &nbsp; ''a''
| &nbsp; ''d''
| &nbsp; ''d''
|-
| style="background:#87CEEB"| &nbsp; ''' ''c'' '''
| &nbsp; ''a''
| &nbsp; ''c''
| &nbsp; ''b''
| &nbsp; ''d''
| &nbsp; ''d''
|-
| style="background:#87CEEB"| &nbsp; ''' ''b'' '''
| &nbsp; ''a''
| &nbsp; ''b''
| &nbsp; ''c''
| &nbsp; ''d''
| &nbsp; ''d''
|-
| style="background:#87CEEB"| &nbsp; ''' ''d'' '''
| &nbsp; ''d''
| &nbsp; ''d''
| &nbsp; ''d''
| &nbsp; ''a''
| &nbsp; ''a''
|-
| style="background:#87CEEB"| &nbsp; ''' ''e'' '''
| &nbsp; ''d''
| &nbsp; ''e''
| &nbsp; ''e''
| &nbsp; ''a''
| &nbsp; ''a''
|}
</center>
 
==Algorithm for  Light's associativity test==
[[Computer software]] can be written to carry out Light's associativity test. Kehayopulu and Argyris have developed such an [[algorithm]] for [[Mathematica]].<ref>{{cite journal|last=Kehayopulu|first=Niovi|coauthors=Philip  Argyris|year=1993|title=An algorithm for Light's associativity test using  Mathematica|journal=J. Comput. Inform.|volume=3|issue=1|pages=87&ndash;98|issn=11803886}}</ref>
 
==Extension of Light's associativity test==
 
Light's associativity test can be extended to test associativity in a more general context.<ref>{{cite journal|doi=10.2307/2314731|last=Bednarek|first=A R|year=1968|title=An extension of Light's associativity test|journal=[[American Mathematical Monthly]]|volume=75|issue=5|pages=531&ndash;532|jstor=2314731}}</ref><ref>{{cite journal|doi=10.1007/BF02572966|last=Kalman|first=J A|year=1971|title=Bednarek's extension of Light's associativity test|journal=Semigroup Forum|volume=3|issue=1|pages=275–276}}</ref>
 
Let ''T'' = { ''t''<sub>1</sub>, ''t''<sub>2</sub>, <math>\ldots</math>, ''t''<sub>''m''</sub> } be a [[Magma (algebra)|magma]] in which the operation is denoted by [[Juxtaposition (literary)|juxtaposition]]. Let ''X'' = { ''x''<sub>1</sub>, ''x''<sub>2</sub>, <math>\ldots</math>, ''x''<sub>''n''</sub> } be a set. Let there be a mapping from the [[Cartesian product]] ''T'' &times; ''X'' to ''X'' denoted by  (''t'', ''x'') {{mapsto}} ''tx'' and let it be required to test whether this map has the property
 
:(''st'')''x'' = ''s''(''tx'')  for all ''s'', ''t'' in ''T''  and all ''x'' in ''X''.
 
A generalization of Light's associativity test can be applied to verify whether the above property holds or not. In mathematical notations, the generalization runs as follows: For each ''t'' in ''T'', let ''L''(''t'') be the ''m'' &times; ''n'' matrix of elements of ''X'' whose ''i'' - th row is
 
:( (''t''<sub>''i''</sub>''t'')''x''<sub>1</sub>, (''t''<sub>''i''</sub>''t'')''x''<sub>2</sub>, <math>\ldots</math> , (''t''<sub>''i''</sub>''t'')''x''<sub>n</sub> )  for ''i'' = 1, <math>\ldots</math>, ''m''
 
and let ''R''(''t'') be the ''m'' &times; ''n'' matrix of elements of ''X'', the elements of whose ''j'' - th column are
 
:( ''t''<sub>1</sub>(''tx''<sub>''j''</sub>), ''t''<sub>2</sub>(''tx''<sub>''j''</sub>), <math>\ldots</math> ,  ''t''<sub>''m''</sub>(''tx''<sub>j</sub>) )  for ''j'' = 1, <math>\ldots</math>, ''n''.
 
According to the generalised test (due to Bednarek), that the property to be verified holds if and only if ''L''(''t'') = ''R''(''t'') for all ''t'' in ''T''. When ''X'' = ''T'', Bednarek's test reduces to Light's test.
 
==More advanced algorithms==
There is a randomized algorithm by Rajagopalan and Schulman to test associativity in time proportional to the input size. (The method also works for testing certain other identities.) Specifically, the runtime is <math>O(n^2 \log \frac1\delta)</math> for an <math>n\times n</math> table and error probability <math>\delta</math>.
The algorithm can be modified to produce a triple <math>\langle a,b,c\rangle</math> for which <math>(ab)c\ne a(bc)</math>, if there is one, in time <math>O(n^2 \log n \cdot\log \frac1\delta)</math>.<ref>Rajagopalan and Schulman, FOCS 1996, SICOMP 2000  http://dx.doi.org/10.1137/S0097539797325387</ref>
 
==See also==
*[[Cayley table]]
*[[Associativity]]
 
==References==
{{reflist}}
* {{Cite book | last1=Clifford | first1=Alfred Hoblitzelle | author1-link=Alfred H. Clifford | last2=Preston | first2=Gordon Bamford | author2-link=Gordon Preston | title=The algebraic theory of semigroups. Vol. I | publisher=[[American Mathematical Society]] | location=Providence, R.I. | series=Mathematical Surveys, No. 7 | isbn=978-0-8218-0272-4 | id={{MathSciNet | id = 0132791}} | year=1961}} (pp.&nbsp;7&ndash;9)
 
[[Category:Abstract algebra]]
[[Category:Semigroup theory]]
[[Category:Binary operations| ]]
[[Category:Elementary algebra]]

Revision as of 10:02, 11 May 2013

Template:Expert-subject In mathematics, Light's associativity test is a procedure invented by F W Light for testing whether a binary operation defined in a finite set by a Cayley multiplication table is associative. Direct verification of the associativity of a binary operation specified by a Cayley table is cumbersome and tedious. Light's associativity test greatly simplifies the task.

Description of the procedure

Let a binary operation ' · ' be defined in a finite set A by a Cayley table. Choosing some element a in A, two new binary operations are defined in A as follows:

x y = x · ( a · y )
x y = ( x · a ) · y

The Cayley tables of these operations are constructed and compared. If the tables coincide then x · ( a · y ) = ( x · a ) · y for all x and y. This is repeated for every element of the set A.

The example below illustrates a further simplification in the procedure for the construction and comparison of the Cayley tables of the operations ' ' and ' '.

It is not even necessary to construct the Cayley tables of ' ' and ' ' for all elements of A. It is enough to compare Cayley tables of ' ' and ' ' corresponding to the elements in a proper generating subset of A.

Example

Consider the binary operation ' · ' in the set A = { a, b, c, d, e } defined by the following Cayley table (Table 1):

Table 1
· a b c d e
  a   a   a   a   d   d
  b   a   b   c   d   d
  c   a   c   b   d   d
  d   d   d   d   a   a
  e   d   e   e   a   a

The set { c, e } is a generating set for the set A under the binary operation defined by the above table, for, a = e · e, b = c · c, d = c · e. Thus it is enough to verify that the binary operations ' ' and ' ' corresponding to c coincide and also that the binary operations ' ' and ' ' corresponding to e coincide.

To verify that the binary operations ' ' and ' ' corresponding to c coincide, choose the row in Table 1 corresponding to the element c :

Table 2
· a b c d e
  a   a   a   a   d   d
  b   a   b   c   d   d
  c   a   c   b   d   d
  d   d   d   d   a   a
  e   d   e   e   a   a

This row is copied as the header row of a new table (Table 3):

Table 3
      a   c   b   d   d
   
   
   
   
   

Under the header a copy the corresponding column in Table 1, under the header b copy the corresponding column in Table 1, etc., and construct Table 4.

Table 4
      a   c   b   d   d
  a   a   a   d   d
  a   c   b   d   d
  a   b   c   d   d
  d   d   d   a   a
  d   e   e   a   a

The column headers of Table 4 are now deleted to get Table 5:

Table 5
                       
  a   a   a   d   d
  a   c   b   d   d
  a   b   c   d   d
  d   d   d   a   a
  d   e   e   a   a

The Cayley table of the binary operation ' ' corresponding to the element c is given by Table 6.

Table 6
  (c)   a   b   c   d   e
  a   a   a   a   d   d
  b   a   c   b   d   d
  c   a   b   c   d   d
  d   d   d   d   a   a
  e   d   e   e   a   a

Next choose the c column of Table 1:

Table 7
· a b c d e
  a   a   a   a   d   d
  b   a   b   c   d   d
  c   a   c   b   d   d
  d   d   d   d   a   a
  e   d   e   e   a   a

Copy this column to the index column to get Table 8:

Table 8
                       
  a
  c
  b
  d
  e

Against the index entry a in Table 8 copy the corresponding row in Table 1, against the index entry b copy the corresponding row in Table 1, etc., and construct Table 9.

Table 9
                       
  a   a   a   a   d   d
  c   a   c   b   d   d
  b   a   b   c   d   d
  d   d   d   d   a   a
  e   d   e   e   a   a

The index entries in the first column of Table 9 are now deleted to get Table 10:

Table 10
                       
      a   a   a   d   d
      a   c   b   d   d
      a   b   c   d   d
      d   d   d   a   a
      d   e   e   a   a

The Cayley table of the binary operation ' ' corresponding to the element c is given by Table 11.

Table 11
(c)   a   b   c   d   e
  a   a   a   a   d   d
  b   a   c   b   d   d
  c   a   b   c   d   d
  d   d   d   d   a   a
  e   d   e   e   a   a

One can verify that the entries in the various cells in Table 6 agrees with the entries in the corresponding cells of Table 11. This shows that x · ( c · y ) = ( x · c ) · y for all x and y in A. If there were some discrepancy then it would not be true that x · ( c · y ) = ( x · c ) · y for all x and y in A.

That x · ( e · y ) = ( x · e ) · y for all x and y in A can be verified in a similar way by constructing the following tables (Table 12 and Table 13):

Table 12
 (e)   a   b   c   d   e
  a   d   d   d   a   a
  b   d   d   d   a   a
  c   d   d   d   a   a
  d   a   a   a   d   d
  e   a   a   a   d   d
Table 13
 (e)   a   b   c   d   e
  a   d   d   d   a   a
  b   d   d   d   a   a
  c   d   d   d   a   a
  d   a   a   a   d   d
  e   a   a   a   d   d

A further simplification

It is not necessary to construct the Cayley tables (Table 6 and table 11) of the binary operations ' ' and ' '. It is enough to copy the column corresponding to the header c in Table 1 to the index column in Table 5 and form the following table (Table 14) and verify that the a -row of Table 14 is identical with the a -row of Table 1, the b -row of Table 14 is identical with the b -row of Table 1, etc. This is to be repeated mutatis mutandis for all the elements of the generating set of A.

Table 14
      a   c   b   d   d
  a   a   a   a   d   d
  c   a   c   b   d   d
  b   a   b   c   d   d
  d   d   d   d   a   a
  e   d   e   e   a   a

Algorithm for Light's associativity test

Computer software can be written to carry out Light's associativity test. Kehayopulu and Argyris have developed such an algorithm for Mathematica.[1]

Extension of Light's associativity test

Light's associativity test can be extended to test associativity in a more general context.[2][3]

Let T = { t1, t2, , tm } be a magma in which the operation is denoted by juxtaposition. Let X = { x1, x2, , xn } be a set. Let there be a mapping from the Cartesian product T × X to X denoted by (t, x) Template:Mapsto tx and let it be required to test whether this map has the property

(st)x = s(tx) for all s, t in T and all x in X.

A generalization of Light's associativity test can be applied to verify whether the above property holds or not. In mathematical notations, the generalization runs as follows: For each t in T, let L(t) be the m × n matrix of elements of X whose i - th row is

( (tit)x1, (tit)x2, , (tit)xn ) for i = 1, , m

and let R(t) be the m × n matrix of elements of X, the elements of whose j - th column are

( t1(txj), t2(txj), , tm(txj) ) for j = 1, , n.

According to the generalised test (due to Bednarek), that the property to be verified holds if and only if L(t) = R(t) for all t in T. When X = T, Bednarek's test reduces to Light's test.

More advanced algorithms

There is a randomized algorithm by Rajagopalan and Schulman to test associativity in time proportional to the input size. (The method also works for testing certain other identities.) Specifically, the runtime is for an table and error probability . The algorithm can be modified to produce a triple for which , if there is one, in time .[4]

See also

References

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    The data offered is for normal info purposes only and isn't supposed to be personalised investment or monetary advice. Motley Fool Singapore contributor Stanley Lim would not personal shares in any corporations talked about. Singapore private home costs increased by 1.eight% within the fourth quarter of 2012, up from 0.6% within the earlier quarter. Resale prices of government-built HDB residences which are usually bought by Singaporeans, elevated by 2.5%, quarter on quarter, the quickest acquire in five quarters. And industrial property, prices are actually double the levels of three years ago. No withholding tax in the event you sell your property. All your local information regarding vital HDB policies, condominium launches, land growth, commercial property and more

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  4. Rajagopalan and Schulman, FOCS 1996, SICOMP 2000 http://dx.doi.org/10.1137/S0097539797325387