Semigroups of order n
Tim
are there tables of the semigroups of order n or maybe even some
classification ?
Thanks
Jan Kristian Haugland
Tim wrote:
From http://mathworld.wolfram.com/Semigroup.html
"The total number of semigroups of order n are 1, 4, 18,
126, 1160, 15973, 836021, ..."
Somehow I don't think a table of the 836021 semigroups of
order 7 would be fun reading...:-)
--
J K Haugland
http://hjem.sol.no/neutreeko
Nico Benschop
Interesting, and surprising: are there only 4 semigroups of order 2 ?
I thought there are 5 non-isomorphic semigroups of order two:
2 with one generator:
cyclic group ~ addition mod 2 ~ Boolean XOR
nilpotent sgrp ('monotone' counter) ~ {2,0} \in Z4(.)
3 with two generators, all idempotent ( aa=a, bb=b):
one commutative ~ multiply mod 2 ~ Boolean AND
two non-commutative:
left-copy sgrp ('L2'): ab = a (model: '2-branch')
right-copy sgrp ('R2'): ab = b (model: set/reset FF)
PS: Any sgrp requiring at least 3 generators has necessarily a proper
subsemigroup (of order >1, with two generators). The above five
are the only ones without a proper subsemigrouop (hence 'basic').
Nico Benschop
>
> Jan Kristian Haugland wrote:
> >
> > Tim wrote:
> >
> > > Similiar to tables of finite groups of order n for small values
> > > of n are there tables of the semigroups of order n or maybe even
> > > some classification ?
> >
> > From http://mathworld.wolfram.com/Semigroup.html
> >
> > "The total number of semigroups of order n
> > are 1, 4, 18, 126, 1160, 15973, 836021, ..."
> >
> > Somehow I don't think a table of the 836021 semigroups
> > of order 7 would be fun reading...:-) -- J K Haugland
>
> Interesting, and surprising: are there only 4 semigroups of order 2 ?
>
> I thought there are 5 non-isomorphic semigroups of order two:
>
> 2 with one generator:
> nilpotent sgrp (monotone counter 'U2') ~ {2,0} in Z_4(.)
> 3 with two generators, all idempotent ( aa=a, bb=b):
> left-copy sgrp ('L2'): ab = a (model: '2-branch')
> right-copy sgrp ('R2'): ab = b (model: set/reset FF)
>
> PS: Any sgrp requiring at least 3 generators has necessarily a proper
> subsemigroup (of order >1, with two generators). The above five
> are the only ones without a proper subsemigrouop (hence 'basic')
>
Robin Chapman
news:3CA1A44B...@chello.nl...
> Jan Kristian Haugland wrote:
> >
> > From http://mathworld.wolfram.com/Semigroup.html
> >
> > "The total number of semigroups of order n
> > are 1, 4, 18, 126, 1160, 15973, 836021, ..."
> >
> > Somehow I don't think a table of the 836021 semigroups
> > of order 7 would be fun reading...:-) -- J K Haugland
>
> Interesting, and surprising: are there only 4 semigroups of order 2 ?
>
> I thought there are 5 non-isomorphic semigroups of order two:
>
You might like to learn some day to follow hyperlinks.
Weisstein refers to Sloane's
http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=001423
which states
Semigroups of order n, considered to be equivalent when they are
isomorphic or anti-isomorphic (by reversal of the operator).
Under this equivalence there are indeed four semigroups of
order 2.
With a little intelligence on your part you may be able
to find a sequence of Sloane's detailing your
favoured method of counting semigroups.
Robin Chapman
--
Posted via Mailgate.ORG Server - http://www.Mailgate.ORG
Jan Kristian Haugland
On Wed, 27 Mar 2002, Nico Benschop wrote:
> Jan Kristian Haugland wrote:
> >
> > Tim wrote:
> >
> > > Similiar to tables of finite groups of order n for small values
> > > of n are there tables of the semigroups of order n or maybe even
> > > some classification ?
> >
> > From http://mathworld.wolfram.com/Semigroup.html
> >
> > "The total number of semigroups of order n
> > are 1, 4, 18, 126, 1160, 15973, 836021, ..."
> >
> > Somehow I don't think a table of the 836021 semigroups
> > of order 7 would be fun reading...:-) -- J K Haugland
>
> Interesting, and surprising: are there only 4 semigroups of order 2 ?
>
> I thought there are 5 non-isomorphic semigroups of order two:
>
> 2 with one generator:
> cyclic group ~ addition mod 2 ~ Boolean XOR
> nilpotent sgrp ('monotone' counter) ~ {2,0} \in Z4(.)
>
> 3 with two generators, all idempotent ( aa=a, bb=b):
> one commutative ~ multiply mod 2 ~ Boolean AND
> two non-commutative:
> left-copy sgrp ('L2'): ab = a (model: '2-branch')
> right-copy sgrp ('R2'): ab = b (model: set/reset FF)
Just one naive question: Aren't the last two isomorphic?
Nico Benschop
>
> Jan Kristian Haugland wrote:
> >
> > Tim wrote:
> >
> > > Similiar to tables of finite groups of order n for small values
> > > of n are there tables of the semigroups of order n or maybe even
> > > some classification ?
> >
> >
> > "The total number of semigroups of order n
> > are 1, 4, 18, 126, 1160, 15973, 836021, ..."
> >
> > Somehow I don't think a table of the 836021 semigroups
> > of order 7 would be fun reading...:-) -- J K Haugland
>
> Interesting, and surprising: are there only 4 semigroups of order 2 ?
> I thought there are 5 non-isomorphic semigroups of order two:
>
> 2 with one generator:
> cyclic group ~ addition mod 2 ~ Boolean XOR
> nilpotent sgrp ('monotone' counter) ~ {2,0} \in Z4(.)
>
> 3 with two generators, all idempotent ( aa=a, bb=b):
> one commutative ~ multiply mod 2 ~ Boolean AND
> two non-commutative:
> left-copy sgrp ('L2'): ab = a (model: '2-branch')
> right-copy sgrp ('R2'): ab = b (model: set/reset FF)
>
> PS: Any sgrp requiring at least 3 generators has necessarily a proper
> subsemigroup (of order >1, with two generators). The above five
> are the only ones without a proper subsemigrouop (hence 'basic').
>
> -- NB - http://home.iae.nl/users/benschop/c-ranksm.dvi
PS2: [1] defines equivalent semigroups as:
isomorphic or anti-isomorphic (by reversal of the operator)."
which equivalences (below) L2 and R2, explaining 4 (vs. 5) of order 2.
It took me a while to realize that as closures of state machines
they both model a 'memory' type component:
short-term memory: set/reset-FF R2 stores (and copies) its last input,
long-term memory: branch L2, or 'if-then-else', stores its first inp.
NB: three states are required - including one initial state -
for fathful representation of L2 by state transformations.
PS3: [1] defines:
"A semigroup with an identity is called a commutative monoid, "
^^^^^^^^^^^^^^^^^^^^^^^^^^ ^^^^^^
which is unusual: 'commutative' is normally omitted, isn't it? -- NB
Nico Benschop
>
> On Wed, 27 Mar 2002, Nico Benschop wrote:
>
> > Jan Kristian Haugland wrote:
> > >
> > > Tim wrote:
> > >
> > > > Similiar to tables of finite groups of order n for small values
> > > > of n are there tables of the semigroups of order n or maybe
> > > > even some classification ?
> > >
> > >
> > > "The total number of semigroups of order n
> > > are 1, 4, 18, 126, 1160, 15973, 836021, ..."
> > >
> > > Somehow I don't think a table of the 836021 semigroups
> > > of order 7 would be fun reading...:-) -- J K Haugland
> >
> >
> > I thought there are 5 non-isomorphic semigroups of order two:
> >
> > 2 with one generator:
> > cyclic group ~ addition mod 2 ~ Boolean XOR
> > nilpotent sgrp ('monotone' counter) ~ {2,0} \in Z4(.)
> >
> > 3 with two generators, all idempotent ( aa=a, bb=b):
> > one commutative ~ multiply mod 2 ~ Boolean AND
> > two non-commutative:
> > left-copy sgrp ('L2'): ab = a (model: '2-branch')
> > right-copy sgrp ('R2'): ab = b (model: set/reset FF)
>
> Just one naive question: Aren't the last two isomorphic?
Well, they are 'anti-isomorphic':
ref [1] counts "isomorphic or anti-isomorphic" as equivalent,
which explains the 4 (vs. 5) sgrps of order two.
See my correction in this thread.
For engineering purposes, L2 and R2 are the two essentially different
memory components: L2 models a 2-branch FSM, storing its first
input, leaving the initial (third/extra) state 'forever' after
taking
either one or the other branch (if-then-else);
and R2 stores its last input (set/reset flipflop). - NB
Nico Benschop
>
> "Nico Benschop" <n.ben...@chello.nl> wrote in message
> news:3CA1A44B...@chello.nl...
>
> > Jan Kristian Haugland wrote:
> > >
> > > From http://mathworld.wolfram.com/Semigroup.html
> > >
> > > "The total number of semigroups of order n
> > > are 1, 4, 18, 126, 1160, 15973, 836021, ..."
> > >
> > > Somehow I don't think a table of the 836021 semigroups
> > > of order 7 would be fun reading...:-) -- J K Haugland
> >
> >
> > I thought there are 5 non-isomorphic semigroups of order two:
> >
>
> You might like to learn some day to follow hyperlinks.
> Weisstein refers to Sloane's
>
http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=001423
>
> which states
>
> Semigroups of order n, considered to be equivalent when they are
> isomorphic or anti-isomorphic (by reversal of the operator).
> Under this equivalence there are indeed four semigroups of order 2.
>
> With a little intelligence on your part you may be able
> to find a sequence of Sloane's detailing your
I'll ignore your condescending tone (as usual), Robin.
In fact I made already a correction in this thread, answering Jan.
-- NB
Nico Benschop
>
> Robin Chapman wrote:
> >
> > "Nico Benschop" <n.ben...@chello.nl> wrote in message
> > news:3CA1A44B...@chello.nl...
> >
> > > Jan Kristian Haugland wrote:
> > > >
> > > >
> > > > "The total number of semigroups of order n
> > > > are 1, 4, 18, 126, 1160, 15973, 836021, ..."
> > > >
> > > > Somehow I don't think a table of the 836021 semigroups
> > > > of order 7 would be fun reading...:-) -- J K Haugland
> > >
> > >
> > > I thought there are 5 non-isomorphic semigroups of order two:
> > >
> >
> > You might like to learn some day to follow hyperlinks.
> >
> > Semigroups of order n, considered to be equivalent when they are
> > isomorphic or anti-isomorphic (by reversal of the operator).
> > Under this equivalence there are indeed four semigroups of order 2.
> >
> > With a little intelligence on your part you may be able to find
> > a sequence of Sloane's detailing your
> > favoured method of counting semigroups. -- Robin Chapman
>
> I'll ignore your condescending tone (as usual), Robin.
> In fact I made already a correction in this thread, answering Jan.
> -- NB
BTW: calling left- and right- copy idempotent semigroup 'equivalent',
as apparently mathematicians in [1] do, is declaring past equivalent
to future - in the practical FSM / engineering interpretation of
these closures as dual memory functions:
branch vs. set/reset FF, that is: long vs. short memory.
Making the word 'intelligence' strongly dependent on context, indeed:
"In theory there is no difference between theory and practice,
but in practice there is..."
-- NB - not surprised anymore about the mismatch of understanding
between mathematicians and engineers (of what is 'essential').
The latter will have to go their own way,
in applying and developing essential discrete_math concepts.
Robin Chapman
>
> BTW: calling left- and right- copy idempotent semigroup 'equivalent',
> as apparently mathematicians in [1] do, is declaring past equivalent
> to future - in the practical FSM / engineering interpretation of
> these closures as dual memory functions:
> branch vs. set/reset FF, that is: long vs. short memory.
> Making the word 'intelligence' strongly dependent on context, indeed:
> "In theory there is no difference between theory and practice,
> but in practice there is..."
You don't half talk some pretentious bollocks, Mr Benschop.
If you examine Sloane's site, you'll find enumerations both
regarding non-isomorphic anti-isomorphic semigroups as
equivalent and as inequivalent. It's just mathematicians'
natural curiosity regarding how many structures of a certain
kind there are under as many notions of equivalence
as they can find.
> -- NB - not surprised anymore about the mismatch of understanding
> between mathematicians and engineers (of what is 'essential').
Yes. there seems to be a mismatch between mathematicians
and at least one engineer. All mathematicians regard
understanding as essential, while at least one engineer
regards understanding with contempt.
David C. Ullrich
They are also _isomorphic_. Mapping a to a and b to b gives
an anti-isomorphism. There is _another_ way to map one semigroup
to the other that gives an _isomorphism_.
(It's not hard to find the isomorphism, since there are not
all that many maps from {a,b} 1-1 onto {a,b} in the first
place - write down all the possible mappings, and check each
one to see whether it's an isomorphism.)
> ref [1] counts "isomorphic or anti-isomorphic" as equivalent,
> which explains the 4 (vs. 5) sgrps of order two.
> See my correction in this thread.
>For engineering purposes, L2 and R2 are the two essentially different
>memory components: L2 models a 2-branch FSM, storing its first
>input, leaving the initial (third/extra) state 'forever' after
>taking
> either one or the other branch (if-then-else);
>and R2 stores its last input (set/reset flipflop). - NB
David C. Ullrich
Dik T. Winter
> They are also _isomorphic_. Mapping a to a and b to b gives
> an anti-isomorphism. There is _another_ way to map one semigroup
> to the other that gives an _isomorphism_.
They are not isomorphic. Consider:
Semigroup 1, with elements a and b:
aa = a, ab = a, ba = b and bb = b.
Semigroup 2, with elements a and b:
aa = a, ab = b, ba = a and bb = b.
How is semigroup 1 isomorphic with semigroup 2?
You have also to reverse the operation.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
David C. Ullrich
wrote:
>In article <3ca1d1b4...@nntp.sprynet.com> ull...@math.okstate.edu writes:
> > They are also _isomorphic_. Mapping a to a and b to b gives
> > an anti-isomorphism. There is _another_ way to map one semigroup
> > to the other that gives an _isomorphism_.
>
>They are not isomorphic. Consider:
>Semigroup 1, with elements a and b:
> aa = a, ab = a, ba = b and bb = b.
>Semigroup 2, with elements a and b:
> aa = a, ab = b, ba = a and bb = b.
>
>How is semigroup 1 isomorphic with semigroup 2?
Oops. Never mind...
>You have also to reverse the operation.
>--
>dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
>home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
David C. Ullrich
David C. Ullrich
Ullrich) wrote:
>On Wed, 27 Mar 2002 13:18:32 +0100, Nico Benschop
><n.ben...@chello.nl> wrote:
>
>>Jan Kristian Haugland wrote:
>>>
[...]
>>
>>Well, they are 'anti-isomorphic':
>
>They are also _isomorphic_. Mapping a to a and b to b gives
>an anti-isomorphism. There is _another_ way to map one semigroup
>to the other that gives an _isomorphism_.
>
>(It's not hard to find the isomorphism, since there are not
>all that many maps from {a,b} 1-1 onto {a,b} in the first
>place - write down all the possible mappings, and check each
>one to see whether it's an isomorphism.)
Thanks to JKH for pointing out I was babbling nonsense here.
Sorry.
David C. Ullrich
James Dolan
robin chapman <r...@maths.ex.ac.uk> wrote:
|"Nico Benschop" <n.ben...@chello.nl> wrote in message
|news:3CA1C201...@chello.nl...
|
|>
|> BTW: calling left- and right- copy idempotent semigroup 'equivalent',
|> as apparently mathematicians in [1] do, is declaring past equivalent
|> to future - in the practical FSM / engineering interpretation of
|> these closures as dual memory functions:
|> branch vs. set/reset FF, that is: long vs. short memory.
|> Making the word 'intelligence' strongly dependent on context, indeed:
|> "In theory there is no difference between theory and practice,
|> but in practice there is..."
|
|You don't half talk some pretentious bollocks, Mr Benschop.
|If you examine Sloane's site, you'll find enumerations both
|regarding non-isomorphic anti-isomorphic semigroups as
|equivalent and as inequivalent. It's just mathematicians'
|natural curiosity regarding how many structures of a certain
|kind there are under as many notions of equivalence
|as they can find.
actually benschop has a point in this case; it _is_ rather asinine to
take "isomorphism or anti-isomorphism" as the most basic kind of
equivalence between semigroups. for example, consider the two
two-element semigroups mis-classified as one; for an illustration of
how quite different they are in some important ways, consider their
actions on sets. a set with a pair of constant maps on it can be
construed as the edges of a certain sort of directed graph, whereas a
set with a pair of co-constant maps on it is a rather different sort
of combinatorial object.
Jan Kristian Haugland
"David C. Ullrich" wrote:
> Thanks to JKH for pointing out I was babbling nonsense here.
> Sorry.
Still confusing me with Dik T. Winter? You
said on August 2nd that you did that...:-)
--
J K Haugland
http://hjem.sol.no/neutreeko
David C. Ullrich
<avai...@web.page> wrote:
>
>"David C. Ullrich" wrote:
>
>> Thanks to JKH for pointing out I was babbling nonsense here.
>> Sorry.
>
>Still confusing me with Dik T. Winter? You
>said on August 2nd that you did that...:-)
Aargh. Yes, sorry. I don't know what the problem is
here, it's not like the names are similar or anything.
>--
>
>J K Haugland
>http://hjem.sol.no/neutreeko
David C. Ullrich
Robin Chapman
news:a7snmb$8a4$1...@glue.ucr.edu...
> in article <a54d3230330a0667e30...@mygate.mailgate.org>,
> robin chapman <r...@maths.ex.ac.uk> wrote:
>
> |"Nico Benschop" <n.ben...@chello.nl> wrote in message
> |news:3CA1C201...@chello.nl...
> |
> |>
> |> BTW: calling left- and right- copy idempotent semigroup 'equivalent',
> |> as apparently mathematicians in [1] do, is declaring past equivalent
> |> to future - in the practical FSM / engineering interpretation of
> |> these closures as dual memory functions:
> |> branch vs. set/reset FF, that is: long vs. short memory.
> |> Making the word 'intelligence' strongly dependent on context, indeed:
> |> "In theory there is no difference between theory and practice,
> |> but in practice there is..."
> |
> |You don't half talk some pretentious bollocks, Mr Benschop.
> |If you examine Sloane's site, you'll find enumerations both
> |regarding non-isomorphic anti-isomorphic semigroups as
> |equivalent and as inequivalent. It's just mathematicians'
> |natural curiosity regarding how many structures of a certain
> |kind there are under as many notions of equivalence
> |as they can find.
>
> actually benschop has a point in this case;
No he doesn't.
> it _is_ rather asinine to
> take "isomorphism or anti-isomorphism" as the most basic kind of
> equivalence between semigroups.
No one is asserting that this is "the most basic kind of equivalence".
Sloane's database contains both counts of semigroups under this
equivalence and under isomorphism. (And also several
other related counts).
James Dolan
james dolan <jdo...@math-lw-n01.math.ucr.edu> wrote:
|actually benschop has a point in this case; it _is_ rather asinine to
|take "isomorphism or anti-isomorphism" as the most basic kind of
|equivalence between semigroups. for example, consider the two
|two-element semigroups mis-classified as one; for an illustration of
|how quite different they are in some important ways, consider their
|actions on sets. a set with a pair of constant maps on it can be
|construed as the edges of a certain sort of directed graph, whereas a
|set with a pair of co-constant maps on it is a rather different sort
|of combinatorial object.
hmm, that turned out _at least_ as bad as the terminology i was
objecting to.
in certain contexts, an element x of an abstract monoid is aptly and
usefully called "constant" iff performing any element y followed by
performing x always yields x. however, in the above post i applied
this terminology not in the case of an abstract monoid but rather in
the case of a concrete semigroup, with bad results. unlike
semigroups, monoids always have an identity element, and letting the
identity element take the role of y shows that the value of y*x being
independent of the value of y is equivalent to the law y*x = x; this
nice feature doesn't work for semigroups. even worse, elements of
concrete semigroups (or monoids) are maps, and describing a map as a
"constant map" of course has a different meaning. actually it's not
_that_ different, but a concrete representation of an abstract monoid
or semigroup generally doesn't preserve the meaning of "constant".
James Dolan
robin chapman <r...@maths.ex.ac.uk> wrote:
|> actually benschop has a point in this case;
|
|No he doesn't.
sure he does.
|> it _is_ rather asinine to
|> take "isomorphism or anti-isomorphism" as the most basic kind of
|> equivalence between semigroups.
|
|No one is asserting that this is "the most basic kind of equivalence".
that's about as relevant as the fact that no one's asserting that
anyone's asserting that this is "the most basic kind of equivalence".
the website:
http://mathworld.wolfram.com/Semigroup.html
that was quoted in this thread baldly declares "the total number of
semigroups of order n are 1, 4, 18, 126, 1160, 15973, 836021, ...",
which is an implicit suggestion that the "default" and most basic kind
of equivalence between semigroups is the asinine kind.
|Sloane's database contains both counts of semigroups under this
|equivalence and under isomorphism. (And also several
|other related counts).
completely irrelevant.
by the way, your bizarre crusade against benschop makes you look like
an unpleasant moron. why not just shut up unless you have something
interesting to say?
Robin Chapman
> in article <333690008b025e2d19b...@mygate.mailgate.org>,
> robin chapman <r...@maths.ex.ac.uk> wrote:
>
> |> actually benschop has a point in this case;
> |
> |No he doesn't.
>
> sure he does.
Certainly not.
> |> it _is_ rather asinine to
> |> take "isomorphism or anti-isomorphism" as the most basic kind of
> |> equivalence between semigroups.
> |
> |No one is asserting that this is "the most basic kind of equivalence".
>
> that's about as relevant as the fact that no one's asserting that
> anyone's asserting that this is "the most basic kind of equivalence".
> the website:
>
> http://mathworld.wolfram.com/Semigroup.html
>
> that was quoted in this thread baldly declares "the total number of
> semigroups of order n are 1, 4, 18, 126, 1160, 15973, 836021, ...",
It doesn't baldly declare that, but provides a link to Sloane's
database. Sloane's page describes exactly what these numbers
count, while Weisstein's doesn't.
(Why Weisstein should want to link to that particular sequence is
another question.)
> which is an implicit suggestion that the "default" and most basic kind
> of equivalence between semigroups is the asinine kind.
We're down to "implicit"s now. Oh dear :-(
> |Sloane's database contains both counts of semigroups under this
> |equivalence and under isomorphism. (And also several
> |other related counts).
>
> completely irrelevant.
Relavent in that it refutes your attempted argument.
> by the way, your bizarre crusade against benschop makes you look like
> an unpleasant moron. why not just shut up unless you have something
> interesting to say?
Tu quoque.
Lee Rudolph
>> an unpleasant moron. why not just shut up unless you have something
>> interesting to say?
>
>Tu quoque.
I've always wanted to see someone challenged to a dual.
Lee Rudolph
Dik T. Winter
> "David C. Ullrich" wrote:
>
> > Thanks to JKH for pointing out I was babbling nonsense here.
> > Sorry.
>
> Still confusing me with Dik T. Winter? You
> said on August 2nd that you did that...:-)
For some Americuns we are both Swedes ;-).
Nico Benschop
>
> in article <333690008b025e2d19b...@mygate.mailgate.org>,
> robin chapman <r...@maths.ex.ac.uk> wrote:
>
> |> actually benschop has a point in this case;
> |
> |No he doesn't.
>
> sure he does.
>
> |Sloane's database contains both counts of semigroups under
> |this equivalence and under isomorphism. (And also several
> |other related counts).
>
> completely irrelevant.
>
> by the way, your bizarre crusade against benschop
> makes you look like an unpleasant moron.
> why not just shut up unless you have something interesting to say?
You take the words from my mouth
(too aggressive for print, I find; yet possibly effective if
coming from from someone else, for a change;-)
PS: I think I've explained why, in a context where *both* left- and
right- copy idempotent semigroups occur (as in FSM closures
modeling a sequential logic circuit with memory functions like
S/R-FFs and Branches - which have anti-isomorphic semigroups),
it is most essential to count them as distinct types of semigroups!
In fact this holds even in case of isomorphic AND(.) ~ OR(+) viewed
as order=2 semigroups: normally *both* are required in combinational
Boolean logic functions, like (a+b).c -- which btw is one of the
two causes of asymmetry in Boolean functions.
-- NB
Jan Kristian Haugland
"Dik T. Winter" wrote:
>
> In article <3CA1E240...@web.page> Jan Kristian Haugland <avai...@web.page> writes:
> > "David C. Ullrich" wrote:
> >
> > > Thanks to JKH for pointing out I was babbling nonsense here.
> > > Sorry.
> >
> > Still confusing me with Dik T. Winter? You
> > said on August 2nd that you did that...:-)
>
> For some Americuns we are both Swedes ;-).
Yeah. I'm actually 25% Swedish...
Nico Benschop
>
> >> By the way, your bizarre crusade against Benschop makes you
> >> look like an unpleasant moron. Why not just shut up unless
> >> you have something interesting to say? -- James Dolan
> >
> > Tu quoque. -- Robin Chapman
>
> I've always wanted to see someone challenged to a dual.
Isn't that a jewel ? -- NB
(my Dutch/English dictionary has no English word for 'duel', other
than 'single combat': one man fighting himself - could that be RC ?-)
Nico Benschop
>
> "Nico Benschop" <n.ben...@chello.nl> wrote in message
> news:3CA1C201...@chello.nl...
>
> >
> > BTW:
> > as apparently mathematicians in [1] do, is declaring past
> > equivalent to future - in the practical FSM / engineering
> > interpretation of these closures as dual memory functions:
> > branch vs. set/reset FF, that is: long vs. short memory.
> > Making the word 'intelligence' strongly dependent on context,
> > indeed:
> > "In theory there is no difference between theory and practice,
> > but in practice there is..."
>
> You don't half talk some pretentious bollocks, Mr Benschop.
(which I presume is in English, not my mother tongue)
And while you're at it: what is "tu quoque" - from another post
by this illustrious and erudite person, as answer to James Dolan's:
>| by the way, your bizarre crusade against benschop makes you look
>| like an unpleasant moron. why not just shut up unless you have
>| something interesting to say?
> > -- NB - not surprised anymore about the mismatch of understanding
> > between mathematicians and engineers (of what is 'essential').
>
> Yes. there seems to be a mismatch between mathematicians
> and at least one engineer. All mathematicians regard
> understanding as essential, while at least one engineer
> regards understanding with contempt. -- Robin Chapman
I wonder, who could that one engineer be? Moreover, from
the posts in this thread, how does the conclusion follow
that he "regards understanding with contempt" ?
According to my analysis, and recently acquired math and
latin knowledge, this is a non sequitur.
But then, twisting meanings is not foreign to you, I noticed.
Only a mismatch of understanding between mathematicians and
engineers was observed, regarding "what is essential". Hardly a
controversial opinion, I presume. But a rather important one to
realize, for practical purposes. In fact this is my personal
experience of at least 20 years (with several mathematicians;
'breek me de bek niet open', meaning: there's a tale to that),
shared with several of my engineering colleagues.
Also, your 'at least' tirade reminds me:
Some mathematicians, as I read here, concluded from limited
observations: in Scotland there are at least 20 sheep which are
black on at least one side (where 'side' was not quite well defined).
Although interesting for some, and showing the math'ns open mind (for
the other side possibly being orange;-) the concept of 'essential'
does seem to have another connotation for them. Just good to know...
(e.g: 90 years old Boolean algebra discovered as an essential model of
binary logic networks [C.Shannon 1938] is a trivial isomorphism;-)
-- NB
John R Ramsden
>
> robin chapman <r...@maths.ex.ac.uk> wrote:
> >
> > [...]
>
> by the way, your bizarre crusade against benschop makes you look like
> an unpleasant moron. why not just shut up unless you have something
> interesting to say?
Hey don't knock it. Many sci.math readers, probably all I'd say, with
any interest in number theory must have found Robin's summaries of
Benschopera ("the works of Benshop") invaluable, and will certainly
have filed them away even if they haven't got round to studying them
yet.
Robin is one of the few people with the skill and patience to provide
them, the only person to my recollection. Needless to say Nico himself
has never been able or willing.
As for the more, um, robust exchanges, well maybe I have a warped sense
of humour (or perhaps I'm an unpleasant moron), but they're always good
for a giggle. The funniest aspect is Nico's pained replies, as if he
thinks it's a genuine injustice to have his persistent evasions and
errors pointed out despite rarely admitting them and never remedying
them or, as far as I know, withdrawing his claims.
Cheers
---------------------------------------------------------------------------
John R Ramsden (j...@adslate.com)
---------------------------------------------------------------------------
Life is what happens to you while you're planning your next move.
Mick Jagger
---------------------------------------------------------------------------
John R Ramsden
> >
> >"David C. Ullrich" wrote:
> >
> > > Thanks to JKH for pointing out I was babbling nonsense here.
> > > Sorry.
> >
> > Still confusing me with Dik T. Winter? You
> > said on August 2nd that you did that...:-)
>
> Aargh. Yes, sorry. I don't know what the problem is
> here, it's not like the names are similar or anything.
Hmm, maybe the following diagram will help. Actually it almost
looks commutative, so maybe we can prove they are isomorphic:
Jan Kristian Haugland
/ \
|/_ _\|
January ->- Christmas (Russian Orthodox)
\ /
_\| |/_
Dik T Winter
d...@cmtzone.net
<n.ben...@chello.nl> wrote:
>Robin Chapman wrote:
>> You don't half talk some pretentious bollocks, Mr Benschop.
>
>Can someone explain the syntax of this sentence to me?
>(which I presume is in English, not my mother tongue)
It's not in English, it's in British. ;-)
>And while you're at it: what is "tu quoque" - from another post
Latin for "you also."
-don
Lee Rudolph
>
>Latin for "you also."
But in context more idiomatically rendered (in American English;
I'm sure there's an equivalent British English idiom) as "same
to you!", sometimes elaborated to, for example, "same to you,
buddy!, "same to you, Buster!", "same to you with bells on!",
etc., etc.
Lee Rudolph
David C. Ullrich
wrote:
>jdo...@math-lw-n01.math.ucr.edu (James Dolan) wrote:
>>
>> robin chapman <r...@maths.ex.ac.uk> wrote:
>> >
>> > [...]
>>
>> by the way, your bizarre crusade against benschop makes you look like
>> an unpleasant moron. why not just shut up unless you have something
>> interesting to say?
>
>Hey don't knock it. Many sci.math readers, probably all I'd say, with
>any interest in number theory must have found Robin's summaries of
>Benschopera ("the works of Benshop") invaluable, and will certainly
>have filed them away even if they haven't got round to studying them
>yet.
>
>Robin is one of the few people with the skill and patience to provide
>them, the only person to my recollection. Needless to say Nico himself
>has never been able or willing.
>
>As for the more, um, robust exchanges, well maybe I have a warped sense
>of humour (or perhaps I'm an unpleasant moron), but they're always good
>for a giggle. The funniest aspect is Nico's pained replies, as if he
>thinks it's a genuine injustice to have his persistent evasions and
>errors pointed out despite rarely admitting them and never remedying
>them or, as far as I know, withdrawing his claims.
Precisely. Well, I think that he _sometimes_ withdraws a claim or
two, but his complaints about people nitpicking regarding things
which are evidently actual errors is indeed the funniest part.
>Cheers
>
>---------------------------------------------------------------------------
>John R Ramsden (j...@adslate.com)
>---------------------------------------------------------------------------
>Life is what happens to you while you're planning your next move.
> Mick Jagger
>---------------------------------------------------------------------------
David C. Ullrich
David C. Ullrich
wrote:
>ull...@math.okstate.edu (David C. Ullrich) wrote:
>>
>> Jan Kristian Haugland <avai...@web.page> wrote:
>> >
>> >"David C. Ullrich" wrote:
>> >
>> > > Thanks to JKH for pointing out I was babbling nonsense here.
>> > > Sorry.
>> >
>> > Still confusing me with Dik T. Winter? You
>> > said on August 2nd that you did that...:-)
>>
>> Aargh. Yes, sorry. I don't know what the problem is
>> here, it's not like the names are similar or anything.
>
>Hmm, maybe the following diagram will help. Actually it almost
>looks commutative, so maybe we can prove they are isomorphic:
>
> Jan Kristian Haugland
> / \
> |/_ _\|
> January ->- Christmas (Russian Orthodox)
> \ /
> _\| |/_
> Dik T Winter
Hey, that's exactly it, that's why I get the two names
confused so often! Thanks.
>Cheers
>
>---------------------------------------------------------------------------
>John R Ramsden (j...@adslate.com)
>---------------------------------------------------------------------------
>Life is what happens to you while you're planning your next move.
> Mick Jagger
>---------------------------------------------------------------------------
David C. Ullrich
Nico Benschop
>
> >>And while you're at it: what is "tu quoque", from another post.
> >
> >Latin for "you also." -- don
>
> But in context more idiomatically rendered (in American English;
> I'm sure there's an equivalent British English idiom) as
> "same to you!", sometimes elaborated to, for example,
> "same to you, buddy!,
> "same to you, Buster!",
Thanks, I get the picture...
So can I now pose as an erudite person, by using this Latin phrase?
(re someones signature: On internet noone knows you're a jerk;-)
-- NB
Nico Benschop
>
> [on NB vs RC]
> As for the more, um, robust exchanges, well maybe I have a warped
> sense of humour (or perhaps I'm an unpleasant moron), but they're
> always good for a giggle. The funniest aspect is Nico's pained
> replies, as if he thinks it's a genuine injustice to have his
> persistent evasions and errors pointed out despite rarely admitting
> them and never remedying them or, as far as I know, withdrawing his
You're pretty out of date, John!
I complemented RC and thanked him several times for his effort,
forgave him his bad language (which he really can't help, the poor
fellow - something like Tourette's syndrome, I suspect;-) Moreover,
I admitted what was wrong, corrected and re-installed most of it,
and even wrote an extended result) separate paper on one of them:
http://www.iae.nl/users/benschop/fst-root.dvi
"On divisors of q = p \pm 1 and the idempotents of Z(.) mod q
for odd prime p, and their p-th power residues mod p^3."
(with one slip in thm3.2 still to be corrected)
So, to follow RCs style: "Stop your excremental drivel,
your bogus arguments, and get lost."-)
-- NB - http://www.iae.nl/users/benschop
Dik T. Winter
> James Dolan wrote:
> > in article <333690008b025e2d19b...@mygate.mailgate.org>,
> > robin chapman <r...@maths.ex.ac.uk> wrote:
> > |Sloane's database contains both counts of semigroups under
> > |this equivalence and under isomorphism. (And also several
> > |other related counts).
> >
> > completely irrelevant.
Oh, but it is.
> PS: I think I've explained why, in a context where *both* left- and
> right- copy idempotent semigroups occur
There are situations where you have to distinguish semigroups that are
equivalent and situations where there is no need to do so. For instance
when you study at the lowest level the subject of semigroups the
distinction is only a difference in notation. At that level the
semigroups are identical for all purposes. However, you can get in
situations where the order becomes important, and in that case you
have to distinguish the two. But you now also that the distinction
will only be there if the semigroup is not commutative.
Something similar occurs in other fields. Consider the polynomial
x^7 - 1. We know it has one root equal to 1 and 6 other roots.
Analytically we can describe those other root and distinguish them.
However, algebraically we are not able to distinguish those 6 roots.
So when we talk about the number field Q(rho) where rho is one of
those six roots, no distinction is made. Or at a lower level, in
some fields you can not distinguish i from -i, and it is just a
notational question, in other fields the distinction becomes
important.
So the Sloane series that gives numbers of semigroups noting as
equivalent also those that are that under an anti-isomorfism
caters for the people that study semigroups (or even higher at
an algebraic level). When the numbers are given for groups
under isomorfisms only, they cater for other people. So it is
quite relevant that both series are given.
Nobuo Saito
> in article <333690008b025e2d19b...@mygate.mailgate.org>,
> robin chapman <r...@maths.ex.ac.uk> wrote:
> by the way, your bizarre crusade against benschop makes you look like
> an unpleasant moron. why not just shut up unless you have something
> interesting to say?
I don't say this far, but I think he's a bit strange about Mr Benschop.
Could anyone(or Robin himself) tell me why he's so, uh, against Mr Benschop?
Pertti Lounesto
>jdo...@math-lw-n01.math.ucr.edu (James Dolan) wrote:
>
>>robin chapman <r...@maths.ex.ac.uk> wrote:
>>
>[snip]
>
>>by the way, your bizarre crusade against benschop makes you look like
>>an unpleasant moron. why not just shut up unless you have something
>>interesting to say?
>>
>
>I don't say this far, but I think he's a bit strange about Mr Benschop.
>Could anyone(or Robin himself) tell me why he's so, uh, against Mr Benschop?
>
Chapman aspires to become a mathematician. He knows
that an essential ingredient is learning the scientific process:
falsification of results of peers. He has decided to start
from an easy target, a non-mathematician like Benschop.
Later, he will try to proceed to falsification of results of
advanced mathematicians. I suggest as a model my site
http://www.teli.stadia.fi/~lounesto/counterexamples.htm.
Dik T. Winter
> jdo...@math-lw-n01.math.ucr.edu (James Dolan) wrote in message news:<a7ssdl$9ms$1...@glue.ucr.edu>...
> > an unpleasant moron. why not just shut up unless you have something
> > interesting to say?
>
> I don't say this far, but I think he's a bit strange about Mr Benschop.
> Could anyone(or Robin himself) tell me why he's so, uh, against Mr Benschop?
I think I can as I have followed it quite a bit. On numerous occasions
Robin Chapman read through Benschop's manuscripts and gave detailed
analysis of errors and problems in those manuscripts. In many cases
Benschop just ignored the critisisms or provided an answer that was not
an answer at all. In one case Robin even provided an explicit
counterexample to a theorem of Benschop, Benschop just waved it away...
Pertti Lounesto
>genki...@hotmail.com (Nobuo Saito) writes:
> > jdo...@math-lw-n01.math.ucr.edu (James Dolan) wrote in message news:<a7ssdl$9ms$1...@glue.ucr.edu>...
> > > by the way, your bizarre crusade against benschop makes you look like
> > > an unpleasant moron. why not just shut up unless you have something
> > > interesting to say?
> >
> > I don't say this far, but I think he's a bit strange about Mr Benschop.
> > Could anyone(or Robin himself) tell me why he's so, uh, against Mr Benschop?
>
>I think I can as I have followed it quite a bit. On numerous occasions
>Robin Chapman read through Benschop's manuscripts and gave detailed
>analysis of errors and problems in those manuscripts. In many cases
>Benschop just ignored the criticisms or provided an answer that was not
>an answer at all. In one case Robin even provided an explicit
>counterexample to a theorem of Benschop, Benschop just waved it away...
>
That seems a strange reason to be against somebody
(to provide a counterexample, which is waved away).
I have provided several counterexamples to theorems
proved by prominent mathematicians. Usually, they
have only been grateful to me for showing them the way,
see http://www.teli.stadia.fi/~lounesto/counterexamples.htm.
There must be something else in the strange relation
between Benschop and Chapman. Maybe a woman?
Nico Benschop
>
> Dik T. Winter wrote:
>
> >genki...@hotmail.com (Nobuo Saito) writes:
> > > jdo...@math-lw-n01.math.ucr.edu (James Dolan) wrote in message news:<a7ssdl$9ms$1...@glue.ucr.edu>...
> > > > by the way, your bizarre crusade against benschop makes you look like
> > > > an unpleasant moron. why not just shut up unless you have something
> > > > interesting to say?
> > >
> > > I don't say this far, but I think he's a bit strange about Mr Benschop.
> > > Could anyone(or Robin himself) tell me why he's so, uh, against Mr Benschop?
> >
> >I think I can as I have followed it quite a bit. On numerous occasions
> >Robin Chapman read through Benschop's manuscripts and gave detailed
> >analysis of errors and problems in those manuscripts. In many cases
> >Benschop just ignored the criticisms or provided an answer that was not
> >an answer at all. In one case Robin even provided an explicit
> >counterexample to a theorem of Benschop, Benschop just waved it away...
> >
> That seems a strange reason to be against somebody
> (to provide a counterexample, which is waved away).
Moreover, it was NOT waved away, but answered appropriately - showing
precisely that it was computed to satisfy a condition which, as first
line mentioned in the proof of a thm, was obviously missing from the
theorem statement as a necessary condition. And instead of noting this
easily corrected oversight on my part, Chapman went out of his way to
compute the counter example. How desperate must he be to go this
far in order to discredit me in such backhanded fashion, I ask you.
I'd call it pathetic, really. And this is not the first time he
'blew up' a rather simple-to-correct mistake in a text of mine, even
to the point of a typo, claimed to be a gross mistake in notation!
> I have provided several counterexamples to theorems
> proved by prominent mathematicians. Usually, they
> have only been grateful to me for showing them the way,
> see http://www.teli.stadia.fi/~lounesto/counterexamples.htm.
> There must be something else in the strange relation
> between Benschop and Chapman. Maybe a woman?
No, I just have no idea what it could be. Possibly a bet of his (with
A.v.d.P?) that he could 'stamp Benschop into the ground' in no time...
Which of course didn't happen, since he (RC) does not know the
difference between main line (the 'approach') and details of a proof.
Details, of course, should be correct as well, but can be corrected
without any loss to the main line, which in the FLT case is that:
"No additive p-th power residue equivalence mod p^k,
that depend on exponent p distributing over a sum,
can be extended to integer equality."
.. what I'd call the "remaining (Chinese) carry theorem" - since the
carries (produced by the p-th powers < p^{pk} of k-digit residues)
make the difference!
Nobuo Saito
> Dik T. Winter wrote:
[snip]
> >
> >I think I can as I have followed it quite a bit. On numerous occasions
> >Robin Chapman read through Benschop's manuscripts and gave detailed
> >analysis of errors and problems in those manuscripts. In many cases
> >Benschop just ignored the criticisms or provided an answer that was not
> >an answer at all. In one case Robin even provided an explicit
> >counterexample to a theorem of Benschop, Benschop just waved it away...
> >
> That seems a strange reason to be against somebody
> (to provide a counterexample, which is waved away).
> I have provided several counterexamples to theorems
> proved by prominent mathematicians. Usually, they
> have only been grateful to me for showing them the way,
> see http://www.teli.stadia.fi/~lounesto/counterexamples.htm.
> There must be something else in the strange relation
> between Benschop and Chapman. Maybe a woman?
I tend to agree with Mr Lounesto.
My hunch is that there is a subtle but important reason only a few people
or none(even not Robin himself) know.
Interesting.
Nico Benschop
> Robin himself) know. Interesting.
Interesting indeed, this concerns (often unconscious) motivations,
and relates to psychology - rather then mathematics... Although
somewhat off topic, let me offer some reflections of mine (being
involved with this and similar math/engineering controversy for say
the past 8 years, at least;-)
1. Being no mathematician by training, but a researcher in industry,
on formal design/spec'n/synthesis methods for sequential logic
VLSI (control logic FSM's, better methods to implement binary
arithmetic) - with finite semigroups = finite function composition
= associative algebra as central concept, I stumbled over 3 as
semi_primitive root of 1 mod 2^k (any k>2): not generally considered
very important by mathematicians ('trivial';-) to my surprise.
re US patent ( http://home.iae.nl/users/benschop/pat3star.dvi )
2. During that research I noticed (also trivial) that each subgroup
of a units group (mod p^k, for instance) sums to zero, hence also
the three cubic roots of 1 mod p^k (any prime p == 1 mod 6, at any
precision k>0) ... which is a 'counter example' to FLT (case_1) but
only for residues, of course, yet (a+b)^p == a^p + b^p (mod p^k)
for such cubic roots --> EDS property of each FLT solution "in core":
Exponent p Distributes over a Sum
(core A_k = units subgrp of order p-1, so n^p == n, as extended FST)
--> Main Question: can a solution of FLT for residues, with the EDS
property (that does NOT hold for integers, re the Pascal Triangle),
be extended to equality for integer p-th powers ( pk digits base p)?
I figured not (summer 1994), and I decided to give it a try at
proving this (to attract attention to finite semigroups for
FSM synthesis purposes, not really explored since
Krohn/Rhodes' decomposition restricted to reset/permutation
machines,
hence using only groups & Jordan/Hoelder)
With a first trial submission to AMM probelem section, using
the (rhetoric) question:
Given this (cubic root) approach to FLT via FST, where is the fault?
(which by consensus MUST be there somewhere, since FLT has
no simple proof, not even its first case - coprime to p).
http://home.iae.nl/users/benschop/scimat98.htm
More than half a year passed, with emails bouncing back-and-forth,
and dragging of feet by the AMM, until I finally withdrew by lack of
response. Next, going into the 'lions den': a math conference on
semigroups (Prague '96) where my short paper on this approach was
excepted. All to no avail: no response, cold shoulder, no discussion
(not even after prompting of the audience by one of the organizing
professors;-(
Reason for me to analyse such attitude, deriving e.g. a formula for
the probability of a self respecting math journal to place a short and
direct proof of any well known (but 'stigmatized') hard math problem;
see http://home.iae.nl/users/benschop/inertia.htm
By lack of better feedback, as I mentioned in "Goldbach's" thread
last week (on a cynical note):
" There's nothing like an angry opponent, to be well motivated
for checking your program (or proof;-)" -- NB
Which RC does for me, with some positive residue, after all - when
passing his replies through a sharp filter. I'm grateful, really.
(een kinderhand is gauw gevuld;-)
And regarding some possible psychological/ motivational reasons for
such - not uncommon- violent behaviour of mathematicians against any
suggestion of a direct approach:
1. They are bombarded by too many simplistic stuff, and get angry.
2. Even if a suggestion makes it to a math conference, the
principle: "Don't stick your neck out if not absolutely
necessary" holds!
Avoid even to be seen, by colleagues, near such a person, since
reputations are built-up slowly through the years, but are
destroyed in a flick of a second (as by a contagious disease...)
3. So a serious mathematician prudently 'keeps distance'.
To which rule RC did not succumb, since he is - apparently -
an ardent self-appointed defender of the Faith, and protector
of the Holy Grail (= unsolvability of FLT by direct means)...
Even pointing at the Grail is punishable by severe ridiculization.
Re non-discussion in the NMBRTHRY list mar99 (yes, I made it that far,
much to the chagrin of the unfortunate/inexperienced moderator,
who got no doubt reprimanded by colleagues for such gross error;-):
in http://home.iae.nl/users/benschop/nr-th.htm
(AFAIK RC was just on sabbatical year at Alf's group in Sydney when
I wrote my request to A.v.d.Poorten for his opinion on my cubic root
and EDS approach, using an extension of FST).
Just some background ('voer voor psychologen'). -- NB
Nico Benschop
>
> [.. on the essential difference between main line and details ..]
>
> Details, of course, should be correct as well, but can be corrected
^^^^^^^^
> "No additive p-th power residue equivalence mod p^k,
> can be extended to integer equality." (e.g: FLT mod p^2)
>
> .. what I'd call the "remaining 'Chinese' carry theorem" - since
> carries, produced by the p-th powers < p^{pk} of k-digit residues
> (base p), make the difference!http://home.iae.nl/users/benschop/carry.htm