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location  Germany  
age  48  
visits  member for  5 years 
seen  48 mins ago  
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My interests:
abstract structures
(e.g. graphs, groups, polytopes, spaces, ...)
^{in the course of this}category theory
model theory
presentations and representations
(e.g. of abstract structures by or inside other abstract structures)
^{next to this}philosophy
(esp. of mathematics, science, and mind)
^{especially concerned with}atomism
(i.e. reductionistic or other theories referring to some kind of "atoms") ancient atomism
 set theories (with and without urelements)
 logical atomism
 particle physics
 neuroscience
^{and}
their refutations
2d

awarded  Yearling 
Oct 14 
awarded  SelfLearner 
Oct 14 
answered  Supervenience in mathematics 
Sep 29 
comment 
Rationale behind an requirement on Turing machines
Notice further, that $b$ can not only be read but also be written. 
Sep 29 
comment 
Rationale behind an requirement on Turing machines
Rereading your answer I am still not satisfied: In Hopcroft/Ullman's definition of a Turing machine as a 7tupel the blank symbol $b$ is distinguished (among the tape symbols), but there seems to be no requirement on a Turing machine involving $b$. Especially the transition function $\delta$ can be defined on the whole tape alphabet (including $b$). The same holds  by the way  for the initial state $q_0$. 
Sep 27 
awarded  Nice Answer 
Sep 24 
awarded  Autobiographer 
Sep 3 
awarded  Peer Pressure 
Aug 30 
comment 
When are two algorithms essentially the same?
I didn't want to ask for a single equivalence, sorry if it sounded like this. I am looking for any (sensible) equivalence. One came to my mind only today: $T$ is equivalent to $T'$ when they have the same symbols and blank symbol, compute the same function, have the same number of internal states and there is a bijection $f$ between the states that preserves  among other things  starting and halting, next state $S(f(x),s)$ is $f(S(x),s)$, and so on. One may find this trivial, but it is a sensible equivalence, isn't? And there may be others. 
Aug 29 
comment 
When are two algorithms essentially the same?
If the question is read "which definition of equivalence between unary addition Turing machines is sensible?" there is an answer (and you gave it): none but the trivial one: all such machines (defined by their inputoutput behaviour) are equivalent. And indeed they are: they compute the same function. 
Aug 29 
comment 
When are two algorithms essentially the same?
Why not did you simply answer by "none" (= one might want to apply no definition of equivalence). A real answer to a real question. 
Aug 28 
asked  When are two algorithms essentially the same? 
Aug 26 
comment 
Rationale behind an requirement on Turing machines
Yet another question: Does your answer imply that my $T_{bindouble}$ computes nothing, thus is no Turing machine at all? And what about $T_{id}$ which rewrites the first symbol and halts? $T_{id}$ seems to work equally well for unary and binary and arbitrary $k$ary (interpreted) inputs. 
Aug 26 
comment 
Rationale behind an requirement on Turing machines
You seem to imply that one does not really need to require a blank symbol? So what is your specific answer? "In fact one does not need to require a blank symbol because there are encoding schemes that allow to neglect them?" How can this metamathematical statement be made totally precise? 
Aug 26 
comment 
Rationale behind an requirement on Turing machines
Yes, this is what I mean. But you put the head initially on a specific square (left or equal to the first $1$). 
Aug 26 
comment 
Rationale behind an requirement on Turing machines
One question: you write "input is padded with infinitely many additional $0$s". How can this be, when we have only finitely many $1$s? 
Aug 26 
comment 
Rationale behind an requirement on Turing machines
@Joel: Thank you very much! (Now, that I know and understand your answer, I see that my question was not really "research level". Do you find it inappropriate for MO, too?) 
Aug 26 
accepted  Rationale behind an requirement on Turing machines 
Aug 26 
revised 
Rationale behind an requirement on Turing machines
edited tags 
Aug 26 
comment 
Rationale behind an requirement on Turing machines
Is it "unmathematical thinking" to think about the reasons why a condition in a definition was chosen? Maybe it's "metamathematical thinking", but it's "mathematical" nevertheless, isn't it? Should I ask the question at meta.mathoverflow.net? 