Changes

6,205 bytes removed , 15:30, 11 October 2013

update

Line 117: Line 117:

3 & 1 & 0 & 0 \\

4 & 1 & 0 & 0 \\

−

5 & {}^{\shortparallel} & {}^{\shortparallel} & {}^{\shortparallel}

+

5 & {}^\shortparallel & {}^\shortparallel & {}^\shortparallel

\end{array}</math>

|}

Line 134: Line 134:

3 & 1 & 0 & 0 \\

4 & 1 & 0 & 0 \\

−

5 & {}^{\shortparallel} & {}^{\shortparallel} & {}^{\shortparallel}

+

5 & {}^\shortparallel & {}^\shortparallel & {}^\shortparallel

\end{array}</math>

|}

Line 1,622: Line 1,622:

===Computation Summary for Logical Equality===

−

Figure 2.1 shows the expansion of <math>g = \texttt{((u, v))}</math> over <math>[u, v]\!</math> to produce the expression:

+

Figure 2.1 shows the expansion of <math>g = \texttt{((} u \texttt{,~} v \texttt{))}\!</math> over <math>[u, v]\!</math> to produce the expression:

{| align="center" cellpadding="8" width="90%"

−

| <math>\~~texttt~~{uv~~} ~~~+~ \texttt{(u)(v)}</math>

+

|

+

<math>\begin{matrix}

+

uv & + & \texttt{(} u \texttt{)(} v \texttt{)}

+

\end{matrix}</math>

|}

−

Figure 2.2 shows the expansion of <math>\mathrm{E}g = \texttt{((u + du, v + dv))}</math> over <math>[u, v]\!</math> to produce the expression:

+

Figure 2.2 shows the expansion of <math>\mathrm{E}g = \texttt{((} u + \mathrm{d}u \texttt{,~} v + \mathrm{d}v \texttt{))}\!</math> over <math>[u, v]\!</math> to produce the expression:

−

{| align="center" cellpadding="8" width="90%"

+

{| align="center" cellpadding="8" width="90%"

−

| <math>\~~texttt~~{uv} \cdot \texttt{((du, dv))} + \texttt{u(v)} \cdot \texttt{(du, dv)} + \texttt{(u)v} \cdot \texttt{(du, dv)} + \texttt{(u)(v)} \cdot \texttt{((du, dv))}</math>

+

|

+

<math>\begin{matrix}

+

uv \cdot \texttt{((} \mathrm{d}u \texttt{,~} \mathrm{d}v \texttt{))} & + &

+

u \texttt{(} v \texttt{)} \cdot \texttt{(} \mathrm{d}u \texttt{,~} \mathrm{d}v \texttt{)} & + &

+

\texttt{(} u \texttt{)} v \cdot \texttt{(} \mathrm{d}u \texttt{,~} \mathrm{d}v \texttt{)} & + &

+

\texttt{(} u \texttt{)(} v \texttt{)} \cdot \texttt{((} \mathrm{d}u \texttt{,~} \mathrm{d}v \texttt{))}

+

\end{matrix}</math>

|}

−

<math>\mathrm{E}g</math> tells you what you would have to do, from ~~where~~ you are in the universe <math>[u, v],\!</math> if you want to end up in a place where <math>g\!</math> is true. In this case, where the prevailing proposition <math>g\!</math> is <math>\texttt{((u, v))},</math> the component <math>\texttt{uv} \~~cdot~~ \texttt{~~((du~~, dv))}</math> of <math>\mathrm{E}g</math> tells you this: If <math>u\!</math> and <math>v\!</math> are both true where you are, then change either both or neither of <math>u\!</math> and <math>v\!</math> at the same time, and you will attain a place where <math>\texttt{((du, dv))}</math> is true.

+

In general, <math>\mathrm{E}g\!</math> tells you what you would have to do, from wherever you are in the universe <math>[u, v],\!</math> if you want to end up in a place where <math>g\!</math> is true. In this case, where the prevailing proposition <math>g\!</math> is <math>\texttt{((} u \texttt{,~} v \texttt{))},\!</math> the component <math>uv \cdot \texttt{((} \mathrm{d}u \texttt{,~} \mathrm{d}v \texttt{))}\!</math> of <math>\mathrm{E}g\!</math> tells you this: If <math>u\!</math> and <math>v\!</math> are both true where you are, then change either both or neither of <math>u\!</math> and <math>v\!</math> at the same time, and you will attain a place where <math>\texttt{((} u \texttt{,~} v \texttt{))}\!</math> is true.

−

Figure 2.3 shows the expansion of <math>\mathrm{D}g</math> over <math>[u, v]\!</math> to produce the expression:

+

Figure 2.3 shows the expansion of <math>\mathrm{D}g\!</math> over <math>[u, v]\!</math> to produce the expression:

−

{| align="center" cellpadding="8" width="90%"

+

{| align="center" cellpadding="8" width="90%"

−

| <math>\~~texttt~~{uv} \cdot \texttt{(du, dv)} ~+~ \texttt{u(v)} \cdot \texttt{(du, dv)} ~+~ \texttt{(u)v} \cdot \texttt{(du, dv)} ~+~ \texttt{(u)(v)} \cdot \texttt{(du, dv)}</math>

+

|

+

<math>\begin{matrix}

+

uv \cdot \texttt{(} \mathrm{d}u \texttt{,~} \mathrm{d}v \texttt{)} & + &

+

u \texttt{(} v \texttt{)} \cdot \texttt{(} \mathrm{d}u \texttt{,~} \mathrm{d}v \texttt{)} & + &

+

\texttt{(} u \texttt{)} v \cdot \texttt{(} \mathrm{d}u \texttt{,~} \mathrm{d}v \texttt{)} & + &

+

\texttt{(} u \texttt{)(} v \texttt{)} \cdot \texttt{(} \mathrm{d}u \texttt{,~} \mathrm{d}v \texttt{)}

+

\end{matrix}</math>

|}

−

<math>\mathrm{D}g</math> tells you what you would have to do, from ~~where~~ you are in the universe <math>[u, v],\!</math> if you want to bring about a change in the value of <math>g,\!</math> that is, if you want to get to a place where the value of <math>g\!</math> is different from what it is where you are. In the present case, where the ruling proposition <math>g\!</math> is <math>\texttt{((u, v))},</math> the term <math>\texttt{uv} \~~cdot~~ \texttt{~~(du~~, dv)}</math> of <math>\mathrm{D}g</math> tells you this: If <math>u\!</math> and <math>v\!</math> are both true where you are, then you would have to change one or the other but not both <math>u\!</math> and <math>v\!</math> in order to reach a place where the value of <math>g\!</math> is different from what it is where you are.

+

In general, <math>\mathrm{D}g\!</math> tells you what you would have to do, from wherever you are in the universe <math>[u, v],\!</math> if you want to bring about a change in the value of <math>g,\!</math> that is, if you want to get to a place where the value of <math>g\!</math> is different from what it is where you are. In the present case, where the ruling proposition <math>g\!</math> is <math>\texttt{((} u \texttt{,~} v \texttt{))},\!</math> the term <math>uv \cdot \texttt{(} \mathrm{d}u \texttt{,~} \mathrm{d}v \texttt{)}\!</math> of <math>\mathrm{D}g\!</math> tells you this: If <math>u\!</math> and <math>v\!</math> are both true where you are, then you would have to change one or the other but not both <math>u\!</math> and <math>v\!</math> in order to reach a place where the value of <math>g\!</math> is different from what it is where you are.

−

Figure 2.4 approximates <math>\mathrm{D}g</math> by the linear form <math>\mathrm{d}g</math> that expands over <math>[u, v]\!</math> as follows:

+

Figure 2.4 approximates <math>\mathrm{D}g\!</math> by the linear form <math>{\mathrm{d}g}\!</math> that expands over <math>[u, v]\!</math> as follows:

{| align="center" cellpadding="8" width="90%"

|

−

<math>\begin{array}{~~lll~~}

+

<math>\begin{array}{*{9}{l}}

\mathrm{d}g

−

& = & \texttt{uv}\!\~~cdot~~\!\texttt{~~(du, dv~~)} + \texttt{u(v)}\!\~~cdot~~\!\texttt{~~(du~~, dv)} + \texttt{(u)v}\!\~~cdot~~\!\texttt{~~(du, dv~~)} + \texttt{(u)(v)}\!\cdot\!\texttt{~~(du~~, dv)}

+

& = & uv \cdot \texttt{(} \mathrm{d}u \texttt{,~} \mathrm{d}v \texttt{)}

−

\\ \\

+

& + & u \texttt{(} v \texttt{)} \cdot \texttt{(} \mathrm{d}u \texttt{,~} \mathrm{d}v \texttt{)}

−

& = & \texttt{(du, dv)}

+

& + & \texttt{(} u \texttt{)} v \cdot \texttt{(} \mathrm{d}u \texttt{,~} \mathrm{d}v \texttt{)}

+

& + & \texttt{(} u \texttt{)(} v \texttt{)} \cdot \texttt{(} \mathrm{d}u \texttt{,~} \mathrm{d}v \texttt{)}

+

\\[8pt]

+

& = & \texttt{(} \mathrm{d}u \texttt{,~} \mathrm{d}v \texttt{)}

\end{array}</math>

|}

−

Figure 2.5 shows what remains of the difference map <math>\mathrm{D}g</math> when the first order linear contribution <math>\mathrm{d}g</math> is removed, namely:

+

Figure 2.5 shows what remains of the difference map <math>\mathrm{D}g\!</math> when the first order linear contribution <math>{\mathrm{d}g}\!</math> is removed, namely:

{| align="center" cellpadding="8" width="90%"

|

−

<math>\begin{~~matrix~~}

+

<math>\begin{array}{*{9}{l}}

\mathrm{r}g

−

& = & ~~\texttt{~~uv} \cdot ~~\texttt{~~0} & + & \texttt{u(v)} \cdot ~~\texttt{~~0} & + & \texttt{(u)v} \cdot ~~\texttt{~~0} & + & \texttt{(u)(v)} \cdot ~~\texttt{~~0}

+

& = & uv \cdot 0

−

\\ \\

+

& + & u \texttt{(} v \texttt{)} \cdot 0

−

& = & ~~\texttt{~~0}

+

& + & \texttt{(} u \texttt{)} v \cdot 0

−

\end{~~matrix~~}</math>

+

& + & \texttt{(} u \texttt{)(} v \texttt{)} \cdot 0

+

\\[8pt]

+

& = & 0

+

\end{array}</math>

|}

Line 1,711: Line 1,732:

</pre>

|}

−

~~ ~~

{| align="center" border="0" cellpadding="10"

Line 1,757: Line 1,776:

</pre>

|}

−

~~ ~~

{| align="center" border="0" cellpadding="10"

Line 1,803: Line 1,820:

</pre>

|}

−

~~ ~~

{| align="center" border="0" cellpadding="10"

Line 1,849: Line 1,864:

</pre>

|}

−

~~ ~~

{| align="center" border="0" cellpadding="10"

Line 1,907: Line 1,920:

==Visualization==

−

In my work on [[Differential Logic and Dynamic Systems 2.0|Differential Logic and Dynamic Systems]], I found it useful to develop several different ways of visualizing logical transformations, indeed, I devised four distinct styles of picture for the job. Thus far in our work on the mapping <math>F : [u, v] \to [u, v],\!</math> we've been making use of what I call the ''areal view'' of the extended universe of discourse, <math>[u, v, du, dv],\!</math> but as the number of dimensions climbs beyond four, it's time to bid this genre adieu, and look for a style that can scale a little better. At any rate, before we proceed any further, let's first assemble the information that we have gathered about <math>F\!</math> from several different angles, and see if it can be fitted into a coherent picture of the transformation <math>F : (u, v) \mapsto ( ~\texttt{((u)(v))}~, ~\texttt{((u, v))}~ ).</math>

+

In my work on [[Differential Logic and Dynamic Systems 2.0|Differential Logic and Dynamic Systems]], I found it useful to develop several different ways of visualizing logical transformations, indeed, I devised four distinct styles of picture for the job. Thus far in our work on the mapping <math>F : [u, v] \to [u, v],\!</math> we've been making use of what I call the ''areal view'' of the extended universe of discourse, <math>[u, v, \mathrm{d}u, \mathrm{d}v],\!</math> but as the number of dimensions climbs beyond four, it's time to bid this genre adieu and look for a style that can scale a little better. At any rate, before we proceed any further, let's first assemble the information that we have gathered about <math>F\!</math> from several different angles, and see if it can be fitted into a coherent picture of the transformation <math>F : (u, v) \mapsto ( ~ \texttt{((} u \texttt{)(} v \texttt{))} ~,~ \texttt{((} u \texttt{,~} v \texttt{))} ~ ).\!</math>

In our first crack at the transformation <math>F,\!</math> we simply plotted the state transitions and applied the utterly stock technique of calculating the finite differences.

Line 1,916: Line 1,929:

|

<math>\begin{array}{c|cc|cc|}

−

t & u & v & du & dv \\[8pt]

+

t & u & v & \mathrm{d}u & \mathrm{d}v \\[8pt]

−

0 & 1 & 1 & 0 & 0 \\

+

0 & 1 & 1 & 0 & 0 \\

−

1 & '' & '' & '' & ~~'' \~~\

+

1 & {}^\shortparallel & {}^\shortparallel & {}^\shortparallel & {}^\shortparallel

\end{array}</math>

|}

−

A quick inspection of the first Table suggests a rule to cover the case when <math>~~\texttt{~~u~=~v~=~1},</math> namely, <math>\~~texttt~~{~~du~~~=~~~dv~~~=~0}.</math> To put it another way, the Table characterizes Orbit 1 by means of the data: <math>(u, v, du, dv) = (1, 1, 0, 0).\!</math> Another way to convey the same information is by means of the extended proposition: <math>\texttt{u~~~v~(du~~)(dv)}.</math>

+

A quick inspection of the first Table suggests a rule to cover the case when <math>u = v = 1,\!</math> namely, <math>\mathrm{d}u = \mathrm{d}v = 0.\!</math> To put it another way, the Table characterizes Orbit 1 by means of the data: <math>(u, v, \mathrm{d}u, \mathrm{d}v) = (1, 1, 0, 0).\!</math> Another way to convey the same information is by means of the extended proposition: <math>u v \texttt{(} \mathrm{d}u \texttt{)(} \mathrm{d}v \texttt{)}.\!</math>

{| align="center" cellpadding="8" style="text-align:center"

Line 1,929: Line 1,942:

|

<math>\begin{array}{c|cc|cc|cc|}

−

t & u & v & du & dv & d^2 u & d^2 v \\[8pt]

+

t & u & v & \mathrm{d}u & \mathrm{d}v & \mathrm{d}^2 u & \mathrm{d}^2 v \\[8pt]

−

0 & 0 & 0 & 0 & 1 & 1 & 0 \\

+

0 & 0 & 0 & 0 & 1 & 1 & 0 \\

−

1 & 0 & 1 & 1 & 1 & 1 & 1 \\

+

1 & 0 & 1 & 1 & 1 & 1 & 1 \\

−

2 & 1 & 0 & 0 & 0 & 0 & 0 \\

+

2 & 1 & 0 & 0 & 0 & 0 & 0 \\

−

3 & '' & '' & '' & '' & '' & ~~'' \~~\

+

3 & {}^\shortparallel & {}^\shortparallel & {}^\shortparallel & {}^\shortparallel & {}^\shortparallel & {}^\shortparallel

\end{array}</math>

|}

−

A more fine combing of the second Table brings to mind a rule that partly covers the remaining cases, that is, <math>\~~texttt~~{~~du~~~=~v}, ~\texttt{~~dv~=~~~(u)}.</math> This much information about Orbit 2 is also encapsulated by the extended proposition, <math>\texttt{(uv)((du, v))(dv, u)},</math> which says that <math>u\!</math> and <math>v\!</math> are not both true at the same time, while <math>du\!</math> is equal in value to <math>v\!</math> and <math>dv\!</math> is opposite in value to <math>u.\!</math>

+

A more fine combing of the second Table brings to mind a rule that partly covers the remaining cases, that is, <math>\mathrm{d}u = v, ~ \mathrm{d}v = \texttt{(} u \texttt{)}.\!</math> This much information about Orbit 2 is also encapsulated by the extended proposition <math>\texttt{(} uv \texttt{)((} \mathrm{d}u \texttt{,} v \texttt{))(} \mathrm{d}v, u \texttt{)},\!</math> which says that <math>u\!</math> and <math>v\!</math> are not both true at the same time, while <math>\mathrm{d}u\!</math> is equal in value to <math>v\!</math> and <math>\mathrm{d}v\!</math> is opposite in value to <math>u.\!</math>

==Turing Machine Example==

+

☞ See [[Theme One Program]] for documentation of the cactus graph syntax and the propositional modeling program used below.

By way of providing a simple illustration of Cook's Theorem, namely, that “Propositional Satisfiability is NP-Complete”, I will describe one way to translate finite approximations of turing machines into propositional expressions, using the cactus language syntax for propositional calculus that I will describe in more detail as we proceed.

Line 1,955: Line 1,970:

A turing machine for computing the parity of a bit string is described by means of the following Figure and Table.

−

~~ ~~

+

{| align="center" border="0" cellspacing="10" style="text-align:center; width:100%"

−

+

| [[Image:Parity_Machine.jpg|400px]]

−

{| align="center" border="0" ~~cellpadding~~="10"

+

|-

−

|

+

| height="20px" valign="top" | <math>\text{Figure 3.} ~~ \text{Parity Machine}\!</math>

−

~~<pre>~~

−

o-~~------------------------------------------------o~~

−

| |

−

| ~~1/1/+1 |~~

−

~~| -~~-~~------> |~~

−

| ~~/\ / \ /\~~ |

−

~~| 0/0/+1 ^ 0 1 ^ 0/0/+1 |~~

−

| \~~/|\ /|\/ |~~

−

~~| | <-------- | |~~

−

~~| #/#/-1 | 1/1/+1 | #/#/-1 |~~

−

~~| | | |~~

−

~~| v v |~~

−

~~| # * |~~

−

~~| |~~

−

~~o-------------------------------------------------o~~

−

Figure ~~21-a~~. Parity Machine

−

</~~pre~~>

|}

Line 1,983: Line 1,981:

|

<pre>

−

Table ~~21-b~~. Parity Machine

+

Table 4. Parity Machine

o-------o--------o-------------o---------o------------o

Line 2,881: Line 2,879:

The output of <math>\mathrm{Stunt}(2)</math> being the symbol that rests under the tape head <math>\mathrm{H}</math> when and if the machine <math>\mathrm{M}</math> reaches one of its resting states, we get the result that <math>\mathrm{Parity}(1) = 1.</math>

−

~~==Work Area==~~

−

~~<pre>~~

−

~~DATA 20. http://forum.wolframscience.com/showthread.php?postid=791#post791~~

−

~~Let's see how this information about the transformation F,~~

−

~~arrived at by eyeballing the raw data, comports with what~~

−

~~we derived through a more systematic symbolic computation.~~

−

~~The results of the various operator actions that we have just~~

−

~~computed are summarized in Tables 66-i and 66-ii from my paper,~~

−

~~and I have attached these as a text file below.~~

−

~~Table 66-i. Computation Summary for f<u, v> = ((u)(v))~~

−

~~o--------------------------------------------------------------------------------o~~

−

~~| |~~

−

~~| !e!f = uv. 1 + u(v). 1 + (u)v. 1 + (u)(v). 0 |~~

−

~~| |~~

−

~~| Ef = uv. (du dv) + u(v). (du (dv)) + (u)v.((du) dv) + (u)(v).((du)(dv)) |~~

−

~~| |~~

−

~~| Df = uv. du dv + u(v). du (dv) + (u)v. (du) dv + (u)(v).((du)(dv)) |~~

−

~~| |~~

−

~~| df = uv. 0 + u(v). du + (u)v. dv + (u)(v). (du, dv) |~~

−

~~| |~~

−

~~| rf = uv. du dv + u(v). du dv + (u)v. du dv + (u)(v). du dv |~~

−

~~| |~~

−

~~o--------------------------------------------------------------------------------o~~

−

~~Table 66-ii. Computation Summary for g<u, v> = ((u, v))~~

−

~~o--------------------------------------------------------------------------------o~~

−

~~| |~~

−

~~| !e!g = uv. 1 + u(v). 0 + (u)v. 0 + (u)(v). 1 |~~

−

~~| |~~

−

~~| Eg = uv.((du, dv)) + u(v). (du, dv) + (u)v. (du, dv) + (u)(v).((du, dv)) |~~

−

~~| |~~

−

~~| Dg = uv. (du, dv) + u(v). (du, dv) + (u)v. (du, dv) + (u)(v). (du, dv) |~~

−

~~| |~~

−

~~| dg = uv. (du, dv) + u(v). (du, dv) + (u)v. (du, dv) + (u)(v). (du, dv) |~~

−

~~| |~~

−

~~| rg = uv. 0 + u(v). 0 + (u)v. 0 + (u)(v). 0 |~~

−

~~| |~~

−

~~o--------------------------------------------------------------------------------o~~

−

~~o---------------------------------------o~~

−

~~| |~~

−

~~| o |~~

−

~~| / \ |~~

−

~~| / \ |~~

−

~~| / \ |~~

−

~~| o o |~~

−

~~| / \ / \ |~~

−

~~| / \ / \ |~~

−

~~| / \ / \ |~~

−

~~| o o o |~~

−

~~| / \ / \ / \ |~~

−

~~| / \ / \ / \ |~~

−

~~| / \ / \ / \ |~~

−

~~| o o o o |~~

−

~~| / \ / \ / \ / \ |~~

−

~~| / \ / \ / \ / \ |~~

−

~~| / \ / \ / \ / \ |~~

−

~~| o o o o o |~~

−

~~| |\ / \ / \ / \ /| |~~

−

~~| | \ / \ / \ / \ / | |~~

−

~~| | \ / \ / \ / \ / | |~~

−

~~| | o o o o | |~~

−

~~| | |\ / \ / \ /| | |~~

−

~~| | | \ / \ / \ / | | |~~

−

~~| | u | \ / \ / \ / | v | |~~

−

~~| o---+---o o o---+---o |~~

−

~~| | \ / \ / | |~~

−

~~| | \ / \ / | |~~

−

~~| | du \ / \ / dv | |~~

−

~~| o-------o o-------o |~~

−

~~| \ / |~~

−

~~| \ / |~~

−

~~| \ / |~~

−

~~| o |~~

−

~~| |~~

−

~~o---------------------------------------o~~

−

~~</pre>~~

−

~~==Discussion==~~

−

~~<pre>~~

−

~~PD = Philip Dutton~~

−

~~PD: I've been watching your posts.~~

−

~~PD: I am not an expert on logic infrastructures but I find the posts~~

−

~~interesting (despite not understanding much of it). I am like~~

−

~~the diagrams. I have recently been trying to understand CA's~~

−

~~using a particular perspective: sinks and sources. I think~~

−

~~that all CA's are simply combinations of sinks and sources.~~

−

~~How they interact (or intrude into each other's domains)~~

−

~~would most likely be a result of the rules (and initial~~

−

~~configuration of on or off cells).~~

−

~~PD: Anyway, to be short, I "see" diamond shapes quite often in~~

−

~~your diagrams. Triangles (either up or down) or diamonds~~

−

~~(combination of an up and down triangle) make me think~~

−

~~soley of sinks and sources. I think of the diamond to~~

−

~~be a source which, during the course of progression,~~

−

~~is expanding (because it is producing) and then starts~~

−

~~to act as a sink (because it consumes) -- and hence the~~

−

~~diamond. I can't stop thinking about sinks and sources in~~

−

~~CA's and so I thought I would ask you if there is some way~~

−

~~to tie the two worlds together (CA's of sinks and sources~~

−

~~together with your differential constructs).~~

−

~~PD: Any thoughts?~~

−

~~Yes, I'm hoping that there's a lot of stuff analogous to~~

−

~~R-world dynamics to be discovered in this B-world variety,~~

−

~~indeed, that's kind of why I set out on this investigation --~~

−

~~oh, gee, has it been that long? -- I guess about 1989 or so,~~

−

~~when I started to see this "differential logic" angle on what~~

−

~~I had previously studied in systems theory as the "qualitative~~

−

~~approach to differential equations". I think we used to use the~~

−

~~words "attractor" and "basin" more often than "sink", but a source~~

−

~~is still a source as time goes by, and I do remember using the word~~

−

~~"sink" a lot when I was a freshperson in physics, before I got logic.~~

−

~~I have spent the last 15 years doing a funny mix of practice in stats~~

−

~~and theory in math, but I did read early works by Von Neumann, Burks,~~

−

~~Ulam, and later stuff by Holland on CA's. Still, it may be a while~~

−

~~before I have re-heated my concrete intuitions about them in the~~

−

~~NKS way of thinking.~~

−

~~There are some fractal-looking pictures that emerge when~~

−

~~I turn to "higher order propositional expressions" (HOPE's).~~

−

~~I have discussed this topic elswhere on the web and can look~~

−

~~it up now if your are interested, but I am trying to make my~~

−

~~e-positions somewhat clearer for the NKS forum than I have~~

−

~~tried to do before.~~

−

~~But do not hestitate to dialogue all this stuff on the boards,~~

−

~~as that's what always seems to work the best. What I've found~~

−

~~works best for me, as I can hardly remember what I was writing~~

−

~~last month without Google, is to archive a copy at one of the~~

−

~~other Google-visible discussion lists that I'm on at present.~~

−

~~</pre>~~

==Document History==

Line 3,056: Line 2,910:

* http://forum.wolframscience.com/archive/topic/228-1.html

* http://forum.wolframscience.com/showthread.php?threadid=228

−

* http://forum.wolframscience.com/printthread.php?threadid=228&perpage=33

+

* http://forum.wolframscience.com/printthread.php?threadid=228&perpage=50

# http://forum.wolframscience.com/showthread.php?postid=664#post664

# http://forum.wolframscience.com/showthread.php?postid=666#post666

Jon Awbrey

12,089

edits

Changes

Directory:Jon Awbrey/Papers/Differential Analytic Turing Automata (view source)

Revision as of 15:30, 11 October 2013