Slightly revised Nov. 16 2011

Explosive traces in 9/11 dust?
Media ducked claim by experts
July 14, 2011
The "criminal-media nexus" under fire in London draws attention to major news events that have gone unreported by the mainstream press on both sides of the Atlantic.

In 2009, for example, a team of scientists reported that evidence of an advanced explosive, favored by the Pentagon, had been found in the dust of the collapsed World Trade Center towers. This paper came after an admission by the National Institute for Standards and Technology that it had done no forensic tests in its investigation of the collapses.

Yet a search today of Google News for the name of one of the investigators, Niels Harrit, a chemistry professor at the University of Copenhagen, turned up no mainstream news organizations, other than Berliner Umschau, which cited Harrit in an opinion piece.

In the paper, published by the Open Chemical Physics Journal, the scientists told of finding red-gray flakes in various samples of dust and determining that the flakes were from a material similar to that in advanced TBX weaponry.

The team had sought submissions of samples of dust from the public and received containers submitted by people who had decided to save such samples. The only samples used in the study came from the five persons who agreed to let themselves be identified publicly.

There has been no public statement from the FBI on the work of the scientists. However, it may be assumed that the evidence can be ignored based on the fact that the chain of custody is broken. There is no way to be sure that the samples weren't doctored.

And yet, such tampering would seem to have required a technically advanced conspiracy, wherein volunteers working for conspirators either submit doctored material or are able to intercept and switch samples. In other words, a tampering conspiracy would require a sophisticated intelligence operation.

But, the question then arises: if intelligence units were behind the 9/11 attacks, why didn't they intercept the samples and switch them for non-incriminating dust. It seems plausible that honest agents had made such a switch too risky, and that conspirators counted on what Britain's former prime minister, Gordon Brown, denounced as a "criminal-media nexus" that includes not only the Murdoch press, but other news organizations as well.

The report, Active Thermitic Material Discovered in Dust from the 9/11 World Trade Center Catastrophe, told of igniting the chips and watching them flame. "The evidence for active, highly energetic thermitic material in the WTC dust is compelling," the scientists wrote.
http://www.diexx88blog.com/wordpress/wp-content/uploads/2011/05/activethermitic_911.pdf

[Also see 9/11 probers skipped key forensic tests
http://www.angelfire.com/ult/znewz1/trade7.html]

They said the residues from ignited chips were "strikingly similar" to the chemical signature of commercial thermite. The scientists believed that the thermite residues were consistent with "super-thermite," also known as "nano-thermite," They cited a 2001 report on Defense Department research into "nano-energetics" and thermobaric (TBX) weapons.
Here is such a report:
http://www.p2pays.org/ref/34/33115.pdf

"Super-thermite electric matches" had been devloped by Los Alamos National Laboratory for such applications as triggering explosives for demolitions, the experts noted.

The authors said their tests ruled out the possibility that the red chips were flakes of ordinary paint.

However, one of the authors, physicist Steven E. Jones, had already been given the Murdoch treatment for having raised questions over the reliability of official accounts, and it is quite possible that journalists and politicians alike shrank from covering the report out of fear of the "criminal-media nexus" blackball.

Harrit works alongside Thomas Bjorn Holm, head of the Nano-Science Center at the Department of Chemistry at the University of Copenhagen and may have been able to consult with Bjorn Holm, whose name does not appear on the report.

Harrit's Facebook page:
http://www.facebook.com/pages/Niels-Harrit/32509864153

Jones was a professor at Brigham Young University before being pressured to retire as a result of his 9/11 criticism,
http://www.physics.byu.edu/research/energy/

The scientists used advanced technology, including scanning electron microscopy, X-ray energy dispersive spectroscopy and differential scanning calorimetry.

One of the authors, Jeffrey Farrer, manages the electron microscopy facility for Brigham Young University's Department of Physics and Astronomy. His research includes nano-particle and thermitic reactions.
http://www.physics.byu.edu/directory.aspx?personID=23

Another author is Kevin R. Ryan, terminated by Underwriters Laboratory after raising technical issues concerning the official 9/11 narrative.
http://www.mindfully.org/Reform/2004/Kevin-R-Ryan22nov04.htm

Physicist Daniel E. Farnsworth, as with a number of experts critical of the government claims about 9/11, is retired and presumably beyond the reach of career retribution incited by the mendacious, criminalized press. Like Jones, Farnsworth taught at Brigham Young.
http://www.physics.byu.edu/research/energy/currvitaApril09.htm

Another author, Frank Legge, is an Australian chemist who serves as a co-editor at the Journal of 9-11 Studies.
http://www.scientistsfor911truth.org/mempages/Legge.html

Gregg Roberts, a 9/11 activist, is also listed, but a Google search gives no inkling of his scientific background.
http://world911truth.org/tag/gregg-roberts/

James R. Gourley identifies himself as a chemical engineer in an extensive criticism he submitted to the National Institute of Standards and Technology 9/11 investigation.
http://911research.wtc7.net/letters/nist/WTC7Comments.html

Bradley R. Larsen is another author. His firm, S&J Scientific, does not appear to have a web page trackable by Google and his scientific background did not show up in a Google search.*

The authors acknowledge conversations with a number of 9/11 critics, including retired naval physicist David L. Griscom
http://www.impactglassresearchinternational.com/
and former University of Iowa physicist Crockett Grabbe.
http://www.sealane.org/speak/index1.html
-----------------------------------------------------
*CORRECTION: A previous version of this page linked to an incorrect web site for Larsen.

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A general continuing fraction recursion algorithm for square roots

A very minor result that happens to please me:

The continuing real fraction

J + 1/(J + 1/(J + 1/(J + 1/ ...

= J + [(J^2 + 4)^.5]/2

is a special case of a recursion function yielding that limit. That general function is

Xsub(n+1) = (Xsub(n) + C)^(-1) + C

Setting Xo = 0 and C = J, we see (where sub(-1) is not an initial value but a designation for the constant prior to application of the function):

---

Convergent ; Our function; Continued fraction

0 ; Xsub(-1) = J; J

1 ; Xsub(1) = (J^-1) + J; J + J^-1

2 ; Xsub(2) = (J + J^-1)^-1 + J; as above

---

Of course, we needn't set Xo = 0. In fact, the curious thing is that this recursion function arrives at the same limit no matter what real initial value is chosen (other than Xo = -C, which must be excluded).

That is, (lim n-->inf)Xsub(n+1) = (lim n-->inf)Ysub(n+1)

when Xsub(1) = (Xo + C)^-1 + C and Ysub(1) = (Yo + C)^-1 + C. It is the constant C that determines the limit, which is the limit of the continuing fraction

1 + 1/C...

That is, beginning with any real but -C for Xo and any real but -C for Yo, we obtain the limit above because we find that

(lim n-->inf)(Xsub(n) - Ysub(n)) = 0,

where (Xsub(n) - Ysub(n)) alternates sign by n.

A bit of perfunctory algebra, which I omit, establishes these facts.

So, this algorithm yields an infinity of approaches to any square root. That is, Xsub(n) =/= Ysub(n) for finite n.

An example: (lim n-->inf)X(sub n) = (2 + 8^.5)/2 = 1 + 2^.5

---

For Xo = 1 and C = 2, some recursive (calculator) values are:

3

2.333...

2.428571429

2.411764706

2.414634146

For Xo = 1/2 and C = 2

2.5 2.4 2.416...6...

2.413793103

2.414285714

For Xo = -31 and C = 2

-29.0

1.965517241

2.50877193

2.398601399

2.416909621

For Xo = 31 and C = 2

33.0

2.03...03...

2.492537313

2.401197605

2.416458853

For Xo = 1/31 and C = 2 2.032258065

2.492063492

2.401273885

2.416445623

2.413830955

For Xo = -1/31 and C = 2

1.967741935

2.508196721

2.39869281

2.416893733

---

Note the pattern of alternately too high--too low.

The Monty Hall problem -- over easy

The 9/11 collapses -- an unlikely sequence
Tests for divisibility by 9 and 11

The Monty Hall problem

A player is shown three closed doors. One hides a nice prize and the other two hide booby prizes. The player picks a door. Monty then opens another door, to reveal a booby prize. The player is asked whether he or she would like to stand pat or switch his choice to the remaining closed door. Should he or she switch?

The reason this problem is so popular is that, for most of us, the answer is counterintuitive. Of course, in the original TV game show, Let's Make a Deal, the player was not given the option to switch.

My mathematician son, Jim Conant, almost instinctively knew the right answer: switch! He saw immediately that in 1/3 of cases standing pat fails so that in 2/3 of cases, switching must succeed.

But to the duller-witted of us, that reasoning doesn't seem to account for the apparent point that once the set is whittled down to two choices, switching gives a 50-50 chance of success.

However, probabilities are about information, and the information the player obtains by Monty's opening of the door affects the probabilities.

I confess I found this very difficult to grasp (and retain!) until I was able to come up with the neat proof below.

I arrived at the proof by thinking: Suppose we divide the doors into subsets {A} and {B,C} and Monty doesn't open a door once a player selects A. There is a 1/3 probability of success if he chooses set A and 2/3 for {B,C}. Now if someone tells him he can't choose, say, element B, that doesn't affect the 2/3 probability for {B,C}. So if he switches to {B,C} he must take C, which then has a 2/3 probability of success.

However, here's the proof over easy:

           A     B    C   case 1.    0    (0    1) case 2.    0    (1    0) case 3.    1    (0    0)   case 1: Monty opens B, switch succeeds case 2: Monty opens C, switch succeeds case 3: Monty opens B or C, switch fails  

In 2/3 of cases, switch succeeds.

I suggest that a source of confusion is the "or" in case 3. Despite there being two options, Monty opens only one door, as in cases 1 and 2.

---

100-door Monty

If you're still unhappy, Jim Conant suggests a light bulb might go on if you consider this scenario:

We have 100 doors, with a prize behind only one. You choose a door. Monty now opens 98 doors, none of which hides a prize. Should you switch? Of course, since the probability that the other door hides the prize is influenced by the knowledge that 98 other doors hid nothing. What is your chance of having chosen the winner? Still, 1 percent. What is the chance that the prize is behind one of the other 99 doors? Still 99 percent. So the knowledge you are given squeezes that probability onto the other closed door.

---

Enter information theory

Let's have a bit of fun with information theory.

The base-2 information content, or value, of the three closed doors is simply log₂3 = 1.58 bit. Each door's information value is of course 1/3 log 3 = 0.528 bit. When one of the doors is opened, the remaining two doors still have information values of 0.528 bit each.

On the other hand, in the case of two closed doors, the information content of each door is 1/2 log 2 = 0.5, which corresponds to maximum uncertainty. Hence the 0.028 difference in information corresponds with your awareness of an asymmetric change in probabilities that occurs once a door is opened. Because one door is opened, the information content of that door, once you get past your surprise, becomes 0. In this case, the information content of the unopened door then becomes 1.06 bit.

Now if we have 100 doors, the scenario opens with a total information content of log 100 = 2 log 10 = 6.64 bit. A single door has an information content of 0.07 bit.

The information content of the 98 doors Monty opens is 6.51, which is far greater than 0.07. Indeed, I suspect you feel much more confident that the prize is behind the other remaining closed door in the 100-door scenario.

We also have that the information content of the set of 99 doors is 6.57. Once 98 of those doors are opened, the information content of those doors becomes 0 and hence the information content of the remaining door in the set is 6.57 versus the measly 0.07 content of the door you chose.

---

The Monty Hall problem is easily solved via the rules of conditional probability. I recommend Afra Zomorodian's proof, which is best accessed by Googling.

---

This page has been deleted from Wikipedia and blacklisted from the Wiki system. Not sure why.

---

Null set uniqueness theorem

John Allen Paulos. Probability and politics
D.J. Velleman. Set theory and logic
Herbert B. Enderton. Set theory and logic
Axioms of Zermelo-Fraenkel set theory

Note: This page was corrected in August 2006 to include a missing
not-symbol and a missing parenthesis. Also, a redundant statement was
deleted and some other minor changes were made.
   

Comment: Some are puzzled by the standard set theoretic fact that even when set A does not equal set B, A - A equals B - B (also written A\A and B\B). To beginners, it is sometimes counterintuitive that the empty subset of A is indistinct from the empty subset of B.

In the Zermelo-Fraenkel version of set theory, the unique empty set is given as an axiom. However, using the rules of logic, it is possible to derive the empty set from other axioms of ZF. See the link above for all the ZF axioms.

The theorem below is not strictly necessary, but hopefully may still prove of use.

---

We accept these rules of inference, letting "~" stand for "not":

   The expression i)    x ® y       means "x implies y", which is equivalent to  ii)   if x is true, then y is true        which is equivalent to  iii)  ~x or y       which "means either x is false, or y is true".  Example:  i)    A person's being alive implies that he or she has       a beating heart  ii)   If a person is alive, then he or she has a       beating heart  iii)  Either a person has a beating heart, or       he or she is dead      

This permits us to write "x ® y" as "~x or y" (although English idiom often has "y or ~x" but the statements are equivalent as a truth table check shows).

We take "x « y" to mean "x is true only if y is true and y is true only if x is true" which can be written (x ® y) and (y ® x)

---

We accept these definitions and conditions a priori:

A set is defined by its elements, not by its descriptions. For example, let A = {r|r is a real root of x² -5x + 6 = 0} and B = {2,3}. In that case, A = B.

Every element is unique. That is, if A = {2} and B = {2,2}, then A = B.

A set is permitted to be an element of another set, (but in standard theory it can't be a member of itself). By this we see that every set is unique.

A subset of a set is itself a set.

B subset of A means (using "e" for "element of"):

x e B ® x e A

Equality means

x e B « x e A, meaning B is a subset of A and A is a subset of B.

---

First, let us see that every set has a an empty subset, meaning you cannot remove the null subset from a set.

Let us provisionally call B a subset of A and then require that B have no elements.

By definition, x e B ® x e A, which is equivalent to

x ~e B or x e A

which is true, since x is not a member of B. Even if x is not a member of A (perhaps A is empty), the statement remains true.

That is, a subset with no elements satisfies the definition of subset.

Now suppose our presumption that any set A has a null subset is false. Using { }_A to mean null subset of A, we have:

i. ~(x e { }_A ® x e A), or,

ii. ~(x ~e { }_A or x e A), or,

iii. x e { }_A and x ~e A,

This last is false simply because x ~e { }_A. That is, "x e { }_A or x ~e A" is true, but the "and" makes the statement wrong.

Hence our suggestion is correct.

Now to prove uniqueness we simply need show that { }_A = { }_B for arbitrary sets A and B. We have

x e { }_A « x e { }_B, or,

(x ~e { }_A or x e { }_B) and (x ~e { }_B or x e { }_A).

To the left of the "and" we have the truthfulness of x not being an element of { }_Aand similarly to the right.

Hence { }_A = { }_B

Remembering that any subset is a set, our claim that only one null set exists is verified.

---

It is also instructive to verify the uniqueness of the null set by considering complement sets.

Adopting the notation for the complement set, we have

x e A\B ® x ~e B

We know that A = A and so A is a subset of A. This fact allows us to write:

x e A\A ® x ~e A

which is true. Look at the rewritten statement:

~(x ~e A) ® ~(x e A\A), or

x e A ® x ~e A\A.

Now this holds for any variable. Hence A\A is a set with no elements.

Now suppose B ~= A. Nevertheless,

x e B\B « x e A\A

as we see from

(x ~e B\B or x e A) and (x e B\B or x ~e A),

which is true.

So let us denote A\A with an empty set symbol { } and consider the cases { }\{ }, ~{ }\{ }, and { }\~{ }.

i.    x e { }\{ } --> x e { } and x e { }           which holds vacuously [note transformations above]  ii.   x e ~{ }\{ } --> x e ~{ } and x ~{ }           which is true.  iii.  x e { }\~{ } --> x e { } and x ~e ~{ }           which means       x e {}\~{} --> x e {} and x e {}           which is false, meaning that the expression           {}\~{} is not defined in standard set 

theory.

Now we have that A\A = B\B = { }\{ } = { }, establishing that the null subset of any set is indistinct from the null subset of any other set, meaning that there is exactly one null set.

Note on 'exclusive or' symbolism

Today's preferred symbol for the exclusive or operation is "XOR". Yet even today the plus sign "+" is used to denote exclusive or. This symbol derives from "<--/-->", meaning "does not strictly imply." The truth table for nonexclusive or is

 i)     A  B     ----     T  F   T     F  T   T     T  T   T     F  F   F 

whereas the truth table for exclusive or is

ii)     A  B     ----     T  F   T     F  T   T     T  T   F     F  F   F 

By transforming the statement A <--/--> B, we will arrive at table ii. To wit: A <--/--> B

= not-((A --> B) & (B --> A))

= not-((not-A v B) & (not-B v A))

applying de Morgan's negation law, we get

not-(not-A v B) v not-(not-B v A)

A second application of that law yields

(A v not-B) v (not-A v B)

This formalism includes the possible expression:

(A & not-B) and (B and not-A), which is equavalent to

(A & not-A) & (B & not-B)

which, as a contradiction, must be disregarded, leaving the two remaining possibilities:

A & not-B

B & not-A

, conforming to table ii.

---

---

Thanks to John Peterson, an alert reader who caught a point of confusion that was due to two typographical errors, since corrected.

Tests for divisibility by 9 and 11

If one adds the digits of a number and the sum is divisible by 9, then the number is divisible by 9. Similarly, if one alternates the signs of a multi-digit number's digits and the sum is divisible by 11, then the number is divisible by 11.

Example

The sum of the digits of 99 is 18, which is divisible by 9. Likewise, the sum 1+8 is also divisible by 9.

The sum of the alternately signed digits of 99 is 9 + (-9) = 0, and 0 is divisible by 9.

And, of course, 99 = 9(11).

How do these tests of divisibility work?

This description is for people with no background in number theory.

Proposition I

A number is divisible by 9 (with a remainder of 0) if and only if the sum of its digits, in base-10 notation, is divisible by 9.

That is, if n is divisible by 9, the sum of its digits is divisible by 9 and if the sum of its digits is divisible by 9, n is divisible by 9.

So our method of proof can either begin with the assumption that n is divisible by 9 or with the assumption that the sum of n's digits is divisble by 9. Below we have chosen to assume n is divisible by 9. But, first some background.

What is base 10?

It is customary to use a place system for numbers of any base. The base-10 system, with its 10 digits, uses digit position to tell us what multiple of 10 we have.

When we write, say 231, this tells us that we are to add 200 + 30 + 1. Each place represents a power of 10. That is, 200 + 30 + 1 = 2 · 10² + 3 · 10¹ + 1 · 10⁰ (where any number with exponent 0 is defined as equal to 1).

If we wish to write 23 in binary, or base-2, notation, we first write:

1 · 2⁴ + 0 · 2³ + 1 · 2² + 1 · 2¹ + 1 · 2⁰.

Then, limiting ourselves to the digit set {0,1}, we write 10111, knowing that the place signifies a power of 2.

Sets of numbers differing by multiples of q

Now we want to think about some sets of numbers, each of which is divisible by some number n.

For example, let's consider the series

{...-17, -10, -3, 4, 11...} where any two members differ by some multiple of 7.

As we see,

(-17) - (-10) = -7

(-17) - 11 = -28

Another such series is

{...-15, -8, -1, 6, 13, 20...}

where subtraction of any two numbers in the series also yields a number divisible by 7.

Once we know one member of such a set, we know them all. That is, the set is writable:

{-3 + 7k|k e K}, with K the set of integers.

It is customary to express such a set thus:

[-3]₇. We can also express this set as [-10]₇. In fact, in this notation,

[-3]₇ = [-10]₇

In general, a series denoted [n]_q is identical to the series denoted [n + qk]_q, where k is any integer.

The set [n]_q is known as congruence class n modulo q.

Note that

[0]_q = [0 + qk]. Every member of this series is divisible by q for all k. That is, every element of this series, when divided by q, equals an integer k.

[q]_q = [q + qk]. Every member of this series is divisible by q. That is, every member, when divided by q, equals k+1.

What happens if we add elements of [m]_q and [n]_q?

We have m + qk + n + qj = m+n + q(k+j), letting j be an integer.

If we set k+j to 0, this gives m+n, which we use to establish the series [m+n]_q = [m+n + qk]_q.

For example, we have [-3]₇ + [-1]₇ = [-4]₇,

which means,

{...-17,-10,-3,4,11...} + {...-15,-8,-1,6,13,20...} = {...-18,-11,-4,3,10,17...}

We see that -18 - 3 = -21, which is indeed divisible by 7.

By kindred reasoning we can show that it is possible to multiply an element from each of two similar series to obtain a third series similar to the other two. By 'similar' is meant a series in which elements differ by an integer multiple of q.

That is, [p]_q · [r]_q = [pr]_q.

Proof of Proposition I

Let d_n, d_n-1 ... d₁, d₀ represent the decimal expansion of some number N.

Then, by definition, we have N = d_n · 10ⁿ + d_n-1 · 10^n-1+...+d₀ · 10⁰

[See explanation above.]

Since both sides of the expression are equal, each must belong to the same congruence class.

That is, putting N as a member of the series [N]_q means that the right side of the equation is also a member of [N]_q.

That is, [N]_q = [d_n · 10ⁿ + d_n-1 · 10^n-1 +...+ d₀ · 10⁰]_q.

Now our condition is that N be divisible by 9. In that case N is a member of the congruence class [0]₉. That is, [N]₉ = [0]₉. Likewise, the decimal expansion is also a member of [0]₉.

That is, [N]₉ = [0]₉ = [N's decimal expansion]₉.

Because [p+r]_q = [p]_q+[r]_q and [pr]_q = [p]_q · [r]_q, we may write:

[N]₉ = [0]₉ = [d_n]₉ · [10]₉ⁿ + [d_n-1] · [10]₉^n-1 + ... + [d₁]₉ · [10]₉⁰

Now we know that [10]₉ = [1]₉, since 10-9 = 1.

Hence we can substitute the number 1 for the number 10, obtaining

[0]₉ = [d_n]₉ · [1]₉ⁿ + [d_n-1]₉ · [1]₉^n-1+...+ [d₀]₉ · [1]₉⁰ ...

Obviously 1^m = 1. So we may write

[N]₉ = [d_n]₉ + [d_n-1]₉+...+[d₀]₉

which equals

[d_n+d_n-1+...+d₀]₉, which equals [N]₉, which equals [0]₉.

Since the sum of the digits is equal to congruence class [0]₉, it must be divisible by 9.

QED

Proposition II

A number is divisible by 11 only if the alternating sum of its digits is divisible by 0.

That is, d_n - d_n-1 + d_n-2 etc. must equal a sum divisible by 11.

The proof for 9 can be used for 11, noting that [10]₁₁ = [-1]₁₁, since 10 - (-1) = 11. And it should be remembered that (-1)²ⁿ is positive, while (-1)^{2n ± 1}

is negative.

First published ca. 2002

First published Monday, June 14, 2010

Freaky facts about 9 and 11

It's easy to come up with strange coincidences regarding the numbers 9 and 11. See, for example,

http://www.unexplained-mysteries.com/forum/index.php?showtopic=56447

How seriously you take such pecularities depends on your philosophical point of view. A typical scientist would respond that such coincidences are fairly likely by the fact that one can, with p/q the probability of an event, write (1-p/q)ⁿ, meaning that if n is large enough the probability is fairly high of "bizarre" classically independent coincidences.

But you might also think about Schroedinger's notorious cat, whose live-dead iffy state has yet to be accounted for by Einsteinian classical thinking, as I argue in this longish article:

http://www.angelfire.com/ult/znewz1/qball.html


Elsewhere I give a mathematical explanation of why any integer can be quickly tested to determine whether 9 or 11 is an aliquot divisor.

http://www.angelfire.com/az3/nfold/iJk.html

Here are some fun facts about divisibility by 9 or 11.

# If integers k and j both divide by 9, then the integer formed by stringing k and j together also divides by 9. One can string together as many integers divisible by 9 as one wishes to obtain that result.

Example:

27, 36, 45, 81 all divide by 9

In that case, 27364581 divides by 9 (and equals 3040509)

# If k divides by 9, then all the permutations of k's digit string form integers that divide by 9.

Example:

819/9 = 91

891/9 = 99

198/9 = 22

189/9 =21

918/9 = 102

981/9 = 109

# If an integer does not divide by 9, it is easy to form a new integer that does so by a simple addition of a digit.

This follows from the method of checking for factorability by 9. To wit, we add all the numerals, to see if they add to 9. If the sum exceeds 9, then those numerals are again added and this process is repeated as many times as necessary to obtain a single digit.

Example a.:

72936. 7 + 2 + 9 + 3 + 6 = 27. 2 + 7 = 9

Example b.:

Number chosen by random number generator:

37969. 3 + 7 + 9 + 6 + 9 = 34. 3 + 4 = 7

Hence, all we need do is include a 2 somewhere in the digit string.

372969/9 = 4144

Mystify your friends. Have them pick any string of digits (say 4) and then you silently calculate (it looks better if you don't use a calculator) to see whether the number divides by 9. If so, announce, "This number divides by 9." If not, announce the digit needed to make an integer divisible by 9 (2 in the case above) and then have your friend place that digit anywhere in the integer. Then announce, "This number divides by 9."

In the case of 11, doing tricks isn't quite so easy, but possible.

We check if a number divides by 11 by adding alternate digits as positive and negative. If the sum is zero, the number divides by 11. If the sum exceeds 9, we add the numerals with alternating signs, so that a sum 11 or 77 or the like, will zero out.

Let's check 5863.

We sum 5 - 8 + 6 - 3 = 0


So we can't scramble 5863 any way and have it divide by 11.

However, we can scramble the positively signed numbers or the negatively signed numbers how we please and find that the number divides by 11.

6358 = 11*578

We can also string numbers divisible by 11 together and the resulting integer is also divisible by 11.

253 = 11*23, 143 = 11*13

143253 = 11*13023

Now let's test this pseudorandom number:

70517. The sum of digits is 18 (making it divisible by 9).

We need to get a -18. So any digit string that sums to -18 will do. The easiest way to do that in this case is to replicate the integer and append it since each positive numeral is paired to its negative.

7051770517/11 = 641070047

Now let's do a pseudorandom 4-digit number:

4556. 4 - 5 + 5 - 6 = - 2. Hence 45562 must divide by 11 (obtaining 4142).

Sometimes another trick works.

5894. 5 - 8 + 9 - 4 = 2. So we need a -2, which, in this case can be had by appending 02, ensuring that 2 is found in the negative sum.

Check: 589402/11 = 53582

Let's play with 157311.

Positive digits are 1,7,1
Negative digits are 5, 3, 1

Positive permutations are

117, 711, 171

Negative permutations are

531, 513, 315, 351, 153, 135

So integers divisible by 11 are, for example:

137115 = 11*12465

711315 = 11*64665

Sizzlin' symmetry
There's just something about symmetry...

To form a number divisible by both 9 and 11, we play around thus:

Take a number, say 18279, divisible by 9. Note that it has an odd number of digits, meaning that its copy can be appended such that the resulting number 1827918279 yields a pattern pairing each positive digit with its negative, meaning we'll obtain a 0. Hence 1827918279/11 = 166174389 and that integer divided by 9 equals 20312031. Note that 18279/9 = 2031,

We can also write 1827997281/11 = 166181571 and that number divided by 9 equals 203110809.

Suppose the string contains an even number of digits. In that case, we can write say 18271827 and find it divisible by 9 (equaling 2030203). But it won't divide by 11 in that the positives pair with positive clones and so for negatives. This is resolved by using a 0 for the midpoint.

Thence 182701827/11 = 16609257. And, by the rules given above, 182701827 is divisible by 9, that number being 20300203.

Ah, wonderful symmetry.

A set of 'gamma' constants

The Euler constant designated g is defined:

lim _b->inf (å_(1,b) 1/i - S_(1,b) 1/x dx) =

lim _b->inf (å_(1,b) 1/i - In b)

We say that a constant is a member of the "gamma" set if there is a formula

lim _b->inf (å_(a,b) f(i) - S_(a,b) f(x) dx) = c, with c non-zero.

A curious thing about this limit is that it differs from lim_(b->inf)å_(a,b)f(i) - lim_(b->inf)S_(a,b)f(x) dx.

That last formula always goes to zero in the limit. Visualize a series of bar graphs with h = 1. So, for i^-1 we put each value side by side. The continuous curve x^-1 intersects the bar graph at integer values. So what we have is the area under the bar graph curve minus the area under the smooth curve. The difference is a sequence of ever-smaller, by percentage, pieces of the bars. Their total area equals gamma.

There is an infinity of formulas whereby the difference between a pair of converging curves yields a constant.

For example, lim _(b->inf) (å_(0,b) 1/i² - S_(0,b) 1/x² dx) = constant less than or equal to p²/6. And in general, for r > 1, lim _(b->inf) (å_(0,b) 1/i^r - S_(0,b) 1/x^r dx = constant. That is, there is a set of left-over pieces of the bars. This area is less than or equal to z(r). We do not know whether any particular r, including r = 2s, corresponds to a rational number.

However, for divergent curves other than the euler constant curves, I have not found a formula which produces a nonzero constant.

The family of Euler constants -- also known as Stieltjes constants -- follows this pattern:

c = lim _(n->inf) (Sum _(1,n) (In m)^r/r - Integral_(1,n) (In x)^m /x dx)

Other cases:

lim _(b->inf) (å_(0,b) i - S_(0,b) x dx). Now at every value of b, we have the b(b-1)/2 - b²/2 = -b/2 -- and lim_(0,b)-b/2 diverges.

Also:

lim_(b->inf) (å_(0,b) i² - S_(0,b) x² dx) = -b²/4, which diverges. Negative divergence also holds for f(i) = i³ and f(x) = x³ and also for exponent 4.

We also have

lim_(b->inf)(å_(0,b) eⁱ - S_(0,b) e^x)dx) =

lim_(b->inf)(å eⁱ - (e^x - 1)) =

lim_(b->inf)(å_(0,b) e^(i-1)) - 1 -- which diverges.

An interesting zero?

Posted Oct. 16 2009 by Paul Conant     We choose arbitrarily A and B as positive reals, neither of which is proportional to the number e.  We have     Ax + Bx - Cx = 0  We rewrite this as     exlnA + exlnB - exlnC  This gives     1 + xlnA + (xlnA)2/2! + (xlnA)3/3! + ...     1 + xlnB + (xlnB)2/2! + (xlnB)3/3! + ...   - 1 - xlnC - (xlnC)2/2! - (xlnC)3/3! - ...  which equals     1 + x(lnAB/C) + x2[(lnA)2 + (lnB)2 - (lnC)2]/2! + ...  That is,

0 = 1 + å¥j=0 xj/j![(lnA)j + (lnB)j - (lnC)j]  So we see the sigma sum equals -1 = eip  So we see that -1 can be expressed by an infinite family of infinite series.  Further, we may write

x = -åxj+1/j![(lnA)j + (lnB)j - (lnC)j]

and so any x may be so expressed.

This also holds for z = x + iy = reiu  And of course, we also have     c = 2/x + [lnAB] + x/2![(lnA)2 + (lnB)2] + x2/3![(lnA)3 + (lnB)3] ...       = 2/x + åj=1¥  xj-1/j![(lnA)j + (lnB)j]

An objection to proposition 1 of Wittgenstein's 'Tractatus'

From Wittgenstein's 'Tractatus Logico-Philosophicus,' proposition 1:

1. The world is all that is the case.

1.1 The world is the totality of facts, not of things.

1.11 The world is determined by the facts, and by their being ALL the facts.

1.12 For the totality of facts determines what is the case, and also whatever is not the case.

1.13. The facts in logical space are the world.

1.2 The world divides into facts.

1.21 Each item can be the case or not the case while everything else remains the same.

We include proposition 2.0, which includes a key concept:

2.0 What is the case -- a fact -- is the existence of states of affairs [or, atomic propositions].

According to Ray Monk's astute biography, 'Ludwig Wittgenstein, the Duty of Genius' (Free Press division of Macmillan, 1990), Gottlob Frege aggravated Wittgenstein by apparently never getting beyond the first page of 'Tractatus' and quibbling over definitions.

However, it seems to me there is merit in taking exception to the initial assumption, even if perhaps definitions can be clarified (as we know, Wittgenstein later repudiated the theory of pictures that underlay the 'Tractatus'; nevertheless, a great value of 'Tractatus' is the compression of concepts that makes the book a goldmine of topics for discussion).

Before doing that, however, I recast proposition 1 as follows:

1. The world is a theorem.

1.1 The world is the set of all theorems, not of things [a thing requires definition and this definition is either a 'higher' theorem or an axiom]

1.12 The set of all theorems determines what is accepted as true and what is not.

1.13 The set of theorems is the world [redundancy acknowledged]

2. It is a theorem -- a true proposition -- that axioms exist.

This world view, founded in Wittgenstein's extensive mining of Russell's 'Principia' and fascination with Russell's paradox is reflected in the following:

Suppose we have a set of axioms (two will do here). We can build all theorems and anti-theorems from the axioms (though not necessarily solve basic philosophical issues).

With p and q as axioms (atomic propositions that can't be durther divided by connectives and other symbols except for vacuous tautologies and contradictions), we can begin:

1. p, 2. ~p

3. q, 4. ~q

and call these 4 statements Level 0 set of theorems and anti-theorems. If we say 'it is true that p is a theorem' or 'it is true that ~p is an anti-theorem' then we must use a higher order system of numbering. That is, such a statement must be numbered in such a way as to indicate that it is a statement about a statement.

We now can form set Level 1:

5. p & q [theorem]

6. ~p & ~q [anti-theorem]

7. p v q

8. ~p & ~q

9. p v ~q

10. ~p & q

11. ~p v q

12. p & ~q

Level 2 is composed of all possible combinations of p's, q's and connectives, with Level 1 statements combined with Level 2 statements, being a subset of Level 2.

By wise choice of numbering algorithms, we can associate any positive integer with a statement. Also, the truth value of any statement can be ascertained by the truth table method of analyzing such statements. And, it may be possible to find the truth value of statement n by knowing the truth value of sub-statement m, so that reduction to axioms can be avoided in the interest of efficiency.

I have no objection to trying to establish an abstract system using axioms. But the concept of a single system as having a priori existence gives pause.

If I am to agree with Prop 1.0, I must qualify it by insisting on the presence of a human mind, so that 1.0 then means that there is for each mind a corresponding arena of facts. A 'fact' here is a proposition that is assumed true until the mind decides it is false.

I also don't see how we can bypass the notion of 'culture,' which implies a collective set of beliefs and behaviors which acts as an auxiliary memory for each mind that grows within that culture. The interaction of the minds of course yields the evolution of the culture and its collective memory.

Words and word groups are a means of prompting responses from minds (including one's own mind). It seems that most cultures divide words into noun types and verb types. Verbs that cover common occurrences can be noun-ized as in gerunds.

A word may be seen as an auditory association with a specific set of stimuli. When an early man shouted to alert his group to imminent danger, he was at the doorstep of abstraction. When he discovered that use of specific sounds to denote specific threats permitted better responses by the group, he passed through the door of abstraction.

Still, we are assuming that such men had a sense of time and motion about like our own. Beings that perceive without resort to time would not develop language akin to modern speech forms.

In other words, their world would not be our world.

Even beings with a sense of time might differ in their perception of reality. The concept of 'now' is quite difficult to define. However, 'now' does appear to have different meaning in accord with metabolic rate. The smallest meaningful moment of a fly is possibly below the threshold of meaningful human perception. A fly might respond to a motion that is too short for a human to cognize as a motion.

Similarly, another lifeform might have a 'now' considerably longer than ours, with the ultimate 'now' being, theoretically, eternity. Some mystics claim such a time sense.

The word 'deer' (perhaps it is an atomic proposition) does not prove anything about the phenomenon with which it is associated. Deer exist even if a word for a deer doesn't.

Or does it? They exist for us 'because' they have importance for us. That's why we give it a name.

Consider the eskimo who has numerous words for phenomena all of which we English-speakers name 'snow.' We assume that each of these phenomena is an element of a class named 'snow.' But it cannot be assumed that the eskimo perceives these phenomena as types of a single phenomenon. They might be as different as sails and nails as far as he is concerned.

These phenomena are individually named because they are important to him in the sense that his responses to the sets of stimuli that 'signal' a particular phenomenon potentially affect his survival. (We use 'signal' reservedly because the mind knows of the phenomenon only through the sensors [which MIGHT include unconventional sensors, such as spirit detectors].

Suppose a space alien arrived on earth and was able to locomote through trees as if they were gaseous. That being might have very little idea of the concept of tree. Perhaps if it were some sort of scientist, using special detection methods, it might categorize trees by type. Otherwise, a tree would not be part of its world, a self-sevident fact.

What a human is forced to concede is important, at root, is the recurrence of a stimuli set that the memory associates with a pleasure-pain ratio. The brain can add various pleasure-pain ratios as a means of forecasting a probable result.

A stimuli set is normally, but not always, composed of elements closely associated in time. It is when these elements are themselves sets of elements that abstraction occurs.

Much more can be said on the issue of learning. perception and mind but the point I wish to make is that when we come upon logical scenarios, such as syllogisms, we are using a human abstraction or association system that reflects our way of learning and coping with pleasure and pain. The fact that, for example, some pain is not directly physical but is 'worry' does not materially affect my point.

That is, 'reality' is quite subjective, though I have not tried to utterly justify the solipsist point of view. And, if reality is deeply subjective, then the laws of form which seem to describe said reality may well be incomplete.

I suggest this issue is behind the rigid determinism of Einstein, Bohm and Deutsch (though Bohm's 'implicate order' is a subtle and useful concept).

Deutsch, for example, is correct to endorse the idea that reality might be far bigger than ordinarily presumed. Yet, it is his faith that reality must be fully deterministic that indicates that he thinks that 'objective reality' (the source of inputs into his mind) can be matched point for point with the perception system that is the reality he apprehends (subjective reality).

For example, his reality requires that if a photon can go to point A or point B, there must be a reason in some larger scheme whereby the photon MUST go to either A or B, even if we are utterly unable to predict the correct point. But this 'scientific' assumption stems from the pleasure-pain ratio for stimuli sets in furtherance of the organism's probability of survival. That is, determinism is rooted in our perceptual apparatus. Even 'unscientific' thinking is determinist. 'Causes' however are perhaps identified as gods, demons, spells and counter-spells.

Determinism rests in our sense of 'passage of time.'

In the quantum area, we can use a 'Russell's paradox' approach to perhaps justify the Copenhagen interpretation.

Let's use a symmetrical photon interferometer. If a single photon passes through and is left undetected in transit, it reliably exits only in one direction. If, detected in transit, detection results in a change in exit direction in 50 percent of trials. That is, the photon as a wave interferes with itself, exiting in a single direction. But once the wave 'collapses' because of detection, its position is irrevocably fixed and so exits in the direction established at detection point A or detection point B.

Deutsch, a disciple of Hugh Everett who proposed the 'many worlds' theory, argues that the universe splits into two nearly-identical universes when the photon seems to arbitrarily choose A or B, and in fact follows path A in Universe A and path B in Universe B.

Yet, we might use the determinism of conservation to argue for the Copenhagen interpretation. That is, we may consider a light wave to have a minimum quantum of energy, which we call a quantum amount. If two detectors intercept this wave, only one detector can respond because a detector can't be activated by half a quantum unit. Half a quantum unit is effectively nothing. Well, why are the detectors activated probablistically, you say? Shouldn't some force determine the choice?

Here is where the issue of reality enters.

From a classical standpoint, determinism requires ENERGY. Event A at time(0) is linked to event B at time(1) by an expenditure of energy. But the energy needed for 'throwing the switch on the logic gate' is not present.

We might argue that a necessary feature of a logically consistent deterministic world view founded on discrete calculations requires that determinism is also discrete (not continuous) and hence limited and hence non-deterministic at the quantum level.

[The hit counters on all of Paul Conant's pages have been behaving erratically, going up and in numbers with no apparent rhyme or reason.]

[This page first posted January 2002]

Most Hamiltonian circuit families grow polynomially

Draft
    

Statement (i) A method exists for obtaining a set of optimal 'traveling salesman' routes that is a member of a family that grows no faster than 0(In2)2ⁿ. (ii) However, even for large n, the method yields a polynomial rate for most sets of fields of nodes.

Remark 1

We operate on a euclidean plane using nodes (cities) that can be specified with cartesian coordinates. We ignore graphs where all points lie on a line.

We adopt the convention that a route begins and ends at an origin city.

Remark 2

We assume that every road linking two cities is straight. This assumption is justified by the fact that a graph composed of curvilinear roads can be deformed into another one that preserves distances -- provided the roads don't cross outside the cities. This proviso is one of the criteria adopted below.

Proof

(i) Our aim is to show that a pretzel circuit can always be replaced by a shorter simple loop (Hamiltonian circuit).

We begin by showing that route backtrackings and crossings are unnecessary.

It is straightforward that if a route covers the same link twice, a shorter route can be drawn.

Suppose a node p is inside a field of nodes and the route enters p from the left, proceeds to x and then backtracks to p exiting toward the right.

                     x                                         n                  p                  m 

For non-trivial cases, there must exist some node m to the immediate right of p and some node n to the immediate left. Since mpx is not straight, it is longer than mx. Hence mxpn is shorter than mpxpn.

Similar reasoning holds if p is outside a field.

We show that a route that intersects at a city can be replaced by a shorter route.

Consider any two two-dimensional fields of n nodes and have them intersect at point p.

       [F  I  E  L  D  1]          m          n                p         q         r        [F  I  E  L  D  2] 

The optimal route for the new field F₁ U F₂ will not cross at p because of the option mpq linking F₁ and F₂ and nr linking the fields, with nr < npr; or because of the option npr and mq.

So then no route drawn on a graph need cross at a city.

We now show that a route that intersects outside a city can be replaced by a shorter route.

We have some field of four nodes.

        a                        d    b      x                  c 

The route acdba crosses at a non-nodal point we call x. Because bxc is not straight, bxc is longer than bc and hence abxcdxa is longer than abcda.

Now we take another four-node field F₂ and attach it to the graph above. But we consider only two-point attachments, as we already know that one-point attachments are not optimal. So it is evident that if F₂'s route does not intersect x, the composite graph's route is shorter than if the alternative holds. An induction proof can be used for any number of fields, of course.

Now suppose we have a ray segment that holds more than two nodes (ignoring origin) adjacent to another ray with at least two.

                     a                       b                                           c                                               O       f       d   

How might we link these rays with an un-loop? Suppose cfdba. We will find that more than one link between two adjacent rays is wasteful. The proof is left for the reader.

Also, we know that backtracking is wasteful and so we would not enter the vertical ray at b. In fact our route enters a ray at top or bottom node and leaves at bottom or top. Intermediate nodes can be ignored.

We will not always require that our rays have more than one non-origin node.

(ii)

Our aim is to show a means of obtaining the set of simple loops on any (non-trivial) field of nodes.

We take a field of n nodes and select any node as origin of a set of rays, with each ray drawn through one or more nodes in the field such that all nodes are intersected by rays.

Now our route must begin at origin and link every ray before returning to origin. At origin, we pick any ray to begin and (b = bottom node, t = top node) our path on that ray is determined: 0bt. However, after that there exists a choice. For ray 2 our path may be bt or tb and likewise for subsequent rays. Now if there are 2 nodes per ray, there are n/2 rays and there are about 2^n/2 acceptable paths.

Repeating this procedure for every node in the field, it will be seen that we have established the set of all n-gons, regular or irregular, that can be drawn on the field. Since any route that is not an n-gon is not a simple loop, such a route intersects itself at least once, which we know can be replaced by a shorter circuit.

The set of simple loops would then have an upper limit of n(2^n/2), though it is evident that n is much too high since changes of reference frame will slide nodes out of ray alignment and there is no reason to expect that an equal number of new alignments will occur.

Now suppose we have a field with all rays intersecting two nodes. If we add a point to that field but place it on a ray, the ray's interior point drops out and there is no significant change.

This tends to show that though a set of 0 2ⁿ may occur, it is very unlikely to occur on the next step. But, at this point, we do not rule out the possibility of a set of fields corresponding to an exponential growth rate 0(In2)2^m for m < n.

However, in most cases, a field's upper limit of acceptable routes is given by the number of rays per origin times the number of nodes in the field, or n²(n-1), which gives 3n² - 2n, or 0n², as the rate of change.

Remark 3

It seems apparent that a computer program needs to identify the nodes that form edges of the regular or irregular polygons associated with each node as origin and then to measure only edge distances. Hence we can expect that, for a random field of n nodes, the program will very likely, but perhaps not certainly, run efficiently.

Remark 4

It may be that, for some families of loops, a hard rate continues from n through infinity; or that a hard rate alternates with an efficient rate over infinity; or that the hard rate always zeroes out. The third statement would prove that NP=P.

---

Assuming that for most sets of fields P-time computer programs can be derived from our method, it may be that encypherment experts may wish to consider their options.

---

Primes up to 50,000

When algorithms collide: An infinite sum that isn't (or is it?)

This page went online Dec. 26, 2001

Peano, Hilbert and others devised space-filling curves about a century ago, helping to initiate the field of topology. So, from a topological standpoint, the result below is unremarkable.

Yet, philosophically it seems curious that the same structure can yield two different values.

---

It seems quite reasonable that an area under a curve is ever better approximated by summing areas defined by rectangular strips that fit the area ever more closely. At the limit of width w = 0, the area is defined as the infinite sum of vertical heights f(x) between x=a and x=b. This calculus approximation algorithm is standard.

Yet it is possible to devise an algorithm which seems to yield a sum of y heights that clashes with the area algorithm.

We create a sawtooth path between some curve f(x) and a finite x axis interval by this means: Divide the interval by h, keeping h a positive odd integer. Then trace the path a to f(a) to f(a+1/h) to f(a+2/h) to a+2/h to a+3/h to f(a+3/h) ... to f(a+n/h) to a+n/h = b. The length of this path is always longer than the path connecting the discrete points of f(x). [There are more than two points between f(a) and f(a+1/h) for the sawtooth path but only two points for the other path.]

As h approaches infinity, the number of sawteeth approaches infinity. At the limit of h=infinity, the sawteeth have narrowed to the infinitesimal area f(x), where the route is presumably twice f(x). But the infinite sum of f(x) cannot be infinite if the area is finite (fits inside a finite square). Quite often, the infinite sum (the integral) is less than the continuous arc length for f(x).

So this would seem to indicate that the sawtooth function must be considered undefined at h = infinity.

The special case of the unit square makes this plain:

Use the unit square lying on the positive x axis between origin and x=1. Divide 1 by the odd (for convenience) positive integer h. Trace a sawtooth path from 0,0 to 0,1 to 1/h,1 to 1/h,0 to 2/h,0 to 2/h,1 and so on to 1,0. The sawtooth path length is h+2. As h approaches infinity, the sawtooth path length also nears infinity.

It would seem that at the limit, the 'infinitesimal entity' is twice the height f(x). The sum is the integral of f(x), which, by the fundamental theorem of calculus, is the area quantified as 1.

So that at infinity, if the sawtooth function exists, the sawtooth route length collapses to distance 1.

But would a long-lived or very speedy salesman traveling through all the points of a sawtooth curve find that the sawtooth route is optimal at infinity?

The area is rarely directly proportional to some x interval or radius. For example, take any finite radius circle and define its radius as 1. Then the area is less than the circumference. But take exactly the same circle and define its radius as 3, and now the area is greater than the circumference. Now make a sawtooth path through half of the same circle.

The salesman will find, as h increases, that his path length exceeds the area quantity for every h after some h=k. If the sawtooth algorithm is completable (an 'actual infinity' is agreed), then he would appear to find that the optimal route is the sawtooth route.

But, there is an infinity of areas that can be assigned to the semicircle. For every area quantity there exists a k after which h means that the sawtooth path length is longer.

This observation can be generalized.

So we conclude that the sawtooth function is either unbounded or undefined at h's infinite limit.

If we say it is unbounded, however, we would need to explain why we are forbidden to sum the f(x) heights.

As my son Jim, a Cornell topologist, points out, sawtooth functions are well-known for producing curves with infinite perimeters that contain a finite area, the Koch snowflake being a famous example.

Of course the issue here is not that such functions occur but that the logical conclusion of this particular sawtooth algorithm yields an anomolous result.

Disjoint nondenumerable sets of irrationals

[This page went online Feb. 28, 2002.]

I wish to thank Amherst logician Dan Velleman, author of 'How to Prove It,' for not permitting any slack or insecure arguments. He has not vouched for the final version of this theorem and proof. The theorem itself is nothing special. But the proof is amusing in that it gives an inkling of how vast is the sea of irrationals with respect to the rationals.

Theorem: There exists a denumerable infinite family of sets such that each family member is a nondenumerable set of irrationals that is disjoint from every other member.

Proof:

We establish a monotonically climbing differentiable real function h(x), where x is continuous, and where the first derivative exists and is neither a constant nor a periodic (as in cos(ax)). We use only the integer values h(n) in our proof.

We denote a rational q's infinite decimal digit string by S_0 and its recurring finite substring by s. We see that the first digit of s, which has k digit places, recurs every (nk)th digit place. We denote the initial digit place of s by p. We pair p with d_2, which symbolizes a specific numeral digit. Initially, we are assuming a base 10 system.

We then establish the erase-and-replace function f by locating the h(n)th (p, d_2) and replacing d_2 with d_3.

The altered string, S_x, is forever aperiodic.

Example:

s = (01), k = 2, h = n^2

S_0 = 0.010101010101010101...

S_x = 0.210101210101010121...

We know the set of h functions is infinite and we presume that that set, and hence the set of f functions, is denumerable. So we form the indexed set U f_i.

However, this permits us to form a 'diagonal' function g, such that g is formed by the sequence f_1(1), f_2(2), f_3(3) ... ad infinitum. We then alter the g(n)th (p,d_1) by use of a third digit symbol, writing (p,d_3).

Since g is aperiodic and cannot be indexed, we have established a nondenumerable set of irrationals, which we denote R.

In general, we will write R_a to mean the set R derived from the rational q_a.

Indexing the denumerable set of rationals by j, we question whether R_1 intersection R_2 is nonempty. If empty, we proceed consecutively until we find -- if ever -- R_1 intersection R_j is nonempty.

In that event, we form a new function t_i and require use of (p, d_4) on the subset of R_j strings, denoted R'_j, that are identical to R_1 strings. R_j U R'_j we call T_j (which still holds if R_j U R'_j = R_j). We then proceed to the next instance where R_j intersection R_(j+n) is nonempty. Again, we insert digit d_5, obtaining T_(j+n).

Obviously, with a base 10 system, this procedure can be repeated 9 times.

As the number of digit places in s is irrelevant, we can increase our set of available digits with a higher base number system. In that case, we apply induction: If a base m system yields m-1 replacement digits, then a base m+1 system, yields m replacement digits.

With m e M = N, and N denumerable, we find that the family U( T_j) exists, where J = N, requiring that U( T) is denumerable and that for all j e J, n e N(T_j is disjoint from T_(j+n).

Proof concluded.

Remark: It is curious that the set of writable functions is considered denumerable because each expression is composed of a finite string of symbols. For example, 2 + log x, contains 9 symbols, including spaces. Hence, each such expression can be counted, and the entire set of such expressions must also be countable. Since our diagonal function is writable as g(n) = f_n(n), we have shown a nondenumerable subset of writable functions.

A geometric note on Russell's paradox

A lengthier discussion of Russell's paradox was found on my web page on vacuous truth before it vanished into the vacuum left by the demise of Geocities. It is there that I discuss banishing 'actual infinity' and replacing it with construction algorithms. However, such an approach, though countering other paradoxes, does not rid us of Russell's, which is summed up with the question: If sets are sorted into two types: those that are elements of themselves ('S is the name of a set that contains sets named for letters of the alphabet' is an example) and those that are not, then what type of set is the set that contains sets that are elements of themselves?

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Here we regard the null set as the initial set and build sets from there, as in:

Step 0: { }

Step 1: { { { } },{ } }

Using an axiom of infinity, we can continue this process indefinitely, leading directly to a procedure for building an abstraction of all countable sets; indirectly, noncountable sets can also be justified.

In Russell's conundrum, some sets are elements of themselves.

So let us regard a plane as pre-existent (null space or some such) and regard Set sub 0, the empty set, as a circle surrounding nothing. Set sub 1 we picture as a concentric circle around set 0. Now those two circles express the element represented as {{ }}.

We might also say the two circles express the set that contains the element { }.

In that case, we may wish to say that the set {{ }} = the element {{ }}.

Suppose we establish two construction algorithms for sets of concentric circles. In algorithm A, the outer circle is a solid line, so that the null space container is not construed to be part of the set. In algorithm B, the outer circle is a dashed line, so that the null space container is construed to be part of the set.

At each step we add a new enclosing circle and require that it be either solid or dashed.

So as step n is taken to infinity, we find that the null space container vanishes. That is, the circles fill the plane.

In that case, the container is non-existent or undefined, so that the rule that requires that it be either part of the set or not part of the set is not applicable.

We have assumed that the initial circle surrounds something that is not an element -- as with { }. But we really must give a rule that says the inmost area is not an element. We could have a rule that permits not only the outmost but the inmost circle to be dashed. In that case the inner void would be considered a part (element) of the set. Though the void in a null set is, for sound reasons, not permitted to be construed as an element, it is still useful to see the relationship of admitting the inner void to Russell's paradox.

---

Russell's paradox is usually disposed of by resort to an axiom prohibiting a set from being an element of itself. But we can see that sets and elements can be identical except for the rule prohibiting a set belonging to itself.

For example, element {{ }} in a set-building algorithm X is identical to the set containing the element { } in set-building algorithm Y.

It seems that the axiom while necessary remains questionable from the standpoint of logical consistency.

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First published ca. 2000

An algorithm to imply the set of reals

I have reconsidered my initial idea that the algorithm I give proves non-denumerability of the reals, an idea which I based on the fact that there is no n for which a cellular diagonal exists that intercepts every row of cells. However, despite intuition, it does not immediately follow that no such diagonal exists for the infinite set.

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An algorithm to imply the set of reals

We present an algorithm for "constructing" the entire set of reals and find that our intuition suggests that no diagonal can exist, which accords with Cantor's proof.

Cantor's diagonalization proof says that if it were possible that all reals were listed, then one could create an anti-diagonal number from the diagonal that intersects all the listed numbers. Hence, no such diagonal exists and the set of reals is not 1-to-1 with the set of naturals.

However, we are assuming that there is a set N containing all naturals. We are also assuming that because a finite n x n square always has a (non-Pythagorean diagonal of n cells) that intersects every row, then the same holds for the tiled quarter-plane.

However, the infinity axiom says that N exists, in which case we may say that the set of X axis integers is 1-to-1 with the set of Y axis integers, and further that the cellular diagonal with a cell vertex at (0,0) has cardN and intersects every horizontal string of cells bounded by y and y+1. In that case, we may use Cantor's argument to say that the reals can't be mapped 1-to-1 onto this tiling.

Interestingly, our algorithm for constructing all the reals intuitively says that such a set has no diagonal. But, that intuition is simply an intuition until we apply Cantor's proof and the infinity axiom.

We write an algorithm for determining the set of reals, thus:

We have the set of 1x1 square cells in a quadrant of the Cartesian grid. Each square is eligible to contain a digit. We consider only the horizontal strings of cells.

Suppose we use a base 2 system. We begin at step 0 using 2 consecutive spaces and obtaining 4 possible combinations, to wit: 00, 11, 10, 01. For step 1 (3 spaces), the number of combinations is 2*4 and for step n the number of combinations is 2ⁿ⁺².

At lim_n->inf. 2ⁿ⁺², we have established all base 2 infinite digit strings, expressing the entire set of reals greater than an arbitrary j e Z and less than or equal to j+1.

Remark: Our algorithm does not compute reals sequentially. It only computes 'pre-reals,' since a real is defined here as an infinite digit string. No element "materializes" prior to the entire set's establishment at (denumerable) infinity.

The algorithm above requires that for any step n, the set of digit strings is a rectangle n2ⁿ⁺².

But lim_n->inf. n/2ⁿ = 0, meaning the rectangle elongates and narrows toward a limiting form of a half-line.

For an integer diagonal to include an element of all horizontal digit strings, we must have a square of n columns and n rows. But such a square of the reals is never attainable. It would then seem safe to say that the set of reals, expressed as infinite digit strings, has no diagonal, which is equivalent to saying the set of reals is non-denumerable.

However, our intuition that the set of reals so constructed should have no diagonal is provable by agreement to the infinity axiom, which permits the cardN diagonal, and by Cantor's anti-diagonal result.

It also follows that no exponentially defined set of tiles n x kⁿ has a cellular diagonal at the tiling algorithm's infinite limit.

On the other hand, a tiling defined by n^k can be said to have k cellular diagonals, such that collectively the k diagonals intersect every horizontal cellular string. It then can be shown that such a tiling is one-to-one with N.

Interestingly, the power set of any n e N has card2ⁿ, which corresponds to step n of our algorithm, in which we have a set of 2ⁿ⁺² pre-reals.

Additional remark:

Lemma: Any set with lim_n->inf kⁿ elements has the cardinality of the reals, with k =/= 0, -1, 1 and k e Z.

Proof:

The set of reals is 1-to-1 with a set that has lim_n->inf 2ⁿ elements. Hence the cardinality of each set is identical. Similarly, the algorithm above can be rewritten as ckⁿ, with c a nonzero integer constant, meaning that all real digit strings are established at lim_n->inf ckⁿ.

Theorem: Some non-enumerable reals can be approximated with explicit rationals to any degree of accuracy in a finite number of steps.

Proof:

Construct n Turing machines consecutively, truncating the initial integer, and compute each one's output as a digit string n digits long. Use some formula to change each diagonal digit.

The infinite diagonal cannot be encoded as a Turing machine number, so it is not enumerable. Yet a computer can compute its approximation as a rational up to n. (The accuracy of this approximation is the same as the accuracy obtainable, in principle, for an enumerable irrational.)

Comment: The denumerable set of computables implies an extension of the concept of denumerability.

Justification:

We give these instructions for diagonalizing Turing computables:

Up to and including the nth diagonal space, follow this rule: if a digit is not 0, replace it with 0; if 0, replace it with 1. After the nth diagonal space, follow this rule: if a digit is not 2, replace it with 2; if it is 2, replace it with 3.

None of these diagonals is enumerable with respect to the Turing numbers. Yet we have a countably infinite set of diagonals. Hence, non-denumerability implies the existence of two denumerable sets of reals which are not denumerable with respect to each other.

If we diagonalize the diagonals, it is not apparent to me that this real is not a member of the computables.

Definition of choice sequence:

If f(n) gives a_n of cauchy sequence {a_n}, then {a_n} is a choice sequence if f(n) --> a_on+1 or a_1n+1 or . . . or a_mn+1.

Note i.Since a choice sequence is cauchy |a_m - a_n| <= 1/k for all m and n after some n_o. However, the rule for determining step n+1 means that more than one choice sequence is possible for every n after some n_o. That is, a choice sequence's limiting value must fall within an upper and lower bound.

Note ii: It may be that a_xn+1 is non-randomly determined. Yet, there exists an infinity of choice sequences such that the limiting value of {a_n} is an effectively random element of some infinite subset of reals (known as a 'spread') bounded by a least upper bound and a greatest lower bound.

Remark: Though choice sequences are primarily of interest to intuitionists, here we require that they be governed by ZFC.

Theorem: The question of whether the set of choice sequences contains an element with a non-enumerable limiting value is not decidable.

Proof:

We first prove (Lemma i) that within a spread (x,y), with x the GLB and y the LUB, a non-enumerable exists.

Use a diagonalization formula on the digit string outputs from the set of Turing machines, obtaining one non-enumerable real. Prefix, in turn, every rational digit string to this real and then move the decimal point to the front of each new string. Lemma i is proved.

So then suppose x and y are irrational enumerables. Rationals arbitrarily close to x from above and to y from below can be found.

Let x < p and q < y. So calling the choice sequence limit L, we have

Case i: (x < p < L < q < y).

Case ii: The possibility x < L < p exists, but then a smaller rational can be found between x and L.

Case iii: Likewise for the possibility q < L < y.

It is now straightforward that if L is a choice function limit, there is an effectively random possibility that L is enumerable or non-enumerable. This possibility is in principle undecidable.

Though probability laws suggest that the set of choice sequences includes a sequence with a non-enumerable limit, this suggestion is undecidable.

The cosmos cannot be fully modeled as a Turing machine or Turing computation

Dave Selke, an electrical engineer with a computer background, has made a number of interesting comments concerning this page, spurring me to redo the argument in another form. The new essay is entitled "On Hilbert's sixth problem" and may be found at

http://paulpages.blogspot.com/2011/11/first-published-tuesday-june-26-2007-on.html

Draft 3 (Includes discussion of Wolfram cellular automata)
Comments and suggestions welcome  

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Note: The word "or" is usually used in the following discussion in the inclusive sense.

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Many subscribe to the view that the cosmos is essentially a big machine which can be analyzed and understood in terms of other machines.

A well-known machine is the general Turing machine, which is a logic system that can be modified to obtain any discrete-input computation. Richard Feynman, the brilliant physicist, is said to have been fascinated by the question of whether the cosmos is a computer -- originally saying no but later insisting the opposite. As a quantum physicist, Feynmen would have realized that the question was difficult. If the cosmos is a computer, it certainly must be a quantum computer. But what does that certainty mean? Feynmen, one assumes, would also have concluded that the cosmos cannot be modeled as a classical computer, or Turing machine [see footnote below].

Let's entertain the idea that the cosmos can be represented as a Turing machine or Turing computation. This notion is equivalent to the idea that neo-classical science (including relativity theory) can explain the cosmos. That is, we could conceive of every "neo-classical action" in the cosmos to date -- using absolute cosmic time, if such exists -- as being represented by a huge logic circuit, which in turn can be reduced to some instance (computation) of a Turing algorithm. God wouldn't be playing dice.

A logic circuit always follows if-then rules, which we interpret as causation. But, as we know, at the quantum level, if-then rules only work (with respect to the observer) within constraints, so we might very well argue that QM rules out the cosmos being a "classical" computer.

On the other hand, some would respond by arguing that quantum fuzziness is so miniscule on a macroscopic (human) scale, that the cosmos can be quite well represented as a classical machine. That is, the fuzziness cancels out on average. They might also note that quantum fluctuations in electrons do not have any significant effect on the accuracy of computers -- though this may not be true as computer parts head toward the nanometer scale. (My personal position is that there are numerous examples of the scaling up or amplification of quantum effects. "Schrodinger's cat" is the archetypal example.)

Of course, another issue is that the cosmos should itself have a wave function that is a superposition of all possible states -- until observed by someone (who?). (I will not proceed any further on the measurement problem of quantum physics, despite its many fascinating aspects.)

Before going any further on the subject at hand, we note that a Turing machine is finite (although the set of such machines is denumerably infinite). So if one takes the position that the cosmos -- or specifically, the cosmic initial conditions (or "singularity") -- are effectively infinite, then no Turing algorithm can model the cosmos.

So let us consider a mechanical computer-robot, A, whose program is a general Turing machine. A is given a program that instructs the robotic part of A to select a specific Turing machine, and to select the finite set of initial values (perhaps the "constants of nature"), that models the cosmos.

What algorithm is used to instruct A to choose a specific cosmos-outcome algorithm and computation? This is a typical chicken-or-the-egg self-referencing question and as such is related to Turing's halting problem, Godel's incompleteness theorem and Russell's paradox.

If there is an algorithm B to select an algorithm A, what algorithm selected B? -- leading us to an infinite regression.

Well, suppose that A has the specific cosmic algorithm, with a set of discrete initial input numbers, a priori? That algorithm, call it T_c, and its instance (the finite set of initial input numbers and the computation, which we regard as still running), imply the general Turing algorithm T_g. We know this from the fact that, by assumption, a formalistic description of Alan Turing and his mathematical logic result were implied by T_c. On the other hand, we know that every computable result is programable by modifying T_g. All computable results can be cast in the form of "if-then" logic circuits, as is evident from Turing's result.

So we have

Tc <--> Tg

Though this result isn't clearly paradoxical, it is a bit disquieting in that we have no way of explaining why Turing's result didn't "cause" the universe. That is, why didn't it happen that T_g implied Turing who (which) in turn implied the Big Bang? That is, wouldn't it be just as probable that the universe kicked off as Alan Turing's result, with the Big Bang to follow? (This is not a philisophical question so much as a question of logic.)

Be that as it may, the point is that we have not succeeded in fully modeling the universe as a Turing machine.

The issue in a nutshell: how did the cosmos instruct itself to unfold? Since the universe contains everything, it must contain the instructions for its unfoldment. Hence, we have the T_c instructing its program to be formed.

Another way to say this: If the universe can be modeled as a Turing computation, can it also be modeled as a program? If it can be modeled as a program, can it then be modeled as a robot forming a program and then carrying it out?

In fact, by Godel's incompleteness theorem, we know that the issue of T_c "choosing" itself to run implies that the T_c is a model (mathematically formal theory) that is inconsistent or incomplete. This assertion follows from the fact that the T_c requires a set of axioms in order to exist (and hence "run"). That is, there must be a set of instructions that orders the layout of the logic circuit. However, by Godel's result, the Turing machine is unable to determine a truth value for some statements relating to the axioms without extending the theory ("rewiring the logic circuit") to include a new axiom.

This holds even if T_c = T_g (though such an equality implies a continuity between the program and the computation which perforce bars an accurate model using any Turing machines).

So then, any model of the cosmos as a Boolean logic circuit is inconsistent or incomplete. In other words, a Turing machine cannot fully describe the cosmos.

If by "Theory of Everything" is meant a formal logico-mathematical system built from a finite set of axioms [though, in fact, Zermelo-Frankel set theory includes an infinite subset of axioms], then that TOE is either incomplete or inconsistent. Previously, one might have argued that no one has formally established that a TOE is necessarily rich enough for Godel's incompleteness theorem to be known to apply. Or, as is common, the self-referencing issue is brushed aside as a minor technicality.

Of course, the Church thesis essentially tells us that any logico-mathematical system can be represented as a Turing machine or set of machines and that any logico-mathematical value that can be expressed from such a system can be expressed as a Turing machine output. (Again, Godel puts limits on what a Turing machine can do.)

So, if we accept the Church thesis -- as most logicians do -- then our result says that there is always "something" about the cosmos that Boolean logic -- and hence the standard "scientific method" -- cannot explain.

Even if we try representing "parallel" universes as a denumerable family of computations of one or more Turing algorithms, with the computational instance varying by input values, we face the issue of what would be used to model the master programer.

Similarly, one might imagine a larger "container" universe in which a full model of "our" universe is embedded. Then it might seem that "our" universe could be modeled in principle, even if not modeled by a machine or computation modeled in "our" universe. Of course, then we apply our argument to the container universe, reminding us of the necessity of an infinity of extensions of every sufficiently rich theory in order to incorporate the next stage of axioms and also reminding us that in order to avoid the paradox inherent in the set of all universes, we would have to resort to a Zermelo-Frankel-type axiomatic ban on such a set.

Now we arrive at another point: If the universe is modeled as a quantum computation, would not such a framework possibly resolve our difficulty?

If we use a quantum computer and computation to model the universe, we will not be able to use a formal logical system to answer all questions about it, including what we loosely call the "frame" question -- unless we come up with new methods and standards of mathematical proof that go beyond traditional Boolean analysis.

Let us examine the hope expressed in Stephen Wolfram's New Kind of Science that the cosmos can be summarized in some basic rule of the type found in his cellular automata graphs.

We have no reason to dispute Wolfram's claim that his cellular automata rules can be tweaked to mimic any Turing machine. (And it is of considerable interest that he finds specific CA/TM that can be used for a universal machine.)

So if the cosmos can be modeled as a Turing machine then it can be modeled as a cellular automaton. However, a CA always has a first row, where the algorithm starts. So the algorithm's design -- the Turing machine -- must be axiomatic. In that case, the TM has not modeled the design of the TM nor the specific initial conditions, which are both parts of a universe (with that word used in the sense of totality of material existence).

We could of course think of a CA in which the first row is attached to the last row and a cylinder formed. There would be no specific start row. Still, we would need a CA whereby the rule applied with aribitrary row n as a start yields the same total output as the rule applied at arbitrary row m. This might resolve the time problem, but it is yet to be demonstrated that such a CA -- with an extraordinarily complex output -- exists. (Forgive the qualitative term extraordinarily complex. I hope to address this matter elsewhere soon.)

However, even with time out of the way, we still have the problem of the specific rule to be used. What mechanism selects that? Obviously it cannot be something from within the universe. (Shades of Russell's paradox.)

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Footnote

Informally, one can think of a general Turing machine as a set of logic gates that can compose any Boolean network. That is, we have a set of gates such as "not", "and," "or," "exclusive or," "copy," and so forth. If-then is set up as "not-P or Q," where P and Q themselves are networks constructed from such gates. A specific Turing machine will then yield the same computation as a specific logic circuit composed of the sequence of gates.

By this, we can number any computable output by its gates. Assuming we have less than 10 gates (which is more than necessary), we can assign a base-10 digit to each gate. In that case, the code number of the circuit is simply the digit string representing the sequence of gates.

Note that circuit A and circuit B may yield the same computation. Still, there is a countable infinity of such programs, though, if we use any real for an input value, we would have an uncountable infinity of outputs. But this cannot be, because an algorithm for producing a real number in a finite number of steps can only produce a rational approximation of an irrational. Hence, there is only a countable number of outputs.

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*************************

Thanks to Josh Mitteldorf, a mathematician and physicist, for his incisive and helpful comments. Based upon a previous draft, Dr. Mitteldorf said he believes I have shown that, if the universe is finite, it cannot be modeled by a subset of itself but he expressed wariness over the merit of this point.

Alternate proof

of the Schroeder-Bernstein theorem

PlanetMath's standard proof

Wikipedia's standard proof

Draft 2   The following proof of the Schroeder-Bernstein theorem is, as far as I know, new as of April 2006. The standard proof can be found at the links above.   Notation:  "cardA" denotes the cardinal number of a set A. "<=" means "less than or equal to." "~" means "equinumerous to." 

Schroeder-Bernstein theorem: If, and only if, A and B are sets such that card A <= cardB and card B <= card A, then cardA = cardB.

Proof:

Remark: We grant that the transitivity of subsets is established from axioms. That is, A Í B Í C implies A Í C. Further, for nonempty sets, we accept that if A Ì B, then B Ì A is false. This follows from the definition of subset: for all x, x element of X implies x element of Y. But, if A is a proper subset of A, then, for some b, b element of B implies b is not an element of A. Hence B does not fit the definition of subset of A.

We rewrite cardA <= cardB and card B <= card A, thus:

A ~ C Í B and B ~ D Í A

If

Suppose A ~ C = B. Then we are done. Likewise for B ~ D = A. So we test the situation for nonempty sets with: A ~ C Ì B and B ~ D Ì A. (We note that if C ~ D, then A ~ B, spoiling our interim hypothesis. So we assume C is not equinumerous to D.)

We now form the sets A U C and B U D, permitting us to write (A U C) Ì (A U B) and (B U D) Ì (A U B).

We now have two options:

i) (A U C) Ì (A U B) Ì (B U D) Ì (A U B)

This is a contradiction since A U B cannot be a proper subset of itself.

ii) (A U C) Ì (A U B) É (B U D) Ì (A U B)

Also a contradiction, since B U D cannot be a proper subset of A U B and also properly contain A U B.

Only if

We let B = C and D = A.

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This proof, like the standard modern proof, does not rely on the Axiom of Choice.

---

In search of a blind watchmaker

Richard Dawkins' web site
Wikipedia article on Dawkins
Wikipedia article on Francis Crick
Abstract of David Layzer's two-tiered adaptation
Joshua Mitteldorf's home page
Do dice play God? A book review

A discussion of The Blind Watchmaker: Why the Evidence of Evolution Reveals a Universe without Design by Richard Dawkins  First posted Oct. 5, 2010 and revised as of Oct. 8, 2010  Please notify me of errors or other matters at "krypto78...at...gmail...dot...com"   

By PAUL CONANT

Surely it is quite unfair to review a popular science book published years ago. Writers are wont to have their views evolve over time.¹ Yet in the case of Richard Dawkins' The Blind Watchmaker: Why the Evidence of Evolution Reveals a Universe without Design (W.W. Norton 1986), a discussion of the mathematical concepts seems warranted, because books by this eminent biologist have been so influential and the "blind watchmaker" paradigm is accepted by a great many people, including a number of scientists.

Dawkins' continuing importance can be gauged by the fact that his most recent book, The God Delusion (Houghton Mifflin 2006), was a best seller, and by the links above. In fact, Watchmaker, also a best seller, was re-issued in 2006.

I do not wish to disparage anyone's religious or irreligious beliefs, but I do think it important to point out that non-mathematical readers should beware the idea that Dawkins has made a strong case that the "evidence of evolution reveals a universe without design."

There is little doubt that some of Dawkins' conjectures and ideas in Watchmaker are quite reasonable. However, many readers are likely to think that he has made a mathematical case that justifies the theory(ies) of evolution, in particular the "modern synthesis" that combines the concepts of passive natural selection and genetic mutation.

Dawkins wrote his apologia back in the eighties when computers were becoming more powerful and accessible, and when PCs were beginning to capture the public fancy. So it is understandable that, in this period of burgeoning interest in computer-driven chaos, fractals and cellular automata, he might have been quite enthusiastic about his algorithmic discoveries.

However, interesting computer programs may not be quite as enlightening as at first they seem.

Cumulative selection

Let us take Dawkins' argument about "cumulative selection," in which he uses computer programs as analogs of evolution. In the case of the phrase, "METHINKS IT IS LIKE A WEASEL," the probability -- using 26 capital letters and a space -- of coming up with such a sequence randomly is 27^-28 (the astonishingly remote 8.3 x 10^-41). However, that is also the probability for any random string of that length, he notes, and we might add that for most probability distributions. when n is large, any distinct probability approaches 0.

Such a string would be fantastically unlikely to occur in "single step evolution," he writes. Instead, Dawkins employs cumulative selection, which begins with a random 28-character string and then "breeds from" this phrase. "It duplicates it repeatedly, but with a certain chance of random error -- 'mutation' -- in the copying. The computer examines the mutant nonsense phrases, the 'progeny' of the original phrase, and chooses the one which, however slightly most resembles the target phrase, METHINKS IT IS LIKE A WEASEL."

Three experiments evolved the precise sentence in 43, 64 and 41 steps, he wrote.

Dawkins' basic point is that an extraordinarily unlikely string is not so unlikely via "cumulative selection."

Once he has the readers' attention, he concedes that his notion of natural selection precludes a long-range target and then goes on to talk about "biomorph" computer visualizations (to be discussed below).

Yet it should be obvious that Dawkins' "methinks" argument applies specifically to evolution once the mechanisms of evolution are at hand. So the fact that he has been able to design a program which behaves like a neural network, really doesn't say much about anything. He has achieved a proof of principle that was not all that interesting, although I suppose it would answer a strict creationist, which was perhaps his basic aim.

But which types of string are closer to the mean? Which ones occur most often? If we were to subdivide chemical constructs into various sets, the most complex ones -- which as far as we know are lifeforms -- would be farthest from the mean.(Dawkins, in his desire to appeal to the lay reader, avoids statistics theory other than by supplying an occasional quote from R.A. Fisher.)

Let us, like Dawkins, use a heuristic analog². Suppose we take the set of all grammatical English sentences of 28 characters. The variable is an English word rather than a Latin letter or space. What would be the probability of any 28-character English sentence appearing randomly?

My own sampling of a dictionary found that words with eight letters appear with the highest probability of 21%. So assuming the English lexicon to contain 500,000 words, we obtain about 105,000 words of length 8.

Now let us do a Fermi-style rough estimate. For the moment ignoring spaces, we'll posit average word length of 2 to 9 as covering virtually all combinations. That is, we'll pretend there are sentences composed of only two-letter words, only three-letter and so on up to nine letters. Further, we shall put an upper bound of 10⁵ on the set of words of any relevant length (dropping the extra 5,000 for eight-letter words as negligible for our purposes).

This leads to a total number of combinations of (10⁵)² + 10⁸ + ... + 10¹⁴, which approximates 10¹⁴.

We have not considered spaces nor (directly) combinations of words of various lengths. It seems overwhelmingly likely that any increases would be canceled by the stricture that sentences be grammatical, something we haven't modeled. But, even if the number of combinations were an absurd 10 orders of magnitude higher, the area under the part of some typical probability curve that covers all grammatical English sentences of length 28 would take up a miniscule percentage of a tail.

Analogously, to follow Dawkins, we would suspect that the probability is likewise remote for random occurrence of any information structure as complex as a lifeform.

To reiterate, the entire set of English sentences of 28 characters is to be found far out in the tail of some probability distribution. Of course, we haven't specified which distribution because we have not precisely defined what is meant by "level of complexity." This is also an important omission by Dawkins.

We haven't really done much other than to underscore the lack of precision of Dawkins' analogy.

Dawkins then goes on to talk about his "biomorph" program, in which his algorithm recursively alters the pixel set, aided by his occasional selecting out of unwanted forms. He found that some algorithms eventually evolved insect-like forms, and thought this a better analogy to evolution, there having been no long-term goal. However, the fact that "visually interesting" forms show up with certain algorithms again says little. In fact, the remoteness of the probability of insect-like forms evolving was disclosed when he spent much labor trying to repeat the experiment because he had lost the exact initial conditions and parameters for his algorithm. (And, as a matter of fact, he had become an intelligent designer with a goal of finding a particular set of results.)

Again, what Dawkins has really done is use a computer to give his claims some razzle dazzle. But on inspection, the math is not terribly significant.

It is evident, however, that he hoped to counter Fred Hoyle's point that the probability of life organizing itself was equivalent to a tornado blowing through a junkyard and assembling from the scraps a fully functioning 747 jetliner, Hoyle having made this point not only with respect to the origin of life, but also with respect to evolution by natural selection.

So before discussing the origin issue, let us turn to the modern synthesis.

The modern synthesis

I have not read the work of R.A. Fisher and others who established the modern synthesis merging natural selection with genetic mutation, and so my comments should be read in this light.

Dawkins argues that, although most mutations are either neutral or harmful, there are enough progeny per generation to ensure that an adaptive mutation proliferates. And it is certainly true that, if we look at artificial selection -- as with dog breeding -- a desirable trait can proliferate in very short time periods, and there is no particular reason to doubt that if a population of dogs remained isolated on some island for tens of thousands of years that it would diverge into a new species, distinct from the many wolf sub-species.

But Dawkins is of the opinion that neutral mutations that persist because they do no harm are likely to be responsible for increased complexity. After all, relatively simple life forms are enormously successful at persisting.

And, as Stephen Wolfram points out (A New Kind of Science, Wolfram Media 2006), any realistic population size at a particular generation is extremely unlikely to produce a useful mutation because the ratio of possible mutations to the number of useful ones is some very low number. So Wolfram also believes neutral mutations drive complexity.

We have here two issues:

1. If complexity is indeed a result of neutral mutations alone, increases in complexity aren't driven by selection and don't tend to proliferate.

2. Why is any species at all extant? It is generally assumed that natural selection winnows out the lucky few, but does this idea suffice for passive filtering?

Though Dawkins is correct when he says that a particular mutation may be rather probable by being conditioned by the state of the organism (previous mutation), we must consider the entire chain of mutations represented by a species.

If we consider each species as representing a chain of mutations from the primeval organism, then we have a chain of conditional probability. A few probabilities may be high, but most are extremely low. Conditional probabilities can be graphed as trees of branching probabilities, so that a chain of mutation would be represented by one of these paths. We simply multiply each branch probability to get the total probability per path.

As a simple example, a 100-step conditional probability path with 10 probabilities of 0.9 and 60 with 0.7 and 30 with 0.5 yields a cumulative probability of 1.65 x 10^-19.

In other words, the more mutations and ancestral species attributed to an extanct species, the less likely it is to exist via passive natural selection. The actual numbers are so remote as to make natural selection by passive filtering virtually impossible, though perhaps we might conjecture some nonlinear effect going on among species that tends to overcome this problem.

Dawkins' algorithm demonstrating cumulative evolution fails to account for this difficulty. Though he realizes a better computer program would have modeled lifeform competition and adaptation to environmental factors, Dawkins says such a feat was beyond his capacities. However, had he programed in low probabilities for "positive mutations," cumulative evolution would have been very hard to demonstrate.

Our second problem is what led Hoyle to revive the panspermia conjecture, in which life and proto-life forms are thought to travel through space and spark earth's biosphere. His thinking was that spaceborne lifeforms rain down through the atmosphere and give new jolts to the degrading information structures of earth life. (The panspermia notion has received much serious attention in recent years, though Hoyle's conjectures remain outside the mainstream.)

From what I can gather, one of Dawkins' aims was to counter Hoyle's sharp criticisms. But Dawkins' vigorous defense of passive natural selection does not seem to square with the probabilities, a point made decades previously by J.B.S. Haldane.

Without entering into the intelligent design argument, we can suggest that the implausible probabilities might be addressed by a neo-Lamarkian mechanism of negative feedback adaptations. Perhaps a stress signal on a particular organ is received by a parent and the signal transmitted to the next generation. But the offspring's genes are only acted upon if the other parent transmits the signal. In other words, the offspring embryo would not strengthen an organ unless a particular stress signal reached a threshold.

If that be so, passive natural selection would still play a role, particularly with respect to body parts that lose their role as essential for survival.

Dawkins said Lamarkianism had been roundly disproved, but since the time he wrote the book molecular biology has shown the possibility of reversal of genetic information (retroviruses and reverse transcription). However, my real point here is not about Lamarkianism but about Dawkins' misleading mathematics and reasoning.

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Joshua Mitteldorf, an evolutionary biologist with a physics background and a Dawkins critic, points out that an idea proposed more than 30 years ago by David Layzer is just recently beginning to gain ground as a response to the cumulative probabilities issue. Roughly I would style Layzer's proposal a form of neo-Lamarckianism. The citation3 is found at the bottom of this essay and the link is posted above.

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On origins

Dawkins concedes that the primeval cell presents a difficult problem, the problem of the arch. If one is building an arch, one cannot build it incrementally stone by stone because at some point, a keystone must be inserted and this requires that the proto-arch be supported until the keystone is inserted. The complete arch cannot evolve incrementally. This of course is the essential point made by the few scientists who support intelligent design.

Dawkins essentially has no answer. He says that a previous lifeform, possibly silicon-based, could have acted as "scaffolding" for current lifeforms, the scaffolding having since vanished. Clearly, this simply pushes the problem back. Is he saying that the problem of the arch wouldn't apply to the previous incarnation of "life" (or something lifelike)?

Some might argue that there is a possible answer in the concept of phase shift, in which, at a threshold energy, a disorderly system suddenly becomes more orderly. However, this idea is left unaddressed in Watchmaker. I would suggest that we would need a sequence of phase shifts that would have a very low cumulative probability, though I hasten to add that I have insufficient data for a well-informed assessment.

Cosmic probabilities

Is the probability of life in the cosmos very high, as some think? Dawkins argues that it can't be all that high, at least for intelligent life, otherwise we would have picked up signals. I'm not sure this is valid reasoning, but I do accept his notion that if there are a billion life-prone planets in the cosmos and the probability of life emerging is a billion to one, then it is virtually certain to have originated somewhere in the cosmos.

Though Dawkins seems to have not accounted for the fact that much of the cosmos is forever beyond the range of any possible detection as well as the fact that time gets to be a tricky issue on cosmic scales, let us, for the sake of argument, grant that the population of planets extends to any time and anywhere, meaning it is possible life came and went elsewhere or hasn't arisen yet, but will, elsewhere.

Such a situation might answer the point made by Peter Ward and Donald Brownlee in Rare Earth: Why Complex Life Is Uncommon in the Universe (Springer 2000) that the geophysics undergirding the biosphere represents a highly complex system (and the authors make efforts to quantify the level of complexity), meaning that the probability of another such system is extremely remote. (Though the book was written before numerous discoveries concerning extrasolar planets, thus far their essential point has not been disproved. And the possibility of non-carbon-based life is not terribly likely because carbon valences permit high levels of complexity in their compounds.)

Now some may respond that it seems terrifically implausible that our planet just happens to be the one where the, say, one-in-a-billion event occurred. However, the fact that we are here to ask the question is perhaps sufficient answer to that worry. If it had to happen somewhere, here is as good a place as any. A more serious concern is the probability that intelligent life arises in the cosmos.

The formation of multicellular organisms is perhaps the essential "phase shift" required, in that central processors are needed to organize their activities. But what is the probability of this level of complexity? Obviously, in our case, the probability is one, but, otherwise, the numbers are unavailable, mostly because of the lack of a mathematically precise definition of "level of complexity" as applied to lifeforms.

Nevertheless, probabilities tend to point in the direction of cosmically absurd: there aren't anywhere near enough atoms -- let alone planets -- to make such probabilities workable. Supposing complexity to result from neutral mutations, probability of multicellular life would be far, far lower than for unicellular forms whose speciation is driven by natural selection. Also, what is the survival advantage of self-awareness, which most would consider an essential component of human-like intelligence?

Hoyle's most recent idea was that probabilities were increased by proto-life in comets that eventually reached earth. But, despite enormous efforts to resolve the arch problem (or the "jumbo jet problem"), in my estimate he did not do so.

(Interestingly, Dawkins argues that people are attracted to the idea of intelligent design because modern engineers continually improve machinery designs, giving a seemingly striking analogy to evolution. Something that he doesn't seem to really appreciate is that every lifeform may be characterized as a negative-feedback controlled machine, which converts energy into work and obeys the second law of thermodynamics. That's quite an "arch.")

The intelligent design proponents, however, face a difficulty when relying on the arch analogy: the possibility of undecidability. As the work of Godel, Church, Turing and Post shows, some theorems cannot be proved by tracking back to axioms. They are undecidable. If we had a complete physical description of the primeval cell, we could encode that description as a "theorem." But, that doesn't mean we could track back to the axioms to determine how it emerged. If the "theorem" were undecidable, we would know it to be "true" (having the cell description in all detail), but we might be forever frustrated in trying to determine how it came to exist.

In other words, a probabilistic argument is not necessarily applicable.

The problem of sentience

Watchmaker does not examine the issue of emergence of human intelligence, other than as a matter of level of complexity.

Hoyle noted in The Intelligent Universe (Holt, Rhinehart and Winston 1984) that over a century ago, Alfred Russel Wallace was perplexed by the observation that "the outstanding talents of man... simply cannot be explained in terms of natural selection."

Hoyle quotes the Japanese biologist S. Ohno:

"Did the genome (genetic material) of our cave-dwelling predecessors contain a set or sets of genes which enable modern man to compose music of infinite complexity and write novels with profound meaning? One is compelled to give an affirmative answer...It looks as though the early Homo was already provided with the intellectual potential which was in great excess of what was needed to cope with the environment of his time."

Hoyle proposes in Intelligent that viruses are responsible for evolution, accounting for mounting complexity over time. However, this seems hard to square with the point just made that such complexity doesn't seem to occur as a result of passive natural winnowing and so there would be no selective "force" favoring its proliferation.

At any rate, I suppose that we may assume that Dawkins in Watchmaker saw the complexity inherent in human intelligence as most likely to be a consequence of neutral mutations.

Another issue not addressed by Dawkins (or Hoyle for that matter) is the question of self-awareness. Usually the mechanists see self-awareness as an epiphenomenon of a highly complex program (a notion Roger Penrose struggled to come to terms with in The Emperor's New Mind (Oxford 1986) and Shadows of the Mind (Oxford 1994).)

But let us think of robots. Isn't it possible in principle to design robots that multiply replications and maintain homeostasis until they replicate? Isn't it possible in principle to build in programs meant to increase probability of successful replication as environmental factors shift?

In fact, isn't it possible in principle to design a robot that emulates human behaviors quite well? (Certain babysitter robots are even now posing ethics concerns as to an infant's bonding with them.)

And yet there seems to be no necessity for self-awareness in such designs. Similarly, what would be the survival advantage of self-awareness for a species?

I don't suggest that some biologists haven't proposed interesting ideas for answering such questions. My point is that Watchmaker omits much, making the computer razzle dazzle that much more irrelevant.

Conclusion

In his autobiographical What Mad Pursuit (Basic Books 1988) written when he was about 70, Nobelist Francis Crick expresses enthusiasm for Dawkins' argument against intelligent design, citing with admiration the "methinks" program.

Crick, who trained as a physicist and was also a panspermia advocate (see link above), doesn't seem to have noticed the difference in issues here. If we are talking about an analog of the origin of life (one-step arrival at the "methinks" sentence), then we must go with a distinct probability of 8.3 x 10^-41. If we are talking about an analog of some evolutionary algorithm, then we can be convinced that complex results can occur with application of simple iterative rules (though, again, the probabilities don't favor passive natural selection).

One can only suppose that Crick, so anxious to uphold his lifelong vision of atheism, leaped on Dawkins' argument without sufficient criticality. On the other hand, one must accept that there is a possibility his analytic powers had waned.

At any rate, it seems fair to say that the theory of evolution is far from being a clear-cut theory, in the manner of Einstein's theory of relativity. There are a number of difficulties and a great deal of disagreement as to how the evolutionary process works. This doesn't mean there is no such process, but it does mean one should listen to mechanists like Dawkins with care.

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1. In a 1996 introduction to Watchmaker, Dawkins wrote that "I can find no major thesis in these chapters that I would withdraw, nothing to justify the catharsis of a good recant."  2. My analogy was inadequately posed in previous drafts. Hopefully, it makes more sense now.  3. Genetic Variation and Progressive Evolution David Layzer The American Naturalist Vol. 115, No. 6 (Jun., 1980), pp. 809-826 (article consists of 18 pages) Published by: The University of Chicago Press for The American Society of Naturalists Stable URL: http://www.jstor.org/stable/2460802

Note: An early draft contained a ridiculous mathematical error that does not affect the argument but was very embarrassing. Naturally, I didn't think of it until after I was walking outdoors miles from an internet terminal. It has now been put right.