Author Topic: The shortest-chain-of-the-World ( problem )  (Read 6016 times)

X1

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Re: The shortest-chain-of-the-World ( problem )
« Reply #15 on: January 13, 2013, 09:06:50 PM »
   
he inflates his response with gratuitous insults.

 
(edited, when my blood pressure came down to normal levels again... :))

  ... It is a tricky problem, indeed, and the people who did not get it right at the first place always feel somehow insulted / betrayed - and then they try to mud the issue, by inventing various / numerous "loopholes" and "variations". It has been done many times in the past, as one can read in the article in Wikipedia, and it will be done again in the future - human narure does not change onernight !  In fact, that problem is widely known exactly because of this childish/wild behaviour : Riding-whatever mathematicians and/or would-like-to-be-omniscient people, tried to formulate their own versions, to trap back the people they feel they had trapped them...
    However, I had no intention to trap anybody. I was talking about the problem of the two bends, placed the one after the other in the same line, and I was arguing that the "average" of the strenghts is meaningless, because it will be pushed lower, towards the weakest "link". I just thought that the two problems had something in common : the idea that "adding" averages and calculating probabilities of not-performed experiments (1), or not-taken choices, can be misleading...
   It was proved to be a wrong idea !   :)

1. A. Peres, 1978
    Unperformed experiments have no results
   
« Last Edit: January 14, 2013, 01:33:00 AM by X1 »

X1

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Re: The shortest-chain-of-the-World ( problem )
« Reply #16 on: January 13, 2013, 10:27:57 PM »
   When I thought it would be better to compare the two-links-with-three-value-strengths problem,  with the infamous choose-one-door three-doors-two-outcomes "Monty Hall" problem, I was not aware of the problems this problem generates to the people who had not solved it at the first place...  :) and then try hopelessly to discover some "loophole" of the correct argument, or some "variation" of the simple initial problem, that could lead to another solution. I underestimated the fact that I had spent my New Year evening trying to explain it to a friend of mine, that has a PhD in engineering, in vein ! So, I should perhaps expect a similar behaviour from some members here - but I have to confess that I had hoped for something else.
   Now I am reading the Wikipedia article ( I was not even aware of the name of the problem till now... ), I see that the behaviour of Dan Lehman is rather typical ... so I should not blame his "expected" behaviour, as I did. People all over the world try to escape by teeth and nails - because, somehow, this problem offends their ego, their self-respect, in a deeper way one would have thought. There are even many studies of psychologists who try to explain this wild behaviour - so we are talking about something rather common.
   Reading this article, I was amazed by the number of "traps" people had invented, to "revenge" the ones they feel they had trapped them in the first place. If I knew this, I would never have thrown more fuel to the fire we have already in this Forum. My original "two-in-line-knots", or "two-links" problem was more than enough !  :)
   Somebody publishes a series of experiments, where one set of knots ( the eyeknots, the loop knots ) is shown to be stronger, on average, than another ( the eye-to-eye knots / the bends ). However, those results do not tell that if we put in line TWO eyeknots, two loops, the one linked to the other by the linkage of their bights, to form a compound bend, they will still be stronger than ONE eye-to-eye knot, one bend. I pointed it right away (1), and I have repeated it in other words (2),(3)... but Dan Lehman had not understood the issue (4)...and have chosen to insist defending his mistaken view ever since (5). So, I should have not be surprized he has chosen a similar escape in the more tricky Monty Hall problem - where, at least, he seems that he is in good company. The interested reader is advised to read the Wikipedia article, to learn how we all can persuade ourselves that the other guy is always "not big"... but WE are !  :)

   1.  http://igkt.net/sm/index.php?topic=4150.msg25822#msg25822
   2.  http://igkt.net/sm/index.php?topic=4150.msg25830#msg25830
   3.  http://igkt.net/sm/index.php?topic=4150.msg25879#msg25879   
   4.  http://igkt.net/sm/index.php?topic=4150.msg25828#msg25828
   5.  http://igkt.net/sm/index.php?topic=4150.msg25864#msg25864
« Last Edit: January 14, 2013, 01:37:55 AM by X1 »

Dan_Lehman

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Re: The shortest-chain-of-the-World ( problem )
« Reply #17 on: January 14, 2013, 08:51:00 PM »
   
he inflates his response with gratuitous insults.

  ... It is a tricky problem, indeed, and the people who did not get it right at the first place
always feel somehow insulted / betrayed - and then they try to mud the issue,
by inventing various / numerous "loopholes" and "variations".
...
In fact, that problem is widely known exactly because of this childish/wild behaviour :
... to trap back the people they feel they had trapped them...
However, I had no intention to trap anybody.
...

A heap of deprecations serving no good purpose,
and which suggest refusing to follow reason and
concede an argument, fearing a loss of pride.

There was no trapping, nor on my part any misunderstanding;
the "Monty Hall" problem has been understood by me for some
time.  But the problem statement is key to distinguishing it
from seeming equal problems that however differ in result:
e.g., to say that the host "opens a door" is ambiguous as to
a key ingredient --randomness or not.

And so I asked the similar problem wherein the door opening
was done by chance --something that X1 denied making any
difference, when in fact it is a critical point : there is no benefit
to switching (nor penalty from doing so), in this case.  One
can read about this via Wikipedia, if not "running out the
probabilities" oneself, on some paper, say.

--to wit:
Quote
... even if you choose to open the one of the two
other-than-the-one-I-had- already-chosen doors
( the B or the C, as I had chosen the A ),
and you do this in random ( your choice was decided by flipping a coin,
between the option door B and the option door C ),
and it happened that this door was the door B,
that contains the knot tyer and not the rare ABoK copy ),

even then, it would be better to me to alter my initial selection,
and select C.

Well, you sank into it, trap or not.  Because I know that
you can run out the *random* possibilities here, and see
that there are equal chances for all --for B being void,
and the prize being A or C.  On only 1/3 of the cases
is C the winner (as for the other two doors' probabilities);
that you got a peek at B, by random chance, doesn't
change this --or enhance A's chances--, but it does
leave these two as candidates, of now equal standing (50/50).
(You'll be no worse off by changing.)

Or, better for your pride, but not your chances of winning
the prize copy of ABOK --but maybe you'll meet some
"big" knot tyer, and learn something!   ;D

But of course you won't heed the repeated argument, from me;
you can read about it elsewhere, though, via URLinks from the
Wikipedia page you now know about.  E.g.,
http://probability.ca/jeff/writing/montyfall.pdf
Quote
Monty Fall Problem: In this variant, once you have selected one of the three doors,
the host slips on a banana peel and accidentally pushes open another door, which just
happens not to contain the car.
Now what are the probabilities that you will win
the car if you stick with your original selection, versus if you switch to the remaining
door?
In this case, it is still true that originally there was just a 1/3 chance that your original
selection was correct.  And yet, ...
the probabilities of winning if  you stick or switch are both 1/2, not 1/3 and 2/3.


All of this --and indeed the considerable, continued discussion of the
Monty Hall issues-- should illustrate how some seemingly subtle,
minor ingredients in a problem statement can be critical to the
solution (or, put another way, to actually describing the intended problem)!


.:.  QED , at last?!   ::)




Now, how did we get into all this statistical debate, at all?
--from this [nb: I've corrected X1's "eye-to-eye" to be "end-2-end",
by which is meant other's "bend", i.e. a single, ends-joining knot]:

Somebody publishes a series of experiments, where one set of knots
( the eyeknots, the loop knots ) is shown to be stronger, on average,
than another ( the [end-2-end] knots / the bends ).
However, those results do not tell that if we put in line TWO eyeknots, two loops,
the one linked to the other by the linkage of their bights, to form a compound bend,
they will still be stronger than ONE [end-2-end] knot, one bend.

I pointed it right away (1), and I have repeated it in other words (2),(3)...
but Dan Lehman had not understood the issue (4)...
and have chosen to insist defending his mistaken view ever since (5).
So, I should have not be surprized he has chosen a similar escape in the more tricky Monty Hall problem
--where, at least, he seems that he is in good company.


And this gratuitous misstatement of my thinking and repeated
denial of reason again confronts us with an apparent need
to (further) explain --although the explanations are ignored.
Here is what was said, in part.

Given these results, it would seem that the strongest means of bending two lines together
should be with two end-of-line loops.

  No ! When you will use two loops, the reduction in the strength of the compound bend
will be larger than the average you have reported. From any pair of loops / links,
you will have to take into account only the weaker, and then to calculate the average
of the weakest links, and the weakest links only.

Yes, not "no"!  Look at the values : except for the fig.8
end-2-end knot
, the other such knots are way below
the strengths of the eyeknots --weaker values or not!


To emphasize, "the other such knots are way below the strengths
of the eyeknots"
!!


Which points to our having more than mere average values
from the testing, but sD values, which give a predicted shaping
to the distribution of expected results.  And given the large
differences that were observed, even taking into consideration
the value-diminishing effect of 2-link sampling, the eye-in-eye
(or also, we should believe, twin eye --CLDay's term) knots
come out on average stronger than single-knot end-2-end joints,
in these observations, by statistical reasoning.


That reasoning uses an expected bell curve,
as referred to ...

Quote
My view is that you should have ignored that single test,
just follow the expected bell curve, and so keep the standard deviation low.

Good, let's do that, indeed --and make the reasoning afforded
by such statistical analysis via the standarDeviation (sD).


Of a normal bell curve, one finds 95% of the results within
2 sD of the mean; one sD contains 34%, the 2nd 14%, and
at a 3rd and beyond are 2% = 50%, reflected on the other
side of the mean.  As in our situation we're concerned about
only the divide between a lower range and the rest, we can
put our divide at 2 sD below the mean and have 98% vs 2%
--quite some diminution of the lower values!


But look as where 2 sD below the observed AVG values puts
us : at eyeknot strengths of the overhand & LazyDog of about
80% & 75%, respectively.  In comparison, the end-2-end knots
corresponding to these have means (not 2 sD below!) of about
70%.  .:.  To my thinking, this supports the proposition that
the stronger means to joining lines is by interlinked (one way
or another) eyeknots, in some cases, at least.  Notably NOT
among those cases by this current data is the fig.8 knot
which tests high in both forms (which surprises me somewhat!?).




And let's take this more refined sD/Guassian-distribution thinking
to the 2-Links problem in the OP.  Note that it assumes values
that are quite different (1 is half of 2 and a third of 3), and and
equal distribution of these values.  I showed, in my answer
to this problem, what happens if one simply doubles the middle
value (2) in population.  But let's now try to model the implications
of a bell curve and it sD implications.


To model a 98% distribution above the low value,
we can use 39 of 40 links, with 38 at value 2, one each at 3 & 1.
To assess the effects of this on expected average strength of
a 2-link joint, we have:


1+1 =1
1+2 =1   x38
1+3 =1
 -------------------> sums to 40x1 =40kg
((
2+1 =1
2++2 =2  x38
2++3 =2
 -------------------> sums to (1 + (38x2) + 2) =79kg
)) x38 !

 -------------------> sums to 38 x 79 =3002kg



3+1 =1
3+2 =2   x38
3+3 =3

 -------------------> sums to (1 + (38x2) + 3) =80kg


total kg = (40 + 3002 + 80) = 3122


There are (40 + 1520 + 40)=1600 cases
dividing the total kg sum of 3,122
. . .
for an average of 1.95kg.


So, yes, taking into consideration the effect of 2-links
alters the expected average, but only slightly, once
the distribution of such values is figured in.




--dl*
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X1

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Re: The shortest-chain-of-the-World ( problem )
« Reply #18 on: January 14, 2013, 09:46:10 PM »
fearing a loss of pride.

Yes, FEAR is the correct word, I agree.  :)
Here comes a unicornman !  :)

Or the unicorn man fears to stand on his own feet ? That is the question...
1. If you have chosen somebody that would fear something, anything - except time -, you have chosen poorly. Look elsewhere, or into the mirror.
2. If you think that all people are feeling like you, remember it the next time you will insult somebody.
3. If you are fearing your loss of pride, relax : your pride is so great, that any, however great that might be, loss of a portion of it, will leave the net sum unchanged.

the "Monty Hall" problem has been understood by me for some
time. 
   
   I do not doubt this...However, we do not know how much time you were trying to understand it, do we ?  :)
   Apparently, you have not understood my solution, and that is why, wisely perhaps, you have not said one word about it - but many words about/against me, as always...
   Speak about what I do, about my knots, not about me... because, apparently, you choose people poorly.

maybe you'll meet some  "big" knot tyer, and learn something!   ;D

  Was this your great final one-liner ?  :) Because if it was, either you underestimate yourself, or you underestimate Ashley. I leave to the reader to guess, judging from your behaviour, which is the most probable, the "expected" outcome !  :)
   To be anle to teach something about anything, you should be a teacher . Frankly, you are the worse teacher I had met in my life !  :)
   However, this will not disturb your loneliness - it will secure it.

   The childish trap with the two players, where the one is choosing after the other, was easily detected - but I did not believe you were so naive to place this one, and not any one of the many much better you could think of - or, as you knew the problem "for some time ", could have copied from the Wikipedia article !
   So, I had excluded it , and I was very strict about it :
 
   The "example" DL got out of his sheaves was specific about it. When the host ( the host, NOT another player !) will open the door, this door will (always) contain a goat.

   That was my hope, that you had not placed that childish trap... although, judging from your behaviour since then, It seems that I should have "expected" this...
   Timeo Danaos et dona ferentes.
   
   Stand on your own feet, Dan Lehman, do not fear the expected outcome - because unperformed experiments have no results.

   Q.E.D.

   P.S. I would hope that you could spent a small percentege of this verbiage about any of the knots I publish in this Forum...but , frankly, I do not expect it any more. And we should never add or subtract hopes with expectations, as the Monty Hall problem has taught us.
« Last Edit: January 14, 2013, 09:52:05 PM by X1 »

X1

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Re: The shortest-chain-of-the-World ( problem )
« Reply #19 on: January 16, 2013, 07:55:10 AM »
The childish trap with the two players,

   My statement that the variation of the original game DL pulled out of his sheaves ( to "trap back" whom he had imagined had tried to trap him at the first place), was a "two players" variation, was not accurate. In fact, it is a "three players" variation - the first player is not allowed to alter his choice, the second is allowed to alter his choice, but he always decides to not actualize this option ( because he decides it will not matter ), and the third is also allowed to alter his choice, and he does so, every time. 
   One need not "run out the probabilities" :), as DL did, in those long pages, to solve this problem !
   The first player chooses right from the beginning, and he retains his choice till the end, because he is not allowed to do anything else. There are 2 goats and one car behind those closed doors, and even a child knows that his chances are 1 in 3. The second player does exactly what the first did - although for a different reason ( he decides to do it, while the first player was obliged to do it ). However, the reason he does whatever he does, can not alter his chances !  :). So, his chances are the same as those of the first player, 1 in 3.
   So, the first player has 1 in 3 chances (=1/3), and the second has 1 in 3 chances (=1/3). What are the chances of the third player ?  :) One need not be a rocket scientist, riding a spacecraft ( this time... :) ), to conclude that the remaining player will have the remaining chances, and 1 - 1/3 - 1.3 = 1/3 ! So, in this childish variation of the original game, every player has the same chances, in all possible three cases, ( just as it should have been expected ) : 1) choosing and not be allowed to alter one s choice, 2) choosing, allowed to alter, but not deciding to alter one s choice, and, 3) choosing, allowed to alter, and deciding to alter one s choice.
   Why should we "run the probabilities', as DL, did, to see this ? I suppose that, if one does not understand something, he finds it "safer" to rely to some numbers, even if they are not needed. However, this use of "statistics" can be misleading, and hide under the rug the lack of deeper understanding. I have seen people in panic to count with their fingers, just like children do, just to be sure they will not be mistaken...
« Last Edit: January 16, 2013, 08:04:45 AM by X1 »

Dan_Lehman

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Re: The shortest-chain-of-the-World ( problem )
« Reply #20 on: January 17, 2013, 05:39:18 PM »
X, there is no connection of all your words about some
"two", now "three players" situation to anything I wrote:
I simply answered to your false assertion that in the case
where whichever of the two unchosen doors was revealed
was selected by chance --a coin toss, say; or as that cited
article puts it, a slip on a banana peel ("Monty Fall")--
there was still favor in switching : no, odds then are equal.


Then my presented running-out-numbers addressed
your continued warnings of dire consequences of using
two eyeknots in tandem to join ropes vs. a single knot,
that one must consider the weaker strength only.

That I did, using the assumptions coming with a bell
curve and standard deviations, enlarging the sample
count to a unit of 40 which allows 39/40 to model
the 98% of the population within 2sD of the Mean
plus the top end (strongest).  There, I showed that
with your 2-links situation, given this appropriately
proportioned population of 1kg/2kg/3kg links, one
would see the Mean only diminished from 2kg to
1.95kg --which is not so bad, and, esp. in light of
the large differences seen in the test data between
the eyeknots & end-2-end knots, supports J.P.'s
proposition.


--dl*
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X1

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Re: The shortest-chain-of-the-World ( problem )
« Reply #21 on: January 18, 2013, 12:18:49 AM »
 
there is no connection of all your words about some "two", now "three players" situation to anything I wrote:

   There is, but you just do not see it ( I can not be sure about why is this happening...).  However, apparently you bite hard into this problem, for whatever reason :)..., so I suggest you read the variation of the problem as I had described it in my last post, and see that it is the same variation you had presented by yourself, at your "contre-attaque"  :). I do not doubt that one can describe this variation in fewer words, in an ever more condensed form - but the essence will remain the same. Your variation is about two players ( two people that, in principle, can not know which door hides which object ), not one host and one player. I had transformed it in the "three players variation", because I believe that helps one understand it better/deeper, without the use of "counting" - by fingers or otherwise  :). It helps one understand that each player has the same chances than anybody else, so that those chances are 1 in 3, and, lastly, answer your specific question, about the chances of the second and the third player ( the player who does not change his choice, and the player who does ). And it does it without even having to "run the probabilities".

I simply answered to your false assertion that in the case where whichever of the two unchosen doors was revealed was selected by chance ... there was still favour in switching


   The false assertion was that I will fall into your trap - and it was proven false, because I had not, and I said it so, word by word :  Read my lips :

   The "example" DL got out of his sheaves was specific about it. When the host ( the host, NOT another player !) will open the door, this door will (always) contain a goat.

   THAT was what I had read out of your text - and that was what I expected you to do, because I expected your trap to lie elsewhere, where it would be more concealed ( and clever/cunning): in the use of the vague word "next".  I thought that you had tried to use a variation of the problem which would contain conditional probabilities, that would relate the initial choice of the player(s) with the conditions the host will place after he is informed about them. Conditional probabilities are always harder to evaluate. However, it turned out you had changed the original problem in a less complex / more childish way - that I had not expected. Perhaps you thought such a variation would be better for you, because a more obvious trap would be ignored... and you were right on this.

   I repeat, for the last time, and I hope that you will get it this time !  :)
   If there is one host, who knows what is behind which door, and one only player, who does not, the player should alter his original choice after the host opens a door that contains a goat. He (the player) can not know how this had happened, and he should not suppose that the host oppened this door by chance, of course - however, he should think that, even if that is what had happened, indeed, he would be no worse if he alters his choice, and he would be better if the host had opened one of the two doors he knew it contains a goat - so he should alter his initial choice. So, ONE player - a door which is opened by choice or chance and reveals a goat - this player should always alter his first choice.
   If there is one host and two players, and those two players have different/opposite strategies ( the first would choose at random, and would open the door he has chosen, and the other will also choose at random, but he has offered, and exercised, the option to alter his original choice after the first player has opened a door, and this door happens to contain a goat ), at the long run, the two players will win about the same number of times : 1 in 3.
   If there is one host, who does not even know which door hides a car and which hides the car, and three players , and the first player choses at random, and opens the door, then the second chooses at random, but he does not open the door he has chosen, and then the third player chooses at random, and he always alters his first choice if the first player has chosen a goat - at this variation of the original ptoblem, all players will have the same probability to win the car, 1 in 3.
   I believe my re-formulating of the variation you had posted into this three-player form, is more "balanced", and explains your two-player variation without the need to "run the probabilities", i.e., without the need of "counting".

[re. the ] 2-links situation, given this appropriately proportioned population of 1kg/2kg/3kg links

  Here we come again !  ( I should have expected that ! ) Your beloved method of twisting the given data as if they were ropes  :), is in action here, again. Would you, ever, be able to address the problem which is posed to you, without trying to figure out "loopholes" and "variations" ?  You did it in the Monty Hall problem, and you do it again here... , because, apparently, that is always your favourite escape route.
   WHO ON EARTH had spoken about this "appropriately proportioned population"  :) of the 1kg/2/kg/3kg/ links ? Is it something like this "abbreviated topological correspondence" of yours ? Read my lips :

 
we know ... that one third of them can withstand a pull of only 1 kilogram, one third of them can withstand a pull of only 2 kilograms, and the last third of them can withstand a pull of only 3 kilograms.
   So, the average weight one link can withstand is 2 kilograms, right ? [ "Yes"... ].

   Of course, another distribution with the same "average" strength of the links, would lead to another result. We do not know which is the expected or the actual distribution of a large sample of similar tests. I had supposed it would be a Gaussian ( normal, bell shaped ) distribution, but this was only a wild guess - so we can not depend on it ! It may well be a more "square" or a more" flattened" one, we just do not know. However, the whatever different result will not always be closer to the 18/9 than the 14/9 is... More weaker-than-average links will tend to "push" the value even lower - and the stronger-than-average links will not help in this : it is the weaker links that count and determine the strength of each compound chain/knot. That is what I had repeated over and over again - the "average" does not mean much - and we should not add and subtract "averages of strengths", like they were weights !  :)
   
 
To my thinking, this supports the proposition that the stronger means to joining lines is by interlinked (one way or another) eyeknots, in some cases, at least. 

   Reading again the looong previous posts, just to check again if I had missed something, I had noticed that "loophole", which was left as an insurance premium !  :) :"in some cases, at least." (sic). How convenient, this "some" and this "at least"... ( For the "one way or another", that hides the escape route of the "shared eye legs" "variation", I had already responded elsewhere (1)) .I could easily say the exact opposite, that "the stronger means to joining lines is by eye-to-eye knots, in some cases, at least" - but I will not, because :
 
 I fear no jamming knot
 or any loss of pride
 or spill by the unicorn
 of which I do not ride.


   Music from the movie " All that Jazz" ( 1979)
   Any other inguistic improvements and rhythm s ideas are welcomed... :)

   An advice : do not be so "sure" about J.P.s conjecture - or, if you have chosen to show persistence on your (wrong) first idea, prepare to figure out a plausible "loophole" or "variation", like you did in my 1-2-3 strength link problem, or in the Monty Hall problem. With your experience on knots, I hope you will find a much better escape route than the excuses you managed to pronounce on the simple, well stated problems of this thread.

 1)  http://igkt.net/sm/index.php?topic=4150.msg25955#msg25955   
« Last Edit: January 18, 2013, 07:12:25 AM by X1 »

X1

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Knot tyer s song.
« Reply #22 on: January 18, 2013, 06:01:32 PM »
I fear no jamming knot
or lose the long-lost pride
or spill from the unicorn
on which I do not ride.

« Last Edit: January 20, 2013, 03:08:08 AM by X1 »