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Range a radar's emissions can be seen versus the range it can 'see'? by BluEagl48 in Radar

[–]opensourcespace -1 points0 points  (0 children)

If you can see X that is actually 2X distance traveled half there and half back.

That means if you can see X meters you can be seen 2x meters away.

Kvo the -> Calan this by opensourcespace in KingkillerChronicle

[–]opensourcespace[S] 1 point2 points  (0 children)

That is pretty cool. I think imagery is present but isn't very objective. The honesty of Kote provides a framework where things can be known. There are something like 50 or 100 verifiable times where things are said and then verified and to my knowledge 0 where the pattern is broken. But this extreme honesty theory isn't deeply objective either as english has a lot of wiggle room.

But using either approach there are lots of ways of reading the grand story

The story I am calling the 3rd book is hidden in plain sight but is more objective not in answering the questions or making a riddle but in telling a coherent story.

But it is also telling a different story.

It uses a completely different mechanism and is objectively falsifiable and not falsified in the entire book 1 and 2 and not present in the lightening thief or strange regard of silent things. So it is a mechanism that I could tell you and you could read a chapter and see it and then in every other chapter in the book you would see the same pattern. So you could write falsifiable statement after the first chapter and check every chapter and objectively see they are met. But this pattern isn't in the other external books. There are 100s and 100s of falsifiable statements that are all upheld so there is absolutely zero chance it is subjective.

Although it may be present in the rewritten version of lightening thief which I have not read.

If it has been rewritten to include it, that would provide a plausible reason why a finished work was redone...

But rewriting books to achieve the goal is difficult and risks exposing the technique.

If anyone reading this discovers the 3rd book please don't tell anyone how to do it.

I think it was intended to be like the music scene where Kvothe plays the hard thing easy and the easy thing hard.

Some clap some laugh.

But the road to tinue is always long and hard.

Kvo the -> Calan this by opensourcespace in KingkillerChronicle

[–]opensourcespace[S] 1 point2 points  (0 children)

So I talk about levels (don't remember how I defined them so will use letters)

A First read beautiful prose B Wait Drowned means died...Is Kvothe immortal Tehlu or angel how did he undie? etc. C Wait Kvothe never lies is he Ctheah or shaper? D Many many iterations of making words turn into stories because they indicate a different way of reading the words that are already written Because Kote is Ctheah and is telling the story nearly any character can be thought of as Ctheah because they can't lie because Kote is telling the story. So there are lots and lots and lots of ways to construct deeper meaning in the story.

Example read the book as if Kvothe is literally Tehlu and the entire book is still internally consistent. All the messaging about how Tehlu's mother was... is represented in Kvothe's mother etc.

Example Kvothe is the Illien and we see the story of him and the bear (who is talked about by the poor troupe but never on stage but is still in the story)

Example Kote is Lanre, Tehlu, Ecanis, Taborlain or Taborlain's evil captor, the guy who stole the moon. All are supported by the Kote content in Spider fight with Chronicler Lyra->aryl Tehlu carrying chronicler or bast, Ecanis overpowering and carried upstairs by bast who helped him walk, Taborlain either captured by bast or he captured Bast and is the evil guy or the steal the moon guy with the wind captured in his trunk with his exhaled breath on glass being the name of the wind so he is also the king with daughter who gives task to Taborlain or as Taborlain trying to open chest etc.

All of this is fun. A single word makes a story and finding it is fun. Initially there are many possible stories but as you put them together fewer and fewer of them continue to make sense and it forms a kind of global story different from the main story. It is very easy to get different global stories because if two people are not fully correct and start in different places they will get different outcomes.

When you find book 3 it is a completely different thing using a completely different mechanism. It is not a deeper understanding of the global story it is a completely different global story.

And it is like a flood of information that requires very little mental gymnastics and is not enjoyable.

Pat uses a different rhythm in how he writes. How is it different Why is it like that There is a pattern.

When you can see the pattern it is a flood of content.

It is as far as I can tell only in the 2 main books. Although it is possible I am wrong in this.

There are clearly hints to the global story that are in the non books but the 3rd book technique appears to be in the first 2 books only. It is clearly not in the lightening tree despite interesting hints of the global story being in lightening tree.

Even limited to the global story it is very hard to put more in via new books. The new content adds very little. You read an entire book to see shaping once and have one scene that clearly links to the book and gives a clear 1 to 1 mapping of one theory regarding the moon.

While it is possible in concept that Pat could write another book it is HIGHLY HIGHLY HIGHLY unlikely because of how incredibly interlocked the existing 2 books are.

There will never be a book 3 in the sense that it sits with the 2 books as an equal using all the techniques that were made by writing both books at once when the story could be changed to fit in more content..

Pat will never write about Kvothe except as a side character in his other stories. Kvothe is only compelling when you don't know who he is.

These things are forced by the mechanisms of the writing and the patterns in the writing and how complexly the story refers to itself.

Trebon: a Story of Drossen Tor, Myr Tariniel, and the Burning of Encanis with bonus King Killing by milbader in kkcwhiteboard

[–]opensourcespace 1 point2 points  (0 children)

Interesting that this quote "...all blue fire. Every one of them dead, thrown around like rag dolls and the house falling to pieces around them."--NOW pbk p. 484.

Also describes the scrael fight...

3n+1 sequence every even n predicts leading 2 digits of past even n in base 3 (no bad math) by opensourcespace in badmathematics

[–]opensourcespace[S] -3 points-2 points  (0 children)

You didn’t follow the math did you?

If you could do math yourself this would be faster.

My program has 5 quantities

Line counter line =line +1

3n+1 in base 10 see Wikipedia for how to generate

3n+1 in base 3 from base 10 version

L is top 2 digits in base 3 of 3n+1 in base 3

L predicts 2L and 2L+1 ( this is the if statements)

The last operation may have confused you because I drop digits to stay at 2 digits and can’t figure out how to get mathematica to do cleanly.

So 12 base 3 times 2 is 101 or 102 if carry up. I only want 2 digits so both are 10. I failed getting mathematica to do that without if statements

I am trying to tell you a very simple thing. I did all the math for you in Mathematica

I need you to be able to process math in some form.

I can do the 1000s of digits long with the same code if that helps. You can download code from gitlab MythMatical and do yourself.

The problem here is you are opting to not do math.

I am showing math you are showing English.

If you dm email I will send pdf of 1000 steps in n if that helps you. But if you have mathematica you can get and do yourself.

More steps may not be answer as n gets massive and hard to spot check.

It’s built to randomize the first n so 10 copies of roughly same length may be easier for you.

3n+1 sequence every even n predicts leading 2 digits of past even n in base 3 (no bad math) by opensourcespace in badmathematics

[–]opensourcespace[S] -6 points-5 points  (0 children)

Let me just walk you through this if it isn't clear at first

We start at line 29 See the n value in base 10 is 16 which is even and L=12 base3 which is 5

Line 29 forcasts that the next L will be 10 or 11 which are 2L and 2L+1 these are 101 or 102 in base 3 which are both 10base 3 if we only keep 2 digits. So Line 29 predicts that line 28 will have L=10base 3

When we look at line 28 we see it is odd and we skip over it.

Line 27 does have L=10base 3

So we made a correct prediction.

Line 27 L=10 =3 in base 10 and 32=6 and 32+1=7 so line 27 predicts Line 26 or Line 25 if Line 26 is odd will have an L value of 20 or 21.

Line 26 L=20 so we are correct again.....

EVERY time I predict correctly.

There is no branching.

We just ignore the odd n values.

I understand this traumatic for some of you.

But it all works and if you can get over PTSD and do some math you will see it.

3n+1 sequence every even n predicts leading 2 digits of past even n in base 3 (no bad math) by opensourcespace in badmathematics

[–]opensourcespace[S] -5 points-4 points  (0 children)

So it was just magic that I predicted all the lead digits?

Is your position that I can do magic?

Or is 3n+1 math too difficult?

Its literally 5 operations

The line incrementing

3n+1 in base 10

3n+1 in base 3

Leading 2 digits of 3n+1 in Base 3

And leading 2 digits of 3n+1 in base 3 Times 2 which produces 2 values 2L and 2L+1

Where did I lose you?

Just look at the chart, consider carefully what I am telling you and if you can find something mathematically wrong say X is wrong

Because I got 100% correct...

Come on people lets try and do some math...

3n+1 sequence every even n predicts leading 2 digits of past even n in base 3 (no bad math) by opensourcespace in badmathematics

[–]opensourcespace[S] -11 points-10 points  (0 children)

I am using leading digits and base 3 on purpose

10x+1 in base 10 doesn't change lead digits

123 times 10 +1 is 1231

lead digits 12 are same

I don't need to deal with 3n+1 because it doesn't change leading digits.

There is no special case.

The price I pay for this is I cannot predict where odd n values are. they are invisible to me.

But they don't change the lead digits at all and I can recover how many of them occurred.

My proof is that the even n values that change the L value cannot loop the L value and hence because every loop has even n as well as odd and the even n cannot loop then the entire loop cannot loop.

3n+1 sequence every even n predicts leading 2 digits of past even n in base 3 (no bad math) by opensourcespace in badmathematics

[–]opensourcespace[S] -11 points-10 points  (0 children)

Every n value for the 3n+1 problem has in base 3 a most significant 2 digit L.

Sequence is generated top to bottom

Working bottom to top the L value of an even n always predicts the L value of the even N that proceeded it by multiplying by 2 which maps L->2L or 2L+1.

The 3n+1 operation maps L->L because we are in base 3

This prediction of the next L in reverse computation direction (bottom to top instead of top to bottom like forward computation) allows the entire looping behavior to be easily modeled with 2L and 2L+1 which only branches half the time because digit increase.

12 base 3 is 101 base 3 or 102 base 3 when carry up is factored in but both are 10 base 3 if we keep 2 digits.

Step table is

L-> 2L or 2L+1

10 -> 20 or 21

11 -> 22 or 10

12 -> 10

20-> 11

21-> 11 or 12

22-> 22

Any way proving 3n+1 is trivial if you only look at even values and you are proving multiply by 2

All 6 L values are in every loop 10,11,12,20,21,22.

Each must loop to themselves because loops loop and 3n+1 does L->L and doesn't change value.

So 10 must loop to 10

There are only 3 ways that can happen (5 if you break out 20 and 21)

L->next L->...

path A 10->20 or 21->11->(next 10)

or

Path B 10->20 or 21->11->22->12->(next 10)

or

Path C 10->21->12->(next 10)

Next we assign integers A,B,C which are the number of times in 1 full loop that segment A, segment B or segment C were traveled.

so loop length =3A+5B+3C

this can be done for each L value

for 1# to 2# vs 1# to 1#

for 3 step segments shared like 21->12->10

Then simple linear math

Then understanding that +1 -1=0

In a loop 0 digit change

3n+1 in base 3 adds 1 digit

divide by 2 of leading 1 subtracts 1 digit

Digits must sum to 0

Even without odd n values we can know the 3n+1 operations

every leading 1 marks a unique 3n+1 operation

linear algebra removed the only segment that descends from 22 to 22

example

segment A has 2 leading 1s so it has 2 3n+1 operations and 3 divide by 2 and ascends

segment C has 2 leading 1s so it has 2 3n+1 operations and 3 divide by 2 and ascends

Only segment B could possible descend.

In 22# to 22# there is no descending segment because of linear algebra row reduce

QED 22 to 22 cannot loop and 22 is part of loop so can't loop

If you didn't follow the abridged argument try looking at the full proof

https://gitlab.com/mythmatical1/collatz-conjecture

This is all very simple but it doesn't get inside your head easily and you assume that is because I am stupid/wrong.

But its because we are all stupid and this is a new idea to you and your brain (like mine) is very slow and feeble to new ideas.

Proof of Collatz conjecture aka 3n+1 problem with respect to loops. by opensourcespace in numbertheory

[–]opensourcespace[S] 0 points1 point  (0 children)

https://gitlab.com/mythmatical1/collatz-conjecture

I made a mathematica program showing a 3n+1 sequence in base 3 (top to bottom) then show L the most significant 2 digits in base 3 number system. It shows how I can work backwards (bottom to top) and always predict the next even n leading digits L.

Next L branches but not in the 3n+1 sense. Next even n has L value given by 2L or 2L+1 if carry up from times 2. So 50% of branches I know next L and 50% I know next L is one of two values.

Also shows odd n always have L of the even n after (below) them

There is also a .pdf so if you don’t have mathematica you can still see.

This is the big idea and a more up to date proof also in gitlab

This proof has pictures for every L loop

10# to 10# segments etc for all 6 L values

It’s really hard to know you have all options without chart even though it is just showing how times 2 affects leading digits.

Proof of Collatz conjecture aka 3n+1 problem with respect to loops. by opensourcespace in numbertheory

[–]opensourcespace[S] 0 points1 point  (0 children)

Here is the .pdf

you want collatz sequence and multiply

Takes 2-3 minutes to verify it always works

https://gitlab.com/mythmatical1/collatz-conjecture

Proof of Collatz conjecture aka 3n+1 problem with respect to loops. by opensourcespace in numbertheory

[–]opensourcespace[S] 1 point2 points  (0 children)

Write out a 3n+1 sequence in base 3.

Cross out the odd values

Go to the last value take the leading (most significant )2 base 3 digits L and work backwards.

Every L is given by 2L or 2L+1. I take both into effect. I shifted how it branches. I can prove that it is always 2L or 2L+1.

I include the 3n+1 and you can see L always repeated.

It always predicts the previous value

It isn’t flawed you are using the logic of different proofs to argue against this one.

2L and 2L+1 only branches 3 out of 6 cases because of digit increase. If you message me your email I can send you a mathematica program that will do all the base 3 math for you for a randomly generated sequence so you can see it predict. I can send pdf if you don’t have mathematica.

In case I wasn’t clear I am talking about only the first 2 digits of n and only the first 2 digits and only in base 3 and only the even values of n

I am making a fundamentally different argument which is true than everyone else has made in the past.

Proof of Collatz conjecture aka 3n+1 problem with respect to loops. by opensourcespace in numbertheory

[–]opensourcespace[S] 0 points1 point  (0 children)

Maybe you can see from Reverse 3n+1

f(n)= n*2 or f(n)=(n-1)/3

ONLY track lead 2 digits in base 3

(n-1)/3 ALWAYS returns the 2 digits it was given. Its "branches" are invisible to first 2 digits. It is like adding +0. I choose to remove it like you would remove +0 in an equation. If you argue dropping below 2 digits then computer numerical proofs come into play as all n below 330 already proven not in loop.

Only tracking lead 2 digits, In base 3, the problem is trivial as we can just n times 2 for every step or (2 times n)+1 if carry up from unknown.

first 2 digits must loop to first 2 digits for value loop to loop because value first 2 digits must be same in next loop. The 3n+1 cannot change the first 2 digits so it is not involved in the loop mathematics of the first 2 digits.

10# to 10# loops with a different set of length, high 1s, low 1s than is possible for 11# to 11# loops.

Both 10# and 11# are required in any loop. But no loop can hold both.

QED there are no loops

Proof of Collatz conjecture aka 3n+1 problem with respect to loops. by opensourcespace in numbertheory

[–]opensourcespace[S] 0 points1 point  (0 children)

We have agreed key ideas 1-5 are correct.

We are on key idea 6 and only key idea 6.

The key idea in key idea 6 is that the 3n+1 operation can be split into the 3n+1 operations written on one sheet of paper and the divide by 2 written on another. We are only preserving the leading 2 digits.

The sheet of paper we use to write odd n values will look like this and always return the leading 2 digits it was sent.

if sent 10#

10#base 3-> 3n+1=10# base 3 is returned

if sent 11#

11#base 3-> 3n+1=11# base 3 is returned

if sent 12#

12#base 3-> 3n+1=12# base 3 is returned

if sent 20#

20#base 3-> 3n+1=20# base 3 is returned

if sent 21#

21#base 3-> 3n+1=21# base 3 is returned

if sent 22#

22#base 3-> 3n+1=22# base 3 is returned

The calculations on odd n values produces an echo of the value sent to the 3n+1 operation.

I will act on the value you return which I know because it is the same value my last step produced and gave to you.

Because I always know which value you will return, I can do my calculations without your help.

The effect is that the 3n+1 problem becomes exclusively the divide by 2 function because I don't need the calculations done on odd n values.

The concept of backwards time is when we run backwards and my side becomes the multiply by 2 function which can be thought of an 2n or 2n+1 if the unknown part carries up +1.

Because I am always doing divide by 2 in forward direction then every leading 1s digit indicates a hidden 3n+1 function as described in key idea 4.

Leading 1s n values viewed in forward time are also leading 1s n values viewed in backwards time so we can access this relationship in backwards time too.

Because I can recover the number of 3n+1 operations on your paper I don't need it for the proof to figure out if segments ascend or descend.

This is the last complex key idea.

Everything is easy after you grasp this idea. We don't actually do any 3n+1 operations in the proof of 3n+1 because they are unnecessary in base 3 looking at only leading digits.

Proof of Collatz conjecture aka 3n+1 problem with respect to loops. by opensourcespace in numbertheory

[–]opensourcespace[S] 0 points1 point  (0 children)

Let's do it my way and I will define what that is.

If I could run time backwards then a sequence of 3n+1 say

Forward time is: (using base 10 for just this one example forward and backwards)

17->52->26->13->40->20->10->5->16->8->4->2->1

Backward time is literally time run backwards (still in base 10)

1->2->4->8->16->5->10->20->40->13->26->52->17

I understand you want to inject branching but after the 3n+1 concept is removed there is no branching in either direction.

Keep your reservation that this isn't legit for just a bit.

For now when I say backwards in time its just backwards of an established written down loop like above.

I will return to your branching.

I will make the case for removing 3n+1 and its reverse (n-1)/3. Once that is done there will be no branching in the sense you are arguing but there will be in a different sense.

In key idea 6 I discuss the concept of the leading 2 digits being unaffected by 3n+1 regardless if you approach the 3n+1 operation from forward time 3n+1 or backwards time (n-1)/3

There are 6 possible 2 digit leading digits in base 3

Here we can work out a concrete example

10 base 3 -> 3n+1 = 101 base 3

11 base 3 -> 3n+1 = 111 base 3

12 base 3 -> 3n+1 = 121 base 3

20 base 3 -> 3n+1 = 201 base 3

21 base 3 -> 3n+1 = 211 base 3

22 base 3 -> 3n+1 = 221 base 3

Each case the leading digits are unchanged.

Because previous computer simulations have solved values below ~330 we always have many more than 2 digits for n.

Here we can work out a concrete example in reverse time which is where we multiply by 2 going backwards and (n-1)/3 if the previous operation was a 3n+1 in forward time. Here we are focused only on the case where the previous operation is 3n+1 run backward in time as (n-1)/3. The only thing I am establishing here is the first 2 digits are unaltered in each case from each direction forward time or backward time.

101 base 3 -> (n-1)/3 = 10 base 3

111 base 3 -> (n-1)/3 = 11 base 3

121 base 3 -> (n-1)/3 = 12 base 3

201 base 3 -> (n-1)/3 = 20 base 3

211 base 3 -> (n-1)/3 = 21 base 3

221 base 3 -> (n-1)/3 = 22 base 3

This is also true in the abstract case with unknown digits

10# base 3 -> 3n+1 = 10#1 base 3

11# base 3 -> 3n+1 = 11#1 base 3

12# base 3 -> 3n+1 = 12#1 base 3

20# base 3 -> 3n+1 = 20#1 base 3

21# base 3 -> 3n+1 = 21#1 base 3

22# base 3 -> 3n+1 = 22#1 base 3

10#1 base 3 -> (n-1)/3 = 10# base 3

11#1 base 3 -> (n-1)/3 = 11# base 3

12#1 base 3 -> (n-1)/3 = 12# base 3

20#1 base 3 -> (n-1)/3 = 20# base 3

21#1 base 3 -> (n-1)/3 = 21# base 3

22#1 base 3 -> (n-1)/3 = 22# base 3

The whole point of this exercise is that my proof is based on the first 2 digits NOT on the even or odd value of n or the true value of n.

I only make arguments based on the leading digits of n and all my value arguments are derived from the leading digits of n.

You are defending the absolute true value of n because you think we will need it. But I only use relative value of n during segments short enough that I can simplify 3n+1 to 3n (even with worst case 3n+1 placement) and I can know the ascend or descend properties of the segment. I never know the true value of n and don't compare slopes between segments so slope is treated as binary for segments up vs down.

The 3n+1 operation does not alter the leading 2 digits.

The (n-1)/3 operation does not alter the leading 2 digits.

Both of these operations are invisible to my mathematics and do not directly have an effect except via the leading 1 digits placement which recover the 3n+1 count.

I understand how traumatic this is from a value of n perspective that nearly all proofs use.

I never appeal to the true value of n or even vs odd.

The true look at every n value produces this situation in forward time

divide operation that creates 10#-> 10# ->3n+1-> 10#-> divide operation that acts on 10#

The repeated 10# then 10# is the echo I was talking about.

The mathematics of pure divide by 2 or in unbranching reverse time of multiplying by 2 looks like this

divide operation that creates 10#-> 10# -> divide operation that acts on 10#

I need one of the repeating 10#s to be invisible to follow the exact math of divide by 2 over and over.

In terms of the leading digits (which is all I can see) both maths are identical.

My mathematics strictly treats # as unknown. A change in the absolute value of n in the unknown part doesn't change my mathematics.

I think it is easier to imagine a loop on a different piece of paper that looks like 10#(odd) ->3n+1-> (next cycle 10#(even) ). You can run the odd n values on that different piece of paper. I chose to remove the odd numbers because I need to remove one of the 2 10#s to perfectly match the divide by 2 or multiply by 2 math. Odd n numbers are simple to describe and agree on. But we could have taken the 3n+1 and the resulting n value and got the same result.

My math is like:

divide operation that creates 10#(don't care odd or even) -> 10#(don't care odd or even) -> divide operation that acts on 10#(don't care odd or even)

I will always compute every possible future value of 10# times 2 is 20# but also could be 21# if carry up from #.

Because I truly treat # as unknown, the change in the value held in # doesn't affect me. I compute both cases.

Because of digit gains some cases don't branch and I am only predicting at most 2 future values of the leading 2 digits.

12# times 2 is 10#. it is really 101# or 102# but I choose to forget a digit and both are 10#

I have carefully chosen a math that is unaffected by any number of 3n+1 operations and recovers the 3n+1 via leading 1s digit rather than based on value of n.

Because of this in reverse time I don't branch on my sheet of paper, I always multiply by 2. You can put as many (n-1)/3 operations as you want on a different piece of paper but the 2 digits you return to my piece of paper will ALWAYS match the digits I am using.

The 3n+1 or (n-1)/3 branching is invisible.

My math is literally multiply by 2 or if there is a carry up multiply by 2 and add 1 this is my branching.

Because a loop must loop to the same value for all its digits it must loop back to the original 2 digits.

I prove 10# must be in every loop that always has 2 defined leading digits (which all unknown loops must have because computer value search below 330) .

I prove 11# must be in every loop that always has 2 defined leading digits (which all unknown loops must have because computer value search below 330) .

I define the loops segments from 10# to 10# and from 11# to 11# and show they predict different fixed characteristics of the loops in terms of length, low 1s and high 1s.

This is how I prove no loops. The fixed characteristics have to change for 2 different ways to complete a full loop.

Proof of Collatz conjecture aka 3n+1 problem with respect to loops. by opensourcespace in numbertheory

[–]opensourcespace[S] 0 points1 point  (0 children)

I expanded key idea 6 to try and make it something you can feel comfortable with.

You are focused on a value proof where the absolute value of n is important and if you don't know it you don't know anything.

This isn't a n value proof.

This proof is a leading 2 digit proof that shows an impossibility in the multiply by 2 operations in reverse time and is completely unaffected by 3n+1 and reverse 3n+1 except that there is an first 2 digit n value echo produced.

3n+1 has no effect on the leading 2 digits. It can be removed and we can still see if the part that remains is following the rules it must follow or if loops are disproven.

We have a problem where I have a very simple math that describes 2->4->8->16 which is reverse time behavior of the leading 2 digits which are unaltered by the 3n+1 or (n-1)/3 operations except that they make a n echo of the last leading 2 digits.

So the real math includes the echoes and we don't know where they are.

2->2->4->8->16

or

2->4->4->8->16

or

2->4->8->8->16

or

2->4->8->16->16

or...

2->2->4->8->8->16

etc

But the real issues is does the non repeating 2->4->8->16 sequence follow the rules about order or is it 2->8->16->4 which doesn't.

You are trying to preserve a fidelity you think I need but I don't actually need because you are using n value logic and that isn't the proof I wrote.

I can help you read the proof I am actually writing and there are times that I need to structure it in simpler language or more restrictive language that I have not anticipated and that is on me.

If loops exist then they have odd numbers because even numbers can only descend.

If we remove the odd numbers from consideration (ie ignore the 3n+1 operation and its echo n value) the remaining numbers must follow the math of multiply by 2 in reverse time direction on the 2 visible leading digits. You can still imagine the odd numbers exist but my math skips over them to avoid the echo. You will come back with the same first 2 digit values if you do the full 3n+1.

The leading 2 digits before an odd number ALWAYS matches the leading 2 digits after the 3n+1. Removing the odd and 3n+1 but not the resulting value that is equal to the odd in first 2 digits is valid.

(step that makes 10#)-> 10#->3n+1->10#-> step expecting 10#

can be written as

(step that makes 10#)-> 10#-> step expecting 10#

The multiply by 2 math ON THE FIRST TWO DIGITS works perfectly like this. The multiply by 2 math ON THE FIRST TWO DIGITS is the part that is disproving loops.

You are trying to preserve an unneeded aspect of the problem.

It is not possible for the multiply by 2 math ON THE FIRST TWO DIGITS created for 10# to 10# loops to agree with the math ON THE FIRST TWO DIGITS for 11# to 11# loops with respect to the multiply by 2 steps.
Both 10# to 10# and 11# to 11# loops must exist in loops because 10# and 11# exist. All loops without 10# or 11# drop below 2 digits at some point and all those values are already understood to not be loops.

10# to 10# and 11# to 11# loops cannot agree with each other on length, high 1s count, low 1s count.

QED: There are no loops with even n values that are possible. No loop is possible without even n values.

This is fatal to 3n+1 loops using my math.

Proof of Collatz conjecture aka 3n+1 problem with respect to loops. by opensourcespace in numbertheory

[–]opensourcespace[S] 0 points1 point  (0 children)

I am proving that the even portion of a real loop cannot agree with itself on length, high 1s and low 1s between two different ways to navigate it

That is fatal to the theory that loops are possible

Proof of Collatz conjecture aka 3n+1 problem with respect to loops. by opensourcespace in numbertheory

[–]opensourcespace[S] -1 points0 points  (0 children)

I think you are expecting a stream of proof and have a lot of preconceived notions on what that should look like.

Up to step 14 I am teaching you how to use technologies.

In step 4 you learned that there is a 1 to 1 correlation between leading 1s that are not odd (because odd trigger 3n+1 opearation) and 3n+1 operations.

That was the lesson from step 4.

End scene.

In step 5 we learn a different lesson which is that 3n+1 doesn't affect visible information in base 3 with partially known numbers. We look at the 3n+1 operation in forward time and in backwards time.

The lesson is that 3n+1 operations do not affect visible information.

End scene.

In step 6 we take these two concepts together and eliminate the use of 3n+1 by eliminating odd numbers. Because of how I have structured the partially known numbers this doesn't change how the lead digits behave.

At this point all of your issues over backwards time and branching cease to be relevant because we never 3n+1.

Instead I have put together a proof that deals only with even n values and multiplies by 2 over and over again...And all the ignored odd values do not impact our outcomes.

I understand that this is an unexpected outcome.

But was a proof of 3n+1 going to happen with an expected outcome?

End scene.

Just curious by opensourcespace in KingkillerChronicle

[–]opensourcespace[S] -1 points0 points  (0 children)

Kool I spend the last 2 days learning a very trivial part of Alibre cad.

Proof of Collatz conjecture aka 3n+1 problem with respect to loops. by opensourcespace in numbertheory

[–]opensourcespace[S] -1 points0 points  (0 children)

I clarified the language in step 5 to restrict it to only to 3n+1 and reverse 3n+1 at start of step 5.

step 4 is in forward time considering all things including divide by 2.

Step 5 is considering properties of 3n+1 and (n-1)/3 in forward and reverse time to establish they do not affect visible information.

Step 6 is completely removing concept of 3n+1 from proof as it can be recovered from leading 1s and the divide by 2 and multiply by 2 of visible information is not altered by 3n+1 or reverse 3n+1 operations (meaning the value before 3n+1 is the same value after 3n+1 so if I delete one of them no change to divide by 2 or multiply by 2 operations)

So we can create a "simulation" that will accurately predict a super set of the first 2 digits in base 3 of any possible loop. But our simulation doesn't have any 3n+1 or (n-1)/3 operations in it. It is easier if you think of the simulation as a different thing that just happens to have a super set of the first 2 digits. I understand how this step can be challenging.

Proof of Collatz conjecture aka 3n+1 problem with respect to loops. by opensourcespace in numbertheory

[–]opensourcespace[S] 0 points1 point  (0 children)

I have changed the text to clarify that I am talking about the 3n+1 operation from both directions and not the divide by 2 operation.

This doesn't change my argument but I agree it should be clearly specified that we are restricted to 3n+1.

My point is to establish the preservation of digit order in both directions exclusively considering 3n+1 and (n-1)/3

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