text:
Announcing: New 4x4x4 Brute Force Solver

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rokicki

Member
What you don't is precisely what you mentioned: execute hash_number modulo some_number = index, then visit database[index] and get your info. At database[index] you have multiple pieces of info, one of which is the FULL 32- bit or 64-bit number. So there is no 1-byte/entry version of a hash table.
For doing an IDA* search, where we just need an admissible heuristic, this approach actually works
*perfectly*; you ensure the single entry contains the minimum distance-to-solved from all values
that collide here, and your table then functions admirably. Further, if you use a fast hash function
(it doesn't have to be cryptographically secure), the elimination of collision detection, etc. cuts the
code down and the performance (with no inessential branching) can be excellent.

You can even use this idea with only one *bit* per entry that says a position is either at a distance
> d for some d, or "no information". I frequently use it with two bits per entry, giving me information
that the position is >d, >d+1, >d+2, and "no information" for a given d. Using 1.6 bits per entry also
works well, in which case you get >d, >d+1, and "no information".

Story time:
Great writeup, Lucas; thanks! That clarifies things tremendously. I think the bulk of this prose should
be on a wiki somewhere for others to profit from.

Everyone on this thread: should we follow the sage advice of Lucas and go with SiGN exclusively, or
should we continue to use whatever we are using? In particular, if we plan to exchange positions
and compare solves it would be nice if we could just copy and paste.

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unsolved

Member
Of course, I can't control what notation people use. But we should have a standard, and SiGN is as good as any. It's simple, and I believe it makes the most sense: r can only stand for one of 2R or Rw, and if we're going to have a standard we need to pick on...
As an outsider, what I am about to offer may be unpopular, but here is my input regarding notation.

The very first cube book that I ever saw in 1980-1981 featured upper case letters for the sides and + for clockwise and - for counterclockwise. Just my opinion, I like this best. How many times have you had to squint and see if a ' was or was not in the middle of a sequences of moves? Then you check over the moves over and over until you see what you missed, etc. But with a + and - after the moves, you get a nice delimeter and your eye does not lose track as easily.

When the 4x4x4 came out, lower case meant slice moves. No problem, I like that.

But 5x5x5 and larger presented problems, of course. Personally, I find E and M and S counterintuitive. I use notation such as {f/r} to indicate a middle slice turn from the front in the direction of the right face. In the opposite direction, it is {r/f}.

I have not given larger cubes any thought, or how to describe multiple slice turns.

Also, I use a right- handed rotational vector notation. R+ and L+ both turn in the same direction, so your L- is my L+. Ditto for top and bottom, your U = my T+ and your D' = my B+, since I use K for the back slice. B = bottom to me, T = top.

Why should r'l and l'r be opposite in direction when you turn them both the same way? In my notation it would be (lr)+ when both turn like r and (lr)- when both spin like r'.

[Ff]+ would be F f and [Tt]- would be U' u'. Slice and face turns use square brackets, pairs of slice turns use round parenthesis, and middle slices use curly braces.

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Christopher Mowla

While most people use SiGN, some people like cmowla will still use old WCA notation. I'm not blaming them, but it makes me sad that as a community we still haven't fully settled on a single notation.
I'm set in my ways! No, I understand the significance of unity (that was the core idea of my single commutator project!), as I made an effort to make my derivation of the "Holy Grail" alg in SiGN notation. I did this because I assumed that in the future, we would all have agreed on SiGN by then. However, for 2 layer wide turns, I used a 2 coefficient in front of the lowercase letter.

Speaking of which,
In addition, R simply meant the outer slice, and r was the simplest multi-slice outer block turn (two slices) -- which was consistent with one common conventions.
It was easy to learn, consistent, and at least *something* we would hopefully all agree on.
Having r = Rw is logically flawed based on the structure of SiGN. r (in SiGN) = 1r = a 1 layer outer block turn. That is, r = 1r = R in SiGN. But in SiGN also wants to make capital letters mean a single slice...so here is a conflict in SiGN, and thus if you omit the "convention" that r = Rw, then SiGN will be consistent.

With the same flawed logic that r = Rw for the sake that r = (R M') on the 3x3x3, then why isn't x = (Rw l') on the 4x4x4 in SiGN? I of course know the answer, but how obvious you think that answer is, I feel it's just as obvious that you shouldn't say that since r = (R M') on the 3x3x3 (which is really equivalent to doing a single slice turn which obviously also has a cube rotation), it's equal to Rw on the 4x4x4.

Since SiGN was claimed to be "simple" and based on "logic", making the exception that 1r = Rw is illogical. Get rid of it, and you will have a good notation. I know it makes things more compact if you keep it, but should that one advantage make it logically inconsistent, and worse, make SiGN cause confusion?

Some day, people would forget r had ever had another meaning. (Slight dramatization, but that's basically how I felt.)
That's never going to happen. r, as I have said before, has no place in SiGN, so omit it (including give a syntax error in alg.cubing.net if someone types r,l,u,d,f,b without a 2 or larger number coefficient, that is, if you eventually make selectable options to use WCA and SiGN separately as is on alg.garron.us), and all of your problems are solved for all time. Until then, I will continue to use WCA whenever I can because there is no mistake about what moves I mean when I post algorithms because almost everyone knows WCA, but I still can't say as many people know SiGN.

In any case, alg.garron.us will start redirecting to alg.cubing.net in a few months (whenever I take a breath from working on the WRC and fix a few alg.cubing.net details first), and all "old WCA" algs will be translated to SiGN. That is, r will be redirected to 2R.
It's good to see that you will be making r become 2R, but I'm sorry to hear that you will be getting rid of alg.garron.us. Oh well, I'm sure for those who prefer alg.garron can always still use CubeTwister on their computers.

EDIT:
I will modify template Alg4 in the wiki to go to alg.cubing.net so that there are not so many redirects because of the 4x4x4 parity algorithm page, but this is of course after you redirect r to 2R. So maybe let me know by pm when this gets done (or you can do it whenever it gets done, if that makes things easier for you).

EDIT2:
And Lucas, what are you going to do about capital M,E,S at alg.cubing.net? It will be a mess unless you make a button like you did in alg.garron.us to use WCA or SiGN.

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cuBerBruce

Member
SiGN notation may be a good idea for a consistent notation across arbitrary size cubes, but for the 4x4x4, it is a bit clumsy, and less consistent with standard mathematical notation. Singmaster notation, in its "pure" form, is actually just a use of mathematical notation (albeit it adds the prime/apostrophe to represent inverses, which alternatively can be expressed as raising to a power of -1). Singmaster extended his notation to 4x4x4 in1982 with the lower case letters representing inner layer turns, and has a long-standing history of being the de facto standard notation. I think SiGN enthusiasts have somewhat of an uphill battle to convince everyone one to use this clumsier notation (as a 4x4x4 notation) in place of the long-standing Singmaster standard.

As far as I know the use of lower case letters for wide turns is a relatively recent addition for 3x3x3. Can anyone show that this convention was in use before 2000?

Using +/- has had some use in cubing, but is also not consistent with standard mathematical notation.

I suppose programs should give the user the choice of notation, because I think a lot of people still prefer the long-standing Singmaster convention as a 4x4x4 notation.

rokicki

Member
I have no beef with anything you say, Bruce. But for our efforts here in this thread it would be great
if we can agree on a notation and use that consistently. If that notation uses "l" for an inner slice turn,
I'm okay with that, and that seems to be what people are doing.

Long term, big picture, I do plan a 4x4x4 contest, and I need to pick a notation for that, and I doubt
I can make everyone happy there. I personally prefer a notation that extends naturally to bigger cubes
and for this reason I prefer the SiGN notation.

It looks to me like the solver created by unsolved uses single-slice-turn metric (SSTM), and your two
solvers use outer-block-turn metric (OBTM) and block-turn metric (BTM). Is this accurate? In that
case I guess there is no direct comparison between the solvers. I am doing my experiments with OBTM
but plan to also support BTM; I find it hard to be interested in SSTM. Is there a motivation for using
SSTM that I'm not seeing?

cuBerBruce

Member
I have no beef with anything you say, Bruce. But for our efforts here in this thread it would be great
if we can agree on a notation and use that consistently. If that notation uses "l" for an inner slice turn,
I'm okay with that, and that seems to be what people are doing.

Long term, big picture, I do plan a 4x4x4 contest, and I need to pick a notation for that, and I doubt
I can make everyone happy there. I personally prefer a notation that extends naturally to bigger cubes
and for this reason I prefer the SiGN notation.

It looks to me like the solver created by unsolved uses single-slice-turn metric (SSTM), and your two
solvers use outer-block-turn metric (OBTM) and block-turn metric (BTM). Is this accurate? In that
case I guess there is no direct comparison between the solvers. I am doing my experiments with OBTM
but plan to also support BTM; I find it hard to be interested in SSTM. Is there a motivation for using
SSTM that I'm not seeing?
If you go to the first post of my thread, you'll find 3 versions, one for each of these 3 metrics.

In SSTM, each of the standard moves affect a minimal number of pieces; and the basic moves correspond directly with the Singmaster letters and their powers. If you have multiple turns in a row on the same axis, it's more straightforward to see what the effect on a single piece or single layer is for those turns. These properties are perhaps of more interest to mathematicians than speedcubers. Like on the 15 puzzle, mathematicians seem to be mostly interested in single-tile metric, while the way normal people manipulate the physical puzzle corresponds more with multi-tile metric.

cmhardw

I think Chris's arguments that the change r /= Rw should be added to the rules for SiGN very convincing. In particular the following is very convincing to me:

Having r = Rw is logically flawed based on the structure of SiGN. r (in SiGN) = 1r = a 1 layer outer block turn. That is, r = 1r = R in SiGN. But in SiGN also wants to make capital letters mean a single slice...so here is a conflict in SiGN, and thus if you omit the "convention" that r = Rw, then SiGN will be consistent.
I would try very hard to use SiGN exclusively if there is consensus to use it as a community standard, but I would prefer that r=Rw be done away with if this happens. Those are my thoughts.

bobthegiraffemonkey

Member
thoughts on r=(1)r = R
Since lowercase letters are multiple layer turns, why would it make sense to think of the default as a single layer turn? 'r' is a shortcut for '2r', the simplest multiple layer turn. I realise it may be confusing since it conflicts with another notation, but having r=1r=R would be even worse IMO since 1r contradicts itself (it can't be both a single layer turn and a double layer turn).

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Lucas Garron

Member
SiGN notation may be a good idea for a consistent notation across arbitrary size cubes, but for the 4x4x4, it is a bit clumsy, and less consistent with standard mathematical notation. Singmaster notation, in its "pure" form, is actually just a use of mathematical notation (albeit it adds the prime/apostrophe to represent inverses, which alternatively can be expressed as raising to a power of -1). Singmaster extended his notation to 4x4x4 in1982 with the lower case letters representing inner layer turns, and has a long-standing history of being the de facto standard notation. I think SiGN enthusiasts have somewhat of an uphill battle to convince everyone one to use this clumsier notation (as a 4x4x4 notation) in place of the long-standing Singmaster standard.

As far as I know the use of lower case letters for wide turns is a relatively recent addition for 3x3x3. Can anyone show that this convention was in use before 2000?

Using +/- has had some use in cubing, but is also not consistent with standard mathematical notation.
I'm somewhat sympathetic to the historical argument, but what you call "relatively recent" is practically all of modern speedcubing.
I also don't think it's so much of an uphill battle. If it weren't for the 2008 Regs, I'm confident we would already be there.

While I care about the math, too, I don't think we should be too beholden to all arbitrary choices.
For example, alg composition is the reverse order of regular group theory.

I suppose programs should give the user the choice of notation, because I think a lot of people still prefer the long-standing Singmaster convention as a 4x4x4 notation.
Programs should not give the user the choice of notation if it is contradictory (but I'm totally fine with allowing non-conflicting aliases like r=Rw, as on alg.cubing.net).

I would rather use Rw and r=2R than have two contradicting definitions for an indefinite period of time. Perhaps that's a discussion for elsewhere.

qqwref

Member
How many times have you had to squint and see if a ' was or was not in the middle of a sequences of moves?
This may be a font issue. For the fonts I use, I see 's very clearly. On your computer, do you also have trouble visually seeing the difference between its and it's?

I use notation such as {f/r} to indicate a middle slice turn from the front in the direction of the right face. In the opposite direction, it is {r/f}.
Ewwwwww. For methods/algs that use a lot of these slices, this would get annoying extremely quickly. (For instance: here's a popular Z perm: {f,u}2 U {f,u}2 U {f,u} U2 {f,u}2 U2 {f,u} U2)

Why should r'l and l'r be opposite in direction when you turn them both the same way?
Because R and L are on opposite sides of the cube With SiGN, sure, you could write it 2R' 2L, but you could also write it 2R' 3R' (or 2-3r', or M) and it would be very clear that they go the same direction.

Having r = Rw is logically flawed based on the structure of SiGN. r (in SiGN) = 1r = a 1 layer outer block turn.
You only think it's flawed because your mental definition is wrong. A lack of number does not mean "1 goes here", it means "the move of this type that is furthest from the center of the puzzle". So r = 2r because 1r wouldn't be lowercase, and M = all but the outermost two layers. And in fact, it's perfectly valid to write 2r - but nobody would, because it wastes time.

With the same flawed logic that r = Rw for the sake that r = (R M') on the 3x3x3, then why isn't x = (Rw l') on the 4x4x4 in SiGN? I of course know the answer, but how obvious you think that answer is, I feel it's just as obvious that you shouldn't say that since r = (R M') on the 3x3x3 (which is really equivalent to doing a single slice turn which obviously also has a cube rotation), it's equal to Rw on the 4x4x4.
Huh? x is a rotation of the whole cube, which will never be equivalent to (r 2L'). The idea that "lowercase letter with no number = two layer turn" is in fact a generalization of the 3x3x3 notation you mentioned, whereas the Singmaster notation of a lowercase letter being a slice is not. Again, the notation only doesn't make sense to you because you have the wrong idea of what things mean.

SiGN notation may be a good idea for a consistent notation across arbitrary size cubes, but for the 4x4x4, it is a bit clumsy, and less consistent with standard mathematical notation.
Actually I find SiGN a lot less clumsy for algs or reconstructions which use very few slice moves. And compared to the WCA 4x4x4/5x5x5 notation, which really cannot be extended nicely to bigger cubes, SiGN makes writing algs on larger cubes very easy. WCA 6x6x6+ notation is also pretty clumsy and really makes no accommodation for people who need slice turns or inner block turns, such as anyone working on pretty pattern algs.

And IMO there is no "standard mathematical notation" for what letters go to what turns - Singmaster chose one convention, but math is not known for forcing people to use particular letters. Math does define what the suffixes should be, but both SiGN and WCA notation use an easy-to-type version of the standard (' = ^-1, 2 = ^2).

As far as I know the use of lower case letters for wide turns is a relatively recent addition for 3x3x3. Can anyone show that this convention was in use before 2000?
Maybe this is my young-person bias showing, but I don't consider any non-theoretical writing about cubes before 2003 to be important in deciding what we do now. And besides, this is a new sport, and we can make new rules - almost everything we do now was decided, by the community, since the founding of the WCA.

Anyway, unrelated to all this stuff: the optimal algs people have been finding for ELL cases (such as u2 b' U' b' l D2 f2 D2 l' b U b u2) are extremely interesting, and generally unknown. Keep it up!

cmhardw

You only think it's flawed because your mental definition is wrong. A lack of number does not mean "1 goes here", it means "the move of this type that is furthest from the center of the puzzle". So r = 2r because 1r wouldn't be lowercase, and M = all but the outermost two layers. And in fact, it's perfectly valid to write 2r - but nobody would, because it wastes time.
That makes sense to me. Lowercase letters denote multiple layer turns, the smallest of which is a turning of two layers. I retract my earlier statement. SiGN notation does appear to be internally consistent.

rokicki

Member
Here are some distance-14 (OBTM) optimal solves, I believe.

(There could be a bug in my program).

These are for the ELL positions posted earlier.

First line is the position, second is the solution.

Sorry for all the three-wide turns; I haven't yet added code
to "untwist" solutions from my optimal solver (which always
holds the DBL corner fixed).

If there is interest I will post some more. Similarly, if anyone
can confirm these that would be great too.

Code:
F2 U' 2L' 2R' U F' L U' F 2L 2R F' U L' F'
Uw2 Fw 3Uw2 3Rw' 3Uw' 3Fw Uw2 3Uw2 3Fw' 3Uw 3Rw 3Uw2 Fw' Uw2
2R2 U F2 U B' R B 2R2 B' R' B U' F2 U' 2R2
R U' Fw2 3Fw U2 Fw2 R Fw2 U2 Fw2 R 3Fw' U R'
F' R' 2D R U R' 2D' 2U' R U' R' 2U R F
F2 Rw2 U R2 U' R2 3Rw2 U R2 U' R2 Rw2 3Rw2 F2
R2 2R2 U2 R2 U' 2R U2 2L 2R U' R2 U 2L' 2R' U2 2R' U' R2 2R2
F2 Rw2 U' F2 U F2 3Fw2 U' F2 U F2 3Fw2 Rw2 F2

Christopher Mowla

If there is interest I will post some more. Similarly, if anyone
can confirm these that would be great too.
They are all correct and amazing!
Uw2 Fw 3Uw2 3Rw' 3Uw' 3Fw Uw2 3Uw2 3Fw' 3Uw 3Rw 3Uw2 Fw' Uw2 =
[Uw2 Fw D2 R' B' L: u2](13 btm, 17 ltm, 14 ftm, 20 qtm)

R U' Fw2 3Fw U2 Fw2 R Fw2 U2 Fw2 R 3Fw' U R' =
R U' B Fw2 L2 Fw2 U Fw2 L2 Fw2 U B' U R' (14 btm, 18 ltm, 14 ftm, 20 qtm)

F2 Rw2 U R2 U' R2 3Rw2 U R2 U' R2 Rw2 3Rw2 F2 =
F2 Rw2 U R2 U' M2 U R2 U' R2 l2 F2 (12 btm, 14 ltm, 14 ftm, 24 qtm)

F2 Rw2 U' F2 U F2 3Fw2 U' F2 U F2 3Fw2 Rw2 F2 =
F2 Rw2 U' F2 U S2 U' F2 U S2 Rw2 F2 (12 btm, 16 ltm, 14 ftm, 24 qtm)

Yes there is a position I would like you to find the optimal OBTM (ftm) solution for.

I'm not sure if 18 OBTM is out of reach for your solver, but 18 OBTM is the briefest I have found (and currently exists) for this case (which I have studied extensively).
(Rw2 Uw U S Lw2 u2) r (u2 Lw2 S' U' Uw' Rw2)
(Rw2 Uw' U' S Lw2 u2) r (u2 Lw2 S' U Uw Rw2)

unsolved

Member
They are all correct and amazing!
Uw2 Fw 3Uw2 3Rw' 3Uw' 3Fw Uw2 3Uw2 3Fw' 3Uw 3Rw 3Uw2 Fw' Uw2 =
[Uw2 Fw D2 R' B' L: u2](13 btm, 17 ltm, 14 ftm, 20 qtm)

R U' Fw2 3Fw U2 Fw2 R Fw2 U2 Fw2 R 3Fw' U R' =
R U' B Fw2 L2 Fw2 U Fw2 L2 Fw2 U B' U R' (14 btm, 18 ltm, 14 ftm, 20 qtm)

F2 Rw2 U R2 U' R2 3Rw2 U R2 U' R2 Rw2 3Rw2 F2 =
F2 Rw2 U R2 U' M2 U R2 U' R2 l2 F2 (12 btm, 14 ltm, 14 ftm, 24 qtm)

F2 Rw2 U' F2 U F2 3Fw2 U' F2 U F2 3Fw2 Rw2 F2 =
F2 Rw2 U' F2 U S2 U' F2 U S2 Rw2 F2 (12 btm, 16 ltm, 14 ftm, 24 qtm)

Yes there is a position I would like you to find the optimal OBTM (ftm) solution for.

I'm not sure if 18 OBTM is out of reach for your solver, but 18 OBTM is the briefest I have found (and currently exists) for this case (which I have studied extensively).
(Rw2 Uw U S Lw2 u2) r (u2 Lw2 S' U' Uw' Rw2)
(Rw2 Uw' U' S Lw2 u2) r (u2 Lw2 S' U Uw Rw2)
I forgot to post your solved scrambles.

Code:
Solutions To Scramble = F2 U' l' r' U F' L U' F l r F' U L' F'

TOP                     FRONT                   RIGHT
---------------------   ---------------------   ---------------------
|####|####|####|####|   |OOOO|OOOO|####|OOOO|   |XXXX|####|XXXX|XXXX|
---------------------   ---------------------   ---------------------
|^^^^|####|####|####|   |OOOO|OOOO|OOOO|OOOO|   |XXXX|XXXX|XXXX|XXXX|
---------------------   ---------------------   ---------------------
|^^^^|####|####|OOOO|   |OOOO|OOOO|OOOO|OOOO|   |XXXX|XXXX|XXXX|XXXX|
---------------------   ---------------------   ---------------------
|####|####|XXXX|####|   |OOOO|OOOO|OOOO|OOOO|   |XXXX|XXXX|XXXX|XXXX|
---------------------   ---------------------   ---------------------

BOTTOM                  BACK                    LEFT
---------------------   ---------------------   ---------------------
|~~~~|~~~~|~~~~|~~~~|   |&&&&|&&&&|&&&&|&&&&|   |^^^^|####|####|^^^^|
---------------------   ---------------------   ---------------------
|~~~~|~~~~|~~~~|~~~~|   |&&&&|&&&&|&&&&|&&&&|   |^^^^|^^^^|^^^^|^^^^|
---------------------   ---------------------   ---------------------
|~~~~|~~~~|~~~~|~~~~|   |&&&&|&&&&|&&&&|&&&&|   |^^^^|^^^^|^^^^|^^^^|
---------------------   ---------------------   ---------------------
|~~~~|~~~~|~~~~|~~~~|   |&&&&|&&&&|&&&&|&&&&|   |^^^^|^^^^|^^^^|^^^^|
---------------------   ---------------------   ---------------------

Solution [001] =  d2 f  D  F2 U' F2 r2 F2 [inside 5-TFS] ---> U  F2 D' f' d2  @ 0128707314012 nodes with 1010604690894 5-TFS probes

Solutions To Scramble = r2 U' B2 U' F L F' r2 F L' F' U B2 U r2

TOP                     FRONT                   RIGHT
---------------------   ---------------------   ---------------------
|####|####|####|####|   |OOOO|####|####|OOOO|   |XXXX|XXXX|&&&&|XXXX|
---------------------   ---------------------   ---------------------
|####|####|####|####|   |OOOO|OOOO|OOOO|OOOO|   |XXXX|XXXX|XXXX|XXXX|
---------------------   ---------------------   ---------------------
|####|####|####|####|   |OOOO|OOOO|OOOO|OOOO|   |XXXX|XXXX|XXXX|XXXX|
---------------------   ---------------------   ---------------------
|####|OOOO|OOOO|####|   |OOOO|OOOO|OOOO|OOOO|   |XXXX|XXXX|XXXX|XXXX|
---------------------   ---------------------   ---------------------

BOTTOM                  BACK                    LEFT
---------------------   ---------------------   ---------------------
|~~~~|~~~~|~~~~|~~~~|   |&&&&|&&&&|XXXX|&&&&|   |^^^^|^^^^|^^^^|^^^^|
---------------------   ---------------------   ---------------------
|~~~~|~~~~|~~~~|~~~~|   |&&&&|&&&&|&&&&|&&&&|   |^^^^|^^^^|^^^^|^^^^|
---------------------   ---------------------   ---------------------
|~~~~|~~~~|~~~~|~~~~|   |&&&&|&&&&|&&&&|&&&&|   |^^^^|^^^^|^^^^|^^^^|
---------------------   ---------------------   ---------------------
|~~~~|~~~~|~~~~|~~~~|   |&&&&|&&&&|&&&&|&&&&|   |^^^^|^^^^|^^^^|^^^^|
---------------------   ---------------------   ---------------------

Solution [001] =  u2 b' U' b' l  D2 f2 D2 [inside 5-TFS] ---> l' b  U  b  u2  @ 0090736443045 nodes with 0750247004859 5-TFS probes
Solution [002] =  u2 b' U' b' l' D2 f2 D2 [inside 5-TFS] ---> l  b  U  b  u2  @ 0090736464321 nodes with 0750247150267 5-TFS probes
Solution [003] =  d' r2 d  B2 D2 F' R2 D2 [inside 5-TFS] ---> L2 B  l2 r' u'  @ 0110414034303 nodes with 0885175449290 5-TFS probes
Solution [004] =  d2 r' U' B2 D' r' f2 r  [inside 5-TFS] ---> D  B2 U  r  d2  @ 0123089215000 nodes with 0972086306647 5-TFS probes
Solution [005] =  d2 b' U' b' l  D2 f2 D2 [inside 5-TFS] ---> l' b  U  b  d2  @ 0130611366693 nodes with 1023666280309 5-TFS probes
Solution [006] =  d2 b' U' b' l' D2 f2 D2 [inside 5-TFS] ---> l  b  U  b  d2  @ 0130611387969 nodes with 1023666424889 5-TFS probes
Solution [007] =  r2 b  U  r  D  l' b2 l  [inside 5-TFS] ---> D' r' U' b' r2  @ 0259062671872 nodes with 1904439161555 5-TFS probes

Solutions To Scramble = l2 U' B2 U' F L F' l2 F L' F' U B2 U l2

TOP                     FRONT                   RIGHT
---------------------   ---------------------   ---------------------
|####|####|####|####|   |OOOO|####|####|OOOO|   |XXXX|&&&&|XXXX|XXXX|
---------------------   ---------------------   ---------------------
|####|####|####|####|   |OOOO|OOOO|OOOO|OOOO|   |XXXX|XXXX|XXXX|XXXX|
---------------------   ---------------------   ---------------------
|####|####|####|####|   |OOOO|OOOO|OOOO|OOOO|   |XXXX|XXXX|XXXX|XXXX|
---------------------   ---------------------   ---------------------
|####|OOOO|OOOO|####|   |OOOO|OOOO|OOOO|OOOO|   |XXXX|XXXX|XXXX|XXXX|
---------------------   ---------------------   ---------------------

BOTTOM                  BACK                    LEFT
---------------------   ---------------------   ---------------------
|~~~~|~~~~|~~~~|~~~~|   |&&&&|XXXX|&&&&|&&&&|   |^^^^|^^^^|^^^^|^^^^|
---------------------   ---------------------   ---------------------
|~~~~|~~~~|~~~~|~~~~|   |&&&&|&&&&|&&&&|&&&&|   |^^^^|^^^^|^^^^|^^^^|
---------------------   ---------------------   ---------------------
|~~~~|~~~~|~~~~|~~~~|   |&&&&|&&&&|&&&&|&&&&|   |^^^^|^^^^|^^^^|^^^^|
---------------------   ---------------------   ---------------------
|~~~~|~~~~|~~~~|~~~~|   |&&&&|&&&&|&&&&|&&&&|   |^^^^|^^^^|^^^^|^^^^|
---------------------   ---------------------   ---------------------

Solution [001] =  u2 f  U' f  r  D2 b2 D2 [inside 5-TFS] ---> r' f' U  f' u2  @ 0088691387205 nodes with 0736229771312 5-TFS probes
Solution [002] =  u2 f  U' f  r' D2 b2 D2 [inside 5-TFS] ---> r  f' U  f' u2  @ 0088691410749 nodes with 0736229934429 5-TFS probes
Solution [003] =  d2 l  U' B2 D' l  b2 l' [inside 5-TFS] ---> D  B2 U  l' d2  @ 0124135640777 nodes with 0979264921448 5-TFS probes
Solution [004] =  d2 f  U' f  r  D2 b2 D2 [inside 5-TFS] ---> r' f' U  f' d2  @ 0128566310853 nodes with 1009645389864 5-TFS probes
Solution [005] =  d2 f  U' f  r' D2 b2 D2 [inside 5-TFS] ---> r  f' U  f' d2  @ 0128566334397 nodes with 1009645555635 5-TFS probes

Solutions To Scramble = r2 U l r U' F R' U F' l' r' F U' R F' r2

TOP                     FRONT                   RIGHT
---------------------   ---------------------   ---------------------
|####|####|^^^^|####|   |OOOO|OOOO|OOOO|OOOO|   |XXXX|####|####|XXXX|
---------------------   ---------------------   ---------------------
|####|####|####|XXXX|   |OOOO|OOOO|OOOO|OOOO|   |XXXX|XXXX|XXXX|XXXX|
---------------------   ---------------------   ---------------------
|&&&&|####|####|XXXX|   |OOOO|OOOO|OOOO|OOOO|   |XXXX|XXXX|XXXX|XXXX|
---------------------   ---------------------   ---------------------
|####|####|####|####|   |OOOO|OOOO|OOOO|OOOO|   |XXXX|XXXX|XXXX|XXXX|
---------------------   ---------------------   ---------------------

BOTTOM                  BACK                    LEFT
---------------------   ---------------------   ---------------------
|~~~~|~~~~|~~~~|~~~~|   |&&&&|####|&&&&|&&&&|   |^^^^|^^^^|####|^^^^|
---------------------   ---------------------   ---------------------
|~~~~|~~~~|~~~~|~~~~|   |&&&&|&&&&|&&&&|&&&&|   |^^^^|^^^^|^^^^|^^^^|
---------------------   ---------------------   ---------------------
|~~~~|~~~~|~~~~|~~~~|   |&&&&|&&&&|&&&&|&&&&|   |^^^^|^^^^|^^^^|^^^^|
---------------------   ---------------------   ---------------------
|~~~~|~~~~|~~~~|~~~~|   |&&&&|&&&&|&&&&|&&&&|   |^^^^|^^^^|^^^^|^^^^|
---------------------   ---------------------   ---------------------

Solution [001] =  u2 f' D  B2 U' B2 r2 B2 [inside 5-TFS] ---> U  B2 D' f  u2  @ 0089361202383 nodes with 0740808355065 5-TFS probes
Solution [002] =  d2 f' D  B2 U' B2 r2 B2 [inside 5-TFS] ---> U  B2 D' f  d2  @ 0129236126031 nodes with 1014223718305 5-TFS probes
Solution [003] =  l2 f' U  R2 U' L2 b2 L2 [inside 5-TFS] ---> U  R2 U' f  l2  @ 0297890262609 nodes with 2170654243169 5-TFS probes
Solution [004] =  l2 f' U  R2 D' B2 r2 B2 [inside 5-TFS] ---> D  R2 U' f  l2  @ 0297890471079 nodes with 2170655663948 5-TFS probes
Solution [005] =  l2 f' U' b2 D' B2 r2 B2 [inside 5-TFS] ---> D  b2 U  f  l2  @ 0297922559391 nodes with 2170875725249 5-TFS probes
Solution [006] =  l2 f' L2 U  R2 U' b2 U  [inside 5-TFS] ---> R2 U' L2 f  l2  @ 0298295606800 nodes with 2173433925721 5-TFS probes
Solution [007] =  l2 f' L2 D  B2 U' l2 U  [inside 5-TFS] ---> B2 D' L2 f  l2  @ 0298301859982 nodes with 2173476807387 5-TFS probes

Solutions To Scramble = R2 r' F' L R F r' F' R' F r2 F' L' F R2

TOP                     FRONT                   RIGHT
---------------------   ---------------------   ---------------------
|####|####|OOOO|####|   |OOOO|OOOO|####|OOOO|   |XXXX|XXXX|####|XXXX|
---------------------   ---------------------   ---------------------
|XXXX|####|####|^^^^|   |OOOO|OOOO|OOOO|OOOO|   |XXXX|XXXX|XXXX|XXXX|
---------------------   ---------------------   ---------------------
|####|####|####|####|   |OOOO|OOOO|OOOO|OOOO|   |XXXX|XXXX|XXXX|XXXX|
---------------------   ---------------------   ---------------------
|####|####|&&&&|####|   |OOOO|OOOO|OOOO|OOOO|   |XXXX|XXXX|XXXX|XXXX|
---------------------   ---------------------   ---------------------

BOTTOM                  BACK                    LEFT
---------------------   ---------------------   ---------------------
|~~~~|~~~~|~~~~|~~~~|   |&&&&|####|&&&&|&&&&|   |^^^^|####|^^^^|^^^^|
---------------------   ---------------------   ---------------------
|~~~~|~~~~|~~~~|~~~~|   |&&&&|&&&&|&&&&|&&&&|   |^^^^|^^^^|^^^^|^^^^|
---------------------   ---------------------   ---------------------
|~~~~|~~~~|~~~~|~~~~|   |&&&&|&&&&|&&&&|&&&&|   |^^^^|^^^^|^^^^|^^^^|
---------------------   ---------------------   ---------------------
|~~~~|~~~~|~~~~|~~~~|   |&&&&|&&&&|&&&&|&&&&|   |^^^^|^^^^|^^^^|^^^^|
---------------------   ---------------------   ---------------------

Solution [001] =  r  F2 r  U  b' L  B' L' [inside 5-TFS] ---> l  B' U2 f  l2  @ 0227762813531 nodes with 1689809785046 5-TFS probes
Solution [002] =  r  B' R  L' B  r  B' L  [inside 5-TFS] ---> B  r2 B' R' B   @ 0231706464436 nodes with 1716850086155 5-TFS probes
Solution [003] =  r' u2 r  u2 b' l  U  F2 [inside 5-TFS] ---> F  r  B  L2 R2  @ 0235278692736 nodes with 1741344225356 5-TFS probes
Solution [004] =  r' F' R' L  F  r' F' R  [inside 5-TFS] ---> F  r2 F' L' F   @ 0241864066531 nodes with 1786499735871 5-TFS probes
Solution [005] =  L  F' L' b2 L  B  L' b' [inside 5-TFS] ---> L  B' F  L' b'  @ 0308623479380 nodes with 2244260809189 5-TFS probes

Solutions To Scramble = r2 U F2 U B' R B r2 B' R' B U' F2 U' r2

TOP                     FRONT                   RIGHT
---------------------   ---------------------   ---------------------
|####|####|####|####|   |OOOO|OOOO|&&&&|OOOO|   |XXXX|####|XXXX|XXXX|
---------------------   ---------------------   ---------------------
|####|####|####|####|   |OOOO|OOOO|OOOO|OOOO|   |XXXX|XXXX|XXXX|XXXX|
---------------------   ---------------------   ---------------------
|XXXX|####|####|^^^^|   |OOOO|OOOO|OOOO|OOOO|   |XXXX|XXXX|XXXX|XXXX|
---------------------   ---------------------   ---------------------
|####|####|####|####|   |OOOO|OOOO|OOOO|OOOO|   |XXXX|XXXX|XXXX|XXXX|
---------------------   ---------------------   ---------------------

BOTTOM                  BACK                    LEFT
---------------------   ---------------------   ---------------------
|~~~~|~~~~|~~~~|~~~~|   |&&&&|&&&&|OOOO|&&&&|   |^^^^|^^^^|####|^^^^|
---------------------   ---------------------   ---------------------
|~~~~|~~~~|~~~~|~~~~|   |&&&&|&&&&|&&&&|&&&&|   |^^^^|^^^^|^^^^|^^^^|
---------------------   ---------------------   ---------------------
|~~~~|~~~~|~~~~|~~~~|   |&&&&|&&&&|&&&&|&&&&|   |^^^^|^^^^|^^^^|^^^^|
---------------------   ---------------------   ---------------------
|~~~~|~~~~|~~~~|~~~~|   |&&&&|&&&&|&&&&|&&&&|   |^^^^|^^^^|^^^^|^^^^|
---------------------   ---------------------   ---------------------

Solution [001] =  R2 f' D  F2 f' u2 l2 u2 [inside 5-TFS] ---> f  F2 D' f  R2  @ 0214056276801 nodes with 1595827407974 5-TFS probes
Solution [002] =  R2 f' D  F2 f' d2 l2 d2 [inside 5-TFS] ---> f  F2 D' f  R2  @ 0214056279072 nodes with 1595827423505 5-TFS probes
Solution [003] =  R2 f' D  b' B2 u2 r2 u2 [inside 5-TFS] ---> B2 b  D' f  R2  @ 0214059607653 nodes with 1595850227820 5-TFS probes
Solution [004] =  R2 f' D  b' B2 d2 r2 d2 [inside 5-TFS] ---> B2 b  D' f  R2  @ 0214059609924 nodes with 1595850243100 5-TFS probes
Solution [005] =  R2 f' D  B2 d2 b' r2 b  [inside 5-TFS] ---> d2 B2 D' f  R2  @ 0214061510578 nodes with 1595863321622 5-TFS probes
Solution [006] =  R2 f' D' F2 u2 b' l2 b  [inside 5-TFS] ---> u2 F2 D  f  R2  @ 0214071048247 nodes with 1595928696941 5-TFS probes
Solution [007] =  R2 f' D' F2 b' u2 l2 u2 [inside 5-TFS] ---> b  F2 D  f  R2  @ 0214071539037 nodes with 1595932069181 5-TFS probes
Solution [008] =  R2 f' D' F2 b' d2 l2 d2 [inside 5-TFS] ---> b  F2 D  f  R2  @ 0214071541308 nodes with 1595932084901 5-TFS probes
Solution [009] =  R2 f' D' f' B2 u2 r2 u2 [inside 5-TFS] ---> B2 f  D  f  R2  @ 0214072943817 nodes with 1595941686238 5-TFS probes
Solution [010] =  R2 f' D' f' B2 d2 r2 d2 [inside 5-TFS] ---> B2 f  D  f  R2  @ 0214072946088 nodes with 1595941701785 5-TFS probes
Solution [011] =  R2 f' D' B2 u2 f' r2 f  [inside 5-TFS] ---> u2 B2 D  f  R2  @ 0214076638621 nodes with 1595967056929 5-TFS probes
Solution [012] =  F  L  f2 D' B  r2 d2 r2 [inside 5-TFS] ---> B' D  f2 L' F'  @ 0348708026223 nodes with 2519103241100 5-TFS probes
Solution [013] =  F  L  f2 D' B  l2 d2 l2 [inside 5-TFS] ---> B' D  f2 L' F'  @ 0348708028575 nodes with 2519103256699 5-TFS probes

Solutions To Scramble = L U F L' U' L U L F' L2 U L U L' U' L U' L'

TOP                     FRONT                   RIGHT
---------------------   ---------------------   ---------------------
|####|####|####|####|   |^^^^|&&&&|&&&&|OOOO|   |XXXX|XXXX|XXXX|XXXX|
---------------------   ---------------------   ---------------------
|####|####|####|####|   |OOOO|OOOO|OOOO|OOOO|   |XXXX|XXXX|XXXX|XXXX|
---------------------   ---------------------   ---------------------
|####|####|####|####|   |OOOO|OOOO|OOOO|OOOO|   |XXXX|XXXX|XXXX|XXXX|
---------------------   ---------------------   ---------------------
|####|####|####|####|   |OOOO|OOOO|OOOO|OOOO|   |XXXX|XXXX|XXXX|XXXX|
---------------------   ---------------------   ---------------------

BOTTOM                  BACK                    LEFT
---------------------   ---------------------   ---------------------
|~~~~|~~~~|~~~~|~~~~|   |&&&&|OOOO|OOOO|^^^^|   |OOOO|^^^^|^^^^|&&&&|
---------------------   ---------------------   ---------------------
|~~~~|~~~~|~~~~|~~~~|   |&&&&|&&&&|&&&&|&&&&|   |^^^^|^^^^|^^^^|^^^^|
---------------------   ---------------------   ---------------------
|~~~~|~~~~|~~~~|~~~~|   |&&&&|&&&&|&&&&|&&&&|   |^^^^|^^^^|^^^^|^^^^|
---------------------   ---------------------   ---------------------
|~~~~|~~~~|~~~~|~~~~|   |&&&&|&&&&|&&&&|&&&&|   |^^^^|^^^^|^^^^|^^^^|
---------------------   ---------------------   ---------------------

Solution [001] =  U  B2 L2 F  B2 U' F  R2 B' [inside 5-TFS] ---> D  F' R2 B  F'  @ 0001119126581 nodes with 0017853193809 5-TFS probes
Solution [002] =  U' R  L' B2 R  D' L  B2 R' [inside 5-TFS] ---> U  L2 R' F2 L2  @ 0001371388154 nodes with 0019582866730 5-TFS probes
Solution [003] =  U' F2 L2 F2 B' U  B' R2 F  [inside 5-TFS] ---> D' B  R2 B  F'  @ 0002020120411 nodes with 0024031075167 5-TFS probes
Solution [004] =  U2 F  R' B  U2 F' L  F' B2 [inside 5-TFS] ---> D2 B2 L' B' F   @ 0002951816961 nodes with 0030420382639 5-TFS probes
Solution [005] =  U2 F  B' R  F2 D2 F2 B  R' [inside 5-TFS] ---> B  U2 F' L  B'  @ 0002982652928 nodes with 0030631752492 5-TFS probes
Solution [006] =  U2 F  B' R' B2 D2 F' B2 R  [inside 5-TFS] ---> F' U2 B  L' F   @ 0002983217161 nodes with 0030635615750 5-TFS probes
Solution [007] =  U2 B' R  F' U2 B  L' F2 B  [inside 5-TFS] ---> D2 F2 L  B' F   @ 0003265166068 nodes with 0032568908988 5-TFS probes
Solution [008] =  R2 F  R  B' R' F' R' D2 L' [inside 5-TFS] ---> F' L  D2 R' B   @ 0009056298908 nodes with 0072277389472 5-TFS probes
Solution [009] =  R2 F  R' D2 L  B' L' D2 R' [inside 5-TFS] ---> B' R' F' R  B   @ 0009059132519 nodes with 0072296853649 5-TFS probes
Solution [010] =  R2 B' R  D2 L' F  L  D2 R  [inside 5-TFS] ---> F  R  B  R' F'  @ 0009373437574 nodes with 0074452290522 5-TFS probes
Solution [011] =  R2 B' R' F  R  B  R  D2 L  [inside 5-TFS] ---> B  L' D2 R  F'  @ 0009381341449 nodes with 0074506439427 5-TFS probes
Solution [012] =  L2 U2 D  R' L  F2 R  D' L  [inside 5-TFS] ---> B2 R' U  R  B2  @ 0011473651735 nodes with 0088853541481 5-TFS probes
Solution [013] =  L2 U2 D' R  L' B2 R' D  L' [inside 5-TFS] ---> F2 R  U' R' F2  @ 0011478354296 nodes with 0088885796272 5-TFS probes
Solution [014] =  L2 D  R  U2 R' L  F2 D' L  [inside 5-TFS] ---> B2 R' U  R  B2  @ 0011558045923 nodes with 0089432096854 5-TFS probes
Solution [015] =  L2 D' R' U2 R  L' B2 D  L' [inside 5-TFS] ---> F2 R  U' R' F2  @ 0011632890326 nodes with 0089945242796 5-TFS probes
Solution [016] =  L2 F2 R  L2 U' R  B2 L' D  [inside 5-TFS] ---> R' B2 L  R' U   @ 0011968390258 nodes with 0092245688140 5-TFS probes
Solution [017] =  L2 B2 R' L2 U  R' F2 L  D' [inside 5-TFS] ---> R  F2 L' R  U'  @ 0012195816719 nodes with 0093805095305 5-TFS probes
Solution [018] =  F  R  B' R' F' R' D2 L' F' [inside 5-TFS] ---> L  D2 R' B  R2  @ 0012755309015 nodes with 0097641491153 5-TFS probes
Solution [019] =  F  R' D2 L  B' L' D2 R' B' [inside 5-TFS] ---> R' F' R  B  R2  @ 0012793133387 nodes with 0097900948987 5-TFS probes
Solution [020] =  F  R' L2 D2 L  B' L' D2 R2 [inside 5-TFS] ---> F' R' F  L2 F'  @ 0012809096856 nodes with 0098010381424 5-TFS probes
Solution [021] =  F  L2 F' R  F  R2 D2 L  B  [inside 5-TFS] ---> L' D2 L2 R  F'  @ 0013110831019 nodes with 0100079073946 5-TFS probes
Solution [022] =  F  B' R2 F  D' B  R2 F' U  [inside 5-TFS] ---> B2 F' L2 B2 U'  @ 0013255306225 nodes with 0101069575163 5-TFS probes
Solution [023] =  F  B' R2 B' D  F' R2 B  U' [inside 5-TFS] ---> B  F2 L2 F2 U   @ 0013257071300 nodes with 0101081687503 5-TFS probes
Solution [024] =  F' U  D2 R2 D' F  D  R2 U2 [inside 5-TFS] ---> B  U  B' D2 F   @ 0013353083421 nodes with 0101740007997 5-TFS probes
Solution [025] =  F' D2 B  U' B' U2 R2 D' F' [inside 5-TFS] ---> D  R2 D2 U' F   @ 0013784940047 nodes with 0104701441777 5-TFS probes
Solution [026] =  F' L  B' U2 F  R' F  B2 D2 [inside 5-TFS] ---> B2 R  B  F' U2  @ 0014111006466 nodes with 0106937420192 5-TFS probes
Solution [027] =  F' B  L  B2 D2 F  B2 L' F  [inside 5-TFS] ---> U2 B' R  F' U2  @ 0014315314555 nodes with 0108338340723 5-TFS probes
Solution [028] =  F' B  L' F2 D2 F2 B' L  B' [inside 5-TFS] ---> U2 F  R' B  U2  @ 0014319216425 nodes with 0108365054083 5-TFS probes
Solution [029] =  F2 R  U  R' F2 L  D' R  B2 [inside 5-TFS] ---> L  R' D  U2 L2  @ 0014922710151 nodes with 0112502821614 5-TFS probes
Solution [030] =  F2 R  U  R' F2 L  D' B2 R' [inside 5-TFS] ---> L  U2 R  D  L2  @ 0014922710294 nodes with 0112502822550 5-TFS probes
Solution [031] =  F2 R  U  R' F2 L  D' B2 L  [inside 5-TFS] ---> R' U2 R  D  L2  @ 0014922710296 nodes with 0112502822551 5-TFS probes
Solution [032] =  B  U' D2 R2 D  B' D' R2 U2 [inside 5-TFS] ---> F' U' F  D2 B'  @ 0015679560453 nodes with 0117692608142 5-TFS probes
Solution [033] =  B  D2 F' U  F  U2 R2 D  B  [inside 5-TFS] ---> D' R2 D2 U  B'  @ 0016014245689 nodes with 0119987577690 5-TFS probes
Solution [034] =  B  L' F  U2 B' R  F2 B' D2 [inside 5-TFS] ---> F2 R' B  F' U2  @ 0016398204444 nodes with 0122620173492 5-TFS probes
Solution [035] =  B' R  D2 L' F  L  D2 R  F  [inside 5-TFS] ---> R  B  R' F' R2  @ 0016988624704 nodes with 0126668356753 5-TFS probes
Solution [036] =  B' R  L2 D2 L' F  L  D2 R2 [inside 5-TFS] ---> B  R  B' L2 B   @ 0017004358800 nodes with 0126776299331 5-TFS probes
Solution [037] =  B' R' F  R  B  R  D2 L  B  [inside 5-TFS] ---> L' D2 R  F' R2  @ 0017094129355 nodes with 0127392092332 5-TFS probes
Solution [038] =  B' L2 B  R' B' R2 D2 L' F' [inside 5-TFS] ---> L  D2 L2 R' B   @ 0017403264353 nodes with 0129512031676 5-TFS probes
Solution [039] =  B2 R' U' R  B2 L' D  R' F2 [inside 5-TFS] ---> L' R  D' U2 L2  @ 0017964864801 nodes with 0133362816543 5-TFS probes
Solution [040] =  B2 R' U' R  B2 L' D  F2 R  [inside 5-TFS] ---> L' U2 R' D' L2  @ 0017964864892 nodes with 0133362817287 5-TFS probes
Solution [041] =  B2 R' U' R  B2 L' D  F2 L' [inside 5-TFS] ---> R  U2 R' D' L2  @ 0017964864896 nodes with 0133362817359 5-TFS probes
That makes sense to me. Lowercase letters denote multiple layer turns, the smallest of which is a turning of two layers. I retract my earlier statement. SiGN notation does appear to be internally consistent.

A Move Notation Proposal: Lodino = Logigcal Direction Notation

Background: In reviewing the various notations used by cubers,
and offering a fresh perspective as an outsider (I am no speed
cuber and can't even solve a 5x5x5 without constantly referencing
cmowla's numerous PDFs on the subject), I decided to start with a
tabula rasa.

A summary first for those who can't read this in its entirety
without falling asleep.

Goals for a hypothetical new system:

1. A notation system that reads in such a way that the cubies
to which a formula applies is obvious, with an intuitive
direction indicator.

2. The system should be telescoping and inclusive of all
cubes from 2x2x2 through NxNxN (with N currently at 11 max)
with no need to add terms as the size of the cube grows.

3. No dependence on the knowledge of mathematical notation
or symbology to ease the burden on the newly initiated.

4. System should resolve the present ambiguity concerning the
ambiguous rotation notation governing the middle slices for
all NxNxN cubes with odd N >= 5.

5. The system should be neither cumbersome for the human
to learn, nor difficult for the programmer to implement.

With these goals in mind, the system below accomplishes all
of the above.

========================================================

Logical Direction Notation (Lodino) has the following
good features.

1. Easy to understand direction notation. It eliminates
the arbitrary prime mark (') assignment altogether,
and the unitiated can understand which way to turn the
cube at once. Older systems may even have a right-handed
bias in that clockwise turns (no prime notation) are more
easily performed with a right hand on the R and U faces
than a left hand on the L and D faces.

2. Rotation notation is also "global" and not "relative."
There is a rotation indicator for every direction, not
merely the absence of one for the arbitrary unprimed move.
In this way, lodino supports positive reinforcement with
a standard number of descriptors per move of the same
cube size.

3. Moves are of the form: cubie_subset.direction

4. Lodino corrects the defect where larger cubes have
some move designations with no bearing or resemblance
to smaller cubes. All moves on larger cubes look like
those already seen before on smaller cubes.

5. Lodino corrects larger cube notations that complicate
situations where:

A. Multiple slices with or without the center slice
are involved in the move.
B. Slice turns + face turns are required for a move.
C. Moves involve only the center slice, which, by definition,
cannot have a rotation indicator that is unambiguous.

6. Lodino smooths out the 4x4x4 move description defect
where 2 slices in motion being turned at once have
opposing direction indicators. For example, ud' moves
the up and down slices in the same direction, yet the
move notation is inconsistent and not natural.

========================================================

Sample of Moves

========================================================

2X2X2 and larger

U.L = turn the top towards the left = U presently
U.R = turn the top towards the right = U' presently
U.2 = turn the top twice

F.R = F
F.L = F'
F.2 = F2

R.B = R
R.F = R'
R.2 = R2

D.R = D
D.L = D'
D.2 = D2

B.L = B
B.R = B'
B.2 = B2

L.F = L
L.B = L'
L.2 = L2

========================================================

4x4x4 single slice turns

Only time to use lower case for the direction is when
all moves are slices.

u.l = turn the top slice towards the left = u presently
u.r = turn the top slice towards the right = u' presently
u.2 = turn the top slice twice

f.r = f
f.l = f'
f.2 = f2

r.b = r
r.f = r'
r.2 = r2

d.r = d
d.l = d'
d.2 = d2

b.l = b
b.r = b'
b.2 = b2

l.f = l
l.b = l'
l.2 = l2

4x4x4 double slice turns

ud.l = ud'
ud.r = u'd

fb.r = fb'
fb.l = f'b

rl.b = rl'
rl.f = r'l

4x4x4 face and slice turns

Uu.L = U u
Uu.R = U' u'
Uu.2 = U2 u2

Ff.R = F f
Ff.L = F' f'

Rr.B = R r
Rr.F = R' r'

Dd.R = D d
Dd.L = D' d'

Bb.L = B b
Bb.R = B' b'

Ll.F = L l
Ll.B = L' l'

========================================================

5x5x5 middle slice turns

Enclose the moves in round parenthesis to demarcate a middle edge.

(uf) = turn the middle slice towards the front face from the top face.
When stand alone, it is both a cube selector and a direction selector.

(fu) = the opposite direction of (uf). Move from the front towards
the top

(uf).2 = move the slice twice.
(fu).2 = same as (uf).2

l.(fu) = adjoin the l slice to the middle slice and turn it from f to u.

lr.(uf) = turn 3 slices at once; l, the middle, and r, all from u to f.
This is essentially like moving R and L' on separate moves.

L.(uf).2 = turn the L face and the middle edge at the same time, from u to f.

d.(fr).2 = Turn the d and middle slices twice, in the direction of f to r.

u.(fr) = adjoin the middle slice to the top slice and it them from f to r.

========================================================

NxNxN very large cubes with multiple slice turns for N > 5

"Build" the moves from the perspective of looking either from UFR with the
middle edge rotating vertically or from DBL, likewise. Pad moves to the
left or right of the middle edge in the order in which they appear.
Enclose the moves in round parenthesis to demarcate a middle edge.

Add a padded 0 for any ommited slice not in a series of consecutive slices.
Repeat the same slice letter as many times as needed if it is in relative
motion. Always use all N terms when describing a move sequence.

There should never be more than N/2 terms in motion for a move sequence.

11x11x11 examples:

F00ff(u.r)00000
0000f(u.r)bb000

000ll(u.f)0r00R

Last edited:

Carrot

Member
========================================================

NxNxN very large cubes with multiple slice turns for N > 5

"Build" the moves from the perspective of looking either from UFR with the
middle edge rotating vertically or from DBL, likewise. Pad moves to the
left or right of the middle edge in the order in which they appear.
Enclose the moves in round parenthesis to demarcate a middle edge.

Add a padded 0 for any ommited slice not in a series of consecutive slices.
Repeat the same slice letter as many times as needed if it is in relative
motion. Always use all N terms when describing a move sequence.

There should never be more than N/2 terms in motion for a move sequence.

11x11x11 examples:

F00ff(u.r)00000
0000f(u.r)bb000

000ll(u.f)0r00R
I prefer this:
5F 3F' F
5F 4F' 5B' 3B

or this:
5f 4f F
5f 5b' 4b'

assuming f is understood as short for 2f. (reason why I would write F and not f)

one could however however argue for this notation (working for nxnxn for n<22):
145f = F 4f 5f = 5F 3F' F
124f = F f 4f = 4F 3F' 2F

and this (but this would give problems with consistency as 4F=4f=/=4F, so bad idea, let's keep 4F=/=4f):
145F = F 4f 5f = 5F 3F' F
124F = F f 4f = 4F 3F' 2F

so in your cases you could write them as:
F00ff(u.r)00000 = 145f
0000f(u.r)bb000 = 5f 45b'

and since this shouldn't ruin the consistency too much of what things mean, you can write things like(completely made up conjugated commutator, I doubt it's for ANY use except it shouldn't scramble too much):
[f: [24u 5b, 2L' U 2L]] = f 24u 5b 2L' u 2L 5b' 24u' 2L' U' 2L f'

Last edited:

unsolved

Member
I prefer this:
5F 3F' F
5F 4F' 5B' 3B

or this:
5f 4f F
5f 5b' 4b'

assuming f is understood as short for 2f. (reason why I would write F and not f)

one could however however argue for this notation (working for nxnxn for n<22):
145f = F 4f 5f = 5F 3F' F
124f = F f 4f = 4F 3F' 2F

and this (but this would give problems with consistency as 4F=4f=/=4F, so bad idea, let's keep 4F=/=4f):
145F = F 4f 5f = 5F 3F' F
124F = F f 4f = 4F 3F' 2F

so in your cases you could write them as:
F00ff(u.r)00000 = 145f
0000f(u.r)bb000 = 5f 45b'

and since this shouldn't ruin the consistency too much of what things mean, you can write things like(completely made up conjugated commutator, I doubt it's for ANY use except it shouldn't scramble too much):
[f: [24u 5b, 2L' U 2L]] = f 24u 5b 2L' u 2L 5b' 24u' 2L' U' 2L f'
I don't have any experience with these larger cubes, so I defer to you. The idea was to toss something out there that made sense. I gave it a fair amount of thought.

In other news, the first implementation of the massively parallel version of OO_4x4x4 was not a success, but for the opposite reason you might suspect. It is too parallel!

I've never seen my 32-threaded computer (16 physical cores) get pegged at 100% before, but then again, the software thinks there are 1.3 million cores (2.6 million threads).

Last edited:

Herbert Kociemba

Member
A Move Notation Proposal: Lodino = Logigcal Direction Notation
I must admit that I do not like this notation at all. Why we can't just accept these simple rules:

1. A letter *without* a number as prefix means a *single* slice move, upper cases like U , F etc. mean the outer slice, lower cases like u , f etc mean the adjacent slice.

2. With a number prefix, with upper case letters we denote a single slice move like 3U, 3F etc. With lower case letters we denote outer block turns, like 3u, 3f etc.

Last edited:

Carrot

Member
I must admit that I do not like this notation at all. Why we can't just accept these simple rules:

1. A letter *without* a number as prefix means a *single* slice move, upper cases like U , F etc. mean the outer slice, lower cases like u , f etc mean the adjacent slice.

2. With a number prefix, with upper case letters we denote a single slice move like 3U, 3F etc. With lower case letters we denote outer block turns, like 3u, 3f etc.
Why not:
• lowercase letter: single slice turns
• uppercase letter: outer block turns
• prefixes: defines "how deep" the turn should be (3f = 3rd layer single slice turn of the f-face in a clockwise direction etc.)
• postfixes: defines the direction of the turn
• special case: since f=F we denote all f's as F. We can then denote all 2f's as f for more compactness.

Your notation gives unconsistency in when uppercase and lower cases means what, it shouldn't. Why should U=OBT and U=SS, u=SS => 3U=SS, 3u=OBT? that makes no sense.

rokicki

Member
Here's another six optimal OTBM solves; some are 14, some are 15. Still all
from the ELL list.

Code:
2R2 U' B2 U' F L F' 2R2 F L' F' U B2 U 2R2
Rw2 U 3Rw 3Uw' Fw Rw2 Uw2 3Uw2 Rw2 Fw' 3Uw 3Rw' U' Rw2
R2 2R' F' L R F 2R' F' R' F 2R2 F' L' F R2
U F 3Fw' Rw 3Fw U' 3Fw' Rw2 3Rw U' 3Rw' F Rw F' 3Fw
2L2 2R2 U 2R2 F2 2R2 F2 U' 2L2 2U2 2R2 2U2
F2 3Fw2 U Uw2 3Fw2 R2 Fw2 U2 Uw2 3Fw2 R2 Fw2 U F2 3Fw2
2L2 2R2 U' 2R2 F2 2R2 F2 U 2L2 2U2 2R2 2U2
F2 3Fw2 U' Uw2 F2 R2 Fw2 U2 Uw2 F2 R2 Fw2 U' F2 3Fw2
R2 D' F2 2F' U 2F U2 2F' U F2 U' 2F U2 2F' U' 2F D R2
Fw2 R2 F' Fw Rw2 F 3Rw U' Rw2 U Uw' R2 Uw 3Rw' Fw
2L2 2R2 U' 2L' 2R U2 F2 2L 2R F2 D2 F2 2L2 F2 2L2 D2 2R2 U' 2L2 2R2
Fw2 U2 Rw2 3Uw R2 3Fw2 Rw2 3Uw2 R2 Rw2 3Uw' Fw2 U2 3Fw2 Rw2
I may not be able to solve all of them with my current program.

Working on the double-edge flip; I've found no solution through
length 14 in the outer block metric and am now looking on a length
15 solution.

I'm not sure if 18 OBTM is out of reach for your solver, but 18 OBTM is the briefest I have found (and currently exists) for this case (which I have studied extensively).
(Rw2 Uw U S Lw2 u2) r (u2 Lw2 S' U' Uw' Rw2)
(Rw2 Uw' U' S Lw2 u2) r (u2 Lw2 S' U Uw Rw2)
That one was easy; 13 outer block turns:

Code:
Rw2 Uw U S Lw2 2U2 2R 2U2 Lw2 S' U' Uw' Rw2
Rw2 U2 R2 Rw' U2 3Uw2 R' Rw U2 3Uw2 Rw' U2 Rw2
And another ELL just popped out.

Code:
2R2 U' 2R2 F2 2R2 F2 U2 F2 2R2 F2 2R2 U' 2R2
R2 Fw2 Rw2 U' R2 F2 R2 U2 R2 F2 R2 U' Rw2 Fw2 R2

Last edited by a moderator:

Christopher Mowla

That one was easy; 13 outer block turns:

Code:
Rw2 Uw U S Lw2 2U2 2R 2U2 Lw2 S' U' Uw' Rw2
Rw2 U2 R2 Rw' U2 3Uw2 R' Rw U2 3Uw2 Rw' U2 Rw2
You just lowered the BHTM for this case by 1 move, the BQTM by 1 move, and the OBTM by 5 moves! It also works on the nxnxn! How did I miss this?!
Rw2 U2 R r' E2 r E2 Rw' U2 Rw2

Now, I'm wondering...

Single Dedge Flip (Current minimum, 19 OBTM)
Rw2 B r B U2 B' r' B U2 Lw' B2 r' B2 Lw B2 Rw2
Rw2 B' r B' U2 B r' B' U2 Lw' B2 r' B2 Lw B2 Rw2

Double Parity (Current minimum, 21 OBTM)
Rw2 U B2 U' r' U B2 U' Rw r U B2 U r' U' B2 U' Rw

Adjacent Double Parity (Current minimum, 22 OBTM)
r2 F Rw' F' r2 U2 r' U2 r' U2 r' U2 F Rw F' r2

Adjacent checkerboard case (Current minimum, 22 OBTM)
Rw' F2 M2 F U2 l U2 l2 U2 l U2 l F' M2 F2 Rw

Adjacent hourglass case (Current minimum, 22 OBTM)
(Lw' U2 M2 U' x) r' U2 r' U2 r2 U2 r' U2 (x' U M2 U2 Lw)

Lastly, algs for all 6 unique "oriented" 3-cycle and 2-cycle last layer cases:

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text:
## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

He paid 75 percent of the original value of the jacket. Thus, the original price must have been: $$\frac{34.5}{.75} = 46\ Dollars$$

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text:
2

On the robustness of noise-blind low-rank recovery from rank-one measurements

We prove new results about the robustness of well-known convex noise-blind optimization formulations for the reconstruction of low-rank matrices from an underdetermined system of random linear measurements. Specifically, our results address random …

On the geometry of polytopes generated by heavy-tailed random vectors

We study the geometry of centrally-symmetric random polytopes, generated by $N$ independent copies of a random vector $X$ taking values in ${\mathbb{R}^n}$. We show that under minimal assumptions on $X$, for $N \gtrsim n$ and with high probability, …

Dictionary-Sparse Recovery From Heavy-Tailed Measurements

The recovery of signals that are sparse not in a given basis, but rather sparse with respect to an over-complete dictionary is one of the most flexible settings in the field of compressed sensing with numerous applications. As in the standard …

Harmonic Mean Iteratively Reweighted Least Squares for Low-Rank Matrix Recovery

We propose a new iteratively reweighted least squares (IRLS) algorithm for the recovery of a matrix $X \in \mathbb{C}^{d_1 \times d_2}$ of rank $r \ll \min(d_1,d_2)$ from incomplete linear observations, solv- ing a sequence of low complexity linear …


text:
## On shattering, splitting and reaping partitions.(English)Zbl 0896.03036

The author deals with cardinal invariants for the family of all partitions of $$\omega$$. Especially he investigates dualizations of some well-known cardinal invariants for the power set of $$\omega$$. So he considers the dual-shattering, dual-splitting and dual-reaping numbers. He shows that it is consistent with ZFC that the dual-shattering cardinal is greater than $$\omega_1$$. This is done using dual-Mathias forcing.
Reviewer: M.Weese (Potsdam)

### MSC:

 03E05 Other combinatorial set theory 03E35 Consistency and independence results 03C25 Model-theoretic forcing 05A18 Partitions of sets
Full Text:

### References:

 [1] Balcar, Fund. Math. 110 pp 11– (1980) [2] Balcar, Topology and its Applications 41 pp 133– (1991) [3] and , Set Theory: The structure of the real line. A.k. Peters, Wellesley 1995. [4] Bell, Fund. Math. 114 pp 149– (1981) [5] Carlson, Adv. in Math. 53 pp 265– (1984) [6] , and , Dualizations of van Douwen diagram. Preprint. [7] Multiple Forcing. Cambridge University Press, Cambridge 1987. [8] Set Theory. Academic Press, London 1978. [9] Set Theory, an Introduction to Independence Proofs. North-Holland Publ. Comp., Amsterdam 1983. · Zbl 0534.03026 [10] Matet, J. Symbolic Logic 51 pp 12– (1986) [11] Piotrowski, Proc. Amer. Math. Soc. 101 pp 156– (1987) [12] Plewnc, Fund. Math. 127 pp 127– (1986) [13] Shelah, Contemporary Math. 31 pp 183– (1984) [14] Partition numbers. Preprint. [15] The integers and topology. In: Handbook of Set-theoretic Topology ( and , eds.), North-Holland Publ. Comp., Amsterdam 1990, pp. 111–167. [16] Small uncountable cardinals and topology. In: Open Problems in Topology ( and , eds.) North-Holland Publ. Comp., Amsterdam 1990, pp. 195–218.
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
# Suppose that quantum mechanics (or QFT), like Newtonian physics and other past theories, turns out to have a limited domain of applicability, beyond which it is replaced by a deeper theory.

## What is your best guess as to what physical parameter will delimit the boundary between quantum mechanics and its successor? Is it a length scale, an energy or energy density, a number of degrees of freedom, or something else?

Radical idea that there might be a new fundamental constant analogous to Planck scale /energy/length where some new theory might take over. What would you look for to observe this?

Is this a large scale problem, or small scale problem?

If QM breaks down it could be at high energies at small scales (high energy density)

What about intermediate / mesoscopic scale?

Can you abandon the wave function (non-ontic?) and build up a new theory that is consistent.

Is QM the problem? We haven’t seen any solid experimental evidence for QFT/QM breaking down?

Future  Ligo / optomechanical experiments to test QM with macroscopic systems, Superconducting experiments currently test in mesoscopic regime.

Could action be a relevant constant in a theory that underlies QM and gravity. Could Planck constant be something that is fixed or something that is emergent. If these things are emergent then length isn’t fundamental but emergent.

Planck Area (in 2 dimensions), in fractional quantum hall effect area of magnetic flux constant, or some Planck surface in higher dimensions.

What regime would we have to go to to see QM/QFT break down? LHC (high energy), LIGO, optomechanics, etc haven’t observed anything.

Does anyone really expect it to break down? So far it has survived every experiment thrown at it.

Perhaps quantum foundations and interpretations is a more interesting direction than what breaks down.

Combination of length scale and energy scale at different limits

Quantum mechanics has been a very successful theory that describes several natural phenomena that can be tested experimentally. This proves up to some extent the validity of quantum mechanics. Until today, most of the challenges that quantum mechanics faces are mostly leaned towards the different interpretations of the concepts of the theory. As a mathematical theory, quantum mechanics is consistent and makes predictions that are experimentally verifiable. With this one can argue that quantum mechanics has no successor theory. However there is no guarantee the there is broader theory that reduces to quantum mechanics. In fact, recent works on String Field Theory suggest that the 'rules' of quantum mechanics  can be derived from physical processes between extended degrees of freedom. However, this does not constitute a bigger theory containing quantum mechanics, but an equivalent formulation. Furthermore, there has been works addressing the fate of quantum mechanics under certain deformations that shows that quantum mechanics is a very robust theory.  In summary, quantum mechanics is a robust theory that as a mathematical framework is very strong, but there is still an active debate on what the right interpretation of it is. The outcome of such debate will probably shed light on the domain of applicability of the theory and the so-called physical parameter that will delimit the boundary between quantum mechanics and its successor (if there is any).

It is known QFT breaks down at a very high-energy scale. Planck scale is definitely beyond QFT’s applicability. At such a scale, quantum gravity effects sit in and it is not clear how to define basic quantities like the number of degrees of freedom and energy density. In another scenario, if a system involves many particles, huge quantum gravity effects may also take place and such system is not well understand.

May be the formulism of QFT needs to be changed. As advocated by Nathan Seiberg, the local Lagrangian formulism with gauge groups may be not the ultimate language to describe quantum physics. On one hand side, we see Seiberg duality, which indicates the gauge symmetry is not fundamental at UV scale. On the other hand, many theories, which do not have a Lagrangian description, can be calculated in the S-matrix approach.

To answer this question we should look at areas that unitarity is breaking.

Obvious scale could be planck mass (Quantum gravity scale).

another candidate would be dark energy scale.

Maybe the question is wrong. We should ask what would be the physical  quantity that we have not probed well and could potentially leads to new physics.  For quantum mechanics it was the action, for special relativity it was the speed.  What would be the physical quantity that leads to the new physics (beyond QM)?

We currently know that physics breaks at low energy (dark energy) and high  energy (QG).

it is probably not a length scale.

Degrees of freedom

Isn’t what we need to explain measure ment problem – how qm goes to classical

Need something past the Copenhagen interpretation “difficult problem”

Is there evidence that we need a successor? Length scale is the planck scale?

Black holes?

Dark energy – doesn’t make sense from the classical  point of view – most of the universe can’t interact – classical part can’t

Multiverse – regions out of causal contact

We know that the speed of light is something we can compare velocities to in order to determine if special relativity is applicable. Similarly, the (reduced) Planck constant is something we compare the action to in order to determine if quantum mechanics is applicable. A natural guess for the scale of a more fundamental theory such as quantum gravity is the Planck scale. The number of degrees of freedom doesn’t seem to be relevant. However, it might be that “the question is wrong”; a boundary between regimes is not actually applicable here. It’s not as simple as a cutoff at the Planck scale. Physics seems to be nonlinear in a sense, so if something breaks down on one scale, it’s going to affect all scales. For example, in inflation quantum fluctuations can have large-scale classical consequences.

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text: If $$\displaystyle I = \int_1^2 \left( \cot^{-1} \sqrt{x-1} \right)^2 \, dx$$, find the value of $$\dfrac { { \pi }^{ 2 }+8I-8\ln { 2 } }{ \pi }.$$


text:
# Chromatic number of associahedron

 Importance: Medium ✭✭
 Author(s): Fabila-Monroy, Ruy Flores-Penaloza, David Huemer, Clemens Hurtado, Ferran Urrutia, Jorge Wood, David R.
 Subject: Geometry
 Keywords: associahedron, graph colouring, chromatic number
 Posted by: David Wood on: June 2nd, 2015
Conjecture   Associahedra have unbounded chromatic number.

An associahedron is the convex polytope in which each vertex corresponds to a way of correctly inserting opening and closing parentheses in a fixed word and the edges correspond to single application of the associativity rule. Equivalently, the vertices of an associahedron correspond to the triangulations of a convex polygon and the edges correspond to edge flips in which a single diagonal is removed from a triangulation and replaced by a different diagonal.

The chromatic number of (the 1-skeleton of) associahedra was first considered by Fabila-Monroy et al [FFHHUW]. They proved that for the associahedron corresponding to edge flips in triangulations of a convex -gon, the chromatic number is at most and at most . The best known lower bound is for [private communication, Ruy Fabila-Monroy].

Update (2019): Addario Berry et al. [ARSW] proved an upper bound of .

## Bibliography

*[FFHHUW] Ruy Fabila-Monroy, David Flores-Penaloza, Clemens Huemer, Ferran Hurtado, Jorge Urrutia, David R. Wood. On the Chromatic Number of some Flip Graphs, Discrete Mathematics and Theoretical Computer Science Vol 11, No 2 (2009).

[ARSW] Louigi Addario Berry, Bruce Reed, Alex Scott, David R. Wood. A logarithmic bound for the chromatic number of the associahedron, 2018.

* indicates original appearance(s) of problem.


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# Introduction and key concepts

 Page 2 / 3
Amplitude

The amplitude is the maximum displacement of a particle from its equilibrium position.

## Investigation : amplitude

Fill in the table below by measuring the distance between the equilibrium and each peak and troughs in the wave above. Use your ruler to measure the distances.

 Peak/Trough Measurement (cm) a b c d e f
2. Are the distances between the equilibrium position and each peak equal?
3. Are the distances between the equilibrium position and each trough equal?
4. Is the distance between the equilibrium position and peak equal to the distance between equilibrium and trough?

As we have seen in the activity on amplitude, the distance between the peak and the equilibrium position is equal to the distance between the trough and the equilibrium position. This distance is known as the amplitude of the wave, and is the characteristic height of wave, above or below the equilibrium position. Normally the symbol $A$ is used to represent the amplitude of a wave. The SI unit of amplitude is the metre (m).

If the peak of a wave measures $2\phantom{\rule{2pt}{0ex}}m$ above the still water mark in the harbour, what is the amplitude of the wave?

1. The definition of the amplitude is the height of a peak above the equilibrium position. The still water mark is the height of the water at equilibrium and the peak is $2\phantom{\rule{2pt}{0ex}}m$ above this, so the amplitude is $2\phantom{\rule{2pt}{0ex}}m$ .

## Investigation : wavelength

Fill in the table below by measuring the distance between peaks and troughs in the wave above.

 Distance(cm) a b c d
2. Are the distances between peaks equal?
3. Are the distances between troughs equal?
4. Is the distance between peaks equal to the distance between troughs?

As we have seen in the activity on wavelength, the distance between two adjacent peaks is the same no matter which two adjacent peaks you choose. There is a fixed distance between the peaks. Similarly, we have seen that there is a fixed distance between the troughs, no matter which two troughs you look at. More importantly, the distance between two adjacent peaks is the same as the distance between two adjacent troughs. This distance is called the wavelength of the wave.

The symbol for the wavelength is $\lambda$ (the Greek letter lambda ) and wavelength is measured in metres ( $m$ ).

The total distance between $4$ consecutive peaks of a transverse wave is $6\phantom{\rule{2pt}{0ex}}m$ . What is the wavelength of the wave?

1. From the sketch we see that 4 consecutive peaks is equivalent to 3 wavelengths.

2. Therefore, the wavelength of the wave is:

$\begin{array}{ccc}\hfill 3\lambda & =& 6\phantom{\rule{0.166667em}{0ex}}\mathrm{m}\hfill \\ \hfill \lambda & =& \frac{6\phantom{\rule{0.166667em}{0ex}}\mathrm{m}}{3}\hfill \\ & =& 2\phantom{\rule{0.166667em}{0ex}}\mathrm{m}\hfill \end{array}$

## Investigation : points in phase

Fill in the table by measuring the distance between the indicated points.

 Points Distance (cm) A to F B to G C to H D to I E to J

What do you find?

In the activity the distance between the indicated points was the same. These points are then said to be in phase . Two points in phase are separate by an integer (0,1,2,3,...) number of complete wave cycles. They do not have to be peaks or troughs, but they must be separated by a complete number of wavelengths.

how do you translate this in Algebraic Expressions
why surface tension is zero at critical temperature
Shanjida
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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# Let A = {1, 2, 3} and B = {2, 1, 5, 6}. Find (A $\cap$ B)

A $\cup$ B = {1, 2, 3, 5, 6}


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# Tag Info

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I am an experimentalist. For me the beauty of the Standard Model of particle physics lies in that it encompasses successfully, in a mathematical manner, the innumerable experimental data of the last sixty years and more. The pros for both proposed theories for gravity are the quantization of gravity, a long sought after holy grail. I am biased by ...

2

I think it may have been Witten (perhaps somebody can correct me) who suggested that a TEO might not actually exist. Just as we might never, even in principle, posses the mathematics that can interpolate between the strongly coupled and free limits of QCD, it might not be possible to write down one set of equations that describes one corner of a TEO, say, ...

21

One thing that stops us from having a theory of everything is actually quite simple. Gravity as we understand it, thanks to the strong equivalence principle, is not a force. It is entirely geometrizable because there is actually no coupling constant between a physical object and the "gravitational field". This means that there is no a priori way to ...

3

A "fifth force" can be given by a scalar field. This field arises in many theories, for example from the dilaton scalar field of string theory, that is non minimally coupled to $R$ in the action. In the low energy limit we can phenomenologically treat string theory as a Scalar Tensor Theories of Gravitation (see the Wikipedia Link for details) in which you ...

0

some theories propose that the infalton field is actually the same a dark energy,, if that were correct, then we would have a fifth field (force might not be correct in this context) that ineracts with the other four. And who knows what else, dark matter might have lots of forces that are invisible to us because they just do not inetract with the four we do. ...

1

Yes, you can simulate non-locality in celular automata, even if the rules are local. In such a case, contiguous cells do not represent adjacent points in our space. In these models space is an approximate, macroscopic emergent property (it is discussed in detail in NKS). This emergent space can have, at microspcopic scales, functional links that resemble the ...

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#### Volume 17, issue 1 (2013)

 Recent Issues
 The Journal About the Journal Subscriptions Editorial Board Editorial Interests Editorial Procedure Submission Guidelines Submission Page Ethics Statement Author Index To Appear ISSN (electronic): 1364-0380 ISSN (print): 1465-3060 Other MSP Journals
Spherical subcomplexes of spherical buildings

### Bernd Schulz

Geometry & Topology 17 (2013) 531–562
##### Abstract

Let $\Delta$ be a thick, spherical building equipped with its natural CAT(1) metric and let $M$ be a proper, convex subset of $\Delta$. If $M$ is open or if $M$ is a closed ball of radius $\pi ∕2$, then $\Lambda$, the maximal subcomplex supported by $\Delta \setminus M$, is $dim\Lambda$–spherical and non-contractible.

##### Keywords
spherical building, Cohen–Macaulay, connectivity
Primary: 51E24
Secondary: 11F75


text:
# NMDS anomaly - data does not support point placement

My data:

Tracking forest communities (via species abundances) in various forest plots across time.

My approach: Non-metric Multidimensional Scaling ordination

I performed NMDS (using nmds() in the R package Ecodist), and then added change vectors to my ordination.

• The ordination was performed using 1000 separate ordinations (i.e., argument nits was set to 1000), so I would think the results should be fairly robust

• Note: each point in my resulting ordination represents a forest plot in a given year, and change arrows connect the points of each consecutive sampling year for each given plot.

I also performed a hierarchical cluster analysis (w/ flex beta), and subsequently colored my points in the ordination based on the cluster analysis results.

The resulting colored NMDS ordination (with change vectors) looks like this:

The point

So if you'll notice, the forest plot in the bottom right corner changes minimally in all but a single year. In that year, the NMDS point is drastically different than in all other years -- it's located all the way toward the top of the NMDS output plot.

Normally, I would see this trend and assume something drastic happened that year. I would assume that the plot's overall species characteristics matched the plots represented by red dots on the top more so than the plots associated with blue dots toward the bottom of the ordination graph in that one given year.

However...

When I actually examine the raw data for this forest plot in the years leading up to, including, and after the massive shift in NMDs space, I find that basically nothing changed between years (especially between the drastic change year and the sampling years immediately before and after).

• In other words, the same individual trees (of the same species and with the same relative distribution of sizes) existed in this massively different year as in the years before and after it.

My Question(s):

So I know that NMDS isn't perfect and can get stuck in "local" minima etc etc... However, after running 1000 ordinations, I would think my results would be robust enough to avoid such issues. But maybe it's still possible?

• So, is this anomaly possibly still due to some random chance issue in the algorithm itself?

• If so, is there any way for me to determine the cause??

Lastly, I have re-run this ordination a hand-full of times (each time with slightly different input), but each time some plot seems to have a weird "outlier " point that doesn't make sense like above. So I'm beginning to think maybe this is something that I won't be able to eliminate or account for.

• If that's the case...

• Is there a way to describe the phenomenon?

• (the phenomenon that a random ordination point just won't cooperate for no apparent reason)
• More importantly, is there precedence for simply excusing an anomaly like this as an artifact of the NMDS approach and just moving forward with the present result?

• examples or suggestions of how to word such an explanation would be appreciated.

Thanks!

• The ordination was performed using 1000 separate ordinations (i.e., argument nits was set to 1000) What's that - for somebody not using same software as you do? What is the model of your NMDS? One-matrix or repeated model or weighted model (INDSCAL)? – ttnphns Aug 23 '17 at 20:55
• Also, I feel (maybe wrongly) that your questions would be hard to answer without being in your research shoes. – ttnphns Aug 23 '17 at 20:59
• @ttnphns so essentially it's 1000 different starts of the NMDS calculation to help avoid the NMDS result finding local minima. I do not know what you mean in your second question (One-matrix or repeated model or weighted model (INDSCAL), so I cannot answer that). – theforestecologist Aug 23 '17 at 21:07
• @ttnphns I don't think you need to be in my shoes to answer it. I've told you the "science" side of it: there is no data/scientific difference b/w the points in question, so the resulting "anomaly" must be due to some oddity in the NMDS algorithm itself and not due to the data. Because the data do not appear to be the problem, I've covered my bases as the researcher. Also, I'm more curious what others do with this situation: do they simply excuse it as a regular occurrence when using NMDS, or do they know the cause/solution to finicky anomaly points? – theforestecologist Aug 23 '17 at 21:10
• UPDATE: By changing the seed (i.e., using the seed() function in R) and rerunning the code I was able to get the point to "correct" itself. So, for me, the solution was finding a seed that allowed the randomizations to sort things out on their own... – theforestecologist Jun 8 '18 at 14:30


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# Solve Limit Problems Online

Tags: What Is Creative Writing ClassHigh School Essay S 2011Free Basic Business Plan Template DownloadThe Adventure Of The Speckled Band Essay QuestionsAn Essay On Man Great Chain Of BeingCritical Thinking Book PdfShort Essays Gender Equality

Byju's Limit Calculator is a tool which makes calculations very simple and interesting.

If an input is given then it can easily show the result for the given number.

Factoring this one is just a little harder, but the idea is exactly the same.

\begin \lim_\dfrac &= \text \\ \\ &= \lim_\dfrac \\ \\ &= \lim_\dfrac \\ \\ &= \lim_[x 1] \\ \\ &= 6 \quad \cmark\end You probably have some problems that look like $$\lim_ \dfrac = ? Instead, there’s probably a polynomial you can expand. For example, here’s a problem a student asked us via Twitter:$$\lim_\dfrac = ?

Textbooks are good sources of graphs showing continuous and discontinuous functions.

If you have a smartboard you may want to try an online interactive gallery for graphing functions which can be found at: Maths Online from the University of Vienna, Austria.

From the left and right side of the limiting x value 2, the y value is approaching -4. If the result of direct substitution results in a function that is 0/0, the answer is undefined at this point, yet the limit still exists.

To evaluate the limits students need to be shown several examples.

## Comments Solve Limit Problems Online

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• ###### Solving limits by factoring - Calculus - YouTube

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In this video, we solve for the limit of various functions, and go over some tricks and tips. TimeStamp ! - Que 1 Starts ** Solution for Que 1.…

• ###### A Guide to Understanding Calculus Limit Problems

Teachers will find that teaching high school calculus limit problems is a concept that can. limits is finding the value of a function as x approaches a certain value, a. which can be found at Maths Online from the University of Vienna, Austria.…

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First of four screencasts on using the algebraic laws of limits to calculate limits. This screencast looks at a rational function and two limits.…


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}

## Definition

Let $\mathbf {Set}$ be the category of sets.

Let $\mathbf C$, $\mathbf D$ be locally small categories.

An adjunction between $\mathbf C$ and $\mathbf D$ is a triple $(F,G,\alpha)$, where

$F : \mathbf D \to \mathbf C$ is a functor.
$G : \mathbf C \to \mathbf D$ is a functor.
$\alpha : \mathrm{Hom}_{\mathbf C}(F-,-) \to \mathrm{Hom}_{\mathbf D}(-,G-)$ is a natural isomorphism between the functors
$\mathrm{Hom}_{\mathbf C}(F-,-) : \mathbf D^{\mathrm{op}} \times \mathbf D \to \mathbf{Set}$
$\mathrm{Hom}_{\mathbf D}(-,G-) : \mathbf D^{\mathrm{op}} \times \mathbf D \to \mathbf{Set}$


text:
Chapter 12

### Objectives

After completion of this chapter, the physical therapist should be able to do the following:

• Explain the neurophysiologic basis of proprioceptive neuromuscular facilitation (PNF) techniques.

• Discuss the rationale for use of PNF techniques.

• Identify the basic principles of using PNF in rehabilitation.

• Demonstrate the various PNF strengthening and stretching techniques.

• Describe PNF patterns for the upper and lower extremity, for the upper and lower trunk, and for the neck.

• Discuss the concept of muscle energy technique and explain how it is similar to PNF.

Proprioceptive neuromuscular facilitation (PNF) is an approach to therapeutic exercise based on the principles of functional human anatomy and neurophysiology.10 It uses proprioceptive, cutaneous, and auditory input to produce functional improvement in motor output and can be a vital element in the rehabilitation process of many conditions and injuries.

The therapeutic techniques of PNF were first used in the treatment of patients with paralysis and various neuromuscular disorders in the 1950s. Originally the PNF techniques were used for strengthening and enhancing neuromuscular control. Since the early 1970s, the PNF techniques have also been used extensively as a technique for increasing flexibility and range of motion.8,9,16,17,18,30,34,36,45,54,67,71

This discussion should guide the therapist in using the principles and techniques of PNF as a component of a rehabilitation program.

### Proprioceptive Neuromuscular Facilitation as a Technique for Improving Strength and Enhancing Neuromuscular Control

#### Original Concepts of Facilitation and Inhibition

Most of the principles underlying modern therapeutic exercise techniques can be attributed to the work of Sherrington,63 who first defined the concepts of facilitation and inhibition.

According to Sherrington, an impulse traveling down the corticospinal tract or an afferent impulse traveling up from peripheral receptors in the muscle causes an impulse volley that results in the discharge of a limited number of specific motor neurons, as well as the discharge of additional surrounding (anatomically close) motor neurons in the subliminal fringe area. An impulse causing the recruitment and discharge of additional motor neurons within the subliminal fringe is said to be facilitatory. Any stimulus that causes motor neurons to drop out of the discharge zone and away from the subliminal fringe is said to be inhibitory.40 Facilitation results in increased excitability, and inhibition results in decreased excitability of motor neurons.75 Thus, the function of weak muscles would be aided by facilitation, and muscle spasticity would be decreased by inhibition.26

Sherrington attributed the impulses transmitted from the peripheral stretch receptors via the afferent system as being the strongest influence on the alpha motor neurons.63 Therefore, the therapist should be able to modify the input from the peripheral receptors and thus influence the excitability of the alpha motor neurons. The ...

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text:
# Merge columns of coordinates and construct sequence

I have to following data frame.

    x1     x2          x3
0   58.33546    15.14554
0   58.33604    15.14571
0   58.33645    15.14581
0   58.33681    15.14591
0   58.33722    15.14600
0   58.33810    15.14604
0   58.33900    15.14599
0   58.33966    15.14578

Consisting of long and lat, each pair belongs to a string (0,1,3,4 and so on, only the string 0 is displayed above).

I would like to first construct a new column with the pairs (x2,x3). How do I do that?

The coordinates above corresponding to 0 is actually a journey (say X), if there is another journey (Y) whose first coordinate is the same as X last, I would like to merge the sequence X and Y coordinates together and call that journey X (the continuation of X simply).

My data set is called data, for some reason merging them together wont work with

data[, x4 := paste(data$x2, data$x3)]

This will create a new column, only isplaying x2.

it probably depends on the use of your new generated column. Here I try to reproduce your example:

x1 <- c(0, 0)
x2 <- c(58.33546, 58.33604)
x3 <- c(15.14554, 15.14571)
df <- data.frame(x1, x2, x3)
df$x4 <- paste("(", as.character(df$x2), ", ", as.character(df$x3), ")")  And df$x4 will look like:

[1] "( 58.33546 ,  15.14554 )" "( 58.33604 ,  15.14571 )"

If you want to show the coordinates, the above format should be fine. If you want to do further analysis with that column, some other transformations are required.

data.table supports list type columns. When doing assignments you need to put them in an additional list() or use :=(newcol=listvalue) syntax instead of newcol:=listvalue to avoid simplification when lists have a single element.

library(data.table)
data<-as.data.table(data)
data[, :=(x2x3=mapply(c, x2,x3, SIMPLIFY=F))]
data[, .(path=list(x1x2)), by=x1]


text:
## anonymous one year ago help!

1. anonymous

@Michele_Laino

2. anonymous

3. anonymous

@Michele_Laino

4. Rushwr

Hey that is going to be another rectangle right?

5. Michele_Laino

hint: we can have this cross section: |dw:1438412145605:dw|

6. Michele_Laino

|dw:1438412291460:dw|

7. Rushwr

oh I missed out the part 'at an angle to the base'

8. Rushwr

I agree with @Michele_Laino

9. anonymous

i think it would be a rectangle

10. Michele_Laino

thanks! @Rushwr

11. anonymous

am i right? D?

12. Michele_Laino

I'm pondering...

13. anonymous

ok

14. Rushwr

oh would it be a parallelogram ?

15. Astrophysics

I don't think it is a rectangle if you look at the original shape |dw:1438412707037:dw|, you notice given the sides are squares, it would be a rectangle if it was cut horizontally, it would be a square if it was cut vertically, so at an angle it must be something in between ;)

16. Astrophysics

I'm debating between square and parallelogram

17. anonymous

ohhh! i see :) its a parellelogram

18. Michele_Laino

and if I'm cutting a cube? @Astrophysics

19. anonymous

i think so im not sure

20. Astrophysics

Well if it was a cube then it would be all squares right

21. anonymous

yea

22. Astrophysics

OR depends on the slice, you could get a parallelogram

23. Michele_Laino

I'm not sure, since my question is: |dw:1438412930882:dw|

24. anonymous

im confused i dont know whether its a parellogram or a square :(

25. Michele_Laino

yes! I think it is a rectangle!

26. anonymous

what! now its a rectangle .. im so confued :((((((((((((((((((((((

27. imqwerty

no its a parellogram

28. Astrophysics

I think it's a parallelogram

29. Rushwr

no its a parallelogram. We can't say it is a square cuz one side is inclined.

30. Michele_Laino

the relative position of these sides, is not changed |dw:1438413189752:dw|

31. Astrophysics

Yeah, it's definitely not a square, because of the angle

32. Michele_Laino

yes! @Astrophysics

33. imqwerty

it can be a square..it can also be a rectangle...wait m posting the reason

34. anonymous

so what is the final answer everyone!

35. anonymous

36. imqwerty

wait :)

37. Astrophysics

The original image Michele drew, makes it seem as if it's a rectangle, but when I look at the question itself, and with the angle it should have opposite sides parallel and equal to the length itself...aaah!

38. Michele_Laino

I think it is a rectangle, since the relative position of these sides is unchanged going from a vertical cross section to an angle to the base cross section: |dw:1438413658506:dw|

39. anonymous

|dw:1438413775838:dw|

40. Rushwr

final answer will be a parallelogram

41. anonymous

I get rectangle

42. Michele_Laino

these angles: |dw:1438413858093:dw| are all right angles, am I right?

43. imqwerty

ok done

44. imqwerty

45. Astrophysics

@jayzdd But how do you know if it's cut like that? I mean it could be like this |dw:1438413969465:dw|

46. Rushwr

I agree @imqwerty

47. imqwerty

:)

48. Astrophysics

I think it's a parallelogram, but I'm not the best at geometry xD

49. imqwerty

@geny55

50. Michele_Laino

how do you do those drawings? @Astrophysics

51. Astrophysics

Do you use chrome?

52. Michele_Laino

yes!

53. Astrophysics

Awesome, I'll send you a few links through message, as I don't want to spam in this question xD

54. Michele_Laino

ok! thanks! :)

55. imqwerty

56. Astrophysics

Ah what ever, I don't think geny really cares about the process anyways, I'll just post it here haha.

57. Astrophysics

You guys can do this stuff right? |dw:1438414519358:dw|

58. imqwerty

hw did u do that!!!

59. Michele_Laino

no, I'm not able to do that

60. Astrophysics

|dw:1438414605710:dw| aaaaah! Ok, I'll teach you guys, it's very helpful as you can see above I took the question and posted it here, without attaching the file!

61. Michele_Laino

please open a question in feedback section @Astrophysics

62. Astrophysics

63. Michele_Laino

thanks! :) @Astrophysics

64. Astrophysics

By the way the draw tool with different colors, I got that from google app store, just search "openstudy" and you will be able to find the applications for that, how ever for the image and gif follow that tutorial as that's not in the applications, from the chrome app store you should be able to get stuff like quote, as such $$\color{blue}{\text{Originally Posted by}}$$ @Michele_Laino please open a question in feedback section @Astrophysics $$\color{blue}{\text{End of Quote}}$$

65. Michele_Laino

ok! :)

66. Astrophysics
67. imqwerty

@Astrophysics is there any way to do this in firefox?

68. Astrophysics

I'm not sure qwerty I only use chrome :\

69. Rushwr

Guys I think its high time to stop commenting or else it will be hard for genny55 to find the answer !

70. Astrophysics

Lol well Rush the answer isn't important, long as he/she follows the process they will be fine.

71. imqwerty

( ͡° ͜ʖ ͡°)

72. Rushwr

okai then u may proceed @Astrophysics

73. Astrophysics

Haha, well I think I gave out all the information on the stuff, so we should be good now :)

74. Rushwr

okai !!!!!!!!!! I'm sorry if I interrupted ur conversation !!!!! @Astrophysics @imqwerty

75. Astrophysics

It's ok!! You did nothing wrong :D, thank you!

76. anonymous

you can also produce a trapezoid http://prntscr.com/7zj8xv

77. anonymous

i think the answer depends on what you mean by 'at an angle to the base'


text:
• ### To Do Lessons

Chemistry » Chemistry 111: Electrochemical Reactions » Standard Electrode Potentials

# Standard Conditions

This is a lesson from the tutorial, Chemistry 111: Electrochemical Reactions and we encourage you to log in or register before you continue, so that you can track your progress.

## Standard conditions

Standard electrode potentials are a measurement of equilibrium potentials. The position of this equilibrium can change if you change some of the conditions (e.g. concentration, temperature). It is therefore important that standard conditions be used:

• pressure = $$\text{101.3}$$ $$\text{kPa}$$ ($$\text{1}$$ $$\text{atm}$$)

• temperature = $$\text{298}$$ $$\text{K}$$ ($$\text{25}$$ $$\text{℃}$$)

• concentration = $$\text{1}$$ $$\text{mol.dm^{-3}}$$


text:
Eddy viscosity ratio

(Difference between revisions)
 Revision as of 16:25, 17 April 2006 (view source)Jola (Talk | contribs)m (Eddy-viscosity ratio moved to Eddy viscosity ratio)← Older edit Revision as of 16:31, 17 April 2006 (view source)Jola (Talk | contribs) mNewer edit → Line 1: Line 1: - The eddy-viscosity ratio, $\frac{\mu_t}{\mu}$, is the ratio between the turbulent viscosity, $\mu_t$, and the dynamic molecular viscosity, $\mu$. + The eddy-viscosity ratio, $\frac{\mu_t}{\mu}$, is the ratio between the turbulent viscosity, $\mu_t$, and the molecular dynamic viscosity, $\mu$. The eddy-viscosity ratio is often also called turbulent viscosity ratio or simply viscosity ratio. The eddy-viscosity ratio is often also called turbulent viscosity ratio or simply viscosity ratio.
The eddy-viscosity ratio, $\frac{\mu_t}{\mu}$, is the ratio between the turbulent viscosity, $\mu_t$, and the molecular dynamic viscosity, $\mu$.


text:
# If the real part of analytic function satisfies $u_x=u_y$, then the function is linear

Let f(z) be analytic function and $\forall z=x+iy\in\mathbb C, u_x=u_y$ ($u_x=\frac{\partial f}{\partial x},u_y=\frac{\partial f}{\partial y}$. Prove that $f(z)=az+b$.

I thought using Cauchy Riemann equations with the given equation: $$\begin{cases}u_x=u_y \\ u_x=v_y\\u_y=-v_x\end{cases}$$ $\Rightarrow u_x=-v_x\Rightarrow u(x,y)=v(x,y)+c(y)$ (where c(y) is a function of $\Im(z)$).As follows $$v_x=u_x=u_y=v_y + c_y(y)$$ and in C.R $u_y=-v_x\Rightarrow v_x=-v_x\Rightarrow v_x=0$ so also $v(x,y)$ is a function only of $y=\Im z$. So I got $f(x,y)=v(y)+c(y)+iv(y)=v(y)(1+i)+c(y)$. I can't find any way to prove that $c(y)\equiv c\in\mathbb C$ or $v(y)$ is a linear function. How can I do it?

• You for $v_x=0$. What does that imply about $u_x,u_y,v_y$? – N. S. Jul 9 '14 at 12:58
• Ouch. So $u=c_1,v=c_2\quad (c_{1,2}\in\mathbb{C})$ ? but that's a constant function (not sure whether they asked for a linear function with incline $\neq 0$). Thanks. – user65985 Jul 9 '14 at 13:05

From $u_x=u_y$ and $u_x=v_y$, you have $u_y=v_y$ which implies $$u=v+2c_1(x).$$ Similarly from $u_x=u_y$ and $u_y=-v_x$, you have $u_x=-v_x$ which implies $$u=-v+2c_2(y).$$ Thus $$u=c_1(x)+c_2(y), v=c_2(y)-c_1(x).$$ Now use $u_x=u_y$ to get $c_1'(x)=c_2'(y)$. Since $c_1(x)$ and $c_2(y)$ are functions of $x$ and $y$, respectively, you have $$c_1'(x)=c_2'(y)\equiv k$$ and hence $c_1(x)=kx+k_1$ and $c_2(y)=ky+k_2$. So $$f(z)=u+iv=k(x+y)+ki(y-x)+b=(1-i)k(x+yi)+k_3=az+b$$ where $a=(1-i)k$ and $b=k_1+ik_2$.
From Cauchy-Riemann we have $u_{xx} + u_{yy} = 0$, if now $u_x = u_y$, we have $$u_{xx} = (u_x)_x = (u_y)_x = u_{xy} = (u_x)_y = u_{yy}$$ so $2u_{xx} = 0$, that is $u_{xx} = u_{yy} = u_{xy} = 0$. For $v$, we have $$v_{xx} = (v_x)_x = -(v_y)_x = -v_{xy} = -(v_x)_y = v_{yy}$$ which again gives $v_{xx} = v_{xy} = v_{yy} = 0$ from Cauchy-Riemann. So both $u$ and $v$ are affine-linear, hence, $f$ is also. As $f$ is analytic, $f(z) = az+b$ for some $a,b \in \mathbb C$.
• why $u_{xx}+y_{yy}=0$? – user65985 Jul 9 '14 at 13:36