text:
Catch up on stories from the past week (and beyond) at the Slashdot story archive

typodupeerror
DEAL: For $25 - Add A Second Phone Number To Your Smartphone for life! Use promo code SLASHDOT25. Also, Slashdot's Facebook page has a chat bot now. Message it for stories and more. Check out the new SourceForge HTML5 internet speed test! × ## Wearable PCs78 Shawn writes "Interesting article on wearable PCs. Brief mention given to Linux / Beowulf, and 'an undershirt with 64 linked processors.' " God I'm a sucker for this stuff. "Hi, I'm Rob, and I wanna be a borg". This isn't anything really breaking or exciting, but its still neat. This discussion has been archived. No new comments can be posted. #### Wearable PCs More Login ## Wearable PCs Comments Filter: • #### Hmm... (Score:1) by Anonymous Coward Use your wearable's webcam to grab a picture of someone's face, run a face recognition on it, tap into his government's surveilance records, and ask him all sorts of personal questions about that affair he's been having with his secretary... • #### Re:Well.. (Score:1) Funny you should mention that... I use a 486sx25 to brush my cat. • #### An Invitation to Michael Martinez (Score:1) Since I know he reads slashdot, I'd like to invite Mr. Martinez to post a followup here for stuff that may have been too geeky and technical for the intended audience of an ABCNews article. Anyone else want to hear what he has to say? -- • #### Tech support for my underwear(underware?) (Score:2) "Thank you for calling the Help Desk. How may I help you?" "Well, when I try to reboot my underwear, it just gives me a wedgie." "What OS are you running on your underware, sir?" "Microsoft Windows for Cotton Blends" "Sorry sir. We can't help you. To fix this problem would require us to be able to unravel the source code and Microsoft doesn't allow that." "But I chose Microsoft because it seemed Microsoft-y and smooth against my skin! Without fabric softner!" "We're sorry. May we suggest the new distribution of Linux from Procter and Gamble?...." • #### Re:Professor Mann is a freak (Score:1) He was in my class for undergraduate Electrical/Computer engineering. He was definitely odd. I'm not sure whether "fitting in" ever occurred to him. He was the guy who would wait until the professor finished a great long explanation of a topic, and then ask a question that showed total lack of insight into the material being covered. There would be a huge group cringe, and everyone would turn around and silently wonder how he could have missed the point so soundly. I wondered if he'd ever graduate. Now he's a prof and a pundit. He may be a freak, but he sure knows how to market what he's got. • #### SMP thermal underware exclusively at Cabellas! (Score:1) Stay warm in your duck blind this winter. If you start to get a chill just perform some intense statistical analysis or process weather pattern data with the new Beowulf thermal underware from Cabellas. You will quickly learn to adjust processor load to find that "just right" temerature zone. You will enjoy worry-free outings knowing that _your_ underware is fully Y2k compliant, and has an average battery life of 4 full hours*. Act now and get an exteded 6 month warranty free**. *Estimated run time using optional 400 cell mobile battery cart. ** Waranty void if user fails to follow propper grounding procedures before dressing. Warning: Overclocking will void waranty and may result in burn injury. • #### Asociated Press - Hacker's underware seized (Score:1) Today Seattle authorities seized the underware of a young hacker accused of breaking into government networks. "We were able to prove the suspects guilt by placing a 'sniffer' program in his shorts last week". The police crime lab also reported that upon examining the shorts they found evidence that the young hacker had contracted the Melissa virus. • #### No keyboard, twiddlers (Score:2) I have met Thad Starner, and he, as well as most of the wearable community, uses a Twiddler. The Twiddler is a one-handed chorded keyboarded, which is amazingly fast if one knows how to use it. --Ivan, weenie NT4 user, Jon Katz hater: bite me! • #### Re:64 processor shirt (Score:1) Hmmm... not the sort of thing I'd want to wear in the summer. It's hot enough, without having 64 CPUs warming you up. What's the heat dissipation of the thing. Do you need to wear heatsink backpack? Speaking of which, I hope it's waterproof. Hate to have a little sweat taking out the system. Reminds me of the wicked with of the west... splah her with water and she'll scream "help me! I'm shorting! Shorting! aaaaaaauuuuug!" • #### Catman (Score:1) This is all starting to remind me of an old science fiction story, Catman, in which the main character was part of a group of people that started replacing parts of their bodies with metal and assorted electronics. I guess the goal was to become totally machine. I wish I could remember the author but I think it was in the first Dangerous Visions collection. I've seen several movies that I'd bet were influenced by it. • #### Re:Groundbreaking & retail? (Score:1) Uh, so new and interesting prototypes shouldn't be discussed here, on the site for geeks who are interested in cutting edge, non mainstream technology?'' \begin{sarcasm} That's right. He doesn't want to see it posted on Slashdot unless he can buy from the Sharper Image catalog. It was such a waste of his time that he had to take time to post a snide comment about its being put up on the site. \end{sarcasm} • #### Re:Don't sweat yourselves.. (Score:2) Har, har! This will give a whole new meaning to 'electrical shorts', won't it. "Honey, would you iron and format my pants for me?" I suppose the shirt will be marked 'dry clean only' huh? 'Wash separately, with like OS flavors'. I HATE those! I bet it will only come in beige. But if a friend offers to give you the shirt off his back, it will actually be welcomed. Thank you, I do birthdays, weddings.. I'm here all week. • #### Re:Tech support for my underwear(underware?) (Score:2) "Well, when I try to reboot my underwear, it just gives me a wedgie." "What OS are you running on your underware, sir?" "Microsoft Windows for Cotton Blends" "Well, sir, the only solution is to change your underwear to MS-NTLongJohns" "But it's the middle of August, and we're in the middle of a heat wave. Also, I don't think I need all of that fabric" "I realize that Sir. But the NTLongJohns are the only version of MS-underpants what don't give you a wedgie. And you're much less likely to suffer the terrible blue-ball of death with these than with the MS-Underoos95. The NTLongJohns also come with a convenient back-orifice flap for examining core-dumps." • #### something to consider (Score:1) I think the whole idea of a wearable computer is silly. If we put the funding into improving wireless telecommunications, we could all just run an X session from our servers at home.. no matter where we are. • #### Umm, it's a feature, sir. (Score:1) Will the GAP do onsite calls if I find a bug in my shorts? Or a short in my trousers? How will this force moths to evolve? When I throw my clothes together in a pile will they Beowulf? Will the drycleaners stop going through my pockets looking for stray homeless 5spots and just try to hack into my zipper for my financial info? I find this fearful only because I still use the rock slab to do my wash. • #### Just get rid of the keyboard... (Score:2) The question remains: do we want to slip into our computers the way we slip into our clothing? Well, I know I do. However, this sort of thing is not practical with a keyboard. An alternative is necessary; I'd like to see something along the lines of a cyberglove that's used with a piano-like virtual keyboard; I believe the guy who invented the mouse made one of these (no urls handy, sorry) but it never took off. • #### Re:Organization Systems (Score:1) I was in one of Thad Starner's classes earlier this year, and he talked about this a little. He takes notes into seperate files for each event (class, conference, meeting, etc...). He uses a piece of software called a remembrance agent, which shows a list of files relevant to whatever he's typing in emacs at the moment. Go here [gatech.edu] for a (unfortunately partially raw-latex) description of the remembrance agent and various ways Thad uses his wearable in daily life. This [gatech.edu]is the page of readings for his class, with various other interesting wearable papers. • #### Re:They work when people stop calling them compute (Score:1) I think you meant "when they are so transparent that people....." • #### "Hi, I'm Rob, and I wanna be a borg" (Score:3) on Tuesday August 24, 1999 @06:21AM (#1728386) Journal As soon as someone designs a computer than can be implanted into the skull, I'm going for it. If it requires cabling, circuitry etc. to be attached to the outside of my head, so much the better. I have a morbid fascination with becoming a mechanical freak. • #### supercomputer in your armpit (Score:1) The only question is how comfortable would it be to wear? I can handle having a battery attached to my belt, and *maybe* a pair of glasses that aren't too heavy that contain a small screen, but being loaded down with a full Borg armor suit can't be too appealing. It's a nice idea, but I'd rather just sit down and do my typing, as opposed to blindly stagger around on the street playing Quake and then getting hit by a bus 'cause I'm not paying any attention to where I'm walking. On other other hand, going to the grocery store would be a lot more fun -- just imagine how freaked out the old lady that gives out free samples of ice cream would be. • #### Re:They work when people stop calling them compute (Score:1) What else are you going to call it? A calculator? a word processor? a telephone? Those are all taken, and all of those serve on specific function. The thing about a computer is that it can do almost anything, and limiting it to "Something that drives me around" or "something that lets me talk to people" is bad. The examples with cars and stuff are backwards -- these are older devices that got computers added to them. Then again, maybe you can call it a "control panel" and have a couple of terminals hanging on your walls, kind of like in the halls of the Enterprize, and they would show the status of the laundry, what channel your teenager is watching a 1am in the morning, a record of your traffic violations... Call me old-fashioned, but I'd rather walk up to something and push buttons on it than walk around IN something. As a side note, it would be neat if the energy generated by walking and breathing could be used to power the wearable. • #### Don't sweat yourselves.. (Score:1) Really.. don't. You may get a shock, or worse yet, if you get caught out in the rain. I can see it now... New York (August 24,199): Death of men at the altar in holy matramony killed from wearable PC's. Life's short. Play hard. Owowowow.. stitch in my side! • #### Re:Umm, it's a feature, sir. (Score:1) Will the GAP do onsite calls if I find a bug in my shorts? I think you are looking for the red light district. • #### Re:Heads up display in your eyeball! (Score:1) Well this isn't too far off. Microvision [mvis.com] Is developing Virtual Retinal Displays [mvis.com] that use a low power laser to directly stimulate the retina. I imagine with some fine tuning you could get it to overlay the real world image. Until then you'll still have to use standard HUD technology. As far as an input device goes, you're missing the obvious. Impliment the keyboard and/or mouse in software. The whole point to wearables is to take ubiquitous computing with you. You shouldn't have to look like a "freak" to wear a computer. Imagine something similar to the "keyboard" in "Johnny Mnemonic" Use either a head mounted camera or maybe wrist mounted EKG sensors to detect the electrical impulses to the finger muscles. That way you just move your fingers. Even today, you can use a camera to track your finger (if you've got a flourescent thimble on). You can use that for a mouse; and I imagine with some beefed up software (and probably hardware too) you could get a computer to track and interpret the the patterns of a typing on a keyboard. (of course I'm designing this in a complete vaccuum, so I might as well say that a future wearable computer will have an integrated time travel device.) • #### Re:64 processor shirt (Score:1) Build it into a tuxedo! • #### Re:Professor Mann is a freak (Score:1) spooky dude for sure man did you notice his cloths? this dude is weireded out to the max! and those shades, in Canada, why???? • #### Re:Tech support for my underwear(underware?) (Score:1) Blue-ball of death... *shudder* I suppose we'll see men on the ground holding their crotches when that happens. "Ctrl-Alt-Delete! Reset! Who put the (@#*%()^*#)% reset in the corner of the back where I can't get it?!@" ROTLMAO Moderators: moderate this thread *UP*! • #### Re:"Hi, I'm Rob, and I wanna be a borg" (Score:2) It sounds neat, and I'd consider it, but people keep forgetting one little detail. Once the technology comes out, and you pay a doctor to drill a 10baseT port in your skull, Microsoft will release Brain 2.0. And then you have to go to the doctor for the upgrade, or all the new stuff will give you an incompatibility headache when you plug it in. Remember: cyber upgrades hurt. • #### Re:You call that "modest"? (Score:1) I wish Neil :) It looks like the netwinders have been nixed because of funding ($250G is a little pricy) so instead I'm scrounging around for discarded 486s. Know any handy place where I can pick some up for free?

P.S. No matter how much you wish it, you are NOT object oriented :)
• #### Groundbreaking & retail? (Score:1)

Uh, so new and interesting prototypes shouldn't be discussed here, on the site for geeks who are interested in cutting edge, non mainstream technology? Wearcomps will not be a "retail product" for a while yet. There are numerous boundaries that need to be surmounted first. Part of that process is discussing the limitations of the technologies that we have now, and how to get around those limitations. It's what Steve Mann does with his cyborg group at U of T every Friday afternoon, and I don't see why people on Slashdot can't contribute their thoughts and feelings too.

If you're not interested in indepth discussion of new ideas and technology, then go read CNN.
• #### Laser eye tap (Score:1)

You don't have to refocus your eyes. The laser eyetap device projects the image right on to your retina, so it has an infinite depth of field and is always in focus, no matter where you're looking.

• #### Anonymous Coward (Score:1)

Posting anon I see. Nuff said.
• #### Re:Organization Systems (Score:1)

You'd be absolutly amazed by how much info just sits in the history file. Got a URL to save? Dump it at the command prompt. Phone number to remember? Dump it at the command prompt. I'm not kidding here. :)
• #### Re:Just get rid of the keyboard... (Score:1)

The obvious answer is to use voice recognition.

A-and move to a big city, where you can stagger down the street talking to yourself and no one will think it's unusual.

"Seedy slash home slash george"
"man chat"
"finger"

George
• #### How do you scrub the hard disk (Score:3)

on Tuesday August 24, 1999 @06:00AM (#1728403) Homepage
on the undershirt cluster? Do you use fdisk or Tide?

George

It has occured to me that there is such a thing as "to much information".
Too much information [unimi.it], perhaps. Too much knowledge, doubtful - information is not knowledge [musicweb.net].

Our external storage - first cave paintings, then clay tablets, scrolls, books, recordings, and now computers - lets us store the information and put our brains to work on the knowledge.

We don't have to remember the endless details (wish I could find a link to Feynman's "map of the cat" story!) and can spend more time on the knowledge rather than the easily recorded and retrieved information.

In net discussions, I often pause to go seach for some little fact. I would love to be able to do this in real time conversations.

(For instance, can you imagine political debates where the candidates could instantly call up, say, the federal budget figures, or their opponent's voting record, or any statistic they needed? And the folks watching at home could instantly check them on it?)

• #### Eyeglasses (Score:1)

The reporter mentions a pair of eye glasses that one of the dudes from MIT has that look normal except for a glow from one lens but they don't show a picture of em.. perhaps if the reporter was wearing a web cam he could have took some decent pictures.
• #### Re:Go right to the source, and ask the horse... (Score:1)

cool thanks.. those glasses look kick-arse.. how long will it be before we get 640x480x256 resolution on a prescription lens with encrypted wireless transmission built in. :)

cool.. check this url: http://computer.org/conferen/proceed/8192/81920048 abs.htm
• #### Re:Just get rid of the keyboard... (Score:1)

What about a cyberglove that recognizes sign language?

... just a thought
• #### Re:Dammit (Score:1)

and I'm sick of looking for worthwhile comments on technology that I'm interested in and finding an abundance of complaints about slashdot. It's a free service that's incredibly easy to avoid completely or read selectively. A little respect never hurt anyone.
• #### You call that "modest"? (Score:1)

eh, Mike :)

I'd hardly call a cluster of 64 netwinders "modest"!

--

• #### Re:supercomputer in your armpit (Score:1)

I think the new iterations are pretty inobtrusive... note that Steve Mann's kind of glasses don't actually contain a display - they just have a laser that draws directly on your retina :)
--
• #### Re:Just get rid of the keyboard... (Score:1)

check out the the twiddler [handykey.com]
• #### Re:How do you scrub the hard disk (Score:1)

from the article:
The question remains: do we want to slip into our
computers the way we slip into our clothing?

for me that would mean slipping into my wearable computer at a dead run on the way to the subway.

and then someone would have to tell me:
hey man you misbuttoned your computer.
• #### Yup (Score:1)

Thad's wearable is unobtrusive from my point of view (I'm a student at Ga Tech), and pretty cool.

There are some interesting rumors about what he got to do with it over at MIT.

• #### Re:Dammit (Score:1)

But wearables are *cool*! There should be *more* stories on wearables, because dammit, that's how cool they are. Cool is good! Need more cool!

---
I'm not sure if I'm being facetious or not, either...
--
"HORSE."

• #### me 2 (Score:1)

Hi I'm Stu, I want to be a borg too.

He calls his computer a third hemisphere of his brain, capable of storing more data and crunching more numbers than his organic brain ever could.
It has occured to me that there is such a thing as "to much information". Though I get jazzed learning and aquiring new information, I think there is something to be said for ignorance. Maybe that means I'm not a geek/nerd after all. DS
• #### That looks like it might chafe... (Score:1)

Just be careful not to wear those boards over anything personal...pin-ends probably wouldn't feel too good on delightfully tender nipples.

I'll just wait for the skull-jacks, my self.
• #### Re:something to consider (Score:1)

Umm, you at least need the visual component of the wearable comp, and something to use as an input device.
Heh, and you should probably have a 300 disc CD changer hooked up to your home box, otherwise you might be stuck on the subway with the wrong CD in at home.

Kintanon
• #### IO (Score:1)

Really current IO is pretty clunky for this kind of proposition. To have to refocus your eyes or have a free hand in order to interface with a computer greatly reduces the utility of always-present devices.

Recently, speech recognition and synthesis have been approaching 'adequate' - HERE is a niche for these technologies.
• #### Re:Tech support for my underwear(underware?) (Score:1)

"Well, when I try to reboot my underwear, it just gives me a wedgie."

"What OS are you running on your underware, sir?"

"Microsoft Windows for Cotton Blends"

"Well, sir, the only solution is to change your underwear to MS-NTLongJohns"

"But it's the middle of August, and we're in the middle of a heat wave. Also, I don't think I need all of that fabric"

"I realize that Sir. But the NTLongJohns are the only version of MS-underpants what don't give you a wedgie. And you're much less likely to suffer the terrible blue-ball of death with these than with the MS-Underoos95. The NTLongJohns also come with a convenient back-orifice flap for examining core-dumps."

"Wow, that does sound... oh ohhhh ohhh S*&T!!!"

"Sir? Sir? Are you still there, Sir?"

"Je-ZUS! My underware were still connected to the phone line... I think my wife just faxed my willie to Canada...."

"Oh dear..."

"What the fsck?? Now my underwear's crashed and my nuts have turned blue!"

• #### Re:You call that "modest"? (Score:1)

Terry told me that Mann got $14M...$250G should be nothing... By the way, this is Lee...
• #### Re:Professor Mann is a freak (Score:1)

I've noticed that too... One time, the CEO of Nortel was giving a speech and he asked a really stupid, non-sequitur question. I could see several profs shaking their heads...

Genius takes on many, weird forms...

• #### Organization Systems (Score:1)

I have followed wearable computing for some time now. I look forward to when the MicroOpticals are available via retail.

What I'm more interested in is how to folks like Starner & Mann actually organize their data. They must collect and manipulate an enormous amount of data. I also know most wearable folks are Emacs fans. What are folks doing to keep all this info accessible?
• #### Powering wearables (Score:2)

...
it would be neat if the energy generated by walking and breathing could be used to power the wearable.
That seems easy enough, but you'd have to get used to having your clothing tight so that it could pull against your movements. Especially in warm places like SoCal, I could see this being unpopular. The alternative would be to have spandex-like clothes, which might be unsightly for many geeks.

I have given some thought to powering a laptop from muscle motion, and I concluded that reflexively bouncing your knee up and down (with your toe on the floor) could be tapped to generate enough power to run a modest computer. This does, however, makes it difficult to put a laptop on top of your lap. As long as you're walking around it would be trivial to generate some tens or hundreds of milliwatts from the flexing of the soles of your shoes; I put over a kilonewton of force on my foot when I step down, and 5 mm of motion is 5 joules right there. I don't see similar possibilities from breathing.

• #### Sign me up!! (Score:1)

I keep threatening to build one, but my wife either
a) doesn't want to be seen in public with a synth-pop cyborg
b) is wary of any project of mine that involves soldering irons
or c) all of the above.

Considering all the electronics I'd have to learn, this might be a good thing...
• #### Well.. (Score:1)

I use a soket 4 pentium 60 to comb my hair... does that count for anything?

bortbox

#### Related LinksTop of the: day, week, month.

When all else fails, read the instructions.

Working...


text:
# Question

You are given two objectives and two eyepieces and are instructed to make a telescope with them. The focal lengths of the objectives are 60.0 cm and 40.0 cm, and the focal lengths of the eyepieces are 0.90 cm and 0.80 cm.
(a) Which lens combination would you pick if you want to have maximum magnification? How about minimum magnification? Explain.
(b) Calculate the maximum and minimum magnifications.

Sales0
Views23

format: plain


text:
# Question #a5995

Jun 16, 2015

Furan is a heterocyclic organic compound.

#### Explanation:

Furan is simply a five-membered aromatic compound with the molecular formula ${C}_{4} {H}_{4} O$. More specifically, furan is a heterocycli compound because its ring contains two types of atoms - carbon and oxygen.

Furan's five-membered ring is actually comprised of four carbon atoms and one oxygen atom. The oxygen is bonded to the ring via two single bonds. Two of the four ring carbons are bonded via double bonds.

In addition to this, each carbon atom has one hydrogen atom attached. Furan's Lewis structure looks like this

Because the lone pairs of electrons are allylic, i.e. they are located next to the double bonds, furan can exhibit resonance

As you an see, one of oxygen's lone pairs of electrons is actually delocalized in the ring, which is why furan is considered an aromatic compound (I won't go into too much detail about why that is). This implies that the actual structure of furan can best be described using a hybrid model that looks like this


text:
# Can't solve a sequence (to determine if a given value is a member)

#### wishmaster

##### Active member
Can somebody help me with this problem? I have to proof if 41/81 is a part of attached sequence. Step by step guide would be useful. Thank you!

#### Amer

##### Active member
Re: Cant solve a sequence

Can somebody help me with this problem? I have to proof if 41/81 is a part of attached sequence. Step by step guide would be useful. Thank you!
$$a_n = \frac{n^2 +1}{2n^2}$$

$$\frac{n^2 +1}{2n^2} = \frac{n^2}{2n^2} + \frac{1}{2n^2} = \frac{1}{2} + \frac{1}{2n^2}$$

can we find "n" such that

$$\frac{1}{2} + \frac{1}{2n^2} = \frac{41}{81}$$ ?
solve it for n

#### wishmaster

##### Active member
Re: Cant solve a sequence

$$a_n = \frac{n^2 +1}{2n^2}$$

$$\frac{n^2 +1}{2n^2} = \frac{n^2}{2n^2} + \frac{1}{2n^2} = \frac{1}{2} + \frac{1}{2n^2}$$

can we find "n" such that

$$\frac{1}{2} + \frac{1}{2n^2} = \frac{41}{81}$$ ?
solve it for n
thank you,but i still dont get it.......

#### MarkFL

Staff member
Re: Cant solve a sequence

I would write:

$$\displaystyle \frac{n^2+1}{2n^2}=\frac{41}{81}$$

Now, cross-multiply and then see if you can find a natural number solution.

#### wishmaster

##### Active member
Re: Cant solve a sequence

I would write:

$$\displaystyle \frac{n^2+1}{2n^2}=\frac{41}{81}$$

Now, cross-multiply and then see if you can find a natural number solution.
Can you continiue?
And many thanks

#### MarkFL

Staff member
Re: Cant solve a sequence

Can you continiue?
And many thanks

What do you get when you cross-multiply? If you are unfamiliar with this method, if you have:

$$\displaystyle \frac{a}{b}=\frac{c}{d}$$

then cross-multiplying, which means to multiply the numerators on each side by the denominators on the other, and then equating both products, will give you:

$$\displaystyle ad=bc$$

It is a kind of shortcut for multiplying both sides by the common denominator of $bd$:

$$\displaystyle \frac{a}{b}bd=\frac{c}{d}bd$$

Canceling, or reducing the fractions, we get:

$$\displaystyle ad=bc$$

Can you do this with the equation I gave?

#### wishmaster

##### Active member
Re: Cant solve a sequence

What do you get when you cross-multiply? If you are unfamiliar with this method, if you have:

$$\displaystyle \frac{a}{b}=\frac{c}{d}$$

then cross-multiplying, which means to multiply the numerators on each side by the denominators on the other, and then equating both products, will give you:

$$\displaystyle ad=bc$$

If it a kind of shortcut for multiplying both sides by the common denominator of bd:

$$\displaystyle \frac{a}{b}bd=\frac{c}{d}bd$$

Canceling, or reducing the fractions, we get:

$$\displaystyle ad=bc$$

Can you do this with the equation I gave?
Yes,i understand this!

So i think thats my result: 81*(n^2+1) = 82(n^2)
So 41/81 is not a member of my sequence. Am i right?
And again,thank you!

#### MarkFL

Staff member
Re: Cant solve a sequence

You have the correct equation:

$$\displaystyle 81\left(n^2+1 \right)=41\left(2n^2 \right)$$

Now, distribute on both sides, then see if you can solve for $n$...

And no, I do not know of any online resources for studying sequences specifically.

#### wishmaster

##### Active member
Re: Cant solve a sequence

You have the correct equation:

$$\displaystyle 81\left(n^2+1 \right)=41\left(2n^2 \right)$$

Now, distribute on both sides, then see if you can solve for $n$...

And no, I do not know of any online resources for studying sequences specifically.
81n2+1 = 82n2
1 = 82n2-81nn
1= n2

I dont know i im right.....or what to do.....

#### MarkFL

Staff member
Re: Cant solve a sequence

81n2+1 = 82n2
1 = 82n2-81nn
1= n2

I dont know i im right.....or what to do.....
When you distribute the $81$ on the left side, you need to do so to both terms within the parentheses, as follows:

$$\displaystyle 81n^2+81=82n^2$$

Next, subtract $81n^2$ from both sides, and what are we left with?

#### wishmaster

##### Active member
Re: Cant solve a sequence

When you distribute the $81$ on the left side, you need to do so to both terms within the parentheses, as follows:

$$\displaystyle 81n^2+81=82n^2$$

Next, subtract $81n^2$ from both sides, and what are we left with?

81 = n2 ??

#### MarkFL

Staff member
Re: Cant solve a sequence

81 = n2 ??
Yes, good!

So, what is the positive root? What is the positive value of $n$?

#### wishmaster

##### Active member
Re: Cant solve a sequence

Yes, good!

So, what is the positive root? What is the positive value of $n$?
9 of course.

But what have i proved?

That 41/81 is a 9th element of sequence?

#### MarkFL

Staff member
Re: Cant solve a sequence

9 of course.

But what have i proved?

That 41/81 is a 9th element of sequence?
Yes, you have shown that:

$$\displaystyle a_{9}=\frac{41}{81}$$

#### wishmaster

##### Active member
Re: Cant solve a sequence

Yes, you have shown that:

$$\displaystyle a_{9}=\frac{41}{81}$$
Ok,thank you.

i still have three questions:

1. How can i explore the monotonoty of this sequence?
2. How to draw this sequence in coordinate system?
3. Can you show me another sequence solution(similar to mine)? Of course,if you have time.....

#### MarkFL

Staff member
Without using calculus, to explore the monotonicity of the sequence, I would write the $n$th term as suggested above by Amer:

$$\displaystyle a_{n}=\frac{1}{2}+\frac{1}{2n^2}$$

Can you see that each succeeding term is smaller than the last, and that as $n$ grows without bound $a_{n}$ asymptotically approaches $$\displaystyle \frac{1}{2}$$?

You could plot the first few terms, along with the asymptote.

You could make up another such problem just as easily as I can.

#### wishmaster

##### Active member
Without using calculus, to explore the monotonicity of the sequence, I would write the $n$th term as suggested above by Amer:

$$\displaystyle a_{n}=\frac{1}{2}+\frac{1}{2n^2}$$

Can you see that each succeeding term is smaller than the last, and that as $n$ grows without bound $a_{n}$ asymptotically approaches $$\displaystyle \frac{1}{2}$$?

You could plot the first few terms, along with the asymptote.

You could make up another such problem just as easily as I can.

Seems im to stupid for sequences......

#### MarkFL

Staff member
Seems im to stupid for sequences......
No, that's not the case at all...let's look at the $n$th term:

$$\displaystyle a_{n}=\frac{1}{2}+\frac{1}{2n^2}$$

And let's for now just look at this part of the $n$th term:

$$\displaystyle \frac{1}{2n^2}$$

Would you agree that when the numerator of a fraction is fixed, that is remains constant, and the denominator gets bigger and bigger, then the value of the fraction is getting smaller and smaller?

#### wishmaster

##### Active member
No, that's not the case at all...let's look at the $n$th term:

$$\displaystyle a_{n}=\frac{1}{2}+\frac{1}{2n^2}$$

And let's for now just look at this part of the $n$th term:

$$\displaystyle \frac{1}{2n^2}$$

Would you agree that when the numerator of a fraction is fixed, that is remains constant, and the denominator gets bigger and bigger, then the value of the fraction is getting smaller and smaller?
Yes,i agree!
I think that this sequence is limited too 0,5, but i dont know how to prove it.....

i have started a master degree with 38,my math was very good at high school,but now...now is all at higher level,i really dont know how would i master this all things......

#### MarkFL

Staff member
Yes,i agree!
I think that this sequence is limited too 0,5, but i dont know how to prove it.....

i have started a master degree with 38,my math was very good at high school,but now...now is all at higher level,i really dont know how would i master this all things......
What does $$\displaystyle \frac{1}{2n^2}$$ approach as $n$ grows without bound?

#### wishmaster

##### Active member
What does $$\displaystyle \frac{1}{2n^2}$$ approach as $n$ grows without bound?
1/2 ?

I really dont know how to get it......

And i really apreciate that you have time for me,thank you

#### MarkFL

Staff member
1/2 ?

I really dont know how to get it......

And i really apreciate that you have time for me,thank you
Let's look at the first few term of this fraction...let:

$$\displaystyle b_{n}=\frac{1}{2n^2}$$

We find:

$$\displaystyle b_1=\frac{1}{2(1)^2}=\frac{1}{2}$$

$$\displaystyle b_2=\frac{1}{2(2)^2}=\frac{1}{8}$$

$$\displaystyle b_3=\frac{1}{2(3)^2}=\frac{1}{18}$$

$$\displaystyle b_4=\frac{1}{2(4)^2}=\frac{1}{32}$$

$$\displaystyle \vdots$$

$$\displaystyle b_{1000}=\frac{1}{2(1000)^2}=\frac{1}{2000000}$$

$$\displaystyle \vdots$$

$$\displaystyle b_{1000000}=\frac{1}{2(1000000)^2}=\frac{1}{2000000000000}$$

Do you see that as $n$ gets bigger and bigger, $b_{n}$ gets smaller and smaller? If we could let $n$ go to infinity, then what would $b_{n}$ be?

#### wishmaster

##### Active member
Let's look at the first few term of this fraction...let:

$$\displaystyle b_{n}=\frac{1}{2n^2}$$

We find:

$$\displaystyle b_1=\frac{1}{2(1)^2}=\frac{1}{2}$$

$$\displaystyle b_2=\frac{1}{2(2)^2}=\frac{1}{8}$$

$$\displaystyle b_3=\frac{1}{2(3)^2}=\frac{1}{18}$$

$$\displaystyle b_4=\frac{1}{2(4)^2}=\frac{1}{32}$$

$$\displaystyle \vdots$$

$$\displaystyle b_{1000}=\frac{1}{2(1000)^2}=\frac{1}{2000000}$$

$$\displaystyle \vdots$$

$$\displaystyle b_{1000000}=\frac{1}{2(1000000)^2}=\frac{1}{2000000000000}$$

Do you see that as $n$ gets bigger and bigger, $b_{n}$ gets smaller and smaller? If we could let $n$ go to infinity, then what would $b_{n}$ be?
I think that bn would be zero,or very close.
but beacuse of + 1/2 bn cant be smaler then 0,5. Am i right?

And sorry,i have to learn to write fractions etc. in right form.

#### MarkFL

Staff member
Yes, $b_{n}$ gets closer and closer to zero, and since:

$$\displaystyle a_n=\frac{1}{2}+b_n$$

then $$\displaystyle a_{n}$$ gets closer and closer to $$\displaystyle \frac{1}{2}$$.

Does this make sense?

#### wishmaster

##### Active member
Yes, $b_{n}$ gets closer and closer to zero, and since:

$$\displaystyle a_n=\frac{1}{2}+b_n$$

then $$\displaystyle a_{n}$$ gets closer and closer to $$\displaystyle \frac{1}{2}$$.

Does this make sense?
Yes,it makes sense ,of course. Only problem si,how to write this in a proper way for my profesor?

And why is $$\displaystyle \frac{n^2+1}{2n^2}$$ equal to $$\displaystyle \frac{1}{2} + \frac{1}{2n^2}$$ ?


text:
## Precalculus (6th Edition) Blitzer

The solutions of the equation are $\left( -\frac{1}{2},\frac{1}{2} \right),\left( 2,8 \right)$
Let us consider the system of equations as \begin{align} & 3x-y=-2 \\ & 2{{x}^{2}}-y=0 \end{align}. Also, consider the equation $2{{x}^{2}}-y=0$: \begin{align} & 2{{x}^{2}}-y=0 \\ & y=2{{x}^{2}} \end{align} Now, put $y=2{{x}^{2}}$ in $3x-y=-2$ and compute the values as given below: \begin{align} & 3x-y=-2 \\ & 3x-2{{x}^{2}}=-2 \\ & 2{{x}^{2}}-3x-2=0 \\ & \left( 2x+1 \right)\left( x-2 \right)=0 \end{align} Solve further the equation: \begin{align} & \left( 2x+1 \right)\left( x-2 \right)=0 \\ & x=-\frac{1}{2},2 \end{align} Substitute the value $x=-\frac{1}{2}$ in the equation $y=2{{x}^{2}}$, to obtain the value of y: \begin{align} & y=2{{x}^{2}} \\ & y=2\times {{\left( -\frac{1}{2} \right)}^{2}} \\ & y=\frac{1}{2} \\ \end{align} Substitute the value $x=2$ in the equation $y=2{{x}^{2}}$, to obtain the value of y : \begin{align} & y=2{{x}^{2}} \\ & y=2\times {{2}^{2}} \\ & y=8 \\ \end{align} Thus, the solution of equations are $\left( -\frac{1}{2},\frac{1}{2} \right),\left( 2,8 \right)$


text:
09
ENE
2021

# disconnected graph dfs

Posted By :

But before returning, we have to check that min(arrival(a), arrival(b), arrival(c), arrival(d)) is less than the arrival(v). To view disconnected members, select a replicated folder from the Replicated folder list, and then expand the Disconnected Members. Why battery voltage is lower than system/alternator voltage. by a single edge, the vertices are called adjacent.. A graph is said to be connected if every pair of vertices in the graph is connected. Mark vertex uas gray (visited). You would get, [3, 5, 1, 4, 13, 2, 6, 17, 9, 11, 12, 10, 18]. Enter your email address to subscribe to new posts and receive notifications of new posts by email. Is it possible to know if subtraction of 2 points on the elliptic curve negative? On a graph of n vertices and m edges, this algorithm takes Θ(n + m), i.e., linear, time.. Uniqueness. Note on Graph Properties. There are a few things to note about how BFS and DFS work on graphs with different properties: BFS and DFS work on both directed and undirected graphs, as shown in the figures above.. Use the Queue. dep := a list of the size of the graph initialized with −1. Compare prices for Dfs Nyse Share Price And Dfs On Disconnected Graph You can order Dfs Nyse Share Price And Dfs On Disconnected Graph after check, compare the Dog likes walks, but is terrified of walk preparation. Then if there is an edge out of the sub tree rooted at v, it’s to something visited before v & therefore with a smaller arrival value. ... Now, from the main function call the function dfs(). Dfs Deferred Compensation And Dfs Disconnected Graph Earlier we have seen DFS where all the vertices in graph were connected. in the above disconnected graph technique is not possible as a few laws are not accessible so the following … When we say subtree rooted at v, we mean all v’s descendants including the vertex itself. Asking for help, clarification, or responding to other answers. So our goal is to petition the vertices into connected components. DFS(G, u) u.visited = true for each v ∈ G.Adj[u] if v.visited == false DFS(G,v) init() { For each u ∈ G u.visited = … For most algorithms boolean classification unvisited / visitedis quite enough, but we show general case here. The tree edges are solid and non-tree edges are dashed. Normally, running DFS (by taking the left-most node first) would stop after visiting node 6. rev 2021.1.8.38287, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, *vertex is the singular of vertices. Do NOT follow this link or you will be banned from the site. Thanks for contributing an answer to Mathematics Stack Exchange! DFS can be used to solve the connectivity problem. Create an unfilled stack ‘S’ and do DFS crossing of a diagram. /*take care for disconnected graph. Help modelling silicone baby fork (lumpy surfaces, lose of details, adjusting measurements of pins). All the vertices may not be reachable from a given vertex (example Disconnected graph). So, for above graph simple BFS will work. Why was there a "point of no return" in the Chernobyl series that ended in the meltdown? This array will help in avoiding going in loops and to make sure all the vertices are visited. Given G = (V, E) and all v in V are marked unvisited, a depth-first search (dfs) (generalisation of a pre-order traversal of tree)is one way of navigating through the graph. Algorithm for finding the longest path in a undirected weighted tree (positive weights). This is demonstrated below in C++, Java and Python: The time complexity of above solutions is O(n + m) where n is number of vertices and m is number of edges in the graph. Breadth First Search (BFS) If min (arrival (a), arrival (b), arrival (c), arrival (d)) is less than the arrival (v), then that means that at-least one back-edge or cross edge is going out of the sub tree rooted at v. If not, then we can stop the procedure and say that the graph is not strongly connected. for undirected graph there are two types of edge, span edge and back edge. DFS(v) returns min of arrival(a), arrival(b), arrival(c) and arrival(d). Moreover, a leaf is not an articulation point. Ultimately DFS is called once for each connected component, and each time it is called again from a new start vertex the componentID increments. How can a Z80 assembly program find out the address stored in the SP register? To learn more, see our tips on writing great answers. Given a directed graph, check if it is strongly connected or not. // flag to determine if graph is strongly connected. Use MathJax to format equations. You continue to run it on different components until the entire graph is "discovered". Two of them are bread-first search (BFS) and depth-first search (DFS), using which we will check whether there is a cycle in the given graph.. Detect Cycle in a Directed Graph using DFS. Now, the Simple BFS is applicable only when the graph is connected i.e. There are several algorithms to detect cycles in a graph. In a connected undirected graph, we begin traversal from any source node S and the complete graph network is visited during the traversal. Suppose there are four edges going out of sub-tree rooted at v to vertex a, b, c and d and with arrival time arrival(a), arrival(b), arrival(c) and arrival(d) respectively. In an undirected graph, a connected component is a set of vertices in a graph that are linked to each other by paths. This link should answer your question. Biconnected components v is a cut vertex if removing v makes G disconnected. How to implement an algorithm for specific kinds of search in a graph. Under any case, it does not take longer than $V+E$. The degree of a vertex in a directed graph is the same,but we distinguish between in- degree and out-degree. How to find connected components using DFS? Call DFS once for each unvisited vertex so far, with a parameter passed to keep track of the connected component associated with vertices reachable from the given start vertex. If The Graph Is Disconnected, Your Algorithm Will Need To Display The Connected Components. all vertices of the graph are accessible from one node of the graph. whether the resulting graph is still connected or not (say by DFS). Algorithm L for computing lowpoint numbers: Do a DFS on the graph starting from an arbitrary vertex called v 0. It starts at a given vertex (any arbitrary vertex) and explores it and visit the any of one which is connected to the current vertex and start exploring it. The idea is to traverse the graph along a particular route and check if the vertices of that route form a loop. Test Your Algorithm With Your Own Sample Graph Implemented As Either An Adjacency List Or An Adjacency Matrix. Making statements based on opinion; back them up with references or personal experience. if none of the edges are connected, then you will simply run DFS on every vertice until you discover your graph is disconnected. if none of the edges are connected, then you will simply run DFS on every vertice until you discover your graph is disconnected. Remember for a back edge or cross edge u -> v,arrival[u] > arrival[v]. Now re-run DFS. In previous post, we have discussed a solution for that requires two DFS traversals of a Graph. Please note that O(m) may vary between O(1) and O(n2), depending on how dense the graph is. When a microwave oven stops, why are unpopped kernels very hot and popped kernels not hot? So let's look at the implementation. All vertices are reachable. Forward edge cannot be going out of the sub tree as they can only be coming in to the sub tree or if it starts from within the sub tree it will go within the sub tree only. The edges which are going out of the sub tree will either be a back edge or a cross edge. # Do DFS traversal starting from first vertex. If the edge is removed, the graph becomes disconnected. Barrel Adjuster Strategy - What's the best way to use barrel adjusters? it is still set to arrival time of, # vertex v, the graph is not strongly connected, # Check if given Graph is Strongly Connected or not, # List of graph edges as per above diagram, # flag to determine if graph is strongly connected or not. // If DFS traversal doesn’t visit all vertices, // Factory method for creating a Edge immutable instance, // A List of Lists to represent an adjacency list, // terminate the search if graph is not strongly connected, // List of graph edges as per above diagram, // flag to determine if graph is strongly connected or not, # A List of Lists to represent an adjacency list, # Perform DFS on graph starting from vertex v, # terminate the search if graph is not strongly connected, # initialize list to arrival time of vertex v, # If the vertex is w is already discovered, that means there is, # either a cross edge or a back edge starting from v. Note that, # the arrival time is already defined for w, # if v is not root node and value of list didn't, # change i.e. if two nodes exist in the graph such that there is no edge in between those nodes. In fact, DFS is often used to determine whether or not a graph is disconnected or not - if we run DFS and do not reach all of the nodes in the graph, the graph must be disconnected. How to apply DFS on a disconnected graph. A directed graphs is said to be strongly connected if every vertex is reachable from every other vertex. In this article, we will extend the solution for the disconnected graph. // Do DFS traversal starting from first vertex. However, the BFS traversal for … A more elegant algorithm always starts at simple ob-servations. When we visit a We can say that the graph is strongly connected if and only if for every edge u->v in the graph, there is at-least one back-edge or cross-edge that is going out of subtree rooted at v. We can modify DFS such that DFS(v) returns the smallest arrival time to which there is an out-edge from the sub tree rooted at v. For example, let arrival(v) be the arrival time of vertex v in the DFS. A path from u to v is and (u,w1)(w1,w2)(w2,w3)…(w This is because the graph might have two different disconnected parts so to make sure that we cover every vertex, we can also run the DFS algorithm on every node. The DFS numbers are shown on each vertex, and the lowpoint numbers are shown in parentheses. What is the policy on publishing work in academia that may have already been done (but not published) in industry/military? Arrival and Departure Time of Vertices in DFS, Types of edges involved in DFS and relation between them. The visiting order that you describe, [5, 1, 4, 13, 2, 6, 17, 9, 11, 12, 10, 18], would happen if the two trees where connected through a root. How can I keep improving after my first 30km ride? DFS starts in arbitrary vertex and runs as follows: 1. March 11, 2018 by Sumit Jain. Repair the topology by performing any of the following procedures, as appropriate: Here’s simple Program for traversing a directed graph through Breadth First Search (BFS), visiting all vertices that are reachable or not reachable from start vertex. select each unvisited vertex w adjacent to v - dfs(w) (recursive!) Under any case, it does not take longer than $V+E$. For every unmarked vertex, we'rere going to run DFS to … You continue to run it on different components until the entire graph is "discovered". re := 0. dfs(0, −1, 0) return re. The BFS traversal of the graph above gives: 0 1 2 5 3 4 6. August 31, 2019. If the root has two or more children, it is an ar-ticulation point. A disconnected graph…. Create a boolean array, mark the vertex true in the array once visited. "Vertice" is not a word. It only takes a minute to sign up. Question: Write And Implement An Algorithm In Java That Modifies The DFS Algorithm Covered In Class To Check If A Graph Is Connected Or Disconnected. Hope that helps! span edge construct spanning tree and back edge connect two node in the same chain(lca of two node is one of them) forms a cycle. MathJax reference. DFS can be used to solve the connectivity problem. Imagine a new node (let's call it 3) which is the parent of 5 and 17. If you use DFS for path-finding reasons, then it makes no sense to try to connect the two components. Write a C Program to implement DFS Algorithm for Connected Graph. Breadth first Search (BFS) traversal for Disconnected Directed Graph is slightly different from BFS traversal for Connected undirected graph. I was wondering how to go about solving a problem with disconnected graphs and depth-first search. site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. Graph – Depth First Search in Disconnected Graph. (14 votes, average: 4.71 out of 5)Loading... You need to spend more on advertising, many people don’t know about these blogs.Such good content should reach everyone. How true is this observation concerning battle? We can check if graph is strongly connected or not by doing only one DFS traversal of the graph. In DFS, each vertex has three possible colors representing its state: white: vertex is unvisited; gray: vertex is in progress; black: DFS has finished processing the vertex. A graph is said to be disconnected if it is not connected, i.e. Following is definite Kosaraju’s calculation. The running time is . Here is an example of a disconnected graph. they are not connected. Consider the example given in the diagram. DFS from e Characterizing cut vertices: Claim The root is … Degree = in-degree + out-degree. Reference: Dr. Naveen garg, IIT-D (Lecture – 30 Applications of DFS in Directed Graphs). And so what we're going to do is for a general graph. Let us take a look at the article to understand the directed graph with strongly connected components. The gure below shows a graph which has been explored by DFS. Should the stipend be paid if working remotely? Depth First Search is a traversing or searching algorithm in tree/graph data structure.The concept of backtracking we use to find out the DFS. Here’s simple Program for traversing a directed graph through Depth First Search(DFS), visiting only those vertices that are reachable from start vertex. Now the DFS cannot send it to any other node hence, it moves out of the DFS () to the parent function which is connected components (). Why would the ages on a 1877 Marriage Certificate be so wrong? select one v in V and mark as visited. Piano notation for student unable to access written and spoken language. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Then you can visit (and apply any transformations on) all nodes just by traversing that list or by using the integers successively to refer to all of your nodes. // construct a vector of vectors to represent an adjacency list, // resize the vector to N elements of type vector, // Perform DFS on graph starting from vertex v, // terminate the search if graph is not strongly, // initialize arr to arrival time of vertex v. // If the vertex is w is already discovered, // that means there is either a cross edge, // or a back edge starting from v. Note that, // the arrival time is already defined for w, // if v is not root node and value of arr didn't, // change i.e. While (any unvisited vertex exist) Add the vertex to the queue. Dfs Deferred Compensation And Dfs Disconnected Graph. When we do a DFS from a vertex v in a directed graph, there could be many edges going out of its sub tree. Description Additional Information Reviews(1). We look at their four arrival times & consider the smallest among them and that will be the value returned by DFS(v). Disconnected graph is a Graph in which one or more nodes are not the endpoints of the graph i.e. If a topological sort has the property that all pairs of consecutive vertices in the sorted order are connected by edges, then these edges form a directed Hamiltonian path in the DAG.If a Hamiltonian path exists, the topological sort order is unique; no other order respects the edges of the path. NB. If you use DFS for traversal reasons, say because you want to make some transformation to each node of the graph, since node 3 is a superficial one that you added, you have to handle that node exceptionally. If min(arrival(a), arrival(b), arrival(c), arrival(d)) is less than the arrival(v), then that means that at-least one back-edge or cross edge is going out of the sub tree rooted at v. If not, then we can stop the procedure and say that the graph is not strongly connected. Recall: DFS to nd 2-connected components This graph is con-nected but removing one vertex b or e dis-connects it. Initially all vertices are white (unvisited). it is still set to arrival time of, // vertex v, the graph is not strongly connected, // Check if given Graph is Strongly Connected or not, // vector of graph edges as per above diagram. i.e. BFS Algorithm for Disconnected Graph Write a C Program to implement BFS Algorithm for Disconnected Graph. // array to store arrival time of vertex. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Why do electrons jump back after absorbing energy and moving to a higher energy level? What is the right and effective way to tell a child not to vandalize things in public places? Along a particular route and check if the vertices of the sub tree will Either be a edge... Degree and out-degree stops, why are unpopped kernels very hot and popped kernels not hot [ v ] function! Best way to use BFS or DFS to … BFS is used as a traversal algorithm graph! The degreeof a vertex, we'rere going to use barrel adjusters ( recursive! in related fields it no. L for computing lowpoint numbers are shown on each vertex, and expand! For nearby vertices of a graph which has been disconnected graph dfs by DFS.... Between those nodes the two components of edges involved in DFS crossing, subsequent to calling recursive for!: DFS to nd 2-connected components this graph is strongly connected or not say. Do the depth first Search is a set of vertices in graph were connected can check if graph is to! Vertex and runs as follows: 1 left-most node first ) would stop after node! Is applicable only when the graph is strongly connected or not by disconnected graph dfs only DFS. Only one DFS traversal of the graph to make sure all the vertices of the along. To solve the connectivity in a graph which has been explored by DFS the connected components nd 2-connected this! Exist in the graph becomes disconnected ( ) for every vertex is reachable from every other vertex more,. $V+E$ link or you will simply run DFS on every vertice until you discover graph. To connect the two components assembly Program find out the address stored in the graph initialized −1! By DFS two nodes exist in the graph is the parent of 5 and.! From the replicated folder list, and the lowpoint numbers are shown in parentheses to new posts and receive of... Was there a  point of no return '' in the Chernobyl series that ended the. Vertices may not be reachable from a given source vertex solve the connectivity in graph! If graph is the same, but we show general case here . Adjacency Matrix the longest path in a connected undirected graph with disconnected graphs depth-first! Enough, but we distinguish between in- degree and out-degree but if use... Push the vertex adjusting measurements of pins ) subscribe to this RSS,... Determine the connectivity in a directed graph, a connected undirected graph disconnected graph dfs! Seen DFS where all the vertices may not be reachable from every other vertex DFS traversal such! Or DFS to … BFS is applicable only when the graph such that there is no edge between... Are accessible from one node of the graph such that there is no in. About solving a problem with disconnected graphs and depth-first Search to vandalize things in places. Remember for a back edge into your RSS reader back edge or cross.! Positive weights ) or responding to other answers sub tree will Either be a back.! First traversal back edge the degreeof a vertex in an undirected graph there are several algorithms to cycles... V in v and mark as visited, clarification, or responding to other answers out-degree. Will be banned from the site children, it is strongly connected every. ( recursive! the vertices are visited on, we begin traversal from source. Little tricky but if you use DFS in marking show general case.! Dfs where all the vertices are visited cookie policy the edge is removed, the simple BFS work. Unmarked vertex, we'rere going to run DFS on every vertice until you discover your graph disconnected graph dfs disconnected do. First Search ( BFS ) if the root has two or more vertices are visited other by paths visited! Shown in parentheses a replicated folder from the site or more vertices are disconnected, do depth... Every vertice until you discover your graph is  discovered '' do electrons jump back after absorbing energy moving!, see our tips on writing great answers visited during the traversal general graph for computing lowpoint numbers are on... Parent of 5 and 17 > arrival [ u ] > disconnected graph dfs [ u ] > [! Tree ( positive weights ), push the vertex true in the Chernobyl series that ended in meltdown... Problem with disconnected graphs and depth-first Search to handle disconnected graph do is for a back or. How can a Z80 assembly Program find out the DFS not connected, i.e it in disconnected graph ) we. To view disconnected members, select a replicated folder from the site user contributions licensed under cc by-sa Inc!, −1, 0 ) return re  point of no return '' the... To solve the connectivity in a non-connected graph will Either be a disconnected graph dfs edge or cross edge u - v! Once visited will Either be a back edge disconnected graph dfs cross edge ) would stop after visiting 6! Which one or more children, it does not take longer than $V+E$ first ) would stop visiting. Is used as a traversal algorithm for graph first Search ( BFS traversal! Wondering how to use DFS in marking form a loop first Search ( BFS traversal... Connect the two components it possible to know if subtraction of 2 points on the elliptic curve negative leave/enter... … BFS is applicable only when the graph such that there is no in... To new posts by email paste this URL into your RSS reader every other vertex an arbitrary vertex v... To determine if graph is strongly connected or not by doing only one DFS traversal of the is... References or personal experience on opinion ; back them up with references or personal.! And cookie policy for graph 's call it 3 ) which is the policy publishing. Subsequent to calling recursive DFS for path-finding reasons, then it is strongly connected or not by doing one. Running DFS ( w ) ( recursive! up with references or personal experience and then expand the members! // flag to determine the connectivity problem structure.The concept of backtracking we use to find out the numbers... Time of vertices in a graph that are linked to each other by paths a microwave oven stops, are... Modelling silicone baby fork ( lumpy surfaces, lose of details, adjusting measurements of )... Distinguish between in- degree and out-degree discovered '' follows: 1 and Departure of! Exchange Inc ; user contributions licensed under cc by-sa to connect the two components kinds of Search in a graph... Display the connected components Need to Display the connected components 2021 Stack is... – 30 Applications of DFS in marking, privacy policy and cookie policy,... Tips on writing great answers to nd 2-connected components this graph is connected. Graph becomes disconnected will help in avoiding going in loops and to make all! Recall: DFS to determine the connectivity problem v ’ S descendants including the vertex Stack... Graphs ) 's the best way to use BFS or DFS to … BFS is as... Policy and cookie policy makes G disconnected is strongly connected or not doing. To learn more, see our tips on writing great answers to learn more, our. A new node ( let 's call it 3 ) which is the policy publishing! In parentheses algorithm in tree/graph data structure.The concept of backtracking we use to find the... … BFS is applicable only when the graph use to find out the address stored in graph! List or an Adjacency list or an Adjacency list or an Adjacency Matrix have already been (... Removed, the simple BFS will work array will help in avoiding going in loops and to make sure the! For student unable to access written and spoken language leaf is not an articulation.. Is connected i.e for that requires two DFS traversals of a graph are. Requires two DFS traversals of a vertex in a connected undirected graph, check if graph is con-nected removing. Members, select a replicated folder list, and then expand the disconnected members discovered.. Illustration for an undirected graph is strongly connected which are going out the. There are several algorithms to detect cycles in a graph in which one or more vertices are disconnected do... All vertices of that route form a loop Write a C Program to implement an algorithm specific... Remember for a general graph receive notifications of new posts by disconnected graph dfs there no! By taking the left-most node first ) would stop after visiting node 6 disconnected graph dfs understand BFS then it no... Vertices are disconnected, your algorithm will Need to Display the connected components solution... Complete DFS traversal of such graphs, we have seen DFS where all vertices! Search is a set of vertices in a graph that are linked to each other paths... Kinds of Search in disconnected graph dfs graph in which one or more children, it does take... To mathematics Stack Exchange under any case, it does not take longer than \$ V+E.! Studying math at any level and professionals in related fields how can i keep improving after my 30km! This link or you will simply run DFS to … BFS is applicable only when the graph along a route... So what we 're going to run it on different components until entire... How can i keep improving after my first 30km ride along a particular route check. We'Rere going to use barrel adjusters ) would stop after visiting node 6 traverses only the vertices DFS... Stop after visiting node 6 IIT-D ( Lecture – 30 Applications of in... And out-degree Either be a back edge or cross edge u - >,.


text:
Heisenberg-limited phase estimation without adaptive measts

# Demonstrating Heisenberg-limited unambiguous phase estimation without adaptive measurements

## Abstract

We derive, and experimentally demonstrate, an interferometric scheme for unambiguous phase estimation with precision scaling at the Heisenberg limit that does not require adaptive measurements. That is, with no prior knowledge of the phase, we can obtain an estimate of the phase with a standard deviation that is only a small constant factor larger than the minimum physically allowed value. Our scheme resolves the phase ambiguity that exists when multiple passes through a phase shift, or NOON states, are used to obtain improved phase resolution. Like a recently introduced adaptive technique [Higgins et al2007 Nature 450 393], our experiment uses multiple applications of the phase shift on single photons. By not requiring adaptive measurements, but rather using a predetermined measurement sequence, the present scheme is both conceptually simpler and significantly easier to implement. Additionally, we demonstrate a simplified adaptive scheme that also surpasses the standard quantum limit for single passes.

###### pacs:
03.67.-a, 42.50.St, 03.65.Ta

,

## 1 Introduction

The interferometric measurement of an optical phase is an indispensable tool in metrology, used in length measurement, optical characterization and a myriad of other applications. Phase measurement is closely related to the more general concept of quantum parameter estimation, which is important for characterizing Hamiltonians for applications in quantum information technology, for example. In practice, there exist many different types of phase measurement problems, each with its own set of requirements and limitations, which has led to a diverse range of interferometric techniques.

In this paper, we consider the estimation of a completely unknown phase in the situation where the total number of photon-passes through the system being probed is the crucial resource, and there is no inherent limitation to the number of times each photon can pass through the phase shift [1, 2, 3, 4] (see Appendix). This corresponds to the case of a single measurement of a fixed phase. The number of photon-passes is important because, for example, the light may cause damage to a sensitive sample, or each sampling of the phase may be difficult to achieve (such as in probing the strength of a Hamiltonian coupling for a quantum information application [5]). In this regime, the aim is to obtain an estimate with minimal uncertainty for a given number of applications, , of the phase shift [3, 4]. In the nomenclature of quantum information, this restriction equates to having a limited number of queries of the system.

The regime we consider may be contrasted with the “phase sensing” regime, such as would be found in the context of gravity-wave detection or atomic clocks. In that case the aim is to measure changes in a time-varying phase. There is an inherent limitation to the number of passes of each photon, because additional passes will eventually lead to less accurate results due to the varying phase. In addition, because the phase shifts are small, the phase is already approximately known, allowing any phase ambiguities to be resolved. In that regime, particular techniques and states of light, such as -photon path-entangled “NOON” states [6, 7], are desirable due to their improved phase resolution from a single pass through the sample. The phase ambiguity in the NOON state (which only allows the phase to be estimated modulo ) is resolved using the pre-existing approximate knowledge of the phase.

In phase measurement, there are two well-known limits to measurement accuracy. One is the standard quantum limit (SQL)1, which is the limit obtained by “standard” techniques, such as coherent states and photon counters. The other is the Heisenberg limit, which is the fundamental limit using arbitrary quantum states and measurements. When there is an inherent limitation to the number of passes that may be used, the SQL and Heisenberg limit are and , respectively, where is the number of photons [8, 9]. To attain the Heisenberg limit in this case it is necessary to create path-entangled -photon states, such as NOON states [6, 7] or other non-classical states [8, 9, 10, 11, 12, 13, 14]. However, such states are very difficult to produce experimentally even for moderate  [15, 16, 17, 18, 19, 20, 21, 22].

In this paper, we follow some recent papers [1, 2] in using the term Heisenberg limit, as defined above, for the general situation, regardless of whether there is a limitation on the number of passes of any individual photon. In terms of the total number of photon-passes, , the Heisenberg limit is  [2, 3]. In order to give a meaningful basis for comparison, we take the SQL to be the maximum accuracy possible with classical states and single passes, which is . When there is no inherent limitation to the number of passes for each photon, interference fringes with a width scaling as may be obtained simply by using passes of a single photon. The difficulty is that this also yields an ambiguous phase estimate, in exactly the same way as using NOON states. When there is no additional knowledge of the phase, resolving this ambiguity is a major problem.

Recently, a number of approaches have been proposed to resolve the above ambiguity, both in the context of phase estimation [1, 3] and in related contexts [2, 23, 24]. These are closely related to the quantum phase estimation algorithm (QPEA) of Cleve et al [25, 26] (which is the heart of Shor’s factorization algorithm [27]), as well as Kitaev’s phase estimation algorithm [28]. None of these protocols were rigorously shown, either numerically or analytically, to give Heisenberg-limited scaling for the standard deviation of the phase estimate (see section 2). Another scheme, not related to the QPEA, was shown numerically to be able to achieve Heisenberg-limited operation [29].

Heisenberg-limited phase estimation was realized experimentally for the first time in [30], using a QPEA-inspired adaptive phase estimation scheme proposed in that work. That this protocol does achieve Heisenberg-limited scaling was also shown using rigorous numerics. In fact, two schemes were considered in [30]. Firstly, the QPEA itself was implemented, showing that this scheme fails to achieve a phase uncertainty below due to the high wings on the distribution. Secondly, a generalization of the QPEA2 was presented and demonstrated to achieve an uncertainty within a factor of of the Heisenberg limit, which approaches for large  [11, 13, 31, 32]. Subsequently, a different QPEA-inspired adaptive phase estimation scheme was analytically shown to give Heisenberg-limited scaling by Boixo and Somma [4].

Higgins et al [30] demonstrated the QPEA and its generalization using multiple passes of a single photon through the phase shift. However they also noted that the same schemes could instead be implemented equivalently using single passes of NOON states. An advantage of using NOON states is that the time taken for a measurement does not scale with , thus allowing the Heisenberg limit to be attained in the phase sensing regime. A disadvantage of using NOON states, or related multiphoton entangled states, is that the size of the state scales as . Another disadvantage is that the NOON-state protocols require photon-resolving detectors with inefficiency at most . By contrast the multipass protocols can achieve Heisenberg-limited scaling for any non-zero detection efficiency. See the end of section 5 for further discussion.

Here we present and experimentally demonstrate two new phase estimation schemes that offer significant simplifications compared with the generalized QPEA of [30], while still giving accuracy better than . Both can be applied to either NOON states or multiple passes of single photons, as with the QPEA and its generalization. As in [30], we have performed the experimental demonstration using multiple passes of single photons, as shown in figure 1. Whereas the Heisenberg-limited scaling of the generalized QPEA of [30] was shown numerically, we here provide analytical proofs bounding the scaling of our schemes.

The first scheme, which we shall call the QPEA–standard hybrid or simply hybrid scheme, combines the original QPEA with standard (single-photon, single-pass) interferometry to achieve a phase uncertainty or better. The second, non-adaptive scheme uses a predetermined sequence of single-photon measurements. Remarkably, this non-adaptive scheme achieves a phase uncertainty , although with a higher overhead than the adaptive scheme of [30]. As well as being of fundamental interest, the reduction in complexity offered by these new schemes makes the practical adoption of multipass phase estimation techniques more attractive.

## 2 Existing schemes

In order to compare the performance of phase estimation schemes, it is necessary to define an appropriate measure of uncertainty. Uncertainty measures are not interchangeable. For example it may be the case that although an accurate phase estimate is obtained the majority of the time, a small proportion of measurement errors are large [13]. This will lead to a large standard deviation in the phase estimate, although other measures (e.g. a confidence interval) may fail to capture this uncertainty. We use the standard deviation to unambiguously quantify phase uncertainty—if the standard deviation of the phase estimate scales as , so will other measures of uncertainty [33]. The converse is not generally true.

When considering a single measurement of an unknown phase, it is necessary for the measurement procedure to be self-contained in order to fairly evaluate it. That is, the procedure should not use additional phase information from some other source without counting the resources used to obtain that information. There are two main issues to be considered here:

1. NOON states or multiple passes through a phase shift provide an ambiguous phase estimate, and extra measurement steps are required to resolve the ambiguity.

2. Additional information about the phase may also be useful during the measurement procedure to design an optimal or near-optimal measurement, as is the case for measurements based on squeezed states, for example [10].

One approach to obtain a self-contained, unambiguous measurement is to use the QPEA [25, 26]. The QPEA involves two steps. In the encoding step, qubits, each originally in state , control the application of a sequence of phase shifts . This prepares the state , where is a state in the logical basis, and . In the measurement step, the inverse quantum Fourier transform is used to perform a measurement in an entangled basis on the encoded state. This measurement is the optimal one for obtaining an estimate of the unknown phase encoded in the state [26, 25]. Remarkably, the measurement step can be implemented without entangling gates by using a Markovian sequence of local adaptive measurements on each of the qubits. At each step the measurement basis depends only on the previous measurement and its result [34].

Although the measurement step is optimal for phase estimation, the encoding step is not. As a consequence, the phase uncertainty (where is the Holevo variance [32]) is dominated by the high wings of the distribution of phase estimates, and the QPEA does not achieve a phase estimate with uncertainty scaling at the Heisenberg limit . Rather, the uncertainty actually scales as  [30].

To achieve scaling at the Heisenberg limit, the QPEA can be modified by repeating each application of the phase shift operator , , on each of independently prepared qubits [30]. Bayesian adaptation (that is, controlling the measurement made on each qubit based on the distribution for conditioned on all preceding measurements [13]) can replace the single qubit measurements to achieve scaling for .

As explained above, a phase shift of can be achieved by using an interferometer and either a single photon with passes through the unknown phase shift, or a NOON state with photons and a single pass. In the case of multiple passes, the probability distribution for the phase based on the measurement is , where the sign depends on which output the photon is detected in. In the case of NOON states, all information about is again contained in a binary outcome , this time corresponding to the parity of the photon number at one of the outputs. Moreover, the probability distribution for the phase is identical to that obtained in the single-photon case. It is evident that both cases only yield information about modulo ; as explained, the role of the QPEA, or the generalized QPEA scheme in [30], is to resolve this phase ambiguity. The experimental demonstration in [30] was achieved with multiple passes.

The total resources used is quantified by the number of applications of the phase shift, and is . When using multiple passes of a photon this is equal to the total number of photon-passes, and it is equal to the total photon number for the equivalent scheme using NOON states. As discussed in [3] and the Appendix, this is the appropriate way to quantify resources for problems of this type.

## 3 QPEA–standard hybrid scheme

For the Bayesian adaptive scheme of [30] reduces to the QPEA. However, as noted above, the control protocol in the case has a much simpler formulation as a Markovian scheme. If the problem of outliers of (due to the high wings of the distribution) could be solved, the QPEA would be an attractive scheme due to its simplicity.

Here we present a solution, resulting in an uncertainty that beats the SQL, in which the QPEA is supplemented by some additional simple measurements. These additional measurements are those of standard interferometry, comprising single phase shift applications on many single qubits without adaptive measurement. A related scheme was explored in [29], using standard interferometry to supplement adaptive NOON state inputs with adaptive measurements; numerical simulations showed Heisenberg-limited scaling with an overhead factor of  [29]. We now analytically determine the scaling for the phase uncertainty using the adaptive scheme proposed above.

Consider using resources to implement a standard phase measurement, in addition to the QPEA using resources. First we consider the uncertainty in the standard phase measurement, and for simplicity alternate the controllable phase shift (see figure 1) between and . There are thus two sets of measurement results governed by probabilities and respectively. For even , we obtain estimates of and , denoted and , each based on measurements. The estimate of is then determined from the phase of . Using the Chernoff bound [35] (or Hoeffding’s inequality [36]), the probability that the errors in these estimates exceed are bounded as

 P(|νk−Pk|≥ϵ)≤2exp(−2NSϵ2), (1)

for . The probability that either of the estimates differs from the by at least is then upper-bounded by . Using simple geometry, it is easily seen that if , then the error in the angle can not exceed . Let us take

 ϵ=√3f(NS)/32NS, (2)

where satisfies for . With this choice, we find that

 arcsin(2√2ϵ) =arcsin(√3f(NS)/4NS) (3) ≤(π/3)√f(NS)/NS.

In the second line we have used the fact that , so , and the convexity of the function over the range . Therefore, the probability of the phase error being at least is no greater than .

We now wish to place bounds on the phase uncertainty. We quantify the phase uncertainty by the square root of the Holevo variance,  [37], where

 Missing or unrecognized delimiter for \right (4)

When small, the Holevo variance is well approximated by [13]

 V≡4⟨sin2[(ϕ−ϕest)/2]⟩. (5)

To place an upper bound on this variance, we consider a suboptimal method of determining the phase estimate. Consider combining the standard phase estimate, and the QPEA phase estimate, , by using when the estimates differ by less than , and using if the two estimates differ by more than . There are then three different alternatives which need to be considered.

1. If the standard phase estimate is inaccurate, in the sense that , then the contribution to can not exceed . Choosing ensures that this contribution is .

2. For the case where the standard phase estimate is accurate (), if the QPEA estimate differs from the standard estimate (so ), then the QPEA estimate must be inaccurate (so ). Provided the QPEA uses a randomized initial setting of the controllable phase, the probability distribution is

 Missing or unrecognized delimiter for \right (6)

Using this probability distribution, the probability of being greater than does not exceed , and the overall contribution to then does not exceed

 (1/NQ)√f(NS)/NS. (7)
3. Next consider the case that the standard phase estimate is again accurate () and the standard and QPEA estimates agree (so ). This means that the QPEA estimate must be accurate, in the sense that . With this limit, we find that the possible contribution to is bounded by

 (2/NQ)√f(NS)/NS+2/(πN2Q). (8)

Collecting together all three cases gives the upper bound on as

 V≤3NQ√f(NS)NS+\Or(N−2S)+\Or(N−2Q). (9)

Ignoring the slow variation of , the minimum of (9) as a function of will be obtained for and , which gives

 V=\Or(√lnNN−3/2). (10)

We have implemented the above hybrid scheme both experimentally (see below) and numerically. We do not consider special cases as in the above derivation, but rather determine the best estimate from the normalized product of the distributions conditioned by Bayesian updating. Also, we increment the controllable phase by after each measurement, rather than alternating between and . Numerically we find that choosing , so that , is close to the optimal choice, as expected. Moreover, we find that for this choice is nearly constant, also as expected (it increases from at to at ).

It is useful to contrast our hybrid scheme with the generalized QPEA of [30]. In the latter, (the number of independent iterations for each -fold application of the phase shift), did not depend on . In our hybrid scheme depends on , as . The hybrid scheme also differs in using mostly non-adaptive measurements ( non-adaptive measurements versus adaptive ones). These observations suggest that, by considering other dependencies , good schemes may be found that dispense with adaptivity altogether.

In the adaptive case, it is possible to obtain scaling at the Heisenberg limit for  [30]. In contrast, for non-adaptive measurements, using small values of such as these results in much larger variances that do not decrease below some fixed amount (see figure 2). Rather than there being a threshold value of above which Heisenberg scaling is obtained, these results suggest that must increase as a function of to avoid results with large error. But this can not yield Heisenberg scaling because it leads to an increasingly large overhead.

On the other hand, as we will now show, with a suitable choice of it is indeed possible to obtain scaling at the Heisenberg limit without adaptive measurements. Specifically, we choose , where is the value of used for the -fold application of the phase shift, and . The optimal performance is obtained by numerically optimizing both of these parameters, which we consider later. First, we prove rigorously that scaling at the Heisenberg limit is guaranteed for a particular choice of and .

Since our scheme is non-adaptive, it can be performed in any order; we consider using independently prepared qubits undergoing the -fold application of the phase shift in the sequence . Consider the results for ; applying the analysis using the Chernoff bound as in the hybrid scheme, for even choosing localizes the true phase to an open segment of angular size with probability at least .

Next, using this analysis on the measurements with a two-fold phase shift (), we see that the phase is localized to two diametrically opposite segments of size . As the separation of these segments is , and they are open, only one of them overlaps with the segment obtained for . Hence, combining the results of these two sets of measurements localizes the system phase to within a segment of size with high probability. This analysis is continued up to the -fold phase shift. At stage the phase is localized to a segment of size , with probability of error bounded above by .

Consider the maximum error for each case where the phase may lie outside the segment. The maximum error in the case for is , and occurs with probability no greater than . The contribution to therefore does not exceed . Then, for , the error can not exceed (for example, if the phase is outside the segment, the error can be as large as the size of the segment). The probability of this error does not exceed . Thus the contribution to from these errors does not exceed

 4\rme−3M(K,0)/16+4K∑k=1sin2[(2π/3)/2k]\rme−3M(K,k)/16. (11)

Finally, if the phase is correctly localized in the segment, the maximum error is . Including this contribution we obtain the upper bound on as

 V≤Vmax= 16\rme−3M(K,0)/16+(2π/3)22−2K (12) +16(2π/3)2K∑k=12−2k\rme−3M(K,k)/16.

The total number of phase shift applications is . It is evident that, for independent of and , , in agreement with the numerical results. If we allow to be a function of but not of , then the best scaling of is obtained for , which yields . To determine a variation of that does yield scaling at the Heisenberg limit, we may take , which gives . To satisfy this, should have the linear variation . With this choice, it is easy to show that

 Vmax≤(2π/3)2(1+32\rme−3MK/16)2−2K. (13)

Since , this proves that the phase uncertainty scales at the Heisenberg limit when we use linear variation of with . However, has a substantial overhead (about 54 for ).

The above result represents an upper bound on the phase variance. However, using numerical methods we find that the optimal values of and are smaller, and yield much smaller overhead. We performed numerical simulations (up to ), incrementing the controllable phase by after each measurement, instead of alternating between and . These indicate that values of and consistently and robustly give a low overhead, less than 2.03 times the Heisenberg limit. This is only marginally above the overhead of 1.56 for the adaptive scheme () of [30], and very close to the scaling constant for the adaptive scheme of [29].

## 5 Experiment

We demonstrate both the hybrid and non-adaptive schemes introduced above using photonic qubits in a common-spatial-mode polarization interferometer, as in figure 1, similar to that of [30]. As with the QPEA and its generalization, these new schemes may be equivalently performed either using multiple passes of single photons or single passes of NOON states—here we use the former. The two arms of the interferometer are the right- and left-circular polarization modes, with phase shifts applied using half-wave plates. Computer-controlled motorized stages adjust the configuration of passes through the system phase and the setting of the control phase in the other arm of the interferometer. Pairs of 820 nm single photons are generated by a type-I spontaneous parametric down-conversion source, pumped by a continuous-wave laser diode. One photon of the pair is detected immediately, the other is sent through the experiment, and a coincidence detection heralds a successful measurement.

The experimental data and numerical predictions for both schemes are shown in figure 3. We vary the number of resources by varying the maximum number of passes . The limited aperture of our optics (50 mm) lead to a limit of maximum 32 passes (maximum ). For this reason, the hybrid scheme had a maximum of 95, and the non-adaptive scheme (with and ) a maximum of 297.

The results for the hybrid scheme agree with our theoretical prediction of an asymptotic scaling of close to . This shows the interesting result that the SQL can be surpassed by combining two measurements that do not do so individually. The data for the non-adaptive scheme are also consistent with numerical predictions, clearly surpassing the SQL and exhibiting scaling at the Heisenberg limit . For the maximum number of resources used, the data demonstrate the predicted overhead factor of 1.91 (slightly below the asymptotic factor of 2.03) relative to the Heisenberg limit.

Finally, we make some observations about the role of photon loss in the techniques described. With either NOON states or multipass single photons, the interferometric scheme is relatively sensitive to loss in the sample [39], with efficiency , where is the efficiency for a pass of a single photon through the sample. However, the multipass scheme is less sensitive to loss at the detectors, where , as opposed to for NOON states. To reduce the effect of loss, it may be valuable to employ phase-sensitive states that are loss-tolerant, such as those described in [38, 39]. It should be possible to configure both adaptive [30] and non-adaptive schemes for use with these states.

## 6 Conclusions

When using an -photon NOON state, or passes of a single photon, in a two-mode interferometer with an unknown phase in one mode, the output probabilities vary sinusoidally with rather than . Although this gives phase resolution of in principle, it does not give a phase estimate with this accuracy, because it is sensitive only to changes modulo . This ambiguity poses a fundamental problem, because the accuracy of the prior information about the phase required to resolve the ambiguity is of the same order as the phase resolution obtained by the measurement. In this paper we have devised two simple yet highly efficient techniques for eliminating the phase ambiguity. For the second technique, we obtain an unambiguous phase estimate with accuracy scaling as in the total number of photon-passes . This is the Heisenberg limit scaling, in that it is the best possible under the restriction of the uncertainty principle for and . Moreover, we have experimentally demonstrated our schemes using multiple passes of single photons.

In contrast to the methods presented in [4, 30], the Heisenberg-limited scheme introduced here does not require adaptive measurements, but instead uses a predetermined sequence of measurements. Reducing or removing adaptive measurements should make practical implementation of precision phase estimation far easier. The crucial feature which enables scaling with nonadaptive measurements is the use of larger numbers of repetitions for the measurements with fewer passes. Using this, we have analytically proven that scaling is obtained. In contrast, it is not possible to obtain the Heisenberg limit scaling with nonadaptive measurements if the number of repetitions is held constant.

This work was supported by the Australian Research Council and Spanish MEC projects QOIT (Consolider-Ingenio 2010) and ILUMA (MWM).

## Appendix A Quantifying resources

In this work we quantify the resources by the total number of applications of the phase shift, regardless of whether they are applied in parallel (to different photons) or in series (repeatedly to the same photon). This allows the resources to be treated in a consistent way, yielding a theoretical limit scaling as  [3]. One might argue that the resources could be quantified by a pair of numbers: the number of photons and the number of passes. Then an -photon NOON state with a single pass and passes of a single photon would be regarded as using different resources. However, this distinction cannot be maintained rigorously, because the experiment can be designed such that exactly the same quantity is being measured in either case. We now demonstrate this fact.

Consider the case that the NOON state is detected by time-resolving photon counters. For a two-mode state of sufficient duration, the timings of the detections will typically be distinct. The time that each photon is detected can be used to determine the time that it passed through the phase shift. (For the experiment we consider superpositions of left- and right-circular polarization, so the photon always passes through the phase shift, in contrast to the case where different spatial modes are used.) Given a set of arrival-time data from a NOON-state experiment, one can design a single-photon experiment (using a set of mirrors and electro-optic switches), such that the single photon passes through the sample at the same times . In each case the phase shift is sampled in exactly the same way, by photons passing through at times . Because the photons are indistinguishable, it makes no difference whether it is photons in a NOON state, or the same photon passing times. Similar ideas have been explored in [1, 3, 4, 23, 24, 29]. Interestingly, it is not difficult to devise a series of thought experiments that interpolate between the NOON state experiment and the multiple-pass experiment. For example, consider the case where the photon passes through the sample times, but it is duplicated between passes. This duplication happens in a coherent way, so the state becomes , where 0 indicates left-circularly polarized, and 1 indicates right-circularly polarized (see figure 4). The interpretation is that the lower photon in the split is the “same” photon, so the photon passes times through the sample. On the other hand, because the photons are indistinguishable, we could equally interpret the upper photon in the split as the “same” photon. Then we have different photons, each passing through the phase shift once. This experiment can then be transformed to the NOON state experiment simply by moving all duplications of the photons to the beginning.

The difference between passes of a single photon, and single passes of different photons is simply due to the arbitrary interpretation of which photon is the “same” photon. This therefore makes it ambiguous as to how the resources should be quantified if the photons and passes are quantified as different resources. Another difficulty with that way of quantifying resources is that in a complicated measurement involving different photons with different numbers of passes, it is ambiguous as to which number of passes should be used. In contrast, the number of passes of single photons through the phase shift is always unambiguous, and has a rigorous mathematical foundation [3]. We emphasize that, although non-classical states and multiple passes may be regarded as equivalent, they present different experimental challenges, with different advantages and disadvantages.

## References

### Footnotes

1. In some areas this is called the shot-noise limit, with the term “standard quantum limit” reserved for a limit set by additional constraints which are not relevent to the situation we consider.
2. Note that in [30], the (generalized) QPEA was called the (generalized) Kitaev algorithm.

### References

1. Giovannetti V, Lloyd S and Maccone L 2006 Phys. Rev. Lett.96 010401
2. Boixo S, Caves C M, Datta A and Shaji A 2006 Laser Physics 16 1525–32
3. van Dam W, D’Ariano G M, Ekert A, Macchiavello C and Mosca M 2007 Phys. Rev. Lett.98 090501
4. Boixo S and Somma R D 2008 Phys. Rev.A 77 052320
5. Cole J H, Schirmer S G, Greentree A D, Wellard C J, Oi D K and Hollenberg L C 2005 Phys. Rev.A 71 062312
6. Bollinger J J, Itano W M, Wineland D J and Heinzen D J 1996 Phys. Rev.A 54 R4649–52
7. Lee H, Kok P and Dowling J P 2002 J. Mod. Opt. 49 2325–38
8. Yurke B, McCall S L and Klauder J R 1986 Phys. Rev.A 33 4033–54
9. Holland M J and Burnett K 1993 Phys. Rev. Lett.71, 1355–8
10. Caves C M 1981 Phys. Rev.D 23 1693–708
11. Summy G S and Pegg D T 1990 Opt. Comm. 77 75–9
12. Sanders B C and Milburn G J 1995 Phys. Rev. Lett.75 2944–7
13. Berry D W and Wiseman H M 2000 Phys. Rev. Lett.85 5098–101
14. Hofmann H F 2006 Phys. Rev.A 74 013808
15. Walther P, Pan J-W, Aspelmeyer M, Ursin R, Gasparoni S and Zeilinger A 2004 Nature 429 158–61
16. Mitchell M W, Lundeen J S and Steinberg A M 2004 Nature 429 161–4
17. Eisenberg H S, Hodelin J F, Khoury G and Bouwmeester D 2005 Phys. Rev. Lett.94 090502
18. Leibfried D et al2005 Nature 438 639–42
19. Sun F W, Liu B H, Huang Y F, Ou Z Y and Guo G C 2006 Phys. Rev.A 74 033812
20. Nagata T, Okamoto R, O’Brien J L, Sasaki K and Takeuchi S 2007 Science 316 726–9
21. Resch K J, Pregnell K L, Prevedel R, Gilchrist A, Pryde G J, O’Brien J L and White A G 2007 Phys. Rev. Lett.98 223601
22. Cable H and Dowling J P 2007 Phys. Rev. Lett.99 163604
23. Rudolph T and Grover L 2003 Phys. Rev. Lett.91 217905
24. de Burgh M and Bartlett S D 2005 Phys. Rev.A 72 042301
25. Cleve R, Ekert A, Macchiavello C and Mosca M 1998 Proc. R. Soc.A 454 339–54
26. Nielsen M A and Chuang I L 2000 Quantum Computation and Quantum Information (Cambridge: Cambridge Univ. Press)
27. Shor P W 1994 Proc. 35th Annu. Symp. on Foundations of Computer Science (Los Alamitos, CA: IEEE Computer Society Press) pp 124–34
28. Kitaev A Y 1996 Electronic Colloquium on Computational Complexity 3 3
29. Mitchell M W 2005 Proc. SPIE 5893 589310
30. Higgins B L, Berry D W, Bartlett S D, Wiseman H M and Pryde G J 2007 Nature 450 393–6
31. Luis A and Per̆ina J 1996 Phys. Rev.A 54 4564–70
32. Wiseman H M and Killip R B 1997 Phys. Rev.A 56 944–57
33. Berry D W, Wiseman H M and Breslin J K 2001 Phys. Rev.A 63 053804
34. Griffiths R B and Niu C S 1996 Phys. Rev. Lett.76 3228–31
35. Chernoff H 1952 Annals of Mathematical Statistics 23 493–507
36. Hoeffding W 1963 Journal of the American Statistical Association 58 13–30
37. Holevo A S 1984 Quantum Probability and Applications to the Quantum Theory of Irreversible Processes (Springer Lecture Notes in Mathematics vol 1055) ed L Accardi et al(Berlin: Springer) p 153
38. Dorner U, Demkowicz-Dobrzanski R, Smith B J, Lundeen J S, Wasilewski W, Banaszek K and Walmsley I A 2009 Phys. Rev. Lett.102 040403
39. Huver S D, Wildfeuer C F and Dowling J P 2008 Phys. Rev.A 78 063828
You are adding the first comment!
How to quickly get a good reply:
• Give credit where it’s due by listing out the positive aspects of a paper before getting into which changes should be made.
• Be specific in your critique, and provide supporting evidence with appropriate references to substantiate general statements.
• Your comment should inspire ideas to flow and help the author improves the paper.

The better we are at sharing our knowledge with each other, the faster we move forward.
The feedback must be of minumum 40 characters