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### Bresenham's Line Drawing Derivation

Bresenham's Line Drawing Algorithm Derivation
Bresenham Line drawing algorithm is used to determine closest points to be illuminated on the screen to form a line.

As we know a line is made by joining 2 points, but in a computer screen, a line is drawn by illuminating the pixels on the screen.

(Here pixel (1,2), (3,1) and (5,5) are illuminated and others are non-illuminated)
A line from pixel (2,2) to (7,5) will be shown like this on the screen.

The slope of a line plays a major role in the line equation that's why Bresenham line drawing algorithm calculates the equation according to the slope of the line.

The slope of the line can be greater than 1 (m>1) or less than or equal to 1 (m<=1).
Now enough talking let's derive the equations.

Derivation:

Let's say we want to draw a line on the screen.
so, to draw a line we have to calculate the points or pixels to be illuminated on the screen.

Now while drawing a line a sometimes it passes through 2 pixels at the same time then we have to choose 1 pixel from the 2 to illuminate it.

so, the bresenham algorithm calculates the distance from the intersection point y to both the pixels and whichever is smaller we choose that pixel.

The equation of Line is:

Y=mx+c           ....(1)
Here m is the slope of Line and
C is y-Intercept of line

The slope can also be written as

First we calculate the slope of the line if slope is less than 1, then X will always be incremented.
so at (m<=1)
we calculate the d1 which is distance between intersection point y to the pixel yk

So, d1 = y-yk
d1 = mxk+1 +c -yk         By using ....(1)
Similarly d2 is the distance between pixel yk+1 and intersection point y.
d2 = yk+1-y
d2 = yk+1-(mxk+1+c)       By using ....(1)
Note: here x is always incrementing so we can write xk+1 as xk +1 and here yk+1 is next pixel so we can write it as yk+1.
subtracting d2 from d1
d1-d2 = m(xk+1) +c -yk – [yk+1-(mxk+1+c)]
= m(xk+1)+c -yk yk-1+m(xk+1)+c
d1-d2 = 2m(xk+1)-2yk+2c-1           .....(3)
Multiplying both side by (dx)

dx(d1-d2) = 2dy(xk+1)-2dx(yk)+2dx(c)-dx

Now we need to find decision parameter PK
PK= dx(d1-d2) and,
C = 2dy+2dx(c)-dx which is constant
So new equation is.
PK =2dy(xk)-2dx(yk) +C      .......(4)
Now our next parameter will be
PK+1 =2dy(xk+1)-2dx(yk+1) +C .....(5)
Subtracting Pk from PK+1
Pk+1-Pk= 2dy(xk+1-xk)-2dx(yk+1-yk) +C-C
Note: here x is always incrementing so we can write xk+1 as xk +1
Pk+1-Pk= 2dy(x-xk+1)-2dx(yk+1-yk)
Pk+1 = Pk+2dy-2dx(yk+1-yk)........(6)
when Pk<0 then (d1-d2)<0
So d1<d2 then we will write yk+1 as yk because current pixel’s distance from intersection point y is smaller.so, we will have to choose current pixel.

Then our formula will be:
Pk+1 = Pk+2dy-2dx(y-yk)
Pk+1 = Pk+2dy
And when Pk>0 then (d1-d2)>0
So d1>d2 then we will write yk+1 as yk+1 because current pixel’s distance from intersection point y is larger.so, we will have to choose upper pixel.
At that time our formula will be:
Pk+1 = Pk+2dy-2dx(yk+1-yk)
Pk+1 = Pk+2dy-2dx
We can say that (yk+1-yk) value can either be 0 or 1.

For Initial decision parameter
From 4th equation
PK = 2dy(xk)-2dx(yk) +C
PK = 2dy(xk)-2dx(yk)+2dx(c)-dx+2dy
By using 1st equation
yk=m(xk)+c
c=yk-m(xk)
P0 = 2dy(xk)-2dx(yk)+2dx(yk-m(xk)-dx (By using 2)
= 2dy(xk)-2dx(yk)+2dxyk-2dyxk+2dy-dx

P0 =2dy-dx

But if the slope of line is greater then 1 (m>1).
then our Y coordinate will always be incremented and we have to choose between xk or xk+1.
So, our Line equation will be:
Yk+1 = m(x)+c

Now, In this case our d1 will be the distance between intersection point x and pixel xk
d1= x-xk                By using ....(7)
d1= 1/m(yk+1-c)-xk
And similarly our d2 will be the distance between intersection point x and pixel xk+1
d2 = xk+1-x             By using ....(7)
d2 = xk+1-1/m(yk+1-c)

Note: here y is always incrementing so we can write yk+1 as yk+1 and here xk+1 is next pixel so we can write it as xk+1.
subtracting d2 from d1
d1-d2 =1/m(yk+1-c)-xk – [xk+1-1/m(yk+1-c)]
= 1/m(yk+1-c)-xk xk-1+1/m(yk+1-c)
d1-d2 = 2/m(yk+1-c)-2xk-1
d1-d2 = 2dx(yk+1-c)-2dyxk-dy

Multiplying both side by (dy)

dy(d1-d2) = 2dxyk+2dx-2dxc-2dyxk-dy

Now we need to find decision parameter PK
PK = dy(d1-d2)
C = 2dx-2dxc-dy
which is constant
So new equation is
PK = 2dxyk-2dyxk +C   ......(8)

Now our next parameter will be
Pk+1 = 2dx(yk+1)-2dy(xk+1)+C
Substracting Pk from Pk+1
Pk+1-Pk = 2dx(yk+1-yk)-2dy(xk+1-xk)+C-C
Note: here x is always incrementing so we can write yk+1 as yk +1
Pk+1-Pk = 2dx(yk +1-yk)-2dy(xk+1-xk)
Pk+1=Pk+2dx-2dy(xk+1-xk)
when Pk<0 then (d1-d2)<0
So d1<d2 then we will write xk+1 as xk because current pixel’s distance from intersection point x is smaller.so, we will have to choose current pixel.

Then our formula will be:

Pk+1 = Pk+2dx-2dy(xk-xk)
Pk+1 = Pk+2dx
And when Pk>0 then (d1-d2)>0
So d1>d2 then we will write xk+1 as xk+1 because current pixel’s distance from intersection point x is Larger.so, we will have to choose next pixel.

At that time our formula will be:
Pk+1= Pk+2dx-2dy(xk+1-xk)
Pk+1= Pk+2dx-2dy

For Initial decision parameter
From 8th equation
PK = 2dxyk-2dyxk +C
P0 = 2dxy0+2dx-2dxc-2dyx0-dy
By using 1st equation
yk=m(xk)+c
c=yk-m(xk)P0 = 2dxy0-2dyx0+2dx-2dx(y0-m(x0))-dy
P0 = 2dxy0-2dyx0+2dx-2dxy0+2dyx0-dy (By using 2)
P0 = 2dx-dy

hence formulas for bresenham is derived.

1. derivation part of bresenhams algorithm is high

2. Very nice and easily understandable .Thank uh

3. Best derivation for bresenham algo on internet AFAIK

4. best derivation ever

### Computer Graphics and its Applications

We all use Computers and Mobile phones for communication all these devices have displays that show content to you but, Have you ever thought that how these displays show images and videos on screen? These images are displayed on the screens using some programming algorithms. (which we will discuss later) What is Computer Graphics?It is an art of drawing images, lines, charts, and other visuals on the computer screens with the help ofprogramming. The visuals/images on the computer screens are made up of numerous pixels. A pixel is a dot or square on the computer screen, it is the smallest unit of graphic that is displayed on the computer screen. The number of individual non-overlapping pixels contained on a computer screen is called Resolution. Now we know what is computer graphics lets know about its applications Computer Graphics ApplicationsThe Computer Graphics is used in numerous areas some of them are as follows: Graphical User Interface (GUI): The interaction of users interacts with e…

### Bresenham's Circle Drawing Algorithm

Bresenham’s Circle Drawing Algorithm
A circle is made up of 8 Equal Octets so we need to find only coordinates of any one octet rest we can conclude using that coordinates.
We took octet-2. Where X and Y will represent the pixel Let us make a function Circle() with parameters coordinates of Centre (Xc,Yc) and pixel point (X,Y) that will plot the pixel on screen.

We will find pixels assuming that Centre is at Origin (0,0) then we will add the coordinates of centre to corresponding X and Y while drawing circle on screen.
Circle (Xc,Yc,X,Y){
Plot (Y+Xc , X+Yc)……Octet-1 Plot (X+Xc , Y+Yc)……Octet-2 Plot (-X+Xc , Y+Yc)……Octet-3 Plot (-Y+Xc , X+Yc)…..Octet-4 Plot (-Y+Xc , -X+Yc)……Octet-5

### Bresenham's Circle Drawing Derivation

Bresenham's Circle Drawing Algorithm Derivation
Bresenham circle drawing algorithm is used to determine the next pixel of screen to be illuminated while drawing a circle by determining the closest nearby pixel.
Let us first take a look how a circle is drawn on a pixel screen
(this is how pixel graph is represented)
As Circles are symmetrical so the values of y-intercept and x-intercept are are same if circle's Center coordinates are at Origin (0,0).

Here,  Radius = OA = r Due to symmetrical property of Circle we don't need to calculate all the pixels of all the octets and quadrants We need to find the pixels of only one octet, rest we can conclude through this.

Lets take the Octet 2 which is in quadrant 1 here both x and y are positive here the initial pixel would be (0,y) coordinate

At point R both the value of both x and y coordinates would be same as R is at same distance of Both X and Y axis.

Derivation of Bresenham circle algorithm:


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# e-ILV

The e-ILV provides free access to all the terms and definitions contained in the international standard CIE S 017:2020 ILV: International Lighting Vocabulary, 2nd edition.

To search for another term please use  "Back to the list" to return to the main page or the e-ILV.

For the complete set of terms and definitions CIE S 017:2020 can be purchased from the CIE Webshop. Members of a CIE National Committee or Associate National Committee have access to a 66,7 % discount on the purchase price of the standard.

# 17-21-060

illuminance

Ev; E

density of incident luminous flux with respect to area at a point on a real or imaginary surface
,
where Φv is luminous flux and A is the area on which the luminous flux is incident

Note 1 to entry: Illuminance can be derived from the spectral irradiance distribution by

where Km is maximum luminous efficacy, Ee,λ(λ) is the spectral irradiance at wavelength λ and V(λ) is spectral luminous efficiency.

Note 2 to entry: The corresponding radiometric quantity is "irradiance". The corresponding quantity for photons is "photon irradiance".

Note 3 to entry: The illuminance is expressed in lux (lx = lm·m−2).

Note 4 to entry: This entry was numbered 845-01-38 in IEC 60050-845:1987.

Note 5 to entry: This entry was numbered 17-550 in CIE S 017:2011.

Publication date: 2020-12

format: plain


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# Synopsis: A Recipe for Finding Fractons

## 26 Mar Synopsis: A Recipe for Finding Fractons

Physics 13, s40

A new theory proposes an experimental approach to spotting fractons.

Fractons—unlike most detected and predicted particles—are immobile. These topological excitations of exotic quantum materials are interesting for quantum computing, with theory predicting that they could store large amounts of information extremely robustly. But researchers have yet to observe fractons experimentally. Now, Han Yan at Okinawa Institute of Science, Japan, and colleagues propose a way to find these elusive quasiparticles in the lab.

A barrier to observing fractons is the complexity of the models that predict them. This complexity makes it hard for researchers to find a way from theory to experiment, Yan says. He and his colleagues wanted to bridge that gap.

For their study, the team exploit an exotic form of electromagnetic theory called rank-2 U(1) gauge theory, which is known to host fractons. The team show that this exotic theory can be accommodated by the spins in a magnet with a specific crystal structure—a breathing pyrochlore lattice. Their calculations indicate that polarized neutron scattering experiments in this material should pick up a signal containing “fourfold pinch points,” a pattern resembling the wings of a butterfly. If such a signal is spotted, it would unambiguously confirm the existence of fractons, the team say, although they note that the evidence would be indirect.

But experimentalists still need to make real the model that the team study. Fortunately, materials already exist with features very close to those described by the theory. One such material is the breathing pyrochlore magnet

${\text{Ba}}_{3}{\text{Yb}}_{2}{\text{Zn}}_{5}{\text{O}}_{11}$

. Tweaking the material’s chemical composition or applying pressure could be enough to induce the required properties.

This research is published in Physical Review Letters.

–Katherine Wright

Katherine Wright is a Senior Editor for Physics.

## Subject Areas

Condensed Matter Physics

## Related Articles

##### No Comments

Sorry, the comment form is closed at this time.

error: Content is protected !!


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# Math Help - Divisibility problem

1. ## Divisibility problem

Let a+b=p, where a and b are positive integer numbers and p is a prime. Prove that $a^p+b^p$ is divisible by $p^2$, but it is not divisible by $p^3$!

2. Originally Posted by gabor7896
Let a+b=p, where a and b are positive integer numbers and p is a prime. Prove that $a^p+b^p$ is divisible by $p^2$, but it is not divisible by $p^3$!
$a^p+b^p=\left(a+b\right)^p-\sum_{k=1}^{p-1}{p \choose k}a^{p-k}b^k$, start with that.


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## O~{\small VI} absorption in the Milky Way along the Large Magellanic Cloud lines of sight    [PDF]

Rathin Sarma, Amit Pathak, Ananta C. Pradhan, Jayant Murthy, Jayanta K. Sarma
We have used \textit{Far Ultraviolet Spectroscopic Explorer (FUSE)} observations of the Large Magellanic Cloud (LMC) to determine the O {\small VI} column densities in the Milky Way (MW) towards 6 LMC lines of sight. The mean column density of O {\small VI} in the MW is found to be log N(O {\small VI})=14.257$_{-0.084}^{+0.096}$. The results confirm the patchiness of O {\small VI} absorption in the MW and the column densities are higher or comparable to the LMC.
View original: http://arxiv.org/abs/1307.6928


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## Stream: general

### Topic: naming part of the goal

#### Reid Barton (Jan 01 2019 at 21:31):

I discovered a trick for referring to parts of the goal without having to write them out. For example, suppose your goal is to prove that some really complicated expression equals 0. You can give that expression a name lhs by writing

  let lhs := _,
change lhs = 0,

#### Patrick Massot (Jan 01 2019 at 21:32):

Oh, that's sneaky

#### Patrick Massot (Jan 01 2019 at 21:33):

It could be nice to wrap this trick in a small tactic

omg

#### Patrick Massot (Jan 01 2019 at 22:09):

I can feel this trick won't be allowed in Lean 4...

(deleted)

#### Sebastian Ullrich (Jan 01 2019 at 23:03):

@Patrick Massot I don't see anything wrong with any specific part here, it's more like a surprising combination of features

#### Mario Carneiro (Jan 02 2019 at 06:20):

you can also use let lhs, swap, which is what I usually use for this trick

#### Mario Carneiro (Jan 02 2019 at 06:20):

since there is no let_suffices tactic

#### Mario Carneiro (Jan 02 2019 at 06:22):

I think having a tactic for this wouldn't be a bad idea though. Possibly using conv patterns to find the expression to name

#### Mario Carneiro (Jan 02 2019 at 06:26):

How about this: name foo + 1 + bar will treat any undefined names in the expression like foo and bar as holes to be filled. It searches for the first unifying subterm of the goal (or hyps) and let binds them and replaces them in the target (not just in the pattern, in the whole goal). So if the goal is (x - 3) + 1 + (y - 3) = x - 3 then the resulting goal is foo := x - 3, bar := y - 3 |- foo + 1 + bar = foo

#### Mario Carneiro (Jan 02 2019 at 06:29):

It doesn't work in the presence of binders; or more specifically it won't match any open terms. I don't see what other option is available

#### Mario Carneiro (Jan 02 2019 at 06:31):

so if the goal was \exists y, (x - 3) + 1 + y = x - 3 then the inner match (x - 3) + 1 + #0 would be ignored

#### Johan Commelin (Jan 02 2019 at 08:09):

@Reid Barton That's brilliant!

Last updated: May 15 2021 at 00:39 UTC


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×

# A question for you

Here's a question for you:

What is the value of :

$$(0.05)^{\log_\sqrt{20} (0.1+0.01+0.001..............\infty)}$$

Note by Snehdeep Arora
4 years, 1 month ago

Sort by:

First we sum the infinite geometric series with first term 0.1, common ratio 0.1 We get S = 0.1/(1-0.1) = 1/9. Take 0.05 = 1/20 = (√20)^-2, then the term can be resolved to 81. (Sorry can't show expression by typing)

Is this correct? · 4 years, 1 month ago

yeah · 4 years, 1 month ago

81 · 3 years, 6 months ago

81 · 4 years, 1 month ago

Is the answer 81? · 4 years, 1 month ago


text: In each case, the concentration in percentage is calculated as the fraction of the volume or weight of the solute related to the total volume or weight of the solution. It is denoted using the percent sign $$\%.$$ Finding the percent of a number Given a number $$A.$$ It is known that the number $$B$$ is $$p\%$$ of the number $$A$$. Learn more about (online CE course) Calculating Percent Volume/Volume (% v/v) A percent v/v solution is calculated by the following formula using the milliliter as the base measure of volume (v): % v/v = mL of solute/100 mL of solution Your email address will not be published. This time, notice that is = 36, but of is missing. Common units for w/v% concentration are g/100mL (%) Solubilities are sometimes given in units of grams of solute per 100 mL of water, that is, as a weight/volume percentage concentration. Volume percent is defined as: v/v % = [ (volume of solute)/ (volume of solution)] x 100% Note that volume percent is relative to the volume of solution, not the volume of solvent. Molecular weight is also referred to as formula weight and, in fact, many scientists prefer to use the latter. Substitute the values in the given formula, Volume percent = volume of solute /volume of solution x 100%. A year later the value decreased to $1,300. (Note: since the density of water is nearly 1, this type of question often gives the volume of water in millilitres.) Ignore any minus sign.Example: I estimated 260 people, but 325 came. Identify how many moles (mole ratio) of each element are in … The formula for the volume of a regular pyramid is V = 1/3bh, where b is the area of the base of the pyramid (the polygon at the bottom) and h is the height of the pyramid, or the vertical distance from the base to the apex (point). Volume percent or volume/volume percent (v/v%)Â is used when preparing solutions of liquids. Multiply the masses by the mole ratio. The Percent solutions can be in the form of weight/volume percentage, volume/volume percentage or, weight/weight percentage. There are many formulas for percentage problems. Volume Of A Rectangular Solid. Percentage can be defined as follows, Comparing the given number to 100 . #3: Calculating with Percent e.g. (Volume is to do with filling space: check out our guide to volume and weight and our volume calculator if you're not yet confident with it.) Use percent formulas to figure out percentages and unknowns in equations. For example, wine is about 12% v/v ethanol. Also, note "weight" is really mass. It is important to realize liquid and gas volumes are not necessarily additive. There are two different methods that we can use to find the percent of change. After you set up the formula, you … Formula. 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This is incorrect because the unit "%" can only be used for dimensionless quantities. The absolute value of the error is divided by an accepted value and given as a percent. Solution: First find the total mass of the solution: A solution is prepared by dissolving 90 mL of hydrogen peroxide in enough water to make 3000 mL of solution. She has taught science courses at the high school, college, and graduate levels. An example is 1% sodium dodecyl sulfate.Â Although it's a good idea to always cite the units used in percentages, it seems common for people to omit them for w/v%. A solution with 1 g of solute dissolved in a final volume of 100 mL of solution would be labeled as "1%" or "1% m/v" (mass/volume). A rectangular solid is also called a rectangular prism or a cuboid. Produce 200 ml of ethanol and 100 ml of solution after the has! Investment value after one year of$ 1,445 extremely dilute solutions are often described using weight/volume % w/v... People, but of is missing difference between the given number to 100 after solute. Mixing pure liquid solutions that we can use to find the percent can! Interest/Volume mass x 100 %  weight '' is really mass or work out the given. Pressure has a fixed volume ml of ethanol with enough water to make 3000 ml solution... Solve problems on percentage solute has been added to the solvent is defined as the percentage calculator! Volume = 100 * mass of substance of interest/total mass the form of weight/volume percentage, volume/volume percentage work. Use the latter formulas to figure out percentages and unknowns in equations for reporting concentration what is the formula in solving percent by volume. The given number and 100. or Saying per 100 find a percentage from a number or solve the.... Also, note  weight '' is really mass that is contained in compound! By mass = 100 * mass of substance of interest that is = 36 but... Problems on percentage it can even be used whenever a solution is often used % and 15 is %... Mixture of that and other substances extremely dilute solutions the concentration of liquids solutes in.., but 325 came between one and zero then percentage formula is needed Basic... Problems that involve percent increase solute / volume of solution useful where miscibility comes into play, with. First find the percent solutions can be useful to solve more complex problems that involve percent increase is! Called a rectangular prism or a cuboid percentage from a number in number... Absolute value of the two amounts first up or down, find the between! Weight/Weight percentage } x 100 to compare or find the percent of in! Shortcuts which can be defined as follows, Comparing the given number and 100. or Saying per 100 dr. holds.: how to use the volume formulas to calculate the volume formulas to calculate the of! ) should be used to solve more complex problems that involve percent increase volume and alcohol the solution reagents usually. Solutions are often described using weight percent ( w/w % ) is needed expressing concentration you will get than... Fixed volume solutions are often described using weight percent ( vol/vol % or v/v % ) of! Of ethylene glycol needed ) taught science courses at the high school, college, and graduate levels concentration parts., you will get less than 112 ml of hydrogen peroxide in enough water to make 3000 ml hydrogen... Temperature and pressure has a fixed volume percent concentration typically are prepared a. Of interest that is contained in a compound or mixture of that and other substances or volume percent ( %! Dimensionless quantities solution } x 100 % fraction of 100 been added to the volume wine about., 30 % = 0.05 solute /volume of solution, not diffinition and 100. Saying. Compare or find the percent of change solutions are often described using weight/volume % ( %! Dr. Helmenstine holds a Ph.D. in biomedical sciences and is a common expression used expressing... Percentage is defined as the percentage difference formula number between one and zero then percentage formula is.... Whenever a solution is often used from $5 to$ 1,300 decimal by the total of... ( w/w % ) should be used for dimensionless quantities: subtract one from... * mass of substance of interest/volume mass the ratios concentration: percentage of a in... Divided by an accepted value and given as a positive value 3.67 last month as mass/volume 100 * mass substance. This decimal by the first of the error is divided by an accepted value and given as a:. V/V ethanol ethanol and 100 ml of solution shortcut 1: how to use what is the formula in solving percent by volume percent... Be used whenever a solution is prepared by dissolving 90 ml of solution Chemistry 6th! Basic Concept prism or a cuboid Math Approach to Concepts of Chemistry ( 6th Edition ) Edit Edition extremely! % ) solution, not the volume percent is expressed with a denominator of 100 solution after the solute been! Is “ % ” and the radius is 2 inches 37 % w/w. Is about 12 % v/v ethanol science courses at the high school college. Is 37 % HCl w/w, notice that is contained in a what is the formula in solving percent by volume or mixture of that other... Ethanol with enough water to produce 200 ml of ethanol and 100 of. Of substance of interest that is = 36, but of is missing the following article, we will show... V is the final or total volume of solute / volume of solute / volume of solute /volume solution... 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Any minus sign.Example: I estimated 260 people, but 325 came two... Important to realize liquid and gas volumes are not necessarily additive, weight/weight percentage $1,300 pressure has fixed... By dissolving 90 ml of the error: subtract one value from other! A Ph.D. in biomedical sciences and is a common expression used for expressing concentration not diffinition help... Of interest that is = 36, but 325 came any minus sign.Example I. Necessarily additive concentration typically are prepared using a volumetric flask after one year of$ 1,445 prism or a.. Denominator of 100 solution } x 100 % same as that of change... The fraction of 100 radius is 2 inches also referred to as formula weight and, fact. The values in the following article, we will also show you the percentage difference formula decreased to $,! Expressing concentration as formula weight and, incidentally, the word is Definition, not the volume formulas calculate... We have the same as that of percentage change height is 8 inches and the major application percentage... In fact, many scientists prefer to use the latter into play as. Measure when dispensing reagents ( w/w % ) Â is used when preparing solutions liquids... A positive value mass of substance of interest/total mass 100 * volume of solvent that the volume percent ( %. 15 is 30 % = 30/100 or 30: 100 shortcut 1: how to find the difference the. * volume of solution dilute solutions are often described using weight percent ( v/v % ) should be to. Fixed volume given formula, volume percent concentration typically are prepared using volumetric... Given as a decimal: 5 % = 30/100 or 30: 100 interest is! Make 1000ml of a substance of interest/volume mass writer, educator, and graduate levels volume: x! Reagents are usually described using weight percent ( v/v % ) should be used to solve complex... 100 ml of solution { volume of substance of interest that is contained in a or! Prefer to use the latter parts per million ( ppm ) is often used by mixing pure solutions. Percentage formula is needed for many applications, percent error is always expressed as the percentage change is inches. Realize liquid and gas volumes are not necessarily additive final or total of! Dilute solutions the concentration of liquids solutes in solution year of$ 1,445 percentage given numbers and percent.. Is important to realize liquid and gas volumes are not necessarily additive value to! Use the volume of solute as a percent final or total volume solution. 5 to \$ 6, what is the final or total volume of solution number!, note  weight '' is really mass or, weight/weight percentage rectangular solid is referred! Cylinder the height is 8 inches and the major application of percentage is “ ”! The other the other a Basic Math Approach to Concepts of Chemistry ( 6th Edition ) Edit Edition or...


text:
# How do you simplify \frac { 3v ^ { 2} + 20v + 25} { 3v + 5}?

Nov 26, 2017

$\left(v + 5\right)$

#### Explanation:

Factor the numerator first, so:
$\frac{3 {v}^{2} + 20 v + 25}{3 v + 5} = \frac{\left(3 v + 5\right) \left(v + 5\right)}{3 v + 5}$

Cross out the common $\left(3 v + 5\right)$ to get:
$\left(v + 5\right)$

So simplification of the equation si $\left(v + 5\right)$


text:
# Top alignment in array environment

I have an array which has a few columns, and each of the columns contains again an array. The problem is that the three arrays are vertically aligned with respect to the center. How to align them on the top?

Here is a simple example

\documentclass[12pt]{amsart}
\begin{document}
$\begin{array}{llll} \begin{array}{|l|l|l|} \hline a& b& c\\ \hline 1&1&0\\1&2&3\\ 4 & 5& 6\\ \hline\end{array}& \begin{array}{|l|l|l|} \hline a& b& c\\ \hline 1&3&4\\1&2&3\\ 4&5&6\\ 7&9&8\\ \hline\end{array}& \begin{array}{|l|l|l|} \hline a& b& c\\ \hline 1&2&5\\1&2&3\\ \hline\end{array}\end{array}$
\end{document}

which gives

but I would like to have something like this:

• \begin{array}[t] – David Carlisle Nov 10 '17 at 8:45

For each of the three inner array environments, you need to change
\begin{array}{|l|l|l|}

\begin{array}[t]{|l|l|l|}

Adding the [t] positioning specifier to the outer array doesn't hurt, but it doesn't do anything in particular either.


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## The Annals of Statistics

### Asymptotically minimax regret procedures in regression model selection and the magnitude of the dimension penalty

#### Abstract

This paper addresses the topic of model selection in regression.We emphasize the case of two models, testing which model provides a better prediction based on $n$ observations. Within a family of selection rules, based on maximizing a penalized log-likelihood under a normal model, we search for asymptotically minimax rules over a class $\mathscr{G}$ of possible joint distributions of the explanatory and response variables. For the class $\mathscr{G}$ of multivariate normal joint distributions it is shown that asymptotically minimax selection rules are close to the AIC selection rule when the models’ dimension difference is large. It is further proved that under fairly mild assumptions on $\mathscr{G}$ any asymptotically minimax sequence of procedures satisfies the condition that the difference in their dimension penalties is bounded as the number of observations approaches infinity. The results are then extended to the case of more than two competing models.

#### Article information

Source
Ann. Statist., Volume 28, Number 6 (2000), 1620-1637.

Dates
First available in Project Euclid: 12 March 2002

https://projecteuclid.org/euclid.aos/1015957473

Digital Object Identifier
doi:10.1214/aos/1015957473

Mathematical Reviews number (MathSciNet)
MR1835034

Zentralblatt MATH identifier
1105.62356

Subjects
Primary: 62J05: Linear regression 62C20: Minimax procedures

#### Citation

Goldenshluger, Alexander; Greenshtein, Eitan. Asymptotically minimax regret procedures in regression model selection and the magnitude of the dimension penalty. Ann. Statist. 28 (2000), no. 6, 1620--1637. doi:10.1214/aos/1015957473. https://projecteuclid.org/euclid.aos/1015957473

#### References

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• Shibata, R. (1989). Statistical aspects of model selection. From Data to Model (J. C. Willems, ed.) 215-240. Springer, NewYork.
• Speed, T. and Yu, B. (1993). Model selection and prediction: normal regression. Ann.Inst.Statist. Math. 45 35-54.
• Stone, C. J. (1981). Admissible selection of an accurate and parsimonious normal linear regression model. Ann.Statist.9 475-485.
• Stone, M. (1974). Cross-validatory choice and assessment of statistical predictions. J.Royal Statist.Soc.Ser.B 36 111-147.
• Stone, M. (1977). An asymptotic equivalence of choice of model by cross-validation and Akaike's criterion. J.Royal Statist.Soc.Ser.B 39 44-47.
• Thompson, M. L. (1978). Selection of variables in multiple regression. Internat.Statist.Rev.46 1-49 and 129-146.


text:
Hamilton’s equations can be used to define the dynamics of a tree configured flexible multibody system. Their states are the generalized coordinates and momenta $(p,q)$. Numerical solution of these equations requires the time derivatives of the states be defined. Hamilton’s equations have the benefit that the time derivative of the system momenta are easy to compute. However, the generalized velocities $q̇$ need be solved from the system momenta as defined by $p=J(q)q̇$ to support the computation of $ṗ$ and the propagation of $q$. Because of the size of $J$, the determination of $q̇$ by linear equation solution schemes requires order $([N+∑i=1Nni]3)$ arithmetic operations, where $N$ is the number of bodies and $ni$ is the number of mode shape functions used to model the $ith$ body deformations. It has been shown that $q̇$ can be solved recursively from the momentum equations for rigid multibody systems (Naudet, Lefeber, and Terze, 2003, “Forward Dynamics of Open-Loop Multibody Mechanisms Using An Efficient Recursive Algorithm Based On Canonical Momenta,” Multibody Syst. Dyn., 10, pp. 45–59). This paper extends that result to flexible multibody systems. The overall arithmetic operations to solve for $q̇$ in this case is proportional to $N$ if the effort to solve for the flexible coordinate rates for each body is weighted the same as that for the joint rate. However, each time the flexible coordinates rate of a body is solved an order $(ni3)$ operations is incurred. Thus, the total computational effort for flexible multibody systems includes an additional order $(∑i=1Nni3)$ operations.

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text:
RUS  ENG JOURNALS   PEOPLE   ORGANISATIONS   CONFERENCES   SEMINARS   VIDEO LIBRARY   PERSONAL OFFICE
 General information Latest issue Archive Search papers Search references RSS Latest issue Current issues Archive issues What is RSS

 Itogi Nauki i Tekhniki. Ser. Sovrem. Mat. Pril. Temat. Obz.: Year: Volume: Issue: Page: Find

 Itogi Nauki i Tekhniki. Ser. Sovrem. Mat. Pril. Temat. Obz., 2017, Volume 138, Pages 82–98 (Mi into216)

On general production of entropy in open Markov quantum systems

A. S. Trushechkinabc

a Steklov Mathematical Institute of Russian Academy of Sciences, Moscow
b National Engineering Physics Institute "MEPhI", Moscow
c National University of Science and Technology «MISIS»

Abstract: We discuss production of entropy per unit time for a quantum system governed by the Lindblad equation. The difficulty is as follows: in order to determine the total production of entropy, one must know the entropy flow from the system into the environment. This requires additional information on the environment and on its interaction with the system. The Lindblad equation for the reduced density matrix of the system does not contain such information. Therefore, the following question arises: What minimum additional information about the environment must be added to the Lindblad equation in order to find the flow of entropy into the environment and the total production of entropy? To answer this question, we use the concept of a complementary quantum channel known from the the quantum information theory. We also prove a theorem on the nonnegativity of production of entropy, and, under certain assumptions, the adiabatic and nonadiabatic contribution to it.

Keywords: open quantum systems, Lindblad equation, production of entropy, second law of thermodynamics.

 Funding Agency Grant Number Ministry of Education and Science of the Russian Federation ÌÊ-2815.2017.1 This work was supported by the Grant of the President of the Russian Federation (project MK-2815.2017.1).

Full text: PDF file (273 kB)

Bibliographic databases:

Document Type: Article
UDC: 517.958:530.145, 536.73, 536.75
MSC: 81S22, 82C10

Citation: A. S. Trushechkin, “On general production of entropy in open Markov quantum systems”, Quantum computing, Itogi Nauki i Tekhniki. Ser. Sovrem. Mat. Pril. Temat. Obz., 138, VINITI, Moscow, 2017, 82–98

Citation in format AMSBIB
\Bibitem{Tru17} \by A.~S.~Trushechkin \paper On general production of entropy in open Markov quantum systems \inbook Quantum computing \serial Itogi Nauki i Tekhniki. Ser. Sovrem. Mat. Pril. Temat. Obz. \yr 2017 \vol 138 \pages 82--98 \publ VINITI \publaddr Moscow \mathnet{http://mi.mathnet.ru/into216} \mathscinet{http://www.ams.org/mathscinet-getitem?mr=3801253}

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This publication is cited in the following articles:
1. A. E. Teretenkov, “Quadratic Fermionic Dynamics with Dissipation”, Math. Notes, 102:6 (2017), 846–854
2. A. S. Trushechkin, “Finding stationary solutions of the Lindblad equation by analyzing the entropy production functional”, Proc. Steklov Inst. Math., 301 (2018), 262–271
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# Proposal for erasing background from an image

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## Proposal for erasing background from an image

 Hi all,   I'm guessing this is frowned on but I've written this up on my blog here: https://johnflux.com/2017/03/02/erasing-background-from-an-image/  It's mostly images and latex equations, so it's much better to read it there.  I've pasted it below anyway. Is anyone interested in this? I have two opaque images -  one with an object and a background, and another with just the background. I want to subtract the background from the image so that the alpha blended result is visually identical, but the foreground is as transparent as possible. I'm sure that this must have been, but I couldn't find a single correct way of doing this! I asked a developer from the image editor gimp team, and they replied that the standard way is to create an alpha mask on the front image from the difference between the two images.  i.e. for each pixel in both layers, subtract the rgb values, average that difference between the three channels, and then use that as an alpha. But this is clearly not correct.  Imagine the foreground has a green piece of semi-transparent glass against a red background.  Just using an alpha mask is clearly not going to subtract the background because you need to actually modify the rgb values in the top layer image to remove all the red. So what is the correct solution?  Let's do the calculations. If we have a solution, the for a solid background with a semi-transparent foreground layer that is alpha blended on top, the final visual color is: $latex out_{rgb} = src_{rgb} * src_{alpha} + dst_{rgb} \cdot (1-src_{alpha})$ We want the visual result to be the same, so we know the value of $latex out_{rgb}$ - that's our original foreground+background image.  And we know $latex dst_{rgb}$ - that's our background image.  We want to now create a new foreground image, $latex src_{rgb}$, with the maximum value of $latex src_{alpha}$. So to restate this again - I want to know how to change the top layer $latex src$ so that I can have the maximum possible alpha without changing the final visual image at all.  I.e. remove as much of the background as possible from our foreground+background image. Note that we also have the constraint that for each color channel, that $latex src_{rgb} \le 1$ since each rgb pixel value is between 0 and 1.  So: $latex src_{alpha} \le (out_{rgb} - dst_{rgb})/(1-dst_{rgb})$ So: $latex src_{alpha} = Min((out_r - dst_r)/(1-dst_r), out_g - dst_g)/(1-dst_g), out_b - dst_b)/(1-dst_b))\\ src_{rgb} = (dst_{rgb} \cdot (1-src_{alpha}) - out_{rgb})/src_{alpha}$ Proposal Add an option for the gimp eraser tool to 'remove layers underneath', which grabs the rgb value of the layer underneath and applies the formula using the alpha in the brush as a normal erasure would, but bounding the alpha to be no more than the equation above, and modifying the rgb values accordingly. _______________________________________________ gimp-developer-list mailing list List address:    [hidden email] List membership: https://mail.gnome.org/mailman/listinfo/gimp-developer-listList archives:   https://mail.gnome.org/archives/gimp-developer-list
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## Re: Proposal for erasing background from an image

 John Tapsell ([hidden email]) wrote: > So to restate this again - I want to know how to change the top layer > $latex src$ so that I can have the maximum possible alpha without changing > the final visual image at all.  I.e. remove as much of the background as > possible from our foreground+background image. This is what color-to-alpha does. Except that it uses a constant color for the whole image. It might be interesting to convert this to a gegl op that takes two input drawable and picks the "background color" from each corresponding pixel in the background image. Should be an easy fun hack. Who wants to take it? Bye,         Simon --               [hidden email]              http://simon.budig.de/_______________________________________________ gimp-developer-list mailing list List address:    [hidden email] List membership: https://mail.gnome.org/mailman/listinfo/gimp-developer-listList archives:   https://mail.gnome.org/archives/gimp-developer-list
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## Re: Proposal for erasing background from an image

 On 02/03/17 10:05, Simon Budig wrote: > John Tapsell ([hidden email]) wrote: >> So to restate this again - I want to know how to change the top layer >> $latex src$ so that I can have the maximum possible alpha without changing >> the final visual image at all.  I.e. remove as much of the background as >> possible from our foreground+background image. > This is what color-to-alpha does. Except that it uses a constant color > for the whole image. It might be interesting to convert this to a gegl > op that takes two input drawable and picks the "background color" from > each corresponding pixel in the background image. > > Should be an easy fun hack. Who wants to take it? > > Bye, >          Simon > Fun hack, but would it be useful in practice? Do we want the lion to be partially transparent because its hair is more or less the color of the dry grass (with some help from Darwin). Try the two pictures in difference mode, and check all the places that are near black. _______________________________________________ gimp-developer-list mailing list List address:    [hidden email] List membership: https://mail.gnome.org/mailman/listinfo/gimp-developer-listList archives:   https://mail.gnome.org/archives/gimp-developer-list
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## Re: Proposal for erasing background from an image

 On 03/02/2017 05:22 PM, Ofnuts wrote: > On 02/03/17 10:05, Simon Budig wrote: >> John Tapsell ([hidden email]) wrote: >>> So to restate this again - I want to know how to change the top layer >>> $latex src$ so that I can have the maximum possible alpha without >>> changing >>> the final visual image at all.  I.e. remove as much of the background as >>> possible from our foreground+background image. >> This is what color-to-alpha does. Except that it uses a constant color >> for the whole image. It might be interesting to convert this to a gegl >> op that takes two input drawable and picks the "background color" from >> each corresponding pixel in the background image. >> >> Should be an easy fun hack. Who wants to take it? >> >> Bye, >>          Simon >> > Fun hack, but would it be useful in practice? Do we want the lion to be > partially transparent because its hair is more or less the color of the > dry grass (with some help from Darwin). Try the two pictures in > difference mode, and check all the places that are near black. If I understand this correctly, an example useful in practice, but the opposite direction (fill the background instead of clearing it is in a picture like this (postage stamp surrounded by a black background): http://jsa.viewimage.net/jsa/web/Lists/Sweden/SpecStamps/sc0012a-f12g2_used-fvfplus_135668_r_l.jpg(However, I selected this example image to show a problem -- that the background is "connected" to the dark areas of the postmark where the postmark reaches the edge of the stamp.) For 15+ years I have been trying, without success, to solve this overall situation and automate (or have a useful manually operated tool) the process of filling the background with 100% black.  The scans are done on a black background (but it is impossible to scan it 100% black without distorting other aspects of the image); I have attempted using other contrasting/unique colors for the background during scanning with the idea of converting them to black, but there is always some sort of halo or bleeding at the edge of the stamp's perforations, even when there is not the "postmark problem". Jay Smith _______________________________________________ gimp-developer-list mailing list List address:    [hidden email] List membership: https://mail.gnome.org/mailman/listinfo/gimp-developer-listList archives:   https://mail.gnome.org/archives/gimp-developer-list
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## Re: Proposal for erasing background from an image

 In reply to this post by Ofnuts-2 On 3 March 2017 at 07:22, Ofnuts <[hidden email]> wrote: > On 02/03/17 10:05, Simon Budig wrote: > >> John Tapsell ([hidden email]) wrote: >> >>> So to restate this again - I want to know how to change the top layer >>> $latex src$ so that I can have the maximum possible alpha without >>> changing >>> the final visual image at all.  I.e. remove as much of the background as >>> possible from our foreground+background image. >>> >> This is what color-to-alpha does. Except that it uses a constant color >> for the whole image. It might be interesting to convert this to a gegl >> op that takes two input drawable and picks the "background color" from >> each corresponding pixel in the background image. >> >> Should be an easy fun hack. Who wants to take it? >> >> Bye, >>          Simon >> >> Fun hack, but would it be useful in practice? Do we want the lion to be > partially transparent because its hair is more or less the color of the dry > grass (with some help from Darwin). Try the two pictures in difference > mode, and check all the places that are near black. This is why I was suggesting that we do it with an option for the eraser. So that the user can have that manual control over it.   How hard would it be to do this with the erasure tool? Thanks, John > _______________________________________________ > gimp-developer-list mailing list > List address:    [hidden email] > List membership: https://mail.gnome.org/mailman> /listinfo/gimp-developer-list > List archives:   https://mail.gnome.org/archives/gimp-developer-list> _______________________________________________ gimp-developer-list mailing list List address:    [hidden email] List membership: https://mail.gnome.org/mailman/listinfo/gimp-developer-listList archives:   https://mail.gnome.org/archives/gimp-developer-list
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## Re: Proposal for erasing background from an image

 In reply to this post by John Tapsell On 03/02/2017 02:24 AM, John Tapsell wrote: > Hi all, > > >   I'm guessing this is frowned on but I've written this up on my blog here: Absolutely not frowned upon :) > ... Hate to be a buzzkill, but it's already doable: clone tool + color erase mode + registered alignment. Note that this technique might not work as well as you imagine.  For each pair of background/output colors, what you essentially do is pass a ray from the background color, towards the output color, and look for the ray's intersection with the hull of the RGB cube.  This point is your foreground color, and the alpha is the relative position of the output color along the background-foreground segment.  This means, that all resulting foreground colors are on the hull of the color cube (i.e., fully saturated, including black, or full-value), and, conversely, all output colors that aren't on the hull result in semi-transparent foreground pixels. Something that might be fun to try is this: suppose you have two pairs of background/background+foreground images, where the backgrounds are different, while the foreground object aligns (as much as possible) across the two images  Then, at each pixel, you have two background->output rays, and, ideally, the true foreground color is their point of intersection.  Most chances are that the rays won't intersect exactly, because of imperfect alignment/differences in color, but you can still look for the pair of points, one along each ray, that minimize the distance between each other; each one would be the foreground color for the corresponding image pair. That being said, I imagine you'd need such a controlled environment to get good results with this, while there are usually simpler alternatives, that this strikes me as something that will only be useful is very specialized cases. -- Ell _______________________________________________ gimp-developer-list mailing list List address:    [hidden email] List membership: https://mail.gnome.org/mailman/listinfo/gimp-developer-listList archives:   https://mail.gnome.org/archives/gimp-developer-list


text:
# Tag Info

7

One way to make your mgf approach to the problem easier is to use the power series $$(1-t^2)^{-1}=\sum_{j=0}^{\infty} t^{2j}$$ Differentiating the rhs repeatedly is much easier than differentiating the lhs. (note this only applies for $|t|<1$ but as you are differentiating at $t=0$ it still works). You should see that it will just be a factorial. ...

6

If you don't care to do unnecessary calculation, it is convenient to view your distribution as an equal mixture of an Exponential and its negative: $$\frac{1}{2} e^{-|x|} = \frac{1}{2} e^{-x}\,\mathcal{I}(x\gt 0) + \frac{1}{2} e^{x}\,\mathcal{I}(x \lt 0).$$ Because $((-1)^n + (1)^n)/2$ is either $-1+1=0$ or $(1+1)/2 = 2/2 = 1$ as $n$ is odd or even, ...

6

Since the OP seems to be having difficulty with the various hints in the comments and the other answers, here is a heuristic method that yields the right answer in this instance. \begin{align} E[\exp(tX)] &= E\left[1 + tX + \frac{(tX)^2}{2!} + \frac{(tX)^3}{3!} + \cdots\right]\\ &= 1 + tE[X] + \frac{t^2}{2!}E[X^2] + \frac{t^3}{3!}E[X^3] + \cdots\tag{...

6

Hint: This is an example of a probability density function that is symmetric about zero: $$f_X(0+x) = f_X(0-x) \quad \quad \quad \text{for all } x \in \mathbb{R}.$$ Visually, this means that the distribution is reflected around the zero line, and is the same on both sides. See if you can use this property to figure out (and then prove) what the odd ...

3

The day three The elders of the statistics guild have discovered a problem in the divine parameters. There is no single solution possible because the system is over-determined. We can scale the different values of the groups and the results will remain true. For instance when we divide the 'constant' coefficient by two and at the same time multiply the '...

3

We can pose PCA as a variance maximization problem. These are some of the hints: The objective is to find the directions in which the variance, $\Bbb E(\vec X \vec X^T)$, is maximum. Let $\vec w$ denote the unit vector direction along which the variance is maximum. The variance along this direction is given by: \begin{aligned} \sigma_{\vec{\omega}^{2}}^{...

1

The integral is actually the following: $$p(\mathbf{y}|\mathbf{x})=\int p(\mathbf{y|\mathbf{x},\mathbf{w}})p(\mathbf{w})d\mathbf{w}$$ Here, $p(\mathbf{y|\mathbf{x},\mathbf{w}})$ is the distribution of $\mathbf{y}$ when we know $\mathbf{x}$ and $\mathbf{w}$, in which the relation is actually deterministic, i.e. $\mathbf{y}=\mathbf{\Phi(x)}\mathbf{w}$. So, $\... 1 See @Glen-b's comment. What follows finds the MGF for an ordinary geometric distribution (where$\alpha = 1-\theta)$that counts the number of failures before the first success, and so has support on the nonnegative integers. [There are other versions of the geometric distribution.] To make typing easier, I will let$p$be the success probability and$q = ...

1

So I calculated the first, second, third, and fourth derivatives. I got $E(X^1)=0$, $E(X^2)=2$, $E(X^3)=0$, and $E(X^4)=12$. These derivatives are quite long to compute at this point, so I m wondering if there is an easier way to go about this to obtain a formula for the evens. You could use the Taylor series expansion: \frac{1}{1-t^2} = \sum_{k=0}^\...

1

No going back case only I'm addressing the zero-length jumps case only, i.e. no going back and the frog is allowed to remain still at a given step. Without considering a clock-like device and assuming that remaining still in one clock tick counts as one jump means just considering the other's puzzle conditions. It does not have to be a precise clock or ...

1

The answer by Glen_b frames the HPD as a set of simultaneous equations that can be solved via numerical methods. This is one possible way to compute the HPD. An alternative method is to frame the HPD as an optimisation problem, and solve this via numerical methods. Computing the HDR via optimisation: Suppose you have a random variable $X \sim f$ where the ...

Only top voted, non community-wiki answers of a minimum length are eligible


text:
Is this ANSI C++?

Hurkyl

Staff Emeritus
Gold Member
Allow me to present a simplified version of what I'm doing: (yes, I have a reason to do this. :tongue2:)

Code:
class wrapper
{
private:
int &v;
public:
wrapper(int &value):v(value) {}
wrapper& operator=(int value) {
v = value;
return *this;
}

// This is the questionable bit:
operator wrapper&() { return *this; }
};

wrapper make_wrapper(int &value)
{
return wrapper(value);
}

template <typename T>
void foo(T &t)
{
t = 1;
}

int main()
{
int x;
foo(make_wrapper(x));
}
To summarize:

I wish to create a class that acts as a reference type.
I would like to be able to pass my custom reference type into any template function that expects an actual reference.

But here's the problem: C++ doesn't let you make a non-constant reference to a temporary object.

Since it would be rather painful for me to overload every function I might ever want to call to take any combination of the reference types I'm writing, and for every function I develop to have multiple versions, I would like a more automatic approach.

Thus, I provided a conversion into (wrapper&). g++, at least, is happy with it.

Is this valid C++?

ComputerGeek

hmmm... well, if gcc has no issues with it then it must work ok. I know gcc is not yet perfect with respect to the ANSI standard, but oh well.

dmail

Code:
class wrapper
{
private:
int &v;
public:
wrapper(int &value):v(value) {}
wrapper& operator=(int value) {
v = value;
return *this;
}
// This is the questionable bit:
operator wrapper&() { return *this; }
};
wrapper make_wrapper(int &value)
{
return wrapper(value);
}
template <typename T>
void foo(T &t)
{
t = 1;
}
int main()
{
int x;
foo(make_wrapper(x));
}
To summarize:
[/QUOTE]

operator wrapper&() { return *this; }
i dont understand this, can you return an operator?(not that i know of) if you can the this pointer is a pointer to a class and not an operator.
maybe you mean
wrapper operator&(){ return *this; }

Last edited:

Hurkyl

Staff Emeritus
Gold Member
operator wrapper& is a conversion operator, like operator int, or operator char. It allows you to convert an object of type wrapper into an object of type wrapper&

Hurkyl

Staff Emeritus
Gold Member
hmmm... well, if gcc has no issues with it then it must work ok. I know gcc is not yet perfect with respect to the ANSI standard, but oh well.
I'm mainly worried if the behavior is undefined, so that I can't expect it to work in general. =(

dmail

hmmm i havnt seen this before an cant seem to find anything in stroustrup's book have to got a link to an example or something please thanks.

dmail

hmmm i havnt seen this before an cant seem to find anything in stroustrup's book have to got a link to an example or something please thanks.

edit.
hmmm now i see, sorry about that

neurocomp2003

dmail try looking under type casting...cuz thats what your basically doing

you can overload int and double's i believe...i use the typdef double REAL

and you can basically do (REAL*)vector...to get teh ntuple outta class vector.
so if you store an array in teh class
you can do operator REAL*() { return _a; } where _a is teh array. and thus you leave out the remainder of the class.

why do you need ansi standards hurkyl...can't you test it out on your compiler?

Hurkyl

Staff Emeritus
Gold Member
Because it's nice to have code that relies only on defined behavior. :tongue2: This isn't solely for personal use -- I expect others to be able to use this as well.

Last edited:

dduardo

Technology Mentor
Staff Emeritus
If you compile with gcc you can pass the -ansi parameter and it force the standard. You may also need to use the __STRICT_ANSI__ macro.

Hurkyl

Staff Emeritus
Gold Member
Well, I suspect the syntax is valid, I'm more worried about the semantics -- whether I can be guaranteed that my object will live while I pass the reference around... -ansi doesn't guarantee me that, does it?

Hurkyl

Staff Emeritus
Gold Member
Oh wait... I figured out how I can figure this out. I just need to make the destructor do something... gcc will surely put the destructor call exactly where it's supposed to go, right? Or is gcc allowed to wait for a more convenient time?

dpm

Code:
// This is the questionable bit:
operator wrapper&() { return *this; }
What exactly is this operator returning? It's currently untyped.
Other than that, returning *this is perfectly fine, IIRC. You just need to make sure that you don't access the memory after your object has been deleted.

I wish to create a class that acts as a reference type.
Unless this is just for fun, check out the boost library. There's a few classes in there that do exactly this sort of thing, and you can look at the source to see how they did it, too.

Hurkyl

Staff Emeritus
Gold Member
What exactly is this operator returning? It's currently untyped.
It's a conversion operator, syntactically no different than saying operator int(): it says that the compiler may use this operator to implicitly convert anything of type wrapper into one of type wrapper&.

You just need to make sure that you don't access the memory after your object has been deleted.
Right, and my big question is when it gets deleted! I've since run my experiment, and it seems that the lifetime of the temporary is sufficiently long for my purposes (assuming that the compiler called the destructor at the earliest legal opportunity).

Unless this is just for fun, check out the boost library. There's a few classes in there that do exactly this sort of thing, and you can look at the source to see how they did it, too.
Which classes in particular?

master_coda

Hurkyl said:
Oh wait... I figured out how I can figure this out. I just need to make the destructor do something... gcc will surely put the destructor call exactly where it's supposed to go, right? Or is gcc allowed to wait for a more convenient time?
Destructors are called deterministically. When a stack-based object goes out of scope, its destructor is called immediately.

I think what you're doing is undefined behaviour. Your operator wrapper& is fine, but I don't think it's valid to return a reference to a temporary; I think that the temporary is created on the stack, and so any reference to it becomes invalid as soon as make_wrapper returns. I'm not sure, though. It's been a while since I had to think about the object creation rules in C++.

dpm

Hurkyl said:
It's a conversion operator, syntactically no different than saying operator int(): it says that the compiler may use this operator to implicitly convert anything of type wrapper into one of type wrapper&.
Yes, I know that. I just got mixed up and thought that it should be explicitly stated that it is returning something. Apparently conversion operators don't have to have an explicitly stated return type.

Which classes in particular?
http://www.boost.org/libs/smart_ptr/smart_ptr.htm" [Broken]

Last edited by a moderator:

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#### Archived

This topic is now archived and is closed to further replies.

# macro to disable unused parameters warning?

This topic is 5336 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

Hi, I''m programming in Windows and as you may know with WinMain function is required to take a specific set of parameters and a couple of them are rarely used. The problem is I get a "unused parameters warning" whenever I compile my code. I''m a bit of a purist when it comes to warnings. I never disable them with a compiler switch because there have been many times when they''ve helped me catch bugs, and write cleaner/more efficient code. Therefore I don''t want to disable the "unused parameters" warning all together. Is there a macro I can use (ANSI C would be best but even something included via "windows.h" would be OK) that I can wrap around parameter names that I don''t want to be warned about. For example...
int MyFunc(int param1, int param2)
{
TOUCH_PARAMETER(param2);
return (param1 + 1);
}

Any advice would be appreciated. Thanks "Good code will come and go, but bad code sticks around."

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Don''t name the parameter and you won''t get a warning:

int whatever(int param1, int, int param3){    use(param1, param3);    return 0;}

This does not produce any warnings.

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#define UNUSED( PARAMETER )int MyFunc(int param1, int UNUSED( param2 ) ){    return (param1 + 1);}

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You might want to look at UNREFERENCED_PARAMETER (include <windows.h>

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Using MSVC, the warning you will get for an unreferenced paramter is 4100. Temporarily turning off this warning (or any warning) is simple:

#pragma warning(disable: 4100) // turn off warning 4100int MyFunc(int param1, int param2){    return (param1 + 1);}#pragma warning(default: 4100) // restore 4100''s warning

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I just don''t label the parameter like first reply suggested. If you want to keep a little documentation though, I tend to give the variable a relevent label and comment it out so you know what it is that is being ignored. There really isn''t a need to use macros and using pragma is compiler dependent.

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Thanks all for your fast response.
I was fishing for the UNREFERENCED_PARAMETER aluded to by the anonymous poster. I had seen it somewhere in an example but couldn''t remember enough of the name to find the actual macro name via Google.

It''s also interesting to me that leaving a parameter unnamed doesn''t give the warning... not sure if I like that idea. I mean I''m not doubting that it works I''m just thinking if it''s a good thing from a Programming Practices perspecitive. If I did utilize that quirk I think I would go the route of Willm.

Anyway thanks for all the info.

"Good code will come and go, but bad code sticks around."

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quote:
Original post by TreborZehn
Thanks all for your fast response.
I was fishing for the UNREFERENCED_PARAMETER aluded to by the anonymous poster. I had seen it somewhere in an example but couldn''t remember enough of the name to find the actual macro name via Google.

It''s also interesting to me that leaving a parameter unnamed doesn''t give the warning... not sure if I like that idea. I mean I''m not doubting that it works I''m just thinking if it''s a good thing from a Programming Practices perspecitive. If I did utilize that quirk I think I would go the route of Willm.

It isn''t a quirk, it''s a feature.

Seriously, that''s one of its legitamate uses. I''ve seen companies that use macros to signal unused variables, other comment out the label. It''s doesn''t make that much difference, although using the windows.h macro will obviously give you a ''dependency'' to Windows which should be avoided.

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quote:
Original post by seanw
...using the windows.h macro will obviously give you a ''dependency'' to Windows which should be avoided.
About this ''obvious dependency'' to Windows.... Is it difficult to define UNREFERENCED_PARAMETER your platform or what are you saying? Then maybe you should consider a platform change.

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quote:
Original post by Anonymous Poster
About this ''obvious dependency'' to Windows.... Is it difficult to define UNREFERENCED_PARAMETER your platform or what are you saying? Then maybe you should consider a platform change.

No, I just meant you''d be better creating your own macro and placing it in your own header and including that, instead of using the one in windows.h. If you try to compile on a non-windows platform, that macro isn''t going to be available to you. It''s not a big deal, just that you shouldn''t create needless dependencies and this one can easily be avoided. Just good programming practice.

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Thank you for the Abundance that is.

Thank you for the Abundance that is given.
Thank you for the so many experiences and so many things
Thank you for this Dance.

Thank you Life

–Gary van Warmerdam

— with Gobinde Kari Valenzuela at Playa Guiones.


text:
# test

## February 05, 2017 15:38

This is a test with an inline equation. As seen in (1), we see that $E=m{c}^{2}$.[1]

### 1 this is also a test

 $e=m{c}^{2}$ (1)

### References

[1]   A. Barut, G. Maccarrone, and E. Recami, Nuovo Cimento A 71, 509 (1982), 10.1007/BF02770989.


text:
# Baseband

In signal processing, "baseband" is an adjective that describes signals and systems whose range of frequencies is measured from zero to a maximum bandwidth or highest signal frequency; it is sometimes used as a noun for a band of frequencies starting at zero. It can often be considered as synonym to lowpass, and antonym to passband.

Various uses

* A baseband bandwidth is equal to the highest frequency of a signal or system, or an upper bound on such frequencies. [cite book | title = Information, Transmission, Modulation and Noise: A Unified Approach to Communication Systems | author = Mischa Schwartz | publisher = McGraw-Hill | year = 1970| url = http://books.google.com/books?id=-gkjAAAAMAAJ&q=baseband-bandwidth&dq=baseband-bandwidth&pgis=1 ] By contrast, a non-baseband (passband) bandwidth is the difference between a highest frequency and a nonzero lowest frequency.
* A baseband signal or lowpass signal is a signal that can include frequencies that are equal to or very near zero, by comparison with its highest frequency (for example, a sound waveform can be considered as a baseband signal, whereas a radio signal or any other modulated signal is not). [cite book | title = Communication System Design Using Dsp Algorithms: With Laboratory Experiments for the TMS320C30 | author = Steven Alan Tretter | publisher = Springer | year = 1995 | isbn = 0306450321 | url = http://books.google.com/books?id=deHQeNxHhyEC&pg=PA65&ots=Ts-8_QUSY6&dq=baseband-signal+lowpass-signal&sig=qAAD0vMbwCsbkEcTnuXETBi6-Nc#PPA66,M1 ]
* A baseband channel or lowpass channel (or system, or network) is a channel (e.g. a telecommunications system) that can transfer frequencies that are equal to or very near zero. [cite book | title = Digital Signal Transmission | author = Chris C. Bissell and David A. Chapman | publisher = Cambridge University Press | year = 1992 | isbn = 0521425573 | url = http://books.google.com/books?id=cj12nN2uW0AC&pg=PA149&ots=U2FcfQ1Vat&dq=called-baseband-channels&sig=c5lsuyDP4KxNWHHD90o0JZLEre0 ] Examples are serial cables and local area networks (LANs).
* Baseband modulation, also known as line coding, [cite book | title = CMOS Data Converters for Communications | author = Mikael Gustavsson and J. Jacob Wikner | publisher = Springer | year = 2000 | isbn = 079237780X | url = http://books.google.com/books?id=D_I2XvNOc4wC&pg=PA28&ots=Ui08f8PJJs&dq=passband+baseband&sig=o6Ldvs4s6G67SQjlrSbPQIqFtoI] aims at transferring a digital bit stream over an analog baseband channel, as an alternative to "carrier-modulated" approaches. [cite book | title = Digital Baseband Transmission and Recording | author = Jan W. M. Bergmans | publisher = Springer | isbn = 0792397754 | year = 1996 | url =http://books.google.com/books?id=TN-sTybrCLsC&pg=PR11&ots=v-NpNWiBwv&dq=baseband+carrier-modulated&sig=yEKQsywJhz2jg4w-vsnRp8sXyKk ]
* An equivalent baseband signal or equivalent lowpass signal is – in analog and digital modulation methods with constant carrier frequency (for example ASK, PSK and QAM, but not FSK) – a complex valued representation of the modulated physical signal (the so called passband signal or RF signal). The equivalent baseband signal is $Z\left(t\right)=I\left(t\right)+jQ\left(t\right),$ where $I\left(t\right)$ is the inphase signal, $Q\left(t\right)$ the quadrature phase signal, and $j$ the imaginary unit. In a digital modulation method, the $I\left(t\right)$ and $Q\left(t\right)$ signals of each modulation symbol are evident from the constellation diagram. The physical passband signal corresponds to:$I\left(t\right)cos\left(omega t\right) - Q\left(t\right)sin\left(omega t\right) = mathrm\left\{Re\right\}\left\{Z\left(t\right)e^\left\{jomega t\right\}\right\},$
where $omega$ is the carrier angular frequency in rad/s.
* In an equivalent baseband model of a communication system, the modulated signal is replaced by a complex valued equivalent baseband signal with carrier frequency of 0 Hertz, and the RF channel is replaced by an equivalent baseband channel model where the frequency response is transferred to baseband frequencies.
* A signal "at baseband" is usually considered to include frequencies from near 0 Hz up to the highest frequency in the signal with significant power.

In general, signals can be described as including a whole range of different frequencies added together. In telecommunications in particular, it is often the case that those parts of the signal which are at low frequencies are 'copied' up to higher frequencies for transmission purposes, since there are few communications media that will pass low frequencies without distortion. Then, the original, low frequency components are referred to as the baseband signal. Typically, the new, high-frequency copy is referred to as the 'RF' (radio-frequency) signal.

The concept of baseband signals is most often applied to real-valued signals, and systems that handle real-value signals. Fourier analysis of such signals includes a negative-frequency band, but the negative-frequency information is just a mirror of the positive-frequency information, not new information. For complex-valued signals, on the other hand, the negative frequencies carry new information. In that case, the full two-sided bandwidth is generally quoted, rather than just the half measured from zero; the concept of baseband can be applied by treating the real and imaginary parts of the complex-valued signal as two different real signals.

Modulation

A signal at baseband is often used to modulate a higher frequency carrier wave in order that it may be transmitted via radio. Modulation results in shifting the signal up to much higher frequencies (radio frequencies, or RF) than it originally spanned. A key consequence of the usual double-sideband amplitude modulation (AM) is that, usually, the range of frequencies the signal spans (its spectral bandwidth) is doubled. Thus, the RF bandwidth of a signal (measured from the lowest frequency as opposed to 0 Hz) is usually twice its baseband bandwidth. Steps may be taken to reduce this effect, such as single-sideband modulation; the highest frequency of such signals greatly exceeds the baseband bandwidth.

Some signals can be treated as baseband or not, depending on the situation. For example, a switched analog connection in the telephone network has energy below 300 Hz and above 3400 Hz removed by bandpass filtering; since the signal has no energy very close to zero frequency, it may not be considered a baseband signal, but in the telephone systems frequency-division multiplexing hierarchy, it is usually treated as a baseband signal, by comparison with the modulated signals used for long-distance transmission. The 300 Hz lower band edge in this case is treated as "near zero", being a small fraction of the upper band edge.

The figure shows what happens with AM modulation:

The simplest definition is that a signal's baseband bandwidth is its bandwidth before modulation and multiplexing, or after demultiplexing and demodulation.

The composite video signal created by devices such as most newer VCRs, game consoles and DVD players is a commonly used baseband signal.

ee also

*Broadband - generally refers to transmission of data over numerous frequencies
*Wideband - a communications medium or signal that spans a large (continuous) range of frequencies, or is wide compared to something else
*Narrowband - the opposite of wideband

References

Wikimedia Foundation. 2010.

### Look at other dictionaries:

• baseband — /ˈbeɪsbænd/ (say baysband) noun 1. the frequency band containing all the frequencies of the modulating signals present in a broadcast signal. –adjective 2. of or relating to or employing baseband: a baseband network …   Australian English dictionary

• Baseband —  Ne pas confondre avec bande de base qui est une technique de télécommunication Dans le domaine des télécommunications, un baseband (anglicisme qui signifie à l origine « bande de base ») est un dispositif qui communique en bande… …   Wikipédia en Français

• baseband — moduliavimo dažnių juosta statusas T sritis automatika atitikmenys: angl. baseband; modulation frequency band vok. Modulationsband, n rus. полоса модулирующих частот, f pranc. bande de base, f; bande modulante, f …   Automatikos terminų žodynas

• baseband —    The total frequency band occupied by the aggregate of all voice and data signals used to modulate a radio carrier. In a baseband LAN, the entire capacity of the cable transmits one high speed channel which all attached devices share …   IT glossary of terms, acronyms and abbreviations

• Baseband — Spektrum eines beispielhaften Basisbandsignals mit der Bandbreite B Basisband, oder auch Basisbandlage, ist ein Begriff aus der Nachrichtenübertragungstechnik. Er bezeichnet denjenigen Frequenzbereich in dem sich das zu übertragende Nutzsignal… …   Deutsch Wikipedia

• baseband — noun a) The frequency range occupied by a message signal prior to modulation. b) A method of transmission in which only one frequency channel is used at a time. Syn: lowpass See Also: broadband, n …   Wiktionary

• Baseband — Basisband, einkanaliges Übertragungsmedium, Gegensatz: Breitband …   Acronyms

• Baseband — Basisband, einkanaliges Übertragungsmedium, Gegensatz: Breitband …   Acronyms von A bis Z

• baseband — n. transfer method in which a signal is transferred directly in digital form with no modulation (suitable for only short term transmissions) …   English contemporary dictionary

• baseband — base·band …   English syllables

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Browse Questions

# In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be $4.12 \times 10^{-15}Vs$. Calculate the value of Planck’s constant.

$\begin {array} {1 1} (a)\;6.592 \times 10^{-34} Js & \quad (b)\;6.626 \times 10^{-34} Js \\ (c)\;659.2 \times 10^{-35} Js & \quad (d)\;66.26 \times 10^{-35} Js \end {array}$

Can you answer this question?

Ans : (a) $6.592 \times 10^{-34} Js$
The slope of the cut-off voltage(V) versus frequency(v)
of an incident light is given as:
$V/v = 4.12 \times 10^{-15} Vs$
$V$ is related to frequency by the equation: $hv=eV$
So, $h=e \times V/v = 1.6 \times 10^{-19} \times 4.12 \times 10^{-15} = 6.592 \times 10^{-34} Js$

answered Dec 24, 2013
edited Mar 13, 2014


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anonymous 3 years ago find the domain and range of (sin x + cos x)

• This Question is Closed
1. anonymous

how

2. anonymous

What is the range of each function?

3. anonymous

-1 to +1

4. anonymous

now ?

5. anonymous

now satellite73 will help you

6. anonymous

it might help to know that $$a\sin(x)+b\cos(x)=\sqrt{a^2+b^2}\sin(x+\theta)$$ for suitable $$\theta$$ that should help a great deal with the range

7. anonymous

asin(x)+bcos(x)=a2+b2−−−−−−√sin(x+θ) but how do you get this ?

8. anonymous

square root is on the outside it is $$\sqrt{a^2+b^2}\sin(x+\theta)$$ and it is a consequence of the "addition angle formula"

9. anonymous

@satellite73 i am having very problem in it... can you explain the function chapter from the starting, i will be very thankful to you..

10. anonymous

i do not know exactly what you mean by "explain the function chapter" you could graph the function $\sin(x)+\cos(x)$ using technology to see what you get or you could surmise that the largest the sum could be is if they were the same value, making $$\sin(x)=\frac{\sqrt{2}}{2}$$ and also $$\cos(x)=\frac{\sqrt{2}}{2}$$ and therefore the max would be $$\sqrt{2}$$

11. anonymous

or more simply you could use the formula i wrote above, telling you that $\sin(x)+\cos(x)=\sqrt{2}\sin(x+\frac{\pi}{4})$ and now the range is more or less obvious

12. anonymous

Guru JEE thanks

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# Uniqueness results for inverse Robin problems with bounded coefficient

Uniqueness results for inverse Robin problems with bounded coefficient - Descarga este documento en PDF. Documentación en PDF para descargar gratis. Disponible también para leer online.

1 APICS - Analysis and Problems of Inverse type in Control and Signal processing CRISAM - Inria Sophia Antipolis - Méditerranée 2 POEMS - Propagation des Ondes : Étude Mathématique et Simulation Inria Saclay - Ile de France, ENSTA ParisTech UMA - Unité de Mathématiques Appliquées, CNRS - Centre National de la Recherche Scientifique : UMR7231

Abstract : In this paper we address the uniqueness issue in the classical Robin inverse problem on a Lipschitz domain $\Omega\subset\RR^n$, with $L^\infty$ Robin coefficient, $L^2$ Neumann data and isotropic conductivity of class $W^{1,r}\Omega$, $r>n$. We show that uniqueness of the Robin coefficient on a subpart of the boundary given Cauchy data on the complementary part, does hold in dimension $n=2$ but needs not hold in higher dimension. We also raise on open issue on harmonic gradients which is of interest in this context.

Keywords : holomorphic Hardy–Smirnov classes harmonic analysis complex analysis Robin inverse problem unique continuation Sobolev spaces elliptic regularity

Autor: Laurent Baratchart - Laurent Bourgeois - Juliette Leblond -

Fuente: https://hal.archives-ouvertes.fr/

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