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# Homework Help: Two Particle Schrodinger Equation in Polar Coordinates

1. Aug 5, 2011

### naharrison

1. The problem statement, all variables and given/known data

Consider two non-identical, non-interacting particles of mass M that are constrained to move on a circle of radius R. Write down the Schrodinger equation for this problem and ﬁnd the eigenfunctions and energy levels of this system.

2. Relevant equations

(see below)

3. The attempt at a solution

I started by defining a coordinate system where the circle is centered at the origin of the x-y plane (z = 0). The particles don't interact and there is no external potential so V = 0. Also, using spherical coordinates, $$r = R, \ \theta = \frac{\pi}{2}$$ . I then wrote down the Schrodinger equation:

$$[-\frac{\hbar^2}{2m_{1}}\nabla^{2}_{1} - \frac{\hbar^2}{2m_{2}}\nabla^{2}_{2} + V]\psi(r,\theta,\phi) = E\psi(r,\theta,\phi)$$

The Laplacian in spherical coordinates is:

$$\frac{\partial^2}{\partial r^2} + \frac{2}{r}\frac{\partial}{\partial r} + \frac{1}{r^2}[\frac{1}{sin\theta}\frac{\partial}{\partial\theta}(sin\theta\frac{\partial}{\partial\theta}) + \frac{1}{sin^2\theta}\frac{\partial^2}{\partial \phi^2}]$$

Since r and theta are fixed, the Schrodinger equation reduces to:

$$\frac{1}{2}(\frac{\hbar}{R})^2(m_1 + m_2)\frac{\partial^2\psi}{\partial\phi^2} + E\psi = 0$$

letting $$a = \sqrt{\frac{2R^2E}{\hbar^2(m_1 + m_2)}}$$ the solution of the Schrodinger equation is:

$$\psi = Ae^{a\phi} + Be^{-a\phi}$$

This is where I get stuck. I assume the next step is to apply boundary conditions which will then give the energy levels. How do you do this in this situation? I tried:

$$\psi(0) = \psi(2\pi)$$
$$\frac{\partial\psi(0)}{\partial\phi} = \frac{\partial\psi(2\pi)}{\partial\phi}$$

but the results didn't make any sense. Any thoughts? Thank you in advance!

2. Aug 5, 2011

### mathfeel

How did you get the mass to the numerator?
You are missing an $i$ in both terms. Also please check the dimension of $a$. With that applying b.c. should give you the quantisation condition on E.

3. Aug 5, 2011

### naharrison

Thank you for your help. Both careless mistakes, I need to be more careful. Now I am trying to determine the degeneracy of the first three energy levels. Here is my logic, I am not sure if it is correct.

Ground state degeneracy = 2. Neither particle is excited, but since they are distinguishable you could swap their positions and get a different state with the same energy. So, state 1 = particle X in ground state at position 1 and particle Y in ground state at position 2; state 2 = particle X in ground state at position 2 and particle Y in ground state at position 1.

First excited state degeneracy = 4. One particle is excited and the other one isn't, and again you can flip their positions to get another state.

Second excited state. This one is more difficult. It seems to depend on which of the following situations has a larger energy, one particle in its second excited state and one particle in its ground state, or both particles in their first excited state. In the first situation the degeneracy would be 4, in the second situation the degeneracy would be 2.

Here is one more thing that is confusing me. After I correct the mistakes from my first post I get the energy to be $$E = \frac{\hbar^2 n^2}{4MR^2} \ , \ n = 0,1,2,3,...$$ Clearly there is a unique value of energy for each value of $n$. How could you determine the degeneracies just from this equation for the energy levels? Thanks again for any feedback.

4. Aug 5, 2011

### naharrison

where $m_1 = m_2 = M \$ (this is given in the original question) sorry for not mentioning that.

5. Aug 10, 2011

### JFuld

fyi im just learning this too

Was V=0 given in the problem?

The particles are confined to a circle of radius R, I would have guessed there would be an infinite potential barrier at r=R.

regardless, from what I have learned thus far,to calculate the eigenfunctions and energy levels, consider a general wave function that describes a system two particles

psi( r1,ms1, r2,ms2) i think you only need polar rather than spherical coordinates

where r1 is the radius of the first particle, ms1 is the internal angular momentum of the first particle and so forth

Since the particles are non interacting, the ms term can be ignored
leaving psi(r1,r2)=psi(r1)psi(r2). I believe you solve psi(r1)psi(r2) separately, but they should be the same wave function

I would solve the energy levels for each particle separately, taking

E(total)=E1n E2n

Also, exchanging coordinates of non-interacting particles won't affect the probability distribution

hence psi(r1,r2)=psi(r2,r1)

6. Aug 10, 2011

### naharrison

V=0 was not explicitly given, but the particles do not interact and and no other external potential is given. I think you are confusing ON a circle with INSIDE a circle. If they were confined to move inside a circle, then yes, I think there would be an infinite potential at r=R. However this just has the particles moving ON the circle. It is like a particle moving in just the x-direction in Cartesian coordinates, there is no need to assume infinite potential barriers keeping it on the x-axis.

I think you are right about solving for the energy levels separately. This would lead to:
$E^{(1)}_n = bn^2$ and $E^{(2)}_{n'} = bn'^2$ where $b = \frac{\hbar^2}{2MR^2}$. Therefore,
$$E^{total}_{n,n'} = b(n^2 + n'^2)$$
Therefore, the ground state is $n = 0 \ , \ n' = 0$ leading to $E = 0$
And the first excited state is $n = 1 \ , \ n' = 0$ OR $n = 0 \ , \ n' = 1$ leading to $E = b$
And the second excited state is $n = 1 \ , \ n' = 1$ leading to $E = 2b$
And the third excited state (which is not asked for in the problem, but I will include it anyway) is $n = 2 \ , \ n' = 0$ or $n = 0 \ , \ n' = 2$ leading to $E = 4b$

If $g_n$ is the degeneracy of the $n^{th}$ excited state then $g_0 = 1, \ g_1 = 2, \ g_2 = 1$. Is this correct?


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# Definition:Vector Cross Product/Complex/Definition 4

## Definition

Let $z_1$ and $z_2$ be complex numbers.

The cross product of $z_1$ and $z_2$ is defined as:

$z_1 \times z_2 := \dfrac {\overline {z_1} z_2 - z_1 \overline {z_2}} {2 i}$

where:

$\overline {z_1}$ denotes the complex conjugate of $z_1$
$\overline {z_1} z_2$ denotes complex multiplication.

format: plain


text:
## PathFeedrateClassType

The feedrate for the axes, or a single axis, associated with a \mtmodel{Path} component a vector. $\frac{MILLIMETER}{SECOND}$ The measurement of the feedrate for the axes, or a single axis, associated with a path component-a vector.

The representation of the PathFeedrateClassType ObjectType in the address space is shown in the following table:

Name Attribute
NodeId ns=1;i=2303
NamespaceUri http://opcfoundation.org/UA/MTConnect/v2/
BrowseName PathFeedrateClassType
NodeClass ObjectType
IsAbstract False
SubtypeOf MTSampleClassType


text:
Browse Questions

# Prove that the perpendicular bisectors of a triangle are concurrent.

Toolbox:
• To prove that the $\perp$ bisectors are concurrent, prove that the third bisector passing through the intersecting point of other two $\perp$ bisectors is also $\perp$ to its corresponding side.
• $\overrightarrow {AB}=\overrightarrow {OB}-\overrightarrow {OA}$
• If $\overrightarrow a.\overrightarrow b=0\:then \:\overrightarrow a\:is\:\perp\:to\:\overrightarrow b$
• Section formula: If D is mid point of AB, then $\overrightarrow {OD}=\frac{\overrightarrow {OA}+\overrightarrow {OB}}{2}$
Let ABC be a triangle. D,E F be mid points of BC,AC and AB respectively
Let O (origin) be the point of intersection of the $\perp$ bisectors
OD and OE
Let $\overrightarrow {OA}=\overrightarrow a,\:\overrightarrow {OB}=\overrightarrow b\:and\:\overrightarrow {OC}=\overrightarrow c$
Since D is mid point of BC, we know from section formula,
$\overrightarrow {OD}=\frac{\overrightarrow {OB}+\overrightarrow {OC}}{2}=\frac{\overrightarrow b+\overrightarrow c}{2}$,
E is mid point of AC
$\Rightarrow\:\overrightarrow {OE}=\frac{\overrightarrow a+\overrightarrow c}{2}$ and
$\overrightarrow {OF}=\frac{\overrightarrow a+\overrightarrow b}{2}$
But $\overrightarrow {OD}$ is $\perp$ bisector to $\overrightarrow {BC}$
$\Rightarrow \overrightarrow {OD}.\overrightarrow {BC}=0$
$\Rightarrow\:\frac{\overrightarrow b+\overrightarrow c}{2}.(\overrightarrow c-\overrightarrow b)=0$
$\Rightarrow\:|\overrightarrow c|^2-|\overrightarrow b|^2=0$............(i)
Similarly $\overrightarrow {OE}$ is $\perp$ bisector to $\overrightarrow {AC}$
$\Rightarrow \overrightarrow {OE}.\overrightarrow {AC}=0$
$\Rightarrow\:\frac{\overrightarrow c+\overrightarrow a}{2}.(\overrightarrow c-\overrightarrow a)=0$
$\Rightarrow\:|\overrightarrow c|^2-|\overrightarrow a|^2=0$................(ii)
Subtrancting (i) - (ii) we get
$\Rightarrow\:(|\overrightarrow c|^2-|\overrightarrow b|^2)-(|\overrightarrow c|^2-|\overrightarrow a|^2)=0$
$\Rightarrow |\overrightarrow a|^2-|\overrightarrow b|^2=0$...........(iii)
To prove that the $\perp$ bisectors are concurrent, we have to prove
that the third bisector $\overrightarrow {OF}$ is $\perp$ to $\overrightarrow {AB}$
That is we have to prove that $\overrightarrow {OF}.\overrightarrow {AB}=0$
$\Rightarrow \overrightarrow {OF}.\overrightarrow {AB}=\frac{\overrightarrow a+\overrightarrow b }{2}.(\overrightarrow b-\overrightarrow a)$
$=\frac{|\overrightarrow b|^2-|\overrightarrow a|^2}{2}=0$ from (iii)
$\Rightarrow\:\overrightarrow {OF}.\overrightarrow {AB}=0$
$\Rightarrow\:\overrightarrow {OF}$ is $\perp$ to $\overrightarrow {AB}$
$\Rightarrow\:\overrightarrow {OF}$ is also a perpendicular bisector.
$\Rightarrow$ All the three $\perp$ bisectors are concurrent.
Hence proved.
edited Apr 24, 2013


text:
# 《Microeconomic Theory》的笔记-第173页

###### gui
• 页码：第173页 2012-02-02 12:41:40

## gui对本书的所有笔记  · · · · · ·

• ###### 第199页

P 6.D.2. From (iii) to (i). Common way to prove is to use two times of integration by part...

• ###### 第175页

E 6.B.2. u, which represents a preference, has the expected utility form. Then the prefere...

• ###### 第197页

D 6.D.2 IF F and G have same mean, but F have smaller variance than G. Does F second-ord...

• ###### 第101页

Exact consumer's surplus and deadweight loss, Hausman(1981). 是这道题的原型。文章本身是非...


text:
## Algebra 1

$2x=4$
Each side of the diagram represents one side of the equation. The bar separating the two sides represents the equal sign. The left side includes two green tiles. Each green tile represents a variable such as x, so 2 green tiles are 2x. The right side contains 4 yellow tiles. Each yellow tile represents +1, so 4 yellow tiles are +4. Combine the parts into an equation. $2x=4$


text:
# Changing pin direction without glitching on LPC800

I'm looking at an LPC812 Cortex M0+ microcontroller and its GPIO functions. On a PIC microcontroller, one would always set the LATch prior to changing the direction of a given pin. This avoids the possibly nasty situation where a pin "gitches". ie: the output state of the pin is a 1 just before you change direction and set it to a 0. This results in a pin output changing from 1 to 0 over a sub-microsecond duration. (related: Why set PORTx before TRISx?)

However, it is not obvious from the user manual (UM10601) whether writing a value to a pin which is currently an input has any logical effect (it will not have a physical effect on the pin state as it's an input). Thus if I write a zero to PIN0 (page 92/93) that should in theory clear all the output bits. If they were latched, which isn't certain.

Section 7.7.2 says that "If either or both of these conditions (ie: set to gpio and set to output) is (are) not met, writing to the pin has no effect.". Which isn't helpful as I know there's no physical effect.

Under the IOCON description (6.3, page 68) is the following Remark:

If the open-drain pins PIO0_10 and PIO0_11 are not available on the package, prevent the pins from internally floating as follows: Set bits 10 and 11 in the GPIO DIR0 register to 1 to enable the output driver and write 1 to bits 10 and 11 in the GPIO CLR0 register to drive the outputs LOW internally.

In my opinion this is an incorrect sequence (unless the GPIO port outputs defaults to low, something else that isn't documented). On a PIC16 I would set the HI/LO state first, then change the direction.

I admit that if the part had arrived I would just test this on the bench. However I'll put this on stackexchange as I'm seeing examples everywhere which set direction and then the value, all of which may be glitching output pins.

You're over-analysing this. There are two registers involved in a GPIO pin, the output data register, and the direction register. Both of these are writable at any time (and I agree your Section 7.7.2 quote does not make this super-clear.)

The remark on p68 doesn't care about a transition, it cares about a power-optimised stable end state. Changing the order makes no real difference, but is cleaner as you describe.

There is no changing from 1 to 0 over a sub-microsecond duration. When you enable the output, it takes the driven value in just the same way as if you drive it from 1 to 0 using the data value.

It is only when the reset data value is 0, the tri-state is pulled up, and after enabling the output you write 1 to the data pin that you would see a pulse.

All peripherals are different, but for example

A read from GPIODATA returns the last bit value written if the respective pins are configured as output, or it returns the value on the corresponding input GPIN bit when these are configured as inputs. All bits are cleared by a reset.

Specifically, the last write and not the last write whilst the pin direction was set as output.

• I over-analyse everything :-) I realise what you are saying about the remark and power saving. But if the pins are physically available, the same advice still stands as there are no PU resistors on those lines. – carveone Mar 2 '18 at 12:48
• Your final paragraph is exactly the problem. I would assume the reset value is 0 and a pulse is what you want to avoid (some of the time at least). I think you are saying that you can change the order: ie: write 1 to the GPIO CLR0 first and then set the GPIO DIR0 register to 1 to enable the output driver and that is effective. I wasn't convinced that it would be effective as the documentation was just a bit too fuzzy there. – carveone Mar 2 '18 at 12:50
• yes, almost certainly that will work. Data is persistent over tri-state en/disable, and is almost certainly writable even when there is no immediate side effect. Writing clear data sheets is hard. – Sean Houlihane Mar 2 '18 at 12:55
• Thanks for clarifying that. And yes, it's even trickier with nit pickers like me. I was in QA before and am always thinking that nothing is obviously so - if it isn't documented than it might not be true! The PIC has LAT which is completely explicit. When it just had PORT, it wasn't so clear, especially on a read-modify-write. – carveone Mar 2 '18 at 13:00

However, it is not obvious from the user manual (UM10601) whether writing a value to a pin which is currently an input has any effect.

It has not. See output enable in schematic below.

In my opinion this is an incorrect sequence (unless the GPIO port outputs defaults to low, something else that isn't documented).

They want you to enable the strong pull-down. The order does not matter, since it was already floating anyway. (see reset value of IOCON/GPIO registers)

• Thanks for including that image for me. When I said "has any effect" I mean "any logical effect at all, including setting some register". My concern is over the "data output" line. If that is enabled, then one of the strong pull up or pull down is enabled. My thinking was that it was unpredictable which one unless "data output" is set by a write to PIN0 or CLR0 or SET0 or whatever. – carveone Mar 2 '18 at 12:45
• @carveone But only if Output Enable is also enabled. And if you enable Open Drain, strong pull-up will never be active. – Jeroen3 Mar 2 '18 at 20:54
• Argh! I don't think I was thinking when I typed the above comment! Yes, I meant to say that when the Output Enable line becomes enabled, it is unclear (from the data sheet) which of pull up or pull down becomes enabled. Nor is it clear whether writing to, say, SET0 will set the data output line high. Now I do know that from reset, changing DIR0 to output will bring the pin low and that SET0/CLR0 will set/clear the data output line (even if an input) but the data sheet doesn't make that explicit. – carveone Mar 2 '18 at 21:40
• And it all seems a bit pedantic but I have found that being suspicious and obsessive up front saves me being confused in the long run. I wonder if NXP will take suggestions on their data sheets :-) – carveone Mar 2 '18 at 21:44


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# If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at

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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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04 Aug 2013, 09:52
2
Stiv wrote:
If two of the four expressions x + y, x + 5y, x - y, and 5x - y are chosen at random, what is the probability that their product will be of the form of x^2 - (by)^2, where b is an integer?
A) 1/2
B) 1/3
C) 1/4
D) 1/5
E) 1/6

the result will be of form : $$x^2 - (by)^2$$ only when we choose x+y and x-y...rest all pair will not satisfy.
therefore favorable case=1
total case = we have to select 2 from 4 ....= $$4C2$$ = 6

probability = favorable/total =$$1/6$$
hence E
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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05 Sep 2013, 07:44
How come the answer is not b) 1/3? The way I approached it was: first chose an expression (any one), then out of the three remaining, one is its "partner". Hence, 1/3. Am I double counting with this approach? Thanks!
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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05 Sep 2013, 07:49
dumptruck wrote:
How come the answer is not b) 1/3? The way I approached it was: first chose an expression (any one), then out of the three remaining, one is its "partner". Hence, 1/3. Am I double counting with this approach? Thanks!

Not all expressions have their appropriate pairs.

Check here: if-two-of-the-four-expressions-x-y-x-5y-x-y-5x-y-are-92727.html#p713823
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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19 Nov 2013, 11:55
1
Hi Bunuel,

Here is how I did it. Since we can choose either (x+y) or (x-y) at first, the probability is 2/4 and for second pick we have only one of these left so the prob is 1/3. Total prob= 2/4*1/3=1/6
Is that right?
Thanks!
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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26 Nov 2013, 18:09
NvrEvrGvUp wrote:
I also got 1/6 but not really understanding why the equation pair is (x+y)(x-y).

My thoughts were that we needed to factor $x^2-(by)^2$ into (x-by)(x+by) where the factored "b" was the squared root of the original "b", or essentially a placeholder for a perfect square.

This led me to think that the only equation pair possible that would work to satisfy the original ask would be (x+5y) and (x-y) since we need two equations with differing signs but also needed an constant infront of the "y" term.

Thanks,
Rich

I thought the same thing. Bunuel why does this not work?
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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27 Nov 2013, 05:11
1
runningguy wrote:
NvrEvrGvUp wrote:
I also got 1/6 but not really understanding why the equation pair is (x+y)(x-y).

My thoughts were that we needed to factor $x^2-(by)^2$ into (x-by)(x+by) where the factored "b" was the squared root of the original "b", or essentially a placeholder for a perfect square.

This led me to think that the only equation pair possible that would work to satisfy the original ask would be (x+5y) and (x-y) since we need two equations with differing signs but also needed an constant infront of the "y" term.

Thanks,
Rich

I thought the same thing. Bunuel why does this not work?

Not sure I understand what you mean but the pair we need is $$(x+y)(x-y)$$ because $$(x+1*y)(x-1*y)$$ is of the form of $$(x-by)(x+by)$$. How is $$(x+5y)(x-y)$$ of the form of $$x^2 - (by)^2$$?
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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14 Oct 2015, 18:49
ronny13 wrote:
If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at random, what is the probability that their product will be of the form of x^2 -(by)^2, where b is an integer?

A. 1/2
B. 1/3
C. 1/4
D. 1/5
E. 1/6

First of all, realize that only two expressions form the given 4, when multiplied will give the form $$x^2 -(by)^2$$. These are $$x+y$$ and $$x-y$$
$$(x+y)(x-y) = x^2 - y^2 = x^2 - (by)^2$$ ; $$b = 1$$

Now, the question becomes: Probability of choosing these 2 expressions from given 4 expressions.

In other words,
What is the probability of choosing a and c from the set {a,b,c,d}??

Solution:
Choose (a AND c) OR (c AND a)
=> $$\frac{1}{4}*\frac{1}{3} + \frac{1}{4}*\frac{1}{3}$$
=> $$2(\frac{1}{4}*\frac{1}{3})$$
=> $$\frac{1}{6}$$
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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25 Oct 2015, 11:08
1
atshy21saraf wrote:
Engr2012 wrote:
atshy21saraf wrote:
Q. If two of the expressions (x+y), (x-y), (x+5y), (5x-y) are chosen then What is the probability that their product will be of the form of x^2-(by)^2, b is an integer.
My doubt: What is the difference between 4c2 and (4c1 x 3c1) in case of total possibilities?

A. 1/2
B. 1/3
C. 1/4
D. 1/5
E. 1/6

Search for a question before you post. This question has already been discussed before. Follow posting guidelines (link in my signatures).

Topics merged. Refer above for the solution.

My bad, I will keep that in mind. I am not very familiar with forum rules.

Although my doubt is still not cleared.

Why did't we do 4C1*3C1 to select 2 out of 4 expression?

Lets say you have 4 objects A,B,C,D and you are asked to select 2 out of the 4. In this case there is no difference whether you take A and then B or you take B and then A. You still are selecting A and B.

But if I ask you that the order of selection matters, i.e. a selection process in which A and then B will be different from B and then A then you can only do 4C1*3C2

Coming back to the original question, a^2-b^2 or (a-b)(a+b) can be formed either by choosing a-b and then a+b or by choosing a+b and then a-b. It does not matter which order you choose a+b and a-b. This is the reason why the number of ways = 4C2.

Hope this helps.
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If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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Updated on: 01 Jun 2017, 09:31
1
Bunuel wrote:
spc11 wrote:
If two of the four expressions x+y, x+5y, x-y, 5x-y are chosen at random, what is the probability that their product will be of the form x [square] - (by) square, where b is a constant.
A.1/2
B.1/3
C. 1/4
D.1/5
E.1/6

First, notice that we are being tested on the difference of squares. We can restate the problem as: What is the probability, when selecting two expressions at random, that the product of those expressions will create a difference of two squares? Remember, the difference of two squares can be written as follows:

a^2 - b^2 = (a + b)(a - b)

So, x^2 - (by)^2 can be written as (x + by)(x - by). Thus, we are looking for two expressions in the form of (x + by)(x - by). Although this problem is attempting to trick us with the expressions provided, the only two expressions that when multiplied together will give us a difference of squares are x + y and x - y. When we multiply x + y and x - y, the result is x^2 - y^2 or x^2 - (1y)^2.

We see that there is just one favorable product, namely (x + y)(x - y). In order to determine the probability of this event, we must determine the total number of possible products. Since we have a total of four expressions and we are selecting two of them to form a product, we have 4C2, which is calculated as follows:

4C2 = (4 x 3)/(2!) = 12/2 = 6 products

Of these 6 products, we have already determined that only one will be of the form x^2 - (by)^2. Therefore, the probability is 1/6.

Note: If you don’t know how to use the combination formula, here is a method that will work equally well:

We are choosing 2 expressions from a pool of 4 possible expressions. That is, there are 2 decisions being made:

Decision 1: Choosing the first expression

Decision 2: Choosing the second expression

Four different expressions are available to be the first decision.

For the second decision, 3 remaining expressions are available because 1 expression was already chosen. We multiply these two numbers: 4 x 3 = 12.

The final step is to divide by the factorial of the number of decisions (2! = 2) because the order in which we multiply the expressions doesn’t matter (for example, (x+y)(x-y) = (x-y)(x+y)). In this case, the two expressions are only considered as one, so we need to divide 12 by 2.

12/2 = 6

Once again the answer would be E.

Alternate Solution:

One other way to solve this problem is to use probability.

Once again, we have determined that the only two expressions that when multiplied together will give us a difference of squares are x + y and x - y. If we select either of those expressions first, since there are 2 favorable expressions and 4 total expressions, there is a 2/4 = 1/2 chance that either x + y or x - y will be selected. Next, since there is 1 favorable expression left and 3 total expressions, there is a 1/3 chance that the final favorable expression will be selected.

Thus, the probability of selecting x - y and x + y is 1/2 x 1/3 = 1/6.

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Originally posted by JeffTargetTestPrep on 13 Feb 2017, 07:41.
Last edited by JeffTargetTestPrep on 01 Jun 2017, 09:31, edited 1 time in total.
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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13 Feb 2017, 11:56
1
Hi All,

This question is based heavily on algebra patterns. If you can spot the patterns involved, then you can save some time; even if you can't spot it though, a bit of 'brute force' math will still get you the solution.

We're given the terms (X+Y), (X+5Y), (X-Y) and (5X-Y). We're asked for the probability that multiplying any randomly chose pair will give a result that is written in the format: X^2 - (BY)^2.

Since there are only 4 terms, and we're MULTIPLYING, there are only 6 possible outcomes. From the prompt, you should notice that the 'first part' of the result MUST be X^2....and that there should be NO 'middle term'....which limits what the first 'term' can be in each of the parentheses....

By brute-forcing the 6 possibilities, you would have...
(X+Y)(X+5Y) = X^2 + 6XY + 5Y^2
(X+Y)(X-Y) = X^2 - Y^2
(X+Y)(5X-Y) = X^2 + 4XY - Y^2
(X+5Y)(X-Y) = X^2 + 4XY - 5Y^2
(X+5Y)(5X-Y)= 5X^2 +24XY - 5Y^2
(X-Y)(5X-Y) = 5X^2 -6XY + Y^2

Only the second option is in the proper format, so we have one option out of six total options.

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Special Offer: Save $75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ *****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***** Manager Joined: 23 Sep 2016 Posts: 217 Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at [#permalink] ### Show Tags 23 Feb 2017, 00:56 Bunuel wrote: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at random, what is the probability that their product will be of the form of x^2 -(by)^2, where b is an integer? A. 1/2 B. 1/3 C. 1/4 D. 1/5 E. 1/6 Four expressions can be paired in $$C^2_4=6$$ ways (choosing 2 out of 4); $$x^2 - (by)^2=(x-by)(x+by)$$ and only one pair is making such expression: $$(x+1*y)(x-1*y)$$, hence probability is 1/6. Answer: E. hey, can you please help me in this question i solved it by knowing out of 2 pairs we need 1 pair so 2c1 and out of 4 no. we need 2 no. so 4c2 so i got probability 2/6 which is 1/3 so how you get 1 in numerator. please explain thankyou EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 12883 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at [#permalink] ### Show Tags 23 Feb 2017, 11:48 Hi rishabhmishra, If you're going to treat this prompt as a Combination question, then that's fine (although it's not necessary). You just have to make sure that you treat the ENTIRE prompt correctly. Using the Combination Formula, you correctly determined that there were 6 possible "products" that could be created. However, you made a mistake in terms of what fraction of those 6 products actually 'fit' what you were looking for. The point of using the Combination Formula is that it is meant to eliminate 'duplicate entries', but it's not correct to use 2C1 here, since there was only ONE product that fit what the question asked for: (X+Y)(X-Y) and since that's the SAME product as (X-Y)(X+Y), we're NOT allowed to count it twice. GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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28 Mar 2017, 09:18
only 2 combination helps us get the answer
we can solve it in 2 ways:
1)
a) we need x+y or x-y out of 4, then 2/4
b) then we need either of them, so 1/3
c)2/4*1/3=1/6

Or let's us combinatorics
there are totally 2C4 combinations of the 4 pairs, when 2 are chose. 2C4=6
there is only one combination that suits us, then 1/6

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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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31 May 2017, 18:23
ronny13 wrote:
If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at random, what is the probability that their product will be of the form of x^2 -(by)^2, where b is an integer?

A. 1/2
B. 1/3
C. 1/4
D. 1/5
E. 1/6

1. The six possible products are:
(x+y)(x+5y)
(x+y)(x-y)
(x+y)(5x-y)
(x+5y)(x-y)
(x+5y)(5x-y)
(x-y)(5x-y)

2. Of the above, we see only the second gives the form required.
Probability is therefore 1/6
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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25 Jun 2017, 17:46
ronny13 wrote:
If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at random, what is the probability that their product will be of the form of x^2 -(by)^2, where b is an integer?

A. 1/2
B. 1/3
C. 1/4
D. 1/5
E. 1/6

Only (x+y)(x-y) or one combination satisfies the condition. There are 4!/(2!)*(2!) total possibilities, since two equations are selected and the other two are not selected (YYNN as an anagram).

The odds are 1/6.
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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25 Jun 2017, 21:26
mbaapp1234 wrote:
ronny13 wrote:
If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at random, what is the probability that their product will be of the form of x^2 -(by)^2, where b is an integer?

A. 1/2
B. 1/3
C. 1/4
D. 1/5
E. 1/6

Only (x+y)(x-y) or one combination satisfies the condition. There are 4!/(2!)*(2!) total possibilities, since two equations are selected and the other two are not selected (YYNN as an anagram).

The odds are 1/6.

Minor correction: it should be should "probability" instead of "odds" (these two things are not the same and GMAT always asks about probability, not odds).
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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25 Jun 2017, 22:45
2 expression can be chosen 4C2 ways=6
only one pair will form the required product so probability =1/6
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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11 Aug 2017, 06:32
Why is the proper approach for this problem to set up a combination 4C2 where as for other problems such as choosing a team you may do 4 * 3. When I solved this problem I almost thought of the four expressions as players for a team. There are four to choose from and I have two slots to put players in. For the first slot, I have four to choose from and in the second slot, I have 3 to choose from so there are 4* 3 = 12 combinations. If there were replacement, that would yield 4 * 4 = 16 combinations.

I guess my question is: would someone be able to explain to me the flaw in my logic between these two different methods of arriving at the number of combinations. Thanks!
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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11 Aug 2017, 11:37
Hi brandmanrocks,

To start, you have to always be sure that you are answering the question that is ASKED (sometimes the question is so specific that there's really only one way to approach it). That having been said, most prompts can be approached in more than one way, so you could certainly approach this prompt in the way that you described. However, you have to make sure to 'adjust' your 'math' to account for how you approached the work.

You can certainly treat this question as a permutation, BUT you then have to keep track of every potential option that 'fits' what is asked for....

(4 choices)(3 choices) = 12 possible options in which the order matters. Of the various 'pairs', there would be TWO that 'fit' what we're looking for:

(X+Y)(X-Y)
(X-Y)(X+Y)

Again, we would have to count that as two options because 'your way' is focused on treating this as a permutation.

2/12 = 1/6 so you end up with the same correct answer as you would if you treated this as a 'Combination' question.

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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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24 Apr 2018, 10:01
Top Contributor
ronny13 wrote:
If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at random, what is the probability that their product will be of the form of x^2 -(by)^2, where b is an integer?

A. 1/2
B. 1/3
C. 1/4
D. 1/5
E. 1/6

Okay, first recognize that x² - (by)² is a DIFFERENCE OF SQUARES.
Here are some examples of differences of squares:
x² - 25y²
4x² - 9y²
49m² - 100k²

In general, we can factor differences of squares as follows:
a² - b² = (a-b)(a+b)

So . . .
x² - 25y² = (x+5y)(x-5y)
4x² - 9y² = (2x+3y)(2x-3y)
49m² - 100k² = (7m+10k)(7m-10k)

----OKAY, ONTO THE QUESTION------------------

So, from the 4 expressions (x+y, x+5y ,x-y and 5x-y), only one pair (x+y and x-y) will result in a difference of squares when multiplied.

So, the question now becomes:
If 2 expressions are randomly selected from the 4 expressions, what is the probability that x+y and x-y are both selected?

P(both selected) = [# of outcomes in which x+y and x-y are both selected]/[total # of outcomes]

As always, we'll begin with the denominator.

total # of outcomes
There are 4 expressions, and we must select 2 of them.
Since the order of the selected expressions does not matter, we can use combinations to answer this.
We can select 2 expressions from 4 expressions in 4C2 ways (= 6 ways)

If anyone is interested, I have added a video (below) on calculating combinations (like 4C2) in your head

# of outcomes in which x+y and x-y are both selected
There is only 1 way to select both x+y and x-y

So, P(both selected) = 1/6

RELATED VIDEO

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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at &nbs [#permalink] 24 Apr 2018, 10:01

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##### Maths - Herons Formula

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NAIPC 2019

#### Start

2019-03-02 09:00 AKST

## NAIPC 2019

#### End

2019-03-02 14:00 AKST
The end is near!
Contest is over.
Not yet started.
Contest is starting in -840 days 22:19:50

5:00:00

0:00:00

# Problem HRocket Powered Hovercraft

You are programming an autonomous rocket powered hovercraft. The vehicle can travel very, very fast, but turning is difficult. Since it’s hovering, it turns by firing directional thrusters, which will turn the vehicle whether it is moving forward or is stationary.

The salt flats on which you’re testing have been mapped with a 2D Cartesian grid. The hovercraft starts at location $(0{,}0)$ on this grid, facing in the positive $X$ direction. Your job is to get the vehicle to another location $(x{,}y)$ on the flats.

The hovercraft has a fixed ground speed $v$ miles per second and a fixed rate of rotation $w$ radians per second. Due to the power of the rockets both for acceleration and deceleration, it can attain its maximum speed virtually instantaneously, and come to a stop from maximum speed just as quickly. Likewise, it can begin rotating at its fixed rate instantaneously, and stop just as quickly. It can rotate either clockwise or counter-clockwise.

You must figure out the minimum amount of time to get the vehicle to its target. The program which controls the vehicle can start forward motion, stop forward motion, start rotating, and stop rotating, each exactly once. Note that starting/stopping forward movement can be independent of starting/stopping rotation.

## Input

Each test case will consist of exactly two lines.

The first line will contain two space-separated integers $x$ and $y$ ($-1\, 000\! \le \! x,y\! \le \! 1\, 000, (x{,}y){\neq }(0{,}0)$), which indicate the location on the grid mapped onto the flats that you are trying to reach, in units of miles.

The second line of input will contain two space-separated real numbers with exactly two decimal places: $v$ ($0.01\! \le \! v\! \le \! 10.00$) and $w$ ($0.01\! \le \! w\! \le \! 10.00$), where $v$ is the fixed speed of travel in miles per second, and $w$ is the fixed rate of rotation in either direction in radians per second.

## Output

Output a single real number, which is the minimum amount of time (in seconds) it would take get the hovercraft from $(0{,}0)$ to $(x{,}y)$ subject to the constraints. Your answer will be accepted if it is within an absolute error of $10^{-3}$.

Sample Input 1 Sample Output 1
20 0
1.00 0.10

20.00000000

Sample Input 2 Sample Output 2
-10 10
10.00 1.00

3.14159265

Sample Input 3 Sample Output 3
0 20
1.00 0.10

28.26445910

Sample Input 4 Sample Output 4
-997 -3
5.64 2.15

177.76915187


text:
# Equation of a Circle Given Two Points and Tangent Line

Consider the general equation circle is given by

If the given circle is passing through two points, say, $A\left( {{x_1},{y_1}} \right)$and $B\left( {{x_2},{y_2}} \right)$, then these points must satisfy the general equation of circle. Now put these two points in the given equation of circle, i.e.

Also the given straight line $ax + by + d = 0\,\,\,{\text{ - - - }}\left( {{\text{iv}}} \right)$ touches the circle at one point as shown in the given diagram and it is clear from the diagram that the distance of a given form the center $\left( { - g, - f} \right)$ must be equal to the radius of the circle. By using the formula for distance between point and the line i.e.

But the radius of the circle in general form is $r = \sqrt {{g^2} + {f^2} - c}$, using this values we get

Squaring both sides of the above equation we have

To evaluate the equation of required circle, we must the find values of $g,f,c$ from the above equations (ii), (iii) and (v). We can solve these three using the method of simultaneous equations, then put all these in the equation (i) to get the required circle.
An alternate way is that we assume that the required equation of circle is

Since the circle passes through the points $A\left( {{x_1},{y_1}} \right)$and $B\left( {{x_2},{y_2}} \right)$, so these points must satisfy the above equation of circle, we have

Since the given line $ax + by + d = 0\,\,\,{\text{ - - - }}\left( {{\text{iv}}} \right)$ touches the circle.
So the distance of the center of the circle from the given line $ax + by + d = 0$ must be equal to its radius, i.e.

Now using the above equation (ii), (iii) and (v), we can find the values of $h,k$ and $r$ in the equation (i), we get the required circle.


text:
5th International Conference on Multiple Comparison Procedures

MCP 2007 Vienna
Vienna, Austria | July 8-11, 2007

## Program

The conference will be held from July 9 to July 11. On July 8 several pre-conference courses are offered.
Keynote:

 Multiple inference in medical research – an experience Peter Bauer Medical University of Vienna The history of applications of multiple comparisons procedures in medical research over the last four decades is given from a personal perspective. Some milestones are sketched. The reaction to new statistical methodology in the medical literature is investigated by comparing how multiplicity issues have been handled in medical journal articles twenty years ago and today. Some of the multiplicity issues in the rapidly developing area of gene expression and gene association studies are discussed. They have provoked new concepts and have helped to establish multiplicity as an important point to consider in the scientific community. Some arguments a consulting medical statistician may expect from his clients will be sketched followed by some pragmatic concluding comments.

Invited Talks:

 Multi-stage gatekeeping procedures with clinical trial applications Alex Dmitrienko; Tamhane, Ajit Eli Lilly and Company This talk introduces a general approach to constructing gatekeeping procedures for multiple testing problems arising in clinical trials with hierarchically ordered objectives (primary/secondary endpoints, dose-control comparisons, etc). The approach is applied to set up gatekeeping procedures based on popular multiple tests (Holm, fallback and Hochberg tests), resampling and parametric tests. The resulting procedures have a straightforward multi-stage structure that facilitates the implementation of gatekeeping procedures and communication of the results to non-statisticians. One can also account for logical restrictions among multiple analyses and improve the power of individual tests by eliminating comparisons that are no longer clinically meaningful. The general approach is illustrated using clinical trial examples.

 Multiple Testing Procedures with Applications to Genomics Sandrine Dudoit; van der Laan, Mark J. University of California In this two-part presentation, we will provide an overview of a general methodology for multiple hypothesis testing and applications to a range of large-scale testing problems in biomedical and genomic research. Specifically, we will describe resampling-based single-step and stepwise multiple testing procedures for controlling a broad class of Type I error rates, defined as tail probabilities and expected values for arbitrary functions of the numbers of Type I errors and rejected hypotheses (e.g., generalized family-wise error rate, tail probability for the proportion of false positives among the rejected hypotheses, false discovery rate). Unlike existing approaches, the procedures are based on a test statistics joint null distribution and provide Type I error control in testing problems involving general data generating distributions (with arbitrary dependence structures among variables), null hypotheses, and test statistics. The multiple testing results are reported in terms of rejection regions, parameter confidence regions, and adjusted $p$-values. A key ingredient of our proposed procedures is the null distribution used in place of the unknown joint distribution of the test statistics. We provide a general characterization for a proper test statistics null distribution, which leads to the explicit construction of two main types of test statistics null distributions. The first null distribution is the asymptotic distribution of a vector of null shift and scale-transformed test statistics, based on user-supplied upper bounds for the means and variances of the test statistics for the true null hypotheses. The most recent proposal defines the null distribution as the asymptotic distribution of a vector of null quantile-transformed test statistics, based on user-supplied marginal test statistics null distributions. We will discuss joint resampling-based empirical Bayes procedures for controlling generalized tail probability and expected value error rates. The approach involves specifying: (i) a null distribution for vectors of null test statistics and (ii) a distribution for random guessed sets of true null hypotheses. By randomly sampling null test statistics and guessed sets of true null hypotheses, one obtains a distribution for an arbitrary guessed function of the numbers of false positives and rejected hypotheses, for any given vector of cut-offs for the test statistics. Cut-offs can then be chosen to control tail probabilities and expected values for this distribution at a user-supplied level. Due to their generality and flexibility, our proposed multiple testing procedures are well-suited to address high-dimensional testing problems arising in different areas of application of statistics. We will conclude with an overview of applications in biomedical and genomic research, including: the identification of differentially expressed and co-expressed genes in high-throughput gene expression experiments, such as microarray experiments; tests of association between gene expression measures and biological annotation metadata (e.g., Gene Ontology); sequence analysis; the genetic mapping of complex traits using single nucleotide polymorphisms. Our forthcoming book provides a detailed account of the theoretical foundations of our multiple testing methodology and discusses its software implementation in the R package(www.bioconductor.org)and applications in biomedical and genomic research (Dudoit and van der Laan, 2007).

 Aesthetics and power in multiple testing – a contradiction? Gerhard Hommel IMBEI, University of Mainz, Germany It seems to be desirable that a multiple testing procedure should be as powerful as possible, given a criterion for type I error control. However, there are important additional aspects to be considered: 1. the pattern of the decisions should be logical; 2. the decisions should be conceivable, e.g., for scientific reasons or within a clinical trial; and 3. the decisions should be taken in such a way that they can be communicated also to non-statisticians. Moreover, there are also aspects of aesthetics that can be considered as relevant. Aesthetics are certainly to some extent subjective, but many people would agree that the closure test or the Bonferroni-Holm procedure (say) have an aesthetic component. I will consider in my talk different concepts related to multiple testing procedures and discuss them under the aspects above. In particular, the following issues are discussed: · Coherence and consonance; · Monotonicity of decisions (dependent on p-values); · Exchangeability; · Criteria for control of type I errors; · Concepts of power. To illustrate the ideas, I will consider the “fallback procedure” (Wiens, 2003; Wiens and Dmitrienko, 2005) as an example and discuss some properties of this procedure. References: Wiens, B.L. (2003). A fixed sequence Bonferroni procedure for testing multiple endpoints. Pharm. Stat. 2, 211-215. Wiens, B.L. and Dmitrienko, A. (2005). The fallback procedure for evaluating a single family of hypotheses. J. Biopharm. Stat. 15, 929-942.

 Modified Weighted Simes Tests in Group Sequential Designs Willi Maurer Novartis Pharma AG Multiple hypothesis testing problems, where the joint distribution between test statistics is not fully known and different weights or priorities are given to hypotheses, will be discussed. If there is no fully hierarchical ordering among the hypotheses, a general approach consists of gatekeeping procedures that usually are based essentially on the Bonferroni inequality with unequal type I error probability allocation, applied to closed testing procedures. A possible improvement with respect to increased power can be achieved by using the Simes inequality. Such scenarios arise, e.g., in trials aiming at investigating the effectiveness of cardiovascular and diabetes treatments in preventing or delaying cardiovascular events. The primary endpoints considered in these cases are often compound variables summarizing events in time, like fatal/nonfatal MI and stroke, revascularization, hospitalization for unstable angina, etc. So called ‘hard’ and ‘soft’ endpoints are built from subsets of these events where ‘hard’ endpoints can be comprised of a subset of the events constituting a ‘soft’ endpoint. Such endpoints are usually highly correlated among themselves but may be of unknown correlation with further endpoints like time to progression of diabetes. An additional complication in such trials is that they are usually of long duration and interim decisions have to be taken in a group sequential or adaptive design setting. We will discuss in this context issues that arise with respect to the validity of the Simes inequality in the case of unequal allocation of the probability of a type I error and show that results for bi-variate test statistics in the two-sided case can be extended to the one sided case if the rejection region is slightly altered. The problem of applying Hochberg-type testing strategies in a group sequential setting is discussed and various options regarding the allocation of spending functions in the arising repeated closed test situation are compared. Much of the work and newer results presented have been done together with Werner Brannath, Frank Bretz and Sanat Sarkar.

 On Consequences of One-Sided Alternative Hypotheses for the Null Hypothesis Joachim Röhmel Germany ONeill (1997) defines the primary endpoint as ‘a clinical endpoint that provides evidence sufficient to fully characterize clinically the effect of a treatment in a manner that would support a regulatory claim for the treatment’. In a clinical trial with an experimental treatment and a reference the situation may occur that two (or even more) primary endpoints may be necessary to describe the experimental treatment’s benefit. Sometimes effects on important secondary endpoints will influence the judgement on the experimental treatment’s value. For these situations multiple testing procedures have been developed to control the rate of false claims on superiority or non-inferiority. Often multiple testing procedures focus on the aim that for at least one primary endpoint the a priori set target is met, and little attention is then given to those endpoints which failed to demonstrate the proposed effect. When taking the above definition of a primary endpoint seriously, however, assurance is needed that none of the “non-significant” endpoints is inferior. Therefore, the focus of interest in a situation with multiple primary endpoints should be the more specific minimal target to demonstrate superiority in one of them given that non-inferiority is observed in the remaining. Several proposals exist in the literature for dealing with this or similar problems, but prove insufficient or inadequate at a closer look (e.g. Bloch et al. (2001, 2006) or Tamhane and Logan (2002, 2004)). In the talk I will focus on the case of two primary endpoints. Many aspects, however, can be transferred easily to the general case. I propose a hierarchical three step procedure, where non-inferiority in both variables must be proven in the first step, superiority has to be shown by a bivariate test (e.g. Holm (1979), O’Brien (1984), Hochberg (1988), a bootstrap (Wang (1998)), or Läuter (1996)) in the second step, and then superiority in at least one variable has to be verified in the third step by a corresponding univariate test. From the above mentioned bivariate superiority tests Läuter’s SS test and the Holm procedure are preferably for the reason that these have been proven to control the type I error strictly, irrespective of the correlation structure among the primary variables and the sample size applied. A simulation study reveals that the performance regarding power of the bivariate test depends to a considerable degree on the correlation and on the magnitude of the expected effects of the two primary endpoints. The part of the talk is based on a joined work with Christoph Gerlinger (Schering), Norbert Benda (Novartis), and Jürgen Läuter (University of Magdeburg). I explore consequences for setting up null hypotheses in situations where similar problems with directional alternative hypotheses might arise, for example in stratified clinical trials. References O’Neill RT (1997). Secondary endpoints cannot validly analyzed if the primary endpoint does not demonstrate clear statistical significance. Contr. Clin. Trials 18, 550-556. Bloch, D.A., Lai, T.L. and Tubert-Bitter, P. (2001) One-sided tests in clinical trials with multiple endpoints. Biometrics 57, 1039-1047. Bloch, D.A., Lai, T.L., Su, Z. and Tubert-Bitter, P. A combined superiority and non-inferiority approach to multiple endpoints in clinical trials. Statistics in Medicine (in preview) DOI: 10.1002/sim.2611 Hochberg, Y. (1988). A sharper Bonferroni procedure for multiple tests of significance. Biometrika 75, 800-802. Holm, S.A. (1979). A Simple Sequentially Rejective Multiple Test Procedure. Scandinavian Journal of Statistics 6, 65-70 Läuter, J. (1996). Exact t and F tests for analyzing clinical trials with multiple endpoints. Biometrics 52, 964-970. O'Brien, P.C. (1984). Procedures for comparing samples with multiple endpoints. Biometrics 40, 1079-1087. Tamhane, A.C. and Logan, B.R. (2002) Accurate critical constants for the one-sided approximate likelihood ratio test for a normal mean vector when the covariance matrix is estimated. Biometrics 58, 650-656. Tamhane, A.C. and Logan, B.R. (2004). A superiority-equivalence approach to one-sided tests on multiple endpoints in clinical trials. Biometrika 91, 715-727. Wang, S.-J.(1998). A closed procedure based on Follmann's test for the analysis of multiple endpoints. Communications in Statistics Theory and Methods 27, 2461-2480.

 Control of Generalized Error Rates in Multiple Testing Joseph P. Romano Stanford University Consider the problem of testing s hypotheses simultaneously. The usual approach restricts attention to procedures that control the probability of even one false rejection, the familywise error rate (FWER). If s is large, one might be willing to tolerate more than one false rejection, thereby increasing the ability of the procedure to correctly reject false null hypotheses. One possibility is to replace control of the FWER by control of the probability of k or more false rejections, which is called the k-FWER. We derive both single-step and stepdown procedures that control the k-FWER in finite samples or asymptotically, depending on the situation. We also consider the false discovery proportion (FDP) defined as the number of false rejections divided by the total number of rejections(and defined to be 0 if there are no rejections). The false discovery rate proposed by Benjamini and Hochberg controls E(FDP). Here, the goal is to construct methods which satisfy, for a given $\gamma$ and $\alpha$, $P\{FDP >\gamma\} \leq \alpha$, at least asymptotically. In contrast to Bonferroni type methods, we construct methods (using resampling) that implicitly take into account the dependence structure of the individual test statistics in order to further increase the ability to detect false null hypotheses. This feature is also shared by related work of van der Laan, Dudoit and Pollard, but our methodology is quite different. Simulations demonstrate improved performance over currently available methods. (This talk is based on joint work with Michael Wolf of the University of Zurich.)

 Stepwise Testing of Multiple Dose Groups Against a Control With Ordered Endpoints James Francis Troendle NIH Hierarchical gatekeeper methods exist for testing multiple dose clinical trials with multiple endpoints. This paper considers the case of ordered endpoints where an endpoint will only be tested at a given dose if all higher endpoints were found significant at that dose. Existing stepwise procedures based on the Bonferroni procedure are compared to new methods that incorporate correlation either through an assumption of Gaussian distribution or through resampling. The methods are compared by simulation for power and control of the familywise error.

 Prospective Strategies and Challenges to Adaptively Designing Genomic Biomarker Targeted Trials in Trials Sue-Jane Wang Center for Drug Evaluation and Research In recent decades, translational research has increasingly aimed to uncover the genomic biomarker signature that has the potential to differentiate the therapeutic effect among patients who are predicted to be good versus poor signatures. The advent of genomic biomarker gradually brings the awareness that phenotypically homogeneous patients may be heterogeneous at the genomic level. In this talk, I will present adaptive designs that prospectively account for genomic heterogeneous patient subpopulations and diagnostics test performance characteristics. The statistics concept of biomarker qualification and the efficiency of the pharmacogenomics clinical trials in view of personalized medicine will be exemplified. Aside from alpha allocation strategies and adaptive multiple hypotheses testing, challenges to adaptively designing pharmacogenomics targeted trials within conventional clinical trials will be elucidated via typical examples.

 Multiple Testing of General Contrasts: Truncated Closure and the Extended Shaffer-Royen Method Peter H. Westfall; Tobias, Randall D. Texas Tech University Powerful improvements are possible for multiple testing procedures when the hypotheses are logically related. Closed testing with alpha-exhaustive tests provides a unifying framework for developing such procedures, but can be computationally difficult and can be “nonmonotonic in p-values“. Royen (1989) introduced a “truncated“ closed testing method for the case of all pairwise comparisons in the ANOVA that is monotonic in p-values. Shaffer (1986) developed a similar truncated procedure for more general comparisons, but using Bonferroni tests rather than alpha-exhaustive tests, and Westfall (1997) extended Shaffer's method to allow alpha-exhaustive tests. This paper extends Royen's method to general contrasts and proves that it is equivalent to the extended Shaffer procedure. For large number k of contrasts, the method generally requires evaluation of O(2^{k}) critical values corresponding to subset intersection hypotheses, and is computationally infeasible for large k. The set of intersections is represented using a tree structure, and a branch-and-bound algorithm is used to search the tree and reduce the O(2^{k}) complexity by obtaining conservative “covering sets“ that retain control of the familywise type I error rate (FWE). The procedure becomes less conservative the deeper the tree search, but computation time increases. In some cases with logical relations, even the more conservative covering sets provide much more power than standard methods. The method is general, computable, and often much more powerful than commonly-used methods for multiple testing of general contrasts, as shown by applications to pairwise comparisons and response surfaces. In particular, with response surface tests, the method is computable with complete tree search, even when k is large, and can make many more “discoveries“ than the standard FDR-controlling method. The Extended Shaffer-Royen method has recently been hard-coded in the SAS/STAT procedure PROC GLIMMIX; syntax and output will be shown. Reference: Westfall, P.H. and Tobias, R.D. (2007). Multiple Testing of General Contrasts: Truncated Closure and the Extended Shaffer-Royen Method, to appear in Journal of the American Statistical Association.

 Ranks of true positives in large scale genetics experiments Russell D. Wolfinger; Zaykin, Dmitri; Zhivotovsky, Lev; Czika, Wendy; Shao, Susan SAS In the context of a large collection of statistical genetics tests in which the number of true positives (TPs) is small, we study the distribution of the ranks of TPs among the false positives (FPs). We investigate the relative efficiency of ranking measures and how many "best" results need to be screened to cover TPs with high probability, using a few different ways of assessing significance and adjusting for multiple testing. This way of looking at the problem can aid in optimally following up on initial significant findings and in planning of future large scale experiments. A whole-genome association scan is a prominent example, where the number of tests, L, is now commonly in the hundreds of thousands. With modern high-throughput genotyping capabilities, L can be large simply from the number of measured genetics markers, which are usually single nucleotide polymorphisms (SNPs). L can then grow exponentially by considering tests of haplotypes constructed from all possible pairs of SNPs, all triplets, etc. We simulate markers that are in linkage disequilibrium, that is, have some correlation structure, typically blocked. The measure of association of the genetic markers with a binary or quantitative trait of interest is usually some kind of p-value, perhaps weighted towards effect size. Multiple testing methods investigated include Sidak and no adjustment at all.

 Bayesian adjusted inference for selected parameters Daniel Yekutieli Tel Aviv University Benjamini and Yekutieli suggested viewing FDR adjusted inference as marginal inference for selected parameters. I will explain this approach – focusing on its weaknesses. I will then argue that overlooking the problem of selective inference is an abuse of Bayesian methodology; and introduce “valid“ Bayesian inference for selected parameters. I will then show that these methods are straightforward generalizations of Bayesian FDR. To make the discussion clearer I will demonstrate use of the Bayesian adjustments on microarray data.

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text:
Yuki726's blog

By Yuki726, history, 12 days ago, ,

In this submission for the problem 31660028 the author used if(ans[ni][nj] < ans[v.fi][v.se] + 1) break; condition. Can anyone guide me in proving that this algorithm with the break condition produces shortest paths and its complexity is O(n*m).


text:
# [bootlin/training-materials updates] master: Remove obsolete boot time labs (87cc20ff)

Michael Opdenacker michael.opdenacker at bootlin.com
Thu May 16 14:17:19 CEST 2019

Repository : https://github.com/bootlin/training-materials
On branch  : master

>---------------------------------------------------------------

commit 87cc20ffbaa23a9f8753231f852348fd15dd7d4a
Author: Michael Opdenacker <michael.opdenacker at bootlin.com>
Date:   Wed May 15 15:04:40 2019 +0200

Remove obsolete boot time labs

Signed-off-by: Michael Opdenacker <michael.opdenacker at bootlin.com>

>---------------------------------------------------------------

87cc20ffbaa23a9f8753231f852348fd15dd7d4a
labs/boottime-install/boottime-install.tex     | 167 -------------
labs/boottime-install/usbdisk-filled.png       | Bin 45232 -> 0 bytes
labs/boottime-install/usbdisk-formated.png     | Bin 37135 -> 0 bytes
labs/boottime-measuring/boottime-measuring.tex | 334 -------------------------
labs/boottime-measuring/videoplayer.dia        | Bin 2322 -> 0 bytes
mk/boot-time.mk                                |   5 +-
6 files changed, 2 insertions(+), 504 deletions(-)

diff --git a/labs/boottime-install/boottime-install.tex b/labs/boottime-install/boottime-install.tex
deleted file mode 100644
index 4625da0a..00000000
--- a/labs/boottime-install/boottime-install.tex
+++ /dev/null
@@ -1,167 +0,0 @@
-\subchapter{Preparing hardware}{Preparing a PC and a USB drive before
-the workshop}.
-
-Make sure the prerequisites are met before the workshop. Make sure you
-do not follow the below instructions on the morning the workshop starts,
-as these preparation steps can take up to several hours, depending
-on the speed of your Internet connection.
-
-Here is the required hardware:
-\begin{itemize}
-\item Microchip SAMA5D3x evaluation kit
-\item 5V power adaptor for the kit
-\item At least one USB-A (male) to micro-USB (male) cable. Two such
-      cables if possible.
-\item Recent PC with at least 2 GB of RAM, a decently fast processor
-      and 20 GB of free contiguous space for installing Linux.
-\item 1 USB flash drive with at least 4 GB of free space
-\end{itemize}
-
-\section{Installing Linux on the PC}
-
-Do this before joining the workshop. This can be either done by
-participants themselves on their own PCs, or on PCs supplied by the
-entity which organizes the event.
-
-For the practical labs in this workshop, we need you to install
-{\bf Ubuntu Desktop 12.04}. We support both 32 and 64 bit variants,
-though 64 bit will probably give you the best performance
-if you have a recent CPU.
-
-Get the Desktop edition of Ubuntu 12.04 at
-instructions.
-
-The way your disk is partition will also matter in the practical labs.
-What matters is that \code{/opt} is in a partition with at least 15 GB
-of disk space. It's still because of the pre-compiled build environment
-which must have a specific path to be useful.
-
-The easiest way to achieve this is to have only one big root (\code{/})
-partition, which would then contain \code{/opt}. If you have enough
-space, you can of course have a dedicated partition for \code{/opt}.
-
-{\bf Important notes}
-\begin{itemize}
-\item We do not support Linux installations in a virtual machine.
-      VMs will suffer from long compile times and will complicate
-\item Using Ubuntu in live mode (from with installing it) will not work
-      either.  Memory may get scarse, compile time will be much longer because
-      of the slow disk, and exporting the root filesystem through NFS
-      won't work.
-\end{itemize}
-
-At the end of installation, we recommend to install the latest updates:
-
-\begin{verbatim}
-sudo apt update
-\end{verbatim}
-
-\section{Preparing a USB disk for the workshop}
-
-The USB disk will be needed by workshop participants:
-
-\begin{itemize}
-\item To access files which could take a long time to download
-      in the workshop venue (if it has a slow connection to the Internet
-      once).
-\item To get the pre-compiled build environment, saving at least
-      one hour of compile time. This file is several hundreds of MB big,
-\end{itemize}
-
-Now take a USB disk with at least 4 GB of free space, and connect it
-to your Ubuntu PC. You should see a window showing its contents.
-
-We are going to reformat this disk, so that every participant
-accesses his disk through the same name and path.
-
-The first thing you need to do is find which device file is
-associated to this newly inserted disk. Type the \code{dmesg} command
-and look at the last lines:
-
-\begin{verbatim}
-...
-[ 6103.687852]  sdb: sdb1
-...
-\end{verbatim}
-
-Let's assume that this disk associated to \code{/dev/sdb}:
-\begin{itemize}
-\item \code{/dev/sdb} represents the whole disk
-\item \code{/dev/sdb1} represents only the first partition on this disk.
-      There could be more partitions.
-\end{itemize}
-
-Now, let's format this first partition for the \code{FAT32} filesystem
-(make sure you replace \code{sdb1} by the right device name, otherwise
-you may destroy data on other disks!):
-
-\begin{verbatim}
-sudo umount /dev/sdb1
-sudo mkfs.vfat -F 32 -n BootTime /dev/sdb1
-\end{verbatim}
-
-Now, remove your USB drive and insert it again. A file explorer window
-should appear:
-\begin{center}
-\includegraphics[width=8cm]{labs/boottime-install/usbdisk-formated.png}
-\end{center}
-
-Now let's fill this disk with data needed during the workshop:
-{\small
-\begin{verbatim}
-cd /media/$USER/BootTime -wget -r -np -nH --cut-dirs=2 --reject "index.html*" https://bootlin.com/labs/boottime/ -sudo umount /media/$USER/BootTime
-\end{verbatim}
-}
-
-take up to several hours.
-
-At the end, your USB drive contents should look like this:
-\begin{center}
-\includegraphics[width=8cm]{labs/boottime-install/usbdisk-filled.png}
-\end{center}
-
-\section{Installing the lab archive}
-
-same directories, your pre-compiled build environment won't work and
-you will waste a lot of time compiling software.
-
-Type the below commands:
-\begin{verbatim}
-cd /opt
-sudo chown -R $USER.$USER .
-wget https://bootlin.com/doc/training/boot-time/boot-time-labs.tar.xz
-sudo tar xf boot-time-labs.tar.xz
-sudo chown -R $USER.$USER .
-\end{verbatim}
-
-\section{Installing software packages needed during the workshop}
-
-This step is recommended for all workshops. It will pre-install all the
-software packages needed during the workshop.
-
-To keep our instructions as close as real life as possible, our labs
-ask to install software packages exactly when they are needed. However,
-if all the participants to a workshop download the same packages at the
-
-This step can be skipped if you are just a few people doing the labs
-from home or from an office with fast Internet access.
-
-Here is the command to pre-install the software packages:
-
-\begin{verbatim}
-/opt/boot-time-labs/setup/install-packages
-\end{verbatim}
-
-
-You are now ready to do the workshop!
diff --git a/labs/boottime-install/usbdisk-filled.png b/labs/boottime-install/usbdisk-filled.png
deleted file mode 100644
index 4d7a0083..00000000
Binary files a/labs/boottime-install/usbdisk-filled.png and /dev/null differ
diff --git a/labs/boottime-install/usbdisk-formated.png b/labs/boottime-install/usbdisk-formated.png
deleted file mode 100644
index cdb6b344..00000000
Binary files a/labs/boottime-install/usbdisk-formated.png and /dev/null differ
diff --git a/labs/boottime-measuring/boottime-measuring.tex b/labs/boottime-measuring/boottime-measuring.tex
deleted file mode 100644
index 0c50b6ce..00000000
--- a/labs/boottime-measuring/boottime-measuring.tex
+++ /dev/null
@@ -1,334 +0,0 @@
-\subchapter{Measuring boot time}{Measuring the various
-components of boot time}
-
-Our first goal is to measure boot time in a coarse way. In a second
-step, we will measure the time spent in the various components of boot
-time.
-
-
-As explained in the lectures, the first thing to do is to choose what do
-measure. In the demo running on this particular board, a natural choice is to
-measure the time taken to start the demo video.
-
-Ideally, we would measure the time elapsed since power-on or since a
-reset event. However, we have no oscilloscope here, and therefore we can only
-rely on messages sent on the serial port.
-
-\begin{itemize}
-\item The earliest message we can have on the serial port is the
-      \code{RomBOOT} one. We will take its time of arrival as the origin of
-      time.
-\item We will implement a way to send a message to the serial line when
-      the video starts playing.
-\end{itemize}
-
-\section{Measuring overall boot time}
-
-We will use \code{grabserial} to measure the time spent while booting.
-It prints the timing information of each message received on the
-serial line\footnote{Beware that grabserial displays the arrival time
-of the beginning of a line on a serial port. If a message was sent
-without a trailing line feed, you could get the impression that
-the following characters have been received much earlier. This is
-particularly true when only a whitespace was first sent to the
-serial line.}, and display the time elapsed since a given event (like
-receiving a special string), used as the origin of time.
-
-You will find \code{grabserial} on \url{http://elinux.org/Grabserial}.
-You can also get it from your USB flash drive:
-
-
-\begin{verbatim}
-cd
-mkdir bin && cd bin
-cp /media/$USER/BootTime/downloads/grabserial . -chmod a+x grabserial -\end{verbatim} - -The command we will be using is: - -\begin{verbatim} -~/bin/grabserial -d /dev/ttyACM0 -m RomBOOT -t -e 30 -\end{verbatim} - -Here are details about this command: - -\begin{itemize} -\item \code{-m "RomBOOT"} specifies the string marking the -origin of time. -\item \code{-t} makes \code{grabserial} show the time when -{\bf the first character of each line is received}. -\item \code{-e 30} tells \code{grabserial} to stop measuring after 30 -seconds. -\end{itemize} - -First, you will have to exit \code{picocom} using \code{C-a C-x} to run -\code{grabserial}. - -Start \code{grabserial} and reset the board using \code{PB1}. You will -see that it takes around 13.5 seconds to reach the Buildroot prompt and -show the demo video: - -\begin{verbatim} -$ ~/bin/grabserial -d /dev/ttyACM0 -m RomBOOT  -t -e 30
-[0.000002 0.000002] RomBOOT
-[0.054440 0.054440]
-[0.054594 0.000154]
-[0.054719 0.000125] AT91Bootstrap 3.5.2 (Wed Jan 30 18:42:28 CET 2013)
-[0.059009 0.004290]
-[0.062645 0.003460] 1-Wire: ROM Searching ... Done, 0x3 1-Wire chips
-found
-[0.110066 0.047421]
-...
-[12.158893 0.064820] Starting portmap: done
-[12.284482 0.125589] Initializing random number generator... done.
-[12.393179 0.108697] ALSA: Restoring mixer settings...
-[12.447447 0.054268] Starting network...
-[13.077587 0.630140] Starting sshd: OK
-[13.467746 0.390159]
-[13.470225 0.002479] Welcome to Buildroot
-\$
-\end{verbatim}
-
-Note that there's no information on exactly when the first user space
-code gets executed. We will fix that in the {\em Add instrumentation}
-section.
-
-You can make a visual check that the \code{buildroot
-login:} message is issued at about the same time as the video
-starts playing. Of course, the accuracy is about 1 or 2 seconds,
-and it's time to measure the video starting time in a more accurate way.
-
-\section{Trick to estimate the video start time}
-
-For this kind of need, finding out when a device starts functioning,
-the ideal solution would be to monitor the LCD screen pins with an
-oscilloscope, and based on the recorded activity, find out when the
-video actually starts playing.
-
-When only software tracing is available, a standard solution is to
-add instrumentation to the application (here the video player) to write a message to
-the serial line.
-
-In our case, however, our video player is based on GStreamer, a very
-complex piece of software. We can write a message on the serial line
-before starting the player application, but the time to load this
-application and actually start playing the video is far from being
-negligible. This time is one of the components of boot time that we want to
-measure.
-
-A trick here is to replace the video by the shortest video possible
-(with the same codecs), containing only one frame. Assuming that
-the time to play the video itself is negligible, we can have a pretty
-accurate measurement of the time to load the video player and start
-playing the video.
-
-\begin{center}
-\includegraphics[width=\textwidth]{labs/boottime-measuring/videoplayer.pdf}
-\end{center}
-
-\section{Reconstitute the build environment for the root filesystem}
-
-In every boot time reduction project, you have to rebuild the root
-filesystem, either by using the original build system, or by compiling
-the software stack with a new build system which could make it
-easier to optimize and simplify the root filesystem.
-
-Another option would have been to make direct changes to the root
-filesystem. However, what you do is difficult to reproduce, and would
-make the system more difficult to maintain. In a real production
-project, it's much better to make changes to the configuration of
-the build system used to generate the root filesystem.
-
-This way, your changes to reduce boot time are always applied,
-whatever the other changes you make on the system.
-
-This is also true when you need to add instrumentation. If you do
-this by hand, you will lose all your instrumentation the next
-time you have to update or optimize a component.
-
-\subsection{Install development software}
-
-To update and recompile the root filesystem and
-components like the Linux kernel, you will need to install several
-software development packages:
-
-\begin{verbatim}
-sudo apt install git libncurses5-dev u-boot-tools g++ bison
-sudo apt install flex gettext texinfo libqt4-dev lzop
-\end{verbatim}
-
-Also install the \code{meld} package. It is a very
-convenient utility to compare different versions of a file, which
-is often useful in an optimization project.
-
-\begin{verbatim}
-sudo apt install meld
-\end{verbatim}
-
-Now, to be able to update the root filesystem, you need to reconstitute
-the Buildroot environment that was used to generate it.
-
-Do it by following the instructions available on
-
-\section{Study how user space boots}
-
-It's important to understand how the various user space programs
-are started in the system. They all originate from the \code{init}
-program, which is the first and only program executed by the Linux
-kernel\footnote{There is one exception: the kernel can also run
-helpers to handle hotplug events.}, when it's done booting the device.
-
-The programs that are executed by the \code{init} program can
-found by reading the \code{/etc/inittab} file, which is a configuration
-file for \code{init}.
-
-Start \code{picocom} if needed and study this file. You will see that
-a \code{/etc/init.d/rcS} script is also executed, and executes more
-scripts at his turn. Take some time to read these scripts, until you
-understand how the video player is started.
-
-
-Make sure you are in the \code{/opt/boot-time-labs/buildroot/buildroot-at91} directory.
-
-Copy the \code{../data/1frame.avi} file\footnote {Here's the command that
-was used to generate a video containing only the first frame of the original video: \\
-\code{avconv -i MPEG2_480_272.avi -vcodec copy -acodec copy -frames 1
-1frame.avi}}
-to \code{board/atmel/sama5d3ek_demo/}, where the original video is
-stored.
-
-Now, modify the \code{board/atmel/sama5d3ek_demo/post-build.sh}
-file\footnote{This script is called after compiling the root filesystem
-components, and is used to add custom scripts and files to the system.}:
-\begin{itemize}
-\item To copy the \code{1frame.avi} video to \code{/root} on the target
-      file system. Do this by looking for the instruction that copies the
-      original video, and by using a similar command.
-\item To modify the contents of the \code{/root/go.sh} script
-      which starts the video player:
-      \begin{itemize}
-      \item To add an \code{echo -e "\nStarting video player"} command
-	    \footnote{We are using a newline (\code{\n}) at the
-beginning of the string because \code{grabserial} only pays attention to
-the arrival time of the first character in each line. Without this, we
-would get the wrong time if the current line already had some
-characters sent earlier. If we didn't use \code{echo -e} (\code{e}:
-suppport for escape characters), \code{\n} wouldn't have been considered
-as a newline.}
-            right before invoking \code{gst-launch}, the video player that the demo
-	    uses.
-      \item To play the \code{1frame.avi} video instead of the original
-  	    one.
-      \item To add an \code{echo -e "\nDone playing video"} right after
- 	    playing the video.
-      \end{itemize}
-\end{itemize}
-
-The next thing we need is to know when the \code{init} program starts to
-execute.
-
-Let's have a look at home this program is built. Run the below command
-to check the configuration of \code{BusyBox}, which implements
-\code{init}:
-
-\begin{verbatim}
-\end{verbatim}
-
-Go to the \code{Init utilities} menu, and disable \code{Be _extra_ quiet
-on boot}. Exit and save your new configuration. This way, we will have
-a message on the serial line when \code{init} starts to execute.
-
-It's now time to rebuild your root filesystem:
-\begin{verbatim}
-make
-\end{verbatim}
-
-The above command should only take a few seconds to run.
-
-Make sure that the \code{output/target/root/go.sh} file
-contains the right modifications and that there are no syntax errors.
-Update the \code{board/atmel/sama5d3ek_demo/post-build.sh} file until
-you get this file right.
-
-If you can't manage to get it right, have a look at the
-\code{../solutions/post-build1.sh} file.
-
-\section{Re-flash the root filesystem}
-
-We will use \code{SAM-BA} again to update the root filesystem.
-
-Let's copy our new root filesystem to it to our \code{SAM-BA} directory:
-
-\begin{verbatim}
-cp output/images/rootfs.ubi \
-/opt/boot-time-labs/flashing/sama5d3xek-demo/
-\end{verbatim}
-
-Put your board in RomBOOT mode (follow the procedure in the previous
-chapter), and assuming that the SAM-BA device is \code{/dev/ttyACM1},
-reflash your board, using the same instructions as before:
-
-\begin{verbatim}
-cd /opt/boot-time-labs/flashing/sama5d3xek-demo/
-sam-ba /dev/ttyACM1 AT91SAMa5d3x-EK sama5d3xek_demo_linux_nandflash.tcl
-\end{verbatim}
-
-Press the reset button and make sure that your board boots as
-expected. You can also check that the new instrumentation messages are
-there. Don't pay attention to timing this time, as SSH keys had to be
-generated from scratch again.
-
-To stop the video player from running in a quick and endless way,
-you can run the below command in the serial console:
-
-\begin{verbatim}
-killall go.sh
-\end{verbatim}
-
-We will remove the endless loop a little bit later.
-
-\section{Measure boot time components}
-
-You can now exit \code{picocom} and run \code{grabserial} again.
-
-We can now time all the various steps of system booting in a pretty
-accurate way. Do it by filling writing down the time stamps in the below
-table.
-
-Then, in each row, compute the elapsed time, which is the difference
-between the time stamp for the next step and the time stamp for the current step.
-For this purpose, you can use the \code{bc -l} command line calculator.
-
-\begin{tabular}{| l | c | c | r |}
-  \hline
-  Boot phase & Start string & Time stamp & Elapsed \\
-  \hline
-  \hline
-  RomBOOT & \code{RomBOOT} & & \\
-  \hline
-  Bootstrap & \code{AT91Bootstrap 3.5.2} & & \\
-  \hline
-  Bootloader & \code{U-Boot 2012.10} & & \\
-  \hline
-  Kernel & \code{Booting Linux on physical CPU 0} & & \\
-  \hline
-  Init scripts & \code{init started: BusyBox v1.21.0} & & \\
-  \hline
-  Critical application & \code{Starting video player} & & \\
-  \hline
-  Critical application ready & \code{Done playing video} & & \\
-  \hline
-  \hline
-  \multicolumn{3}{| c |}{Total} & \\
-  \hline
-\end{tabular}
-
-In the next labs, we will work on reducing boot time for each of the boot phases.
diff --git a/labs/boottime-measuring/videoplayer.dia b/labs/boottime-measuring/videoplayer.dia
deleted file mode 100644
index 5f80c1b6..00000000
Binary files a/labs/boottime-measuring/videoplayer.dia and /dev/null differ
diff --git a/mk/boot-time.mk b/mk/boot-time.mk
index bd72eac3..7ed6b758 100644
--- a/mk/boot-time.mk
+++ b/mk/boot-time.mk
@@ -20,9 +20,8 @@ BOOT_TIME_SLIDES = \
boottime-hardware-init \

-BOOT_TIME_LABS = boottime-install \
-		boottime-getting-started \
-		boottime-measuring \
+BOOT_TIME_LABS = setup \
+		boot-time-board-setup \
boottime-setup \
boottime-init-scripts \
boottime-application \


text:
# The opt-out illusion - Technology - TLS

source
https://www.the-tls.co.uk/articles/the-opt-out-illusion/
tags
Privacy

Opting out is an illusion. The extraction of our behavioural data is non-negotiable.


text:
# 2.4.1.1 X-Functions Overview

X-Functions provide a uniform way to access nearly all of Origin's capabilities from your LabTalk scripts. The best way to get started using X-Functions is to follow the many examples that use them, and then browse the lists of X-Functions accessible from script provided in the LabTalk-Supported X-Functions section.

## Syntax

You can recognize X-Functions in script examples from their unique syntax:

xFunctionName input:=<range> argument:=<name/value> output:=<range> -switch;

General Notes:

• X-Functions can have multiple inputs, outputs, and arguments.
• X-Functions can be called with any subset of their possible argument list supplied.
• If not supplied a value, each required argument has a default value that is used.
• Each X-Function has a different set of input and output arguments.

Notes on X-Function Argument Order:

• By default, X-Functions expect their input and output arguments to appear in a particular order.
• Expected argument order can be found in the help file for each individual X-Function or from the Script window by entering XFunctionName -h.
• If the arguments are supplied in the order specified by Origin, there is no need to type out the argument names.
• If the argument names are explicitly typed, arguments can be supplied in any order.
• You can mix handling of arguments as long as omitted arguments come first in the specified order, followed by explicitly typed arguments in any order.
• The argument name can be shortened by trimming characters from the end of the argument name, but the shortened name needs to be unique.

The following examples use the fitpoly X-Function to illustrate these points.

## Examples

The fitpoly X-Function has the following specific syntax, giving the order in which Origin expects the arguments:

fitpoly iy:=(inputX,inputY) polyorder:=n fixint:=(0=No/1=Yes) intercept:=value coef:=columnNumber oy:=(outputX,outputY) N:=numberOfPoints;

If given in the specified order, the X-Function call,

//Polynomial fit on XY data (X-column 1, Y-column 2),
//Polynomial order = 4
// Not fix intercept value, Initial intercept Value is 0,
//Output Coef in column 3,
//Output X value in column 4, Output Y values in column 5,
//Number of points of output XY =100
fitpoly (1,2) 4 0 0 3 (4,5) 100;

tells Origin to fit a 4th order polynomial with 100 points to the X-Y data in columns 1 and 2 of the active worksheet, putting the coefficients of the polynomial in column 3, and the X-Y pairs for the fit in columns 4 and 5 of the active worksheet.

In contrast, the command with all options typed out is a bit longer but performs the same operation:

fitpoly iy:=(1,2) polyorder:=4 coef:=3 oy:=(4,5) N:=100;

In return for typing out the input and output argument names, LabTalk will accept them in any order, and still yield the expected result:

fitpoly coef:=3 N:=100 polyorder:=4 oy:=(4,5) iy:=(1,2);

It is also permissible to omit some argument names, then follow with other named arguments in any order, as in the following script, which yields the same result as above.

fitpoly (1,2) 4 oy:=(4,5) N:=100 coef:=3;

You can also shorten the argument name if the shortened name is unique in the argument list, as in ...

fitpoly iy:=(1,2) poly:=4 co:=3 o:=(4,5) N:=100;

... where poly is short for polyorder, and co for coef, and o for oy.

However, the following script will produce an error...

fitpoly i:=(1,2) poly:=4 co:=3 o:=(4,5) N:=100;

... because there are two argument names (iy and intercept) that begin with letter i. Here Origin cannot tell which argument i refers to; that is to say i is not unique.

Inputs and outputs can be placed on separate lines from each other and in any order, as long as they are explicitly typed out.

fitpoly
coef:=3
N:=100
polyorder:=4
oy:=(4,5)
iy:=(1,2);

Notice that the semicolon ending the X-Function call comes only after the last parameter associated with that X-Function.

## Option Switches

Option switches such as -h or -d allow you to access alternate modes of executing X-functions from your scripts. They can be used with or without other arguments. The option switch (and its value, where applicable) can be placed anywhere in the argument list. This table summarizes the primary X-Function option switches:

Name Function
-h Prints the contents of the help file to the Script window.
-d Brings up a graphical user interface dialog to input parameters.
-s Runs in silent mode; results not sent to Results log.
-t <themeName> Uses a pre-set theme.
-r <value> Sets the output to automatically recalculate if input changes.

For more on option switches, see the section X-Function Execution Options.

## Generate Script from Dialog Settings

The easiest way to call an X-Function is with the -d option and then configures its settings using the graphical user interface (GUI).

In the GUI, once the dialog settings are done, you can generate the corresponding LabTalk script for the configuration by selecting the Generate Script item in the dialog theme fly-out menu. Then a script which matches the current GUI settings will be output to script window and you can copy and paste it into a batch OGS file or some other project for use.

Note: The Dialog Theme box and corresponding Generate Script fly-out menu item are not available from all X-Function dialogs that you open with the -d option (for instance, compare fitpoly -d with rnormalize -d).


text:
# If a curve $y = f(x)$ passes through the point $(1, -1)$ and satifies the differential equation, $y(1+xy) dx = x ;dy$, then $f(-\frac{1}{2})$ is equal to :
( A ) $\frac{2}{5}$
( B ) $\frac{4}{5}$
( C ) $-\frac{2}{5}$
( D ) $-\frac{4}{5}$


text:
Question 222324
<font face="Garamond" size="+2">

The perimeter of a triangle is the sum of the measures of the three sides, so for your triangle,

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P\ =\ 4(x -3) + 3x + 5x + 4]

Simplify:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P\ =\ 12x - 8]

The perimeter can be no more than 100, so it is less than or equal to 100 and no less than 28, so it is greater than or equal to 28, so:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 28 \leq 12x - 8 \leq 100]

Add 8 to all three parts of the compound inequality:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 36 \leq 12x \leq 108]

Divide all three parts of the compound inequality by 12:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3 \leq x \leq 9]

However, we have to change the inequality sign on the low end of the interval.  That is because *[tex \Large x \neq 3].  If *[tex \Large x = 3] then the measure of the left side of the triangle would be 0, i.e. *[tex \Large 4(3 - 3) = 0], and there would be no triangle.

Hence, the final interval for *[tex \Large x] is:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3 < x \leq 9]

John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>


text:
2 added 3 characters in body

Consider the property of a vertex $v$ of a planar graph $G$ that the circular ordering of its edges is the same (upto orientation) for every graph embedding $\pi$ of $G$ into the plane $\mathbb{R}^2$.

1. Does this property have an official name?

2. (How) can it be defined purely combinatorially?

3. (How) can planar graphs be characterized in which every vertex has this property?

1

# Unique circular ordering of edges around a vertex

Consider the property of a vertex $v$ of a planar graph $G$ that the circular ordering of its edges is the same (upto orientation) for every graph embedding $\pi$ of $G$ into the plane $\mathbb{R}^2$.

• Does this property have an official name?

• (How) can it be defined purely combinatorially?

• (How) can planar graphs be characterized in which every vertex has this property?