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# What are the advantages of the ILC over the LHC?

USA Today has an article on Japan's interest as the site for the $10 billion future International Linear Collider. This accelerator will utilize electron/positron collisions (like CERN's former LEP collider) at energies of 500 GeV and will be 30 km long. The ILC site claims precision capabilities over and above those of the 14 TeV LHC for Higgs and other "new physics" areas. What are the specific advantages of the ILC over the LHC and what new particle physics insites are more likely to be discovered? - BTW, I understand that electron/positron collisions are cleaner than proton/antiproton. There have to be more reasons why the ILC is more advantageous than the LHC. – Michael Luciuk Jan 2 '12 at 21:27 "what new particle physics insites are more likely to be discovered" If we knew that then it wouldn't be new. More interesting questions are things like "what candidate beyond the standard model theories could it distinguish between" or "how much of the string theory phase space could it constrain". Of course, any surprises will be...well, surprising. – dmckee Jan 2 '12 at 21:53 dmckee: I agree. Those are better questions. – Michael Luciuk Jan 2 '12 at 23:39 "There have to be more reasons why the ILC is more advantageous than the LHC." - Well, you're essentially wrong, Michael. There don't have to be any other reasons. jdm's answer is detailed and exhaustive but it confirms the basic thesis you summarized by the word "cleaner". – Luboš Motl Jan 3 '12 at 7:58 ## 1 Answer One often says a hadron collider like the LHC is used for discovering, while an electron collider is rather used for precision measurements. There are a couple of benefits of a high-energy electron collider: • All electrons have roughly the same energy. One can vary the center of mass energy$\sqrt{s}$and map out resonances (think$ee\rightarrow Z$or$J/\psi$). Compare this to the LHC, which is actually in a sense a gluon-gluon collider. The gluons in the protons have wildly varying energies (the probability of a certain energy is given by the parton distribution functions), so you never know with which energy your particles are colliding. You are probing all energies up to 7 TeV at the same time, in a way. Even worse, most of the time your initial state particles have different momenta in beam direction, so your whole system is boosed along the beam. These effects smear out and shift your distributions, e.g. if you want to measure the invariant mass of something. This is the reason that we use transverse variables at hadron colliders - there is no initial boost in transverse direction (perpendicular to the beam line). • Related to the above, in some searches for new particles you can see nice mass edges at a lepton collider. They occur when you have a chain decay, say$\tilde\chi^0_2 \rightarrow \tilde\chi^0_1 + Z \rightarrow \tilde\chi^0_1 + \ell^+ \ell^-$. The invariant mass distribution of the lepton pairs rises up to the point$\Delta m = m(\tilde\chi^0_2) - m(\tilde\chi^0_1)$where all of the energy from the chain decay is carried by the two leptons and none by the neutralino, and then it drops off suddenly. You can look for mass edges at the LHC too, but they will be more smeared out. • As you said, the collisions are 'cleaner'. There are no jets from proton remnants for example. A lot of backgrounds just don't exist there, and if you are running near the energy of an interesting resonance, you will get a lot of signal events. However, at the energies of the ILC there will still be a lot of jets produced. Also the ILC will have a pretty high luminosity, so they will also have to deal with multiple simultaneous events. And then, there are entirely new problems. After the electron beam is focused in the collision point, it diverges quickly behind it because the electrons repel each other. The part of the detector nearest to the beam line will be 'blinded' from all the electrons ('beamstrahlung'). If you look at data vs. background plots from ILC simulations, you see that the beam-induced background spikes at high$\eta\$, at small angles to the beam. It looks like this can be controlled, but it is still challenging for detector designers and analyzers.

It's very probably the LHC experiments will conclusively observe the Higgs Boson soon. But then we still don't know which Higgs Boson it is - the standard model one, or another one with different properties? In the MSSM (a supersymmetric theory), there are five Higgs bosons, two of which are charged. And what's the exact mass (masses)? We will eventually be able to answer these questions at the LHC with a lot of data, but we can do it more precisely with a special tool like the ILC. The same not only applies for the Higgs, but also for other new physics, like SUSY, extra dimensions, or whatever other thing there is to be discovered.

-
I would add one more point, the advantage of closure, that missing energy can be a clean sign of a missing particle in a four pi detector, so important for dark matter candidates. –  anna v Jan 3 '12 at 5:32


text:
How are there pi bonds in B_2 molecule without sigma bonds?

There shouldn't be, and that's why "B"_2 is very unstable; it has two orthogonally-localized half-pi bonds, which is quite a weak bond.

Imagine that each "B" takes the place of a "CH" on "HC"-="CH".

Then, take away the sigma bond, the two "H" atoms, and one electron from each of the two full pi bonds, and you have a structure similar to "B"_2.

Che dire "Li"_2 "molecule"? Is that stable? (Its bond length is "267.3 pm", over twice the length of an average bond.)

The first chemical bond made in a molecule is preferenzialmente a bbsigma legame.

sigma bonds are formed from a direct atomic orbital overlap. In comparison, pi bonds are sidelong overlaps and thus, sigma overlaps are made preferentially because they form the stronger bond.

"B"_2 contains two boron atoms, which each use a basis of a 1s, un 2sE tre 2p orbitali atomici.

Per qualificarti per il "Li"_2 attraverso "N"_2, there exists an orbital mixing effect che rende il sigma molecular orbital for a 2p_z-2p_z overlap (sigma_(g(2p))) superiore in energia rispetto al pi molecular orbitals for a 2p_(x//y)-2p_(x//y) overlaps (pi_(u(2p))).

So, this particular orbital energy ordering takes place in the diagramma orbitale molecolare, simile al "C"_2 molecola:

where |E_(sigma_(2p_z))| > {|E_(pi_(2p_x))| = |E_(pi_(2p_y))|}.

Dal "C" has one more electron, "B"_2 would have a similar MO diagram, EXCEPT for two meno electrons. This, at first glance, seems to suggest that "B"_2 has two half-pi bonds, with a molecular configurazione elettronica di:

color(green)((sigma_(1s))^2(sigma_(1s^"*"))^2(sigma_(2s))^2(sigma_(2s^"*"))^2(pi_(2p_x))^1(pi_(2p_y))^1)

Dal "B"_2 is paramagnetica with two electrons, in order to make that bond, there are indeed two half-bbpi obbligazioni, which form what we represent improperly in line notation as a sigma bond... and it isn't actually a sigma legame.

:"B"-"B":

That isn't a sigma bond, but two half-pi bonds. We would then expect "B"_2 to be very unstable, which it is. (The lone pairs are from the sigma_(2s) e sigma_(2s)^"*", since bonding + antibonding filled = nonbonding.)

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#### [SOLVED] How do I avoid "too" lucky / unlucky streaks in random number generation?

I'm currently dealing with a multiplayer combat system where the damage dealt by the players is always multiplied by a random factor between 0.8 and 1.2.

In theory, a truly random RNG may eventually yield the same number many times (see the Tetris dilemma). This could result in a match where player is always making very high damage while the other always makes very low damage.

What can I do to make sure this doesn't happen? Are some RNGs better than others at avoiding repetition?

# Use a shifting bias

The base generator uses a uniform distribution between $$\0\$$ and $$\1\$$ to generate $$\r\$$. Initially set a bias value, $$\b\$$, to $$\0\$$.

The overall distribution will be biased by the following formula:

$$\ r^{\exp (-b)} \$$

The effect here is that when $$\b\$$ is positive, the resulting number will be biased toward $$\1\$$. When $$\b\$$ is negative, the resulting number will be biased toward $$\0\$$.

Take this number and scale it appropriately to the desired range.

Each time a player rolls favorably, subtract from the bias. Each time the player rolls unfavorably, add to the bias. The amount changed could be scaled by how (un)favorable the roll is or could be a flat amount (or a combination). You will need to adjust specific values to fit the feel you are going for.

#### @Kromster says support Monica 2012-07-16 13:28:52

Sid Meier had an excellent speech on GDC 2010 just about this topic and Civilization games. I will try to find and paste the link later. In essence - perceived randomness is not the same as true randomness. To make things feel fair you need to analyze previous results and pay attention to players psychology.

Avoid streaks of bad luck at all costs (if previous two turns were unlucky next one should be guaranteed to be lucky). Player should be always luckier than AI opponent.

#### @Jonathan Dickinson 2012-07-16 09:40:07

I actually wrote some code to do this. The gist of it is using statistics to correct unlucky streaks. The way you can do this is to keep track of how many times the event has occurred and use that to bias the number generated by the PRNG.

Firstly, how do we keep track of the percentage of events? The naive way of doing this would be to keep all numbers ever generated in memory and average them out: which would work but is horribly inefficient. After a little thinking I came up with the following (which is basically a cumulative moving average).

Take the following PRNG samples (where we proc if the sample is >= 0.5):

``````Values: 0.1, 0.5, 0.9, 0.4, 0.8
Events: 0  , 1  , 1  , 0  , 1
Percentage: 60%
``````

Notice that each value contributes to 1/5 of the final result. Let's look at it another way:

``````Values: 0.1, 0.5
Events: 0  , 1
``````

Notice that the `0` contributes to 50% of the value and the `1` contributes 50% of the value. Taken slightly further:

``````Values: [0.1, 0.5], 0.9
Events: [0  , 1  ], 1
``````

Now the first values contribute 66% of the value and the last 33%. We can basically distil this down to the following process:

``````result = // 0 or 1 depending on the result of the event that was just generated
new_samples = samples + 1

average = (average * samples / new_samples) + (result * 1 / new_samples)
// Essentially:
average = (average * samples / new_samples) + (result / new_samples)

// You might want to limit this to, say, 100.
// Leaving it to carry on increasing can lead to unfairness
// if the game draws on forever.
samples = new_samples
``````

Now we need to bias the result of the value sampled from the PRNG, because we are going for a percentage chance here things are a lot easier (versus, say, random amounts of damage in a RTS). This is going to be hard to explain because it 'just occurred to me'. If the average is lower it means that we need to increase the chance of the event occurring and visa-versa. So some examples

``````average = 0.1
desired = 0.5
corrected_chance = 83%

average = 0.2
desired = 0.5
corrected_chance = 71%

average = 0.5
desired = 0.5
corrected_change = 50%
``````

Now what 'occurred to me' is that in the first example 83% was just "0.5 out of 0.6" (in other words "0.5 out of 0.5 plus 0.1"). In random event terms that means either:

``````procced = (sample * 0.6) > 0.1
// or
procced = (sample * 0.6) <= 0.5
``````

So in order to generate an event you would basically use the following code:

``````total = average + desired
sample = rng_sample() * total // where the RNG provides a value between 0 and 1
procced = sample <= desired
``````

And therefore you get the code that I put in the gist. I am pretty sure this all can be used in the random damage case scenario, but I haven't taken the time to figure that out.

Disclaimer: This is all home-grown statistics, I have no education in the field. My unit tests do pass though.

#### @Kylotan 2012-07-16 11:08:05

Looks like an error in your first example because both a 0.1 and a 0.9 value result in a 0 event. But you're basically describing keeping a cumulative moving average (en.wikipedia.org/wiki/Moving_average#Cumulative_moving_average) and correcting based on that. One risk is that each result would be significantly inversely correlated with the prior result, though this correlation would decrease over time.

#### @Kylotan 2012-07-16 11:11:48

I would be tempted to alter this to use a 'leaky integrator' system instead: start with the average initialised to 0.5 and instead of counting samples pick arbitrary constant value (eg. 10, 20, 50, or 100) which does not get incremented. Then at least the correlation between 2 subsequent values is constant throughout the use of the generator. You can also tweak the constant value - larger values mean slower correction and more apparent randomness.

#### @Jonathan Dickinson 2012-07-16 11:46:03

@Kylotan thanks, thanks for providing the name. I am not sure exactly what you mean with your second comment - maybe provide a new answer?

That's quite clever and doesn't have the limitations of arrays. I understand Kylotan's suggestion, which is to initialize `samples` at its maximum value (in this case, 100) from the start. That way, it doesn't take 99 iterations for the RNG to stabilize. Either way, the one disadvantage I can see with this method is that it does not guarantee fairness, it simply ensures a constant average.

#### @Jonathan Dickinson 2012-07-16 12:04:59

@jSepia - indeed, you would still get runs of fairness/unfairness but they would be followed (usually) by a balanced run. E.g. In my unit test I 'forced' 100 non-procs and was met with ~60 procs when I did the real samples. Under uninfluenced situations (if you look at the code) a 50% proc usually sees, at worst, runs of 2/3 in either direction. But one player could have a run allowing them to defeat the other player. If you want to bias it more strongly to fair: `total = (average / 2) + desired`.

#### @user744 2011-04-17 08:44:22

You can solve it the same way Tetris does, by making a preset list of damage results and shuffling.

Let's say you know the player is going to deal 0.8x to 1.2x damage with a linear distribution. Take the list [0.8, 0.9, 1.0, 1.1, 1.2]. Shuffle it randomly, so you get e.g. [1.2, 1.0, 0.8, 0.9, 1.1].

The first time the player deals damage, they deal 1.2x. Then 1x. Then, etc, to 1.1x. Only when the array is empty should you generate and shuffle a new array.

In practice, you'll probably want do this to 4+ arrays at once (e.g. start with [0.8,0.8,0.8,0.8,0.9,0.9,0.9,0.9,...]). Otherwise the period of the sequence is low enough that players can figure out whether their next hit is "good" or not. (Although that can also add more strategy to the combat, as in Dragon Quest IX's Hoimi table, which people figured out how to probe by looking at healing numbers and tweak until you're guaranteed a rare drop.)

To make it a bit more random you could always have half the list as random numbers, and the other half calculated as (2-x) to get the average correct.

#### @user744 2011-04-17 11:21:54

@Adam: That method really only works for this particular example; if you are dealing out Tetris pieces rather than damage multipliers, what's 2 - S block?

#### @Kylotan 2011-04-17 16:15:59

The usual term for this is sort of system is "random without replacement". It's just analogous to using a deck of cards instead of dice, really.

#### @o0'. 2011-04-17 17:21:44

Even better, you could do half the numbers really random, and only half of them subject to this rule.

#### @user744 2011-04-17 18:59:39

@Lo'oris: I can't see how that's in any way better.

#### @o0'. 2011-04-17 19:04:51

@Joe: because it is more "really random", but still avoids too long streaks (which was what the OP wanted).

#### @user744 2011-04-17 19:07:40

It can still result in the local distribution not resembling the global distribution, which is exactly what the question doesn't want. Terms like "really random" are vague pseudomathematics; the more you define what statistical properties you want, the clearer your intent and game design will be.

#### @John McDonald 2012-07-16 15:38:26

There's a modification to the popular board game Settlers of Catan that uses 36 shuffled cards (one for each dice result), but instead of keeping all 36 cards every round, you discard 4 of the cards off the top to provide a little uncertainty.

#### @Runonthespot 2012-08-01 13:34:58

Also take a look at ShuffleBags which implement this sort of behaviour quite neatly - (kaioa.com/node/53)

#### @coderanger 2011-04-17 02:19:58

What you are asking for is actually the opposite of most PRNGs, a non-linear distribution. Just put in some kind of diminishing returns logic in your rules, Assuming that everything over 1.0x is a "critical hit" of some kind, just say that each round your chances of getting a crit go up by X, until you get one at which point they reset to Y. You then do two rolls each round, one to determine crit or not, and then another for the actual magnitude.

#### @Michael 2011-04-17 02:42:22

This is the general approach I'd take, you use the uniform distribution of the RNG but transform it. You could also use the output of the RNG as an input to your own custom distribution that re-adjusts itself based on recent history, i.e. to force variance in the outputs, so that it looks "more random" in human perception terms.

#### @coderanger 2011-04-17 02:46:00

I actually know of an MMO that does something like this, but the chance of a crit actually increases each time you get one until you don't get one, then it resets to a very low value. This leads to rare streaks of crits which are very satisfying to the player.

#### @Michael 2011-04-17 02:53:16

Sounds like a good alg, long dry spells have always been frustrating, but it doesn't lead to crazy streaks of crits.

#### @user744 2011-04-17 08:53:23

Fixing this doesn't require a nonlinear distribution, it just requires that short time-sequential subsets of the distribution have the same properties as the distribution itself.

#### @dreta 2012-07-16 13:45:29

this is how Blizzard games do it, atleast since Warcraft 3

### [SOLVED] How to improve this random number generation in my context?

• 2016-06-13 16:17:31
• Daniyal Azram
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### [SOLVED] Random number hlsl

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### [SOLVED] How do I produce "enjoyably" random, as opposed to pseudo-random?

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text:
# What is a virtual particle?

1. Sep 25, 2008

### Trip2

what is the differace between a particle and a virtual particle?

2. Sep 25, 2008

### Staff: Mentor

3. Sep 25, 2008

### Tac-Tics

A virtual particle is a particle which is not observed in an experiment, but influences the outcome.

In a Feynman diagram, these are particles that are created during an event, but are destroyed again before a measurement is taken. They are called "virtual", then, because they are never actually seen directly. They only serve to influence the final result, and to say they actually ever existed during the experiment is a question you can't answer.

4. Sep 25, 2008

### clem

A virtual particle is just a useful mathematical step in calculating QFT in perturbation theory. A VP never exists as a physical particle, which is why SP is a suitable candidate.

Last edited: Sep 25, 2008
5. Sep 25, 2008

### Count Iblis

That's not true, because there are then also processes in which SP would be the real particle. The properties of SP as a real particle are not so good. That's why the squared matrix element

|<JMC|WH>|^2

has been declining lately.

Last edited: Sep 25, 2008
6. Sep 25, 2008

### vanesch

Staff Emeritus
To the OP: the last two contributions are jokes where the smiley was forgotten

7. Sep 25, 2008

### clem

I am so conservative, I did not know how to make smileys, but now see its simple. Anyway the part of my previous post up to SP was serious.

8. Sep 25, 2008

### Trip2

so virtual particles are retrocausal?

9. Sep 25, 2008

### vanesch

Staff Emeritus
They can formally be "retrocausal", but as they don't correspond to anything physically well-defined, it's a bit strange to say that. They are part of a quantum field calculation. In that sense, you could almost ask whether the square root of two is retrocausal.
On the other hand, the line between "almost real virtual particles" and real particles is also thin. But these are not retrocausal.

10. Sep 26, 2008

### Trip2

so why is it said that they are causality violaters?

11. Sep 26, 2008

### Staff: Mentor

Who says they violate causality?

12. Sep 26, 2008

### masudr

Virtual particles are not bound by the relation

$$E^2 = p^2 + m^2.$$

This is not a problem however, as spacelike separated field operators have a commutator of zero, and therefore causality is not violated in quantum field theory.

13. Sep 26, 2008

### jnorman

virtual particles are just another of the many ways in which physicists indicate that they do not know what is actually happening.


text:
Publication
Title
Tuning the Fermi level of $SiO_{2}$-supported single-layer graphene by thermal annealing
Author
Abstract
 The effects of thermal annealing in inert Ar gas atmosphere of SiO2-supported, exfoliated single-layer graphene are investigated in this work. A systematic, reproducible change in the electronic properties of graphene is observed after annealing. The most prominent Raman features in graphene, the G and 2D peaks, change in accord to what is expected in the case of hole doping. The results of electrical characterization performed on annealed, back-gated field-effect graphene devices show that the neutrality point voltage VNP increases monotonically with the annealing temperature, confirming the occurrence of excess hole accumulation. No degradation of the structural properties of graphene is observed after annealing at temperatures as high as 400 °C. Thermal annealing of single-layer graphene in controlled Ar atmosphere can therefore be considered a technique to reproducibly modify the electronic structure of graphene by tuning its Fermi level.
Language
English
Source (journal)
The journal of physical chemistry : C : nanomaterials and interfaces. - Washington, D.C., 2007, currens
Publication
Washington, D.C. : 2010
ISSN
1932-7447 [print]
1932-7455 [online]
Volume/pages
114:5(2010), p. 6894-6900
ISI
000276562500002
Full text (Publisher's DOI)
UAntwerpen
 Faculty/Department Publication type Subject


text:
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How many times will the digit 7 be written when listing the integers from 1 to 1000?

A. 110
B. 111
C. 271
D. 300
E. 304
[Reveal] Spoiler: OA

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Official Solution:

How many times will the digit 7 be written when listing the integers from 1 to 1000?

A. 110
B. 111
C. 271
D. 300
E. 304

Many approaches are possible. For example:

Approach #1:

Consider numbers from 0 to 999 written as follows:

1. 000

2. 001

3. 002

4. 003

...

1000. 999

We have 1000 numbers. We used 3 digits per number, hence used total of $$3*1000=3000$$ digits. Now, why should ANY digit have preferences over another? We used each of 10 digits equal # of times, thus we used each digit (including 7) $$\frac{3000}{10}=300$$ times.

Approach #2:

There are several ways to count the number of times 7 appears between 7 and 997. One way is to consider the number of 7's in single, double, and triple digit numbers separately.

One-digit numbers: 7 is the only one-digit number.

Two-digit numbers: 7 could be the tens digit or the units digit. Case 1: 7 is the tens digit. There are 10 ways to place 7 as the tens digit of a two-digit number. Case 2: There are 9 ways to place the units digit. Thus, for two-digit numbers we have: $$10+9=19$$ numbers that contain a 7.

Three-digit numbers: Use the knowledge from the previous two scenarios: each hundred numbers will contain one 7 in numbers such as 107 or 507 and also 19 other sevens in numbers such as 271 or 237. Thus a total of 20 sevens per each hundred and 200 sevens for 1000. Since we have 700's within the range, that adds another 100 times that a seven will be written for a total of 300 times.

Approach #3:

In the range 0-100:

7 as units digit - 10 times (7, 17, 27, ..., 97);

7 as tens digit - 10 time (71, 72, 73, ..., 79);

So in first one hundred numbers 7 is written $$10+10=20$$ times.

In 10 hundreds 7 as units or tens digit will be written $$10*20=200$$ times. Plus 100 times when 7 is written as hundreds digit (700, 701, 702, ..., 799).

Total $$200+100=300$$.

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07 Nov 2014, 10:07
Dear experts,
Approach #1:
What if:
Consider numbers from 1 to 1000 written as follows:

1. 0001

2. 0002

3. 0003

...

1000. 1000

We still have 1000 numbers. However, we used 4 digits per number, hence used total of $$4*1000=4000$$ digits.
$$\frac{4000}{10}=400$$ times.

What's wrong with the above inference?

Thank you so much!

Bunuel wrote:
Official Solution:

How many times will the digit 7 be written when listing the integers from 1 to 1000?

A. 110
B. 111
C. 271
D. 300
E. 304

Many approaches are possible. For example:

Approach #1:

Consider numbers from 0 to 999 written as follows:

1. 000

2. 001

3. 002

4. 003

...

1000. 999

We have 1000 numbers. We used 3 digits per number, hence used total of $$3*1000=3000$$ digits. Now, why should ANY digit have preferences over another? We used each of 10 digits equal # of times, thus we used each digit (including 7) $$\frac{3000}{10}=300$$ times.

Approach #2:

There are several ways to count the number of times 7 appears between 7 and 997. One way is to consider the number of 7's in single, double, and triple digit numbers separately.

One-digit numbers: 7 is the only one-digit number.

Two-digit numbers: 7 could be the first digit or the second digit. Case 1: 7 is the first digit. There are 9 ways to place 7 as the first digit of a two-digit number. Case 2: There are 10 ways to place the second digit, i.e. 0-9. Remember that we have counted 07 already. Thus, for two-digit numbers we have: $$10+9=19$$ numbers that contain a 7.

Three-digit numbers: Use the knowledge from the previous two scenarios: each hundred numbers will contain one 7 in numbers such as 107 or 507 and also 19 other sevens in numbers such as 271 or 237. Thus a total of 20 sevens per each hundred and 200 sevens for 1000. Since we have 700's within the range, that adds another 100 times that a seven will be written for a total of 300 times.

Approach #3:

In the range 0-100:

7 as units digit - 10 times (7, 17, 27, ..., 97);

7 as tens digit - 10 time (71, 72, 73, ..., 79);

So in first one hundred numbers 7 is written $$10+10=20$$ times.

In 10 hundreds 7 as units or tens digit will be written $$10*20=200$$ times. Plus 100 times when 7 is written as hundreds digit (700, 701, 702, ..., 799).

Total $$200+100=300$$.

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04 Feb 2015, 13:34
Why doesn't the following approach work?

Clearly, 7 appears once between 1 and 9. Then, there are 90 integers between 10 and 99. How many times does 7 NOT appear? 8 x 9, which is 72. 90-72 is 18. Therefore 7 appears 18 + 1 = 19 times between 1 and 99. There are 900 integers from 100 to 999. How many times does 7 NOT appear? 8 x 9 x 9 = 648. 900-648 = 252.
Therefore, 7 appears 252 times between 100 and 999.

252 + 19 = 271.

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05 Feb 2015, 03:16
hessen923 wrote:
Why doesn't the following approach work?

Clearly, 7 appears once between 1 and 9. Then, there are 90 integers between 10 and 99. How many times does 7 NOT appear? 8 x 9, which is 72. 90-72 is 18. Therefore 7 appears 18 + 1 = 19 times between 1 and 99. There are 900 integers from 100 to 999. How many times does 7 NOT appear? 8 x 9 x 9 = 648. 900-648 = 252.
Therefore, 7 appears 252 times between 100 and 999.

252 + 19 = 271.

7 is not in 72 numbers (from 10 to 99) but out of 18 numbers with 7, one, 77, has two 7's, so 7 appears from 1 to 99: 18 + 1 + 1 =20 times.

Similarly you are missing that 177, 277, 377, ... contain two 7's and 777 contain three 7's.
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13 Apr 2015, 23:26
possibilities are

X X 7 = 9*9*1
X 7 X = 9*1*9
7 X X = 1*9*9
7 7 X = 1*1*9
7 X 7 = 1*9*1
X 7 7 = 9*1*1
777 = 1
if we add all of them = 81+81+81+9+9+9+1=271
where did i go wrong??
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12 Aug 2015, 03:42
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WaterFlowsUp wrote:
possibilities are

X X 7 = 9*9*1
X 7 X = 9*1*9
7 X X = 1*9*9
7 7 X = 1*1*9
7 X 7 = 1*9*1
X 7 7 = 9*1*1
777 = 1
if we add all of them = 81+81+81+9+9+9+1=271
where did i go wrong??

This is how I did..same approach as you took...this might help you understand what went wrong

_ single digit -> 1 possibility
_ _ two digits -> (1 possibility * 10 possibilities) + (9 possibilities * 1 possibility) -> (9, as we cannot place a 0) = 19
_ _ _ three digits -> (1 possibility * 10 possibilities * 10 possibilities) + (9 possibilities * 1 possibility * 10 possibilities) + (9 possibilities * 10 possibilities * 1 possibility) = 100 + 90 + 90 = 280

add up, 1 + 19 + 280 -> 300

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29 Aug 2015, 02:04
Bunuel wrote:
hessen923 wrote:
Why doesn't the following approach work?

Clearly, 7 appears once between 1 and 9. Then, there are 90 integers between 10 and 99. How many times does 7 NOT appear? 8 x 9, which is 72. 90-72 is 18. Therefore 7 appears 18 + 1 = 19 times between 1 and 99. There are 900 integers from 100 to 999. How many times does 7 NOT appear? 8 x 9 x 9 = 648. 900-648 = 252.
Therefore, 7 appears 252 times between 100 and 999.

252 + 19 = 271.

7 is not in 72 numbers (from 10 to 99) but out of 18 numbers with 7, one, 77, has two 7's, so 7 appears from 1 to 99: 18 + 1 + 1 =20 times.

Similarly you are missing that 177, 277, 377, ... contain two 7's and 777 contain three 7's.

Hi Bunuel
I understand your logic, but what if we count the numbers without 7 such that we have a 3 digit number that can be made by numbers {0,1,2,3,4,5,6,8,9} and then subtract the result from 1000. given this logic, 1000-9*9*9= 271

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29 Aug 2015, 02:25
amirzohrevand wrote:
Bunuel wrote:
hessen923 wrote:
Why doesn't the following approach work?

Clearly, 7 appears once between 1 and 9. Then, there are 90 integers between 10 and 99. How many times does 7 NOT appear? 8 x 9, which is 72. 90-72 is 18. Therefore 7 appears 18 + 1 = 19 times between 1 and 99. There are 900 integers from 100 to 999. How many times does 7 NOT appear? 8 x 9 x 9 = 648. 900-648 = 252.
Therefore, 7 appears 252 times between 100 and 999.

252 + 19 = 271.

7 is not in 72 numbers (from 10 to 99) but out of 18 numbers with 7, one, 77, has two 7's, so 7 appears from 1 to 99: 18 + 1 + 1 =20 times.

Similarly you are missing that 177, 277, 377, ... contain two 7's and 777 contain three 7's.

Hi Bunuel
I understand your logic, but what if we count the numbers without 7 such that we have a 3 digit number that can be made by numbers {0,1,2,3,4,5,6,8,9} and then subtract the result from 1000. given this logic, 1000-9*9*9= 271

This gives numbers with 7's but some of them will have more than one 7.
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30 Aug 2015, 01:47
*** three digits

lets keep right most as 7 this time two can be be filled any 10 digits that includes 0 and 7 also, because we can have to take from 1 to 1000 e.g. 007, 107, 117... so on where 7 at right most.... 10*10 = 100
similarly keep 7 at middle place we more 100
now keep 7 and left most we 100
100+100+100 = 300

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19 Sep 2015, 11:43
My method.

7 can be in units tens or hundres digit.

In units digit - 7 , 17, 27,...87,97 - so 10 times for every 100. so 10*10 for 1000 numbers = 100

In tens digit - 70,71,72,73.....79 - so 10 times for every 100. so 10*10 for 1000 numbers = 100

In hundreds digit - 700,701,702,703....799 - so there are - 100

no other place where 7 can appear.
total = 100+100+100 = 300

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20 Sep 2015, 09:19
So I made the same mistake and counted the number of integers with a 7 (271) but should have counted the number of times a 7 occurs. I understand approach 2 and 3, but not approach #1. If I count the number of digits from 1-1000 inclusive this would be:

single digit numbers: 1, 2, ...9 gives 9 numbers with 1 digit -> 9x1=9 digits
two-digit numbers: 10, 11, ... 99 gives 90 numbers with 2 digits -> 90x2=180 digits
three-digit numbers: 100, 101, ...999 gives 900 numbers with 3 digits -> 900x3=2700 digits
four-digit numbers: 1000. Only a single number with 4 digits-> 4 digits

This sums to 9+180+2700+4=2893 digits and not 3000. What am I missing?

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29 Sep 2015, 10:47
can we solve this problem by using permutations and combinations concept.
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09 Oct 2015, 10:53
Bunuel wrote:
Official Solution:

How many times will the digit 7 be written when listing the integers from 1 to 1000?

A. 110
B. 111
C. 271
D. 300
E. 304

Many approaches are possible. For example:

Approach #1:

Consider numbers from 0 to 999 written as follows:

1. 000

2. 001

3. 002

4. 003

...

1000. 999

We have 1000 numbers. We used 3 digits per number, hence used total of $$3*1000=3000$$ digits. Now, why should ANY digit have preferences over another? We used each of 10 digits equal # of times, thus we used each digit (including 7) $$\frac{3000}{10}=300$$ times.

Approach #2:

There are several ways to count the number of times 7 appears between 7 and 997. One way is to consider the number of 7's in single, double, and triple digit numbers separately.

One-digit numbers: 7 is the only one-digit number.

Two-digit numbers: 7 could be the first digit or the second digit. Case 1: 7 is the first digit. There are 9 ways to place 7 as the first digit of a two-digit number. Case 2: There are 10 ways to place the second digit, i.e. 0-9. Remember that we have counted 07 already. Thus, for two-digit numbers we have: $$10+9=19$$ numbers that contain a 7.

Three-digit numbers: Use the knowledge from the previous two scenarios: each hundred numbers will contain one 7 in numbers such as 107 or 507 and also 19 other sevens in numbers such as 271 or 237. Thus a total of 20 sevens per each hundred and 200 sevens for 1000. Since we have 700's within the range, that adds another 100 times that a seven will be written for a total of 300 times.

Approach #3:

In the range 0-100:

7 as units digit - 10 times (7, 17, 27, ..., 97);

7 as tens digit - 10 time (71, 72, 73, ..., 79);

So in first one hundred numbers 7 is written $$10+10=20$$ times.

In 10 hundreds 7 as units or tens digit will be written $$10*20=200$$ times. Plus 100 times when 7 is written as hundreds digit (700, 701, 702, ..., 799).

Total $$200+100=300$$.

Bunuel - aren't you double counting 77 in approach #3? Answer should be 19 * 9 + 100 = 271

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11 Oct 2015, 06:11
mystseen wrote:
Bunuel wrote:
Official Solution:

How many times will the digit 7 be written when listing the integers from 1 to 1000?

A. 110
B. 111
C. 271
D. 300
E. 304

Many approaches are possible. For example:

Approach #1:

Consider numbers from 0 to 999 written as follows:

1. 000

2. 001

3. 002

4. 003

...

1000. 999

We have 1000 numbers. We used 3 digits per number, hence used total of $$3*1000=3000$$ digits. Now, why should ANY digit have preferences over another? We used each of 10 digits equal # of times, thus we used each digit (including 7) $$\frac{3000}{10}=300$$ times.

Approach #2:

There are several ways to count the number of times 7 appears between 7 and 997. One way is to consider the number of 7's in single, double, and triple digit numbers separately.

One-digit numbers: 7 is the only one-digit number.

Two-digit numbers: 7 could be the first digit or the second digit. Case 1: 7 is the first digit. There are 9 ways to place 7 as the first digit of a two-digit number. Case 2: There are 10 ways to place the second digit, i.e. 0-9. Remember that we have counted 07 already. Thus, for two-digit numbers we have: $$10+9=19$$ numbers that contain a 7.

Three-digit numbers: Use the knowledge from the previous two scenarios: each hundred numbers will contain one 7 in numbers such as 107 or 507 and also 19 other sevens in numbers such as 271 or 237. Thus a total of 20 sevens per each hundred and 200 sevens for 1000. Since we have 700's within the range, that adds another 100 times that a seven will be written for a total of 300 times.

Approach #3:

In the range 0-100:

7 as units digit - 10 times (7, 17, 27, ..., 97);

7 as tens digit - 10 time (71, 72, 73, ..., 79);

So in first one hundred numbers 7 is written $$10+10=20$$ times.

In 10 hundreds 7 as units or tens digit will be written $$10*20=200$$ times. Plus 100 times when 7 is written as hundreds digit (700, 701, 702, ..., 799).

Total $$200+100=300$$.

Bunuel - aren't you double counting 77 in approach #3? Answer should be 19 * 9 + 100 = 271

No, I'm not double counting and the answer is correct. The question asks "How many times will the digit 7 be written when listing the integers from 1 to 1000?" not how many number will have 7.
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25 Nov 2015, 23:29
Bunuel wrote:
mystseen wrote:
Bunuel wrote:
Official Solution:

How many times will the digit 7 be written when listing the integers from 1 to 1000?

A. 110
B. 111
C. 271
D. 300
E. 304

Many approaches are possible. For example:

Approach #1:

Consider numbers from 0 to 999 written as follows:

1. 000

2. 001

3. 002

4. 003

...

1000. 999

We have 1000 numbers. We used 3 digits per number, hence used total of $$3*1000=3000$$ digits. Now, why should ANY digit have preferences over another? We used each of 10 digits equal # of times, thus we used each digit (including 7) $$\frac{3000}{10}=300$$ times.

Approach #2:

There are several ways to count the number of times 7 appears between 7 and 997. One way is to consider the number of 7's in single, double, and triple digit numbers separately.

One-digit numbers: 7 is the only one-digit number.

Two-digit numbers: 7 could be the first digit or the second digit. Case 1: 7 is the first digit. There are 9 ways to place 7 as the first digit of a two-digit number. Case 2: There are 10 ways to place the second digit, i.e. 0-9. Remember that we have counted 07 already. Thus, for two-digit numbers we have: $$10+9=19$$ numbers that contain a 7.

Three-digit numbers: Use the knowledge from the previous two scenarios: each hundred numbers will contain one 7 in numbers such as 107 or 507 and also 19 other sevens in numbers such as 271 or 237. Thus a total of 20 sevens per each hundred and 200 sevens for 1000. Since we have 700's within the range, that adds another 100 times that a seven will be written for a total of 300 times.

Approach #3:

In the range 0-100:

7 as units digit - 10 times (7, 17, 27, ..., 97);

7 as tens digit - 10 time (71, 72, 73, ..., 79);

So in first one hundred numbers 7 is written $$10+10=20$$ times.

In 10 hundreds 7 as units or tens digit will be written $$10*20=200$$ times. Plus 100 times when 7 is written as hundreds digit (700, 701, 702, ..., 799).

Total $$200+100=300$$.

Bunuel - aren't you double counting 77 in approach #3? Answer should be 19 * 9 + 100 = 271

No, I'm not double counting and the answer is correct. The question asks "How many times will the digit 7 be written when listing the integers from 1 to 1000?" not how many number will have 7.

Hi Bunuel,
you have missed counting 70 as a number having digit 7 in tens place and hence the doubt of the person..
7 at tens place=70,71,72...79=10 places
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26 Nov 2015, 11:48
chetan2u wrote:

Hi Bunuel,
you have missed counting 70 as a number having digit 7 in tens place and hence the doubt of the person..
7 at tens place=70,71,72...79=10 places

Its merely a typo. Bunuel 's solution does capture 10 values and 71-79 can not be 10 values. It must have included 70 to make it 10 values with 7 at the ten's place.
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26 Nov 2015, 21:46
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Engr2012 wrote:
chetan2u wrote:

Hi Bunuel,
you have missed counting 70 as a number having digit 7 in tens place and hence the doubt of the person..
7 at tens place=70,71,72...79=10 places

Its merely a typo. Bunuel 's solution does capture 10 values and 71-79 can not be 10 values. It must have included 70 to make it 10 values with 7 at the ten's place.

Hi buddy,
few points for you..
1) i have written it is 'missed out' and not a 'mistake'...
2) counting 7 in ten places is a very average thing to know and i believe if you and i can see that then bunuel must have done a 'typo' error and that is why i have mentioned 'missing out'.
3) for any one counting 7 in tens place from 71 to 79 can count ten taking two 7s in 77 and that is what someone has written earlier.
4) the essence of comment was to include 70 in the list..
Regards
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28 Apr 2016, 22:40
Would this be a possible way of solving this problem?

XX7
We can have 10 various digits = 0-9 instead of X

And there are 3 possible positions for 7: {7XX, X7X, XX7}

P(2,3)*C(1,10)*C(1,10)

Hence 3*10*10=300.

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22 May 2016, 08:58
Dear Bunuel,

I think there is a typo in the explanation of approach 2 below:

I think there are 10 numbers meetign the first case which is

" Case 1: 7 is the first digit. There are 9 ways to place 7 as the first digit of a two-digit number. "

Those numbers are : 70, 71, 72, 73, 74, 75, 76, 77, 78, 79.

So the total number of 2 digit numbers become: 10 + 9 = 19.

Regards,
Ankush

Bunuel wrote:
Official Solution:

How many times will the digit 7 be written when listing the integers from 1 to 1000?

A. 110
B. 111
C. 271
D. 300
E. 304

Many approaches are possible. For example:

Approach #1:

Consider numbers from 0 to 999 written as follows:

1. 000

2. 001

3. 002

4. 003

...

1000. 999

We have 1000 numbers. We used 3 digits per number, hence used total of $$3*1000=3000$$ digits. Now, why should ANY digit have preferences over another? We used each of 10 digits equal # of times, thus we used each digit (including 7) $$\frac{3000}{10}=300$$ times.

Approach #2:

There are several ways to count the number of times 7 appears between 7 and 997. One way is to consider the number of 7's in single, double, and triple digit numbers separately.

One-digit numbers: 7 is the only one-digit number.

Two-digit numbers: 7 could be the first digit or the second digit. Case 1: 7 is the first digit. There are 9 ways to place 7 as the first digit of a two-digit number.Case 2: There are 10 ways to place the second digit, i.e. 0-9. Remember that we have counted 07 already. Thus, for two-digit numbers we have: $$10+9=19$$ numbers that contain a 7.

Three-digit numbers: Use the knowledge from the previous two scenarios: each hundred numbers will contain one 7 in numbers such as 107 or 507 and also 19 other sevens in numbers such as 271 or 237. Thus a total of 20 sevens per each hundred and 200 sevens for 1000. Since we have 700's within the range, that adds another 100 times that a seven will be written for a total of 300 times.

Approach #3:

In the range 0-100:

7 as units digit - 10 times (7, 17, 27, ..., 97);

7 as tens digit - 10 time (71, 72, 73, ..., 79);

So in first one hundred numbers 7 is written $$10+10=20$$ times.

In 10 hundreds 7 as units or tens digit will be written $$10*20=200$$ times. Plus 100 times when 7 is written as hundreds digit (700, 701, 702, ..., 799).

Total $$200+100=300$$.

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QUESTION

# POL 319 Week 3 DQ 1 Reinventing State and Local Governance

This file of POL 319 Week 3 Discussion Question 1 Reinventing State and Local Governance comprises:

State and local governments on average, have grown at a rate much higher than that of the population. However, there does not appear to be an equivalent improvement in government service for the increase in personnel. Discuss whether you think your local and state government service has improved or worsened where you live. New Public Management (NPM) is an approach for improving local government. What is this system and do you think it will improve local governance?Your initial post should be at least 250 words in length. Support your claims with examples from required material(s) and/or other scholarly resources, and properly cite any references. Respond to at least two of your classmates

• $5.19 ANSWER Tutor has posted answer for$5.19. See answer's preview

*** 319 **** * DQ 1 *********** State *** ***** **********


text:
An analysis of the celestial body of beta pictoris system in astronomical research

We have no samples of materials from bodies elsewhere in the solar system for analysis in for astronomical research-that of star beta pictoris. Much of today's research in astronomy does not astronomers examine radiation of all wavelenghts emitted by a celestial body beta pictoris (about. Is an icy small solar system body the first exocomet system detected was around beta pictoris, a very young type a v star, in documents similar to cometdocx. The young beta pictoris is a university of arizona) are seeing in our own solar system their analysis of the more distant planet 9 planet 10 planet x.

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An analysis of the celestial body of beta pictoris system in astronomical research
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1. ## Coins

You have 10 boxes with 10 coins each. All the coins from 9 of those boxes(all 90 of them) all weigh 1g. However the other box has coins which all weigh 1.1g. You have a scale that measures the weight of the load you put on. How can you determin which box has the 1.1g coins just by weighing one load of any combination of these 100 coins?

2. Originally Posted by usagi_killer
You have 10 boxes with 10 coins each. All the coins from 9 of those boxes(all 90 of them) all weigh 1g. However the other box has coins which all weigh 1.1g. You have a scale that measures the weight of the load you put on. How can you determin which box has the 1.1g coins just by weighing one load of any combination of these 100 coins?
Put one coin of the first box, two coins of the second box,..., ten coins of the tenth box on the scale. You should easily see why this works.

3. Sorry can you explain why it works?

4. Originally Posted by usagi_killer
Sorry can you explain why it works?
Can you make in into an equation?

The weight of this pile will be $w=1*x_1+2*x_2+ \ldots + 10*x_{10}$, with $x_i$ being the weight of one coin from the $i^{th}$ box. So, what will the weight be precisely? Substitute in the $x_i$'s, and what do you get?

5. Originally Posted by usagi_killer
You have 10 boxes with 10 coins each. All the coins from 9 of those boxes(all 90 of them) all weigh 1g. However the other box has coins which all weigh 1.1g. You have a scale that measures the weight of the load you put on. How can you determin which box has the 1.1g coins just by weighing one load of any combination of these 100 coins?
I would have thought with one weighing its impossible (except if you get lucky)

CB

6. Originally Posted by CaptainBlack
I would have thought with one weighing its impossible (except if you get lucky)

CB
This is also an interesting question:
The Ultimate Puzzle Site - Brain-Teasers

7. Originally Posted by Laurent
Put one coin of the first box, two coins of the second box,..., ten coins of the tenth box on the scale. You should easily see why this works.
Ahh.. it weighs the coins, I was reading what I expected to read .. that it was a balance.

CB

8. Originally Posted by malaygoel
This is also an interesting question:
The Ultimate Puzzle Site - Brain-Teasers

My solution is a bit different from theirs:

Spoiler:
Split the coins into three sets of 4 coins, and compare two of these on your balance. Either one set which you weigh and is heavier, or the two sets you weigh are equal and so the other set has the odd coin. That is one weigh, and you know the odd coin is in a specific set of 4 coins. Solving the problem for 4 coins and two weighs is arbitrary, and so we are done.

9. ## thanx

__________________

thx


text:
# Homework Help: Wave function for Hydrogen and Probability

1. Apr 4, 2005

### Twigs

Hey, if any one can throw in some thoughts im a little lost. Not sure If I need to integrate, or what. Thanks for any help.

The wave function for a hydrogen atom in the 2 s state is:(attachment)

I need to Calculate the probability that an electron in the 2 s state will be found at a distance less than 3.00 a from the nucleus.

File size:
1.2 KB
Views:
129
2. Apr 5, 2005

### dextercioby

Of course you need to inegrate.The Probability is the integral of the probability density...

$$\mathcal{P}=\iiint_{\mathcal{D}} \left| \Psi \left(\vec{r},t\right) \right|^{2} \ dV$$

is the probability of finding the electron (infinite mass nucleus) in the domain $\mathcal{D}\subseteq \mathbb{R}^{3}$

Daniel.


text:
# Graphs, groups, and more: celebrating Brian Alspach’s 80th and Dragan Marušič’s 65th birthdays

from 28 May 2018 to 1 June 2018
Koper
UTC timezone
Home > Timetable > Contribution details

# On decomposing $3$-uniform hypergraphs into loose $m$-cycles

## Speakers

• Prof. Saad EL-ZANATI

## Content

A loose $m$-cycle is a 3-uniform hypergraph with vertex set ${v_1, v_2, \ldots, v_{2m}}$ edge set ${{v_1,v_2,v_3}, {v_3,v_4,v_5}, \ldots, {v_{2m-1},v_{2m},v_1}}$. We consider the problem of decomposing $K_{v}^{(3)}$, the complete 3-uniform hypergraph of order $v$, into edge-disjoint loose $m$-cycles. We settle the problem in the case $m=4$.


text:
## Can Anyone See The Bug Here?

Discussions on extending SharpCap using the built in Python scripting functionality
1CM69
Posts: 63
Joined: Tue Nov 14, 2017 9:49 pm

### Can Anyone See The Bug Here?

Hi,

I've spent most of the day debugging my script but to no avail.

Lots of errors have been rectified, mostly pertaining to INDENTS but still I cannot see what else I am missing.

It's really strange because I have a working script version but now I have built a GUI around this script, it is failing.

Here's my latest unworking script with GUI:
GUI_2.txt
Here's the working version without GUI:
LRGB_Sequence_Capture.txt
Any pointers would be lovely.

Forgot to mention that the script runs fine but then hangs right after this line:

Code: Select all

self.appendMessageText('Capturing LUMINANCE Channel: (' + str(count) + '/' + str(capSeq) + ')')
SharpCap.Wheels.SelectedWheel.Position = 1

Cheers..,

Kirk

oopfan
Posts: 593
Joined: Sat Jul 08, 2017 2:37 pm
Location: New York
Contact:

### Re: Can Anyone See The Bug Here?

Code: Select all

while True:
if not SharpCap.SelectedCamera.Capturing :
break
time.sleep(0.5)

If you ask me, that 'time.sleep' should be inside the while-loop. The way you have it written that while-loop is bogarting the CPU. By putting the 'time.sleep' inside the while-loop you are giving the capture time to execute.

That was it. In your working code you have the indentation right. This is one thing that I dislike about Python -- indentations.

Brian

1CM69
Posts: 63
Joined: Tue Nov 14, 2017 9:49 pm

### Re: Can Anyone See The Bug Here?

oopfan wrote:
Fri Aug 24, 2018 10:02 pm

Code: Select all

while True:
if not SharpCap.SelectedCamera.Capturing :
break
time.sleep(0.5)

If you ask me, that 'time.sleep' should be inside the while-loop. The way you have it written that while-loop is bogarting the CPU. By putting the 'time.sleep' inside the while-loop you are giving the capture time to execute.

That was it. In your working code you have the indentation right. This is one thing that I dislike about Python -- indentations.

Brian
OK, thanks Brian.

I’ll have a look at that in the morning.

Yes, indentation, so very easy to get lost in whitespace lol

Cheers..,

Kitk

1CM69
Posts: 63
Joined: Tue Nov 14, 2017 9:49 pm

### Re: Can Anyone See The Bug Here?

Well I am stumped...

I definitely thought that it was down to indents causing the issue but I have seemingly tried every combination under the sun with no joy.

The script works perfectly as a bare script with no GUI but as soon as I wrap the code inside a Button Click it hangs.

By trial and error and the insertion of a 'MessageBox.Show' at various places along the code I have narrowed it down to the 'While True:' loop

So upon clicking my GUI Button the first part of the function performs as planned, moving the filterwheel to position1 if not already there.

It displays the message that it's capturing LUMINANCE channel and starts to capture.

Here is my first indication of failure because when the working bare script reaches this point SharpCap shows the capture countdown as per normal in the right hand side of the statusbar and the 'Stop capture' button on the toolbar is enabled.

On my GUI script, neither the countdown is displayed or the the 'Stop capture' button enabled.

So like I have stated, by careful placement of a 'MessageBox.Show' directly after the first 'While.True:' I can see that the code never gets past this point and constantly rewrites the 'MessageBox' to the screen, therefore for some reason the 'While True:' is not stepping in to the 'if not SharpCap.SelectedCamera.Capturing :' line and breaking the loop when the 'CaptureLimitValue' is reached.

Here are the section of code, first the working code:
Working Capture Code.txt
and the non working code:
Non Working Capture Code.txt
Regards..,

Kirk

oopfan
Posts: 593
Joined: Sat Jul 08, 2017 2:37 pm
Location: New York
Contact:

### Re: Can Anyone See The Bug Here?

Kirk,

I can't say if this will fix your problem but if you go back to one of your original posts where I attached my Python code:

viewtopic.php?f=14&t=1029

You will see that I spawn a worker thread for operations that take a long time to execute. In GUI programming you don't want to kick off a lengthy procedure from the UI thread, especially if that procedure also wants to update the UI. This is true for all languages not just Python.

Here are the important bits:

Code: Select all

class MyForm(Form):
def onClickButton(self, sender, event):

def __init__(self, form):
self.form = form

def doWork(self):
self.form.appendMessageText("Working...")
my_function(self.form);
self.form.appendMessageText("Done.")

"my_function" is in global scope in my example. If you want you can inline the code in the doWork method but it is not good practice.

Brian

1CM69
Posts: 63
Joined: Tue Nov 14, 2017 9:49 pm

### Re: Can Anyone See The Bug Here?

oopfan wrote:
Mon Aug 27, 2018 7:17 pm
Kirk,

I can't say if this will fix your problem but if you go back to one of your original posts where I attached my Python code:

viewtopic.php?f=14&t=1029

You will see that I spawn a worker thread for operations that take a long time to execute. In GUI programming you don't want to kick off a lengthy procedure from the UI thread, especially if that procedure also wants to update the UI. This is true for all languages not just Python.

Here are the important bits:

Code: Select all

class MyForm(Form):
def onClickButton(self, sender, event):

def __init__(self, form):
self.form = form

def doWork(self):
self.form.appendMessageText("Working...")
my_function(self.form);
self.form.appendMessageText("Done.")

"my_function" is in global scope in my example. If you want you can inline the code in the doWork method but it is not good practice.

Brian
Thanks Brian,
I was just looking through your earlier code too.

I had to change this line:

Code: Select all

my_function(self.form);
to:

Code: Select all

self.form.my_function()
and all is now working.

Just need to finish off a little and tidy up my script.

Thanks for all the help.

Regards..,

Kirk


text:
High variance in path tracing specular objects

I'm testing my path tracer for specular surfaces. Images rendered with my own code have very noisy appearance, and I think they look like variance.

I use point light and the bunny is of mirror material, walls are matte. If it is the problem with sampling, how should I modify the program. To my understanding, point lights are specially handled and thus the sampling results should be a constant value. So I don't think the noise is caused by light sampling. So could there be any bugs?

When only matte material exists in the scene, there are also variance, but not that bad, and global illumination effects are visible. So what could the problem?

specular bunny:

matte bunny:

For the specular bunny image, spp is 4. Since it is a point light, wi is directly set to the vector between point light and intersection point when calculating direct illumination, so there is actually no "sampling". When computing scattering directions, BRDF is matte when intersecting the wall, and BRDF is specular reflection when the ray hit the bunny.

Matte BRDF is cosine weighted sampled, while specular reflection just use the perfect reflection direction.

• Do the fire flies get darker after more samples or are they stuck at white? That would suggest a NAN has been output. – PaulHK Nov 13 '18 at 7:23
• @PaulHK Yes they get darker but still exists. Does that suggest anything? – jinglei Nov 13 '18 at 14:23

If your bunny is purely specular, then sampling the light directly at the shade point would give no contribution since the specular BSDF is a delta BSDF. It generally evaluates to zero for any direction other than the mirror direction. If it was a glossy BSDF, then it might be possible that the pdf value could be very small so that the monte-carlo estimator $$\frac{f(X)}{P(X)}$$ just explodes (which is generally the source of fireflies i.e having a very small pdf and the function is large).
If it was a mirror direction, then you should just see a black bunny for direct illumination and just the reflection of the surrounding scene for multiple bounces. Multiple bounces from the specular bunny should not be cause fireflies, because if you're sampling based on cosine term at the matte surface, the estimator becomes $$\frac{L_i(X)f_r(X)\cos(\theta_i)}{\text{pdf}(\theta_i)}$$, where $$\theta_i$$ is the sampled direction using BSDF sampling at the matte surface. The pdf and the $$\cos(\theta_i)$$ should cancel out each other perfectly. However since you're using a point light, you can never sample from the BSDF since the light is a delta light and needs an explicit connection. Hence the pdf for the estimator would be 1 all the time. The only way for large values to creep up are in the numerator. So I would suggest you to check the values by increasing bounce one by one to see when the problems start creeping in.


text:
HomeEnglishClass 14ReasoningChapterSolved Paper

B 5 R 1 @ E K 4 F 7 c L A M 2 ...

# B 5 R 1 @ E K 4 F 7 c L A M 2 P 3 % 9 H I W 8 * 6 U J \$ V Q # Which of the following is the fifth to the left of the seventeenth form the left end of the above arrangement?

Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

Updated On: 17-8-2021

Apne doubts clear karein ab Whatsapp par bhi. Try it now.

Get Answer to any question, just click a photo and upload the photo
and get the answer completely free,
UPLOAD PHOTO AND GET THE ANSWER NOW!

Watch 1000+ concepts & tricky questions explained!

Click here to get PDF DOWNLOAD for all questions and answers of this chapter - IBPS & SBI PREVIOUS YEAR PAPER Class 14 SOLVED PAPER

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Text Solution

L W * 4

5^(th) to left of 17^(th) from left =17-5 =12^(th) from left =L


text:
Answered You can buy a ready-made answer or pick a professional tutor to order an original one.

QUESTION

# contemporary world

Below is a discussion question

Brower and Sanders (2014) write that observers of the post-World War I Middle East referred to the region as one of "states without nations, nations without states" (p. 353). Does this characterization still fit the region generally? If so, why? If not, why not?

Below is an answer to the above question need a relpy/response to the given answer below like agreeing or disagree so on. Needs to be only a small paragraph no double spacing or no APA format.

Yes, I believe it does.  This region is consistently battling over borders and/or religions and/or ethnicities.  Despite all of the help from other nations and the UN, these battles continue today.

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Q. 15.9

Design a feedforward-feedback control system for the blending system in Example 15.5, for a situation in which an improved sensor is available that has a smaller time delay of $0.1 min$. Repeat parts (b), (c), and (d) of Example 15.5. For part (c), approximate $G_{v} G_{p} G_{m}$ with a first-order plus time-delay transfer function, and then use a PI controller with ITAE controller tuning for disturbances (see Table 12.4). For the feedforward controller in Eq. 15-34, use $\alpha=0.1$.

$G_{f}(s)=-4.17 \frac{1.0833 s+1}{\alpha(1.0833) s+1}$             (15-34)

Develop a Simulink diagram for feedforward-feedback control and generate two graphs similar to those in Fig. 15.13.

 Table 12.4 Controller Design Relations Based on the ITAE Performance Index and a First-Order-plus-Time-Delay Model (Lipták, 2006)*,† Type of Input Type of Controller Mode A B Disturbance PI P 0.859 −0.977 I 0.674 −0.680 Disturbance PID P 1.357 −0.947 I 0.842 −0.738 D 0.381 0.995 Set point PI P 0.586 −0.916 I 1.03† −0.165† Set point PID P 0.965 −0.85 I 0.796† −0.1465† D 0.308 0.929

${ }^{*}$ Design relation: $Y=A(\theta / \tau)^{B}$ where $Y=K K_{c}$ for the proportional mode, $\tau / \tau_{I}$ for the integral mode, and $\tau_{D} / \tau$ for the derivative mode.

${ }^{†}$For set-point changes, the design relation for the integral mode is $\tau / \tau_{I}=A+B(\theta / \tau)$.

Verified Solution

The block diagram for the feedforward-feedback control system is shown in Fig. $15.12 .$

(a) Not required

(b) Feedforward controllers

From Example 15.5,

\begin{aligned}&G_{I P}=K_{I P}=0.75 psi / mA \\&G_{v}(s)=\frac{K_{v}}{\tau_{v} s+1}=\frac{250}{0.0833 s+1}\end{aligned}

Since the measurement time delay is now $0.1 min$, it follows that:

$G_{t}(s)=G_{m}(s)=K_{t} e^{-\theta s}=32 e^{-0.1 s}$

The process and disturbance transfer functions are:

$\frac{X^{\prime}(s)}{W_{2}^{\prime}(s)}=\frac{2.6 x 10^{-4}}{4.71 s+1}, \quad \frac{X^{\prime}(s)}{X_{1}^{\prime}(s)}=\frac{0.65}{4.71 s+1}$

The ideal dynamic feedforward controller is given by Eq. 15-21:

$G_{f}=-\frac{G_{d}}{K_{I P} G_{t} G_{v} G_{p}}$       (15-21)

Substituting the individual transfer functions into Eq. 15-21 gives,

$G_{f}(s)=-0.417(0.0833 s+1) e^{+0.1 s}$           (1)

The static (or steady-state) version of the controller is simply a gain, $K_{f}$ :

$K_{f}=-0.417$          (2)

Note that $G _{f(s)}$ in (1) is physically unrealizable. In order to derive a physically realizable dynamic controller, the unrealizable controller in (1) is approximated by a lead-lag unit, in analogy with Example 15.5 :

$G_{f}(s)=-0.417 \frac{0.1833 s+1}{0.01833 s+1}$              (3)

Equation 3 was derived from (1) by: (i) omitting the time delay term, (ii) adding the time delay of $0.1 min$ to the lead time constant, and (iii) introducing a small time constant of $\alpha x 0.1833=0.01833$ for $\alpha=0.1$.

(c) Feedback controller

Define $G$ as,

$G=G_{I P} G_{v} G_{p} G m=(0.75)\left(\frac{25}{0.0833 s+1}\right)\left(\frac{2.6 \times 10^{-4}}{4.71 s+1}\right)\left(32 e^{-0.1 s}\right)$

First, approximate $G$ as a FOPTD model, $\tilde{G}$ using Skogestad’s half-rule method in Section $6.3$ :

\begin{aligned}&\tau=4.71+0.5(0.0833)=4.75 min \\&\theta=0.1+0.5(0.0833)=0.14 min\end{aligned}

Thus,

$\tilde{G}=\frac{0.208 e^{-0.14 s}}{4.75 s+1}$

The ITAE controller settings are calculated as:

\begin{aligned}&K K_{c}=0.859\left(\frac{\theta}{\tau}\right)^{-0.977}=0.859\left(\frac{0.14}{4.752}\right)^{-0.977} \quad \Rightarrow K_{c}=134 \\&\frac{\tau}{\tau_{I}}=0.874\left(\frac{\theta}{\tau}\right)^{-0.680}=0.674\left(\frac{0.14}{4.752}\right)^{-0.680} \quad \Rightarrow \tau_{I}=0.642 min\end{aligned}

(d) Combined feedforward-feedback control

This control system consists of the dynamic feedforward controller of part (b) and the PI controller of part (c).

The closed-loop responses to $a +0.2$ step change in $x_{1}$ for the two feedforward controllers are shown in Fig. S15.9a. The dynamic feedforward controller is superior to the static feedforward controller because both the maximum deviation from the set point and the settling time are smaller. Figure S15.9b shows that the combined feedforward-feedback control system provides the best control and is superior to the PI controller. A comparison of Figs. S15.9a and S15.9b shows that the addition of feedback control significantly reduces the settling time due to the very large value of $K_{c}$ that can be employed because the time delay is very small. (Note that $\theta / \tau=0.14 / 4.75=0.0029 .$ )


text:
# Math Help - differentiate e^(-x)/x^2

1. ## differentiate e^(-x)/x^2

Differentiate $H(x)=\frac{e^{-x}}{x^2}$

2. Originally Posted by Minino777
$H(x)=\frac{e^{-x}}{x^2}$
Have you tried anything? Write $H(x)=e^{-x}x^{-2}$ and use the product rule in conjunction with the chain rule.

3. Originally Posted by Minino777
Differentiate $H(x)=\frac{e^{-x}}{x^2}$
The quotient rule immediately comes to mind. Review it. Use it. If you need more help, please show all your work and say where you get stuck.


text:
# ListDensityPlots with different variations of colors are not visible

I have four different lists that show different temperature changes in a wire. The minimum and maximum values of the lists are as follows (since there are 1000 elements in each list, I preferred just to insert the maximums and minimums):

min1 = -2; max1 = 2;
min2 = -10; max2 = 10;
min3 = -20; max3 = 20;
min4 = -40; max4 = 30;

Now, I would like to use ListDensityPlot to plot the temperatures along the wire. My question is very general. How can I set the ColorFunction in such a way that the color legend for all the four plots will be the same but I will be also able to see the temperature change for the first plot (with min1 = -2 , max1 = 2). I tried to use "TemperatureMap" but I couldn't control the range. Is there any suggestion?

The lists are as follows:

S1 = {{0, 0, 0}, {0.01025, 391/40, 0.252497}, {0.01625, 783/40,
0.679141}, {0.02225, 235/8, 0.111459}, {0.029, 183/20,
1.62408}, {0.035, 379/20, 2.08228}, {0.041, 115/4,
0.234905}, {0.04775, 341/40, 0.375804}, {0.058, 733/40,
0.00116476}, {0.064, 225/8, -0.28142}, {0.07075, 79/
10, -0.719961}, {0.07675, 177/10, -1.09275}, {0.08275, 55/
2, -0.4097}, {0.0895, 291/40, -1.21021}, {0.0955, 683/
40, -2.18459}, {0.1, 30, 0.}};

S2 = {{0, 0, 0}, {0.01025, 391/40, 0.323326}, {0.01625, 783/40,
1.77028}, {0.02225, 235/8, 0.697066}, {0.029, 183/20,
9.29734}, {0.035, 379/20, 12.8282}, {0.041, 115/4,
1.59031}, {0.04775, 341/40, 12.3881}, {0.058, 733/40,
8.00715}, {0.064, 225/8, 0.58125}, {0.07075, 79/
10, -3.5893}, {0.07675, 177/10, -3.88711}, {0.08275, 55/
2, -2.08424}, {0.0895, 291/40, -7.60716}, {0.0955, 683/
40, -10.0477}, {0.1, 30, 0.}};

S3 = {{0, 0, 0}, {0.01025, 391/40, 1.3907}, {0.01625, 783/40,
6.65005}, {0.02225, 235/8, 2.79659}, {0.029, 183/20,
14.9936}, {0.035, 379/20, 19.4062}, {0.041, 115/4,
5.48806}, {0.04775, 341/40, 31.2655}, {0.058, 733/40,
27.7984}, {0.064, 225/8, 7.13376}, {0.07075, 79/10,
10.9521}, {0.07675, 177/10, 5.07357}, {0.08275, 55/2,
1.40042}, {0.0895, 291/40, -3.24734}, {0.0955, 683/40,
1.72303}, {0.1, 30, 0.}};

S4 = {{0, 0, 0}, {0.01025, 391/40, 2.90893}, {0.01625, 783/40,
7.22693}, {0.02225, 235/8, 8.96227}, {0.029, 183/20,
16.8355}, {0.035, 379/20, 21.9125}, {0.041, 115/4,
18.6437}, {0.04775, 341/40, 33.9003}, {0.058, 733/40,
30.6217}, {0.064, 225/8, 20.3588}, {0.07075, 79/10,
19.0083}, {0.07675, 177/10, 13.4778}, {0.08275, 55/2,
12.8769}, {0.0895, 291/40, 3.17187}, {0.0955, 683/40,
2.93293}, {0.1, 30, 0.}};

• Have you played around with ColorFunctionScaling? Could you post your code (perhaps with simulated data), so that people can see how it's not working? – aardvark2012 Jun 29 '17 at 8:56
• The lists are too large. Its not possible to copy past them here. – KratosMath Jun 29 '17 at 9:00
• I inserted abstracts of the data. – KratosMath Jun 29 '17 at 9:17

Maybe

Legended[Grid[Partition[
ListDensityPlot[#,
ColorFunction -> ColorData[{"TemperatureMap", {-40, 30}}],
ColorFunctionScaling -> False] & /@ {S1, S2, S3, S4},
2]],
BarLegend[{ColorData[{"Temperature", {-40, 30}}], {-40, 30}}]]

• Thanks for your solution. But is it possible to increase the sensitivity of the colours so that even for the first one we would be able to see the temperature change? One possible solution could be defining many colours for the colorfunction. But I don't know how to do it. – KratosMath Jun 29 '17 at 10:08
• @M.Rock, i am not sure how that can be done. Some scaling of your data may work but i don't know how. – kglr Jun 29 '17 at 10:22
• thanks for your answer by the way. – KratosMath Jun 29 '17 at 10:25
• @M.Rock, my pleasure. Welcome to mma.se. – kglr Jun 29 '17 at 10:26
• @M.Rock. You are over-constraining the problem when you insist on one legend for all the plots as well as being able to discriminate over small temperature changes. Especially, when the temperature range in specify for the legend greatly exceeds the temperature range of your data. You will get somewhat better discrimination by specify temperature range to be {-4, 30} – m_goldberg Jun 29 '17 at 17:30

I solved it. In order to have the same colour legend for all plots so that you can easily compare the results, a new colour function should be defined. By using RGBcolor function I created my own function and then inserted it in the listdensityplot.

mycolorTemp =
Blend[{{-30 , color1} , {-28.75 , color2} , {-27.5 ,
color3} , {-26.25 , color4} , {-25 , color5} , {-23.75 ,
color6} , {-22.5 , color7} , {-21.25, color8} , {-20 ,
color9} , {-19 , color10} , {-18 , color11} , {-17 ,
color12} , {-16 , color13} , {-15 , color14} , {-13.375 ,
color15} , {-11.75 , color16} , {-10.125 , color17} , {-8.5 ,
color18} , {-6.875 , color19} , {-5.25 , color20} , {-3.625 ,
color21} , {-2 , color22} , {-1.6 , color23} , {-1.2 ,
color24} , {-0.8 , color25} , {-0.4 , color26} , {0 ,
color27} , {0.4 , color28} , {0.8 , color29} , {1.2 ,
color30} , {1.6 , color31} , {2 , color32} , {3.6 ,
color33} , {5.2 , color34} , {6.8 , color35} , {8.4 ,
color36} , {10 , color37} , {15 , color38} , {20,
color39} , {25 , color40} , {30, color41} , {35 , color42} , {37,
color43} , {39 , color44} , {41 , color45} , {43 ,
color46} , {45 , color47}} , #1]

where colorn is defined as,

colorn = RGBColor[x1 , x2 , x3].

The intensity of x1, x2 and x3 could be easily obtained according to the colour that you want. The following link is also helpful in creating the colour function.

https://codepen.io/leemark/pen/lpEHr


text:
# How to compare multivariate forecasting methods?

Let $X$ be a multivariate time series of $N$ variables and $T$ observations. Let split $X$ into two separate datasets :

• $X_{train}$ : a train set with $N$ variables and $T_{train}$ observations
• $X_{test}$ : a test set with $N$ variables and $T_{test}$ observations

We train different forecasting methods ($k$-NN, $VAR$, etc.) on $X_{train}$. Then, for each method, we perform a one step forecast on $X_{test}$.

We then want to compare the forecasting performance of the different methods. We could compute the RMSE for each variable, but given the differences in amplitude between variables: we cannot average the RMSE over the variables in order to get a unique score for each method.

I thought of two alternatives:

1. compute MASE instead of RMSE
2. for each forecasting methods, count the number of variables for which it has (strictly or almost) the lowest RMSE.

Do you know other ways to compare forecasting performance of several methods – either in a quantitative manner like I did or in a more qualitative manner? The number of variables is high (~600); this is why I tried to summarize the results into a single evaluation criterion.

• What measure do you use for training error? And why is this not appropriate for test error? (For model-selection, my understanding is that most of the "extra sophistication" is essentially heuristics for comparing models of different complexity ... in the case where a test set is not available? I could be wrong!) – GeoMatt22 Nov 3 '16 at 15:08
• The measure of training error can vary from one method to the other. For example : - for the Vector Autoregressive Model (VAR(p)), I can use an information criterion (AIC, BIC, ...) to choose the best value of meta-parameter p and the corresponding model estimation. - for $k$-NN, the parameter $k$ is usually fitted with cross-validation. The classical error measure is the residual sum of squares. But in my case, it is prone to the same problem of difference of amplitude between variables. Note : maybe I shouldn't use the tag model-selection but rather method comparison ? – Jaewon Nov 4 '16 at 9:44


text:
How to obtain an unbiased post-stratified estimator for survival probabilities?

Blain Waan

New Member
Suppose I have estimated the survival probabilities $\hat{S(t)}_1, \hat{S(t)}_2, \ldots, \hat{S(t)}_H$ from $H$ strata. Now I want to get an unbiased estimate of population survival probability $S(t)$ using these stratum-specific estimates. Then what should be the estimator?

Should it be $\hat{S(t)}_{pst} = \sum_{i=1}^h \dfrac{N_h}{N} \times \hat{S(t)}_h$?

If I want to adjust for under or over representation in the sample, then what should be the estimator (suppose I know the population and sample proportions)?

staassis

Active Member
No easy way to adjust. At any time horizon T, the event of survival up to T is co-dependent with the stratum membership. Not only do conditional probabilities differ from the unconditional probabilities but they change with T.

The easiest way out is simulating a representative sample of the whole population out of your strata multiple times. For each simulated sample, run survival analysis and get the survival curve. Use the survival curves from multiple simulations to draw the confidence band for the true survival curve.

ehsanz

New Member
Hi
i want do stratified cox model in R :
library(survival)
library(foreign)
leuk.data
cox1<-coxph(Surv(t,status)~Rx+logWBC+strata(sex),data=leuk.data,method="breslow")
but it shows this error :

Error in Surv(t, status) : Time variable is not numeric
pleas help me how can numeric time and whats your idea ?
TNX

hlsmith

Less is more. Stay pure. Stay poor.
Not a regular R user, but have you looked to see how time is formatted in the dataset?

Code:
 is.numeric(leuk.data\$t)