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AliPhysics  04715ba (04715ba)
EMCal Embedding Framework

# EMCal Embedding Framework

The EMCal Embedding Framework provides the means to embed an already reconstructed signal event into an existing event. For example, the user may embed a Pythia signal event into a Pb-Pb event in an effort to study the jet energy scale. To do so, a new class, AliAnalysisTaskEmcalEmbeddingHelper, manages access to the embedded event, while users usually access the new input objects via AliEmcalContainer derived classes (of course, direct access to the input objects is possible for those who are really interested - see below). Note that the use of AliEmcalContainer objects does not require use of AliAnalysisTaskEmcal, although using it does simplify the process.

Note that this framework was presented at an Analysis Tutorial in Nov 2016. The presentation is a supplement to the documentation presented here (it is highly recommended to read this documentation first) and provides a practical example of embedding charged and full jets. For more information, see below.

# Basics of embedding via the Embedding Helper

// Create the Embedding Helper
// Set the file pattern. This example uses ptHardBin 4 of LHC12a15e_fix.
// The pT hard bin and anchor run can also be set by adding a printf() wild card to the string (ie %d)
// See the documentation of AliAnalysisTaskEmcalEmbeddingHelper::GetFilenames()
embeddingHelper->SetFilePattern("/alice/sim/2012/LHC12a15e_fix/169838/%d/AOD149");
// If the embedded file is an ESD, then set:
embeddingHelper->SetESD();
// For some examples...
// ... randomly select which file to start from:
embeddingHelper->SetRandomFileAccess(kTRUE);
// ... Start from a random event within each file
embeddingHelper->SetRandomEventNumberAccess(kTRUE);
// ... Set pt hard bin properties
embeddingHelper->SetPtHardBin(4);
embeddingHelper->SetNPtHardBins(11);
// etc..
// As your last step, always initialize the helper!
embeddingHelper->Initialize();

There are a large number of configuration options, including event selection and MC outlier rejection, which are available as options for the embedding helper. See AliAnalysisTaskEmcalEmbeddingHelper. Regardless of the options set, always be certain to call AliAnalysisTaskEmcalEmbeddingHelper::Initialize() when you are done configuring the task!

Once the embedding helper has been created, it will manage access to the file. Then, the basic ideas is that users access the embedded input objects via AliEmcalContainer derived classes (AliClusterContainer, AliParticleContainer, AliTrackContainer, etc). For example, accessing tracks in an embedded AOD file via a track container would look like:

// Create the new track container
AliTrackContainer * trackCont = new AliTrackContainer("tracks");
// Tell the track container that it should get the tracks from the embedded event
trackCont->SetIsEmbedding(kTRUE);

Although there are some details to understand, that's roughly all there is to it from the user perspective - it should just work! (Recall again that AliEmcalContainers are not required, although they substantially simplify the user experience - see below for more information if you don't want to use them.)

With these basics in mind, there are a few main topics to discuss:

# Charged input objects (charged tracks)

To access charged tracks, it is as simple as described above: Simply create a track container and set that it is embedding. To repeat from above:

// Create the new track container
AliTrackContainer * trackCont = new AliTrackContainer("tracks");
// Tell the track container that it should get the tracks from the embedded event
trackCont->SetIsEmbedding(kTRUE);

That is all there is to it - the embedded tracks are now available!

Although this process is quite straightforward, for this to be useful, we need to check on some additional details. In particular, to actually embed the tracks in another event, that means that we will need to add at least two track containers - one corresponding the input event, and one to the embedded event. Thus, it is imperative that whatever task you are adding the containers to actually supports multiple containers explicitly. Many base classes already do. For more information, see below.

# Full (charged + neutral) input objects (cells, clusters, and tracks)

In the case of embedding full objects, everything described above still applies! The Embedding Helper still needs to be configured in the same way, and tracks can be accessed in exactly the same way! We are just building on previous steps described here.

To use full objects, we need to also configure the cells and clusters, which involves configuring the EMCal Corrections Framework. Fortunately, the correction framework will handle all of the steering, so so we will just use advanced configuration techniques to setup the correction task (please review the correction task documentation if you are not familiar with these techniques).

Consider an example where we run the cell corrections separately for the input and embedded events, then combine those cells to run the clusterizer, and then run the cluster level corrections. A sample configuration (with an explanation below) is as follows:

inputObjects:
cells:
cells:
branchName: "usedefault"
cells_embed:
branchName: "usedefault"
embedding: true
cells_combined:
branchName: "emcalCellsCombined"
clusterContainers:
# Used for the clusterizer
baseClusterContainer:
branchName: "caloClustersCombined"
# Used after clusterizer
baseClusterContainer_1:
minE: 0.0
minPt: 0.0
# Used for cluster-track matcher and after
baseClusterContainer_2:
branchName: "caloClustersCombined"
minE: 0.0
minPt: 0.0
clusNonLinCorrEnergyCut: 0.15
trackContainers:
trackContainerName:
# Sets the branch name
branchName: "usedefault"
minPt: 0.15
trackContainerName_embed:
embedding: true
CellEnergy_data:
enabled: true
cellsNames:
- cells
enabled: true
cellsNames:
- cells
enabled: true
cellsNames:
- cells_embed
CellTimeCalib_data:
enabled: true
cellsNames:
- cells
CellCombineCollections_combined:
enabled: true
# Name of the cells branch in the embedded (external) event.
externalCellsBranchName: "usedefault"
# Name of the cells branch to be created for the combined cells in the input event.
combinedCellsBranchName: "emcalCellsCombined"
# Name of the cells to connect from the input event
cellsNames:
- cells
Clusterizer_combined:
enabled: true
cellsNames:
- cells_combined
# By selecting the cluster container here, we set where it will be output
clusterContainersNames:
- baseClusterContainer
ClusterExotics_combined:
enabled: true
cellsNames:
- cells_combined
clusterContainersNames:
- baseClusterContainer_1
ClusterTrackMatcher_combined:
enabled: true
cellsNames:
- cells_combined
clusterContainersNames:
- baseClusterContainer_2
trackContainersNames:
- trackContainerName
- trackContainerName_embed
# ...

// To store the 3 corrections tasks
// Create the 3 needed correction tasks
// These handle the cell corrections to the input data
// Selecting all corrections with "_data" in the name
// These handle the cell corrections to the embedded data
// Selecting all corrections with "_embed" in the name
// These handle the cluster corrections to the combined (input + embedded) data
// Selecting all corrections with "_combined" in the name
// It is important that combined is last!
// Loop over all of the correction tasks to configure them the same
{
// Local configuration
// Initialize the configuration
}

## Cells

The cell corrections are split up into multiple sets of corrections because you often need to treat MC and data cells differently. In particular, you'll often need to disable the energy and time calibration tasks for the MC production. Once these individual cell corrections are completed, the cell collections will need to be combined to be an input to the clusterizer.

To combine the cells, we utilize the task AliEmcalCorrectionCellCombineCollections which will handle the combining the two cell collections. To ensure that each cell branch is corrected before they all of the cells are combined, it is imperative that the correction tasks running the cells corrections on the input and embedded events are run before the combined correction task! It is done properly in the example above.

## Clusters

After the cells have been combined, only one correction task needs to continue. This correction task is labeled as "combined". Since all corrections are disabled by default, the other correction tasks will automatically stop after running the cell corrections. All cluster corrections are then run as normal, using the combined cluster (which are in cluster branch "caloClustersCombined" and accessible in the input objects "baseClusterContainer", "baseClusterContainer_1", and "baseClusterContainer_2").

In the case that the user does not want to run the clusterizer again (this should be fairly uncommon), then multiple cluster containers can simply be added to each correction task, just as one would handle tracks.

## Tracks (in cluster corrections)

Simply add each relevant track container to the relevant task. Make sure to mark the proper track container as embedded for cluster corrections which require tracks. Everything will just work, as in the charged track example above.

Note on the Hadronic Correction: Due to technical limitations when working with AOD files, the track selection of the first particle container is applied to all particles from all particle containers. This should not matter for most users, but it is important to be aware of it. A warning will be thrown when running with such a configuration.

# Multiple Container Support

To actually embed into another event, the user must add at least two EMCal containers to the task: one corresponding the input event, and one to the embedded event. Thus, that task must support multiple containers. There are two parts to this support:

2. Properly utilizing multiple containers.

Properly utilizing multiple containers can be a bit more complicated. There are two relevant groups here:

1. Core framework and embedding relevant classes.
2. The user's own analysis task.

In the case of the core framework and embedding relevant classes, most of them already support multiple containers. Examples include

In the case of the user's own task needing access to multiple containers, it is often as simple as adding just a few lines of code to loop over container collections. For AliAnalysisTaskEmcal derived tasks, it should look something like this:

// Assuming that we want to loop over clusters
AliClusterContainer * clusCont = 0;
TIter nextClusCont(&fClusterCollArray);
while ((clusCont = static_cast<AliClusterContainer*>(nextClusCont()))) {
// A cluster container called cont is available.
// It can be used as normal. That might look like:
for (auto clusterIter = clusCont->all_momentum()) {
// Get the cluster
AliVCluster * cluster = clusterIter.second;
}
// Alternatively, if the index of the cluster is needed, use:
auto clusItCont = clusCont->accepted_momentum();
for (AliClusterIterableMomentumContainer::iterator clusIterator = clusItCont.begin(); clusIterator != clusItCont.end(); ++clusIterator) {
// Get the cluster
AliVCluster * cluster = clusterIter.second;
// Get the cluster index
int clusterIndex = clusIterator.current_index();
}
}

Looping over multiple track containers is extremely similar. Once you've setup a loop for both, your class is ready to go! It is up to you on how to handle it from here.

# Recording of Embedded Events Properties

The embedding helper will automatically store the properties of the embedded events such as cross section and number of trials. These properties are recorded before event selection is applied to either the embedded events or the data. However, it is often helpful to record those same properties after both the embedded and data event selections. For such a case, there is an additional class, AliEmcalEmbeddingQA.

To use this case requires a few simple steps. First, an AliEmcalEmbeddingQA object should be added to your task. Next, it must be initialized via AliEmcalEmbeddingQA::Initialize(). This should usually be done when other histograms are being created and configured, such as in UserCreateOutputObjects(). This histograms that are created in this class should be added to your tasks output list via AliEmcalEmbeddingQA::AddQAPlotsToList(). Lastly, the properties should be recorded by calling AliEmcalEmbeddingQA::RecordEmbeddedEventProperties() after the event selection has been performed in your task. The properties of the embedded events will then be available in your tasks output.

Note that for compatibility reasons, AliAnalysisTaskEmcal::IsPythia() should be set to false if you task inherits from that class. Otherwise, histogram names will conflict.

# Embedding on LEGO trains

Embedding can be used as expected on the LEGO train. If available, it is best to use centralized wagons, while changing the configuration for your particular dataset. The main difficulty with the LEGO train is that a new train must be configured for each pt hard bin. Previously, that meant that a variable must be changed every time, which can be a rather error prone process. Now, the embedding helper can handle this configuration automatically.

The relevant functions are in the function group "pT hard bin auto configuration". Once enabled, the user must set the proper paths to determine where to locate a scratch file, as well as how the particular set of trains will be identified. See the example below for an illustrative example:

// Configure the embedding helper as usual
//...
// Enable pt hard bin auto configuration
embeddingHelper->SetAutoConfigurePtHardBins();
// Tells it the base path of where to write the auto configuration file.
embeddingHelper->SetAutoConfigureBasePath("/alice/cern.ch/user/a/alitrain/");
// Identifies the type of train this train is running on.
embeddingHelper->SetAutoConfigureTrainTypePath("PWGJE/Jets_EMC_PbPb/");
// This value must be unique for your train run. Probably best to use your name!
embeddingHelper->SetAutoConfigureIdentifier("myEmbedding");
// Note that if you've enabled the pt hard bin auto configuration, be certain to not select a particular pt hard bin!
// However, you _must_ set the number of pt hard bins in your production!
embeddingHelper->SetNPtHardBins(11);

Now just request the trains. Each one can be started with the desired settings, and they will automatically coordinate to determine which pt hard bins should be selected for each train. No need to manually change variables for each pt hard bin!

With the above settings

# Note on jets and jet finding

When handling jet finding, a bit more care needs to be applied, especially if apply an artificial tracking efficiency. This is due to the fact that the jet finder keeps track of constituents by index, which can cause some ambiguity in which object applies to which container (if you are interested, see the advanced topics for further details). Thus, the recommend approach is as follows:

To apply an artificial efficiency to the embedded input objects, it is best to do so via the jet finder:

// Set the track efficiency as desired
// Tell it to apply to embedding only

Whether or not an artificial efficiency is applied, it is best to access jet constituents through new, dedicated functions in AliEmcalJet which properly handle returning the appropriate object.

// Assuming that you have an AliEmcalJet called "jet"
// And you want to access the constituent tracks
for (int i = 0; i < jet->GetNumberOfTracks(); ++i) {
// Access the track via the jet
AliVTrack * track = jet->Track(i);
}

Previous functions such as AliEmcalJet::TrackAt(Int_t index, TCloensArray * arr) still work, but with multiple containers, it is impossible to disambiguate which object comes from which container without external help (this is handled by AliEmcalContainerIndexMap). AliEmcalJet::Track(Int_t index) handles such situations properly and is provided as a convenience. And of course the use is still welcome to get the index directly via AliEmcalJet::TrackAt(Int_t index) and then retrieve the object themselves (using AliEmcalContainerIndexMap will be required if there are multiple TClonesArrays or containers available).

The above procedure also applies equally to clusters. The function AliEmcalContainer::Cluster(Int_t index) replaces AliEmcalJet::ClusterAt(Int_t index, TClonesArray * arr) in the same way that AliEmcalJet::Track(Int_t index) replaces AliEmcalJet::TrackAt(Int_t index, TClonesArray * arr).

Note that AliEmcalJet::Track() and AliEmcalJet::Cluster() will work regardless of which track or cluster containers were added to the jet container that was used to access the jet!

# Analysis Tutorial Presentation

The framework was presented at the Analysis Tutorial during the November 2016 ALICE Week. Note that a work in progress version of the framework was presented. However, the slides were updated afterwards, and the interface to the framework was already nearly complete, so the information should be still be useful. Also note the presentation on the EMCal corrections framework. Since it is an integral part of steering the embedding framework in the EMCal, it also worth watching that tutorial if you are unfamiliar with the framework and are interested in embedding cells or clusters.

If users want to access the events directly instead of using AliEmcalContainer derived classes - this is not recommended - then the embedded event can be retrieved via:

From here, the user is then responsible for retrieving the information they are interested in.

## Framework details

These details are intended for experts - users can safely ignore them!

The framework was designed with the goal of making it as separate (ie encapsulated) as possible from all other classes. By leveraging existing code as much as possible and keeping implementation details hidden from the user, it ensures that the framework will maintain compatibility as the code base evolves, as well as keeping each task focused on its particular task.

The implementation details for a variety of classes are as follows:

### AliEmcalContainer

AliEmcalContainer::SetArray() checks whether the container is embedded and automatically selects the proper event. Thus, this works seamlessly for the user. Plus, since it is in the base class, it will work for all derived classes - nothing additional code is required.

### AliEmcalContainerIndexMap

AliEmcalContainerIndexMap is an important class that manages the mapping between all of the input object arrays and the indices that are often stored in classes to identify other input objects. Effectively, it can map from a local index in each array to a global index. For an instance in which this matters, the cluster-track matcher stores the track index number in the cluster and the cluster index number in the track under certain configurations. Without this mapping, it would be impossible to know which input object array the index belongs. Note that although this map is used internally, it should be transparent to most uses. If they only add one TClonesArray of a particular input object type, it will never be apparent that it is used (ie indices will just look as normal).

As a concrete example, consider two input arrays, A and B. A has 200 entries and B has 150 entires. Since the size of the arrays can change between events and we need a consistent mapping for A and B, a large offset is required selected between the two arrays. For example, 100000. Thus, the third element in array A is index 2, while the third element in array B is index 100002. Then, when 100002 is given to AliEmcalContainerIndexMap, it will automatically return the third element in array B.

Note that this mapping can be between TCloensArrays and indices or AliEmcalContainer derived classes and indices. AliClusterContainer and AliParticleContainer automatically store the mapping between accessed TClonesArrays and indices. These mappings can be copied and automatically adapted to AliEmcalContainer derived classes for use in another class by calling AliEmcalContainerMap::CopyMappingFrom().

Additional information is available in the class documentation (AliEmcalContainerIndexMap).

## Appendix

Development of this framework was documented in a JIRA ticket.


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• # question_answer Study the following passage carefully and answer the questions given below: Culture can play a huge role in the shaping of an individual's identity and behavior, this can happen through a theory known as the 'tool kit theory'. The tool kit theory as introduced by Swidler, is a combination of the internal and external elements of culture. An individual seeks out activities that cater to their cultural competences in order to be successful. These competences may be a big part of someone's identity. For example, if an individual considered being a student as a main factor of their identity and they were intellectually developed, using this competence they could have a higher chance of succeeding academically. In addition to this, culture can modify behavior as culture dictates the expectations that have to be met to ensure acceptance into a social group. Individuals will modify their behavior to fit others as this will provide a sense of belonging and protection. But sometimes, this creates moderate level of anxiety in the minds of individuals and further prolonged anxious state creates social abnormal behaviors. How behavior can be modified by the culture? A) It creates moderate level of anxiety B) Caters cultural competences in society C) Through cultural expectation to be met D) By fitting behaviors to others in a group

 Rationale: (c) Reasoning type: Cause-and-Effect Reasoning$\to$showing causes and resulting effect Focus of the question: Argument$\to$strong argument for behavior change by culture Relationship analogy: Group analogy$\to$Where several objects together make a group The question focuses on the argument for the behavioral change through the cultural expectations. The culture can modify behavior as culture dictates the expectations that have to be met to ensure acceptance into a social group.

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# How to determine the continuity of a piecewise function defined respectively on $\mathbb{Q}$ and $\mathbb{I}$

How can I tell the continuity of function $$f(x)=\left\{ \begin{array}{lll} 3x^2 & \text{if} & x\in\mathbb Q\\ 4x^2 & \text{if} & x\in\mathbb I \end{array} \right.$$

I could see it is continuous at x=0 and guess it discontinuous no-where-else. How can I start with the formalised proof?

And also, is $f$ differentiable at 0?

Let $x_0 \in \mathbb{R}$. If $(x_n)$ is a sequence of real numbers such that $x_n \to x_0$ as $n \to \infty$ then $f$ is continuous at $x_0$ if and only if $$\lim_{n\to \infty} f(x_n) = f(x_0)$$

Lets assume $x_0 \in \mathbb{Q}$, and let $(x_n)$ be a sequence of irracional numbers (converging to $x_0$), therefore, $f(x_n)$ is the sequence $4x_n^2$ and then $\lim f(x_n) = \lim 4x_n^2 = 4 x_0^2$ and $f(x_0) = 3x_0^2$. If $x_0 \in \mathbb{R}\setminus\mathbb{Q}$ you construct a rational sequence and will reach the same thing, namely: $f$ is continuous at $x_0$ if, and only if $3x_0^2 = 4x_0^2$ and then $f$ is continuous only in $x_0=0$

Let's see now that $f'(0)=0$. But $$f'(x) = lim_{x_0 \to x} \frac{f(x)-f(x_0)}{x-x_0}$$ And then $$f'(0) = \lim_{x \to 0} \frac{f(x)}{x}$$

But $$\frac{f(x)}{x}=\left\{ \begin{array}{lll} 3x & \text{if} & x\in\mathbb Q^*\\ 4x & \text{if} & x\in\mathbb I \end{array} \right.$$ And again using sequences if necessary, $\lim_{x \to 0} \frac{f(x)}{x} = 0$ and $f'(0) = 0$

• That's a very neat proof making use of sequential continuity, thanks! And I do have another question: is $f$ differentiable at x=0 then? I think yes, since $\lim\limits_{h\to 0} \frac{f(h)-f(0)}{h} =\lim\limits_{h\to 0} \frac{4h^2}{h} = 0$ if $h\in\mathbb Q$ and similarly the limit is also $0$ if $h\in\mathbb I$. – MatheMagic Oct 10 '14 at 2:52
• Yes, $f$ is continuous at $0$ because $3*0 = 4*0$ and $f$ is discontinuous everywhere else – Jonas Gomes Oct 10 '14 at 2:56
• sorry i wasn't clear in the first comment. Now it's corrected...and it's differentiable at x=0 right? – MatheMagic Oct 10 '14 at 3:04
• $f'(0) = \lim_{h \to 0} \frac{f(h)}{h} \leq \lim_{h\to 0} \frac{4h^2}{h} = 0$ – Jonas Gomes Oct 10 '14 at 3:09
• Do you understand now @llqlei? – Jonas Gomes Oct 10 '14 at 3:27


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Question

# Write the conjugate acids for the following Bronsted bases $$\displaystyle { NH }_{ 2 }^{ - },\;{ NH }_{ 3 }$$ and $${ HCOO }^{ - }$$.

Solution

## To convert the Bronsted base into their conjugate base, the base has to gain a proton. $$\underset {\text {Bronsted Base}}{\displaystyle { NH }_{ 2 }^{ - }} + H^+ \rightarrow \underset {\text {Conjugate acid}}{{ NH }_{ 3 }}$$The conjugate acids for the Bronsted bases $$\displaystyle { NH }_{ 2 }^{ - },\;{ NH }_{ 3 }$$ and $${ HCOO }^{ - }$$ are $$\displaystyle { NH }_{ 3 },\ { NH }_{ 4 }^{ + }$$ and $$HCOOH$$ respectively.ChemistryNCERTStandard XI

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# What is 'non-expansion work' and how does ΔG represent the maximum non-expansion work that can be extracted in a reversible process? [duplicate]

I have just learned about Gibbs free energy and how its value indicates the change in entropy of the universe and can thus be used to find the spontaneity of a reaction. However, I found this other interpretation of Gibbs free energy on Wikipedia that I don't get at all.

The Gibbs free energy is the maximum amount of non-expansion work that can be extracted from a thermodynamically closed system (one that can exchange heat and work with its surroundings, but no matter). This maximum can be attained only in a completely reversible process. When a system transforms reversibly from an initial state to a final state, the decrease in Gibbs free energy equals the work done by the system to its surroundings, minus the work of the pressure forces.

What exactly does 'non-expansion work' mean in the context of, say, a gas? How can it even do work other than by expanding or contracting?

Secondly, I thought the extractable energy during a reaction was ΔH( heat transferred during a physical or chemical reaction carried out at constant pressure). If that is actually ΔG=ΔΗ-TΔS instead, what kind of energy does 'TΔS" represent? I know that it is equal to the heat transferred into or out of the system if the process is done reversibly but I don't understand how that's relevant here.

Lastly, Wikipedia mentions that this maximum can only be extracted by a reversible process. How exactly would you extract this? For example, if a reaction has a ΔΗ=-50 and ΔG=-75, how do you get the extra 25 units of energy?


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# Solving such integrals numerically?

1. Apr 11, 2005

### TSN79

Is there any way of solving $$\int_{0}^{1} \frac{1}{\sqrt{100-x^5}} dx$$ by some regular method? If not, how does one go about solving such integrals numerically?

2. Apr 11, 2005

### arildno

1. I don't know about any method to crack this integral analytically.
2. As for numerical methods:
These typically start with some fancy finite Riemann sum to approximate the integral.
You could check up on the Trapezoid rule or Simpson's rule or some other rule..

3. Apr 11, 2005

### dextercioby

$$\int_{0}^{1} \frac{dx}{\sqrt{100-x^{5}}} =\frac{1}{10} \ _{2}F_{1}\left(\frac{1}{2},\frac{1}{5},\frac{6}{5},\frac{1}{100}\right)$$

according to my friend,Maple.

Daniel.

4. Apr 11, 2005

### Zurtex

According to Mathematica:

$$\int_{0}^{1} \frac{dx}{\sqrt{100-x^{5}}} = \frac{1}{10} \text{Hypergeometric2F1} \left[ \frac{1}{5},\frac{1}{2},\frac{6}{5},\frac{1}{100} \right]$$

Hmm wonder why the fractions are different, does it make a difference?

Anyway to 100 s.f:

0.1000836762086654607634453194543998535914445494139945308954127536431450557575506273354667189780577481

Last edited: Apr 11, 2005
5. Apr 11, 2005

### dextercioby

Conventions,on normal basis they should be the same...

Gauss's function is $_{2}F_{1}\left(a,b,c,z\right)$

Daniel.

P.S.Here are 125 sig.digs.

$$\allowbreak .\,10008\,36762\,08665\,46076\,34453\,19454\,39985\,35914\,44549\,41399\,45308\,95412\,$$
$$75364\,31450\,55757\,55062\,73354\,66718\,97805\,77480\,65696\,66047\,63945\,64052\,7984$$

Last edited: Apr 11, 2005
6. Apr 11, 2005

### Zurtex

Ahh so it's going to be like that is it, well in that case, to 2000 sf:

0.1000836762086654607634453194543998535914445494139945308954127536431450557575\
506273354667189780577480656966604763945640527983993080671953023089297911515248\
996013282637274784238797044064349870623646876674724483462693995830072965962405\
190019049666775160232759330205668592578946075338958508523904278007888848833088\
060488556739935166104773509141768174230265551654691550416849059337448757791416\
602630567642386763728600302390370505747803022415391215722818064179823297750442\
434640467815965016423148168846638958221953928239081869470007493153876431418021\
157217678724106639004941740270189408736496188721513628469894013953107138449253\
791957534610058071313307798195976366400486448615041008866156544661543944553681\
807447545511569915551726649431123256963503190869480487316815727122678140209122\
738110692425545202808759680890972292418663402273431929145662980727173557540648\
152030821410027920261779677495993071187486068413623093242779260125263946810946\
539669010663042017697120941744353678783323221747296966263595416822115595315497\
544773344894826763667430935304194135002414853743041716586627006645641572144003\
038754961014790401073888263046103372397829209944018459246323194799313562020937\
625258141676770207971455031095503662945181184143286220172118052176868976940819\
363006570343110478540840140496393128738996412270357183445395020573388989278555\
813556144036563102455032850874625946333692746942852322508699926140544281120662\
532525325402041764035161169228790566576239939498698256797892476673581793593341\
565591033055741801465791277510653241163913554713152060127185501516632153831744\
084485276203982058024033779697637618294711132918471464969716668181939944429129\
590221571580066039538720668949128150132064065578982106579514989803518960288881\
581003421562857862205496312407174042287815539537348909650284394685560662915042\
685107335399595953096758002153118745564316408031460221673554730734019949375781\
303013799317984319741189620338837708872894884720821818310366605811188532339923\
2084126619627384173536035029500741646467671020765461

:tongue: :tongue: :tongue:

7. Apr 11, 2005

### dextercioby

Heh,i use a 10 y.o.Maple.It can't more than 1000 decimals.

Daniel.

8. Apr 11, 2005

### marlon

It always breaks my heart if an integral has no analytical solution. I hate it when those damned Maple or Mathematica-programs are needed in order to crack the problem. Doesn't it disturbe you that we don't have a TOE when it come to integrals. You know, how about a String Theory-variant for integral-solving...i dunno, just wondering...

regards
marlon

9. Apr 11, 2005

### dextercioby

I'm not pissed off.I like when the results come out in terms of "fancy functions" (i call them:"common special functions").

This is not really elementary mathematics...

Daniel.


text:
# The A-B chocolate puzzle!

Imagine you have 2 types of chocolates (A and B). You randomly pick up two chocolates at once from your bag in a specific pattern. If the same type of chocolates come out, you give them both to your sister. If you pick up different type of chocolates, you put 'B' type back in your bag and give the 'A' type to your sister. Initially, you have 101 chocolates of 'A' type in the bag. At the end of this chocolate distribution task, you want one chocolate of 'B' for yourself and also one in the bag. Number of 'B' chocolates can be of the form 101*n (where n>100). What is the smallest number 'n' to fulfil this condition?

I have no clue, so any hint will be appreciated

• Hi, and welcome to Puzzling SE! On this site we have an absolute requirement, that if you post a puzzle you didn’t design yourself, you must tell where it’s from and/or who designed it. Thanks, and have fun! – Bass Mar 14 '18 at 6:35
• This is a Question from Technothlon mock – ami_ba Mar 14 '18 at 6:45

## 1 Answer

So it sounds like you want there to be one left in the bag, which is a B chocolate?

In that case,

N=101
You just need to have an odd number of B chocolates, and the smallest N that fulfills that is 101.

Reasoning:

Let's see how this works again
pick AA = lose 2 A from your bag
pick AB/BA = lose 1 A from your bag
pick BB = lose 2 B from your bag

As you can see, you can only lose 2 B chocolates at a time.
Meanwhile, every time you pick an A chocolate, no matter what happens, you'll end up giving that A chocolate to your sister.

Therefore, as long as you have an odd number of B, you'll eventually be left with one B in the bag. A will run out no matter what, and this last B will never be given to your sister because you can only give B chocolates in pairs.


text:
# Problems with Capitalism?

1. Apr 10, 2015

### lavinia

More Documentation

Here are two articles - one a detailed study of the S&L crisis - which document the effects of deregulation of that industry during the 1980s and sent the country into recession by the early 1990's and cost the US taxpayer large sums to resolve the ruined S&L industry. The FED, in order to stem the recession and finance the resolution of failed banks brought the overnight rate from over 8% to 3% by 1993.

https://www.fdic.gov/bank/historical/history/167_188.pdf

http://www.dailyfinance.com/2010/07/03/financial-meltdown-vs-savings-loan-crisis-recession.

The S&L crisis is parallel in many ways to the recent real estate crisis, particularly in the wide spread of speculation and fraudulent practices in the quest for high returns and in its effect upon the economies of several states. The crisis is commonly called the "S&L Debacle" .

Last edited: Apr 10, 2015
2. Apr 10, 2015

### lavinia

3. Apr 10, 2015

### lavinia

More Documentation

Here are references on the speculative bubble in Japan during the 1980's that triggered an at least one decade recession (The Lost Decade) and some say two decades.

http://www.thebubblebubble.com/japan-bubble/

After World War 2 Japan instituted banking regulations that were stricter even than those in the United States. These regulations were part of a program to nurture the Japanese economy. The policy was considered to be successful but in the mid-1980's significantly because of pressure from the United States and Europe, Japan deregulated its banking system After 1985 the bubble took off.(See the referenced paper below).

There is an interesting paper written in 1992 by Kenusuke Hotta, a Managing Director at Sumitomo Bank, who strongly favors the deregulation. I could not figure out how to copy the link but it is called "Deregulation of Japanese Financial Markets and the Role of Japanese Banks" Occasional Paper no. 7 ,Occasional Paper Series
February 1992

4. Apr 10, 2015

### lavinia

More Documentation

https://www0.gsb.columbia.edu/faculty/fedwards/papers/Hedge_Funds_&_the_Collapse_of_LTCM.pdf

This paper reviews the collapse of Long Term Capital Management
It also has a review of Hedge Funds and their trading strategies.

It also covers the 1998 Russia default on a large amount of its Treasury debt and its cancellation of derivative contracts. This created a cascade of margin calls and asset devaluations. Basically, firms were forced to sell assets in order to meet margin calls thus causing a drop in prices which then triggered more margin calls and so on. Panic led to a flight to quality which widened yield spreads on risky assets generally for instance, B rated corporate bonds. Long Term Capital Management had huge leveraged bets on yields spreads. The spread widening destroyed it and I have heard nearly wiped others hedge funds like D.E. Shaw & Co. Here is a section of the article that discusses the "systemic risk" of its impending bankruptcy and the resulting bailout organized by the Federal Reserve Bank of New York.

" In early September 1998, LTCM advised the Federal Reserve Bank of New York of its impending difficulties. During the next few weeks, Federal Reserve representatives organized meetings of LTCM's creditors and met with LTCM partners to discuss the situation. On the evening of September 22, the president of the Federal Reserve Bank summoned more than a dozen top executives of the firms that had loaned money to LTCM to an 8 p.m. meeting and warned them that "systemic risk posed by LTCM going into default was 'very real'" (Siconolfi,1998). The next day a 16-member creditor consortium agreed to put in additional capital of $3.625 billion in exchange for 90 percent of the remaining equity in LTCM; ...." Last edited: Apr 16, 2015 5. Apr 10, 2015 ### WWGD Correlation, Causation? Didn't socialist countries achieve a similar gain in life expectancy (during the years of communism)? I am not arguing in favor of communism, but, how do you explain, e.g. that Cuba's life expectancy is 80.7 years for female, 75.9 for males (World Almanac and Book of Facts, 2015, p.768), similar for former communist block countries (I assume life expectancy does not turn on a dime, that , if the economic system is changed, it may take a long time for the benefits/consequences of this change to take effect.) ? Besides, capitalism as we know it today has not been in effect for 150 years. Weren't many of the main discoveries made before today's capitalism was in effect? Vaccination, much of the medicine used to prevent childhood death, etc.? How about the scientific discoveries. And there is the additional fact that most of the people who have ever lived, have lived in the last 150 years, so it makes sense that the more people that are alive (living longer), the more discoveries you have. And there were 100's of important scientific discoveries before the 1900's. And the Soviet Union had some of the best scientists in the world. You may even reasonably argue that a critical mass of discoveries had been reached by the time today's capitalism went into effect. Last edited: Apr 10, 2015 6. Apr 10, 2015 ### WWGD How is the oppression of descent the result of abandoning capitalism? Despite the repeated mantra by the far right that freedom, capitalism and economic growth go together, the examples of China and India belie that very clearly. Same with South Korea, which was under repressive regimes for a long time. I think part of the problem with communist countries is that they educated many of their people too well, and the educated tend to ask the difficult questions. Of course, it is also true that there were no jobs for them. Still, capitalism as we know it today has less than 100 years in effect; too soon to decide whether it works in a sustainable way. 7. Apr 10, 2015 ### lavinia More Documentation This 2012 Business Insider piece quotes Bill Gross, CEO of PIMCO which is one of the largest bond funds in America, who says that the economy is at risk of a Japan style "Lost Decade" by which he seems to mean an extended period of "slow growth" despite signs finally of some recovery in the housing market and resumption of bank lending. His arguments focus on the shape of the yield curve and the record low interest rate policy of the Fed which he says has been in place in order to bring down long term yields to stimulate the housing market. His arguments underscore the severity of the current slowdown and also the risk that ensues when possibly unsound attempts at stimulating recovery are implemented. I found his arguments interesting. http://www.businessinsider.com/bill...ed-is-taking-us-on-the-course-of-japan-2012-2 Last edited: Apr 16, 2015 8. Apr 10, 2015 ### lavinia More Documentation The 1987 Stock Market crash While the 1987 stock market crash had a relatively mild after affect, it is one of the early examples of the risk of the new unregulated leveraged markets that took off in the 1980's Here are two articles that discuss it, one by a contributor to Forbes Magazine, the other a Federal Reserve Discussion Paper. http://www.thebubblebubble.com/1987-crash/ http://www.federalreserve.gov/pubs/feds/2007/200713/200713pap.pdf Both of these papers discuss how leveraged trading strategies - in particular Portfolio Insurance and Index Arbitrage - catalyzed the largest one day drop ever in the market. They also indicate how the stock market - during the vaunted Reagan recovery - was bolstered by low interest rates, leveraged buyouts, IPO's, mergers, hostile takeovers, and a general investment euphoria. These were the days of junk bonds, Michael Millikan's$500 million dollar a year bonus, soaring P/E ratios and the advent of computerized trading of equities.

I have included a page of financial data history such as market prices, bond yields. P/E ratios etc for reference.

http://www.sniper.at/stock-market-crash-of-1987.htm

The crash seems to be attributed to panic that ensued after several things happened. Here is the description in the Fedeal Reserve Paper.

First, news organizations reported that the Ways and Means Committee of the U.S. House of Representatives had filed legislation to eliminate tax benefits associated with financing mergers (Securities and Exchange Commission (SEC) Report 1988,p.3-10). Stocks’values were reassessed as investors reduced the odds that certain companies would be take-over targets. Second, the Commerce Department’s announcement of the trade deficit for August was notably above expectations. On this news, the dollar declined and expectations that the Federal Reserve would tighten policy increased (Wall Street Journal 1987b). Interest rates rose, putting further downward pressure on equity prices (see Figure 2).

On Thursday, equity markets continued to decline. Some of this decrease was attributed to anxiety among institutions, especially pension funds, and among individual investors, which led to a movement of funds from stocks into the relative safety of bonds (Wall Street Journal 1987c). There was also heavy selling during the last half hour of the day amid heavier-than-usual activity by portfolio insurers (Brady Report 1988, p. 21).

Markets continued to decline on Friday, as ongoing anxiety was augmented by some technical factors. A variety of stock index options expired on Friday; price movements during the previous two days had eliminated many at-the-money options so that investors could not easily roll their positions into new contracts for hedging purposes. These developments pushed more investors into the futures markets, where they sold futures contracts as a hedge against falling stocks. Increased sales of futures contracts created a price discrepancy between the value of the stock index in the futures market and the value of the stocks on the NYSE. Index arbitrage traders reportedly took advantage of this price discrepancy to buy futures and sell stocks, which transmitted the downward pressures to the NYSE (Brady Report 1988, Study III, p. 12).

By the end of the day on Friday, markets had fallen considerably, with the S&P 500 down over nine percent for the week. This decrease was one of the largest one-week declines of the preceeding couple of decades, and it helped set the stage for the turmoil the following week (Wall Street Journal 1987d). Portfolio insurers were left with an “overhang” as their models suggested that they should sell more stocks or futures contracts (SEC Report 1988, p. 2-10). Mutual funds experienced redemptions and needed to sell shares (Brady Report 1988, p. 29). Further, some aggressive institutions anticipated the portfolio insurance sales and mutual fund redemptions and wanted to pre-empt the sales by selling first (Brady Report 1988, p. 29; SEC Report 1988, p. 3- 12). ...."

This is an interesting account and the entire event deserves careful study because it exhibits the dangers of an unregulated leveraged market. It shows how a price bubble driven by euphoria and leveraged speculation can be rapidly burst when the factors driving the bubble go away. This can be a change in tax law, a rise in interest rates, a bad economic report, and so on. It also shows how the use of financial derivatives - in this case futures contracts and portfolio insurance (which is a "replicated" put option on the stock market) - can magnify a decline into a crash.

An important thing to note is that with such a huge decline in stock prices, the stock exchange itself was at risk of breaking down . This is because stock specialists financed their positions in stock on bank borrowing using the stock as collateral The banks became reluctant to finance these positions because the stocks had lost significant value and there was huge risk still remaining. BTW: This sort of financing is common in securities markets. In Government bonds, the market for this type of short term - usually overnight - borrowing is called the Repo market.

http://en.wikipedia.org/wiki/Portfolio_insurance

Last edited: Apr 16, 2015
9. Apr 18, 2015

### Siv

I agree.
But I think there are plenty of examples of its imperfections.

The bad aspects of human nature need to be suppressed and discouraged. Thats the whole definition of "civilization".
Capitalism does just the opposite.

Going one step further ... and I know there will be cries of "Nazism" in response to this ... but I don't know why we should continue to live with this problematic human nature and struggle with its consequences. Its a taboo subject to even discuss attempting to improve/modify it. Just because it resulted in atrocities once, does not mean it should be a banned subject forever

10. Apr 19, 2015

### lavinia

In my opinion, one problem with capitalism as it is practiced today (and historically), is the way that central banks handle problems of economically weaker countries to pay debt. The solution is to roll debt over in return for "austerity measures" which force these countries into economic hardship. Paul Krugman has this article where he says that austerity measures not only don't work i.e. they prevent the debtor country from paying by slowing its economy but their harshness leads to political radicalism.

http://www.nytimes.com/2013/02/25/opinion/krugman-austerity-italian-style.html?_r=1

To me, the imposition of austerity is not intrinsic to capitalism so I consider Krugman's observations to be criticism of practice rather than theory.

Last edited: Apr 20, 2015
11. Apr 21, 2015

### lavinia

SIV a couple of questions.

What do you mean by "human nature"?
What are the bad aspects of it?

12. Apr 21, 2015

### Siv

Human nature is essentially the behavioural and psychological aspects of human beings, a lot of which helped our evolutionary journey.
There are common aspects to all human behaviour.
This is a very brief description to something that can cover libraries ....

Bad aspects - oh there are so many ... ethnocentrism, xenophobia, short term benefits at the cost of long term benefits, violence, rape, deception .... all these are innate traits.

13. Apr 22, 2015

### Czcibor

May I, by rules of the forum ask for some data to back up your claim? Because it starts sounding quite interesting...

Because I have limited luck being born in a communist country, and consider such claims in the same as a black person consider claims that apartheid wasn't so bad...

14. Apr 22, 2015

### Czcibor

One question concerning intellectual honesty of this topic. (I mean especially Lavinia)

Are we comparing RL capitalism with some idealised, hypothetical system? Because I may already solve your dillema. Any RL system would be worse than any hypothetical idealised system.

No I mean as fair idea - idealised capitalism vs. idealised alternative. But under such capitalism there are no market failures or market crash :D

Or we should talk about RL capitalism and compare its successes with RL "successes" of its contenders. Like the West vs. Communist countries. Or compare India vs. PRC from times of Deng Xiaoping, when communism was effectively started being jettisoned. Or compare India to Asian Tigers.

15. Apr 22, 2015

### Siv

Sure, Czibor. But which claim, precisely?
I had quoted E.O Wilson's famous "Great idea, wrong species."

16. Apr 22, 2015

### Czcibor

"capitalism is really not much better"

And we may try to compare let's say North Korea with South Korea.

17. Apr 22, 2015

### lavinia

I am not sure why I am suspected of intellectual dishonesty. Let's dismiss it as an unfortunate choice of words.

Many of my post were originally in response to a challenge to document the problems with unregulated markets and to show their long term effects. This is not a criticism of capitalism in all of its forms but of completely free unregulated markets. I argued that deregulation of financial markets was a key component of the speculative difficulties of recent decades. The effect of these difficulties is not trivial and in Japan it led to the "lost decade" and the CEO of PIMCO in another article I cited, warned of the risk of a lost decade in the United States. That is my point and I look forward to discussing it with anyone who wishes to take a rigorous view.

Of course, speculative markets are only one problem. I intend to discuss others, and have started to talk about "austerity measures". Paul Krugman's article is worth reading. He is a Nobel Laureate in Economics and a regular contributor to the New York Times.

As far as what systems work or don't work and what systems are better than others, I have no rigorous opinion on that. Nor have I seen a rigorous opinion in this thread. For instance, to say that the Soviet System collapsed ergo it was an inferior system is at worst ideology and at best a hypothesis to be examined.

Last edited: Apr 22, 2015
18. Apr 22, 2015

### Siv

When comparing with communism, it may be better, I agree.
But where is this (unwritten) rule that there should be no other options?

19. Apr 23, 2015

### Czcibor

I'd say that claim that claim "unregulated market" is bit risky. You pointed for example housing bubbles. Anyway why housing happens to be so expensive on those free market? Shouldn't those greedy speculators build something out of bricks and mortar instead of just building up bubbles?

Let me quote Krugman, who grudgingly admitted one interesting thing:

http://www.nytimes.com/2014/08/25/opinion/paul-krugman-wrong-way-nation.html

I fully understand idea - risky behaving financial companies and gov that puts supervision on them to limit risk. However to such picture don't fit well ex. US gov that by allowing mortgage deduction is actually encouraging people to take excessive debt.

The problem is that markets are quite often already strongly regulated, just from time to time in the wrong direction.

Who said that I use such claim?
Concerning SU collapse - the problem that's not one observation (that could have been caused by some random, unobserved factor.
So have you tried to compare GDP of countries on similar development level before and after testing this system you have no rigorous opinion? I mean North vs. South Korea (under Japanese rule north was actually more industrialized); China vs. Taiwan; West vs. East Germany; Spain vs. Poland (Roman Catholic countries, similar size: in 1950 Polish GDP per capita was a bit higher, in 1990 it had 30% of Spanish GDP)
How do you think, maybe a pattern emerge?

20. Apr 26, 2015

I am not defending communism, Czcibor, but didn't people in the Eastern block have most daily needs paid for by the state? Maybe doing a PPP (Purchase Power Parity) do $5,000 buy you the same in Madrid than in Warsaw (back then and now )? So, comparing maybe a$3,000 GDP per capita in the East vs more than $10,000 in the west, but in the East you are provided with housing, stable job, medical care? 21. Apr 26, 2015 ### Czcibor A bit tricky comparison: -$5000 for sure bought a lot in last days of Polish People Republic... according to black market exchange rate an average salary was about \$15
-how do you want to calculate a basket with adjustment for ration stamps? (especially in cases when you had ration stamps for stuff that was NOT accessible, even with ration stamps)
-how do you adjust that in the last decade shops were mostly empty and one was standing in a queue to buy anything? (sometimes it worked, sometimes not) (A realistic approach would involve calculate value of wasting a few hours per day on such staying in queue/shopping)
-how do you adjust for transactions done on black market or through a net of friends?
-how do you adjust for stealing in your place of work? (awful pathology at that time)

I may try to find you some data if you want.

22. Apr 26, 2015

### Czcibor

If data for 1990 (semi-free election was in late 1989), in PPP and adjusted 2005 dollars, are OK, then according to World Bank:

The data are a bit too optimistic for Polish consumer, as GDP was somewhat inflated by ineffective heavy and chemical industry, which was producing equipment for WW3.

But according to them we had 42% of Spanish GDP. (yes, there is a big dispersion in the data, pending on source)

23. Apr 26, 2015

### WWGD

My point is that while capitalism may be overall preferable to communism (at least as commonly-practiced) , it may be a good idea
to try to incorporate o capitalism some positive aspects of communism.

24. Apr 27, 2015

### Czcibor

Which one do you mean?

25. Jun 6, 2015

### lavinia

Some economists point to FDIC instance as a regulatory measure that led to the fall of the S&L's. This is a case of what they is call a Moral Hazard.Their point is that deregulation is not the problem per se.

It is true that FSLIC insurance remained after the S&L's were otherwise deregulated. But in this situation, FSLIC insurance became a free put option written by the US Government. So instead of functioning as a regulatory measure, it transmuted into a derivatives contract.

Before the deregulation of the S&L's FSLIC insurance worked fine in the context of a complete regulatory framework. From the 1930's until deregulation in the 1980's it worked fine. The same holds for Japan as I have described above.

On the other side of the coin, without a safety net such as FSLIC insurance there is risk of runs on financial institutions that lose investor confidence.

Last edited: Jun 6, 2015


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# LBIT Proofs 7: Propositions 48-52

19 min read16th Dec 2020No comments

# 4

AI
Frontpage

Proposition 48: A function  is an infrakernel and fulfills the property:

iff the function  is bounded and continuous when  is equipped with the IKR metric.

So, our first direction is to assume that the relevant function is bounded and continuous, and show that it fulfills the infrakernel properties and the one extra compact b-uniformity property. The second direction is to show that, assuming the b-uniformity property and being an infrakernel, the function is bounded and continuous.

Anyways, so let's assume the function  is bounded and continuous. Ie,

And, if  limits to , then  limits to  in Hausdorff-distance/the infra-KR metric.

First, for purposes of disproof, let's assume  doesn't have bounded Lipschitz constant/amount of measure present. That is,

Then, you can let  be any number you want, no matter how large, and find some  where  and  and . Now, let's look at the norm of the infradistribution . Plug in the constant function  and we have:

This is because  had  or more measure present. Then, making that substitution, we have:

And at this point we observe that the Lipschitz constant of a constant is 0, and the norm of a constant is just its absolute value, and  by assumption, to get:

Now we use our earlier inequalities and the absolute value (everything in our earlier inequalities is a negative quantity) to get:

But  can be unboundedly high, so our maximum norm is infinity, which we stipulated isn't the case. Thus,  does have bounded Lipschitz constant (this was derived from 's bounded IKR-norm and continuity).

Now, we just need to derive the other two conditions on an infrakernel from boundedness and continuity of .

If  is a continuous function , and  is an arbitrary compact subset of , then  must be a compact subset of , which, from our characterization of the iff conditions of compactness in Proposition 46, means that it fulfills the shared compact-almost-support property and the b-uniformity condition. Thus the function  fulfills the compact-shared compact-almost-support property, and the compact b-uniformity condition. Finally, for the pointwise-convergence condition of an infrakernel, continuity of the function  means that if  limits to , then  limits to  in Hausdorff-distance, so by Proposition 45, for all continuous bounded functions f,  limits to , and this is the last condition we need to call  an infrakernel.

So, bounded and continuous functions  imply that the function is an infrakernel and fulfills the compact b-uniformity condition. Time for the reverse direction. Let's assume that  is an infrakernel  and fufills the compact b-uniformity condition. For showing boundedness, there's some Lipschitz constant upper bound , and we can go:

And we have established boundedness. For continuity, it will take a more detailed argument. Let  limit to . Our task is to show that  limits to  in Hausdorff-distance/infra-KR distance.

Now, the set  is compact, so by the conditions for an infrakernel, there's a bound on the  value of the minimal points of the  infradistributions, and the  sets have shared compact-almost-supports. We're also assuming the compact b-uniformity condition, so the  sets have the b-uniformity condition on them. By our necessary and sufficient conditions to be precompact, we have that the set of points  is precompact in , so we can isolate a convergent subsequence of the sequence of infradistributions K.

What we'll do is show that all convergent subsequences of  have the same limit point, namely . Here's the argument for it. Index the convergent subsequence with  and have  be the limiting infradistribution of it. From the pointwise convergence condition on an infrakernel,

Thus, our arbitrarily generated limit point  must perfectly match up with  and be identical to it, so since there's only one limit point, the sequence  actually converges to . So, because convergent sequences in the input induce convergent sequences in the output, the function  must be continuous, and we've shown the other direction, that infrakernels fulfilling the compact b-uniformity condition must be bounded and continuous.

Proposition 49: The space  equipped with the IKR-topology is a Polish space if  is too.

So, we've got a metric, the infra-KR distance ie the Hausdorff-distance. We know that the notion of convergence remains independent of which metric on  these metrics are being induced by, so the topology is unchanged, and they're all metrizations. We also know from Proposition 43 that  is complete under any infra-KR distance ie Hausdorff-distance. So,  is completely metrizable by any infra-KR distance induced by a complete metric on , all that remains is to show that it's a separable space, we need to find a countable dense subset.

We know that a single point is compact, and we also know that compactness implies that for all , there is a  value where  by Proposition 46, b-uniformity.

So, the space of infradistributions with the  values of their minimal points having an upper bound is dense in the space of infradistributions. All we need to do now is to have a dense countable subset of the space of infradistributions with upper bounds on the  values of their minimal points.

The way we'll be doing this is taking a dense countable subset of the space of a-measures. There are only countably many ways to choose finitely many points from this subset, and a choice of finitely many a-measures can be taken to define an infradistribution by convex hull, closure and upper-completion. Assuming the fiddly details work out, we'll have countably many infradistributions which are (hopefully) dense in the space of infradistributions with bounded  values, which are dense in the space of infradistributions, so we've got our dense countable subset.

There's still much to do. We need to show that there's a dense countable subset of the space of a-measures at all, we need to deal with technicalities involving normalization to show that we always get an infradistribution by this process, and we need to show that given any infradistribution with bounded  values, we can make an infradistribution of this form arbitrarily close to it in Hausdorff-distance.

For our first part, the cone of a-measures can be viewed as , and you can also view the space of measures over  as  (a pair of a distribution and a scaling term). This isn't quite right, because there's an identification of everything when the amount of measure is 0, but this doesn't really matter. So, we're trying to find a dense countable subset of

Now, when  is a Polish space,  equipped with the KR-metric is also a Polish space, so there's a dense countable subset. You can take the product of this with the dense countable subset of  consisting of the rational numbers 0 and above, to get a dense countable subset of the cone of a-measures.

For the normalization technicalities, we need to make sure that whenever we pick finitely many points from our dense countable subset, there's a point with , a point with , and all points have , in order to have normalization work and your batch of points define  an actual infradistribution. This can always be done, throwing out the "bad" choices of finitely many points still leaves you with a countable set.

Finally, given an infradistribution  with its minimal points all having , can we make infradistributions arbitrarily close to it in Hausdorff-distance of this form where we just specify finitely many points?

Well, yes. Because of the compact-projection property of infradistribution sets, and truncating at , we can see that all the minimal points of  lie in a compact set. So, given any , we can cover  (well, the part with  values below ) with -sized balls centered at its points, get a finite subcover from compactness, and then use denseness of our countable subset to take our finitely many balls with center points and pick something from the dense countable subset in each ball in order to get: A collection of finitely many points from the dense countable subset where each point in  with  value below  is  distance away from one of these finitely many points, so the closed convex hull of these finitely many points is only  distance away from  (but truncated at ), and then when upper completion is done, they stay only  distance away from each other.

So, given any infradistribution with bounded  values, and any , we can find finitely many points from a dense countable subset of the space of a-measures, which induces an infradistribution that's only  distance away. Because there are only countably many ways to pick finitely many points from a countable set, we have a countable set of infradistributions where any infradistribution has infradistributions of this restricted form arbitrarily close to them, and so we have separability, the last condition we need for Polishness.

Proposition 50: The infra-Vietoris topology and the topology induced by the IKR metrics are the same.

This proof will proceed in three ways. The first is showing that this collection of sets declared-to-be-open in the infra-Vietoris topology is closed under finite intersections, and thus forms a basis (instead of a sub-basis). The second way is taking any basis open set from the basis induced by the IKR metric (we'll be using the strongly equivalent Hausdorff-metric for this) and any point/infradistribution in it, and finding a Infra-Vietoris open set from the basis which is a subset of the open ball in the IKR metric. Finally, we taken any basis open set from the infra-Vietoris topology, and any point/infradistribution in such a set, and find a tiny open ball w.r.t. the IKR metric that fits within the infra-Vietoris open set.

So, here's what we're going to do: First, we're going to show that the intersection of two open sets from the basis is also in the basis. Obviously, we only need to show this for pairwise intersection, as that shows closure under finite intersection.

Consider the open set  of infradistributions to be generated by the finite family of bounded-b open sets , and the open set  of infradistributions to be generated by the finite family of bounded-b open sets .

Claim: The finite family of bounded-b open sets  and , for all the  and , is a suitable batch of open sets in the a-measures to perfectly replicate the  open set of infradistributions.

The properties for  are:

And, the properties for  lying in the set induced by the  and  families of open sets (upper completion of open set is open) is:

So let's get started on showing equivalence of these things. We'll use  for an a-measure, and use a subscript of  or  to denote which sort of set it's in.

First, let's take the conjunction of these two things.

This is equivalent to "for any , it is above some  and also above some ". Since the only notion of upper completion is adding the  term, this is equivalent to "for any , we are in one of two cases, either there is a  such that , or that "

Similarly,

Is saying "for any , we are one of two cases, either there is a  such that , or vice-versa". And so these two conditions are equivalent.

Also, if you have both of:

Then you trivially have

And in the reverse direction, if you have:

then you'll get both of

That just leaves the projection property.

If you have

then, let's say  is your arbitrary point in the closure of the projection of . This is saying  and . Again, by how upper completion is just adding , this is equivalent to "given an arbitary point  in the closure of the projection of , either  or "

And the condition

is saying "given an arbitary point  in the closure of the projection of , either  or ", which is clearly equivalent. So, we've got our bidirectional equivalence, and any binary (and thus finite) intersection of infra-Vietoris basis opens is an infra-Vietoris basis open, so it is indeed a basis.

Now we must show that this infra-Vietoris topology is identical to the topology induced by the infra-KR metric/Hausdorff distance metric. This can be done by taking any ball in any Hausdorff distance metric centered at a point/infradistribution, and finding an infra-Vietoris open set that contains the same point and is a subset of the open ball. Conversely, we should be able to take any infra-Vietoris open set, and any infradistribution/point within it, and find an IKR open ball centered on that point that lies entirely within the infra-Vietoris open set, showing that the two topologies are equal.

So, first, assuming we've got a Hausdorff-distance of  centered at some infradistribution, can we make an infra-Vietoris set that lies entirely within the ball and engulfs the center point?

Well, there's some  cutoff where the projection of the infradistribution  (cut off at that  value) is within Hausdorff-distance  of the projection of the entire infradistribution  to the measure component. Accordingly, we have one of our open sets be a -thickening of the chopped-off infradistribution, as this upper completion engulfs the entire infradistribution  within it with  room to spare. The rest is done by covering the compact chopped-off portion with all possible open balls of size , and using compactness of the chopped-off infradistribution to pick a finite subcover/finitely many open balls.  has the projection of its closure lie entirely within the -thickening of the lower part, has nonempty intersection with each open ball, and is a subset of the upper completion of the union of everything.

Now, our task is to show that any infradistribution that lies in the open set induced by that finite collection of open sets lies in the Hausdorff-distance-of- ball centered at . So, let our arbitrary new infradistribution  be a subset of the union of the upper completion of all these open sets, and have nonempty intersection with each of them, and have its closure lie in the projection of the union of them. Note that the initial open set which engulfs the entire bottom portion has all the other open sets as subsets of it, so the union and upper closure of the opens is just the upper completion of the "big" open set. Any point in  must lie in this "big" open set, and it's made by thickening up  (but chopped off) and taking the upper completion, witnessing that the distance from  to  is  or less.

For the other direction, pick an arbitrary point in . It's within  of the upper completion of one of the finitely many balls used for the open cover. And because those center points are center points of -open balls which  must have nonempty intersection with, and  is upper-complete, it has a point of distance  or less from your starting point in . Thus, anything in this basis open set has Hausdorff-distance of  or less from the center point , so all open balls in the infra-KR metric topology can have their central point engulfed by an infra-Vietoris open set, which is a subset of the initial open ball. So we're half-done, we know that all open sets of the infra-KR metric are open in the infra-Vietoris topology.

In the other direction, we need to take an infra-Vietoris open set, and an infradistribution in it, and find some tiny hausdorff-distance from that arbitrary infradistribution that remains in the infra-Vietoris open set. Let  be your infradistribution, which is a subset of the union of the upper completion of the , and has nonempty intersection with each , and the closure of its projection is a subset of the projections of the .

Now, there's one important fact we have to establish first. It is impossible to have a sequence  that gets arbitrarily close to the complement of . Fix some sequence  and assume that this sequence gets arbitrarily close to the complement of

One of two things can happen. The first thing that can happen is that  has a subsequence with bounded  value. Then, we can pick out a convergent subsequence of that subsequence because infradistributions, when cut off at any finite  value, are compact, and take the limit point , which lies in the set  due to closure, and yet has distance 0 from the exterior of , which, being the complement of an open set, is closed. So, we've got a point in  and also not in , which is impossible.

Therefore, the  sequence must have unboundedly large  value as  increases. This projects down to make an  sequence of measures, in the compact set , and we can isolate a convergent subsequence of those, with limit point . Now, something interesting happens.

For the  sequence of a-measures, eventually the  value of them goes well past the largest  value of one of the finitely many bounded- open sets. And when we go a tiny distance away to a point  which lies outside of , then... well, it's got nothing from any of the  below it. And nothing from any of the  above it either, because the  value of  is too high since it's near . So,  projects down to some measure  which lies outside of . And since  is really close to  (in the closure of the projection of ) is really close to the exterior of the projection of , as witnessed by . And this is a closed set, because the complement of an open set is closed, and the projection of an open set is open.

This lets us show that our  limit point lies in the complement of , which is impossible, because  lies in  which is a subset of .

So our original assumption was wrong, and there is some  where any point in  is always  or more distance away from the exterior of . Further, for the , since , you should be able to pick said point, and find some -sized open ball you can stick around it that lies entirely within the open set .

Now, a Hausdorff-distance of  around , all infradistributions that close must lie in the same Vietoris-open set. Here's why. Since all points in your new infradistribution  are at most  distance away from a point in , and given any point in  you need to travel  distance to get to the exterior of  is a subset of that open set as well.

Also, , because given the central points in the  sets, you can travel  distance to get to a point in , which lies in the -sized open ball around said point which still lies entirely within the . So any  that close must intersect the same collection of open sets.

Finally, the projection thing is taken care of by the same sort of Hausdorff-distance argument we used to argue that said  must lie in the union of upper-completions of the open sets. So, given any Vietoris-open, and any point in it, we can find an IKR-open which engulfs said point and lies entirely in the Vietoris open.

Therefore, the two topologies are the same.

Theorem 3: Given a crisp infrakernel   then if there's a  where, regardless of  and , there's a unique crisp infradistribution  where , and can be found by starting with any crisp infradistribution on , and repeatedly applying .

This theorem is going to require showing a lot of fiddly measure-theoretic stuff, and then once everything's all set up, the actual theorem itself falls pretty fast. So, we're going to start off by noting or proving three things needed to make progress.

The first is: "given any measurable function , and any probability distribution , and any , we have that there is a continuous function  s.t. ".

This is an analogue of the celebrated result that continuous functions are dense in the measurable functions. I'm gonna skip small bits of it, to be reconstructed if you ask. Pretty much, we use the fact that  is separable in the KR-metric, and the results from this pdf to get a few results.  is separable according to the KR-metric, and  has all its output values in that separable space, so it's separably valued. Also s is measurable, so by proposition 1.8 in the paper,  is strongly measurable, so it's the pointwise limit of a sequence of simple functions. The actual proof of this fact goes through Theorem 1.5 in that paper, and looking carefully at Theorem 1.5, the simple functions constructed have their value in , because the proof is like "fix a dense countable subset of your space of interest, then modify  to be a simple function by rounding off the outputs to the nearest of a finite collection of points from that subset". Proposition 1.10 from that paper says that since  is strongly measurable, then given any  is strongly -measurable, and then proposition 1.16 from that paper says that  is Bochner-integrable because , because all the  are probability distributions. Further, part of the proof of proposition 1.16 from that paper shows that the -valued simple functions, due to their bounded norm, converge to  in the  norm by the dominated convergence theorem.

Anyways, at this point, we've shown that any measurable function  has simple -valued functions that limit to it, so the simple functions are dense in the measurable functions . Thus, at this point, we just need to show that we can make continuous functions  that are arbitrarily close to some simple function .

Let's begin. Our simple functions , from looking carefully at the proof of Theorem 1.5 from the linked paper, take the form of partitioning  into finitely many disjoint and covering measurable sets , and mapping each of them to some probability distribution . Let  be how many disjoint sets there are.

Thus, our task is to take a simple function , and find a continuous function  where  If we can do this for any , then it will show that continuous functions are dense in the simple functions which are dense in the measurable functions in

So, here's what you do. Every probability distribution over a Polish space has the property that the measure of any measurable set  can be approximated from below by the measure of a compact subset . We have finitely many disjoint measurable sets , and we associate each of them with a compact subset  where .

Let's abuse notation a bit and have  be the minimum distance from  to a point in . Then, let  be defined as:

Admittedly, the distance may be 0, leading to , however, this will be considered to be 1. The reason this works and is continuous is because all the  are bounded away from each other. This is because they're all subsets of  which are disjoint, so all the  are disjoint, and any two compact sets which are disjoint must be bounded away from each other, because otherwise you could fix a sequence in one compact set getting closer and closer to the other compact set, isolate a convergent subsequence, and it'd limit to be distance 0 from the other compact set, and so in it, contradicting disjointness. Now, we can go:

And then, for the next step, we observe that when  is like "well it's in , I'll return  as my output", and  is like "well, it's distance 0 from , so I turn into  by convention". Further,  always returns probability distributions, and the maximum KR-distance two probability distributions can have (when the original space  has its metric bounded above by 1, which can always be imposed without altering the topology) is 1. So, we get

And we're done. We showed that any measurable function  could be arbitrarily closely approximated in  by a simple function , and that any simple function could be arbitrarily closely approximated in  by a continuous function .

The next fact we'll need is that, when  has its distance metric bounded above by 1, which can always be done without altering the topology, then the KR-distance on probability distributions is exactly equal to the Earthmover distance, the first Wasserstein metric.

The third and final fact we need is that, if you have an function  (function to the space of compact subsets of ), which is continuous w.r.t. the Hausdorff metric, and always produces compact convex sets, then the function  (defined relative to some continuous function  and some ), defined by:

Is lower-hemicontinuous.

The requirement for being lower-hemicontinuous is if you have some sequence  limiting to , and some , then you should be able to find some sequence of points  where  limits to .

First, we'll need that, if  limits to , then  is a compact set. We know that  is continuous and  is compact, so Lemma 3 steps in to show that said set is compact.

The relevant implication of this, when paired with Hausdorff-continuity of , is that you can take any sequence of distributions , and find a convergent subsequence that converges to something in .

The second piece for this argument we'll need is that:

So, first, we observe that since all the  are compact, we can find actual minimizers . We don't know that the limit actually exists, so pass to a subsequence indexed by  where we're taking the liminf. Pass to another subsequence where the  actually converge to a point in , which we can do by the immediately preceding point about compactness. Then,

This was done by  being continuous, so  converges to , and .

Second, fix some minimizer , and by Hausdorff-continuity of , we can find a sequence  which limits to it. Then,

And at this point, we can go:

So, we have

Alright, that's part two of the argument for part 3 of the preliminaries for this theorem down. Time for part 3 of the argument. We'll assume that  limits to  limits to , and pick an arbitrary  s.t. , and

Let  be some point in  that's as close as possible to . Obviously, these limit to , due to Hausdorff-continuity of . Non-obviously, a tail of them (all but finitely many), will lie in . Here's why.

Let  be:

There is some finite  by which you are guaranteed three things forever afterwards.

First, , because  is Hausdorff-continuous and  limits to .

Second, , because  is continuous and  limits to .

Third,

Because we know that this minimum-distance quantity limits to the minimum-distance quantity of the limit, we showed it above. Now, at this point, we can go:

This was splitting the distance up, observing that  was selected to be as close as possible to , so it'd be as close or closer than the Hausdorff distance between the sets they were selected from, using that two of the distances were small, using how  was defined, and using our bound on the minimum-distance quantity. Anyways, this shows that forever after some timestep,

And therefore, since

We have that

Time for our final part! Showing lower-hemicontinuity of  requires showing that, for all sequences  limiting to , and all , we can find a sequence  limiting to . We just showed that for  fulfilling the defining inequality. But we took the closure, didn't we? What if our  was added in the closure? Then we can take a sequence  (but without the closure part) that does fulfill the defining inequality, and do a thing where we make a sequence  which limits to , one sequence for each . Then, we basically proceed for a while in  until we have a guarantee that  will remain close to , and switch to picking from the sequence , and repeat, slowly ratcheting up the second index over time. Said sequence will get arbitrarily close to the  values, which get arbitrarily close to  itself, so this "step up the sequence index when you've waited long enough" sequence can limit to

And with this argument, we know that  is lower-hemicontinuous.

Alright, now we can start embarking on the true proof instead of just setting everything up. Remember, we assumed that there was a crisp infrakernel , all the  have compact sets of minimal points (by CAS), and that there was some fixed  where  regardless of  and , and our goal is to show that there's a unique stationary crisp infradistribution which can be found by selecting any crisp infradistribution and repeatedly iterating .

There's a bit of setup to do first, but thankfully it isn't as demanding as we've done before. First, we'll abuse notation a bit by using  and  and  and  to refer, not to the infradistribution as a whole, but to their compact convex sets of minimal points, ie, sets of probability distributions.

Second, we recall from the proof of proposition what the form for  (the pushforward of ) is. Well, actually, the set of minimal points. It is:

Ie, the closed convex hull of the union of mixtures of the sets of minimal points of . Mixing sets with regards to a probability distribution is done by fixing a measurable selection function  s.t. for all  is a probability distribution, and then you just union all those points together for all the possible selection functions to make your set .

Our final thing to recall (well, actually, look up from Wikipedia), is that since the Earthmover distance/1st Wasserstein metric/KR-distance is all the same, we'll be using an alternate formulation of distance. If  are probability distributions over , then

Ie, of all probability distributions over , restrict to those which make  and  when you project it down to the appropriate coordinates, and try to pick one that makes the integral of the distance function as low as possible. The lowest integral of distance you can get is the earthmover distance. You can check Wikipedia for this formulation. We'll be using this form once.

Let's begin. First, by the form of the set of minimal points of the pushforward,

And then, we observe that taking the closure doesn't affect the Hausdorff distance at all, so we can go:

The reason for that inequality is that convex hull doesn't increase the Hausdorff distance. If you've got a finite mix of points from  in the convex hull, then you could just take each of those, find the closest point from the analogue for , and mix those together to make a point close to your thing added in the convex hull, so distance doesn't go up when you do convex hull.

At this point, we have the inequality:

Our mission is to impose a suitable bound on the second term, and once we've done that, show that  has type  (maps compact sets of probability distributions to cmpact sets of probability distributions), and then we can apply the Banach fixpoint theorem on  to get a fix-set of probability distributions that doesn't alter when you apply  (pushforward), and can be found by iteration.

For our first part, imposing a bound on

We shall do the following. We will pick a point in the first set, and find a nearby point in the second set. A point from the first set must be associated with some , and some measurable selection function  where, for all . Said point is written as , or alternatively, as . Our task is to find a point in the second set.

Construct it as follows: Select  to be as close to  as you can get. Pick an . Then, pick some  s.t. the projection of  to the first coordinate is  and the projection to the second coordinate is , and:

Such a  can always be found. Pick a continuous function  such that

We can always pick such a function because we've already shown that continuous functions are dense in the measurable functions.

At this point, recall our function  defined as:

We know that said function will be lower-hemicontinuous. Further, it always produces closed sets (we explicitly took the closure), and they're nonempty (just look at it a bit), and are convex. Convexity is a bit more involved, and it occurs because a mix of points a certain distance or less from some set point ( in this case), will also be the same or less distance away from said point, and the  are all convex. At this point, we can invoke the Michael selection theorem to get a continuous function  where, for all .

At this point, remember our distribution  which projects down to ? Well, by the disintegration theorem (the usual probability theory version),  can be written as  where  gives the conditional probability distribution over  if you see some .

And now we can finally define our measurable selection function , it is defined as:

And, our final point that we will pick to be close to  will be: .

Uh... Is  even a legit selection function? What we need is that for all . Well, fear not.

And, because  is closed and we took the closure of a batch of points within it, we have:

And, because  is closed and compact, making  by integrating over a bunch of different choices of  to plug in just makes the average of a batch of points within , which lies in , so we're good,  regardless of  and it's a legit selection function. Now, we can get into inequality shuffling!

And then we can notice something really interesting. We can rewrite that final integral as

Look familiar? That's just the semidirect product!! It's the exact same as

And, by our construction, . So, writing this as an integral, we have:

Also, because  projects to  in the first coordinate, we can write our integral over  as an integral over  instead, it doesn't change the value.

And move the norm inside to get:


text:
### Home > CCG > Chapter 10 > Lesson 10.2.2 > Problem10-80

10-80.

A cement block is the shape of a prism with length $1.5$ ft, width $1$ ft and height $1$ ft. Centered on the top of the block and passing all the way through the block are two $0.25$ ft by $0.2$ ft rectangular holes. Homework Help ✎

1. Draw a diagram of the block.

2. What is the volume of the block?

• Subtract the volumes of the two inner prisms from the volume of the large outer prism.