text:
# Windows error 0x00000014, 20

## Detailed Error Information

Message The system cannot find the device specified. winerror.h

This appears to be a raw Win32 error. More information may be available in error 0x80070014.

### CREATE_DELETE_LOCK_NOT_LOCKED[2]

This is a Blue Screen of Death stop code. More information is available in the Knowledge Base article Bug Check 0x14: CREATE_DELETE_LOCK_NOT_LOCKED.

### HRESULT analysis[3]

This is probably not the correct interpretation of this error. The Win32 error above is more likely to indicate the actual problem.
Flags SuccessThis code indicates success, rather than an error. This may not be the correct interpretation of this code, or possibly the program is handling errors incorrectly. false Microsoft false false 0 (0x000) FACILITY_NULL[3][1] The default facility code.[3][1] 20 (0x0014)

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(plain text only)

## Sources

1. winerror.h from Windows SDK 10.0.14393.0
2. https://docs.microsoft.com/en-us/windows-hardware/drivers/debugger/bug-check-code-reference2
3. https://msdn.microsoft.com/en-us/library/cc231198.aspx

User contributions licensed under CC BY-SA 3.0


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• Level: GCSE
• Subject: Maths
• Word count: 1977

# Emma is playing with arrangements of the letters of her name to see how many combinations there are.

Extracts from this document...

Introduction

Year 11 Coursework:

## Emma’s Dilemma

Emma is playing with arrangements of the letters of her name to see how many combinations there are.

EMMA                 AEMM                 MMAE                MAEM

EMAM                 AMEM                 MMEA                MEMA

EAMM                 AMME                 MEAM                MAME

I found that there are 12 combinations of the letters in the name Emma. However, I think that the fact that there is a repeated letter in the name has affected the result. I will try another name with no repeated letter to see if it will affect the number of combinations. I will use the name Lucy.

## LUCY                  UCYL                   CYLU                  YLUC

LUYC                  UCLY                   CYUL                  YLCU

LCUY                  ULCY                   CUYL                  YCUL

LCYU                  ULYC                   CULY                  YCLU

LYUC                  UYLC                   CLYU                  YUCL

LYCU                  UYCL                   CLUY                  YULC

I found 24 different combinations, twice as many for Emma. I will try another couple of four-letter names to see if I get the same results. I will try another name with a repeated letter to see if I find 12 combinations.

## NINA                  NANI                     AINN                   IANN

NIAN                  NNAI                     ANIN                   INAN

NAIN                  NNIA                     ANNI                   INNA

Like Emma, there are 12 combinations for the name Nina.

To make sure I don’t miss out any combinations or repeat any I use a system.

I start with the name then swap around the last two letters.

Eg. NINA

## NIAN

I then change the second letter and repeat this.

NANI

NAIN

Middle

KRYER                EKYRR                 REKYR                 RKRYE

YKERR                YERRK                 YRREK                 YREKR

YKRER                YERKR                 YRRKE                 YRKRE

YKRRE                YEKRR                 YRKER                 YRERK

There are 60 combinations for a five-letter name with one repeated letter.

I will now try a name with two repeated letters I will use the name Hanna.

HANNA                              ANNAH                          NANAH

HANAN                              ANNHA                          NANHA

HAANN                              ANHNA                          NAHNA

HNNAA                              ANHAN                          NAHAN

HNANA                              ANANH                          NAAHN

HNAAN                              ANAHN                          NAANH

AAHNN                          NNAAH

AANHN                          NNAHA

AANNH                          NNHAA

AHNNA                          NHAAN

AHNAN                          NHANA

AHANN                          NHNAA

I have found 30 different combinations for a five-letter name with two repeats.

 No Repeats 1 Repeat 2 Repeats 5 Letters 120 60 30

I can see from my table that the pattern for the results for five-letter names is the same. The number of arrangements is halved each time a letter is repeated.

I am trying to find a formula to find the number of combinations for any given number of letters. I will find results for six-letter names to help me find the formula. I predict that there will be 720 combinations for a six-letter name with no repeats. I think this because there are 6 combinations beginning with NIC;

## NICOLA                          NICALO                           NICLOA

### Therefore there will be 6 combinations each for NIL, NIA and NIO.

Eg. NILCAO                      NILACO                        NILOCA

NILCOA                      NILAOC                         NILOAC etc.

There will therefore be 24 combinations for NI, 24 for NC and so on. Altogether there will be 120 combinations beginning with the letter N.

Conclusion

No. of  x’s

No. of combinations

1

1!   (= 1)

1

1

2

2!   (= 2)

2

1

3

3!   (= 3)

3

1

4

4!   (= 4)

4

1

I know that  n!  equals the number of combinations for any word with no

x!

repeated letters and  n!  is for one repeated letter etc. However, I want to

2!

find out the formula for working out the number of combinations for words with letters repeated more than once.

Eg.   XXXYYY

The formula for this could be written as:

6!           (no.of letters)

3! × 3!                 (no. of x’s > y’s)

Or

720  = 20

36

I will write out all thepossiblecombinations for XXXYYY to check my result.

## XXYYXY

XXYYYX

XYXXYY

XYXYXY

XYXYYX

XYYXXY

XYYXYX

XYYYXX

YXXXYY

YXXYXY

YXYXXY

YXYXYX

YXYYXX

YYXXXY

YYXXYX

YYXYXX

YYYXXX

YXXYYX

My formula worked correctly – I have found 20 possible combinations for XXXYYY.

I can now do a formula for any word with any number of letters and any number of repeats.

For example:

For XXYYZZ it would be

6!

2! 2! 2!

For XXYZABC it wouldbe

7!

2! 1! 1! 1! 1! 1!

My overall formula would be:

n!

a!b!c!d!e!f!g!h!i!j!k!l!m!n!o!p!q!r!s!t!u!v!w!x!y!z!

This can be used to find out the number of combinations for the letters of any word. I would just have to substitute the number of each letter for the letter. So, instead of a!, I would put in the number of times that a appears in the word I was trying to find out the number of combinations for.

For example:

‘Maths’ would be:

5!

1! 1! 1! 1! 1!

Which would equal 120 different combinations.

Whereas ‘classes’ would be:

7!

1! 1! 1! 3! 1!

Which would equal 840 different combinations.

This student written piece of work is one of many that can be found in our GCSE Emma's Dilemma section.

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# Related GCSE Emma's Dilemma essays

1. ## Arrangements for names.

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2. ## Emma's Dilemma

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4

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Pairwise MRS The Marginal Rate of Substitution is usually defined as a pairwise thing. E.g., in case of three goods $x,y$ and $z$, you would have $\text{MRS}_{xy}, \text{MRS}_{xz}$ and $\text{MRS}_{yz}$. (And you could also flip any of these, i.e. $\text{MRS}_{yx}$.) These are defined as the slope of the tangent line of the two dimensional indifference curve ...

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# Geometry/Algebra/Calculus

Why isn't the surface area of a sphere with radius r the following:
2*pi * (pi*r)

That comes from the following flow of logic:
Doesn't it makes sense to think of the surface area of the sphere with radius r as the the circumference of the semi-circle with radius r, pi*r (2*pi*r/2 = pi*r), multiplied by 2*pi, effectively "rotating" the semicircle along an orthogonal angular axis 2*pi radians (360 degrees)? In my mind, that would "trace" the surface area of a sphere, wouldn't it? What am I missing / not seeing?

Thanks

1. 👍 0
2. 👎 0
3. 👁 427
1. the problem is, that to rotate a curve, there has to be a variable radius of rotation. You have the 2pi, but don't specify what the radius of rotation is.

area needs to be in square units, but you only have r once, a linear unit.

1. 👍 0
2. 👎 0
posted by Steve

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3. ### mathematics

What is the solution for this? The surface area of a sphere varies in direct proportion to the square of its radius. If a sphere of radius 6cm has a surface area of 452 square cm, find, to the nearest mm, the radius of a

asked by Richelle on October 29, 2011
4. ### calculus *improper integrals*

A hole of a radius of 1cm is pierced in a sphere of a 4cm radius. Calculate the volume of the remaining sphere. I know that the volume of a sphere with an exterior radius is of: V = pie(R^2 - r^2) dH where R^2 is the exterior

asked by John on September 16, 2012
5. ### physics

Sphere 1 has surface area A1 and volume V1 and sphere 2 has surface area A2 and volume V2. If the radius of sphere 2 is 3.53 times the radius of sphere 1, what is the ratio of the areas A2/A1 ? 1.25×101 & 4.40×101 are correct

asked by bill on December 5, 2017
6. ### Algebra II

12. A sphere has a radius of 4x + 1. Which polynomial in standard form best describes the total surface area of the sphere? Use the formula S = 4 r2 for the surface area of a sphere. (1 point) 64 x2 + 32 x + 4 64 x2 + 16 x + 4 *my

asked by Sandra on October 6, 2014
7. ### calculus

A hole of a radius of 1cm is pierced in a sphere of a 4cm radius. Calculate the volume of the remaining sphere. I know that the volume of a sphere with an exterior radius is of: V = pie(R^2 - r^2) dH where R^2 is the exterior

asked by John on September 15, 2012
8. ### PHYSICS

The surface area A of a sphere of radius R is given by the equation A = 4pi*R2, where pi equals 3.1416 (to 5 digits). Calculate the area of a sphere which has a radius of 2 m. If 2600 grains of sand are uniformly spread out on the

asked by Anonymous on October 2, 2018
9. ### Algebra

A sphere has a radius of 4x + 1. Which polynomial in standard form best describes the total surface area of the sphere? Use the formula S = 4(pi)r2 for the surface area of a sphere.

asked by Sandra on March 7, 2014
10. ### algebra--1 question

A sphere has a radius of 4x + 1. Which polynomial in standard form best describes the total surface area of the sphere? Use the formula S = 4(pi)r^2 for the surface area of a sphere. 16(pi)x^2 + 16(pi) 16(pi)x^2 + 48(pi)x + 16(pi)

asked by SkatingDJ on March 24, 2015

More Similar Questions


text:
QUESTION

# River DeltasWhat differences exist between rivers with deltas and those without? What potential impact does the existence of deltas have on the human population in the U.S.?Your initial post should be

River Deltas

What differences exist between rivers with deltas and those without? What potential impact does the existence of deltas have on the human population in the U.S.?

Your initial post should be at least 250 words and must substantively integrate the assigned readings from the module with proper APA (Links to an external site.)Links to an external site. style formatting// You may use additional sources and materials as long as they are relevant to the discussion and cited properly.

• @
• 95 orders completed
Tutor has posted answer for $15.00. See answer's preview$15.00


text:
MapleSim Hydraulics Library from Modelon - MapleSim Help

Home : Support : Online Help : MapleSim Toolboxes : MapleSim Hydraulics Library from Modelon : Pumps : ModelonHydraulics/Pumps/ControlledPressureSource

MapleSim Hydraulics Library from Modelon

Controlled Pressure Source $—$ Simple pressure controlled pump

The Controlled Pressure Source component describes a pressure-controlled pump with externally commanded pressure, limited flow rate, and internal leakage. The pressure at the outlet port of this component may differ from the commanded value because of the settings of the controller (gain k) and the time constant T of the pump delay. If the boundary pressure should exactly follow the reference pressure model, then Ideal Pressure Source should be used instead.

Connections

 Name Description Modelica ID $\mathrm{oil}$ oil ${\mathrm{port}}_{B}$ Port B, where oil leaves the component ($q<0$, ${p}_{B}<{p}_{A}$ means $0<\mathrm{Δp}$) port_B $\mathrm{commandSignal}$ Connector of input signal used as flow rate commandSignal ${p}_{\mathrm{set}}$ p_set

Parameters

General Parameters

 Name Default Units Description Modelica ID use input $\mathrm{true}$ If true, input port is present useInput $G$ $5.·{10}^{-12}$ $\frac{{m}^{3}}{s\mathrm{Pa}}$ Conductance of pump leakage G use volume B $\mathrm{true}$ If true, a volume is present at port B useVolumeB ${\mathrm{ΔT}}_{\mathrm{system}}$ $0$ $K$ Temperature offset from system temperature dT_system ${p}_{\mathrm{bias}}$ $5.·{10}^{6}$ $\mathrm{Pa}$ Bias of pressure pbias ${q}_{\mathrm{max}}$ $0.05$ $\frac{{m}^{3}}{s}$ Maximum pump flow rate qMax $t$ $0.05$ $s$ Time constant of pressure control system t $k$ ${10}^{-6}$ Gain of pressure controller k

Oil Parameters

 Name Default Units Description Modelica ID ${V}_{B}$ ${10}^{-6}$ ${m}^{3}$ Geometric volume at port B volumeB

Constant Parameters

 Name Default Units Description Modelica ID ${q}_{\mathrm{min}}$ $0$ $\frac{{m}^{3}}{s}$ Minimum pump flow rate qMin


text:
# function of crypt called “ICE” [closed]

Using exeinfo, I have identified a function of crypt called ICE.

What is it? How can I find out the workings of the function?

Some idea's I had for doing this were:

• Perhaps there is a reference
• Could one reverse engineer it?
• If I have examples of (plaintext-ciphertext) pairs, could I construct an equivalent function without knowing the key?
-

## closed as unclear what you're asking by Gilles, DrLecter, e-sushi, AFS, archieJan 6 at 3:23

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

Have you, eehm, tried to google for ICE? –  orlp Jan 3 at 9:27
Yes. I have a look and says that is equal to DES, in is structure. –  user36384 Jan 3 at 9:30
Then what is your question? The source code is freely available. –  orlp Jan 3 at 9:36
@user36384: I have edited the question to reflect what I think you're asking. If not, please could you edit it further to reflect what you're asking. –  figlesquidge Jan 3 at 12:49


text:
×

can you Crack this...!!!!!!!!!

The number of isosceles triangles with sides having integral lengths is equal to....??

Note by Manish Bhargao
2 years ago

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text:
amazonka-redshift-1.4.5: Amazon Redshift SDK.

Copyright (c) 2013-2016 Brendan Hay Mozilla Public License, v. 2.0. Brendan Hay auto-generated non-portable (GHC extensions) None Haskell2010

Network.AWS.Redshift.DescribeEvents

Description

Returns events related to clusters, security groups, snapshots, and parameter groups for the past 14 days. Events specific to a particular cluster, security group, snapshot or parameter group can be obtained by providing the name as a parameter. By default, the past hour of events are returned.

This operation returns paginated results.

# Creating a Request

Creates a value of DescribeEvents with the minimum fields required to make a request.

Use one of the following lenses to modify other fields as desired:

• deStartTime - The beginning of the time interval to retrieve events for, specified in ISO 8601 format. For more information about ISO 8601, go to the ISO8601 Wikipedia page. Example: 2009-07-08T18:00Z
• deSourceType - The event source to retrieve events for. If no value is specified, all events are returned. Constraints: If SourceType is supplied, SourceIdentifier must also be provided. * Specify cluster when SourceIdentifier is a cluster identifier. * Specify cluster-security-group when SourceIdentifier is a cluster security group name. * Specify cluster-parameter-group when SourceIdentifier is a cluster parameter group name. * Specify cluster-snapshot when SourceIdentifier is a cluster snapshot identifier.
• deSourceIdentifier - The identifier of the event source for which events will be returned. If this parameter is not specified, then all sources are included in the response. Constraints: If SourceIdentifier is supplied, SourceType must also be provided. * Specify a cluster identifier when SourceType is cluster . * Specify a cluster security group name when SourceType is cluster-security-group . * Specify a cluster parameter group name when SourceType is cluster-parameter-group . * Specify a cluster snapshot identifier when SourceType is cluster-snapshot .
• deMarker - An optional parameter that specifies the starting point to return a set of response records. When the results of a DescribeEvents request exceed the value specified in MaxRecords , AWS returns a value in the Marker field of the response. You can retrieve the next set of response records by providing the returned marker value in the Marker parameter and retrying the request.
• deMaxRecords - The maximum number of response records to return in each call. If the number of remaining response records exceeds the specified MaxRecords value, a value is returned in a marker field of the response. You can retrieve the next set of records by retrying the command with the returned marker value. Default: 100 Constraints: minimum 20, maximum 100.
• deEndTime - The end of the time interval for which to retrieve events, specified in ISO 8601 format. For more information about ISO 8601, go to the ISO8601 Wikipedia page. Example: 2009-07-08T18:00Z
• deDuration - The number of minutes prior to the time of the request for which to retrieve events. For example, if the request is sent at 18:00 and you specify a duration of 60, then only events which have occurred after 17:00 will be returned. Default: 60

See: describeEvents smart constructor.

Instances

 Source # Methods Source # Methodsgfoldl :: (forall d b. Data d => c (d -> b) -> d -> c b) -> (forall g. g -> c g) -> DescribeEvents -> c DescribeEvents #gunfold :: (forall b r. Data b => c (b -> r) -> c r) -> (forall r. r -> c r) -> Constr -> c DescribeEvents #dataCast1 :: Typeable (* -> *) t => (forall d. Data d => c (t d)) -> Maybe (c DescribeEvents) #dataCast2 :: Typeable (* -> * -> *) t => (forall d e. (Data d, Data e) => c (t d e)) -> Maybe (c DescribeEvents) #gmapT :: (forall b. Data b => b -> b) -> DescribeEvents -> DescribeEvents #gmapQl :: (r -> r' -> r) -> r -> (forall d. Data d => d -> r') -> DescribeEvents -> r #gmapQr :: (r' -> r -> r) -> r -> (forall d. Data d => d -> r') -> DescribeEvents -> r #gmapQ :: (forall d. Data d => d -> u) -> DescribeEvents -> [u] #gmapQi :: Int -> (forall d. Data d => d -> u) -> DescribeEvents -> u #gmapM :: Monad m => (forall d. Data d => d -> m d) -> DescribeEvents -> m DescribeEvents #gmapMp :: MonadPlus m => (forall d. Data d => d -> m d) -> DescribeEvents -> m DescribeEvents #gmapMo :: MonadPlus m => (forall d. Data d => d -> m d) -> DescribeEvents -> m DescribeEvents # Source # Methods Source # MethodsshowList :: [DescribeEvents] -> ShowS # Source # Associated Typestype Rep DescribeEvents :: * -> * # Methods Source # Methods Source # Methodsrnf :: DescribeEvents -> () # Source # Methods Source # Associated Typestype Rs DescribeEvents :: * # Methods Source # Methods Source # Methods Source # Methods Source # type Rep DescribeEvents = D1 (MetaData "DescribeEvents" "Network.AWS.Redshift.DescribeEvents" "amazonka-redshift-1.4.5-2iQeXmFrftS29tYZtVOT2g" False) (C1 (MetaCons "DescribeEvents'" PrefixI True) ((:*:) ((:*:) (S1 (MetaSel (Just Symbol "_deStartTime") NoSourceUnpackedness SourceStrict DecidedStrict) (Rec0 (Maybe ISO8601))) ((:*:) (S1 (MetaSel (Just Symbol "_deSourceType") NoSourceUnpackedness SourceStrict DecidedStrict) (Rec0 (Maybe SourceType))) (S1 (MetaSel (Just Symbol "_deSourceIdentifier") NoSourceUnpackedness SourceStrict DecidedStrict) (Rec0 (Maybe Text))))) ((:*:) ((:*:) (S1 (MetaSel (Just Symbol "_deMarker") NoSourceUnpackedness SourceStrict DecidedStrict) (Rec0 (Maybe Text))) (S1 (MetaSel (Just Symbol "_deMaxRecords") NoSourceUnpackedness SourceStrict DecidedStrict) (Rec0 (Maybe Int)))) ((:*:) (S1 (MetaSel (Just Symbol "_deEndTime") NoSourceUnpackedness SourceStrict DecidedStrict) (Rec0 (Maybe ISO8601))) (S1 (MetaSel (Just Symbol "_deDuration") NoSourceUnpackedness SourceStrict DecidedStrict) (Rec0 (Maybe Int))))))) type Rs DescribeEvents Source #

# Request Lenses

The beginning of the time interval to retrieve events for, specified in ISO 8601 format. For more information about ISO 8601, go to the ISO8601 Wikipedia page. Example: 2009-07-08T18:00Z

The event source to retrieve events for. If no value is specified, all events are returned. Constraints: If SourceType is supplied, SourceIdentifier must also be provided. * Specify cluster when SourceIdentifier is a cluster identifier. * Specify cluster-security-group when SourceIdentifier is a cluster security group name. * Specify cluster-parameter-group when SourceIdentifier is a cluster parameter group name. * Specify cluster-snapshot when SourceIdentifier is a cluster snapshot identifier.

The identifier of the event source for which events will be returned. If this parameter is not specified, then all sources are included in the response. Constraints: If SourceIdentifier is supplied, SourceType must also be provided. * Specify a cluster identifier when SourceType is cluster . * Specify a cluster security group name when SourceType is cluster-security-group . * Specify a cluster parameter group name when SourceType is cluster-parameter-group . * Specify a cluster snapshot identifier when SourceType is cluster-snapshot .

An optional parameter that specifies the starting point to return a set of response records. When the results of a DescribeEvents request exceed the value specified in MaxRecords , AWS returns a value in the Marker field of the response. You can retrieve the next set of response records by providing the returned marker value in the Marker parameter and retrying the request.

The maximum number of response records to return in each call. If the number of remaining response records exceeds the specified MaxRecords value, a value is returned in a marker field of the response. You can retrieve the next set of records by retrying the command with the returned marker value. Default: 100 Constraints: minimum 20, maximum 100.

The end of the time interval for which to retrieve events, specified in ISO 8601 format. For more information about ISO 8601, go to the ISO8601 Wikipedia page. Example: 2009-07-08T18:00Z

The number of minutes prior to the time of the request for which to retrieve events. For example, if the request is sent at 18:00 and you specify a duration of 60, then only events which have occurred after 17:00 will be returned. Default: 60

# Destructuring the Response

Arguments

 :: Int dersResponseStatus -> DescribeEventsResponse

Creates a value of DescribeEventsResponse with the minimum fields required to make a request.

Use one of the following lenses to modify other fields as desired:

• dersEvents - A list of Event instances.
• dersMarker - A value that indicates the starting point for the next set of response records in a subsequent request. If a value is returned in a response, you can retrieve the next set of records by providing this returned marker value in the Marker parameter and retrying the command. If the Marker field is empty, all response records have been retrieved for the request.
• dersResponseStatus - -- | The response status code.

See: describeEventsResponse smart constructor.

Instances

 Source # Methods Source # Methodsgfoldl :: (forall d b. Data d => c (d -> b) -> d -> c b) -> (forall g. g -> c g) -> DescribeEventsResponse -> c DescribeEventsResponse #gunfold :: (forall b r. Data b => c (b -> r) -> c r) -> (forall r. r -> c r) -> Constr -> c DescribeEventsResponse #dataCast1 :: Typeable (* -> *) t => (forall d. Data d => c (t d)) -> Maybe (c DescribeEventsResponse) #dataCast2 :: Typeable (* -> * -> *) t => (forall d e. (Data d, Data e) => c (t d e)) -> Maybe (c DescribeEventsResponse) #gmapT :: (forall b. Data b => b -> b) -> DescribeEventsResponse -> DescribeEventsResponse #gmapQl :: (r -> r' -> r) -> r -> (forall d. Data d => d -> r') -> DescribeEventsResponse -> r #gmapQr :: (r' -> r -> r) -> r -> (forall d. Data d => d -> r') -> DescribeEventsResponse -> r #gmapQ :: (forall d. Data d => d -> u) -> DescribeEventsResponse -> [u] #gmapQi :: Int -> (forall d. Data d => d -> u) -> DescribeEventsResponse -> u #gmapM :: Monad m => (forall d. Data d => d -> m d) -> DescribeEventsResponse -> m DescribeEventsResponse #gmapMp :: MonadPlus m => (forall d. Data d => d -> m d) -> DescribeEventsResponse -> m DescribeEventsResponse #gmapMo :: MonadPlus m => (forall d. Data d => d -> m d) -> DescribeEventsResponse -> m DescribeEventsResponse # Source # Source # Methods Source # Associated Typestype Rep DescribeEventsResponse :: * -> * # Methods Source # Methodsrnf :: DescribeEventsResponse -> () # Source # type Rep DescribeEventsResponse = D1 (MetaData "DescribeEventsResponse" "Network.AWS.Redshift.DescribeEvents" "amazonka-redshift-1.4.5-2iQeXmFrftS29tYZtVOT2g" False) (C1 (MetaCons "DescribeEventsResponse'" PrefixI True) ((:*:) (S1 (MetaSel (Just Symbol "_dersEvents") NoSourceUnpackedness SourceStrict DecidedStrict) (Rec0 (Maybe [Event]))) ((:*:) (S1 (MetaSel (Just Symbol "_dersMarker") NoSourceUnpackedness SourceStrict DecidedStrict) (Rec0 (Maybe Text))) (S1 (MetaSel (Just Symbol "_dersResponseStatus") NoSourceUnpackedness SourceStrict DecidedUnpack) (Rec0 Int)))))

# Response Lenses

A list of Event instances.

A value that indicates the starting point for the next set of response records in a subsequent request. If a value is returned in a response, you can retrieve the next set of records by providing this returned marker value in the Marker parameter and retrying the command. If the Marker field is empty, all response records have been retrieved for the request.

• - | The response status code.


text:
## 2.6Forces at a surface

The next law to deﬁne is that of conservation of momentum, i.e. Newton’s second law of motion () for ﬂuids. It involves forces within the ﬂuid so requires a description of forces at a surface bounding a volume .

The force is in the direction of the traction vector with magnitude of (traction surface area) — compare with Eq. (2.3 )

 (2.16)
The tractions at a surface bounding some volume of ﬂuid (or any continuum, e.g. solid) depend on the orientation of the surface. Consequently, we cannot deﬁne forces within a ﬂuid as traction vectors at points within the ﬂuid.

Instead, we require 3 traction vectors , and , deﬁned in planes perpendicular to one another, i.e. , and . This results in the stress tensor with 9 components, consisting of 3 traction vectors, each containing 3 components.

The traction can be calculated at a surface with any orientation by taking the inner product of the unit normal vector and the stress tensor such that

 (2.17)

### Tensors

We have deﬁned the stress tensor3, an entity with 9 component values, corresponding to our , and axes (or more speciﬁcally base vectors of unit length aligned with our , and axes).

In fact, the term “tensor” describes any entity with multiple component values corresponding to the dimensions of space — here 3. A tensor has rank , such that the number of component values for 3D space = .

In this book we use the term “tensor” to mean “tensor of rank 2” unless otherwise noted. A vector is a tensor of rank 1 and a scalar is rank 0.

The inner product of a vector and tensor produces a vector whose 3 components are evaluated as follows:

 (2.18)
This inner product tensor is commutative only if is symmetric since — see Eq. (2.24 ) for the transpose .
3More precisely the “Cauchy stress tensor”, introduced by Augustin-Louis Cauchy in De la pression ou tension dans un corps solide, 1827.

Notes on CFD: General Principles - 2.6 Forces at a surface


text:
# Graphing Polynomial Functions

To sketch any polynomial function, you can start by finding the real zeros of the function and end behavior of the function .

Steps involved in graphing polynomial functions:

1. Predict the end behavior of the function.

2. Find the real zeros of the function. Check whether it is possible to rewrite the function in factored form to find the zeros. Otherwise, use Descartes' rule of signs to identify the possible number of real zeros.

3. Make a table of values to find several points.

4. Plot the points and draw a smooth continuous curve to connect the points.

5. Make sure that the graph follows the end behavior as found in the above step.

Example:

Graph the polynomial function ${x}^{3}-2{x}^{2}-3x$.

Predict the end behavior of the function.

The degree of the polynomial function is odd and the leading coefficient is positive.

The degree of the polynomial is $3$ and there would be $3$ zeros for the functions.

The function can be factored as $x\left(x+1\right)\left(x-3\right)$. So, the zeros of the functions are $x=-1,0$ and $3$.

Make a table of values to find several points.

 $x$ $f\left(x\right)$ $-3$ $-36$ $-2$ $-10$ $-1$ $0$ $0$ $0$ $1$ $-4$ $2$ $-6$ $3$ $0$ $4$ $20$

Plot the points and draw a smooth continuous curve to connect the points


text:
# Model Card: GroupViT

This checkpoint is uploaded by Jiarui Xu.

## Model Details

The GroupViT model was proposed in GroupViT: Semantic Segmentation Emerges from Text Supervision by Jiarui Xu, Shalini De Mello, Sifei Liu, Wonmin Byeon, Thomas Breuel, Jan Kautz, Xiaolong Wang. Inspired by CLIP, GroupViT is a vision-language model that can perform zero-shot semantic segmentation on any given vocabulary categories.

June 2022

### Abstract

Grouping and recognition are important components of visual scene understanding, e.g., for object detection and semantic segmentation. With end-to-end deep learning systems, grouping of image regions usually happens implicitly via top-down supervision from pixel-level recognition labels. Instead, in this paper, we propose to bring back the grouping mechanism into deep networks, which allows semantic segments to emerge automatically with only text supervision. We propose a hierarchical Grouping Vision Transformer (GroupViT), which goes beyond the regular grid structure representation and learns to group image regions into progressively larger arbitrary-shaped segments. We train GroupViT jointly with a text encoder on a large-scale image-text dataset via contrastive losses. With only text supervision and without any pixel-level annotations, GroupViT learns to group together semantic regions and successfully transfers to the task of semantic segmentation in a zero-shot manner, i.e., without any further fine-tuning. It achieves a zero-shot accuracy of 52.3% mIoU on the PASCAL VOC 2012 and 22.4% mIoU on PASCAL Context datasets, and performs competitively to state-of-the-art transfer-learning methods requiring greater levels of supervision.

### Use with Transformers

from PIL import Image
import requests
from transformers import AutoProcessor, GroupViTModel

model = GroupViTModel.from_pretrained("nvidia/groupvit-gcc-yfcc")
processor = AutoProcessor.from_pretrained("nvidia/groupvit-gcc-yfcc")

url = "http://images.cocodataset.org/val2017/000000039769.jpg"
image = Image.open(requests.get(url, stream=True).raw)

inputs = processor(text=["a photo of a cat", "a photo of a dog"], images=image, return_tensors="pt", padding=True)

outputs = model(**inputs)
logits_per_image = outputs.logits_per_image  # this is the image-text similarity score
probs = logits_per_image.softmax(dim=1)  # we can take the softmax to get the label probabilities

## Data

The model was trained on publicly available image-caption data. This was done through a combination of crawling a handful of websites and using commonly-used pre-existing image datasets such as YFCC100M. A large portion of the data comes from our crawling of the internet. This means that the data is more representative of people and societies most connected to the internet which tend to skew towards more developed nations, and younger, male users.

For more code examples, we refer to the documentation.

### BibTeX entry and citation info

@article{xu2022groupvit,
author    = {Xu, Jiarui and De Mello, Shalini and Liu, Sifei and Byeon, Wonmin and Breuel, Thomas and Kautz, Jan and Wang, Xiaolong},
title     = {GroupViT: Semantic Segmentation Emerges from Text Supervision},
journal   = {arXiv preprint arXiv:2202.11094},
year      = {2022},
}

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Stage: 3 Challenge Level:

Use a single sheet of A4 paper and make a cylinder having the greatest possible volume. The cylinder must be closed off by a circle at each end.

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This is the first article in a series which aim to provide some insight into the way spatial thinking develops in children, and draw on a range of reported research. The focus of this article is the. . . .

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This article looks at levels of geometric thinking and the types of activities required to develop this thinking.

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This task depends on groups working collaboratively, discussing and reasoning to agree a final product.

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Choose a couple of the sequences. Try to picture how to make the next, and the next, and the next... Can you describe your reasoning?

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Framed

Stage: 3 Challenge Level:

Seven small rectangular pictures have one inch wide frames. The frames are removed and the pictures are fitted together like a jigsaw to make a rectangle of length 12 inches. Find the dimensions of. . . .

Two Squared

Stage: 2 Challenge Level:

What happens to the area of a square if you double the length of the sides? Try the same thing with rectangles, diamonds and other shapes. How do the four smaller ones fit into the larger one?

Trice

Stage: 3 Challenge Level:

ABCDEFGH is a 3 by 3 by 3 cube. Point P is 1/3 along AB (that is AP : PB = 1 : 2), point Q is 1/3 along GH and point R is 1/3 along ED. What is the area of the triangle PQR?

Introducing NRICH TWILGO

Stage: 1, 2, 3, 4 and 5 Challenge Level:

We're excited about this new program for drawing beautiful mathematical designs. Can you work out how we made our first few pictures and, even better, share your most elegant solutions with us?

Multiplication Series: Illustrating Number Properties with Arrays

Stage: 1 and 2

This article for teachers describes how modelling number properties involving multiplication using an array of objects not only allows children to represent their thinking with concrete materials,. . . .

Ten Hidden Squares

Stage: 2 Challenge Level:

These points all mark the vertices (corners) of ten hidden squares. Can you find the 10 hidden squares?

Square It

Stage: 3 and 4 Challenge Level:

Players take it in turns to choose a dot on the grid. The winner is the first to have four dots that can be joined to form a square.

Khun Phaen Escapes to Freedom

Stage: 3 Challenge Level:

Slide the pieces to move Khun Phaen past all the guards into the position on the right from which he can escape to freedom.

Sliding Puzzle

Stage: 1, 2, 3 and 4 Challenge Level:

The aim of the game is to slide the green square from the top right hand corner to the bottom left hand corner in the least number of moves.

World of Tan 14 - Celebrations

Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outline of Little Ming and Little Fung dancing?

World of Tan 8 - Sports Car

Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outline of this sports car?

Turning Cogs

Stage: 2 Challenge Level:

What happens when you turn these cogs? Investigate the differences between turning two cogs of different sizes and two cogs which are the same.

World of Tan 15 - Millennia

Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outlines of the workmen?

Making Tangrams

Stage: 2 Challenge Level:

Here's a simple way to make a Tangram without any measuring or ruling lines.

World of Tan 9 - Animals

Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outline of this goat and giraffe?

Cuboids

Stage: 3 Challenge Level:

Find a cuboid (with edges of integer values) that has a surface area of exactly 100 square units. Is there more than one? Can you find them all?

Fred the Class Robot

Stage: 2 Challenge Level:

Billy's class had a robot called Fred who could draw with chalk held underneath him. What shapes did the pupils make Fred draw?

Buses

Stage: 3 Challenge Level:

A bus route has a total duration of 40 minutes. Every 10 minutes, two buses set out, one from each end. How many buses will one bus meet on its way from one end to the other end?

Bands and Bridges: Bringing Topology Back

Stage: 2 and 3

Lyndon Baker describes how the Mobius strip and Euler's law can introduce pupils to the idea of topology.


text:
# Proofs of convergence of random variables

Several results will be established using the portmanteau lemma: A sequence {Xn} converges in distribution to X if and only if any of the following conditions are met:

1. E[f(Xn)] → E[f(X)] for all bounded, continuous functions f;
2. E[f(Xn)] → E[f(X)] for all bounded, Lipschitz functions f;
3. limsup{Pr(XnC)} ≤ Pr(XC) for all closed sets C;

## Convergence almost surely implies convergence in probability

$X_n\ \xrightarrow{as}\ X \quad\Rightarrow\quad X_n\ \xrightarrow{p}\ X$

Proof: If {Xn} converges to X almost surely, it means that the set of points {ω: lim Xn(ω) ≠ X(ω)} has measure zero; denote this set O. Now fix ε > 0 and consider a sequence of sets

$A_n = \bigcup_{m\geq n} \left \{ \left |X_m-X \right |>\varepsilon \right\}$

This sequence of sets is decreasing: AnAn+1 ⊇ ..., and it decreases towards the set

$A_{\infty} = \bigcap_{n \geq 1} A_n.$

For this decreasing sequence of events, their probabilities are also a decreasing sequence, and it decreases towards the Pr(A); we shall show now that this number is equal to zero. Now any point ω in the complement of O is such that lim Xn(ω) = X(ω), which implies that |Xn(ω) − X(ω)| < ε for all n greater than a certain number N. Therefore, for all nN the point ω will not belong to the set An, and consequently it will not belong to A. This means that A is disjoint with O, or equivalently, A is a subset of O and therefore Pr(A) = 0.

Finally, consider

$\operatorname{Pr}\left(|X_n-X|>\varepsilon\right) \leq \operatorname{Pr}(A_n) \ \underset{n\to\infty}{\rightarrow} 0,$

which by definition means that Xn converges in probability to X.

## Convergence in probability does not imply almost sure convergence in the discrete case

If Xn are independent random variables assuming value one with probability 1/n and zero otherwise, then Xn converges to zero in probability but not almost surely. This can be verified using the Borel–Cantelli lemmas.

## Convergence in probability implies convergence in distribution

$X_n\ \xrightarrow{p}\ X \quad\Rightarrow\quad X_n\ \xrightarrow{d}\ X,$

### Proof for the case of scalar random variables

Lemma. Let X, Y be random variables, a a real number and ε > 0. Then

$\operatorname{Pr}(Y \leq a) \leq \operatorname{Pr}(X\leq a+\varepsilon) + \operatorname{Pr}(|Y - X| > \varepsilon).$

(or $\{Y \leq a\} \subset \{X\leq a+\varepsilon\}\cup \{|Y - X| > \varepsilon\}.$)

Proof of lemma:

\begin{align} \operatorname{Pr}(Y\leq a) &= \operatorname{Pr}(Y\leq a,\ X\leq a+\varepsilon) + \operatorname{Pr}(Y\leq a,\ X>a+\varepsilon) \\ &\leq \operatorname{Pr}(X\leq a+\varepsilon) + \operatorname{Pr}(Y-X\leq a-X,\ a-X<-\varepsilon) \\ &\leq \operatorname{Pr}(X\leq a+\varepsilon) + \operatorname{Pr}(Y-X<-\varepsilon) \\ &\leq \operatorname{Pr}(X\leq a+\varepsilon) + \operatorname{Pr}(Y-X<-\varepsilon) + \operatorname{Pr}(Y-X>\varepsilon)\\ &= \operatorname{Pr}(X\leq a+\varepsilon) + \operatorname{Pr}(|Y-X|>\varepsilon) \end{align}

Proof of the theorem: Recall that in order to prove convergence in distribution, one must show that the sequence of cumulative distribution functions converges to the FX at every point where FX is continuous. Let a be such a point. For every ε > 0, due to the preceding lemma, we have:

\begin{align} \operatorname{Pr}(X_n\leq a) &\leq \operatorname{Pr}(X\leq a+\varepsilon) + \operatorname{Pr}(|X_n-X|>\varepsilon) \\ \operatorname{Pr}(X\leq a-\varepsilon)&\leq \operatorname{Pr}(X_n\leq a) + \operatorname{Pr}(|X_n-X|>\varepsilon) \end{align}

So, we have

$\operatorname{Pr}(X\leq a-\varepsilon) - \operatorname{Pr} \left (\left |X_n-X \right |>\varepsilon \right ) \leq \operatorname{Pr}(X\leq a+\varepsilon) + \operatorname{Pr} \left (\left |X_n-X \right |>\varepsilon \right ).$

Taking the limit as n → ∞, we obtain:

$F_X(a-\varepsilon) \leq F_X(a+\varepsilon),$

where FX(a) = Pr(Xa) is the cumulative distribution function of X. This function is continuous at a by assumption, and therefore both FX(a−ε) and FX(a+ε) converge to FX(a) as ε → 0+. Taking this limit, we obtain

$\lim_{n\to\infty} \operatorname{Pr}(X_n \leq a) = \operatorname{Pr}(X \leq a),$

which means that {Xn} converges to X in distribution.

### Proof for the generic case

We see that |XnX| converges in probability to zero, and also X converges to X in distribution trivially. Applying the property proved later on this page we conclude that Xn converges to X in distribution.

## Convergence in distribution to a constant implies convergence in probability

$X_n\ \xrightarrow{d}\ c \quad\Rightarrow\quad X_n\ \xrightarrow{p}\ c,$ provided c is a constant.

Proof: Fix ε > 0. Let Bε(c) be the open ball of radius ε around point c, and Bεc(c)its complement. Then

$\operatorname{Pr}\left(|X_n-c|\geq\varepsilon\right) = \operatorname{Pr}\left(X_n\in B_\varepsilon^c(c)\right).$

By the portmanteau lemma (part C), if Xn converges in distribution to c, then the limsup of the latter probability must be less than or equal to Pr(cBεc(c)),which is obviously equal to zero. Therefore

\begin{align} \lim_{n\to\infty}\operatorname{Pr}\left( \left |X_n-c \right |\geq\varepsilon\right) &\leq \limsup_{n\to\infty}\operatorname{Pr}\left( \left |X_n-c \right | \geq \varepsilon \right) \\ &= \limsup_{n\to\infty}\operatorname{Pr}\left(X_n\in B_\varepsilon^c(c)\right) \\ &\leq \operatorname{Pr}\left(c\in B_\varepsilon^c(c)\right) = 0 \end{align}

which by definition means that Xn converges to c in probability.

## Convergence in probability to a sequence converging in distribution implies convergence to the same distribution

$|Y_n-X_n|\ \xrightarrow{p}\ 0,\ \ X_n\ \xrightarrow{d}\ X\ \quad\Rightarrow\quad Y_n\ \xrightarrow{d}\ X$

Proof: We will prove this theorem using the portmanteau lemma, part B. As required in that lemma, consider any bounded function f (i.e. |f(x)| ≤ M) which is also Lipschitz:

$\exists K >0, \forall x,y: \quad |f(x)-f(y)|\leq K|x-y|.$

Take some ε > 0 and majorize the expression |E[f(Yn)] − E[f(Xn)]| as

\begin{align} \left|\operatorname{E}\left[f(Y_n)\right] - \operatorname{E}\left [f(X_n) \right] \right| &\leq \operatorname{E} \left [\left |f(Y_n) - f(X_n) \right | \right ]\\ &= \operatorname{E}\left[ \left |f(Y_n) - f(X_n) \right |\mathbf{1}_{\left \{|Y_n-X_n|<\varepsilon \right \}} \right] + \operatorname{E}\left[ \left |f(Y_n) - f(X_n) \right |\mathbf{1}_{\left \{|Y_n-X_n|\geq\varepsilon \right \}} \right] \\ &\leq \operatorname{E}\left[K \left |Y_n - X_n \right |\mathbf{1}_{\left \{|Y_n-X_n|<\varepsilon \right \}}\right] + \operatorname{E}\left[2M\mathbf{1}_{\left \{|Y_n-X_n|\geq\varepsilon \right \}}\right] \\ &\leq K \varepsilon \operatorname{Pr} \left (\left |Y_n-X_n \right |<\varepsilon\right) + 2M \operatorname{Pr} \left( \left |Y_n-X_n \right |\geq\varepsilon\right )\\ &\leq K \varepsilon + 2M \operatorname{Pr} \left (\left |Y_n-X_n \right |\geq\varepsilon \right ) \end{align}

(here 1{...} denotes the indicator function; the expectation of the indicator function is equal to the probability of corresponding event). Therefore

\begin{align} \left |\operatorname{E}\left [f(Y_n)\right ] - \operatorname{E}\left [f(X) \right ]\right | &\leq \left|\operatorname{E}\left[ f(Y_n) \right ]-\operatorname{E} \left [f(X_n) \right ] \right| + \left|\operatorname{E}\left [f(X_n) \right ]-\operatorname{E}\left [f(X) \right] \right| \\ &\leq K\varepsilon + 2M \operatorname{Pr}\left (|Y_n-X_n|\geq\varepsilon\right )+ \left |\operatorname{E}\left[ f(X_n) \right]-\operatorname{E} \left [f(X) \right ]\right|. \end{align}

If we take the limit in this expression as n → ∞, the second term will go to zero since {Yn−Xn} converges to zero in probability; and the third term will also converge to zero, by the portmanteau lemma and the fact that Xn converges to X in distribution. Thus

$\lim_{n\to\infty} \left|\operatorname{E}\left [f(Y_n) \right] - \operatorname{E}\left [f(X) \right ] \right| \leq K\varepsilon.$

Since ε was arbitrary, we conclude that the limit must in fact be equal to zero, and therefore E[f(Yn)] → E[f(X)], which again by the portmanteau lemma implies that {Yn} converges to X in distribution. QED.

## Convergence of one sequence in distribution and another to a constant implies joint convergence in distribution

$X_n\ \xrightarrow{d}\ X,\ \ Y_n\ \xrightarrow{d}\ c\ \quad\Rightarrow\quad (X_n,Y_n)\ \xrightarrow{d}\ (X,c)$ provided c is a constant.

Proof: We will prove this statement using the portmanteau lemma, part A.

First we want to show that (Xn, c) converges in distribution to (X, c). By the portmanteau lemma this will be true if we can show that E[f(Xn, c)] → E[f(X, c)] for any bounded continuous function f(x, y). So let f be such arbitrary bounded continuous function. Now consider the function of a single variable g(x) := f(x, c). This will obviously be also bounded and continuous, and therefore by the portmanteau lemma for sequence {Xn} converging in distribution to X, we will have that E[g(Xn)] → E[g(X)]. However the latter expression is equivalent to “E[f(Xn, c)] → E[f(X, c)]”, and therefore we now know that (Xn, c) converges in distribution to (X, c).

Secondly, consider |(Xn, Yn) − (Xn, c)| = |Ync|. This expression converges in probability to zero because Yn converges in probability to c. Thus we have demonstrated two facts:

$\begin{cases} \left| (X_n, Y_n) - (X_n,c) \right|\ \xrightarrow{p}\ 0, \\ (X_n,c)\ \xrightarrow{d}\ (X,c). \end{cases}$

By the property proved earlier, these two facts imply that (Xn, Yn) converge in distribution to (X, c).

## Convergence of two sequences in probability implies joint convergence in probability

$X_n\ \xrightarrow{p}\ X,\ \ Y_n\ \xrightarrow{p}\ Y\ \quad\Rightarrow\quad (X_n,Y_n)\ \xrightarrow{p}\ (X,Y)$

Proof:

\begin{align} \operatorname{Pr}\left(\left|(X_n,Y_n)-(X,Y)\right|\geq\varepsilon\right) &\leq \operatorname{Pr}\left(|X_n-X| + |Y_n-Y|\geq\varepsilon\right) \\ &\leq\operatorname{Pr}\left(|X_n-X|\geq\tfrac{\varepsilon}{2}\right) + \operatorname{Pr}\left(|Y_n-Y|\geq\tfrac{\varepsilon}{2}\right) \end{align}

Each of the probabilities on the right-hand side converge to zero as n → ∞ by definition of the convergence of {Xn} and {Yn} in probability to X and Y respectively. Taking the limit we conclude that the left-hand side also converges to zero, and therefore the sequence {(Xn, Yn)} converges in probability to {(X, Y)}.