text:
# 4.6. Global Ocean Simulation in Pressure Coordinates¶

This example experiment demonstrates using MITgcm to simulate the planetary ocean circulation in pressure coordinates, that is, without making the Boussinesq approximations. The simulation is configured as a near copy of tutorial_global_oce_latlon (Section 4.5). with realistic geography and bathymetry on a $$4^{\circ} \times 4^{\circ}$$ spherical polar grid. Fifteen levels are used in the vertical, ranging in thickness from 50.4089 dbar $$\approx$$ 50 m at the surface to 710.33 dbar $$\approx$$ 690 m at depth, giving a maximum model depth of 5302.3122 dbar $$\approx$$ 5200 m. At this resolution, the configuration can be integrated forward for thousands of years on a single processor desktop computer.

## 4.6.1. Overview¶

The model is forced with climatological wind stress data from Trenberth (1990) [TOL90] and surface flux data from Jiang et al. (1999) [JSMR99]. Climatological data (Levitus and Boyer 1994a,b [LB94a, LB94b]) is used to initialize the model hydrography. Levitus and Boyer seasonal climatology data is also used throughout the calculation to provide additional air-sea fluxes. These fluxes are combined with the Jiang et al. climatological estimates of surface heat flux, resulting in a mixed boundary condition of the style described in Haney (1971) [Han71]. Altogether, this yields the following forcing applied in the model surface layer.

(4.42)${\cal F}_{u} = g\frac{\tau_{x}}{\Delta p_{s}}$
(4.43)${\cal F}_{v} = g\frac{\tau_{y}}{\Delta p_{s}}$
(4.44)${\cal F}_{\theta} = - g\lambda_{\theta} ( \theta - \theta^{\ast} ) - \frac{1}{C_{p} \Delta p_{s}}{\cal Q}$
(4.45)${\cal F}_{s} = + g\rho_{FW}\frac{S}{\rho\Delta p_{s}}({\cal E} - {\cal P} - {\cal R})$

where $${\cal F}_{u}$$, $${\cal F}_{v}$$, $${\cal F}_{\theta}$$, $${\cal F}_{s}$$ are the forcing terms in the zonal and meridional momentum and in the potential temperature and salinity equations respectively. The term $$\Delta p_{s}$$ represents the top ocean layer thickness in Pa. It is used in conjunction with a reference density, $$\rho_{FW}$$ (here set to 999.8 kg m-3), the surface salinity, $$S$$, and a specific heat capacity, $$C_{p}$$ (here set to 4000 J kg-1 K-1), to convert input dataset values into time tendencies of potential temperature (with units of oC s-1), salinity (with units ppt s-1) and velocity (with units m s-2). The externally supplied forcing fields used in this experiment are $$\tau_{x}$$, $$\tau_{y}$$, $$\theta^{\ast}$$, $$\cal{Q}$$ and $$\cal{E}-\cal{P}-\cal{R}$$. The wind stress fields ($$\tau_x$$, $$\tau_y$$) have units of N m-2. The temperature forcing fields ($$\theta^{\ast}$$ and $$Q$$) have units of oC and W m-2 respectively. The salinity forcing fields ($$\cal{E}-\cal{P}-\cal{R}$$) has units of m s-1 respectively. The source files and procedures for ingesting these data into the simulation are described in the experiment configuration discussion in section Section 4.5.3.

## 4.6.2. Discrete Numerical Configuration¶

Due to the pressure coordinate, the model can only be hydrostatic (de Szoeke and Samelson 2002 [dSS02]). The domain is discretized with a uniform grid spacing in latitude and longitude on the sphere $$\Delta \phi=\Delta \lambda=4^{\circ}$$, so that there are 90 grid cells in the zonal and 40 in the meridional direction. The internal model coordinate variables $$x$$ and $$y$$ are initialized according to

\begin{split}\begin{aligned} x=r\cos(\phi),~\Delta x & = r\cos(\Delta \phi) \\ y=r\lambda,~\Delta y & = r\Delta \lambda \end{aligned}\end{split}

Arctic polar regions are not included in this experiment. Meridionally the model extends from 80oS to 80oN. Vertically the model is configured with fifteen layers with the following thicknesses

$$\Delta p_{1}$$ = 7103300.720021 Pa
$$\Delta p_{2}$$ = 6570548.440790 Pa
$$\Delta p_{3}$$ = 6041670.010249 Pa
$$\Delta p_{4}$$ = 5516436.666057 Pa
$$\Delta p_{5}$$ = 4994602.034410 Pa
$$\Delta p_{6}$$ = 4475903.435290 Pa
$$\Delta p_{7}$$ = 3960063.245801 Pa
$$\Delta p_{8}$$ = 3446790.312651 Pa
$$\Delta p_{9}$$ = 2935781.405664 Pa
$$\Delta p_{10}$$ = 2426722.705046 Pa
$$\Delta p_{11}$$ = 1919291.315988 Pa
$$\Delta p_{12}$$ = 1413156.804970 Pa
$$\Delta p_{13}$$ = 1008846.750166 Pa
$$\Delta p_{14}$$ = 705919.025481 Pa
$$\Delta p_{15}$$ = 504089.693499 Pa

(here the numeric subscript indicates the model level index number, $${\tt k}$$; note that the surface layer has the highest index number 15) to give a total depth, $$H$$, of -5200 m. In pressure, this is $$p_{b}^{0}$$ = 53023122.566084 Pa. The implicit free surface form of the pressure equation described in Marshall et al. (1997) [MHPA97] with the nonlinear extension by Campin et al. (2004) [CAHM04] is employed. A Laplacian operator, $$\nabla^2$$, provides viscous dissipation. Thermal and haline diffusion is also represented by a Laplacian operator.

Wind-stress forcing is added to the momentum equations in (4.46) for both the zonal flow, $$u$$ and the meridional flow $$v$$, according to equations (4.42) and (4.43). Thermodynamic forcing inputs are added to the equations in (4.47) for potential temperature, $$\theta$$, and salinity, $$S$$, according to equations (4.44) and (4.45). This produces a set of equations solved in this configuration as follows:

(4.46)$\begin{split}\frac{Du}{Dt} - fv + \frac{1}{\rho}\frac{\partial \Phi^\prime}{\partial x} - \nabla _h \cdot ( A_{h} \nabla _h u )- (g\rho_0)^2\frac{\partial}{\partial p}\left( A_{r}\frac{\partial u}{\partial p}\right) &= \begin{cases} {\cal F}_u & \text{(surface)} \\ 0 & \text{(interior)} \end{cases} \\ \frac{Dv}{Dt} + fu + \frac{1}{\rho}\frac{\partial \Phi^\prime}{\partial y} - \nabla _h \cdot ( A_{h} \nabla _h v) - (g\rho_0)^2\frac{\partial}{\partial p}\left( A_{r}\frac{\partial v}{\partial p}\right) &= \begin{cases} {\cal F}_v & \text{(surface)} \\ 0 & \text{(interior)} \end{cases}\end{split}$
$\frac{\partial p_{b}}{\partial t} + \nabla _h \cdot \vec{\bf u} = 0$
(4.47)$\begin{split}\frac{D\theta}{Dt} - \nabla _h \cdot (K_{h} \nabla _h \theta) - (g\rho_0)^2\frac{\partial}{\partial p}\left( \Gamma(K_{r})\frac{\partial\theta}{\partial p}\right) &= \begin{cases} {\cal F}_\theta & \text{(surface)} \\ 0 & \text{(interior)} \end{cases} \\ \frac{D S}{Dt} - \nabla _h \cdot (K_{h} \nabla _h S) - (g\rho_0)^2\frac{\partial}{\partial p}\left( \Gamma(K_{r})\frac{\partial S}{\partial p}\right) &= \begin{cases} {\cal F}_S & \text{(surface)} \\ 0 & \text{(interior)} \end{cases}\end{split}$
$\Phi_{-H}'^{(0)} + \alpha_{0}p_{b}+ \int^{p}_{0}\alpha' dp = \Phi'$

where $$u=\frac{Dx}{Dt}=r \cos(\phi)\frac{D \lambda}{Dt}$$ and $$v=\frac{Dy}{Dt}=r \frac{D \phi}{Dt}$$ are the zonal and meridional components of the flow vector, $$\vec{\bf u}$$, on the sphere. As described in Section 2, the time evolution of potential temperature $$\theta$$ equation is solved prognostically. The full geopotential height $$\Phi$$ is diagnosed by summing the geopotential height anomalies $$\Phi'$$ due to bottom pressure $$p_{b}$$ and density variations. The integration of the hydrostatic equation is started at the bottom of the domain. The condition of $$p=0$$ at the sea surface requires a time-independent integration constant for the height anomaly due to density variations $$\Phi_{-H}'^{(0)}$$, which is provided as an input field.

## 4.6.3. Experiment Configuration¶

The experiment files

contain the code customizations and parameter settings for these experiments. Below we describe the customizations to these files associated with this experiment.

### 4.6.3.1. Driving Datasets¶

Figure 4.32-Figure 4.37 show the relaxation temperature ($$\theta^{\ast}$$) and salinity ($$S^{\ast}$$) fields, the wind stress components ($$\tau_x$$ and $$\tau_y$$), the heat flux ($$Q$$) and the net fresh water flux ($${\cal E} - {\cal P} - {\cal R}$$) used in equations (4.42) - (4.45). The figures also indicate the lateral extent and coastline used in the experiment. Figure 4.38 shows the depth contours of the model domain.

### 4.6.3.2. File input/data¶

Listing 4.25 verification/tutorial_global_oce_oce_in_p/input/data
 1# ====================
2# | Model parameters |
3# ====================
4#
5# Continuous equation parameters
6 &PARM01
7 tRef = 15*20.,
8 sRef = 15*35.,
9 viscAh =3.E5,
10 no_slip_sides=.TRUE.,
11 viscAr =1.721611620915750e5,
12#viscAz =1.67E-3,
13 no_slip_bottom=.FALSE.,
14 diffKhT=1.E3,
15 diffKrT=5.154525811125000e3,
16#diffKzT=0.5E-4,
17 diffKhS=1.E3,
18 diffKrS=5.154525811125000e3,
19#diffKzS=0.5E-4,
20 cosPower=0.5,
21 implicitDiffusion=.TRUE.,
22 ivdc_kappa=1.030905162225000e9,
23#ivdc_kappa=10.0,
24 gravity=9.81,
25 rhoConst=1035.,
26 rhoConstFresh=1000.,
27 buoyancyRelation='OCEANICP',
28 integr_GeoPot=1,
29 eosType='JMD95P',
30 useNHMTerms=.TRUE.,
31 implicitFreeSurface=.TRUE.,
32 exactConserv=.TRUE.,
33 nonlinFreeSurf=4,
34 hFacInf=0.2,
35 hFacSup=2.0,
36#- to check that it has no impact:
37 doResetHFactors=.TRUE.,
38#useRealFreshWaterFlux=.TRUE.,
40 writeBinaryPrec=64,
41 &
42
43# Elliptic solver parameters
44 &PARM02
45 cg2dMaxIters=200,
46 cg2dTargetResidual=1.E-9,
47 &
48
49# Time stepping parameters
50 &PARM03
51 startTime =           0.,
52 endTime   =     3456000.,
53# after 100 years of intergration, one gets a reasonable flow field
54#endTime   = 3110400000.,
55 deltaTMom      =   1200.0,
56 deltaTtracer   = 172800.0,
57 deltaTFreeSurf = 172800.0,
58 deltaTClock    = 172800.0,
59 abEps     = 0.1,
60 pChkptFreq  = 3110400000.,
61 dumpFreq    = 3110400000.,
62 taveFreq    = 3110400000.,
63 monitorFreq =   1.,
64 periodicExternalForcing=.TRUE.,
65 externForcingPeriod=2592000.,
66 externForcingCycle=31104000.,
67 tauThetaClimRelax=5184000.0,
68 &
69
70# Gridding parameters
71 &PARM04
72 usingSphericalPolarGrid=.TRUE.,
73 dxSpacing=4.,
74 dySpacing=4.,
75#Ro_SeaLevel=53023122.566084,
76 top_Pres=1.E-6,
77 delR=7103300.720021, 6570548.440790, 6041670.010249,
78      5516436.666057, 4994602.034410, 4475903.435290,
79      3960063.245801, 3446790.312651, 2935781.405664,
80      2426722.705046, 1919291.315988, 1413156.804970,
81      1008846.750166,  705919.025481,  504089.693499,
82 ygOrigin=-80.,
83 &
84
85# Input datasets
86 &PARM05
87 bathyFile      ='topog.bin',
88 geoPotAnomFile ='deltageopotjmd95.bin',
89 hydrogThetaFile='lev_t.bin',
90 hydrogSaltFile ='lev_s.bin',
91 zonalWindFile  ='trenberth_taux.bin',
92 meridWindFile  ='trenberth_tauy.bin',
93 thetaClimFile  ='lev_sst.bin',
94#saltClimFile   ='lev_sss.bin',
95 surfQnetFile   ='shi_qnet.bin',
96 EmPmRFile      ='shi_empmr.bin',
97 &

This file specifies the main parameters for the experiment. The parameters that are significant for this configuration are

• Line 9–10,

viscAh=3.E5,
no_slip_sides=.TRUE.

these lines set the horizontal Laplacian frictional dissipation coefficient to $$3 \times 10^{5}$$ m2 s-1 and specify a no-slip boundary condition for this operator, i.e., $$u=0$$ along boundaries in $$y$$ and $$v=0$$ along boundaries in $$x$$.

• Lines 11-13,

 viscAr =1.721611620915750e5,
#viscAz =1.67E-3,
no_slip_bottom=.FALSE.,

These lines set the vertical Laplacian frictional dissipation coefficient to $$1.721611620915750 \times 10^{5}$$ Pa2 s-1, which corresponds to $$1.67\times10^{-3}$$ m2 s-1 in the commented line, and specify a free slip boundary condition for this operator, i.e., $$\frac{\partial u}{\partial p}=\frac{\partial v}{\partial p}=0$$ at $$p=p_{b}^{0}$$, where $$p_{b}^{0}$$ is the local bottom pressure of the domain at rest. Note that the factor $$(g\rho)^2$$ needs to be included in this value.

• Line 14,

diffKhT=1.E3,

this line sets the horizontal diffusion coefficient for temperature to 1000 m2 s-1. The boundary condition on this operator is $$\frac{\partial}{\partial x}=\frac{\partial}{\partial y}=0$$ on all boundaries.

• Line 15–16,

 diffKrT=5.154525811125000e3,
#diffKzT=0.5E-4,

this line sets the vertical diffusion coefficient for temperature to $$5.154525811125 \times 10^{3}$$ Pa2 s-1, which corresponds to $$5\times10^{-4}$$ m2 s-1 in the commented line. Note that the factor $$(g\rho)^2$$ needs to be included in this value. The boundary condition on this operator is $$\frac{\partial}{\partial p}=0$$ at both the upper and lower boundaries.

• Line 17–19,

 diffKhS=1.E3,
diffKrS=5.154525811125000e3,
#diffKzS=0.5E-4,

These lines set the diffusion coefficients for salinity to the same value as for temperature.

• Line 21–23,

 implicitDiffusion=.TRUE.,
ivdc_kappa=1.030905162225000E9,
#ivdc_kappa=10.0,

Select implicit diffusion as a convection scheme and set coefficient for implicit vertical diffusion to $$1.030905162225\times10^{9}$$ Pa2 s-1, which corresponds to 10 m2 s-1.

• Line 24,

gravity=9.81,

This line sets the gravitational acceleration coefficient to 9.81 m s-1.

• Line 25,

rhoConst=1035.,

sets the reference density of sea water to 1035 kg m-3.

• Line 29,

eosType='JMD95P',

Selects the full equation of state according to Jackett and McDougall (1995) [JM95]. Note that the only other sensible choice here could be the equation of state by McDougall et al. (2003) [MJWF03], MDJFW. Other model choices for equations of state do not make sense in this configuration.

• Line 28-29,

implicitFreeSurface=.TRUE.,

Selects the barotropic pressure equation to be the implicit free surface formulation.

• Line 32,

exactConserv=.TRUE.,

Select a more accurate conservation of properties at the surface layer by including the horizontal velocity divergence to update the free surface.

• Line 33–35

nonlinFreeSurf=3,
hFacInf=0.2,
hFacSup=2.0,

Select the nonlinear free surface formulation and set lower and upper limits for the free surface excursions.

• Line 39-40,

readBinaryPrec=64,
writeBinaryPrec=64,

Sets format for reading binary input datasets and writing binary output datasets containing model fields to use 64-bit representation for floating-point numbers.

• Line 45,

cg2dMaxIters=200,

Sets maximum number of iterations the 2-D conjugate gradient solver will use, irrespective of convergence criteria being met.

• Line 46,

cg2dTargetResidual=1.E-13,

Sets the tolerance which the 2-D conjugate gradient solver will use to test for convergence in (2.15) to $$1 \times 10^{-9}$$. Solver will iterate until tolerance falls below this value or until the maximum number of solver iterations is reached.

• Line 51,

startTime=0,

Sets the starting time for the model internal time counter. When set to non-zero, this option implicitly requests a checkpoint file be read for initial state. By default the checkpoint file is named according to the integer number of time steps in the startTime value. The internal time counter works in seconds.

• Line 52–54,

endTime=8640000.,
# after 100 years of intergration, one gets a reasonable flow field
#endTime=3110400000.,

Sets the time (in seconds) at which this simulation will terminate. At the end of a simulation a checkpoint file is automatically written so that a numerical experiment can consist of multiple stages. The commented out setting for endTime is for a 100 year simulation.

• Line 55–57,

deltaTmom      =   1200.0,
deltaTtracer   = 172800.0,
deltaTfreesurf = 172800.0,

Sets the timestep $$\delta t_{v}$$ used in the momentum equations to 20 minutes and the timesteps $$\delta t_{\theta}$$ in the tracer equations and $$\delta t_{\eta}$$ in the implicit free surface equation to 48 hours. See Section 2.2.

• Line 60,

pChkptFreq  =3110400000.,

write a pickup file every 100 years of integration.

• Line 61-63,

dumpFreq    = 3110400000.,
taveFreq    = 3110400000.,
monitorFreq =   1.,

write model output and time-averaged model output every 100 years, and monitor statistics every model time step (this is set for testing purposes; change to a larger number for long integrations).

• Line 64–66,

periodicExternalForcing=.TRUE.,
externForcingPeriod=2592000.,
externForcingCycle=31104000.,

Allow periodic external forcing: set one month forcing period during which a single time slice of data is valid, and the repeat cycle to one year. Thus, external forcing files will contain twelve periods of forcing data.

• Line 67,

tauThetaClimRelax=5184000.0,

Set the restoring timescale to 2 months.

• Line 59,

abEps=0.1,

• Line 72,

usingSphericalPolarGrid=.TRUE.,

Select spherical grid.

• Line 73–74,

dXspacing=4.,
dYspacing=4.,

Set the horizontal grid spacing in degrees spherical distance.

• Line 77–81,

delR=7103300.720021, ...

set the layer thickness in pressure units, starting with the bottom layer.

• Line 87–96,

bathyFile='topog.box'
hydrogThetaFile='lev_t.bin',
hydrogSaltFile ='lev_s.bin',
zonalWindFile  ='trenberth_taux.bin',
meridWindFile  ='trenberth_tauy.bin',
thetaClimFile  ='lev_sst.bin',
surfQFile      ='shi_qnet.bin',
EmPmRFile      ='shi_empmr.bin',

These lines specify the names of the files holding the bathymetry data set, the time-independent geopotential height anomaly at the bottom, initial conditions of temperature and salinity, wind stress forcing fields, sea surface temperature climatology, heat flux, and fresh water flux (evaporation minus precipitation minus runoff) at the surface. See file descriptions in section Section 4.6.3.

Other lines in the file input/data are standard values that are described in the Section 3.8.

### 4.6.3.3. File input/data.pkg¶

This file uses standard default values and does not contain customizations for this experiment.

### 4.6.3.4. File input/eedata¶

This file uses standard default values and does not contain customizations for this experiment.

### 4.6.3.5. File input/topog.bin¶

This file is a 2-D ($$x,y$$) map of depths. This file is assumed to contain 64-bit binary numbers giving the depth of the model at each grid cell, ordered with the $$x$$ coordinate varying fastest. The points are ordered from low coordinate to high coordinate for both axes. The units and orientation of the depths in this file are the same as used in the MITgcm code (Pa for this experiment). In this experiment, a depth of 0 Pa indicates a land point (wall) and a depth of >0 Pa indicates open ocean.

### 4.6.3.6. File input/deltageopotjmd95.box¶

The file contains twelve identical 2-D maps ($$x,y$$) of geopotential height anomaly at the bottom at rest. The values have been obtained by vertically integrating the hydrostatic equation with the initial density field (using input/lev_t.bin and input/lev_s.bin). This file has to be consistent with the temperature and salinity field at rest and the choice of equation of state!

### 4.6.3.7. Files input/lev_t.bin and input/lev_s.bin¶

The files input/lev_t.bin and input/lev_s.bin specify the initial conditions for temperature and salinity for every grid point in a 3-D array ($$x,y,z$$). The data are obtain by interpolating monthly mean values using Levitus and Boyer (1994a,b) [LB94a, LB94b] for January onto the model grid. Keep in mind that the first index corresponds to the bottom layer and highest index to the surface layer.

### 4.6.3.8. Files input/trenberth_taux.bin and input/trenberth_tauy.bin¶

The files input/trenberth_taux.bin and input/trenberth_tauy.bin contain twelve 2-D ($$x,y$$) maps of zonal and meridional wind stress values, $$\tau_{x}$$ and $$\tau_{y}$$, respectively, in 3-D arrays ($$x,y,t$$). These are monthly mean values from Trenberth et al. (1990) [TOL90], units of N m-2.

### 4.6.3.9. File input/lev_sst.bin¶

The file input/lev_sst.bin contains twelve monthly surface temperature climatologies from Levitus and Boyer (1994b) [LB94b] in a 3-D arrays ($$x,y,t$$).

### 4.6.3.10. Files input/shi_qnet.bin and input/shi_empmr.bin¶

The files input/shi_qnet.bin and input/shi_empmr.bin contain twelve monthly surface fluxes of heat (qnet) and freshwater (empmr) from Jiang et al. (1999) [JSMR99] in 3-D arrays ($$x,y,t$$). Both fluxes are normalized so that the total forcing over one year results in no net flux into the ocean (note, the freshwater flux is actually constant in time).

### 4.6.3.11. File code/SIZE.h¶

The file code/SIZE.h is identical to that described in tutorial global ocean simulation, for more specifics see Section 4.5.3.7.

### 4.6.3.12. File code/CPP_OPTIONS.h¶

This file uses standard default values except for:

• #define ATMOSPHERIC_LOADING

enables pressure loading which is abused to include the initial geopotential height anomaly.

• #define EXACT_CONSERV

enables more accurate conservation properties to include the horizontal mass divergence in the free surface.

• #define NONLIN_FRSURF

enables the nonlinear free surface.


text:
# KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.2

KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangle Ex 2.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangle Exercise 2.2.

## Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangle Exercise 2.2

Exercise 2.2 Class 10 Maths Triangles Question 1.
In the following fig. (i) and (ii). DE || BC.
Find EC in (i) and AD in (ii).

Solution:
(i) In ∆ABC, DE || BC, EC =?
$$\quad \frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{AE}}{\mathrm{EC}}$$
$$\frac{1.5}{3}=\frac{1}{\mathrm{EC}}$$
1.5 EC = 1 × 3
$$\mathrm{EC}=\frac{3}{1.5}$$
$$E C=\frac{30}{15}$$
∴ EC = 2 cm

(i) In ∆ABC, DE || BC, AD =?
$$\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{AE}}{\mathrm{EC}}$$
$$\frac{\mathrm{AD}}{7.2}=\frac{1.8}{5.4}$$
$$\quad A D=\frac{1.8}{5.4} \times \frac{7.2}{1}$$
$$=\frac{1.8}{54} \times \frac{72}{1}$$
= 0.6 × 4
∴ AD = 2.4 cm

KSEEB Solutions For Class 10 Maths Triangles Question 2.
E and F are points on the sides PQ and PR respectively of a PQR For each of the following cases, state whether EF || QR:
(i) PE = 3.9cm. EQ = 3 cm
PF=3.6cm. FR = 2.4 cm.
(ii) PE = 4 cm. QE = 4.5 cm.
PF = 8 cm. RF = 9 cm.
(iii) PQ = 1.28 cm. PR = 2.56 cm.
PE = 0.18 cm. PF = 0.36 cm.
Solution:

In ∆PQR, if EF || QR then,
$$\frac{\mathrm{PE}}{\mathrm{QR}}=\frac{\mathrm{PF}}{\mathrm{FR}}$$
$$\frac{3.9}{3}=\frac{3.6}{2.4}$$
1.3 ≠ 1.5
Sides are not dividing In ratio.
∴ EF is not parallel to QR.

(ii)

In ∆PQR, if EF || QR then
$$\frac{\mathrm{PE}}{\mathrm{QR}}=\frac{\mathrm{PF}}{\mathrm{FR}}$$
$$\frac{4}{4.5}=\frac{8}{9}$$
$$\frac{8}{9}=\frac{8}{9}$$
Here sides are dividing in ratio.
∴ EF ||QR

(iii)

PE + EQ = PQ
0.18 + EQ = 1.28
∴ EQ=1.28 – 0.18
EQ = 1.1 cm.
Similarly. PF + FR = PR
0.36 + FR = 2.56
FR = 2.56 – 0.36
FR = 2.2cm.
In ∆PQR, if EF || QR then
$$\frac{\mathrm{PE}}{\mathrm{QE}}=\frac{\mathrm{PF}}{\mathrm{FR}}$$
$$\frac{0.18}{1.1}=\frac{0.36}{2.2}$$
$$\frac{1.8}{11}=\frac{3.6}{22}$$
$$\frac{1.8}{11}=\frac{11.8}{11}$$
Here sides are dividing in ratio.
∴ EF || QR

10th Maths Triangles Exercise 2.2 Question 3.
In the following figure. If LM || CB and LN || CD, prove that $$\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}$$

Solution:
Data: LM || CB and LN || CD then prove that $$\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}$$
Solution: In ∆ACB, LM || CB
$$\quad \frac{\mathrm{AL}}{\mathrm{AC}}=\frac{\mathrm{AM}}{\mathrm{AB}} \quad \ldots \ldots(1)(\text { Theorem } 1)$$
Similarly in ∆ADC, LN || CD
$$\quad \frac{\mathrm{AL}}{\mathrm{AC}}=\frac{\mathrm{AN}}{\mathrm{AD}} \quad \ldots \ldots(2) \quad(\ldots \text { Theorem } 1)$$
From equation (1) and (2) we have
$$\frac{\mathrm{AL}}{\mathrm{AC}}=\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}$$
$$\quad \frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}$$

Triangles Class 10 Exercise 2.2 Question 4.
In the following figure, DE ||AC and DF || AE. Prove that $$\frac{\mathrm{BF}}{\mathrm{FE}}=\frac{\mathrm{BE}}{\mathrm{EC}}$$

Solution:
Data: In this figure, AE || AC and DF || AE, then we have to prove that $$\frac{\mathrm{BF}}{\mathrm{FE}}=\frac{\mathrm{BE}}{\mathrm{EC}}$$
Solution: In ∆ABC, DE || AC.
$$\quad \frac{\mathrm{BD}}{\mathrm{AD}}=\frac{\mathrm{BE}}{\mathrm{EC}} \ldots \ldots(1) \quad(\text { Theorem } 1)$$
Similarly, In ∆ABE, DF ||AE.
$$\quad \frac{\mathrm{BD}}{\mathrm{AD}}=\frac{\mathrm{BF}}{\mathrm{FE}}$$
from equation (1) and (2). we have
$$\frac{\mathrm{BD}}{\mathrm{AD}}=\frac{\mathrm{BE}}{\mathrm{EC}}=\frac{\mathrm{BF}}{\mathrm{FE}}$$
$$\quad \frac{\mathrm{BE}}{\mathrm{ED}}=\frac{\mathrm{BF}}{\mathrm{FE}}$$

Triangle Lesson Exercise 2.2 Question 5.
In the following figure. DE || OQ and DF || OR. Show that EF || QR.

Solution:
Data: In this figure. DE || OQ and DF || OR.
To Prove: EF || QR
Solution: In ∆POQ, DE || OQ.
$$\quad \frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PD}}{\mathrm{DO}} \quad \ldots \ldots \text { (i) }(\text { Theroem } 1)$$
Similarly. In ∆POR. DF || OR.
$$\quad \frac{\mathrm{PD}}{\mathrm{DO}}=\frac{\mathrm{PF}}{\mathrm{FR}} \quad \ldots \ldots \text { (ii) }(\text { Theorem } 1)$$
from equation (1) and (11), we have
$$\frac{P F}{E Q}=\frac{P D}{D O}=\frac{P F}{F R}$$
$$\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PF}}{\mathrm{FR}}$$
In ∆PQR. if $$\frac{P E}{E Q}=\frac{P F}{F R}$$ then EF || QK. (∵ Theorem 1).

Triangles Class 10 Solutions KSEEB Question 6.
In the following figure, A, B an C are points on OP. OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

Solution:
Data: In ∆QPR, AB || PQ and AC || PR. A, B, and C are the points on OP. OQ and OR.
To Prove: BC || QR
Solution: In ∆OPQ, AB || PQ.
$$\quad \frac{\mathrm{OA}}{\mathrm{AP}}=\frac{\mathrm{OB}}{\mathrm{BQ}} \quad \ldots \ldots \text { (i) }(\text { Theorem } 1)$$
Similarly, In ∆OPR. AC|| PR.
$$\quad \frac{\mathrm{OA}}{\mathrm{AP}}=\frac{\mathrm{OC}}{\mathrm{CR}} \quad \ldots \ldots \text { (i) }(\text { Theorem } 1)$$
from equation (i) and (ii), we have
$$\frac{\mathrm{OA}}{\mathrm{AP}}=\frac{\mathrm{OB}}{\mathrm{BQ}}=\frac{\mathrm{OC}}{\mathrm{CR}}$$
$$\quad \frac{\mathrm{OB}}{\mathrm{BQ}}=\frac{\mathrm{OC}}{\mathrm{CR}}$$
∴ BC || QR (∵ Theorem 2)

KSEEB Solutions For Class 10 Maths Question 7.
Using Theorem 2.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Solution:
Data: In ∆ABC. D is the mid-point of AB.
DE Is drawn parallel to BC from D.
To Prove: DE bisects AC side at E.
Solution: In ∆ADE and ∆ABC,
∠D = ∠B (corresponding angles)
∠E = ∠C (corresponding angles)
$$\quad \frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}$$
$$\frac{1}{2}=\frac{\mathrm{AE}}{\mathrm{AC}}$$
∴ AC = AE + EC
∴ E is the mid-point of AC
∴ DE bisects AC at E.

Triangles Class 10 KSEEB Solutions Question 8.
Using Theorem 2.2, prove that the line joining the mid-points of any two sides of a triangle Is parallel to the third side. (Recall that you have done it in Class IX)

Solution:
Solution: In ∆ABC, D and E are mid-points of AB and AC.
∴ AD = DB
AE = EC
AC = 2AE
$$\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{1}{2}$$
As per S.S.S. Postulate.
∴ They are equiangular triangles.
∴ ∠A is common.
These are pair of corresponding angles
∴ DE || BC

Triangles Class 10 Exercise 2.2 Solutions Question 9.
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that $$\frac{\mathrm{AO}}{\mathrm{CO}}=\frac{\mathrm{CO}}{\mathrm{DO}}$$
Solution:
Data : ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O.
To Prove: $$\frac{\mathrm{AO}}{\mathrm{CO}}=\frac{\mathrm{CO}}{\mathrm{DO}}$$
Proof: In Trapezium ABCD. AB || DC.
∴ In ∆AOB and ∆DOC,
∠OCO = ∠OAB (alternate angles)

∠ODC = ∠OBA (alternate angles)
∠DOC = ∠AOB (vertically opposite angles)
∴ ∆AOB and ∆DOC are equiangular triangles.
∴ ∆AOB ||| ∆DOC
Similar triangles divides sides in ratio.
$$\frac{\mathrm{AO}}{\mathrm{CO}}=\frac{\mathrm{CO}}{\mathrm{DO}}$$

10th Maths Triangles Exercise 2.2 Solutions Question 10.
The diagonals of a quadrilateral ABCD intersect each other at the point O such that $$\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}$$ . Show that ABCD is a trapezium.

Solution:
Data: In the quadrilateral ABCD. the diagonals intersect at 0’ such that $$\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}$$
To Prove: ABCD is a trapezium.
Solution: In the qudrilateral ABCD, $$\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}$$
$$\quad \frac{\mathrm{AO}}{\mathrm{CO}}=\frac{\mathrm{OB}}{\mathrm{OD}}$$
It means sides of ∆AOB and ∆DOC divides proportionately.
∴ ∆AOB ||| ∆DOC.
Similarly. ∆AOD ||| ∆BOC.
Now, ∆AOB + ∆AOD = ∆BOC + ∆DOC
∆ABD = ∆ABC.
Both triangles are on the same base AB and between two pair of lines and equal in area.
∴ AB || DC.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangle Ex 2.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangle Exercise 2.2, drop a comment below and we will get back to you at the earliest.

format: plain


text:
# How should I understand path-space measurement equation?

I read the corresponding chapter of PBRT-v3, as well as Veach's thesis which introduced BDPT. However I didn't quite understand why we use the measurement equation. It seems that the measurement equation can be further formulated to obtain different light transport algorithms, such as photon map and bdpt. But why?

I've searched for other literature but didn't find anything really useful. Can anyone give some hint?

• Can you include the actual equation you're asking about, and maybe some more background in your question for the benefit of people who may not have those references handy? – Nathan Reed Jan 22 '18 at 18:52


text:
# Tag Info

30

I will not provide an in-depth comparison, as I only have limited experience with the different plugins. I know a couple of plugins well (LaTeX-Box and LaTeX-Suite), and I know one plugin very well, since I am developing it myself: vimtex. Thus, I will mostly write about vimtex. However, I will first point out some references that might be of interest to ...

12

I can't do comparison, as Vim-LaTeX is the only LaTeX plugin I've used. I have been using Vim-LaTeX for almost a year. So I will talk about Vim-LaTeX alone. There are many features present in Vim-LaTeX. I don't remember all of them. I'll just talk about features that I know and use constantly. Note: These are my limited user experience, which may be very ...

6

Honza's UltiSnips Snippet gentbl<number>x<number> available in the repository https://github.com/honza/vim-snippets/blob/master/UltiSnips/tex.snippets Using this snippet which uses the python interpolation feature of UltiSnips, you can enter e.g. gentbl5x3<tab> which will expand to \begin{tabular}{||||} & & \\ & &...

5

I can not answer your question regarding vim-latex (aka LaTeX-Suite), however, I can recommend two plugins that provide good label and bibliography completion: vimtex Vimtex is a simple and lightweight LaTeX plugin for vim that is initially based on LaTeX-Box. The code is now very much diverged from LaTeX-Box, and the plugin is IMHO better maintained and ...

5

The reason this is happening is because the indentexpr [1] prefers text to be indented inside \items. You can see this by pressing gg=G. This is what it considers correct indenting: \begin{document} \begin{itemize} \item Here is an equation: \begin{align*} x = y \end{align*} \item Here is another item \end{itemize} \end{...

4

You could probably implement this as a snippet by leveraging Ultisnips's Interpolation feature (VimCasts demo). As an alternative, I whipped up a simplistic function to generate a matrix template on the current line per your sample: function! CreateMatrix(rows, ...) abort let cols = a:0 ? a:1 : 3 let matrix = ['\begin{bmatrix}'] call extend(matrix, ...

4

I'll give an answer, but not the one you are looking for. Hopefully, you will find it even better. Note that it does rely on you installing a plugin. First, what you are trying to do is to simplify the process of creating a new theorem environment. You want to apply a template and to fill out some parts automatically. You are in luck, because this is the ...

4

If you use a nore mapping to tie a sequence of keys to <plug>something, even if there is a mapping behind that <plug>something, it'll be ignored. In the documentation, nore is said to prevent recursion. That's what is happening here. That means that: the <plug>something mapping shall indeed be defined with noremap, but the actual ...

4

Installation into a Vim 8 packpath and an example vimrc ❯ git clone https://github.com/vim-latex/vim-latex ~/.vim/pack/my_latex_plugins/start/vim-latex Typically, you want at least following settings in your vimrc ❯ cat ~/.vim/vimrc filetype plugin on filetype indent on let g:tex_flavor='latex' A minimal installation would look like visualized with tree: ...

4

While infuriating, a number of vim-latex's features can be customised without editing the source code. Disable the EEA mapping In .vim/after/ftplugin/tex.vim, add call IMAP('EEA', 'EEA', 'tex') Typing EEA in insert mode will no longer insert the eqnarray environment. F5 prompted environments In .vim/ftplugin/tex/texrc, add: let g:...

3

This is a question that is difficult to write a very rigorous answer to. But I find it easy to state that the prospects are not quite good. At least not if you try to include the corner cases. As you say, you must identify periods and punctuations, but you should also parse many macros, such as \emph{...}, and you should parse stuff like some text {\macro ...

3

Your problem has nothing to do with jedi-vim. It has to do with vim-latex using the conceal feature of gVim, where some parts of text are replaced by (supposedly better) graphic representations when you aren't editing the corresponding lines. In your case a superscript is replaced, but IIRC vim-latex does that with many other things, such as Greek letters ...

3

Google is your friend. Specifically, Google's cache: Make sure that you create a few necessary settings in your ~/.vimrc. " REQUIRED. This makes vim invoke Latex-Suite when you open a tex file. filetype plugin on " IMPORTANT: win32 users will need to have 'shellslash' set so that latex " can be called correctly. set shellslash " IMPORTANT: grep will ...

3

This functionality comes from the internal LaTeX plugin. See :h tex-conceal and :h g:tex_conceal. You can turn it off either by setting conceallevel=0 or by letting g:tex_conceal=''. To be more precise, you can use either of the following lines in your vimrc file: let g:tex_conceal = '' set conceallevel = 0 I propose that you read :h conceallevel before ...

3

This is actually a complicated matter. $...$ is a difficult construct to parse, because it has no clear beginning or end. That is, in math here? $or math here?$ math here?, how do you know which region is really a math region? You need to search further back to know. However, there are some possibilities. In particular, I would take advantage of the fact ...

3

This is a part of standard filetype plugin, see :h tex.vim for a complete info. That particular feature is controlled by :h g:tex_conceal So it's enough to add let g:tex_conceal = '' to your vimrc (or to disable only ligatures and accents: let g:tex_conceal = 'dmgs', i.e. no 'a' in tex_conceal's value).

2

I took time to install vim-latexsuite and I can confirm that my comment was right: you simply have to execute :set iskeyword+=. to make completion work. To avoid executing this function on each latex file you edit you can add the following line setlocal iskeyword+=. To the syntax file .vim/ftplugin/tex.vim

2

vim-latexsuite is a very good addition for writing LaTeX documents and will handle spell checking as you want it to. Meaning, ignore equasions in spell checking and it also does a lot of other great stuff.

2

Your updated findings give it away: The Latex plugin (incorrectly, that's why :inoremap causes the addition of a *) parses the output of the :iab command in s:LookupCharacter(char) from plugin/imaps.vim: " An extremeley wierd way to get around the fact that vim " doesn't have the equivalent of the :mapcheck() function for ...

2

I am not quite sure, however there is something strange with your system() call. You do: let retval = system(latex_path."/outline.py". shellescape(fname) . shellescape (leader) Note, that you do concat the filename to the python script without spaces, so you call ~/vim/ftplugin/latex-suite/outline.pymyfile.texleader which most likely does not make sense. ...

2

This is a shell command run with :silent! !okular [..] in the Tex_ForwardSearchLaTeX() function (ftplugin/latex-suite/compiler.vim). It looks like okular is printing some text to stdout or stderr – the silent! command only silences Vim :echo and :echoerr messages, not stdout from external programs. To fix this you'll need to add something like >/dev/...

2

Disclaimer: I'm the author of vimtex. You should not use both vimtex and vim-latex at the same time. They are filetype plugins that try to implement the same set of features, at least partially. For some more information about the difference between the plugins, you can read some of the answers here: What are the differences between LaTeX plugins? That ...

2

This seems to be a known issue, see the comment in ftplugin/latex-suite/envmacros.vim: " Only include the <M-i> mapping if the user want this. Note that it " conflicts with inserting 'é'. if !hasmapto("<Plug>Tex_InsertItemOnThisLine", "i") && g:Tex_AdvancedMath == 1 imap <buffer> <M-i> <Plug>Tex_InsertItemOnThisLine ...

2

When you read the error message you can see that the problem comes from ~/.vim/bundle/LaTeX-Suite-aka-Vim-LaTeX/plugin/remoteOpen.vim on line 33. This line is the following: com -nargs=1 RemoteOpen :call RemoteOpen('<args>') It uses :h :command to create the user defined command :RemoteOpen. What Vim says is Command already exists: add ! to replace ...

2

Your problem is that you are using inoremap <C-A> <Plug>InsertComment That basically means, that recursive resolution of your mapped keys is prevented which is usually not what you want. Therefore change your mapping to imap <C-A> <Plug>InsertComment See also the extensive help (Start reading from the paragraph that starts with ...

2

All you have to do to install vim-latex manually is download the tarball and extract it in your vimfiles or .vim directory. The .vimrc and the tex.vim files will change some latex settings, but do not effect installation. vim-latex recommends using a plugin manager like pathogen or Vundle to install vim-latex. You can install pathogen and then enter cd ~...

2

I have written a code that is pretty similar to the :only command. But it closes also unsaved buffers. It keeps the only the active window in the current tab. function! CloseAllWindowsButCurrent() let tabnr= tabpagenr() let tabinfo=gettabinfo(tabnr) let windows=tabinfo[0]['windows'] for winid in windows let curwin=winnr() "could change let ...

1

I checked with a minimal example the proposed solution from https://stackoverflow.com/q/16711143 and they work Put the following into ~/.vim/after/ftplugin/tex.vim: setlocal foldlevel=999 Alternatively, you could define in \$MYVIMRC: augroup latex autocmd! autocmd Filetype tex setlocal foldlevel=999 augroup END Question missed mentioning of plugin ...

1

I figured it out, I added a tex.vim file in the ~/../vim-latex/ftplugin directory, the text of the file is: call IMAP("'",'','tex') call IMAP('',"'",'tex') Where the quotes around ' are necessary since it's a special character Edit: now every 3 ' I get ' someone knwos why?

1

In this case, read vim-latex FAQ. The problem is old, and there is a solution, that enables to map <m-i> (which vim confounds with é) to any thing else. PS: your mappings should be <buffer> local. You don't want them triggered in commit messages for instance. BTW, I've been using accentuated characters in my French LaTeX documents for quite ...

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


text:
Find the realand complex solutions to the equation X3(cubed) + 1=0

my choices

a. -1,1,cis(pi/3)
b. 1, cis(pi/3), cis (2pi/3)
c. -1, cis(pi/3), cis(5pi/3)
d. cis (pi/3), cis(2pi/3), cis(4pi/3)

2. Originally Posted by ccdalamp
Find the realand complex solutions to the equation X3(cubed) + 1=0

my choices

a. -1,1,cis(pi/3)
b. 1, cis(pi/3), cis (2pi/3)
c. -1, cis(pi/3), cis(5pi/3)
d. cis (pi/3), cis(2pi/3), cis(4pi/3)
$x^3=-1= cis((2n+1) \pi )\ \ \ n=0,1,2, ...$

so

$
x=cis\left(\frac{2n+1}{3}\pi \right)\ \ \ n=0,1,2, ...
$

which if I am not mistaken is answer (c)

RonL

3. Hello, ccdalamp!

With "cis" in the answer choices, I assume that DeMoivre's Formula is expected.
But it can be solved algebraically . . . with some trig.

Find the real and complex solutions to the equation: . $x^3 + 1\:=\:0$

$a)\;\text{-}1,\:1,\:\text{cis}\left(\frac{\pi}{3}\right)\qqua d\qquad\qquad b)\;1,\:\text{cis}\left(\frac{\pi}{3}\right),\:\te xt{cis}\left(\frac{2\pi}{3}\right)$
$c)\;\text{-}1,\:\text{cis}\left(\frac{\pi}{3}\right),\:\text{ cis}\left(\frac{5\pi}{3}\right)\qquad d)\;\text{cis}\left(\frac{\pi}{3}\right),\:\text{c is}\left(\frac{2\pi}{3}\right),\:\text{cis}\left(\ frac{4\pi}{3}\right)$

Factor: . $(x + 1)(x^2 - x + 1) \:=\:0$

And we have: . $\begin{array}{ccccccc}x + 1 & = & 0 & \;\;\Rightarrow\;\; & x & = & -1 \\
x^2-x+1 & = & 0 & \Rightarrow & x & = & \frac{1\pm i\sqrt{3}}{2} \end{array}$

And the complex roots can be rewritten: . $\begin{array}{ccc}\frac{1}{2} + \frac{\sqrt{3}}{2}i & \Rightarrow & \cos\left(\frac{\pi}{3}\right) + i\sin\left(\frac{\pi}{3}\right) \\
\frac{1}{2} - \frac{\sqrt{3}}{2}i & \Rightarrow & \cos\left(\frac{5\pi}{3}\right) + i\sin\left(\frac{5\pi}{3}\right) \end{array}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

On the other hand, there is an "eyeball" solution.

A cubic equation has three roots, of course.
Either all three are real or there is one real and two complex conjugates.
. . Only (c) meets this requirement.

Moreover, we already know that: . $\sqrt[3]{\text{-}1} = \text{-}1$

Therefore, (c) is the only candidate.


text:
Heteroscedastic model with variance as function of mean magnitude

This isn’t something I’m specifically trying to do with Stan but I’m trying to formulate it as a Bayesian model, which may then later need MCMC.

I have a model where I have observations d_{i,j} where d_i is a 3 dimensional vector. I have a forwards model which can produce modeled data m_{i,j}. Unfortunately, there are multiple sources of noise that produce the observations, one which is likely heteroscedastic and scales with |m_i| and another which is likely constant. I think a simple model I could use to describe this would be something like:

d_{i,j}\sim normal(m_{i,j},(\omega|m_i|)^2+\delta^2)

where \delta and \omega are unknown.

Alternatively, it might be simpler to model the variance as a single parameter \sigma_i^2 where p(\sigma_i)=f(|d_i|) and f is a distribution that has a mean with the properties defined above (tends to a constant at low magnitude, order 1 at high magnitude).

Because the variance is effectively a nuisance parameter, it would also be nice to have some simple/complementary form to the prior so that the variance can be integrated out in the posterior, reducing the number of parameters that have to be computed. I haven’t been able to find much literature on this particular kind of problem which does this- although it’s a frequently used trick with homoscedastic problems.


text:
# pharo-vcs/iceberg

Iceberg is a set of tools that allow one to handle git repositories directly from a Pharo image. Right now we support only git, but Iceberg is designed to allow other code versioning systems in the future. The final aim of Iceberg is to become the default repository manager for Pharo-core, allowing for smoother and faster integration of contribu…
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 Failed to load latest commit information. BaselineOfIceberg.package Aug 13, 2018 Iceberg-ArchitecturalRules.package May 2, 2018 Iceberg-GitCommand.package Dec 14, 2016 Iceberg-Libgit-Filetree.package Jul 17, 2018 Iceberg-Libgit-Tonel.package Jun 27, 2018 Iceberg-Libgit.package Aug 10, 2018 Iceberg-Memory.package Aug 2, 2018 Iceberg-Metacello-Integration.package Aug 7, 2018 Iceberg-Pharo6.package May 3, 2018 Iceberg-Plugin-GitHub.package Aug 10, 2018 Iceberg-Plugin-Metacello.package Aug 1, 2018 Iceberg-Plugin-Migration-Tests.package Aug 2, 2018 Iceberg-Plugin-Migration.package Jul 17, 2018 Iceberg-Plugin-Pharo.package Aug 3, 2018 Iceberg-Plugin.package Jun 5, 2018 Iceberg-Tests-MetacelloIntegration.package Jul 17, 2018 Iceberg-Tests.package Aug 13, 2018 Iceberg-TipUI.package Aug 10, 2018 Iceberg-UI-Tests.package Jun 28, 2018 Iceberg.package Aug 13, 2018 docs Jul 7, 2018 scripts Aug 3, 2018 .filetree Apr 19, 2016 .project Jun 14, 2018 .properties Jun 14, 2018 .smalltalk.metacellotests.ston Jul 31, 2018 .smalltalk.ston Jun 27, 2018 .travis.yml Aug 6, 2018 CONTRIBUTING.md May 24, 2018 LICENSE Apr 28, 2018 README.md Aug 7, 2018 appveyor.yml May 14, 2018 git_cookbook.md Jul 15, 2018 iceberg_test.key.enc Sep 23, 2016

# Git integration for Pharo

Iceberg is a set of tools that allows one to handle git repositories directly from a Pharo image. Right now we only support git, but Iceberg is designed to allow other code versioning systems in the future. The final aim of Iceberg is to become the default repository manager for Pharo-core, allowing for smoother and faster integration of contributions, as well as better branch and version management.

Please be aware that this is still an experimental tool, so you have to be careful. But we already have a few early adopters, so you are invited to try it and provide feedback.

To better understand Iceberg (or even this documentation), I recommend to read the wiki pages which explain (parts of) the Iceberg and Git nomenclature

## FAQ

Q. The Image seems to freeze when doing a clone of a repository.
A. This is because the operation is taking time and Iceberg still does not show feedback properly. You just need to be patient :)

Q. The Image freezes when cloning/commiting and there is nothing I can do to fix it. I used HTTPS as the protocol when doing the clone.
A. There is a known bug around HTTPS and getting the credentials. We will fix this, but while waiting for the fix, you can work around the problem by:

• Adding your credentials before doing any operation (go to System Settings/Tools/Software Configuration Management/Iceberg/Plain credentials)
• Just using the SSH protocol instead (but that option is not as easy on Windows).

Q. I installed Iceberg and try to clone a project, but I'm getting the error message "LGit_GIT_ERROR no ssh-agent suitable credentials found".
A. Currently Iceberg SSH support just handles SSH/ssh-agent. You can fix this problem in two ways:

• In the command line, run this: ssh-add ~/.ssh/id_rsa (macOS users may prefer to execute this: ssh-add -K ~/.ssh/id_rsa)
• Go to Iceberg settings System Settings/Tools/Software Configuration Management/Iceberg/Use custom SSH keys and add the path to your id_rsa.pub and id_rsa files there.

Q. I'm on Linux and when I want to commit I get the message "LGit_GIT_ERROR: no error message set by libgit2."
A. Iceberg for Pharo6 on Linux requires a different version of the VM than the one that is provided by default (this will be the default on Pharo7). The easiest way to get this VM is to execute curl get.pharo.org/vmT60 | bash.

Q. I'm using a server with a different SSH port and I'm getting a "connection timeout" when I'm trying to clone.
A. You need to use a different URL format than the default (the one that is proposed on most sites). You need a URL that looks like this template: ssh://git@url:port/team/project.git.

Q. I'm using Iceberg on Windows - whilst trying to clone a repository I get the error "LGit_GIT_ERROR: error authenticating: failed connecting agent".
A. Prompting for credentials currently doesn't work on Windows (we can't use ssh-agent). You need to setup authentication using SSH keys. Something like this:

IceCredentialsProvider useCustomSsh: true.
IceCredentialsProvider sshCredentials
publicKey: 'c:\path\to\ssh\id_rsa.pub';
privateKey: 'c:\path\to\ssh\id_rsa'

(Your key should not have a passphrase)

Q. I'm using 2FA on github and when I try to to create a PR from Iceberg, I'm asked a login/passwd. I try to login but it doesn't work. A. There is no support for 2FA in Iceberg for the moment. Please create a personal Access Tokens to replace your password in order to avoid this problem: https://github.com/settings/tokens

## Installation (for development)

### Prerequisites

• Latest Pharo 6.1+ image.
• Pharo VM for Pharo 6.1+.

You can get both by downloading it from the Pharo site or in the command line with zeroconf:

wget -O- get.pharo.org | bash

### Install Iceberg

Since Pharo 6.0, Iceberg is included in the Image, so you probably will need to update rather than install (see below).

### Update Iceberg

Pharo 7.0a comes with the latest stable Iceberg version. To update it, just clone Iceberg into your repo and then pull the changes.

#### For Pharo 6.1

MetacelloPharoPlatform select.
#(
'BaselineOfTonel'
'BaselineOfLibGit'
'BaselineOfIceberg'
'MonticelloTonel-Core'
'MonticelloTonel-FileSystem'
'MonticelloTonel-Tests'
'Iceberg-UI'
'Iceberg-TipUI'
'Iceberg-Plugin-Pharo'
'Iceberg-Plugin-Metacello'
'Iceberg-Plugin-GitHub'
'Iceberg-Plugin'
'Iceberg-Metacello-Integration'
'Iceberg-Libgit-Tonel'
'Iceberg-Libgit-Filetree'
'Iceberg-Libgit'
'Iceberg-Tests'
'Iceberg-Memory'
'Iceberg-UI-Tests'
'Iceberg-Core'
'Iceberg-Changes'
'Iceberg'
'Iceberg-GitCommand'
'Iceberg-SmartUI'
'Iceberg-Pharo6'
'LibGit-Core')
do: [ :each | (each asPackageIfAbsent: [ nil ]) ifNotNil: #removeFromSystem ].
"update icons (iceberg needs some new)"
Metacello new
baseline: 'Iceberg';
repository: 'github://pharo-vcs/iceberg:v1.?';
onWarningLog;
"Re-initialize libgit2"
(Smalltalk at: #LGitLibrary) initialize.

"In some case Pharo/Calypso can have a problem with Obsolete classes. If you encounter this problem just execute this command and retry your action:

Smalltalk compilerClass recompileAll

"

NOTE: you need to update Iceberg in a NEW image, otherwise there will be obsolete repositories around.

## 5 minutes tutorial

(the 5 minutes tutorial is no longer valid for this version, we are working on an updated version)

Have a look at the videos [https://github.com/pharo-vcs/iceberg/tree/master/docs/Contribute-to-Pharo-with-Iceberg-0.7.3]


text:
# Quantities of information

A misleading[1] information diagram showing additive and subtractive relationships among Shannon's basic quantities of information for correlated variables ${\displaystyle X}$ and ${\displaystyle Y}$. The area contained by both circles is the joint entropy ${\displaystyle \mathrm {H} (X,Y)}$. The circle on the left (red and violet) is the individual entropy ${\displaystyle \mathrm {H} (X)}$, with the red being the conditional entropy ${\displaystyle \mathrm {H} (X|Y)}$. The circle on the right (blue and violet) is ${\displaystyle \mathrm {H} (Y)}$, with the blue being ${\displaystyle \mathrm {H} (Y|X)}$. The violet is the mutual information ${\displaystyle \operatorname {I} (X;Y)}$.

The mathematical theory of information is based on probability theory and statistics, and measures information with several quantities of information. The choice of logarithmic base in the following formulae determines the unit of information entropy that is used. The most common unit of information is the bit, based on the binary logarithm. Other units include the nat, based on the natural logarithm, and the hartley, based on the base 10 or common logarithm.

In what follows, an expression of the form ${\displaystyle p\log p\,}$ is considered by convention to be equal to zero whenever ${\displaystyle p}$ is zero. This is justified because ${\displaystyle \lim _{p\rightarrow 0+}p\log p=0}$ for any logarithmic base.

## Self-information

Shannon derived a measure of information content called the self-information or "surprisal" of a message ${\displaystyle m}$:

${\displaystyle \operatorname {I} (m)=\log \left({\frac {1}{p(m)}}\right)=-\log(p(m))\,}$

where ${\displaystyle p(m)=\mathrm {Pr} (M=m)}$ is the probability that message ${\displaystyle m}$ is chosen from all possible choices in the message space ${\displaystyle M}$. The base of the logarithm only affects a scaling factor and, consequently, the units in which the measured information content is expressed. If the logarithm is base 2, the measure of information is expressed in units of bits.

Information is transferred from a source to a recipient only if the recipient of the information did not already have the information to begin with. Messages that convey information that is certain to happen and already known by the recipient contain no real information. Infrequently occurring messages contain more information than more frequently occurring messages. This fact is reflected in the above equation - a certain message, i.e. of probability 1, has an information measure of zero. In addition, a compound message of two (or more) unrelated (or mutually independent) messages would have a quantity of information that is the sum of the measures of information of each message individually. That fact is also reflected in the above equation, supporting the validity of its derivation.

An example: The weather forecast broadcast is: "Tonight's forecast: Dark. Continued darkness until widely scattered light in the morning." This message contains almost no information. However, a forecast of a snowstorm would certainly contain information since such does not happen every evening. There would be an even greater amount of information in an accurate forecast of snow for a warm location, such as Miami. The amount of information in a forecast of snow for a location where it never snows (impossible event) is the highest (infinity).

## Entropy

The entropy of a discrete message space ${\displaystyle M}$ is a measure of the amount of uncertainty one has about which message will be chosen. It is defined as the average self-information of a message ${\displaystyle m}$ from that message space:

${\displaystyle \mathrm {H} (M)=\mathbb {E} \left[\operatorname {I} (M)\right]=\sum _{m\in M}p(m)\operatorname {I} (m)=-\sum _{m\in M}p(m)\log p(m).}$

where

${\displaystyle \mathbb {E} [-]}$ denotes the expected value operation.

An important property of entropy is that it is maximized when all the messages in the message space are equiprobable (e.g. ${\displaystyle p(m)=1/|M|}$). In this case ${\displaystyle \mathrm {H} (M)=\log |M|}$.

Sometimes the function ${\displaystyle \mathrm {H} }$ is expressed in terms of the probabilities of the distribution:

${\displaystyle \mathrm {H} (p_{1},p_{2},\ldots ,p_{k})=-\sum _{i=1}^{k}p_{i}\log p_{i},}$ where each ${\displaystyle p_{i}\geq 0}$ and ${\displaystyle \sum _{i=1}^{k}p_{i}=1.}$

An important special case of this is the binary entropy function:

${\displaystyle \mathrm {H} _{\mbox{b}}(p)=\mathrm {H} (p,1-p)=-p\log p-(1-p)\log(1-p).\,}$

## Joint entropy

The joint entropy of two discrete random variables ${\displaystyle X}$ and ${\displaystyle Y}$ is defined as the entropy of the joint distribution of ${\displaystyle X}$ and ${\displaystyle Y}$:

${\displaystyle \mathrm {H} (X,Y)=\mathbb {E} _{X,Y}\left[-\log p(x,y)\right]=-\sum _{x,y}p(x,y)\log p(x,y)\,}$

If ${\displaystyle X}$ and ${\displaystyle Y}$ are independent, then the joint entropy is simply the sum of their individual entropies.

(Note: The joint entropy should not be confused with the cross entropy, despite similar notations.)

## Conditional entropy (equivocation)

Given a particular value of a random variable ${\displaystyle Y}$, the conditional entropy of ${\displaystyle X}$ given ${\displaystyle Y=y}$ is defined as:

${\displaystyle \mathrm {H} (X|y)=\mathbb {E} _{\left[X|Y\right]}[-\log p(x|y)]=-\sum _{x\in X}p(x|y)\log p(x|y)}$

where ${\displaystyle p(x|y)={\frac {p(x,y)}{p(y)}}}$ is the conditional probability of ${\displaystyle x}$ given ${\displaystyle y}$.

The conditional entropy of ${\displaystyle X}$ given ${\displaystyle Y}$, also called the equivocation of ${\displaystyle X}$ about ${\displaystyle Y}$ is then given by:

${\displaystyle \mathrm {H} (X|Y)=\mathbb {E} _{Y}\left[\mathrm {H} \left(X|y\right)\right]=-\sum _{y\in Y}p(y)\sum _{x\in X}p(x|y)\log p(x|y)=\sum _{x,y}p(x,y)\log {\frac {p(y)}{p(x,y)}}.}$

This uses the conditional expectation from probability theory.

A basic property of the conditional entropy is that:

${\displaystyle \mathrm {H} (X|Y)=\mathrm {H} (X,Y)-\mathrm {H} (Y).\,}$

## Kullback–Leibler divergence (information gain)

The Kullback–Leibler divergence (or information divergence, information gain, or relative entropy) is a way of comparing two distributions, a "true" probability distribution ${\displaystyle p}$, and an arbitrary probability distribution ${\displaystyle q}$. If we compress data in a manner that assumes ${\displaystyle q}$ is the distribution underlying some data, when, in reality, ${\displaystyle p}$ is the correct distribution, Kullback–Leibler divergence is the number of average additional bits per datum necessary for compression, or, mathematically,

${\displaystyle D_{\mathrm {KL} }{\bigl (}p(X)\|q(X){\bigr )}=\sum _{x\in X}p(x)\log {\frac {p(x)}{q(x)}}.}$

It is in some sense the "distance" from ${\displaystyle q}$ to ${\displaystyle p}$, although it is not a true metric due to its not being symmetric.

## Mutual information (transinformation)

It turns out that one of the most useful and important measures of information is the mutual information, or transinformation. This is a measure of how much information can be obtained about one random variable by observing another. The mutual information of ${\displaystyle X}$ relative to ${\displaystyle Y}$ (which represents conceptually the average amount of information about ${\displaystyle X}$ that can be gained by observing ${\displaystyle Y}$) is given by:

${\displaystyle \operatorname {I} (X;Y)=\sum _{y\in Y}p(y)\sum _{x\in X}{p(x|y)\log {\frac {p(x|y)}{p(x)}}}=\sum _{x,y}p(x,y)\log {\frac {p(x,y)}{p(x)\,p(y)}}.}$

A basic property of the mutual information is that:

${\displaystyle \operatorname {I} (X;Y)=\mathrm {H} (X)-\mathrm {H} (X|Y).\,}$

That is, knowing ${\displaystyle Y}$, we can save an average of ${\displaystyle \operatorname {I} (X;Y)}$ bits in encoding ${\displaystyle X}$ compared to not knowing ${\displaystyle Y}$. Mutual information is symmetric:

${\displaystyle \operatorname {I} (X;Y)=\operatorname {I} (Y;X)=\mathrm {H} (X)+\mathrm {H} (Y)-\mathrm {H} (X,Y).\,}$

Mutual information can be expressed as the average Kullback–Leibler divergence (information gain) of the posterior probability distribution of ${\displaystyle X}$ given the value of ${\displaystyle Y}$ to the prior distribution on ${\displaystyle X}$:

${\displaystyle \operatorname {I} (X;Y)=\mathbb {E} _{p(y)}\left[D_{\mathrm {KL} }{\bigl (}p(X|Y=y)\|p(X){\bigr )}\right].}$

In other words, this is a measure of how much, on the average, the probability distribution on ${\displaystyle X}$ will change if we are given the value of ${\displaystyle Y}$. This is often recalculated as the divergence from the product of the marginal distributions to the actual joint distribution:

${\displaystyle \operatorname {I} (X;Y)=D_{\mathrm {KL} }{\bigl (}p(X,Y)\|p(X)p(Y){\bigr )}.}$

Mutual information is closely related to the log-likelihood ratio test in the context of contingency tables and the multinomial distribution and to Pearson's χ2 test: mutual information can be considered a statistic for assessing independence between a pair of variables, and has a well-specified asymptotic distribution.

## Differential entropy

The basic measures of discrete entropy have been extended by analogy to continuous spaces by replacing sums with integrals and probability mass functions with probability density functions. Although, in both cases, mutual information expresses the number of bits of information common to the two sources in question, the analogy does not imply identical properties; for example, differential entropy may be negative.

The differential analogies of entropy, joint entropy, conditional entropy, and mutual information are defined as follows:

${\displaystyle h(X)=-\int _{X}f(x)\log f(x)\,dx}$
${\displaystyle h(X,Y)=-\int _{Y}\int _{X}f(x,y)\log f(x,y)\,dx\,dy}$
${\displaystyle h(X|y)=-\int _{X}f(x|y)\log f(x|y)\,dx}$
${\displaystyle h(X|Y)=\int _{Y}\int _{X}f(x,y)\log {\frac {f(y)}{f(x,y)}}\,dx\,dy}$
${\displaystyle \operatorname {I} (X;Y)=\int _{Y}\int _{X}f(x,y)\log {\frac {f(x,y)}{f(x)f(y)}}\,dx\,dy}$

where ${\displaystyle f(x,y)}$ is the joint density function, ${\displaystyle f(x)}$ and ${\displaystyle f(y)}$ are the marginal distributions, and ${\displaystyle f(x|y)}$ is the conditional distribution.


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# Differentiating an unknown solution to a PDE

Sorry if this question is too basic -- I'm not very familiar with Mathematica. I am interested in a way to systematically address the following sort of problem:

Suppose that $u=u(x,y)$ is a function of two variables on $\mathbb{R}^2$ that satisfies a Liouville type equation: $$\Delta u= e^{-2u}$$.

I can check by hand that the function $$v=\partial^2_{xy} u+\partial_x u \cdot \partial_y u$$ is harmonic, i.e. $$\Delta v=0.$$

My question is how can I make Mathematica verify this? I'm interested in more complicated examples I don't want to check by hand, but the principle is the same.

-
Do you explicitly have $u$, or just the PDE? –  Ｊ. Ｍ. Aug 23 '12 at 13:57
Just the PDE. It should be entirely a symbolic manipulation. The point is that for any $u$ satisfying the PDE the function $v$ is harmonic. You can verify this directly by differentiating. –  oberon Aug 23 '12 at 13:59
so you are looking to automatically find a transformation that makes an arbitrary nonlinear PDE into the Laplace equation? –  acl Aug 23 '12 at 14:05
@acl No that would be crazy. I'm looking for a way to start with a fixed PDE (the Liuoville equation say) and a solution $u$ to the PDE. I also start with a $v$ which is an explicit algebraic combination of derivatives of $u$ (say the $v$ I wrote down). I want Mathematica to check that $v$ is harmonic using the rules of differentiation and the PDE. –  oberon Aug 23 '12 at 14:10
it's easy enough to differentiate the v and insert the explicit expression, but that gives a mass of symbols in which you'd need to replace $u_{xx}+u_{yy}$ by $\exp(-u)$ (not to mention the derivatives) and I don't see how to do that; I hope someone else will –  acl Aug 23 '12 at 17:14

The general idea, as best I understand, is to take your differential identities, generate more by differentiating (prolongation), treat all derivatives as algebraically independent variables, and try to show that the putative identity is in the (algebraic) ideal generated by the known polynomials in these variables. The trick is to know how far to prolong.

While there are algorithmic approaches (I believe this goes by the name "Cartan Kuranishi algorithm, or something to that effect), I'll just assume that it will suffice to get all second derivatives of our defining equations. Here is the code.

The independent ariables and initial differential "polynomials".

ivars = {x, y};
dpolys = {D[u[x, y], {x, 2}] + D[u[x, y], {y, 2}] - Exp[-2*u[x, y]],
v[x, y] - (D[u[x, y], x, y] + D[u[x, y], x]*D[u[x, y], y])};

Get the Laplacian of v.

lapv = Total[Flatten[Outer[D[#1, {#2, 2}] &, {v[x, y]}, ivars]]];

Now get all first and second derivatives of these diff polys.

allpolys = Flatten[Join[dpolys, Outer[D, dpolys, ivars],
Outer[D[#1, {#2, 2}] &, dpolys, ivars], D[dpolys, x, y]]];
allvars = Variables[allpolys];

I want to impose an ordering on the "variables". I'm not sure this is necessary but at least it will help to make sure we use the same term ordering when we attempt to reduce the Laplacian of v to zero.

uvars = Select[allvars, ! FreeQ[#, u] &];
duvars = Select[uvars, ! FreeQ[#, Derivative] &];

vvars = Select[allvars, ! FreeQ[#, v] &]; dvvars =
Select[vvars, ! FreeQ[#, Derivative] &];

derivorder[Derivative[a : __][_][__]] := Plus[a]

sortuvars =
Append[Sort[duvars, derivorder[#1] >= derivorder[#2] &], u[x, y]];
sortvvars =
Append[Sort[dvvars, derivorder[#1] >= derivorder[#2] &], v[x, y]];

Small problem: when we go to compute a Groebner basis, the exponentials will not correctly be seen as invertible paramaters. So I replace them up front with a new symbol. We will treat it as invertible. It turns out that this both makes the reduction step viable and also makes the basis computation hugely faster.

allpolys2 = allpolys /. E^(-2 u[x, y]) -> exp2u;

Compute a(n algebraic) Groebner basis for our polynomials. Again, while they involve derivatives, we treat those as being algebraically independent variables.

Timing[
gb = GroebnerBasis[allpolys2, Join[sortvvars, sortuvars],
MonomialOrder -> DegreeReverseLexicographic,
CoefficientDomain -> RationalFunctions];]

(* Out[615]= {0.11, Null} *)

Now check whether the Laplacian of v lives in this (algebraic) ideal. If so, that proves that its vanishing is a consequence of the original differential equations.

PolynomialReduce[lapv, gb, Join[sortvvars, sortuvars],
MonomialOrder -> DegreeReverseLexicographic,
CoefficientDomain -> RationalFunctions][[2]]

(* Out[619]= 0 *)

[Last step: Declare victory and go home.]

-
This sounds like the correct approach. I will investigate further and see if I can get it to work. Thanks! –  oberon Aug 26 '12 at 9:56

I am fairly certain there is a better way to do this, but you can try the following which involves using the VectorAnalysis package:

<<VectorAnalysis

\[CapitalDelta]u[x_, y_] := Exp[-2 u[x, y]]
v[x_, y_] := Derivative[1, 1][u][x,y] + (Derivative[1, 0][u][x,y])*(Derivative[0, 1][u][x, y])

r1 = Laplacian[u[x, y], Cartesian[x, y, z]] -> \[CapitalDelta]u[x,y];
dx = D[Laplacian[u[x, y], Cartesian[x, y, z]], x] -> D[\[CapitalDelta]u[x, y], x];
dy = D[Laplacian[u[x, y], Cartesian[x, y, z]], y] -> D[\[CapitalDelta]u[x, y], y];
dxy = D[Laplacian[u[x, y], Cartesian[x, y, z]], x, y] -> D[\[CapitalDelta]u[x, y], x, y];

FullSimplify[Laplacian[v[x, y], Cartesian[x, y, z]]] //. {dx, dy, dxy,r1}
`
-


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# Tag Info

21

Observation: An individual 1-byte pearson hash behaves like an 8 bit block cipher, encrypting the initial state using the message as key. This means that given a fixed message, each possible initial state produces a different output. This implies that a combined hash will never contain duplicate bytes. Without this property a hash would forget about the ...

18

A cryptographic hash function $f : \{0,1\}^{*} \to \{0,1\}^n$ has three properties: (1) preimage resistance, (2) second-preimage resistance, and (3) collision resistance. Even further, these properties form a hierarchy where each property implies the one before it, i.e., a collision-resistant function is also second-preimage resistant, and a second-preimage ...

15

The difference is in the choice of $m_1$. In the first case (second preimage resistance), the attacker is handed a fixed $m_1$ to which he has to find a different $m_2$ with equal hash. In particular, he can't choose $m_1$. In the second case (collision resistance), the attacker can freely choose both messages $m_1$ and $m_2$, with the only requirement ...

12

With the definitions that a function $F$ is collision-resistant when a [computationally bounded] adversary can't [with sizable odds] exhibit any $(a,b)$ with $a\ne b$ and $F(a)=F(b)$; first-preimage-resistant when, given $f$ determined as $F(a)$ for an unknown random $a$, a [computationally bounded] adversary can't [with sizable odds] exhibit any $b$ with $... 12 Right now, the best published attack against MD5's preimage resistance (first preimage, actually, but it applies to second preimage resistance as well) finds preimages in cost$2^{123.4}$average cost, which is slightly better than the generic attack (average cost of$2^{128}$), but still way beyond the technologically feasible. The attack rebuilds the ... 10 Because it is not secure enough. Hash functions rely a lot on diffusion (a single bit change must change half of the other bits) and confusion (the value of a bit should depend on the value of other bits). This is also known as the avalanche effect. Because it lacks a permutation, my first intuition: it lacks diffusion and has weakness to differential ... 8 Consider the function$H$transforming a message$m$to the SHA-512 hash of the first 1024 bits of$m$(right-padded with$1024-n$zero bits if the bit length$n$of$m$is less than 1024).$H$is first-preimage resistant, but not second-preimage resistant: once you have a first preimage$m_1$, it is trivial to get another$m_2$with the same hash (e.g. ... 8 It is neither pre-image resistant, second pre-image resistant nor collision resistant. It is easy to compute square-roots modulo a prime (assuming, of course, a square root exists, it will half the time). If$p = 3 \bmod 4$, then the simple formula$x^{(p+1)/4} \bmod p$will work; for$p = 1 \bmod 4$, it's a tad more complicated but still sufficiently ... 7 I prefer using definitions that explicitly specify who does what. Weak collision resistance: After Bob creates some message x1, it is "computationally infeasible" for an attacker Mallory to compute some other message x2 such that h(x1) == h(x2). Strong collision resistance: It is "computationally infeasible" for an attacker Mallory to find any two messages ... 7 Yes, it has happened. If you look at the SHA3 hash zoo, there are a number of hashes who has the best attack listed as "2nd preimage". One general place this can occur is if you have a hash function with a weak message compression step, but a fairly strong finalization step. Here, we might not be able to generate first preimages (because we don't know what ... 7 As far as I am aware, there are no practical known second pre-image attacks on MD5, under the conditions you listed. However: if the attacker can control any part of the original, I would worry about using MD5 in this setting. Its security in this setting may be fragile and there may well be cleverer attacks than anything currently in the literature. I ... 7 Let me try to elaborate on their proof. Suppose you had a hash function$H$that was second-preimage resistant but not first-preimage resistant. By showing that this leads to a contradiction, we will be showing that with second-preimage resistance, you must have first-preimage resistance. Namely, we will show that the lack of first-preimage resistance is ... 7 Hash + digital signature If the hash is not collision resistant, the attacker can produce two messages having the same hash. They'll request a signature on the first and present the signature on the second, a forgery. When second pre-image resistance is violated, this attack becomes much more severe, since now the attacker doesn't need control over both ... 6 In their paper Second Preimages on$n$-Bit Hash Functions for Much Less than$2^n$Work, Kelsey and Schneier provide: a second preimage attack on all$n$-bit iterated hash functions with Damgard-Merkle strengthening and$n$-bit intermediate states, allowing a second preimage to be found for a$2^k$-message-block message with about$k\times2^{n/2+1}+ 2^{n-...

5

Actually, to the best of our knowledge, it's computationally infeasible. By the terminology what we use when we discuss cryptographical hash functions, you're not asking for a hash collision (which is "find two different messages that hash to the same value"), but instead you're asking for a hash preimage (which is "for this hash value, find a message that ...

5

Pre-image resistant but not 2nd pre-image resistant? describes the relationship between the three basic hash function security notions: Collision Resistance, Second Preimage Resistance and Preimage Resistance. In short, Collision Resistance implies Second Preimage Resistance (but not vice-versa) - there is a good diagram on page 4 of RogawayShrimpton04 that ...

5

Without the specific reference I can't be sure this is what you are talking about, but generally a "long message" attack is a way to defeat second preimage resistance with less complexity than expected. It uses a time-space tradeoff to find a second preimage with complexity $2^{n/2}$ for a $n$-bit hash function (normally you would expect $2^n$). In the ...

4

Take a function $H:\mathbb S\to\{0,1\}^k$ where $\mathbb S$ is a large finite subset of $\{0,1\}^*$, such that $H$ "compress data" [however this is defined], and $H$ is [conjectured] collision-resistant [thus second-preimage-resistant] and first-preimage-resistant; e.g., SHA-512, for $k=512$. Let $«0»$ and $«1»$ be two public distinct elements of $\mathbb S$....

4

A long message is a message that, when padded, is longer than the block size of the hash function. That means that the hash function has to process the message in parts and keep track of state somehow, which may allow for attacks. Such attacks would not apply to messages shorter than the block size, and may additionally require a large number of blocks to ...

4

Preliminary: Almost the same article is available for free without breaking any law, nor downloading 5GB (formatting is shifted by at most one third of a page). It is also (as well as all other articles of IACR crypto conferences from 2000-2011) in the IACR Online Proceedings, specifically in the FSE 2008 section, but then you need to subtract about 223 from ...

4

We know no practically feasible way to do what you ask for, except if the hash X=acf3602b5eb9a2db3e365d3043682faf or the content of the file wczasp.rb was prepared specially to make that possible. Assuming that the content of file wczasp.rb is arbitrary, what is asked would be a preimage attack. This is further sub-classified as first preimage if only the ...

4

You are not asking for a collision but for a preimage. Collision attack: the attacker computes two messages m and m', distinct from each other, such that m and m' hash to the same value. Preimage: the attacker is given a goal (a hash value h) and finds a message m which hashes to h. MD5 is weak for collisions, but not for preimages: no attack method is ...

4

I really like this question, and have two things to say. First note that CBC-MAC is no good since given the key it's easy to find a collision. Let $t$ be a tag for a message $m=m_1,m_2$ of length $\ell$ bits. Then, in CBC-MAC the input to AES first is $\ell$ and then the output is XORed with $m_1$ and input to AES, and so on. Let $t_1$ be the intermediate ...

3

I have ask as part of my answer, "What problem are you trying to solve?" Do you want a secure unkeyed hash function? If you do, then there are plenty of them around. Even some of the ones that are broken for some uses might be okay for yours (SHA-1 springs to mind -- note the discussion above on HMAC and how broken a hash function has to be). But really, ...

3

Given message $A$, you have to find message $B$, such that the first 64 bits (say, MSB) of their hashes collide: $$MSB_{64}(H(A)) = MSB_{64}(H(B))$$ This problem is called Second Preimage Search for the function $MSB_{64}(H)$, or Partial Second Preimage Search for the hash function $H$ alone. When $H$ is the full round SHA-1, there is no result, ...

3

While collision resistance can be defined for normal hash functions like SHA1, for target collision resistance you need a so called keyed hash function, that is a hash function that additionally to a message $m$ also takes a key $k$. The simplest way to construct a keyed hash function out of a regular one is to prepend the key in front of the message: $f(k,m)... 3 Can we exhibit collisions, or second-preimages (with implies the former), for the ChaCha core? No, likely not. The Salsa20 and ChaCha cores both consist of a large number of "quarter-rounds" each of which is invertible and bijective. The only reason neither core is a bijection (and thus can have collisions) is the final addition of the input elements ... 3 SipHash is a MAC (aka Pseudo Random Function Family) with 64-bit output and 128-bit key, rather than a hash (aka random public member of a Pseudo Random Function Family). It is explicitly designed to be used with a secret random key. Quoting Jean-Philippe Aumasson and Daniel J. Bernstein's SipHash: a fast short-input PRF (in proceedings of Indocrypt 2012): ... 3 The short answer is that for any secure hash it is impossible to generate collisions or second preimages, regardless of message length. If you look at theoretical attacks, however, message length can be a factor. When looking for another message that hashes to the same value as a specific message you are talking about second preimage resistance (or target ... 3 The right terminology is second preimage resistance and preimage resistance. Edited to reflect comment by otus. Preimage attack takes$O(2^{n})$hash function calls on average. Second preimage attack takes just one extra call so the complexity is essentially the same as that of the preimage attack. Given$x$you want$x'\neq x$such that$H(x)=H(x').\$ So ...

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


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# Working alone at a constant rate, machine P produces a widgets in 3...

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08 Jun 2018, 08:38
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Working alone at a constant rate, machine P produces $$a$$ widgets in 3 hours. Working alone at a constant rate, machine Q produces $$b$$ widgets in 4 hours. If machines P and Q work together for $$c$$ hours, then in terms of $$a, b,$$ and $$c$$ how many widgets will machines P and Q produce?

A) $$\frac{3ac + 4bc}{12}$$

B) $$\frac{4ac + 3bc}{12}$$

C) $$\frac{4ac + 3bc}{6}$$

D) $$4ac + 3bc$$

E) $$\frac{ac + 2bc}{4}$$

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Re: Working alone at a constant rate, machine P produces a widgets in 3...  [#permalink]

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08 Jun 2018, 08:55
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Drawing a table to document the information provided in the question will be very useful as follows:-

Attachment:

Kp soln1.jpg [ 29.83 KiB | Viewed 621 times ]

$$\frac{a}{3} + \frac{b}{4} = \frac{x}{c}$$

Solve for x to get

x = $$\frac{4ac + 3bc}{12}$$
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Re: Working alone at a constant rate, machine P produces a widgets in 3...  [#permalink]

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08 Jun 2018, 10:36
CAMANISHPARMAR wrote:
Working alone at a constant rate, machine P produces $$a$$ widgets in 3 hours. Working alone at a constant rate, machine Q produces $$b$$ widgets in 4 hours. If machines P and Q work together for $$c$$ hours, then in terms of $$a, b,$$ and $$c$$ how many widgets will machines P and Q produce?

A) $$\frac{3ac + 4bc}{12}$$

B) $$\frac{4ac + 3bc}{12}$$

C) $$\frac{4ac + 3bc}{6}$$

D) $$4ac + 3bc$$

E) $$\frac{ac + 2bc}{4}$$

Let a=3, implying that A produces 3 widgets every 3 hours, the equivalent of 1 widget per hour.
Let b=4, implying that B produces 4 widgets every 4 hours, the equivalent of 1 widget per hour.
Let c=1 hour.
Since A and B each produce 1 widget per hour, the number of widgets produced when they work together for 1 hour = 1+1 = 2.

Now plug a=3, b=4 and c=1 into the answer choices to see while yields a production of 2 widgets.
Only B works:
$$\frac{4ac + 3bc}{12}$$ = $$\frac{(4*3*1) + (3*4*1)}{12}$$ = 2.

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Re: Working alone at a constant rate, machine P produces a widgets in 3...  [#permalink]

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23 Sep 2018, 07:49
CAMANISHPARMAR wrote:
Working alone at a constant rate, machine P produces $$a$$ widgets in 3 hours. Working alone at a constant rate, machine Q produces $$b$$ widgets in 4 hours. If machines P and Q work together for $$c$$ hours, then in terms of $$a, b,$$ and $$c$$ how many widgets will machines P and Q produce?

A) $$\frac{3ac + 4bc}{12}$$

B) $$\frac{4ac + 3bc}{12}$$

C) $$\frac{4ac + 3bc}{6}$$

D) $$4ac + 3bc$$

E) $$\frac{ac + 2bc}{4}$$

$$P \cup Q\,\,\, - \,\,\,c\,\,{\text{h}}\,\,\, - \,\,\,\boxed{? = f\left( {a,b,c} \right)}\,\,\,{\text{widgets}}$$

$$P\,\,\, - \,\,\,3\,\,{\text{h}}\,\,\, - \,\,\,a\,\,{\text{widgets}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,P\,\,\, - \,\,\,12\,\,{\text{h}}\,\,\, - \,\,\,4a\,\,{\text{widgets}}$$

$$Q\,\,\, - \,\,\,4\,\,{\text{h}}\,\,\, - \,\,\,b\,\,{\text{widgets}}\,\,\,\,\, \Rightarrow \,\,\,\,\,Q\,\,\, - \,\,\,12\,\,{\text{h}}\,\,\, - \,\,\,3b\,\,{\text{widgets}}$$

Now let´s apply the UNITS CONTROL, one of the most powerful tools of our method!

$${\text{?}}\,\,\,\,{\text{ = }}\,\,\,\,{\text{c}}\,\,{\text{h}}\,\,\,\left( {\frac{{4a + 3b\,\,\,{\text{widgets}}}}{{12\,\,\,{\text{h}}}}\,\,\begin{array}{*{20}{c}} \nearrow \\ \nearrow \end{array}\,\,} \right)\,\,\,\,\, = \,\,\,\,\,\,\frac{{\left( {4a + 3b} \right)c}}{{12}}\,\,\,\,\left[ {{\text{widgets}}} \right]$$

Obs.: arrows indicate licit converter.

The right answer is therefore (B).

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: Working alone at a constant rate, machine P produces a widgets in 3... &nbs [#permalink] 23 Sep 2018, 07:49
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## Potassium Carbonate

Updated on 11-Jan-2023 16:09:03
Introduction The primary compound of potash is Potassium Carbonate and pearl ash or the salts of tartar are a more refined form of it. In the making of pearl ash, potash is baked in a kiln and this process removes the impure agents. In the 18th century before the use of baking powder, pearl ash was used to make quick bread. What is Potassium Carbonate? Potassium Carbonate is an inorganic compound that can be represented with the chemical formula- $\mathrm{K_2CO_3}$. It is also known as Carbonate of potash, or Dipotassium carbonate or Pearl ash. It can be defined as ... Read More

## Base Meaning

Updated on 14-Dec-2022 12:39:58
Introduction There are different definitions for the base. But the one definition that is common in all three definitions is bases react with acid to form a neutral compound or salts and water. And is given by G.-F. Rouelle. It is the ion that is present in this compound that makes it a characteristic property. Arrhenius proposed that the presence of hydroxyl ion, $\mathrm{OH^{-}}$ leads to the characteristic property of a base. According to Bronsted Lowry's theory of bases, a base is a substance that can accept protons or Hydrogen ions, $\mathrm{H^{+}}$. And in lewis's base theory base is a ... Read More

## Barium Hydroxide

Updated on 14-Dec-2022 10:13:15
Introduction Barium Hydroxide is made up of the elements Ba, H and O. It seems to be a soft, shiny, alkaline earth metal found within Group 2 of the current periodic table. This is scientifically quite reactive. As a result, it would never be discovered in the environment as a pure element. These derivatives are utilised in the petroleum industry as well as radiography, but it is also useful in metallurgy. It is indeed a transparent white granule having no naturally toxic odour. Because of the hydroxides, $\mathrm{Ba(OH)_{2}}$ is harmful or could harm the skin as well as the eyes. ... Read More

## Barium Carbonate

Updated on 13-Oct-2022 11:19:47
Introduction Barium Carbonate has a molecular formula, (BaCO3). It is an inorganic chemical compound. It is tasteless and odourless. Like or similar to most of the other alkaline earth metal carbonates, it appears as white salt and it is insoluble or poorly soluble in water solution but, it is soluble in most of the acids, except sulphuric acid. It is one of the very important barium compounds commercially or economically. Although it is not soluble in pure water it is partially or slightly soluble in water saturated with carbon dioxide (CO2). It is used for the formation of rat poisons. ... Read More


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Search for a feebly interacting particle X in the decay K+ → π+X
A bstract A search for the K + → π + X decay, where X is a long-lived feebly interacting particle, is performed through an interpretation of the K + → $${\pi}^{+}\nu \overline{\nu}$$ π + ν ν ¯ analysis of data collected in 2017 by the NA62 experiment at CERN. Two ranges of X masses, 0–110 MeV /c 2 and 154–260 MeV /c 2 , and lifetimes above 100 ps are considered. The limits set on the branching ratio, BR( K + → π + X ), are competitive with previously reported searches in the first mass range, and improve on current limits in the second mass range by more than an order of magnitude.
Authors:
; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; more »
Award ID(s):
Publication Date:
NSF-PAR ID:
10272507
Journal Name:
Journal of High Energy Physics
Volume:
2021
Issue:
3
ISSN:
1029-8479
2. A bstract We present a search for the dark photon A ′ in the B 0 → A ′ A ′ decays, where A ′ subsequently decays to e + e − , μ + μ − , and π + π − . The search is performed by analyzing 772 × 10 6 $$B\overline{B}$$ B B ¯ events collected by the Belle detector at the KEKB e + e − energy-asymmetric collider at the ϒ(4 S ) resonance. No signal is found in the dark photon mass range 0 . 01 GeV /c 2 ≤ m Amore »
3. Abstract A search for heavy resonances decaying into a pair of Z bosons leading to $$\ell ^+\ell ^-\ell '^+\ell '^-$$ ℓ + ℓ - ℓ ′ + ℓ ′ - and $$\ell ^+\ell ^-\nu {{\bar{\nu }}}$$ ℓ + ℓ - ν ν ¯ final states, where $$\ell$$ ℓ stands for either an electron or a muon, is presented. The search uses proton–proton collision data at a centre-of-mass energy of 13 TeV collected from 2015 to 2018 that corresponds to the integrated luminosity of 139 $$\mathrm {fb}^{-1}$$ fb - 1 recorded by the ATLAS detector during Run 2 of the Largemore »
4. A bstract The NA62 experiment reports an investigation of the $${K}^{+}\to {\pi}^{+}\nu \overline{\nu}$$ K + → π + ν ν ¯ mode from a sample of K + decays collected in 2017 at the CERN SPS. The experiment has achieved a single event sensitivity of (0 . 389 ± 0 . 024) × 10 − 10 , corresponding to 2.2 events assuming the Standard Model branching ratio of (8 . 4 ± 1 . 0) × 10 − 11 . Two signal candidates are observed with an expected background of 1.5 events. Combined with the result of amore »
5. Abstract The production of $$\pi ^{\pm }$$ π ± , $$\mathrm{K}^{\pm }$$ K ± , $$\mathrm{K}^{0}_{S}$$ K S 0 , $$\mathrm{K}^{*}(892)^{0}$$ K ∗ ( 892 ) 0 , $$\mathrm{p}$$ p , $$\phi (1020)$$ ϕ ( 1020 ) , $$\Lambda$$ Λ , $$\Xi ^{-}$$ Ξ - , $$\Omega ^{-}$$ Ω - , and their antiparticles was measured in inelastic proton–proton (pp) collisions at a center-of-mass energy of $$\sqrt{s}$$ s = 13 TeV at midrapidity ( $$|y|<0.5$$ | y | < 0.5 ) as a function of transverse momentum ( $$p_{\mathrm{T}}$$ p T ) using the ALICE detector at the CERNmore »


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Home > Standard Error > What Is A Standard Error Of The Mean

# What Is A Standard Error Of The Mean

## Contents

Retrieved 17 July 2014. Traditional IRAs & 401(k)s

Guides Stock Basics Economics Basics Options Basics
Exam Prep Series 7 Exam CFA Level 1 Series 65 Exam Simulator Stock Simulator For any random sample from a population, the sample mean will usually be less than or greater than the population mean. For the purpose of this example, the 9,732 runners who completed the 2012 run are the entire population of interest. navigate here

Text is available under the Creative Commons Attribution-ShareAlike License; additional terms may apply. Standard error is a statistical term that measures the accuracy with which a sample represents a population. In cases where the standard error is large, the data may have some notable irregularities.Standard Deviation and Standard ErrorThe standard deviation is a representation of the spread of each of the Notice that s x ¯   = s n {\displaystyle {\text{s}}_{\bar {x}}\ ={\frac {s}{\sqrt {n}}}} is only an estimate of the true standard error, σ x ¯   = σ n https://en.wikipedia.org/wiki/Standard_error

## Standard Error Of The Mean Formula

n is the size (number of observations) of the sample. The standard error is a measure of central tendency. (A) I only (B) II only (C) III only (D) All of the above. (E) None of the above. ISBN 0-521-81099-X ^ Kenney, J.

CliffsAP StatisticsDavid A KayList Price: $16.99Buy Used:$0.01Buy New: $50.85Texas Instruments TI-Nspire TX Handheld Graphing CalculatorList Price:$149.00Buy Used: $51.88Buy New:$170.00Approved for AP Statistics and Calculus About Us Contact Hutchinson, Essentials of statistical methods in 41 pages ^ Gurland, J; Tripathi RC (1971). "A simple approximation for unbiased estimation of the standard deviation". The 95% confidence interval for the average effect of the drug is that it lowers cholesterol by 18 to 22 units. Standard Error Regression Standard error From Wikipedia, the free encyclopedia Jump to: navigation, search For the computer programming concept, see standard error stream.

Edwards Deming. Standard Error Of The Mean Excel The age data are in the data set run10 from the R package openintro that accompanies the textbook by Dietz [4] The graph shows the distribution of ages for the runners. For the age at first marriage, the population mean age is 23.44, and the population standard deviation is 4.72. https://www.khanacademy.org/math/statistics-probability/sampling-distributions-library/sample-means/v/standard-error-of-the-mean Correction for correlation in the sample Expected error in the mean of A for a sample of n data points with sample bias coefficient ρ.

The table below shows formulas for computing the standard deviation of statistics from simple random samples. Standard Error Of Proportion The distribution of these 20,000 sample means indicate how far the mean of a sample may be from the true population mean. In each of these scenarios, a sample of observations is drawn from a large population. Similarly, the sample standard deviation will very rarely be equal to the population standard deviation.

## Standard Error Of The Mean Excel

As an example of the use of the relative standard error, consider two surveys of household income that both result in a sample mean of \$50,000. The sample mean will very rarely be equal to the population mean. Standard Error Of The Mean Formula III. Standard Error Of The Mean Definition It can only be calculated if the mean is a non-zero value.

In an example above, n=16 runners were selected at random from the 9,732 runners. http://3cq.org/standard-error/when-to-report-standard-deviation-and-standard-error.php By using this site, you agree to the Terms of Use and Privacy Policy. The standard deviation is computed solely from sample attributes. The distribution of the mean age in all possible samples is called the sampling distribution of the mean. Standard Error Vs Standard Deviation

To understand this, first we need to understand why a sampling distribution is required. Standard error of mean versus standard deviation In scientific and technical literature, experimental data are often summarized either using the mean and standard deviation or the mean with the standard error. In fact, data organizations often set reliability standards that their data must reach before publication. his comment is here A natural way to describe the variation of these sample means around the true population mean is the standard deviation of the distribution of the sample means.

Perspect Clin Res. 3 (3): 113–116. Difference Between Standard Error And Standard Deviation No problem, save it as a course and come back to it later. This often leads to confusion about their interchangeability.

## Because these 16 runners are a sample from the population of 9,732 runners, 37.25 is the sample mean, and 10.23 is the sample standard deviation, s.

Standard Error of the Estimate A related and similar concept to standard error of the mean is the standard error of the estimate. The standard error is important because it is used to compute other measures, like confidence intervals and margins of error. For any random sample from a population, the sample mean will usually be less than or greater than the population mean. Standard Error Symbol It is the variance -- the SD squared -- that doesn't change predictably, but the change in SD is trivial and much much smaller than the change in the SEM.)Note that

doi:10.4103/2229-3485.100662. ^ Isserlis, L. (1918). "On the value of a mean as calculated from a sample". Scenario 1. Scenario 2. http://3cq.org/standard-error/why-is-standard-error-smaller-than-standard-deviation.php Two data sets will be helpful to illustrate the concept of a sampling distribution and its use to calculate the standard error.

Because these 16 runners are a sample from the population of 9,732 runners, 37.25 is the sample mean, and 10.23 is the sample standard deviation, s. ISBN 0-8493-2479-3 p. 626 ^ a b Dietz, David; Barr, Christopher; Çetinkaya-Rundel, Mine (2012), OpenIntro Statistics (Second ed.), openintro.org ^ T.P. For illustration, the graph below shows the distribution of the sample means for 20,000 samples, where each sample is of size n=16. II.

Roman letters indicate that these are sample values. For the purpose of hypothesis testing or estimating confidence intervals, the standard error is primarily of use when the sampling distribution is normally distributed, or approximately normally distributed. The standard error is an estimate of the standard deviation of a statistic. The sample mean x ¯ {\displaystyle {\bar {x}}} = 37.25 is greater than the true population mean μ {\displaystyle \mu } = 33.88 years.

Wilson Mizner: "If you steal from one author it's plagiarism; if you steal from many it's research." Don't steal, do research. . T-distributions are slightly different from Gaussian, and vary depending on the size of the sample. Notation The following notation is helpful, when we talk about the standard deviation and the standard error.