text:
• Coherent control with a short-wavelength Free Electron Laser(1701.03387)

Jan. 12, 2017 physics.optics
XUV and X-ray Free Electron Lasers (FELs) produce short wavelength pulses with high intensity, ultrashort duration, well-defined polarization and transverse coherence, and have been utilised for many experiments previously possible at long wavelengths only: multiphoton ionization, pumping an atomic laser, and four-wave mixing spectroscopy. However one important optical technique, coherent control, has not yet been demonstrated, because Self- Amplified Spontaneous Emission FELs have limited longitudinal coherence. Single-colour pulses from the FERMI seeded FEL are longitudinally coherent, and two-colour emission is predicted to be coherent. Here we demonstrate the phase correlation of two colours, and manipulate it to control an experiment. Light of wavelengths 63.0 and 31.5 nm ionized neon, and the asymmetry of the photoelectron angular distribution was controlled by adjusting the phase, with temporal resolution 3 attoseconds. This opens the door to new shortwavelength coherent control experiments with ultrahigh time resolution and chemical sensitivity.
• Tunable orbital angular momentum in high-harmonic generation(1608.06907)

Sept. 6, 2016 physics.optics
Optical vortices are currently one of the most intensively studied topics in optics. These light beams, which carry orbital angular momentum (OAM), have been successfully utilized in the visible and infrared in a wide variety of applications. Moving to shorter wavelengths may open up completely new research directions in the areas of optical physics and material characterization. Here, we report on the generation of extreme-ultraviolet optical vortices with femtosecond duration carrying a controllable amount of OAM. From a basic physics viewpoint, our results help to resolve key questions such as the conservation of angular momentum in highly-nonlinear light-matter interactions, and the disentanglement and independent control of the intrinsic and extrinsic components of the photon's angular momentum at short-wavelengths. The methods developed here will allow testing some of the recently proposed concepts such as OAM-induced dichroism, magnetic switching in organic molecules, and violation of dipolar selection rules in atoms.
• CITIUS: an IR-XUV light source for fundamental and applied ultrafast science(1310.3355)

Oct. 12, 2013 physics.optics
We present the main features of CITIUS, a new light source for ultrafast science, generating tunable, intense, femtosecond pulses in the spectral range from IR to XUV. The XUV pulses (about 10^5-10^8 photons/pulse in the range 14-80 eV) are produced by laser-induced high-order harmonic generation in gas. This radiation is monochromatized by a time-preserving monochromator, allowing also to work with high-resolution bandwidth selection. The tunable IR-UV pulses (10^{12}-10^{15} photons/pulse in the range 0.4-5.6 eV) are generated by an optical parametric amplifier, which is driven by a fraction of the same laser pulse that generates high order harmonics. The IR-UV and XUV pulses follow different optical paths and are eventually recombined on the sample for pump-probe experiments. The new light source will become the fulcrum of a new center located at the University of Nova Gorica, active in a wide range of scientific fields, including materials science, catalysis, biochemistry and magnetism. We also present the results of two pump-probe experiments: with the first one, we fully characterized the temporal duration of harmonic pulses in the time-preserving configuration; with the second one, we demonstrated the possibility of using CITIUS for studying of ultra-fast dynamics.
• Two-colour generation in a chirped seeded Free-Electron Laser(1306.1470)

We present the experimental demonstration of a method for generating two spectrally and temporally separated pulses by an externally seeded, single-pass free-electron laser operating in the extreme-ultraviolet spectral range. Our results, collected on the FERMI@Elettra facility and confirmed by numerical simulations, demonstrate the possibility of controlling both the spectral and temporal features of the generated pulses. A free-electron laser operated in this mode becomes a suitable light source for jitter-free, two-colour pump-probe experiments.
• Fully-tunable femtosecond laser source in the ultraviolet spectral range(1110.0709)

Oct. 4, 2011 physics.optics
We demonstrate experimentally the full tunability of a coherent femtosecond source in the whole ultraviolet spectral region. The experiment relies on the technique of high-order harmonic generation driven by a near-infrared parametric laser source in krypton gas. By tuning the drive wavelength in the range between 1100 to 1900 nm, we generated intense harmonics from near to extreme ultraviolet. A number of photons per shot of the order of 10^7 has been measured for the first harmonic orders. Many novel scientific prospects are expected to benefit from the use of such a table-top tunable source.
• Experimental demonstration of frequency pulling in single-pass free-electron lasers(1105.5536)

May 27, 2011 physics.optics
Frequency pulling is a well-known phenomenon in standard laser physics, leading to a shift of the laser frequency when the cavity and maximum gain frequencies are detuned. In this letter we present the first experimental demonstration of frequency pulling in single-pass free-electron lasers. Measurements are performed using the single-pass free-electron laser installed on the Elettra storage ring.
• VUV and X-ray coherent light with tunable polarization from single-pass free-electron lasers(1105.5538)

May 27, 2011 physics.optics
Tunable polarization over a wide spectral range is a required feature of light sources employed to investigate the properties of local symmetry in both condensed and low-density matter. Among new-generation sources, free-electron lasers possess a unique combination of very attractive features, as they allow to generate powerful and coherent ultra-short optical pulses in the VUV and X-ray spectral range. However, the question remains open about the possibility to freely vary the light polarization of a free-electron laser, when the latter is operated in the so-called nonlinear harmonic-generation regime. In such configuration, one collects the harmonics of the free-electron laser fundamental emission, gaining access to the shortest possible wavelengths the device can generate. In this letter we provide the first experimental characterization of the polarization of the harmonic light produced by a free-electron laser and we demonstrate a method to obtain tunable polarization in the VUV and X-ray spectral range. Experimental results are successfully compared to those obtained using a theoretical model based on the paraxial solution of Maxwell's equations. Our findings can be expected to have a deep impact on the design and realization of experiments requiring full control of light polarization to explore the symmetry properties of matter samples.
• Angular Distribution of Nonlinear Harmonic Generation in Helical Undulators: a comparison between experiments and theory(1105.1725)

May 9, 2011 physics.optics
Using harmonic emission from circularly polarized undulator is a procedure that is normally employed on synchrotron beamlines in order to extend the covered spectral range. A similar capability is likewise beneficial for next generation free-electron lasers. In this paper, we perform a first quantitative experimental analysis of the angular distribution of free-electron laser harmonic emission from helical undulators. Experimental results are compared to those obtained by means of a theoretical model based on the paraxial solution of Maxwell's equations.
• Stability of inhomogeneous states in mean-field models with a local potential(1010.4647)

Feb. 23, 2011 cond-mat.stat-mech
The Vlasov equation is well known to provide a good description of the dynamics of mean-field systems in the $N \to \infty$ limit. This equation has an infinity of stationary states and the case of {\it homogeneous} states, for which the single-particle distribution function is independent of the spatial variable, is well characterized analytically. On the other hand, the inhomogeneous case often requires some approximations for an analytical treatment: the dynamics is then best treated in action-angle variables, and the potential generating inhomogeneity is generally very complex in these new variables. We here treat analytically the linear stability of toy-models where the inhomogeneity is created by an external field. Transforming the Vlasov equation into action-angle variables, we derive a dispersion relation that we accomplish to solve for both the growth rate of the instability and the stability threshold for two specific models: the Hamiltonian Mean-Field model with additional asymmetry and the mean-field $\phi^4$ model. The results are compared with numerical simulations of the $N$-body dynamics. When the {\it inhomogeneous} state is stationary, we expect to observe in the $N$-body dynamics Quasi-Stationary-States (QSS), whose lifetime diverge algebraically with $N$.
• Out-of-equilibrium phase transitions in the HMF model: a closer look(1012.1566)

Dec. 7, 2010 cond-mat.stat-mech
We provide a detailed discussion of out-of-equilibrium phase transitions in the Hamiltonian Mean Field (HMF) model in the framework of Lynden-Bell's statistical theory of the Vlasov equation. For two-levels initial conditions, the caloric curve $\beta(E)$ only depends on the initial value $f_0$ of the distribution function. We evidence different regions in the parameter space where the nature of phase transitions between magnetized and non-magnetized states changes: (i) for $f_0>0.10965$, the system displays a second order phase transition; (ii) for $0.109497<f_0<0.10965$, the system displays a second order phase transition and a first order phase transition; (iii) for $0.10947<f_0<0.109497$, the system displays two second order phase transitions; (iv) for $f_0<0.10947$, there is no phase transition. The passage from a first order to a second order phase transition corresponds to a tricritical point. The sudden appearance of two second order phase transitions from nothing corresponds to a second order azeotropy. This is associated with a phenomenon of phase reentrance. When metastable states are taken into account, the problem becomes even richer. In particular, we find a new situation of phase reentrance. We consider both microcanonical and canonical ensembles and report the existence of a tiny region of ensembles inequivalence. We also explain why the use of the initial magnetization $M_0$ as an external parameter, instead of the phase level $f_0$, may lead to inconsistencies in the thermodynamical analysis.
• Experimental perspectives for systems based on long-range interactions(1004.4963)

June 14, 2010 nlin.CD, physics.acc-ph
The possibility of observing phenomena peculiar to long-range interactions, and more specifically in the so-called Quasi-Stationary State (QSS) regime is investigated within the framework of two devices, namely the Free-Electron Laser (FEL) and the Collective Atomic Recoil Laser (CARL). The QSS dynamics has been mostly studied using the Hamiltonian Mean-Field (HMF) toy model, demonstrating in particular the presence of first versus second order phase transitions from magnetized to unmagnetized regimes in the case of HMF. Here, we give evidence of the strong connections between the HMF model and the dynamics of the two mentioned devices, and we discuss the perspectives to observe some specific QSS features experimentally. In particular, a dynamical analog of the phase transition is present in the FEL and in the CARL in its conservative regime. Regarding the dissipative CARL, a formal link is established with the HMF model. For both FEL and CARL, calculations are performed with reference to existing experimental devices, namely the FERMI@Elettra FEL under construction at Sincrotrone Trieste (Italy) and the CARL system at LENS in Florence (Italy).
• Can one measure negative specific heat in the canonical statistical ensemble?(1003.0631)

March 2, 2010 cond-mat.stat-mech
According to thermodynamics, the specific heat of Boltzmannian short-range interacting systems is a positive quantity. Less intuitive properties are instead displayed by systems characterized by long-range interactions. In that case, the sign of specific heat depends on the considered statistical ensemble: negative specific heat can be found in isolated systems, which are studied in the framework of the microcanonical ensemble; on the other hand, it is generally recognized that a positive specific heat should always be measured in systems in contact with a thermal bath, for which the canonical ensemble is the appropriate one. We demonstrate that the latter assumption is not generally true: one can in principle measure negative specific heat also in the canonical ensemble if the system under scrutiny is non-Boltzmannian and/or out-of-equilibrium.
• Lyapunov stability of Vlasov Equilibria using Fourier-Hermite modes(0904.4656)

April 29, 2009 physics.plasm-ph
We propose an efficient method to compute Lyapunov exponents and Lyapunov eigenvectors of long-range interacting many-particle systems, whose dynamics is described by the Vlasov equation. We show that an expansion of a distribution function using Hermite modes (in momentum variable) and Fourier modes (in configuration variable) converges fast if an appropriate scaling parameter is introduced and identified with the inverse of the system temperature. As a consequence, dynamics and linear stability properties of many-particle states, both in the close-to and in the far-from equilibrium regimes can be predicted using a small number of expansion coefficients. As an example of a long-range interacting system we investigate stability properties of stationary states of the Hamiltonian mean-field model.
• Out-of-equilibrium phase re-entrance(s) in long-range interacting systems(0903.5039)

March 29, 2009 cond-mat.stat-mech
Systems with long-range interactions display a short-time relaxation towards Quasi Stationary States (QSSs) whose lifetime increases with system size. The application of Lynden-Bell's theory of "violent relaxation" to the Hamiltonian Mean Field model leads to the prediction of out-of-equilibrium first and second order phase transitions between homogeneous (zero magnetization) and inhomogeneous (non-zero magnetization) QSSs, as well as an interesting phenomenon of phase re-entrances. We compare these theoretical predictions with direct $N$-body numerical simulations. We confirm the existence of phase re-entrance in the typical parameter range predicted from Lynden-Bell's theory, but also show that the picture is more complicated than initially thought. In particular, we exhibit the existence of secondary re-entrant phases: we find un-magnetized states in the theoretically magnetized region as well as persisting magnetized states in the theoretically unmagnetized region.
• Chaotic dynamics in a storage-ring Free Electron Laser(physics/0203083)

March 27, 2002 physics.optics, cond-mat
The temporal dynamics of a storage-ring Free Electron Laser is here investigated with particular attention to the case in which an external modulation is applied to the laser-electron beam detuning. The system is shown to produce bifurcations, multi-furcations as well as chaotic regimes. The peculiarities of this phenomenon with respect to the analogous behavior displayed by conventional laser sources are pointed out. Theoretical results, obtained by means of a phenomenological model reproducing the evolution of the main statistical parameters of the system, are shown to be in a good agreement with experiments carried out on the Super-ACO Free Electron Laser.


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# Math Help - Orthogonal Complement 2

1. ## Orthogonal Complement 2

Dear Colleagues,

If $M\neq \phi$ is any subset of a Hilbert space $H$, show that $M^{\bot\bot}$ is the smallest closed subspace of $H$ which contains M, that is, $M^{\bot\bot}$ is contained in any closed subspace $Y\subset H$ such that $Y\supset M$.

Regards,

Raed.

2. What did you try ?

3. Give me a hint please and I will then continue.

4. Originally Posted by raed
Give me a hint please and I will then continue.
No, that's not how it works.

What have you tried? Where are you stuck? You have to make an effort.

format: plain


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# FAQ#

## Installation#

### Setup fails#

• In case you encounter errors during setup or build process, please make sure to use an appropriate branch for your environment
• ROS 2 Galactic, Ubuntu 20.04: galactic branch
• ROS 2 Humble, Ubuntu 22.04: main branch
• It's also important to delete ./build ./install and ./src directory before rebuilding CARET if you have built CARET using wrong settings

### CLI tool doesn't work#

In case CLI tool execution fails, please make sure to perform CARET environment settings

source /opt/ros/humble/setup.bash
source ~/ros2_caret_ws/install/local_setup.bash

ros2 caret check_caret_rclcpp --workspace <path-to-workspace>

Warning

CARET CLI tool doesn't work properly in Anaconda environment. Please use pure Python.

## Recording#

### LTTng session doesn't start after ros2 caret record#

• Check detailed status for recording sequence by adding -verbose option (e.g. ros2 caret record -v )
• You will see N/M process started recording , where N is the number of processes which have been started recording and M is the number of total processes to be started recording
• If N increases very slowly, add --recording-frequency option with integer greater than 100 (e.g. ros2 caret record -f 500 )
• Please be careful that it increases the possibility of recording failure
• If N remains 0, remove ~/.lttng and start recording again
• Please be careful that the first recorded trace data after removing ~/.lttng tends to lack some events. So, please ignore the data
• It's also important to make sure you don't have another LTTng session running

### So many nodes named /caret_trace_ooooooo created#

• As described in design section, a node to store CARET events is created for each process. Therefore, if a target application is huge and has a lot of processes, the number of CARET nodes also becomes huge

## Visualization#

### Result (plot, message_flow, etc.) is not outputted, or there seems something wrong with the result#

• Please use the following commands for verification
• ros2 caret check_caret_rclcpp to check if a target application is built with CARET/rclcpp
• ros2 caret check_ctf to check if tracing data is recorded properly
• Please make sure the followings:
• A target application is built with CARET/rclcpp
• CARET environment is set properly before running a target application
• export LD_PRELOAD=\$(readlink -f ~/ros2_caret_ws/install/caret_trace/lib/libcaret.so)
• source ~/ros2_caret_ws/install/local_setup.bash
• LTTng trace is started before running a target application
• ros2 trace -s e2e_sample -k -u "ros2*"
• or consider to use launch file
• Trace data is not discarded
• The size of trace data is proper
• If the size of trace data is extremely small (e.g. only few KByte) and a target application has lots of nodes, the maximum number of file descriptors may not be enough. It can be increased by ulimit -n 65536
• See Recording for more details

### Parts of results are not outputted#

• If certain nodes are not traced but some nodes are traced, some packages may be built without CARET/rclcpp. Please make sure <depend>rclcpp</depend> is described in package.xml
• Another possibility is that some nodes can't be analyzed due to CARET's limitations:
• CARET cannot analyze a node which has two or more timer callbacks with the same period time setting
• CARET cannot analyze a node which has two or more subscription callbacks with the same topic name
• Callback information in such nodes are not outputted. Also, message flow will be discontinued at such nodes

### TraceResultAnalyzeError: Failed to find error occurs#

• The error occurs if information in an architecture file and trace data are inconsistent with each other
• Please modify the architecture file or check recording process
• e.g.
• TraceResultAnalyzeError: Failed to find callback_object.node_name: /localization/pose_twist_fusion_filter/ekf_localizer, callback_name: timer_callback_0, period_ns: 19999999, symbol: void (EKFLocalizer::?)()

## Visualization (callback)#

### Callback frequency is smaller than expected value#

• Plot.create_callback_frequency_plot calculates frequency from one second to one second. It counts how many times a callback function is called for a second, and just uses the count as frequency. Therefore, the frequency on the last term tends to small because the last term is usually shorter than one second
• Another possibility is that the frequency of a subscription callback will be small if it receives topics not periodically but infrequently. Also, the frequency of a timer callback will be small if the timer dynamically stops/starts

### Callback latency is bigger than expected value#

• Some nodes may run initialization process. In this case, the latency time calculated by Plot.create_callback_latency_plot is huge on the first execution

## Visualization (message flow)#

### Message flow is discontinued#

• If parts of nodes/communications don't run at all during recording, message flow stops on the way and such nodes/communications are not displayed on y-axis
• Please make sure that all nodes/communications in a target path run during recording, or modify a target path to analyze actually working path
• Another possibility is that a target path includes a node which CARET cannot analyze due to its limitations as explained above

### What is a gray color rectangle in a message flow diagram?#

• A rectangle in a message flow diagram indicates a period from the entry to the exit of a callback function, while a line shows a flow of topics
• Note: a rectangle is not always illustrated

### Huge delay between topic publication and callback start#

• In a message flow diagram, elapsed time from ooo/rclcpp_publish to ooo/callback_start means latency from when a topic is published to when the following callback starts
• It includes the following time:
• Communication (topic) latency
• Wait by ROS scheduler
• Wait by OS scheduler
• In most cases, it doesn't take so much time. In case the time is huge, the followings are possible causes:
• There is a problem in communication
• An executor cannot wake up because other processes occupy CPU
• A callback cannot wake up because other callbacks in the same callback group occupy the executor
• Processing time of a callback is longer than topic subscription period

### Message flow looks split#

• Take the following system for example;
• Node_C publishes a topic when it receives a topic from Node_A
• Node_A publishes a topic with a rate of 50 Hz, while Node_B publishes a topic at 10 Hz
• Message flow (Node_B -> Node_C -> Node_D) looks split at Node_C. It's because Node_C publishes 5 topics while receiving 1 topic from Node_B
• Note: A similar phenomenon will happen even if Callback c0 is a timer callback

### Message flow looks dropped#

• Take the following system for example;
• Node_C publishes a topic when it receives a topic from Node_A
• Node_A publishes a topic with a rate of 10 Hz, while Node_B publishes a topic at 50 Hz
• Message flow (Node_B -> Node_C -> Node_D) looks disconnected at Node_C four times every five messages. It's because Node_C publishes 1 topic while receiving 5 topics from Node_B. So 4 topics don't have a corresponding topic to be published to Node_D
• Note: A similar phenomenon will happen even if Callback c0 is a timer callback

### How response time is calculated?#

• In general, response time is the time a system or functional unit takes to react to a given input (reference). Response time calculated by CARET is the time it takes for input data to arrive at the last node. It doesn't include processing time at the first/last node nor latency of an actuator. It's calculated as the sum of communication latency time (from the time when a node publishes a topic to the time when the following node subscribes the topic) and node latency time (from the time when a node subscribes a topic to the time when it publishes another topic) in a path
• In the following diagram, input data at point A is first reflected with output at point X (ResponseTime_Best)
• ResponseTime_Best can be considered as a path (dataflow) latency time
• ResponseTime_Best can be considered as a happy case, which is contrary to the following worst case scenario
• Assuming that input information is created by a sensor such as an object detection sensor, delay in a sensor should be considered. For instance, if a new object appears at point B, the time from point B to point A should added to the response time. The worst case scenario is that a new object appears just after the previous flow (point C). Response time for the worst case is shown as ResponseTime_Worst
• CARET can calculate both ResponseTime_Best and ResponseTime_Worst using the following APIs:
• response_time.to_best_case_timeseries() , response_time.to_best_case_histogram()
• response_time.to_worst_case_timeseries() , response_time.to_best_worst_histogram()
• CARET also provides response_time.to_histogram() API. It creates histogram assuming a new object appears from point C to point A at intervals of histogram bin size


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## Trigonometry (10th Edition)

Published by Pearson

# Chapter 1 - Trigonometric Functions - Section 1.2 Angle Relationships and Similar Triangles - 1.2 Exercises - Page 18: 59

500 m; 700m

#### Work Step by Step

1. This is the ratio of the two shortest sides of the similar triangle. Reduced to make easier to solve equations $\frac{4}{400}=\frac{1}{100}$ 2. Then construct all three ratios to find unknown sides $\frac{1}{100}=\frac{5}{a}=\frac{7}{b}$ 3. Cross-multiplication --> Solutions : a= 500 b= 700

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


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# HW2, Q.3(a)

Hi all,

I’m a little confused by what the sequences $S_0,S_1,\dots,S_k$ look like. I’m thinking these are cycles (ie. permutations) with elements from $[n]$. And so then we’d be looking for the number of increasing and decreasing sequences of subsets. Any thoughts? Thanks!

### About Anastasia Chavez

I am currently a 3rd year graduate student at UC Berkeley in mathematics. My research interests are in ennumerative combinatorics. I am originally from Santa Rosa, Ca and have been a Bay Area transplant since 2003. My husband, two children, dog and 5 chickens now reside in Berkeley. Besides counting, I enjoy hiking, yoga, playing games of the imagination with my daughters, stand-up comedy and day dreaming about backpacking adventures with my husband. I plan to make those dreams, among others, a reality!

1. I think they are simply a sequence of subsets each of which increases or decreases by one element comparing to the previous subset.

2. Anastasia Chavez

Ah, so we consider subsets of \$\latex [n]\$ as usual (ie. subsets of \$\latex [3]\$ are \$\latex\{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\}, \{1,2,3\}\$). Thanks!

3. plus the empty set, i do believe!

4. (assuming that |{}-{1}|=1, etc. why not?)

5. Hi everyone! For n=2 and k=3 is the following sequence a “valid sequence”?:
$S_0=\emptyset, \ S_1=\{1\}, S_2=\{1,2\}, S_3=\{2\}$
If so… Does n have to be even on part (b)?

6. Sorry, I mean, does k have to be even?
By the way…
Is there a way to edit a comment?

7. yes & yes!

8. & (last question) i don’t know!

9. i mean if k is odd there are zero sequences with this property

10. Thanks!


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# Microbiological assay of vitamins

Microbiological assay of vitamins is a biological assay method performed with the help of microorganisms. Vitamins and amino acids are the essential components for the growth of microorganisms. The basis of this assay is to measure the ability of the test organism to utilize the substance being assayed under a proper nutritional condition. Some examples of microorganisms used for the bioassay of vitamins are listed in Table 1.1.

Materials required

A stock solution, inoculum media, and assay medium are required. Finally, a standard curve is obtained.

## Assay of Vitamin B12

Vitamin B12 is also known as cyanocobalamin. It is a water-soluble vitamin. Its main sources are liver, eggs, milk, meat, and fish. Vitamin B12 deficiency causes Macrolytic anemia, Pernicious anemia.

Principle: The test organism selected is Lactobacillus liechmannii that is capable of utilizing free cyanocobalamin. The assay is performed by using the either titrimetric or turbidimetric method.

### Preparation of Standard Stock Solution

An accurately weighed amount of Cyanocobalamin reference standard is added to sufficient 25% ethanol (resulting in a solution containing 1.0 µ g of cyanocobalamin per ml) and stored in the refrigerator. Further dilutions of this stock solution (1 µg/ ml) are made by adding 1 ml stock solution to 99 ml purified water (1 ml = 10 ng) and further adding 1 ml of the above solution to 199 ml purified water (1 ml = 0.05 ng).

Test Solution: Accurate amount of vitamin to be assayed is taken and dissolved in water, dilute HCl or NaOH is added to adjust pH at 6.0 and make up the volume with water up to the mark.

Inoculum Preparation: Transfer a loop full of Lactobacillus liechmannii from a recent sub-culture into two tubes each containing 10 ml of sterile culture medium.

### Composition of culture media

Yeast extract – 0.75 gm

Peptone – 0.75 gm

Dextrose – 1 gm

Potassium dihydrogen phosphate – 0.2 gm

Tomato juice filtrate – 10 ml

Sorbitan monooleate solution – 1 ml

Water up to – 100 ml

Media is incubated for 18 to 24 hours at 37o C. The culture is centrifuged and decanted the supernatant fluid, under aseptic conditions. These cultured cells are suspended into 10 ml of a sterile suspension of Basal medium stock solution and again centrifuged and decanted off supernatant fluid. The cells are uniformly suspended in 10 ml of sterile medium and then aseptically transferred 1 ml of the cell suspension to 10 ml of sterile medium and mixed. Finally, this resulting cell suspension is taken as inoculum.

### Assay Procedure

In clean ten test tubes, added 0, 0.5, 1.0, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5, and 5 ml of standard cyanocobalamin solution. To each tube added 5 ml of Basal medium solution and the volume of each is adjusted to 10 ml by water. In another four test tubes added 1, 2, 3, 4 ml of test solution which is to be assayed. To each of these also added 5 ml of Basal medium stock solution and adjusted the volume to 10 ml with water. All test tubes are sterilized in an autoclave at 121o C for 15 minutes after that cooled the test tubes at room temperature. Inoculate a drop of inoculum prepared of Lactobacillus liechmannii and incubated the test tubes for 64 to 72 hours at the temperature range of 30 to 37o C. After incubation period titrated contents of each test tube with 0.05 N NaOH using bromothymol blue as indicator until a green color endpoint. Calculate the readings.

The average of titration values of each level of both standard and test solutions are determined and plotted on a graph considered average titration values (in ml) of 0.05 N NaOH against the concentration of standard cyanocobalamin solution. A linear graph is obtained by interpolating the standard curve to determine the concentration as activity per ml of vitamin B12. From the graph, the concentration of the test solution of cyanocobalamin is calculated (Fig. 1.1).

## Assay of Calcium Pantothenate

It refers to the B complex vitamins (or vitamin B complex).

### Standardized Stock Solution

Each ml of the stock solution consists of 50 mcg of calcium pantothenate. It is prepared by dissolving 50 mg of calcium pantothenate in 500 ml of double-distilled water as per British Pharmacopeial Chemical Reference Standard; 10 ml of 0.2 M acetic acid, 100 ml of 1.6% (w/v) sodium acetate; and volume made up to 1 liter with distilled water.

Standard Solution: The standard solution should contain approximately 0.04 mcg of calcium pantothenate in 1 ml, and is duly prepared by diluting the Standard Stock Solution.

Test Solution: The test solution contains nearly the same equivalent amount of calcium pantothenate as present in the Standard Solution i.e., 0.4 mcg/ml prepared in double-distilled water.

### Culture Medium

The culture medium is composed of the following solutions and ingredients:

1. Casein hydrolysate solution: 25 ml
2. Cysteine-tryptophane solution : 25 ml
3. Polysorbate-80 solution: 0.25 ml
4. Dextrose (anhydrous): 10 g
5. Sodium acetate (anhydrous): 5 g
7. Riboflavin-Thiamine hydrochloride-Biotin Solution: 5 ml
8. PABA-Niacin-Pyridoxine hydrochloride solution: 5 ml
9. Calcium pantothenate solution A: 5 ml
10. Calcium pantothenate solution B: 5 ml

The culture medium is usually prepared by dissolving both anhydrous dextrose and sodium acetate in previously mixed solutions and the pH is carefully adjusted to 6.8 with 1 M NaOH solution. The final volume is duly made up to 250 ml with distilled water and mixed thoroughly.

### Stock Culture of Organism

The stock culture of the organism may be prepared to dissolve 2 g water-soluble yeast extract in 100 ml distilled water, 500 mg anhydrous dextrose, 500 mg anhydrous sodium acetate, and 1.5 g agar. The resulting mixture is heated gently so as to dissolve the agar. Now, 10 ml of hot solution is transferred to test tubes and sterilized at 121°C. The stab culture is now prepared duly in three tubes employing Lactobacillus Plantarum, incubated at 30 to 37°C for 16 to 24 hours, and stored in a refrigerator ultimately.

### Preparation of Inoculum

The cells consequently obtained from the stock culture, organism (Lactobacillus Plantarum) are transferred to a sterile tube containing 10 ml of the culture medium and is incubated at 30 to 37°C for a duration of 16–24 hours. Methodology. The various steps involved are namely:

• Standard Solution is added to five test tubes in varying amounts viz., 1, 2, 3, 4, and 5 ml in duplicate.
• To each of the five above test tubes plus another four similar tubes without any standard solution is added 5 ml of culture medium, and the final volume made up to 10 ml with distilled water.
• Now, volumes of test solution corresponding to either three or more of the levels as taken above, are incorporated carefully to similar test tubes, in duplicate.
• To each test tube, 5 ml of the medium solution and volume is made up to 10 ml with distilled water.
• Tubes of both the series are duly heated in an autoclave at 121°C for 5 minutes, cooled to ambient temperature, and added 1 drop of inoculum to each tube except two of the four tubes that specifically has no ‘standard solution’ (i.e., the uninoculated tubes), and mixed thoroughly. The tubes are adequately incubated at 121°C at 30–37°C for 16– 24 hours.
• The absorbance of the various tubes is measured with a spectrophotometer at a wavelength ranging between 540–660 nm. A standard concentration-response curve is plotted between the transmittance Vs log ml (volume) of the standard solution in each tube. Finally, the exact concentration of the calcium pantothenate in the ‘test sample’ is determined with the help of the standard concentration-response curve.
Make sure you also check our other amazing Article on : Principles and Methods of Different Microbiological Assay


text:
# Galois extension

In mathematics, a Galois extension is an algebraic field extension E/F that is normal and separable; or equivalently, E/F is algebraic, and the field fixed by the automorphism group Aut(E/F) is precisely the base field F. The significance of being a Galois extension is that the extension has a Galois group and obeys the fundamental theorem of Galois theory. [1]

A result of Emil Artin allows one to construct Galois extensions as follows: If E is a given field, and G is a finite group of automorphisms of E with fixed field F, then E/F is a Galois extension.

## Characterization of Galois extensions

An important theorem of Emil Artin states that for a finite extension ${\displaystyle E/F,}$ each of the following statements is equivalent to the statement that ${\displaystyle E/F}$ is Galois:

• ${\displaystyle E/F}$ is a normal extension and a separable extension.
• ${\displaystyle E}$ is a splitting field of a separable polynomial with coefficients in ${\displaystyle F.}$
• ${\displaystyle |\!\operatorname {Aut} (E/F)|=[E:F],}$ that is, the number of automorphisms equals the degree of the extension.

Other equivalent statements are:

• Every irreducible polynomial in ${\displaystyle F[x]}$ with at least one root in ${\displaystyle E}$ splits over ${\displaystyle E}$ and is separable.
• ${\displaystyle |\!\operatorname {Aut} (E/F)|\geq [E:F],}$ that is, the number of automorphisms is at least the degree of the extension.
• ${\displaystyle F}$ is the fixed field of a subgroup of ${\displaystyle \operatorname {Aut} (E).}$
• ${\displaystyle F}$ is the fixed field of ${\displaystyle \operatorname {Aut} (E/F).}$
• There is a one-to-one correspondence between subfields of ${\displaystyle E/F}$ and subgroups of ${\displaystyle \operatorname {Aut} (E/F).}$

## Examples

There are two basic ways to construct examples of Galois extensions.

• Take any field ${\displaystyle E}$, any subgroup of ${\displaystyle \operatorname {Aut} (E)}$, and let ${\displaystyle F}$ be the fixed field.
• Take any field ${\displaystyle F}$, any separable polynomial in ${\displaystyle F[x]}$, and let ${\displaystyle E}$ be its splitting field.

Adjoining to the rational number field the square root of 2 gives a Galois extension, while adjoining the cubic root of 2 gives a non-Galois extension. Both these extensions are separable, because they have characteristic zero. The first of them is the splitting field of ${\displaystyle x^{2}-2}$; the second has normal closure that includes the complex cubic roots of unity, and so is not a splitting field. In fact, it has no automorphism other than the identity, because it is contained in the real numbers and ${\displaystyle x^{3}-2}$ has just one real root. For more detailed examples, see the page on the fundamental theorem of Galois theory.

An algebraic closure ${\displaystyle {\bar {K}}}$ of an arbitrary field ${\displaystyle K}$ is Galois over ${\displaystyle K}$ if and only if ${\displaystyle K}$ is a perfect field.

## References

1. ^ See the article Galois group for definitions of some of these terms and some examples.

• Artin, Emil (1998) [1944]. Galois Theory. Edited and with a supplemental chapter by Arthur N. Milgram. Mineola, NY: Dover Publications. ISBN 0-486-62342-4. MR 1616156.
• Bewersdorff, Jörg (2006). Galois theory for beginners. Student Mathematical Library. 35. Translated from the second German (2004) edition by David Kramer. American Mathematical Society. doi:10.1090/stml/035. ISBN 0-8218-3817-2. MR 2251389.
• Edwards, Harold M. (1984). Galois Theory. Graduate Texts in Mathematics. 101. New York: Springer-Verlag. ISBN 0-387-90980-X. MR 0743418. (Galois' original paper, with extensive background and commentary.)
• Funkhouser, H. Gray (1930). "A short account of the history of symmetric functions of roots of equations". American Mathematical Monthly. The American Mathematical Monthly, Vol. 37, No. 7. 37 (7): 357–365. doi:10.2307/2299273. JSTOR 2299273.CS1 maint: ref=harv (link)
• "Galois theory", Encyclopedia of Mathematics, EMS Press, 2001 [1994]
• Jacobson, Nathan (1985). Basic Algebra I (2nd ed.). W.H. Freeman and Company. ISBN 0-7167-1480-9. (Chapter 4 gives an introduction to the field-theoretic approach to Galois theory.)
• Janelidze, G.; Borceux, Francis (2001). Galois theories. Cambridge University Press. ISBN 978-0-521-80309-0.CS1 maint: ref=harv (link) (This book introduces the reader to the Galois theory of Grothendieck, and some generalisations, leading to Galois groupoids.)
• Lang, Serge (1994). Algebraic Number Theory. Graduate Texts in Mathematics. 110 (Second ed.). Berlin, New York: Springer-Verlag. doi:10.1007/978-1-4612-0853-2. ISBN 978-0-387-94225-4. MR 1282723.CS1 maint: ref=harv (link)
• Postnikov, Mikhail Mikhaĭlovich (2004). Foundations of Galois Theory. With a foreword by P. J. Hilton. Reprint of the 1962 edition. Translated from the 1960 Russian original by Ann Swinfen. Dover Publications. ISBN 0-486-43518-0. MR 2043554.
• Rotman, Joseph (1998). Galois Theory (Second ed.). Springer. doi:10.1007/978-1-4612-0617-0. ISBN 0-387-98541-7. MR 1645586.
• Völklein, Helmut (1996). Groups as Galois groups: an introduction. Cambridge Studies in Advanced Mathematics. 53. Cambridge University Press. doi:10.1017/CBO9780511471117. ISBN 978-0-521-56280-5. MR 1405612.CS1 maint: ref=harv (link)
• van der Waerden, Bartel Leendert (1931). Moderne Algebra (in German). Berlin: Springer.CS1 maint: ref=harv (link). English translation (of 2nd revised edition): Modern algebra. New York: Frederick Ungar. 1949. (Later republished in English by Springer under the title "Algebra".)
• Pop, Florian (2001). "(Some) New Trends in Galois Theory and Arithmetic" (PDF).CS1 maint: ref=harv (link)


text:
# A double series with divisibility restrictions

How may I restrict $i,j$ such that they run over $\mathbb{N}$ excepting the numbers divisible by $2$ or $3$ or $7$?

Sum[1/( i j (i + j + 1)), {i, 1, Infinity}, {j, 1, Infinity}]

-
Could you please add a bit of context just to say how you got into this problem ? –  b.gatessucks Sep 13 '13 at 15:20
@b.gatessucks I'm doing some personal research on multiseries. –  Chris's sis the artist Sep 13 '13 at 15:32

Could do explicit inclusion/exclusion as below (I forget what the correct term is but you'll get the idea). First do one index.

s1 = Sum[1/(i j (i + j + 1)), {j, 1, Infinity}];

s2 = Sum[1/(i j (i + j + 1)), {j, 2, Infinity, 2}];
s3 = Sum[1/(i j (i + j + 1)), {j, 3, Infinity, 3}];
s7 = Sum[1/(i j (i + j + 1)), {j, 7, Infinity, 7}];

s23 = Sum[1/(i j (i + j + 1)), {j, 2*3, Infinity, 2*3}];
s27 = Sum[1/(i j (i + j + 1)), {j, 2*7, Infinity, 2*7}];
s37 = Sum[1/(i j (i + j + 1)), {j, 3*7, Infinity, 3*7}];
s237 = Sum[1/(i j (i + j + 1)), {j, 2*3*7, Infinity, 2*3*7}];

stot = s1 - (s2 + s3 + s7) + (s23 + s27 + s37) - s237;

Can now repeat the process with the other index. Below would be the first step.

t1 = Sum[stot, {i, 1, Infinity}]

Unfortunately I cannot get a closed form for this.

If there is a way to rewrite as some form of convolution sum that might be amenable to Fourier series type methods. Offhand I don't see such a rewrite but I could easily be missing something of that sort.

-
OK. Thank you for answer! –  Chris's sis the artist Sep 13 '13 at 15:11

I tried to go the usual tricky route in order to use symmetries :

$$\sum_{i,j} \frac{1}{i\ j (i+j+1)} = \int_{0}^{\infty} dx\ e^{-x} \left(\sum_{i} \frac{e^{-i x}}{i} \right)^2$$

The sum can be performed in closed form :

s = Sum[Exp[-i x]/i Boole[Mod[i, 2] != 0] Boole[Mod[i, 3] != 0] Boole[
Mod[i, 7] != 0] , {i, 1, Infinity}]

but the integral cannot. Numerically :

NIntegrate[Exp[-x] s s, {x, 0, Infinity}]
(* 0.480345 *)

which looks in line with @tintin.

-
Thank you for your answer! –  Chris's sis the artist Sep 13 '13 at 15:41

You can also use Boole as follows:

Sum[1/(i j (i + j + 1)) Boole[! Divisible[i, 2] && !
Divisible[i, 3] && ! Divisible[i, 7] && ! Divisible[j, 2] && !
Divisible[j, 3] && ! Divisible[j, 7]], {i, 1, 1000}, {j, 1,
1000}] // N

0.478819

Which gives the same numerical value as tintin's Mod version.

To speed things up, you can use the new iteration form in Mathematica

First define lis as follows:

lis = Select[Range[1, 1000, 2], Not@Divisible[#, 3] && Not@Divisible[#, 7] &] // N;

Then:

Sum[1/(i j (i + j + 1)), {i, lis}, {j, lis}]

0.478819

-
OK. Thank you for answer! –  Chris's sis the artist Sep 13 '13 at 15:42
2x speedup: add N@ before Select –  ybeltukov Sep 13 '13 at 17:52
@ybeltukov. Yes, good point. And this makes the // N after Sum redundant. Thanks. –  RunnyKine Sep 13 '13 at 18:10
f[i_, j_] := 0;
f[i_, j_] :=
1/(i j (i + j + 1)) /;
Mod[i, 2] != 0 && Mod[i, 3] != 0 && Mod[i, 7] != 0 &&
Mod[j, 2] != 0 && Mod[j, 3] != 0 && Mod[j, 7] != 0

you can sum it with

Sum[f[i,j],{i,1,1000},{j,1,1000}]

which I get the number of 0.478819 I write 1000 because I cant deal with the infinity thing

-
Thank you for answer! –  Chris's sis the artist Sep 13 '13 at 14:59
note you can do Sum[f[i, j], {i, 1, 1000, 2}, {j, 1, 1000, 2}] and omit the Mod[,2] check.. speed it up a tad. –  george2079 Sep 13 '13 at 15:13
@george2079,that is definately a big improvement –  tintin Sep 13 '13 at 15:19

I think you can achieve this way too,

Total[Total[
Table[If[(Divisible[i, #] && Divisible[j, #]) & /@ {2, 3, 7}  /.
List -> Or, 1/(i j (i + j + 1)), 0], {i, 1, 1000}, {j, 1,
1000}]]] // N

0.344854

Sum will achieve the same result, its gives same result as a Table with two iterators as in your case. If you do NSum[1/(i j (i + j + 1)), {i, 1, 1000}, {j, 1, 1000}] you get

SequenceLimit::seqlim: The general form of the sequence could not be determined, and the result may be incorrect. 1.97715

But if you use Total[Total[Table[1/(i j (i + j + 1)), {i, 1, 1000}, {j, 1, 1000}]] // N]

1.98443 (No warnings)

Now to see that same results are generated by Sum and custom loop check these,

Sum[1/(i j (i + j + 1)), {i, 1, 50}, {j, 1, 50}] // N

1.81095

Total[Total[Table[1/(i j (i + j + 1)), {i, 1, 50}, {j, 1, 50}]] // N]

1.81095

'NSum' probably uses approximations hence its result is pretty close but might be wrong if high precision needed.

-


text:
## anonymous one year ago help please

1. anonymous

With?

2. anonymous

3. campbell_st

well you can use the law of cosines to check do you know how to use it..?

4. anonymous

no i dont :(

5. jennyrlz

sec let me see what i can remeber :)

6. anonymous

k

7. anonymous
8. anonymous

im to lazy to draw the abc triangle lol, but that explains the law of cosines well

9. campbell_st

the law of cosines says $Cos(A) = \frac{b^2 = c^2 - a^2}{2bc}$ does that help

10. campbell_st

you have a = 5.3, b = 7 and c = 4... plug them in and post the answer you get

11. campbell_st

oops should read $\cos(A) = \frac{b^2 + c^2 - a^2}{2bc}$

12. jennyrlz

^

13. anonymous

oh wait is it true?

14. jennyrlz

ill leave it to him, he knows what he is doing :)

15. jennyrlz

oh he left...

16. jennyrlz

well why do you think it is true?

17. jennyrlz

he did the "hard" part the rest is algebra

18. campbell_st

well you need to find the angle measure using the law of cosines to see if its true or not

19. campbell_st

an alternative solution is to google a triangle solver put in the 3 sides and then look at the angles that are calculated

20. anonymous

which law of cosine should i use

21. jennyrlz

if only i wouldve thought of this when i was taking pre-calc...

22. campbell_st

use the one I posted above

23. anonymous

how do i use it....do i sub

24. campbell_st

and look for the measures that match the labels a, b and c

25. campbell_st

a = 5.3 b = 7 and c = 4 like this $cos(\theta) = \frac{7^2 + 4^2 - 5.3^2}{2 \times 7 \times 4}$

26. anonymous

doin the math.....

27. anonymous

$\cos \theta \frac{ 49+16-28.09 }{66 }$ thats what i have so far

28. anonymous

is it right

29. anonymous

so far

30. campbell_st

well I think the denominator is 56

31. campbell_st

2 x 4 x 7 = 56

32. anonymous

oh yea

33. anonymous

wait so it is true?

34. campbell_st

not yet, if you do the calculation you'll find $\cos(\theta) = 0.659107$ so to find the angle, using a calculator its $\theta = \cos^{-1}(0.659107)$

35. anonymous

i got 0.79053943 ? how do i punch this equation in a calculator


text:
# Revisiting Deep Learning on CPU, New Optimizations Made to TensorFlow and Caffe

[ ]

Intel has been working hard to optimizing the popular deep learning frameworks, e.g., TensorFlow and Caffe, on (Intel) CPUs. Their work has reduced the performance gaps between CPU and GPU from about 80x to 2x.

# TensorFlow* Optimizations on Modern Intel® Architecture

This is a joint work between Intel and Google.

Optimizing deep learning models performance on modern CPUs presents a number of challenges not very different from those seen when optimizing other performance-sensitive applications in High Performance Computing (HPC):

• Code refactoring needed to take advantage of modern vector instructions. This means ensuring that all the key primitives, such as convolution, matrix multiplication, and batch normalization are vectorized to the latest SIMD instructions (AVX2 for Intel Xeon processors and AVX512 for Intel Xeon Phi processors).
• Maximum performance requires paying special attention to using all the available cores efficiently. Again this means looking at - parallelization within a given layer or operation as well as parallelization across layers.
• As much as possible, data has to be available when the execution units need it. This means balanced use of prefetching, cache blocking techniques and data formats that promote spatial and temporal locality.

To meet these requirements, Intel developed a number of optimized deep learning primitives that can be used inside the different deep learning frameworks to ensure that we implement common building blocks efficiently. In addition to matrix multiplication and convolution, these building blocks include:

• Direct batched convolution
• Inner product
• Pooling: maximum, minimum, average
• Normalization: local response normalization across channels (LRN), batch normalization
• Activation: rectified linear unit (ReLU)
• Data manipulation: multi-dimensional transposition (conversion), split, concat, sum and scale.

The result on Alexnet can be found here.

The installation is also very simple:

# Python 2.7

# Python 3.5

# Python 3.6

# Intel® Optimized Caffe

Intel has been optimizing Caffe as well.

The is a comparison of original Caffe with optimized Caffe on Cifar-10.

The installation is simple as:

git clone https://github.com/intel/caffe

then change the configuration file:

BLAS :=mkl

BLAS_INCLUDE := /opt/intel/mkl/include
BLAS_LIB := /opt/intel/mkl/lib/intel64

The other compilation and installation steps are identical to original Caffe. The optimizations made includes:

• Use ‘mkl’ as BLAS library : Specify ‘BLAS := mkl’ in Makefile.config and configure the location of your MKL’s include and lib location also.
• Set CPU utilization limit :
echo "100" | sudo tee /sys/devices/system/cpu/intel_pstate/min_perf_pct
echo "0" | sudo tee /sys/devices/system/cpu/intel_pstate/no_turbo

• Put ‘engine:”MKL2017” ‘ at the top of your train_val.prototxt or solver.prototxt file or use this option with caffe tool : -engine “MKL2017”
• Current implementation uses OpenMP threads. By default the number of OpenMP threads is set to the number of CPU cores. Each one thread is bound to a single core to achieve best performance results. It is however possible to use own configuration by providing right one through OpenMP environmental variables like KMP_AFFINITY, OMP_NUM_THREADS or GOMP_CPU_AFFINITY. For the example run below , ‘OMP_NUM_THREADS = 64’ has been used.
• Intel® Optimized Caffe* has edited many parts of original BVLC Caffe* code to achieve better code parallelization with OpenMP. Depending on other processes running on the background, it is often useful to adjust the number of threads getting utilized by OpenMP. For Intel Xeon Phi™ product family single-node we recommend to use OMP_NUM_THREADS = numer_of_cores-2.
Written on October 13, 2017


text:
The current GATK version is 3.2-2

#### Howdy, Stranger!

It looks like you're new here. If you want to get involved, click one of these buttons!

Bug Bulletin: The recent 3.2 release fixes many issues. If you run into a problem, please try the latest version before posting a bug report, as your problem may already have been solved.

# java.lang.NullPointerException using BaseRecalibrator

GermanyPosts: 6Member

I became this error and wonder what could be the reason for it:

##### ERROR stack trace

java.lang.NullPointerException at org.broadinstitute.sting.utils.recalibration.covariates.RepeatCovariate.keyForRepeat(RepeatCovariate.java:225) at org.broadinstitute.sting.utils.recalibration.covariates.RepeatCovariate.recordValues(RepeatCovariate.java:100) at org.broadinstitute.sting.utils.recalibration.RecalUtils.computeCovariates(RecalUtils.java:921) at org.broadinstitute.sting.utils.recalibration.RecalUtils.computeCovariates(RecalUtils.java:901) at org.broadinstitute.sting.gatk.walkers.bqsr.BaseRecalibrator.map(BaseRecalibrator.java:263) at org.broadinstitute.sting.gatk.walkers.bqsr.BaseRecalibrator.map(BaseRecalibrator.java:132) at org.broadinstitute.sting.gatk.traversals.TraverseReadsNano$TraverseReadsMap.apply(TraverseReadsNano.java:228) at org.broadinstitute.sting.gatk.traversals.TraverseReadsNano$TraverseReadsMap.apply(TraverseReadsNano.java:216) at org.broadinstitute.sting.utils.nanoScheduler.NanoScheduler$ReadMapReduceJob.run(NanoScheduler.java:471) at java.util.concurrent.Executors$RunnableAdapter.call(Executors.java:471) at java.util.concurrent.FutureTask.run(FutureTask.java:262) at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1145) at java.util.concurrent.ThreadPoolExecutor\$Worker.run(ThreadPoolExecutor.java:615) at java.lang.Thread.run(Thread.java:744)


text:
# Get Equal Substrings Within Budget

You are given two strings s and t of the same length. You want to change s to t. Changing the i-th character of s to i-th character of t costs |s[i] - t[i]| that is, the absolute difference between the ASCII values of the characters.

You are also given an integer maxCost.

Return the maximum length of a substring of s that can be changed to be the same as the corresponding substring of twith a cost less than or equal to maxCost.

If there is no substring from s that can be changed to its corresponding substring from t, return 0.

Example 1:

Input: s = "abcd", t = "bcdf", maxCost = 3
Output: 3
Explanation: "abc" of s can change to "bcd". That costs 3, so the maximum length is 3.

Example 2:

Input: s = "abcd", t = "cdef", maxCost = 3
Output: 1
Explanation: Each character in s costs 2 to change to charactor in t, so the maximum length is 1.

Example 3:

Input: s = "abcd", t = "acde", maxCost = 0
Output: 1
Explanation: You can't make any change, so the maximum length is 1.

Constraints:

• 1 <= s.length, t.length <= 10^5
• 0 <= maxCost <= 10^6
• s and t only contain lower case English letters.

Solution:

Two pointer.

class Solution {
public:
int equalSubstring(string s, string t, int maxCost) {
vector<int> dif(s.size(), 0);
for(int i = 0; i<s.size(); i++){
dif[i] = abs((int)s[i]-t[i]);
}
int a = 0, b=0;
int cur = 0, ans =0 ;
while(b < s.size()){
cur += dif[b];
while(cur > maxCost && a <= b){
cur -= dif[a];
a++;
}
ans = max(ans, b-a+1);
b++;
}
return ans;
}
};


text:
# Why does \marginpar sometimes add an unnecessary newline?

In the following example, the main paragraph gets typeset as 3 lines of text, with the paragraph ending on the third line before it reaches the margin. However, when un-commenting the line with the \marginpar command, the paragraph is reformatted so that the third line extends all the way to the margin, and a blank fourth line is inserted. What is the \marginpar inserting into the main text that doesn't "fit" into the third line, and why?

\documentclass{amsart}
\begin{document}

\noindent
Suppose furthermore that abcde is a regular cardinal such thati abcdefghiji
The collection abcdefg is called a
nice collection of elementary submodels of $H(\theta)$ indexed by $T$ if,
in addition to being a nice collection of sets as above, we have:  Bake
%\marginpar{Bake a cake.}

Next paragraph.
\end{document}

(Note: The \noindent and the choice of document class are probably not essential to the problem, but I couldn't quite get the spacing right when trying to reproduce the problem without them.)

I realize that the practical workaround is to remove the inter-word space before the \marginpar command. However, it is not clear to me why this should be necessary. Furthermore, I discovered this problem when using the \todo command of the todonotes package, which relies on \marginpar. It seems that \todo intentionally gobbles the space following its argument, meaning that it is designed so that one should (in general) leave a space before it, but this case then becomes an exception.

The end of paragraph command (from \par or a blank line) attempts to remove any space at the end of the paragraph (before it does the linebreaking) by doing an \unskip operation. However the insert node for the marginpar protects the interword space so it doesn't get removed. If you are unlucky you get a linebreak and the last line of the paragraph only has the insert node, so gets filled with \parfillskip glue and has the appearance of a blank line. Put a % at the end of the line above and things work better. (You see the same with \footnote or \index)


text:
# Math Help - The Cantor Set

1. ## The Cantor Set

The hint given for the problem is that the set of isolated pts of a countable complete metric space X forms a dense subset of X, and using this i need to prove that the cantor set is uncountable. so i was thinking that we need to use the fact that since cantor set has no isolated points (which was proven in previous exercise), but i keep getting stuck. can someone help me?

2. Originally Posted by squarerootof2
The hint given for the problem is that the set of isolated pts of a countable complete metric space X forms a dense subset of X, and using this i need to prove that the cantor set is uncountable. so i was thinking that we need to use the fact that since cantor set has no isolated points (which was proven in previous exercise), but i keep getting stuck. can someone help me?
Suppose that $C$, the Cantor set, is countable. Since $(\mathbb{R}, | ~ |)$ is a complete metric space it would mean the isolated points of $C$, call this subset $S$, would be dense in $C$. Topologically it means $\bar S = C$. But $S = \emptyset$ by your previous exercise, so the closure of $S$ is the empty set, but that is a contradiction because it has to be $C$. This means the Cantor set is uncountable.

3. thanks so much for the input, but one small question, when you say "Since is a complete metric space " what is the metric you are using? just the ordinary euclidean metric? thanks.

4. Originally Posted by squarerootof2
thanks so much for the input, but one small question, when you say "Since is a complete metric space " what is the metric you are using? just the ordinary euclidean metric? thanks.
Yes. Here it is the same thing as taking the absolute value of the number.


text:
# Evaluate square root of 8^2

Pull terms out from under the radical, assuming positive real numbers.
Evaluate square root of 8^2


text:
# In a triangle ABC, angle ABC = 90 and BD is perpendicular to AC. If BD = 8 cm and AD = 4 cm then find the length of CD?

Feb 1, 2017

$C D = 16$ cm

#### Explanation:

Given $\angle A B C = {90}^{\circ} , B D$ is perpendicular to $A C$, $B D = 8 , \mathmr{and} A D = 4$, as shown in the diagram.
Let $\angle A C B = x , \implies \angle C B D = 90 - x$
$\implies \angle A B D = x , \implies \angle B A D = 90 - x$
$\implies \Delta A B C , \Delta A D B , \mathmr{and} \Delta B D C$ are similar

$\implies \frac{B D}{C D} = \frac{A D}{D B}$
$\implies \frac{8}{C D} = \frac{4}{8}$
$\implies C D = \frac{8 \times 8}{4} = 16$


text:
# Talk:Thalamus

Jump to: navigation, search

## Anatomy requires explanation on a more broader scale

The anatomy section requires more explanation in the sense that on reading the first paragraph, the reader can picturise the structure of thalamus.

## backwards arrow

I think one of the red arrows is backwards?

## Typo in Figure 3C?

I think the legend in Figure 3C is mislabeled -- the green is burst and red is tonic.

## Expanded lead

I just made a small edit (awaiting approval) to expand the lead of the article. I would be interested in adding more material if it can be done in an efficient way. One thing that I think would benefit the article is an "overview" section at the top discussing the diversity of thalamic nuclei and their relationships with one another and with the rest of the brain. If anybody is watching this page, please let me know if that would be, at least in principle, workable. Regards, William Skaggs 16:36, 24 July 2013 (UTC)


text:
# Generating Function for Triangular Numbers/Corollary

## Theorem

Let $\sequence {b_n}$ be the sequence defined as:

$\forall n \in \N_{> 0}: b_n = \dfrac {\paren {n + 1} \paren {n + 2} } 2$

That is:

$\sequence {b_n}_{n \mathop \ge 0} = 1, 3, 6, 10, \ldots, \dbinom {n + 2} 2, \ldots$

Then the generating function for $\sequence {b_n}$ is given as:

$H \paren z = \dfrac 1 {\paren {1 - z}^3}$

## Proof

$H \paren z = \dfrac z {\paren {1 - z}^3}$

is the generating function for $\sequence {a_n}$ as given by:

$\sequence {a_n} = 1, 3, 6, 10, \ldots, \dbinom {n + 1} 2, \ldots$

that is, such that:

$a_1 = 1, a_2 = 3, a_3 = 6, \ldots$

The result follows from Generating Function Divided by Power of Parameter.

$\blacksquare$


text:
• Lesson
6-8
1

Students explore two different methods for dividing the area of a circle in half, one of which uses concentric circles. The first assumption that many students make is that half of the radius will yield a circle with half the area. This is not true, and it surprises students. In this lesson, students investigate the optimal radius length to divide the area of a circle evenly between an inner circle and an outer ring.

Present the problem: A student has brought a giant cookie in for a celebration, and the cookie is as big as a large pizza. How could the cookie be cut so that everyone in the class gets their fair share? Note: You want students to think about cutting the circle in half, so make sure your fair share number is even—if there are an odd number of students in your class, include yourself.

Have students consider the following questions:

• Is the cookie big enough for the whole class to share?
• What fraction of the cookie should each student get?
• How would you divide the cookie?
• What is the first cut you would make? Why?

Suppose there are 22 students in a class. Ask:

[Many students will suggest cutting the cookie in half.]
• How should we divide the circle in half?
[Many students will suggest using the diameter.]

Pass out the Fair Share activity sheet. Direct students to draw lines in the first circle to indicate how they would partition it into 22 equal pieces. Specify that they must start by drawing a diameter. Observe the difficulties they encounter in drawing the rest of their lines—the middle of the circle gets filled with lines and becomes difficult to see.

 Fair Share Activity Sheet

Ask students whether this is the best way to divide up the cookie. Because of the large number of sectors, students will probably agree that the partitioning gets clumsy.

In the next problem, students will explore other ways to make the first partition—ways that will not create so much crowding at the center. The person who cuts the cookie will still need to cut 11 equal pieces from each half, but these pieces will be easier to make and share. Model the fraction problem on the board for students:

1/2 × 1/11 = 1/22

Challenge students to come up with another way to divide the circle’s area in half in Question 2 of the activity sheet. As you observe students working, ask those who have unique solutions to draw their solution on the board. This is an opportunity for differentiation. Some students will create elaborate half-cuts while others may still be struggling to understand how the diameter cuts the circle in half. If a student finishes quickly, ask them to think of other ways to divide the circle in half.

Students will probably have many ideas for ways to divide the area of the circle into two equal pieces, and some may be interesting to look at in the future. If no one mentions the concentric circle cut, draw the figure shown below on the board. Ask students where the inner circle should be drawn in order to get half of the total area in the inner circle and half the total area in the outer ring. This can also be posed as dividing the circle so the inner circle has an area equal to half the area of the outer circle.

Pass out different colors of construction paper and compasses. Have students make a point on the paper that is the center of the circle, construct a circle of any size, and then cut it out. Pass out the Giant Cookie Dilemma activity sheet. To complete Question 1 on the activity sheet, students will measure the radius of their circle, calculate the area, and then find half of the area. Have students estimate a radius for the inner circle that will result in the area of the inncer circle being equal to half the area of the outer circle. For Question 2, they should record this radius and construct it on the circle they cut out. Students should continue filling in the table in Question 3 until they have a good estimate of the inner radius, and then complete the remaining questions. At the end of the activity, students should have a good estimate for the radius of the inner circle and a drawing of the concentric circles based on their best radius estimate.

 The Giant Cookie Dilemma Activity Sheet

Ask students which cuts would make a better piece to eat: their original circle with 22 radial cuts? or the new concentric circles with 11 radial cuts? Have students compare their models and consider what it would be like to eat a cookie of that size and shape.

After students have worked with the paper copy and arrived at a final radius for the inner circle, ask them to calculate the ratio of the inner circle's radius to that of the outer circle. Students can then record their ratios on the Giant Cookie Ratios overhead. As data are written on the overhead, students should begin to see that the decimal ratio is close to 0.71, no matter how big the original circle is. Calculated algebraically, the exact ratio is 1/√2. Middle school students will probably not figure out the answer algebraically, but they should recognize that all the answers are approximately the same.

If time permits, or as an alternative to using construction paper circles, use the Cookie Cutter Applet to give students the opportunity to test different values of the smaller radius. As students change the length of the smaller radius, the applet shows the change in the area of the inner circle and the ratio of the smaller to the larger radius. As with the overhead sheet, students should see that the ratio approaches approximately 0.71.

Assessments

1. Have students construct the two concentric circles on grid paper. They can then verify that the area of the inner circle is equal to that of the outer ring by counting the squares.
2. Present students with the problem using a fixed value for the radius of the cookie. Tell them you baked a cookie on a pizza pan that is 16 inches in diameter. Ask students to write a paragraph explaining how they would use the concentric circles method to divide the cookie in half. They should provide their calculations with their written explanation.
3. Have students create a two-color target game so that there is an equal chance of landing on each color.

Extensions

1. If some students still don't believe that their inner cirlce has the same area as the outer ring, you could make a giant cookie, pressed onto and baked on a pizza tray, and bring it in to do an actual demonstration of the problem. Weigh the inner circle and outer ring separately. If they are the same size, then they will also be the same weight.
2. Present this situation to students: A bathroom tissue company has manufactured a roll of tissue that is "twice as big" as a standard roll. What will the roll look like compared to the original? Will it look twice as big? Students can write an explanation to a friend who doesn't believe that the new roll is really twice the size of the original.
3. Have students consider paper towels, toilet paper, or thread. What percentage of the radius represents halfway through a roll? Is this half the same as the half in the cookie problem?
4. Students who understand the relationship involved in dividing a circle in half can attempt to draw concentric circles that divide the area into 3 or 4 equal parts. Again, it is important for them to be able to generalize the rule for any circle.
5. Have students construct 2 concentric circles, in which the inner radius is 1/√2 time the size of the larger radius. Have them cut out the inner circle and then cut the ring into pieces. Challenge them to fit the pieces into the inner circle so that it is filled up.
6. Have students use algebra to verify the ratio of 1/√2. The area of the inner circle, radius r1 is equal to half of the area of the larger circle r2. That is,

1/2r22 = r12.

Solving this equation leaves 1/√2 = r1/r2.

Questions for Students

1. How many ways are there to divide the area of a circle in half?

[There are limitless ways to do this. However, in the context of further subdividing the cookie so 22 people can have equal shares of it, there are fewer practical ways to make the first cut.]

2. How long must the radius of the inner circle be in order to create an inner circle and an outer ring that have equal areas?

[1/√2]

3. If you wanted to double the area of a circle, how much bigger should you make the radius of a circle?

[The radius should be √2 (or about 1.41) times the size of the original radius. This question could help students understand what is happening mathematically in the cookie problem.]

Teacher Reflection

• Did students connect the number calculations to the geometric model?
• Did they discover the algebraic general rule?
• Did students have the opportunity to address twice as big and half as big in a real world context?
• Are students confident that half of the radius creates less than half the area?
• What difficulties did your students have? How did you resolve them?

### Learning Objectives

Students will:

• Find the areas of circles and rings
• Determine the ratio of the outer ring area to the inner circle area
• Find the optimal radius to use to create the inner circle in order for the areas to be equal

### Common Core State Standards – Mathematics

• CCSS.Math.Content.6.RP.A.1
Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, ''The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak.'' ''For every vote candidate A received, candidate C received nearly three votes.''

• CCSS.Math.Content.7.NS.A.3
Solve real-world and mathematical problems involving the four operations with rational numbers.

• CCSS.Math.Content.7.EE.B.3
Solve multi-step real-life and mathematical problems posed with positive and negative rational numbers in any form (whole numbers, fractions, and decimals), using tools strategically. Apply properties of operations to calculate with numbers in any form; convert between forms as appropriate; and assess the reasonableness of answers using mental computation and estimation strategies. For example: If a woman making $25 an hour gets a 10% raise, she will make an additional 1/10 of her salary an hour, or$2.50, for a new salary of \$27.50. If you want to place a towel bar 9 3/4 inches long in the center of a door that is 27 1/2 inches wide, you will need to place the bar about 9 inches from each edge; this estimate can be used as a check on the exact computation.

• CCSS.Math.Content.7.G.B.4
Know the formulas for the area and circumference of a circle and use them to solve problems; give an informal derivation of the relationship between the circumference and area of a circle.

### Common Core State Standards – Practice

• CCSS.Math.Practice.MP1
Make sense of problems and persevere in solving them.
• CCSS.Math.Practice.MP4
Model with mathematics.
• CCSS.Math.Practice.MP5
Use appropriate tools strategically.
• CCSS.Math.Practice.MP7
Look for and make use of structure.


text:
# Why is Skolem normal form equisatisfiable while the second order form equivalent?

I asked in another question when is it appropriate to de-Skolemize a statement. The answer, I'm not sure I'm satisfied with yet, relies on a second order logical equivelance, but Skolem normal form isn't logically equivalent. I've read first order Skolemization can be defined through a second order logic equivelance:

$\forall$x$\exists$y $\phi$(x,y) $\iff$ $\exists$F$\forall$x $\phi$(x,F(x))

From this we can omit the quantifiers, and treat all variables as implicitly universally quantified:

$\phi$(x,f(x))

I also understand that Skolem normal form is equisatisfiable:

$\forall$x$\exists$y $\phi$(x,y) $\vDash$ $\phi$(x,f(x))

[I apparently incorrectly understood it, see accepted answer]

But apparently this is not equivalent.

What does this mean? Can we freely move between Skolem normal form and the second order sentence even though they aren't equivalent? I only see equisatisfiability with Skolemization used to prove satisfiability or unsatisfiability, and I haven't seen it used to prove logical consequence from first order formulas except through unsatisfiability with searching for a contradiction.

Why is the second order form of a first order sentence that moves existential quantifiers out of scope of universal quantifiers logically equivalent, while Skolem normal form only equisatisfiable?

## 1 Answer

It's perhaps best to think about this first in the context of a particular structure and a particular formula, where you can easily see what's going on. Consider the structure of the natural numbers with the usual ordering relation $<$. The sentence $\forall x\exists y\,(x<y)$ is true in this structure, because for every natural number $x$ there is a larger one $y$, for example $x+1$ (or $x+17$ or $(2x)!$ or lots of other choices). Equivalently, the second-order sentence $\exists F\forall x\,(x<F(x))$ is true in the same structure, witnessed by functions like $F(x)=x+1$ (or $F(x)=x+17$ or $F(x)=(2x)!$, or lots of other functions $F$).

On the other hand, we cannot say (as you did) that $\forall x\exists y\,(x<y)\models\forall x\,(x<F(x))$. To see this, notice first that this "logical consequence" relationship concerns structures for a language that has not only the relation symbol $<$ but also the function symbol $F$. Consider the structure of natural numbers with $<$ as before and some interpretation of the function symbol $F$. Then $\forall x\exists y\,(x<y)$ is true (regardless of the interpretation of $F$ because it doesn't mention $F$), but whether $\forall x\,(x<F(x))$ is true depends on how we interpret $F$. It's true if we take $F(x)=x+1$ (or $F(x)=x+17$ or $F(x)=(2x)!$, or lots of other functions $F$), but it's false if we take $F(x)=0$ (or $F(x)=17$ or $F(x)=\lfloor x/2\rfloor$, or lots of other functions). But $\models$ means that every structure satisfying what's on the left of $\models$ also satisfies what's on the right. So in our case, it would require that $\forall x\,(x<F(x))$ is true regardless of how we interpret $F$ as long as $\forall x\exists y\,(x<y)$ is true. Since that doesn't hold when $F$ is "badly" interpreted ($0$ or $17$ or $\lfloor x/2\rfloor$), we don't have $\forall x\exists y\,(x<y)\models\forall x\,(x<F(x))$. The most we can say is that, having satisfied $\forall x\exists y\,(x<y)$, we can, by suitably interpreting $F$, also satisfy $\forall x\,(x<F(x))$ --- and vice versa. That's what "equisatisfiable" means.

• I take it this means you can always reintroduce the second order existential quantifier of the function to restore logical equivalence... and then move the existential quantifier inside the universal quantifier to restore it to a first order formula. – dezakin Oct 18 '14 at 23:08
• Reintroducing the second-order quantifier (or several of them, in more complicated cases) will convert "satisfiable" to "true". Moving the second-order quantifier inside the universal quantifier to return to first order is OK if the formula you're working with originally came from Skolemizing a first-order formula, but in general, starting with an arbitrary second-order existential formula, there will be no way to reduce it to an equivalent first-order formula. – Andreas Blass Oct 19 '14 at 3:01
• Why can't you convert arbitrary second order formulas? I thought the second order equivalence relation implies both ways. If you can't convert arbitrary second order formulas, does that mean the answer to the linked question (about deskolemizing sentences produced from inferences on Skolemized statements) is incorrect? Or is converting something defined explicitly as a 'Skolem function' always allowed? – dezakin Oct 19 '14 at 6:40
• As far as I can see, the linked answer is only about the case of second-order sentences obtained by Skolemizing first-order ones, not about general second-order sentences. It's not clear to me how you would even try to go about "de-Skolemizing" something that wasn't obtained by Skolemization in the first place. What if, for example, the function symbol $F$ that you want to get rid of has different arguments at different places in your second-order formula? – Andreas Blass Oct 19 '14 at 11:28


text:
Two sides of a triangle have lengths ‘a’ and ‘b’ and the angle between them is θθ.
Question:

Two sides of a triangle have lengths ‘a‘ and ‘b‘ and the angle between them is θθ. What value of θθ will maximize the area of the triangle? Find the maximum area of the triangle also.

Solution:

As, the area of the triangle, $A=\frac{1}{2} a b \sin \theta$

$\Rightarrow A(\theta)=\frac{1}{2} a b \sin \theta$

$\Rightarrow A^{\prime}(\theta)=\frac{1}{2} a b \cos \theta$

For maxima or minima, $A^{\prime}(\theta)=0$

$\Rightarrow \frac{1}{2} a b \cos \theta=0$

$\Rightarrow \cos \theta=0$

$\Rightarrow \theta=\frac{\pi}{2}$

Also, $A^{\prime \prime}(\theta)=-\frac{1}{2} a b \sin \theta$

or, $A^{\prime \prime}\left(\frac{\pi}{2}\right)=-\frac{1}{2} a b \sin \frac{\pi}{2}=-\frac{1}{2} a b<0$

i. e. $\theta=\frac{\pi}{2}$ is point of maxima

Now,

The maximum area of the triangle $=\frac{1}{2} a b \sin \frac{\pi}{2}=\frac{a b}{2}$


text:
Textual Scanning Options for Arrays, Matrices, and Vectors - Maple Programming Help

Textual Scanning Options for Arrays, Matrices, and Vectors

Description

 • The scanf, fscanf, and sscanf family of textual scanning functions (hereinafter referred to only as scanf), can be used to parse objects of type Array, Matrix, and Vector, collectively known as rtables.
 • Parsing specifications for scanf are introduced by the "%" character, as described in the scanf help page. However, the occurrence of "{" in the specification alters the interpretation of the remainder in a way specific to parsing rtables:

 %[*]{dimensions;options}width[modifiers][basefmt][rtfmt]

 • The initial optional "*" means the same as it does for all scanf formats, which is to perform the parse but discard the result. Any other "*" appearing in an rtable scanf specification indicates that the appropriate information will be specified by additional parameters to scanf, much like the meaning of * in printf.
 • Normally, an optional width could follow, but for the rtable formats, this comes after the options, because the width applies to the individual elements being scanned, not the rtable as a whole.
 • The rtable parsing flags are enclosed in braces, "{}", and are broken into two sections, dimensions and options, separated by a semicolon. The semicolon may be left out if no options occur after the dimensions.
 • The optional modifiers are the same as for any scanf format (one of "l", "L", "z", or "Z").
 • The basefmt is any valid scanf format code (for example, "f").
 If the basefmt is one of "a" or "s", the normal behavior of that format is modified slightly. The "a" and "s" formats usually scan only up to the first whitespace character (space, tab, or newline). When used as part of an rtable scanning format, and if the d(chars) option (see below) is being used, the "a" or "s" format will scan until the specified delimiter, or a newline, is encountered. This is illustrated in the examples at the end of this help page.
 • The rtfmt is one of "a", "m", "r", or "c", indicating that the result of scanning should be an Array, Matrix, Vector[row], or Vector[column].
 • The options consist of a combination of one or more of the following.
 t
 Interpret the file as providing the data transposed. In other words, each row of the data represents a column of the rtable. When indices are explicitly provided, the data ordering is unambiguous and this option is ignored.
 F
 The created rtable will be stored in Fortran (column major) order.
 C
 The created rtable will be stored in C (row major) order.
 d(chars)
 Any of the specified characters may appear between values read. This specification allows for the characters to appear as a delimiters, but their appearance is not mandatory. If the ")" character is allowed, it must be the first character specified in chars. At most 64 characters can be specified. You don't have to specify whitespace characters, since these are already allowed to appear. If a whitespace character is specified in chars, it must be the one and only character specified. In that case, fields are expected to be separated by at most one such character, and two consecutive such characters are assumed to refer to an empty entry (assumed to be zero for numeric typed rtables).
 b(b1,b2), b-(b1,b2), or b+(b1,b2),
 Each row of data gives a band of the Matrix, starting with the sub-diagonals (if b- was specified) or super-diagonals (if b+ was specified), or only the necessary-to-specify elements one row at a time (if b was specified). The (b1,b2) part specifies the number of sub- and super-diagonals. The ",b2" part is optional; if omitted, b2 = b1 is assumed. This specification only concerns the format of the parsed data; it does not affect the indexing functions or storage format, which will be rectangular unless otherwise specified using the s option.
 u, u-, or u+
 Each row of data gives only the necessary-to-specify elements of an upper triangular Matrix, strict upper triangular Matrix, or upper Hessenberg Matrix, respectively. Thus, each row is one shorter than the one before it. This specification only concerns the format of the parsed data; it does not affect the indexing functions or the storage format, which will be rectangular unless specified otherwise with the s option.
 l, l-, or l+
 Each row of data gives only the necessary-to-specify elements of a lower triangular Matrix, strict lower triangular Matrix, or lower Hessenberg Matrix. Thus, each row is one longer than the one before it. This specification only concerns the format of the parsed data; it does not affect the indexing functions or the storage format, which will be rectangular unless specified otherwise with the s option.
 e(numElems)
 Each row of data gives one entry, as a set of indices followed by a value, and there are numElems such entries. The number of indices expected is determined from the number of dimensions that were specified. The value of numElems may be specified as *. You can also omit(numElems), in which case numElems is computed from the specified dimensions. This specification only concerns the format of the parsed data; it does not affect the indexing functions or the storage format, which will be rectangular unless specified otherwise with the s option.
 v(numElems)
 The data is a set of vectors, the first of which give the indices (one index per vector), and the last one of which gives the values. There are numElems entries per vector. The number of index vectors expected is determined from the number of dimensions that were specified. The value of numElems may be specified as *. You can also omit (numElems), in which case numElems is computed from the specified dimensions. This specification only concerns the format of the parsed data; it does not affect the indexing functions or storage format, which will be rectangular unless specified otherwise with the s option.
 y(dataType)
 Indicates that the created table will have the specified dataType. See also in, fn, and c n that follow.
 If the specified dataType is a numeric type, and the basefmt is "v" or "V", an "l" modifier is implied (i.e. the basefmt is treated as if it were "lv" or "lV"). This ensures that the CSV-formatted data is interpreted as numbers if possible.
 a
 Indicates that the created table will have dataType anything. This is equivalent to y(anything). This is the default except as specified for in and fn in the following section.
 in where n must be 1, 2, 4, 8, or *.
 Indicates that the created rtable will have a dataType of integer[n]. This is equivalent to y(integer[n]), and it is the default if the basefmt is for an integer format (i.e. "d", "o", or "x").
 fn where n must be 8 or *.
 Indicates that the created rtable will have a dataType of float[8]. This is equivalent to y(float[8]), and it is the default if the basefmt is a floating-point format (i.e. "e", "f", or "g").
 cn where n must be 8 or *.
 Indicates that the created rtable will have a dataType of complex[8]. In this case, the basefmt is applied to the real and the imaginary parts of the input. This is equivalent to y(complex[8]), and it is the default if the basefmt contains a complex format flag ("zc" or "Z").
 r
 The created rtable will be read-only when completed.
 x(indfns) where indfns is a sequence of indexing functions, or *.
 Specifies the indexing functions of the created rtable. If the indfns themselves have parameters (such as band[b]), these can also be specified as *.
 s(storage) where storage is an rtable storage layout, or *.
 Specifies the storage format of the created rtable. If storage itself has parameters (such as band[b]), these can also be specified as *.
 h(numElems)
 Reads a file of hardware floats and produces a dense rectangular rtable with C ordering, and no indexing functions (compatible with the hfarray of Maple V Release 5). The numElems is optional, and specifies the approximate number of elements (which will preallocate the appropriate storage). This format option gives the behavior of the %hX formats of Release 5.
 m(mismatchValue)
 Specifies the value that is to be assigned to an rtable element if the scanned value is not type-compatible with the rtable. The mismatchValue is optional, and if omitted, a suitable value compatible with the rtable is used (zero for numeric tables, $""$ for strings, false for booleans, and NULL for other tables).
 Note that the mismatch value is only used if the scanf format successfully scanned a value, and that value is not compatible with the rtable. If the data in the input doesn't even match the scanf format, an exception will be raised, and the mismatch value will not be used. Thus, this option is primarily useful in conjunction with the "%v" and "%V" formats when reading CSV files into numeric rtables. Any non-numeric values will be converted to the mismatchValue (which then should be numeric).
 The mismatch value option is also useful when reading into rtables of other specific types. For example, if the option y(posint) was specified, an $m\left(99999\right)$ option would cause all non-positive integers (i.e. 0 and negative integers) that were read to be turned into 99999.
 o(postEOFValue)
 Specifies the value that is to be assigned to an rtable element if the end of the input (typically a file) is reached before the rtable has been filled. The postEOFValue is optional, and if omitted, a suitable value compatible with the rtable is used (zero for numeric tables, $""$ for strings, false for booleans, and NULL for other tables).
 • In all of the above, * means that a corresponding extra parameter to scanf is expected, and the * is replaced with the line-printed form of that parameter. Note that this is more like the use of * in printf, and is different from the usual use of * in scanf. Usually, * means to scan but to not return the specified object (that is, to skip over it).
 • For backward compatibility with Maple V Release 5, the %he, %hf, %hg, and %hx will still work as well. In general,

 %widthhcode %hcode

 are equivalent to,

 %{;h(width)}codea %{;h}codea

 for code in {"e", "f", "g", "y"}, and %hx is a synonym for %hy.
 • In the examples below, we use sscanf for the convenience of showing both the input and the parsing specification together. The first argument is the input (which would come from a file if we were using fscanf), and the second is the parsing specification. Line breaks in the input are represented by "\n".

Examples

 > $\mathrm{sscanf}\left("1 2 3 4","%\left\{2,2\right\}fm"\right)$
 $\left[\left[\begin{array}{cc}{1.}& {2.}\\ {3.}& {4.}\end{array}\right]\right]$ (1)
 > $\mathrm{sscanf}\left("1 2 3 4","%\left\{4\right\}fr"\right)$
 $\left[\left[\begin{array}{cccc}{1.}& {2.}& {3.}& {4.}\end{array}\right]\right]$ (2)
 > $\mathrm{sscanf}\left("1 2 3 4","%\left\{;h\right\}fa"\right)$
 $\left[\left[\begin{array}{cc}{1.}& {2.}\\ {3.}& {4.}\end{array}\right]\right]$ (3)
 > $\mathrm{sscanf}\left("1 1 3+2I 2 2 4-5I","%\left\{3,3;e\left(2\right)c8\right\}Zfm"\right)$
 $\left[\left[\begin{array}{ccc}{3.}{+}{2.}{}{I}& {0.}{+}{0.}{}{I}& {0.}{+}{0.}{}{I}\\ {0.}{+}{0.}{}{I}& {4.}{-}{5.}{}{I}& {0.}{+}{0.}{}{I}\\ {0.}{+}{0.}{}{I}& {0.}{+}{0.}{}{I}& {0.}{+}{0.}{}{I}\end{array}\right]\right]$ (4)
 > $\mathrm{sscanf}\left("1 2 abc 123 4","%\left\{2,2;d\left( \right)\right\}sm"\right)$
 $\left[\left[\begin{array}{cc}{"1"}& {"2"}\\ {"abc 123"}& {"4"}\end{array}\right]\right]$ (5)
 > $\mathrm{sscanf}\left("1 2 abc + 123 1 / \left(x+y\right)","%\left\{2,2;d\left( \right)\right\}am"\right)$
 $\left[\left[\begin{array}{cc}{1}& {2}\\ {\mathrm{abc}}{+}{123}& \frac{{1}}{{x}{+}{y}}\end{array}\right]\right]$ (6)
 > $\mathrm{sscanf}\left("1,2,, 3,x,5, 6,y,8,9","%\left\{3,4\right\}vm"\right)$
 $\left[\left[\begin{array}{cccc}{"1"}& {"2"}& {""}& {""}\\ {"3"}& {"x"}& {"5"}& {""}\\ {"6"}& {"y"}& {"8"}& {"9"}\end{array}\right]\right]$ (7)
 > $\mathrm{sscanf}\left("1,2,, 3,x,5, 6,y,8,9","%\left\{3,4;f8m\left(-1\right)\right\}vm"\right)$
 $\left[\left[\begin{array}{cccc}{1.}& {2.}& {0.}& {0.}\\ {3.}& {-}{1.}& {5.}& {0.}\\ {6.}& {-}{1.}& {8.}& {9.}\end{array}\right]\right]$ (8)
 > $\mathrm{sscanf}\left("1 2","%\left\{2,3;o\left(999\right)\right\}dm"\right)$
 $\left[\left[\begin{array}{rrr}{1}& {2}& {999}\\ {999}& {999}& {999}\end{array}\right]\right]$ (9)
 > $\mathrm{sscanf}\left("1 -2 -3 4","%\left\{2,2;y\left(posint\right)m\left(9999\right)\right\}dm"\right)$
 $\left[\left[\begin{array}{rr}{1}& {9999}\\ {9999}& {4}\end{array}\right]\right]$ (10)