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  • wvMaxima Error: "Not Connected To Maxima" on Windows

    - by muntoo
    Maxima gives an error "Not Connected To Maxima" whenever I try to do anything with it. I've looked around, but I couldn't figure out how to fix this on Windows. The only results I got were for Fedora. According to them, I think this may have something to do with the Firewall, but I can't figure out what. I even tried adding wvMaxima to the Windows Firewall exceptions list, but the same error comes up.

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  • Genetic/Evolutionary algorithms and local minima/maxima problem

    - by el.gringogrande
    I have run across several posts and articles that suggests using things like simulated annealing to avoid the local minima/maxima problem. I don't understand why this would be necessary if you started out with a sufficiently large random population. Is it just another check to insure that the initial population was, in fact, sufficiently large and random? Or are those techniques just an alternative to producing a "good" initial population?

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  • Oracle Sparc e Solaris - Performance Máxima para Aplicações de Missão Crítica

    - by Wesley Faria
    Olá pessoal, convido todos a assistirem a entrevista do especialista de Sparc e Solaris da Oracle. Serão abordados temas relevantes como a estratégia da Oracle para essa linha de produtos, roadmap e é claro, os benefícios de se usar a Red Stack.Alem disso terão 3 apresentações que vão detalhar melhor os temas Sparc, Solaris e Integração de Hardware e Software. A entrevista estará disponivel a partir do dia 20 de Setembro de 2012 no link http://www.voit.com.br/NL/Oracle_SPARC/webinar_sparc.htm.

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  • Using Matlab to find maxima for data with a lot of noise

    - by jimbo
    I have noisy data set with three peaks in Matlab and want to do some image processing on it. The peaks are about 5-9 pixels wide at the base, in a 50 x 50 array. How do I locate the peaks? Matlab is very new to me. Here is what I have so far... For my original image, let's call it "array", I tried J = fspecial('gaussian',[5 5], 1.5); C = imfilter(array, J) peaks = imregionalmax(C); but there is still some noise along the baseline between the peaks so I end up getting a ton of local max that are really just noise values. (I tried playing with the size of the filter, but that didn't help.) I also tried peaks = imextendedmax(C,threshold); where the threshold was determined visually... which works but is definitely not a good way to do it since it's not that robust obviously. So, how do I locate these peaks in a robust way? Thanks.

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  • Extended maxima transform in Matlab

    - by garvin
    I use imtophat to apply a filter to an m x n array. I then find the local max using imextendedmax(). I get mostly 0's everywhere except for 1's in the general areas where I am expecting a local max. The weird thing is, though, that I don't get just one local max. Instead in these places I get MANY elements with 1's such as 00011100000 00111111000 00000110000 yet the values there are close but NOT equal so I would expect that there would be one that is higher than all of the rest. So I'm wondering a) if this is a bug and how I might fix it and b) how you would choose choose the element of these 1's with the highest value.

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  • how to match a regulas expresion like (%i1) in python pexpect

    - by mike
    I want to use maxima from python using pexpect, whenever maxima starts it will print a bunch of stuff of this form: $ maxima Maxima 5.27.0 http://maxima.sourceforge.net using Lisp SBCL 1.0.57-1.fc17 Distributed under the GNU Public License. See the file COPYING. Dedicated to the memory of William Schelter. The function bug_report() provides bug reporting information. (%i1) i would like to start up pexpect like so: import pexpect cmd = 'maxima' child = pexpect.spawn(cmd) child.expect (' match all that stuff up to and including (%i1)') child.sendline ('integrate(sin(x),x)') chil.expect( match (%o1 ) ) print child.before how do i match the starting banner up to the prompt (%i1)? and so on, also maxima increments the (%i1)'s by one as the session goes along, so the next expect would be: child.expect ('match (%i2)') child.sendline ('integrate(sin(x),x)') chil.expect( match (%o2 ) ) print child.before how do i match the (incrementing) integers?

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  • image processing toolbox in matlab

    - by yasuhiro89
    I've got a specific question and a related more general one... Why does imextendedmax() not give for example 9 in A(3,3) as a max? Generally... what is the best way for finding multiple maxes/peaks? The nice thing about imextended max is it allows a threshold where presumably everything under that threshold is not counted whereas imregionalmax and findpeaks are more general, less effective. A=round(rand(5)*10) A = 1 5 4 8 3 5 1 8 8 3 9 3 9 1 2 9 7 3 5 9 6 3 5 6 8 B=imextendedmax(A,8) B = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

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  • As an indie game dev, what processes are the best for soliciting feedback on my design/spec/idea? [closed]

    - by Jess Telford
    Background I have worked in a professional environment where the process usually goes like the following: Brain storm idea Solidify the game mechanics / design Iterate on design/idea to create a more solid experience Spec out the details of the design/idea Build it Step 3. is generally done with the stakeholders of the game (developers, designers, investors, publishers, etc) to reach an 'agreement' which meets the goals of all involved. Due to this process involving a series of often opposing and unique view points, creative solutions can surface through discussion / iteration. This is backed up by a process for collating the changes / new ideas, as well as structured time for discussion. As a (now) indie developer, I have to play the role of all the stakeholders (developers, designers, investors, publishers, etc), and often find myself too close to the idea / design to do more than minor changes, which I feel to be local maxima when it comes to the best result (I'm looking for the global maxima, of course). I have read that ideas / game designs / unique mechanics are merely multipliers of execution, and that keeping them secret is just silly. In sharing the idea with others outside the realm of my own thinking, I hope to replicate the influence other stakeholders have. I am struggling with the collation of changes / new ideas, and any kind of structured method of receiving feedback. My question: As an indie game developer, how and where can I share my ideas/designs to receive meaningful / constructive feedback? How can I successfully collate the feedback into a new iteration of the design? Are there any specialized websites, etc?

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  • LyX - Change the fontsize of the compiled math

    - by Soren M
    I have worked with LaTeX in a few years now. My default editor is of course emacs, but I want to try LyX. It seems to work quite well, especially the maxima-calculator intergrated. But I have one problem with it: How do I change the font-size of the run-time compiled math? (It's too small).

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  • Cepstral Analysis for pitch detection

    - by Ohmu
    Hi! I'm looking to extract pitches from a sound signal. Someone on IRC just explain to me how taking a double FFT achieves this. Specifically: take FFT take log of square of absolute value (can be done with lookup table) take another FFT take absolute value I am attempting this using vDSP I can't understand how I didn't come across this technique earlier. I did a lot of hunting and asking questions; several weeks worth. More to the point, I can't understand why I didn't think of it. I am attempting to achieve this with vDSP library. it looks as though it has functions to handle all of these tasks. However, I'm wondering about the accuracy of the final result. I have previously used a technique which scours the frequency bins of a single FFT for local maxima. when it encounters one, it uses a cunning technique (the change in phase since the last FFT) to more accurately place the actual peak within the bin. I am worried that this precision will be lost with this technique I'm presenting here. I guess the technique could be used after the second FFT to get the fundamental accurately. But it kind of looks like the information is lost in step 2. as this is a potentially tricky process, could someone with some experience just look over what I'm doing and check it for sanity? also, I've heard there is an alternative technique involving fitting a quadratic over neighbouring bins. Is this of comparable accuracy? if so, I would favour it, as it doesn't involve remembering bin phases. so questions: does this approach makes sense? Can it be improved? I'm a bit worried about And the log square component; there seems to be a vDSP function to do exactly that: vDSP_vdbcon however, there is no indication it precalculates a log-table -- I assume it doesn't, as the FFT function requires an explicit pre-calculation function to be called and passed into it. and this function doesn't. Is there some danger of harmonics being picked up? is there any cunning way of making vDSP pull out the maxima, biggest first? Can anyone point me towards some research or literature on this technique? the main question: is it accurate enough? Can the accuracy be improved? I have just been told by an expert that the accuracy IS INDEED not sufficient. Is this the end of the line? Pi PS I get SO annoyed (npi) when I want to create tags, but cannot. :| I have suggested to the maintainers that SO keep track of attempted tags, but I'm sure I was ignored. we need tags for vDSP, accelerate framework, cepstral analysis

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  • Oracle Database 11g Implementation Specialist - 14 a 16 Março, 2011

    - by Claudia Costa
    OPN Bootcamp Curso de Especialização em Software OracleCaro Parceiro, O novo programa de parcerias da Oracle assenta na Especialização dos seus seus parceiros. No último ano fiscal muitos parceiros já iniciaram as suas especializações nas temáticas a que estão dedicados e que são prioritárias para o seu negócio. Para apoiar o esforço e dedicação de muitos parceiros na obtenção da certificação dos seus recursos, a equipa local de alianças e canal lançou uma série de iniciativas. Entre elas, a criação deste OPN Bootcamp em conjunto com a Oracle University, especialmente dedicado à formação e preparação para os exames de Implementation, obrigatórios para obter a especialização Oracle Database 11g. Este curso de formação tem o objectivo de preparar os parceiros para o exame de Implementation a realizar já no dia 29 de Março, durante o OPN Satellite Event que terá lugar em Lisboa (outros detalhes sobre este evento serão brevemente comunicados). A sua presença neste curso de preparação nas datas que antecedem o evento OPN Satellite, é fundamental para que os seus técnicos fiquem habilitados a realizar o exame dia 29 de Março com a máxima capacidade e possibilidade de obter resultados positivos. Deste modo, no dia 29 de Março, podem obter a tão desejada certificação, com custos de exame 100% suportados pela Oracle. Contamos com a sua presença! Conteúdo: Oracle Database 11g: 2 Day DBA Release 2 + preparação para o exame 1Z0-154 Oracle Database 11g: Essentials Audiência: - Database Administrators - Technical Administrator- Technical Consultant- Support Engineer Pré Requisitos: Conhecimentos sobre sistema operativo Linux Duração: 3 dias + exame (1 dia)Horário: 9h00 / 18h00Data: 14 a 16 de Março Local: Centro de Formação Oracle Pessoas e Processos Rua do Conde Redondo, 145 - 1º - LisboaAcesso: Metro do Marquês de Pombal Custos de participação: 140€ pax/dia = 420€/pax (3 dias)* - Este preço inclui o exame de Implementation *Custo final para parceiro. Já inclui financiamento da equipa de Alianças e Canal Data e Local do Exame: 29 de Março - Instalações da Oracle University _______________________________________________________________________________________ Inscrições Gratuitas. Lugares Limitados.Reserve já o seu lugar : Email   Para mais informações sobre inscrição: Vítor PereiraFixo: 21 778 38 39 Móvel: 933777099 Fax: 21 778 38 40Para outras informações, por favor contacte: Claudia Costa / 21 423 50 27

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  • Data structure for pattern matching.

    - by alvonellos
    Let's say you have an input file with many entries like these: date, ticker, open, high, low, close, <and some other values> And you want to execute a pattern matching routine on the entries(rows) in that file, using a candlestick pattern, for example. (See, Doji) And that pattern can appear on any uniform time interval (let t = 1s, 5s, 10s, 1d, 7d, 2w, 2y, and so on...). Say a pattern matching routine can take an arbitrary number of rows to perform an analysis and contain an arbitrary number of subpatterns. In other words, some patterns may require 4 entries to operate on. Say also that the routine (may) later have to find and classify extrema (local and global maxima and minima as well as inflection points) for the ticker over a closed interval, for example, you could say that a cubic function (x^3) has the extrema on the interval [-1, 1]. (See link) What would be the most natural choice in terms of a data structure? What about an interface that conforms a Ticker object containing one row of data to a collection of Ticker so that an arbitrary pattern can be applied to the data. What's the first thing that comes to mind? I chose a doubly-linked circular linked list that has the following methods: push_front() push_back() pop_front() pop_back() [] //overloaded, can be used with negative parameters But that data structure seems very clumsy, since so much pushing and popping is going on, I have to make a deep copy of the data structure before running an analysis on it. So, I don't know if I made my question very clear -- but the main points are: What kind of data structures should be considered when analyzing sequential data points to conform to a pattern that does NOT require random access? What kind of data structures should be considered when classifying extrema of a set of data points?

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  • misleading plot issue in mathematica

    - by Qiang Li
    I want to study some "strange" functions by plotting them out in mathematica. One example is the following: mod2[x_] := Which[Mod[x, 2] >= 1, -2 + Mod[x, 2], True, Mod[x, 2]]; f[x_] := Which[-1 <= x <= 1, Abs[x], True, Abs[mod2[x]]]; fn[x_, n_] := Sum[(3/4)^i*f[4^n*x], {i, 0, n}] Plot[{fn[x, 0], fn[x, 1], fn[x, 2], fn[x, 5]}, {x, -2, 2}] However, the plot I got from mma is misleading, in the sense that the maxima and minima of fn[x, 5] should be on the same two levels. But due to high oscillation of the function, and the fact that clearly mma only takes limited number of points to draw the function, you see the plot exhibit strange behavior. Is there any option in plot to remedy for this? Many thanks.

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  • For each element A[i] of array A, find the closest j such that A[j] > A[i]

    - by SamH
    Hi everyone. Given : An array A[1..n] of real numbers. Goal : An array D[1..n] such that D[i] = min{ distance(i,j) : A[j] > A[i] } or some default value (like 0) when there is no higher-valued element. I would really like to use Euclidean distance here. Example : A = [-1.35, 3.03, 0.73, -0.06, 0.71, -0.21, -0.12, 1.49, 1.41, 1.42] D = [1, 0, 1, 1, 2, 1, 1, 6, 1, 2] Is there any way to beat the obvious O(n^2) solution? The only progress I've made so far is that D[i] = 1 whenever A[i] is not a local maxima. I've been thinking a lot and have come up with NOTHING. I hope to eventually extend this to 2D (so A and D are matrices).

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  • Intersection of line and rectangle with maximum segment length

    - by Aarkan
    I have a vector represented by the slope m. Then there is rectangle (assume axis aligned), which is represented by top-left and bottom-right corner. Of course, there may be many lines with slope m and intersecting the given rectangle. The problem is to find out the line whose length of line intercept inside the rectangle is maximum among all such lines. i.e., if the line intersects rectangle at P1 and P2, then the problem is to find the equation of line for which length of P1P2 is maximum. I proceeded like this. Let the line is: y = m*x + c. Then find out the intersection with each side of rectangle and finding out the maxima for distance function between each pair of points. But it will only give me the length of line segment and there seem to be many corner cases to handle. Could anyone please suggest a better way to do this. Thanks in advance.

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  • HPC Server Dynamic Job Scheduling: when jobs spawn jobs

    - by JoshReuben
    HPC Job Types HPC has 3 types of jobs http://technet.microsoft.com/en-us/library/cc972750(v=ws.10).aspx · Task Flow – vanilla sequence · Parametric Sweep – concurrently run multiple instances of the same program, each with a different work unit input · MPI – message passing between master & slave tasks But when you try go outside the box – job tasks that spawn jobs, blocking the parent task – you run the risk of resource starvation, deadlocks, and recursive, non-converging or exponential blow-up. The solution to this is to write some performance monitoring and job scheduling code. You can do this in 2 ways: manually control scheduling - allocate/ de-allocate resources, change job priorities, pause & resume tasks , restrict long running tasks to specific compute clusters Semi-automatically - set threshold params for scheduling. How – Control Job Scheduling In order to manage the tasks and resources that are associated with a job, you will need to access the ISchedulerJob interface - http://msdn.microsoft.com/en-us/library/microsoft.hpc.scheduler.ischedulerjob_members(v=vs.85).aspx This really allows you to control how a job is run – you can access & tweak the following features: max / min resource values whether job resources can grow / shrink, and whether jobs can be pre-empted, whether the job is exclusive per node the creator process id & the job pool timestamp of job creation & completion job priority, hold time & run time limit Re-queue count Job progress Max/ min Number of cores, nodes, sockets, RAM Dynamic task list – can add / cancel jobs on the fly Job counters When – poll perf counters Tweaking the job scheduler should be done on the basis of resource utilization according to PerfMon counters – HPC exposes 2 Perf objects: Compute Clusters, Compute Nodes http://technet.microsoft.com/en-us/library/cc720058(v=ws.10).aspx You can monitor running jobs according to dynamic thresholds – use your own discretion: Percentage processor time Number of running jobs Number of running tasks Total number of processors Number of processors in use Number of processors idle Number of serial tasks Number of parallel tasks Design Your algorithms correctly Finally , don’t assume you have unlimited compute resources in your cluster – design your algorithms with the following factors in mind: · Branching factor - http://en.wikipedia.org/wiki/Branching_factor - dynamically optimize the number of children per node · cutoffs to prevent explosions - http://en.wikipedia.org/wiki/Limit_of_a_sequence - not all functions converge after n attempts. You also need a threshold of good enough, diminishing returns · heuristic shortcuts - http://en.wikipedia.org/wiki/Heuristic - sometimes an exhaustive search is impractical and short cuts are suitable · Pruning http://en.wikipedia.org/wiki/Pruning_(algorithm) – remove / de-prioritize unnecessary tree branches · avoid local minima / maxima - http://en.wikipedia.org/wiki/Local_minima - sometimes an algorithm cant converge because it gets stuck in a local saddle – try simulated annealing, hill climbing or genetic algorithms to get out of these ruts   watch out for rounding errors – http://en.wikipedia.org/wiki/Round-off_error - multiple iterations can in parallel can quickly amplify & blow up your algo ! Use an epsilon, avoid floating point errors,  truncations, approximations Happy Coding !

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  • Compute directional light frustum from view furstum points and light direction

    - by Fabian
    I'm working on a friends engine project and my task is to construct a new frustum from the light direction that overlaps the view frustum and possible shadow casters. The project already has a function that creates a frustum for this but its way to big and includes way to many casters (shadows) which can't be seen in the view frustum. Now the only parameter of this function are the normalized light direction vector and a view class which lets me extract the 8 view frustum points in world space. I don't have any additional infos about the scene. I have read some of the related Questions here but non seem to fit very well to my problem as they often just point to cascaded shadow maps. Sadly i can't use DX or openGl functions directly because this engine has a dedicated math library. From what i've read so far the steps are: Transform view frustum points into light space and find min/max x and y values (or sometimes minima and maxima of all three axis) and create a AABB using the min/max vectors. But what comes after this step? How do i transform this new AABB back to world space? What i've done so far: CVector3 Points[8], MinLight = CVector3(FLT_MAX), MaxLight = CVector3(FLT_MAX); for(int i = 0; i<8;++i){ Points[i] = Points[i] * WorldToShadowMapMatrix; MinLight = Math::Min(Points[i],MinLight); MaxLight = Math::Max(Points[i],MaxLight); } AABox box(MinLight,MaxLight); I don't think this is the right way to do it. The near plain probably has to extend into the direction of the light source to include potentional shadow casters. I've read the Microsoft article about cascaded shadow maps http://msdn.microsoft.com/en-us/library/windows/desktop/ee416307%28v=vs.85%29.aspx which also includes some sample code. But they seem to use the scenes AABB to determine the near and far plane which I can't since i cant access this information from the funtion I'm working in. Could you guys please link some example code which shows the calculation of such frustum? Thanks in advance! Additional questio: is there a way to construct a WorldToFrustum matrix that represents the above transformation?

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  • Oracle OpenWorld 2012. Rueda de prensa de Mark Hurd: pilares de la estrategia de Oracle

    - by Fabian Gradolph
    Estamos en la segunda jornada de OpenWorld 2012. La sesión ha comenzado con la presentación de Mark Hurd, quien después de su intervención, ha ofrecido una rueda de prensa multitudinaria. Es probable que más de 80 o 90 periodistas estuviesen presentes en la sala. El presidente de Oracle nos ha presentado una muy buena síntesis sobre la estrategia de la compañía, que básicamente está basada en cuatro pilares. He aquí un breve resumen: 1. Ofrecer los mejores productos y soluciones en cada una de las capas tecnológicas. Es decir, los mejores productos de almacenamiento de información, los mejores servidores, la mejor base de datos… todos ellos basados en estándares de la industria, lo que favorece la interoperabilidad con los productos de otros fabricantes, si esa es la elección del cliente. 2. Integrar los productos desde la fase de diseño y desarrollo para ofrecer a los clientes soluciones que proporcionen un rendimiento extremo. Los mayores exponentes de esta estrategia son los Engineered Systems (sistemas de ingeniería conjunta), que incluyen productos estrella como Exadata, Exalogic, o Exalitycs. Al combinar las mejores capacidades de los productos Oracle e integrarlos verticalmente, se consiguen los siguientes beneficios. Rendimiento extremo de las soluciones tecnológicas. Por ejemplo, la última versión de Exadata, presentada ahora, ofrece una velocidad hasta 100 veces mayor que la anterior. Reducción de costes. Los sistemas integrados permiten reducir la necesidad de espacio, el tiempo de instalación e integración y los costes de mantenimiento. Su mayor rendimiento también se traduce en una menor necesidad de inversión en infraestructura. Simplificación de la gestión y el mantenimiento. Al integrar diferentes tipos de soluciones en una sola, también se simplifica la contratación de sistemas de soporte, centralizándolos en un único proveedor. 3. 3. Una propuesta completa para entornos cloud. La propuesta de Oracle incluye una cloud pública (Oracle Public Cloud), infraestructura para clouds privadas e infraestructura para sistemas híbridos. Así, la compañía ofrece Aplicaciones como Servicio (AaaS), Plataforma como servicio (PaaS) e infraestructura como servicio (IaaS). Al mismo tiempo, facilita soluciones para que las empresas construyan sus propias infraestructuras cloud. La ventaja de la propuesta de Oracle es que se utiliza la misma tecnología para la cloud pública o la privada, de forma que los clientes tengan facilidad para escoger qué aplicaciones mantendrán en un sistema público, cuáles en un sistema privado, etc. La máxima interoperabilidad permite, además, trasnferir aplicaciones de unas modalidades a otras sin problemas. 4. Aplicaciones sectoriales. Oracle está apostando fuertemente por el desarrollo de aplicaciones específicas para los diferentes sectores de actividad. Así, Oracle proporciona soluciones concretas para Banca y Finanzas, Distribución, Logística, Sector Público, Telecomunicaciones y un largo etcétera. Tras sus palabras preliminares hubo una interesante ronda de preguntas. No es posible hacer un resumen de todas ellas, pero me quedo con un mensaje que Mark ha repetido en un par de ocasiones: Oracle quiere seguir creciendo en todos los mercados en los que opera y se trata de una estrategia para crecer. Así lo esperamos todos.

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  • 2 eventos, 2 países, 1 jornada.

    - by Noelia Gomez
    Normal 0 21 false false false ES X-NONE X-NONE MicrosoftInternetExplorer4 /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0cm 5.4pt 0cm 5.4pt; mso-para-margin:0cm; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-family:"Calibri","sans-serif";} El pasado Martes 23 de Octubre fue un día de gran actividad tanto en España como en Portugal. El Dialogo CxO , organizado por Econique, y en el que participó Oracle, tuvo lugar en Madrid en el Hotel Puerta de Ámerica. Este encuentro tenía como objetivo intercambiar opiniones sobre todos los aspectos relacionados con la gestión estratégica de clientes y el Contact Centre. En este marco, los asistentes tuvieron la oportunidad de realizar reuniones “one to one” con nuestros mejores expertos. Además Oracle presentó dos coloquios relacionados con la visión de las "Nuevas necesidades, estrategias y tendencias en la gestión del Marketing", de la mano de Gema Sebastian, Principal Sales Consultant de Oracle. En dichos coloquios los participantes de empresas, como Caprabo, Carrefour, Endesa, Jaguar Land Rover y Repsol (entre otros) trataron temas de máxima actualidad para los directivos de Marketing. Esta mesa redonda se centró sobre todo en el Marketing en redes sociales, compartiendo entre todos nuestra percepción de que es algo necesario pero que todavía el mercado no sabe muy bien cómo tratar. La escucha activa dentro de las redes y la posibilidad de reaccionar ante determinados factores se veía como un claro punto donde comenzar a trabajar de manera activa y donde Oracle puede ayudar. La experiencia de cliente fue otro de los puntos tratados en esta mesa, donde se dejó claro que ahora es el consumidor el que manda, el que quiere ver las cosas donde quiere y como quiere y que un mensaje de marketing ha de darse en el momento adecuado y aportando un valor real para que el consumidor lo acepte como algo interesante. Igualmente Oracle dispone de herramientas para hacer que esto sea posible. Por otro lado, en Lisboa, tenía lugar el Total Training 2012, una conferencia organizada por el Grupo IFE. En ella participaron más de 100 profesionales de los recursos humanos de las empresas más importantes de Portugal y tuvo como base de partida los conocimientos y experiencias, el intercambio de ideas y la discusión de oportunidades a las que actualmente se enfrentan los profesionales de este área. En este marco Oracle realizó una ponencia sobre “Los nuevos conceptos en RRHH”, de la mano de Julio Rodriguez, Principal Sales Consultant de Oracle, y que puso de manifiesto algunos conceptos tecnológicos relevantes para la gestión del talento que por su novedad, no eran muy conocidos por los profesionales de los RRHH cómo: · Saas (Software as a service) · BI (Business Intelligence) para RRHH · Social Networking y cómo integrarla dentro de la empresa · El mapa del talento, por fin fuera del Excel y en una aplicación · La movilidad en las aplicaciones de RRHH. Sin duda, esta fue una jornada cargada de intercambio de experiencias y de conocimientos para dos grandes áreas: los Recursos Humanos y la Gestión Estratégica del cliente. Si quieres saber más sobre la experiencia del cliente: Customer Concepts Magazine Customer Concepts Exchange in LinkedIn Customer Concepts Web TV Customer Experience @ Oracle.com Customer Experience Facebook Hub Customer Experience YouTube Channel Customer Experience Twitter Puede conocer más sobre HCM (Gestión de RRHH): Oracle Fusion Applications Oracle Fusion Human Capital Management Oracle PartnerNetwork Oracle Consulting Services Oracle Human Capital Management Blog Oracle HCM on Twitter Oracle HCM on Facebook

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  • Why does adding Crossover to my Genetic Algorithm gives me worse results?

    - by MahlerFive
    I have implemented a Genetic Algorithm to solve the Traveling Salesman Problem (TSP). When I use only mutation, I find better solutions than when I add in crossover. I know that normal crossover methods do not work for TSP, so I implemented both the Ordered Crossover and the PMX Crossover methods, and both suffer from bad results. Here are the other parameters I'm using: Mutation: Single Swap Mutation or Inverted Subsequence Mutation (as described by Tiendil here) with mutation rates tested between 1% and 25%. Selection: Roulette Wheel Selection Fitness function: 1 / distance of tour Population size: Tested 100, 200, 500, I also run the GA 5 times so that I have a variety of starting populations. Stop Condition: 2500 generations With the same dataset of 26 points, I usually get results of about 500-600 distance using purely mutation with high mutation rates. When adding crossover my results are usually in the 800 distance range. The other confusing thing is that I have also implemented a very simple Hill-Climbing algorithm to solve the problem and when I run that 1000 times (faster than running the GA 5 times) I get results around 410-450 distance, and I would expect to get better results using a GA. Any ideas as to why my GA performing worse when I add crossover? And why is it performing much worse than a simple Hill-Climb algorithm which should get stuck on local maxima as it has no way of exploring once it finds a local max?

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  • Why does adding Crossover to my Genetic Algorithm give me worse results?

    - by MahlerFive
    I have implemented a Genetic Algorithm to solve the Traveling Salesman Problem (TSP). When I use only mutation, I find better solutions than when I add in crossover. I know that normal crossover methods do not work for TSP, so I implemented both the Ordered Crossover and the PMX Crossover methods, and both suffer from bad results. Here are the other parameters I'm using: Mutation: Single Swap Mutation or Inverted Subsequence Mutation (as described by Tiendil here) with mutation rates tested between 1% and 25%. Selection: Roulette Wheel Selection Fitness function: 1 / distance of tour Population size: Tested 100, 200, 500, I also run the GA 5 times so that I have a variety of starting populations. Stop Condition: 2500 generations With the same dataset of 26 points, I usually get results of about 500-600 distance using purely mutation with high mutation rates. When adding crossover my results are usually in the 800 distance range. The other confusing thing is that I have also implemented a very simple Hill-Climbing algorithm to solve the problem and when I run that 1000 times (faster than running the GA 5 times) I get results around 410-450 distance, and I would expect to get better results using a GA. Any ideas as to why my GA performing worse when I add crossover? And why is it performing much worse than a simple Hill-Climb algorithm which should get stuck on local maxima as it has no way of exploring once it finds a local max?

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  • Save result of for loop in a vector

    - by hendrik
    i think I'm just too tired to see the mistake. i wrote a function to get the maximal value for two data sets from a for loop plot_zu <- function(x) {for (i in 1:x){ z=data_raw[grep(a[i], data_raw$Gene.names),] b=data_raw_ace[grep(a[i], data_raw_ace$Gene.names),] p<-vector("numeric", length(1:length(a))) p[i]<-max(z$t_test_diff) return(p)} } so picture: a is a vector of names and the data set (data_raw(_ace)) are filtered by it. In the end i would like to have all maxima values of column t_test_diff in a vector. After that i want to add the t_test_diff column values from data_raw_ace also. So the problem is, that i get this: [1] 1.210213 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 [8] 0.000000 0.000000 So there is a problem with brackets or something but i cannot see it ( first value fits). Sorry for no good example but i think it is understandable and an easy to solve question. If im to dumb to explain my problem right, i will add an example. !! Thx a lot !! grateful Hendrik

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  • Scared of Calculus - Required to pass Differential Calculus as part of my Computer science major

    - by ke3pup
    Hi guys I'm finishing my Computer science degree in university but my fear of maths (lack of background knowledge) made me to leave all my maths units til' the very end which is now. i either take them on and pass or have to give up. I've passed all my programming units easily but knowing my poor maths skills won't do i've been staying clear of the maths units. I have to pass Differential Calculus and Linear Algebra first. With a help of book named "Linear Algebra: A Modern Introduction" i'm finding myself on track and i think i can pass the Linear Algebra unit. But with differential calculus i can't find a book to help me. They're either too advanced or just too simple for what i have to learn. The things i'm required to know for this units are: Set notation, the real number line, Complex numbers in cartesian form. Complex plane, modulus. Complex numbers in polar form. De Moivre’s Theorem. Complex powers and nth roots. Definition of ei? and ez for z complex. Applications to trigonometry. Revision of domain and range of a function Working in R3. Curves and surfaces. Functions of 2 variables. Level curves.Partial derivatives and tangent planes. The derivative as a difference quotient. Geometric significance of the derivative. Discussion of limit. Higher order partial derivatives. Limits of f(x,y). Continuity. Maxima and minima of f(x,y). The chain rule. Implicit differentiation. Directional derivatives and the gradient. Limit laws, l’Hoˆpital’s rule, composition law. Definition of sinh and cosh and their inverses. Taylor polynomials. The remainder term. Taylor series. Is there a book to help me get on track with the above? Being a student i can't buy too many books hence why i'm looking for a book that covers topics I need to know. The University library has a fairly limited collection which i took as loan but didn't find useful as it was too complex.

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  • C - array count, strtok, etc

    - by Pedro
    Hi... i have a little problem on my code... HI open a txt that have this: LEI;7671;Maria Albertina da silva;[email protected]; 9;8;12;9;12;11;6;15;7;11; LTCGM;6567;Artur Pereira Ribeiro;[email protected]; 6;13;14;12;11;16;14; LEI;7701;Ana Maria Carvalho;[email protected]; 8;13;11;7;14;12;11;16;14; LEI, LTCGM are the college; 7671, 6567, 7701 is student number; Maria, Artur e Ana are the students name; [email protected], ...@gmail are emails from students; the first number of every line is the total of classes that students have; after that is students school notes; example: College: LEI Number: 7671 Name: Maria Albertina da Silva email: [email protected] total of classes: 9 Classe Notes: 8 12 9 12 11 6 15 7 11. My code: typedef struct aluno{ char sigla[5];//college char numero[80];//number char nome[80];//student name char email[20];//email int total_notas;// total of classes char tot_not[40]; // total classes char notas[20];// classe notes int nota; //class notes char situacao[80]; //situation (aproved or disaproved) }ALUNO; void ordena(ALUNO*alunos, int tam)//bubble sort { int i=0; int j=0; char temp[100]; for( i=0;i<tam;i++) for(j=0;j<tam-1;j++) if(strcmp( alunos[i].sigla[j], alunos[i].sigla[j+1])>0){ strcpy(temp, alunos[i].sigla[j]); strcpy(alunos[i].sigla[j],alunos[i].sigla[j+1]); strcpy(alunos[i].sigla[j+1], temp); } } void xml(ALUNO*alunos, int tam){ FILE *fp; char linha[60];//line int soma, max, min, count;//biggest note and lowest note and students per course count float media; //media of notes fp=fopen("example.txt","r"); if(fp==NULL){ exit(1); } else{ while(!(feof(fp))){ soma=0; media=0; max=0; min=0; count=0; fgets(linha,60,fp); if(linha[0]=='L'){ if(ap_dados=strtok(linha,";")){ strcpy(alunos[i].sigla,ap_dados);//copy to struct // i need to call bubble sort here, but i don't know how printf("College: %s\n",alunos[i].sigla); if(ap_dados=strtok(NULL,";")){ strcpy(alunos[i].numero,ap_dados);//copy to struct printf("number: %s\n",alunos[i].numero); if(ap_dados=strtok(NULL,";")){ strcpy(alunos[i].nome, ap_dados);//copy to struct printf("name: %s\n",alunos[i].nome); if(ap_dados=strtok(NULL,";")){ strcpy(alunos[i].email, ap_dados);//copy to struct printf("email: %s\n",alunos[i].email); } } } }i++; } if(isdigit(linha[0])){ if(info_notas=strtok(linha,";")){ strcpy(alunos[i].tot_not,info_notas); alunos[i].total_notas=atoi(alunos[i].tot_not);//total classes for(z=0;z<=alunos[i].total_notas;z++){ if(info_notas=strtok(NULL,";")){ strcpy(alunos[i].notas,info_notas); alunos[i].nota=atoi(alunos[i].notas); // student class notes } soma=soma + alunos[i].nota; media=soma/alunos[i].total_notas;//doesn't work if(alunos[i].nota>max){ max=alunos[i].nota;;//doesn't work } else{ if(min<alunos[i].nota){ min=alunos[i].nota;;//doesn't work } } //now i need to count the numbers of students in the same college, but doesn't work /*If(strcmp(alunos[i].sigla, alunos[i+1].sigla)=0){ count ++; printf("%d\n", count); here for LEI should appear 2 students and for LTCGM appear 1, don't work }*/ //Now i need to see if student is aproved or disaproved // Student is disaproved if he gets 3 notes under 10, how can i do that? } printf("media %d\n",media); //media printf("Nota maxima %d\n",max);// biggest note printf("Nota minima %d\n",min); //lowest note }i++; } } } fclose(fp); } int main(int argc, char *argv[]){ ALUNO alunos; FILE *fp; int tam; fp=fopen(nomeFicheiro,"r"); alunos = (ALUNO*) calloc (tam, sizeof(ALUNO)); xml(alunos,nomeFicheiro, tam); system("PAUSE"); return 0; }

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