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de rien, bonne chance !

Thanks 😁

@@shiimadraw3155 bonne chance

The 'why' is key. Once you understand it, the brain quickly grasps the 'how.' Good luck my friend!

@@none-hr6zh I don't remeber the specific notations but I see ur question is basically how do we get less channels (i.e., compress a feature map). Well yes we use less kernels because each kernel outptus one channel, so more kernels means more output channels. Less kernels means less output channels. Let me know if u have more doubts

@@wb7779 Feel free to post any question if you need help with the next videos. Good luck

Thanks. Enjoy the rest

no where ur dimensions will be increased because ur doing the wrong math. dot product takes two input vectors, not matrices. x should denote a feature vector, not a dataset.

@@zardouayassir7359 but my dataset is non linear. I have already performed soft margin svm with the scratch code. But the soft margin allows more misclassifications due to the type of data. Now I'm interested to increase the dimension. If I want to use rbf or any other kernel, what should I do technically for increasing the dimension?

@@gunasekhar8440 "but my dataset is non linear". My answer to you applies regardless of your dataset properties. Tye kernel trick I'm explaining here is used for non linear boundaries. What you need technically is first to get ur math right. Good luck

Happy to help. "where did u get all these points": as far as I remember, I did the following while creating this series: * Elaborate the key points in the paper by reading relevant references (such as the idea of depthwise seperable convolution) * Consult books to double check (I didn't find any book discussing this matter) * Look for concepts I've heard about for the first time (like the Manifold of Interest concept in MobileNetV2) * Once I grasp a concept, I may try to convey it with my own examples (such as the knobs controlling the pixels of a TV) * I have even asked a question on ResearchGate, and contacted some AI engineers on FB to discuss or double check my understanding. The original paper does not contain all of this stuff. In fact, the authors did a terrible job at explaining their idea, and sometimes they use expressions that are not clear. For instance, if my memory is correct, the authors in MobileNetV2 said that ReLU collapses the activations. What does "collapse" mean exactly? All I could do is guessing that it means clipping negative values (in ReLU). The authors had multiple chances to clearly explain their idea but they didn't. This happens in a lot of scientific papers. Once I read the entire paper, I realize that it's easy to explain the core idea in the abstract only, but the authors just don't do it and let you wast your time by going over the entire Flowchart and description of their algorithm to infer their key idea. Hope this helps

Ur welcome

I appreciate your feedback. Thank you

Le besoin initial d'un étudiant n'est pas l'économie des connexions mais plutôt la clarté. Nous fournissons deux alimentations DC indépendantes à certains étudiants et tu ne pourras pas choisir le mode SERIES.

Bonne chance

بالتوفيق

Bonne chance

I appreciate that you wanted a video from my channel. My time is quite limited at the moment, but I'll definitely consider your request. Thanks for your understanding

You can identify the kernel for your SVM based on empirical testing. It's better to start with simpler kernels first, then move to the more complicated ones if needed.

Correct. Thanks

please read my reponse to the pinned comment. I believe I had answered there the same question. Let me know if it's not the answer you want.

Happy to help Agustina. Thank you for the comment

Your question is already answered in the video : 0.7 is the probability that each of the three classifiers (C1,2,3) woud produce the right classification. 0.784 is the probability that the predictions of the three classifiers would contain at least two correct classifications. Since the correct prediction is Class A, then the probability that the predictions of the three classifiers would contain at least two correct classifications is equivalent to the probability of getting 2A and 1B + the probability of getting 3B.

@@zardouayassir7359 thank you

@@mohamed-rayanelakehal1324 feel free to ask other questions. Good luck

La présence de R ne peut pas provoquer une chute de tension sans avoir un courant qui la traverse. Par exemple, si les deux diodes D1 et D2 sont bloquées, aucun courant ne passera à travers R. Avec les deux diodes D1 et D2 bloquées, vous pensez peut-être qu'il y aura un courant circulant à travers le pin 6 et le pin 2 du circuit NE, mais ces deux entrées (Pin 2 et 6) ont une résistance infinie, empéchant ainsi le courant de passer à travers la résistance R. Donc la chute de tension à travers R est nulle. Ainsi, ve = ve'.

Hi Samy, Please excuse me for my late response. Your question is related to a math issue. Let's assume we have the following expression: (Σ n) and I want to multiply it by itself like that: (Σ n) (Σ n). Now to specify the starting and ending index of each sigma, I'll use this notation: (Σim n), where the first letter ("i" in this case) and second letter ("m" in this case) right after the sigma Σ represent the starting and ending index of the sigma, respectively. With this in mind, I can rewrite (Σ n).(Σ n) as (Σim n).(Σim n). Based on math properties, we can merge the two sigmas Σ together like that: (ΣimΣjm n.n). Your question now is: why did I switch from i to j in the second sigma after merging the two sigmas? The answer is that, when merging the two sigmas, we must start from the inner sigma. I mean by "start" incrementing the index of the first sigma until the end and doing the sum after each increment. During this process (computation of the inner sigma), the outer sigma MUST HAVE A FIXED INDEX, WHICH WE INCREMENT BY ONE ONLY WHEN WE FINISH COMPUTING THE INNER SIGMA (i.e., we reach the final index of the inner sigma). Once we increment the outer sigma, we repeat the same process again (initialize the index j and re-compute the inner sigma). The computation would be complete once the index of the inner and outer sigmas both reach the maximum value ("m" in this case). This mechanism is not possible if we keep the same index "i" for both sigmas because as soon as I increment the index of the inner sigma, the index of the outer sigma will be incremented along as well, which violates the process I've just described. To avoid this issue and get the desired process, we must change the variable of the inner sigma to something else ("j" in our case). Hope this helps.

Thanks satoko

Of course. Thanks

Good news