Introduction to Deep Reinforcement Learning – Hugging Face Deep RL Course

https://huggingface.co/learn/deep-rl-course/en/unit1/introduction

Mapping the Mind of a Large Language Model \ Anthropic

https://www.anthropic.com/news/mapping-mind-language-model

A Complete Guide to BERT with Code | by Bradney Smith | May, 2024 | Towards Data Science

https://towardsdatascience.com/a-complete-guide-to-bert-with-code-9f87602e4a11

残花 – 李清照

花开花落花无悔,缘来缘去缘如水。
花谢为花开,花飞为花悲。
花悲为花泪,花泪为花碎。
花舞花落泪,花哭花瓣飞。
花开为谁谢,花谢为谁悲。

I can fix this PC, boss, but only by playing games • The Register

https://www.theregister.com/2024/05/03/on_call/

Question and Answering over Documents using Langchain and Pinecone

Read this story from Atul Verma on Medium: https://medium.com/@atul.auddy/question-answering-over-documents-using-%EF%B8%8Flangchain-and-pinecone-30250391d6a5

Flaws in Chinese keyboard apps leave 750 million users open to snooping, researchers claim

https://www.theregister.com/2024/04/26/pinyin_keyboard_security_risks/

A Survey of Techniques for Maximizing LLM Performance

Quantum Fourier Transform (QFT) of a Single Qubit is Hadamard Transform

Below is the definition of QFT as illustrated in the YouTube lecture by Abraham Asfaw.

The LaTex code for the equation is as follows and also available here.

Latex
| \tilde{x} \rangle \equiv ~ QFT ~ |x \rangle ~ \equiv \frac{1}{\sqrt{N}}\sum_{y=0}^{N-1}{e^{\frac{2\pi ix y}{N}}} ~| y \rangle

For the one qubit case, N = 21 = 2:

Latex
| \tilde{x} \rangle \equiv ~ QFT ~ |x \rangle ~ \equiv \frac{1}{\sqrt{}N}\sum_{y=0}^{N-1}{e^{\frac{2\pi ix y}{N}}} ~| y \rangle

Latex
\frac{1}{\sqrt{2}}\sum_{y=0}^{1}{e^{\pi ix y}} ~| y \rangle = \frac{1}{\sqrt{2}}[~e^{i \pi x 0}~ | 0 \rangle ~ + ~ e^{i \pi x 1}~| 1 \rangle] = \frac{1}{\sqrt{2}}[~|0\rangle ~+~e^{i \pi x}~|1 \rangle~]

When x = 0:

Latex
QFT~| 0 \rangle = \frac{1}{\sqrt{2}}[~|0\rangle ~+~e^{i \pi 0}~|1 \rangle~] = \frac{1}{\sqrt{2}}[~| 0 \rangle + |1 \rangle~] = |+\rangle

When x = 1:


Latex
QFT~| 1 \rangle = \frac{1}{\sqrt{2}}[~|0\rangle ~+~e^{i \pi 1}~|1 \rangle~] = \frac{1}{\sqrt{2}}[~| 0 \rangle - |1 \rangle~] = |-\rangle

Hence the QFT of a single qubit is essentially the Hadamard transform.

Robotic Packaging

https://www.facebook.com/share/r/3QReDt9KYwQxDuUY/?mibextid=xCPwDs