I have three word lists of different sizes. The smallest one contains the most common English words. The word 'DAD' is apparently not common enough :). The larger two lists are from different machine learning datasets (you can find the sources in GitHub). The easy and hard mode also has different words.
I will add the word 'DAD' to the hand-picked list.
I gave it a photo of my face, and it said that I look approachable and friendly. I asked it to improve my hair, and it replaced me with a completely different person. Oh well! Still very impressive tech.
Just took the interview for the Senior JavaScript Developer position! (I'm not actually a js developer, but this was the closest I could find)
- It was very nice to have "two minutes from now" offered as an interview slot
- Interviewer: "Can you tell about a time you used typescript in a project"
Me: "I've only used typescript in personal projects, writing little games to play remotely with my family."
- Interviewer: "Can you tell about a time you demonstrated excellent collaboration at work?"
Me: "I worked closely with two other people when building product at my last company. I was the data science/ml expert, ..."
- Interviewer: "Can you tell me your English language proficiency?"
Me: "Very proficient."
- Interview over after about 3 minutes! No technical questions at all, sadly.
- A results email was sent to my inbox within a minute or two.
...
RESULTS
Highlights: Good collaboration skills, Experience with AI
Keypoints: Has only used TypeScript in personal projects, Demonstrated effective collaboration and initiative, Comfortable with English
Overall Score: 68%
English Level: B2
Looks pretty good! Agreed that a senior js dev should have some real experience with TypeScript.
But wait, what does B2 mean?
Can interact with a degree of fluency and spontaneity that makes regular interaction with native speakers quite possible without strain for either party.
Hm. I am a native English speaker, and there are two levels above B2. This is a pretty big miss.
But overall, the process was so quick and painless that I'd certainly do it again, just for the chance to practice giving my canned responses to standard behavioral-type questions.
Woods Hole Oceanographic Institution (very shortly) to release footage of the wreck of Titanic that was captured in July 1986 from cameras on the human-occupied submersible Alvin and the newly built, remotely operated Jason Junior. Most of this footage has never been released to the public.
Cycles on (X, A) really can connect parts of A to other parts that look remote if you are forced to stay within A. This is visible in the long exact sequence; when the third map is nonzero, you have cycles in A that become boundaries when you allow chains in X.
Sorry, I do not see what you mean by "dual space," and I do not myself view chains as a analogous to functions.
Thanks for drawing my attention to the long exact sequence*; if I'm interpreting it correctly, the injection at 𝛼 and projection at 𝛽 split, so we have a direct sum: H(X,∅) ≃ H(X,A) ⊕ H(A,∅)?
(never mind the function analogy, I was trying to handwave a quotient in the other direction but that would fall immediately out of the direct sum if I understand correctly: A ≃ (A⊕B)/B and B ≃ (A⊕B)/A?)
* which forms a Möbius band in its own way, because at H(A,∅) we feed it into 𝛼 as ∅ ⊕ Cycles(A) but get it out of 𝛾 as Cycles(A) ⊕ ∅, leading to a "twist"?
The direct sum decomposition you mention doesn't always happen, as you can see in some of the examples in the calculator. It is closer to happening when \gamma is zero, and if \gamma is zero and we switch from integer coefficient to field coefficients, then it always happens!
obviously I need to play with the calculator more (and probably work the exercises) ... hope you don't mind if I get back to you with more Q's once I have them.
well, I love the subject, so I'm glad to discuss whatever! And let me know if all of the exercises are too hard. There should always be a couple of easy ones, but I don't know this audience very well.
Yes, this is true (or nearly) for simplicial complexes. A cycle on (X, A) is the same as a cycle on (X/A, A/A) = (X/A, single point), and this is a useful point of view, since X/A can be much simpler to visualize.
