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2025-10-28 21:26:41: short editorial for all pairs similarity
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.obsidian/plugins/typst-mate/main.js

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{"id":"typst-mate","name":"Typst Mate","version":"2.1.4","minAppVersion":"1.0.0","description":"Render math expressions with Typst instead of MathJax.","author":"azyarashi","isDesktopOnly":false,"fundingUrl":{"Buy me a Coffee":"https://www.buymeacoffee.com/azyarashi"}}
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{"id":"typst-mate","name":"Typst Mate","version":"2.1.7","minAppVersion":"1.0.0","description":"Render math expressions with Typst instead of MathJax.","author":"azyarashi","isDesktopOnly":false,"fundingUrl":{"Buy me a Coffee":"https://www.buymeacoffee.com/azyarashi"}}

content/posts/watchlist.md

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- Attack on Titan
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- Howl's Moving Castle
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- To be Hero X
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- Fullmetal Alchemist
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- Delicious in Dungeon
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- Chainsaw Man
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- Dandadan
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- Re:Zero
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- Violet Evergarden
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- Mob Pyscho 100
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## TV
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- [[frieren|Frieren]]
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---
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title: "USACO Dec 2024: All Pairs Similarity"
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tags:
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- bitmasks
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- sos
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---
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This is a pretty traditional SOS problem. We only care about two values, intersection and union.
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We can have one of these as a DP dimension, and sum the other value over it. In my code, I have union as the DP dimension, and sum intersection over it.
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[Implementation](https://github.com/xyzqm/cp/tree/main/USACO/12-24)

content/thoughts/frieren.md

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- also doesn't really understand helping people because for her, whatever is incoveniencing her will eventually go away, and she has all the time in the world
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- so this really arises again from her temporal disconnect
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- at the end, her just thinking abt helping the person is already big character development
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- frieren to sein is like eisen to stark
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- eisen also ran away from his family, so he wanted to help stark
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## Remembrance + Praise + Religion
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Ideas about remembrance, praise, and religion lie at the core of Frieren. We see a variety of perspectives on these topics throughout the show:
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- **Himmel** epitomizes a pursuit of remembrance. Everything he does, he does so he will be remembered: helping others, building statues of himself, setting out to defeat the Demon King in the first place, etc.

content/thoughts/treaps.md

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---
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title: Treaps
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tags:
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- ds
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- cs
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---
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proof of complexity (let all values be $1...n$):
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$y$ is an ancestor of $x$ iff $y$ is the lowest among all values in the range $[min(x, y), max(x, y)]$. This is with probability $1/(|y - x + 1|)$. Summing this over all $y$ for a fixed $x$ means the expected number of ancestors is just $2 (1 / 2 + 1 / 3 + 1/4 + 1/5...1/n)$, which is $2 log n + cal(O)(1)$

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