HN user

debdut

1,363 karma

github.com/debdut iawaiponly@gmail.com

Posts102
Comments124
View on HN
old.reddit.com 10mo ago

An LLM-Proof Approach to Reinventing Captcha Systems

debdut
2pts1
github.com 1y ago

Entropy Based Sampling and Parallel Cot Decoding

debdut
1pts0
github.com 1y ago

Show HN: Sequelize ORM with SQLite on IndexedDB

debdut
6pts0
github.com 1y ago

Golive – Live Server in Golang

debdut
7pts1
www.avanderlee.com 2y ago

Swift Compile Time Macros

debdut
1pts0
jalammar.github.io 2y ago

The Illustrated Retrieval Transformer

debdut
2pts0
x-dev.pages.jsc.fz-juelich.de 2y ago

Math Behind Transformers and LLMs

debdut
30pts1
jalammar.github.io 2y ago

The Illustrated Transformer (2018)

debdut
162pts11
x-dev.pages.jsc.fz-juelich.de 2y ago

A mathematician's introduction to transformers and large language models

debdut
4pts1
www.minisim.app 2y ago

Show HN: MiniSim – Painless Simulators from Your Mac Menubar

debdut
1pts0
swiftpackageindex.com 2y ago

SwiftUI for HTML

debdut
2pts0
shiningsword.substack.com 3y ago

Thoughts on Cynical Modern Tourism

debdut
1pts1
twitter.com 3y ago

InstructCodeT5: 16B model beats every model in HumanEval

debdut
6pts0
github.com 3y ago

macOS Apps in Rust

debdut
184pts86
twitter.com 3y ago

The Chinchilla Trap

debdut
2pts0
twitter.com 3y ago

Amazon Engineer admits Serverless can't be used in Production

debdut
11pts5
www.primevideotech.com 3y ago

Scaling up the Prime Video audio/video monitoring service and reducing costs

debdut
989pts507
justinpombrio.net 3y ago

Algebraic Types in Rust Explained

debdut
2pts0
www.jitbit.com 3y ago

Now that's what I call an hacker

debdut
36pts10
type.withmarko.com 3y ago

Type Speed test using ChatGPT

debdut
2pts0
github.com 3y ago

Visual ChatGPT

debdut
698pts228
twitter.com 3y ago

Model Converts Benglish to English

debdut
1pts0
www.unum.cloud 3y ago

Unistore: Replace MongoDB, Neo4j, Pinecone

debdut
2pts0
news.ycombinator.com 3y ago

Ask HN: Product Hunt is dead. What happened?

debdut
7pts3
openai-downloads.com 3y ago

Fake OpenAI site w Malicious Intent

debdut
2pts2
twitter.com 3y ago

Companies offering to pass you interviews and do the job for you in US companies

debdut
1pts0
debdut.github.io 3y ago

Generate Project Starter Using ChatGPT

debdut
1pts0
github.com 3y ago

Browser-extension: 17 browsers support, Vite fast builds

debdut
51pts6
www.usecomponent.com 3y ago

Show HN: She built a better Zapier

debdut
7pts1
github.com 3y ago

Fluuid – UUID readable lib in go

debdut
1pts0

This is crazy, you guys basically rebuilt Swift (interpreter instead of compiler) and also the frameworks. This is the level of engineering that does the magic of "it just works". I, personally, am very interested in swift, and would love to work for free for you guys.

51.1999971, 41.1999989, 4.6000039, In this 3 numbers there are total 9+9+8=26 digits, if these 3 numbers are chosen randomly, what's the probability that there are 10 9's with 2 block of 4 consecutive 9's, 5 1's, 4 0's?

Let's approach this step-by-step using combinatorics:

1) First, let's consider the total number of possible arrangements: We have 26 digits in total, and the order matters. So, the total number of arrangements is: 10^26 (as each position can be filled by any digit from 0 to 9)

2) Now, let's count the favorable arrangements:

   a) We need:
      - 10 nines (including two blocks of 4 consecutive nines)
      - 5 ones
      - 4 zeros
      - 7 other digits (26 - 10 - 5 - 4 = 7)

   b) Let's start by placing the two blocks of 4 consecutive nines:
      We have 19 positions to place the first block (26 - 4 - 3 = 19, as we need to leave room for the second block)
      Then we have 15 positions for the second block
      So, there are 19 * 15 = 285 ways to place these blocks

   c) We need to place 2 more nines:
      We have 18 positions left, so there are C(18,2) = 153 ways to do this

   d) Now, we need to place 5 ones in the remaining 16 positions:
      This can be done in C(16,5) = 4368 ways

   e) Next, place 4 zeros in the remaining 11 positions:
      This can be done in C(11,4) = 330 ways

   f) Finally, we need to fill the remaining 7 positions with the other digits:
      There are 7! = 5040 ways to arrange these

   g) For these last 7 digits, we can choose any digit except 0, 1, and 9:
      So we have 7^7 = 823543 possibilities for what these digits could be
3) Putting it all together: The number of favorable outcomes is: 285 * 153 * 4368 * 330 * 5040 * 823543 = 2.51654 × 10^17

4) Therefore, the probability is: (2.51654 × 10^17) / (10^26) = 2.51654 × 10^-9

So, the probability is approximately 0.00000000251654 or about 1 in 397,371,070,190.