James Youn eternal44

Architecture

Server-side rendered monolith.

Summary

Backend:

Frontend:

React via the react-rails ruby library

Synopsis:

There are a couple different ways to design an app but most are centered around the way the data is stored and accessed. This is the reason why going from a one-directional marketplace to a bi-directional marketplace was difficult: everything built "on top" of the data had to change as well.

Goals

Streamline our development and product planning processes.

Goal 1 - finalize data model based on our desired functionality

8 ball problem

Of 8 balls one of them is a different weight. The only way to find the different ball is to use a scale. Find the odd ball (no pun intended) and determine if it's heavier or lighter than the rest.

Solution summary

Use a decision tree to deduce what balls need to be removed.

Max number of weighings: 3 #####Notes:

Finding normal balls for reference in the 2 ball series is trivial so I won't reference finding them.
Numbers in an array represent balls on one side of a scale like "[1, 2, 3]".

	standing
	closing the distance
	double leg takedown
	haymaker
	helmet -> clinch
	- aggressive opponent
	- passive


	from clinch

	import React from 'react'
	import { render } from '@testing-library/react'
	import '@testing-library/jest-dom/extend-expect'

	import ERshow from '../components/ERshow'
	import {
	offer,
	YetAnotherMUAprofileShow,
	MUAoffer,
	exchangeRequest,

	<!DOCTYPE html>
	<html>
	<head>
	<style>
	.flex-container{
	display:flex;
	align-items: center;
	font-family: arial;
	}

	function foo(arg1, arg2, cb) {
	return arg1 + arg2 + cb(4)
	}

	console.log(foo(1, 2, function(num) {
	return num + 2
	}))


	var myArr = [1,2,3]

	function countObjKeys(arr) {
	var keyCount = 0

	function traverseObj(obj) {
	for (var key in obj){
	if(typeof obj[key] === 'object')
	traverseObj(obj[key])

	keyCount++
	}

	function bsearch(ns, n) {
	var startPoint = 0
	var endPoint = ns.length - 1
	var midPoint

	while((startPoint < (endPoint + 1))) {
	midPoint = Math.floor((endPoint - startPoint) / 2) + startPoint

	if(ns[midPoint] === n){
	return midPoint

	// 'AAAABBCDDDEEAA' => 'A4B2C1D3E2A2'

	function countChars (str) {
	var currentChar = str[0];
	var charCount = 1;
	var results = [];

	for (var i = 1; i < str.length; i++) {
	if(str[i] === currentChar) {
	charCount++;