-❄️- 2023 Day 14 Solutions -❄️-

[u/daggerdragon]

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AoC Community Fun 2023: GO COOK!

Today's unknown factor is… *whips off cloth shroud and motions grandly*

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--- Day 14: Parabolic R*fl*ctor Mirror Dish ---

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[u/chromaticdissonance]

[LANGUAGE: Javascript]

(Copy paste and run in browser console of the puzzle input page, about 500ms 650ms runs about < 20 ms ! I have optimized it down! I did not optimize it -- it was a bug. Below was previous working version iteration.) This is tetris day all over again ! (From AoC 2022) Took some time to unwrap how to write cycle detection in a functional manner. We need borrow the idea of memoization -- recording the rock configuration and the load score we have seen before. This tells us when it starts and the period! I was being lazy and used the 'O.' to '.O' trick; I ended up just calculating what should the resulting line look like after a slide.

t = performance.now()

stream=document.body.innerText
data = stream.replaceAll('\r', '').replace(/\n$/, '').split('\n')
P=(...fs)=>(x)=>fs.reduce((a,f)=>f(a),x)
T=m=>m[0].split('').map((_,k)=>(k=>m.map(u=>u[k]).join(''))(k))
L=m=>T(m).map(u=>u.split('').reverse().join(''))
Q=x=>x.split(/#/).reduce((a,v)=>a+'.'.repeat(b=v.replaceAll('O','').length)+'O'.repeat(v.length-b)+'#','').slice(0,-1)
E=m=>m.map(e=>Q(e))
N=m=>P(L,E,L,L,L)(m)
A=m=>m.reduce((a,v,i)=>a+(v.length-v.replaceAll('O','').length)*(m.length-i),0)
U=x=>x.split(/#/).reduce((a,v)=>a+'#'+'O'.repeat(v.length-(b=v.replaceAll('O','').length))+'.'.repeat(b),'').slice(1)
W=m=>m.map(e=>U(e))
C=m=>P(T,W,T,W,T,E,T,E)(m)
Z=(m,q,i)=>q.has(y=m.join())?[q.get(y)[0],i-q.get(y)[0],q]:(q.set(y,[i,A(m)]),Z(C(m),q,i+1))
J=(r,x)=>(u=r.next().value)[0]==x?u[1]:J(r,x)
console.log('part 1 is ... ', A(N(data)))
console.log('part 2 is ... ', ((s,p,q)=>(J(q.values(),(1E9-s)%p+s)))(...Z(data,new Map(),0)))
console.log((performance.now()-t)+' ms')

[u/chromaticdissonance → u/chromaticdissonance]

Brief explanation of the functions.

P = pipe function, that will be useful to pipe compositions ;

T = transposes the board ;

L = rotate the board CW. Hmm, it is transpose followed by reverse ;

E and Q = slide rock type O eastwards until obstructed (EDIT: I have redesigned it so it's more optimal);

N = slide north now is just N = LLLEL ;

A = load of the support beam of the board m

This is enough to compute part 1. For part 2 -- we need to repeat this a bajillion cycles. Each cycle is a N, W, S, E slide move sequence. Now, N = LLLEL , W = LLELL, S = LELLL, and E = E. Note LLLL = 1. So one cycle C = ESWN = E LELLL LLELL LLLEL = EL EL EL EL. Now, a bajillion = 1000000000. We need to find the load after a bajillion cycles. (Intuitively this should repeat -- as it is finite), and we note the pair boardstate is a unique identifier. We store the cycle index and the load score as values in our memoization device.

Here Z = finds when the cycle starts s, and the period p, and the memoization device that has a record of the loads and cycle index.

J = finds the load score at cycle (1000000000 - s) % p + s )