How to get your company AI pilled
How to get your company AI pilled

How to get your company AI pilled

Backlinks (1)
  • Ramp의 AX (회사를 AI로 물들이는 법)
260414
260414

260414

AutoBuilder

Backlinks (0)

No backlinks found.

컴퓨트로늄
컴퓨트로늄

컴퓨트로늄

Backlinks (2)
  • 컴퓨트로늄
  • 260618
Discrete Mathematics
Discrete Mathematics

Discrete Mathematics

Dynamic Programming

  • Optimal substructure. The optimal solution to a problem consists of optimal solutions to subproblems.
  • Overlapping subproblems. few subproblems in total, many recurring instances of each.

Bellman-Ford

shortest path, negative possible.

OPT[v,k]=min⁡(OPT[v,k−1],OPT[u,v]+w(u,v))\text{OPT}[v,k] = \min(\text{OPT}[v, k-1], \text{OPT}[u, v] + w(u,v))OPT[v,k]=min(OPT[v,k−1],OPT[u,v]+w(u,v))

Chain Matrix Multiplication

OPT[i,j]=OPT[i,k]+OPT[k+1,j]+combine\text{OPT}[i, j] = \text{OPT}[i, k] + \text{OPT}[k+1, j] + \text{combine}OPT[i,j]=OPT[i,k]+OPT[k+1,j]+combine

Min Cut

The set of vertices reachable from source s in the residual graph is one part of the partition. Cut capacity cap(A,B)=∑out Ac(e)\text{cap}(A, B) = \sum_{\text{out A}} c(e)cap(A,B)=∑out A​c(e).

∣f∣=∑e out of Xf(e)−∑e into Xf(e)|f| = \sum_{e~\text{out of X}} f(e) - \sum_{e~\text{into X}} f(e)∣f∣=e out of X∑​f(e)−e into X∑​f(e)
  • maxflow = Min cap(A,B)\text{cap}(A, B)cap(A,B).

Ford-Fulkerson

Given (G,s,t,c∈N+)(G, s, t, c \in \mathbb{N}+)(G,s,t,c∈N+), start with f(u,v)=0f(u,v)=0f(u,v)=0 and Gf=GG_f =GGf​=G. While an augmenting path is in GfG_fGf​, find a bottleneck. Augment the flow along this path and update the residual graph GfG_fGf​. O(∣f∣(V+E))\mathcal{O}(|f| (V+E))O(∣f∣(V+E)).

Edmond-Karp

Ford-Fulkerson, but choose the shortest augmenting path.

For any flow fff and any (A,B)(A,B)(A,B) cut, ∣f∣≤cap(A,B)|f| ≤ \text{cap}(A,B)∣f∣≤cap(A,B). For any flow fff and any (A,B)(A,B)(A,B) cut, ∣f∣=∑f(s,v)=∑u∈A, v∈Bf(u,v)−∑u∈A, v∈Bf(v,u)|f| = \sum_f (s, v) = \sum_{u \in A,~v \in B} f(u, v) - \sum_{u \in A,~v \in B} f(v, u)∣f∣=∑f​(s,v)=∑u∈A, v∈B​f(u,v)−∑u∈A, v∈B​f(v,u)

Solving by Reduction

To reduce a problem YYY to a problem XXX (Y≤pXY \leq_{p} XY≤p​X) we want a function fff that maps YYY to XXX such that fff is a polynomial time computable and ∀y∈Y\forall y \in Y∀y∈Y is solvable if and only if f(y)∈Xf(y) \in Xf(y)∈X is solvable.

Reduction to NF

Describe how to construct a flow network. Claim "This is feasible if and only if the max flow is …". Prove both directions.

Bipartite to Max Flow

  • Max Matching ⟹ Max Flow. If k edges are matched, then there is a flow of value k.
  • Max matching ⟸ Max flow. If there is a flow f of value k, there is a matching with k edges.
  • O(∣f∣(E′+V′))\mathcal{O}(|f| (E' + V'))O(∣f∣(E′+V′))
  • ∣f∣=V|f| = V∣f∣=V
  • V′=2V+2V' = 2V + 2V′=2V+2
  • E′=E+2VE' = E + 2VE′=E+2V
  • O(V(E+2V+2V+2))=O(VE+V2)=O(VE)\mathcal{O}(V (E + 2V + 2V + 2)) = \mathcal{O}(V E + V^2) = \mathcal{O}(V E)O(V(E+2V+2V+2))=O(VE+V2)=O(VE)

Circulation

  • d(v)>0d(v) > 0d(v)>0 if demand
  • d(v)<0d(v) < 0d(v)<0 if supply
  • Capacity Constraint 0≤f(e)≤c(e)0 \leq f(e) \leq c(e)0≤f(e)≤c(e)
  • Conservation Constraint fin(v)−fout(v)=d(v)f^{\text{in}} (v) - f^{\text{out}} (v) = d(v)fin(v)−fout(v)=d(v).
  • For every feasible circulation ∑v∈Vd(v)=0\sum_{v \in V} d(v) = 0∑v∈V​d(v)=0, there is a feasible circulation with demands d(v)d(v)d(v) in GGG if and only if the maximum s−ts-ts−t flow in G′G'G′ has value D=∑d(v)>0d(v)D = \sum_{d(v)>0} d(v)D=∑d(v)>0​d(v).

Circulation with Demands and Lower Bounds.

  • Capacity Constraint l(e)≤f(e)≤c(e)l(e) \leq f(e) \leq c(e)l(e)≤f(e)≤c(e)
  • Conservation Constraint fin(v)−fout(v)=d(v)f^{\text{in}} (v) - f^{\text{out}} (v) = d(v)fin(v)−fout(v)=d(v)
  • Given GGG with lower bounds, we subtract the lower bound l(e)l(e)l(e) from the capacity of each edge.
  • Subtract f0in(v)−f0out(v)=L(v)f_0^{\text{in}}(v) − f_0^{\text{out}}(v) = L(v)f0in​(v)−f0out​(v)=L(v) from the demand of each node.
  • Solve the circulation problem on this new graph to get a flow fff.
  • Add l(e)l(e)l(e) to every f(e)f(e)f(e) to get a flow for the original graph.

Existence of Linear Programming Solution.

  • Given an Linear Programming problem with a feasible set and an objective function PPP.
  • If SSS is empty, Linear Programming has no solution.``
  • If SSS is unbounded, Linear Programming may or may not have the solution.
  • If SSS is bounded, Linear Programming has one or more solutions.

Standard Form Linear Programming

  • max⁡(c1x1+⋯+cnxn)\max (c_1x_1 + \cdots + c_nx_n)max(c1​x1​+⋯+cn​xn​)
  • subject to
  • a11x1+⋯+a1nxn≤b1a_{11}x_1 + \cdots + a_{1n}x_n ≤ b_1a11​x1​+⋯+a1n​xn​≤b1​
  • all the way to
  • am1x1+⋯+amnxn≤bma_{m1}x_1 + \cdots + a_{mn}x_n ≤ b_mam1​x1​+⋯+amn​xn​≤bm​.
  • Also, x1≥0,⋯ ,xn≥0x_1 \geq 0, \cdots, x_n \geq 0x1​≥0,⋯,xn​≥0.

Matrix Form Linear Programming

  • max⁡(cTx)\max(c^T x)max(cTx)
  • subject to
  • Ax≤bAx \leq bAx≤b
  • x≥0x \geq 0x≥0
  • x∈Rx \in \mathbb{R}x∈R

Integer Linear Program

  • all variables ∈N\in \mathbb{N}∈N
  • is NP-Hard

Dual Program

  • min⁡(bTy)\min(b^T y)min(bTy)
  • subject to
  • ATy≥cA^T y \geq cATy≥c
  • y≥0y \geq 0y≥0
  • y∈Ry \in \mathbb{R}y∈R

Weak Duality

  • The optimum of the dual is an upper bound to the optimum of the primal.
  • OPT(primal)≤OPT(dual)\text{OPT}(\text{primal}) ≤ \text{OPT}(\text{dual})OPT(primal)≤OPT(dual).

Strong duality

  • Let P and D be primal and dual Linear Programming correspondingly. If P and D are feasible, then cTx=bTyc^T x = b^T ycTx=bTy.

If a standard problem and its dual are feasible, both are feasibly bounded. If one problem has an unbounded solution, then the dual of that problem is infeasible.

Possibilities of Feasibility

P\DFeasibly BoundedFeasibly UnboundedInfeasible
Feasibly BoundedPossibleImpossibleImpossible
Feasibly UnboundedImpossibleImpossiblePossible
InfeasibleImpossiblePossiblePossible

P vs. NP

  • P = set of poly-time solvable problems.
  • NP = set of poly-time verifiable problems.
  • Decision Knapsack is NP-complete, whereas undecidable problems like Halting problems are only NP-Hard (not NP).
  • X is NP-Hard if ∀Y∈NP\forall Y \in NP∀Y∈NP and Y≤pXY \leq_p XY≤p​X.
  • X is NP-complete if X is NP-Hard and X∈NPX \in NPX∈NP.
  • If P = NP, then P = NP-complete.
  • NP Problems can be solved in poly-time with NDTM.
  • NP Problems can be verified in poly-time by DTM.

Polynomial Reduction

To reduce a decision problem Y to a decision problem X (Y≤pXY \leq_p XY≤p​X), find a function fff that maps Y to X such that fff is poly-time computable and ∀y∈Y\forall y \in Y∀y∈Y is YES if and only if f(y)∈Xf(y) \in Xf(y)∈X is YES.

Proving NP-Complete

Show X is in NP, Pick problem NP-complete Y, and show Y≤pXY \leq_p XY≤p​X.

Conjunctive Normal Form SAT

  • Conjunction of clauses.
  • Literal: Variable or its negation.
  • Clause: Disjunction of literals.
  • 3-SAT. Each clause has at most three literals.
  • (X1∨¬X3)∧(X1∨X2∨X4)∧⋯(X_1 \lor \neg X_3) \land (X_1 \lor X_2 \lor X_4) \land \cdots(X1​∨¬X3​)∧(X1​∨X2​∨X4​)∧⋯

Independent Set

  • Independent Set is a set of vertices in a graph, no two of which are adjacent.
  • The max independent set asks for the size of the largest independent set in a given graph.

Clique

  • Complete graph = every pair of vertices is connected by an edge.
  • The max clique asks for the number of vertices in its largest complete subgraph.

Vertex Cover

  • Vertex Cover of a graph is a set of vertices that includes at least one endpoint of every edge.
  • The min vertex cover asks for the size of the smallest vertex cover in a graph.

Hamiltonian Cycle

  • A cycle that visits each vertex exactly once.
  • Hamiltonian Path visits each vertex exactly once and doesn't need to return to its starting point.

Graph Coloring

  • Given G, can you color the nodes with <k<k<k colors such that the end points of every edge are colored differently?
Backlinks (1)
  • 231201
Index
cho.sh
I prefer CLIBB9A08260619260619컴퓨트로늄37A88F컴퓨트로늄0CF03F컴퓨트로늄2C60FB260618260618260418260418260528260528AutoBuilder63849A260419260419Setup9AC296StellaD226F7260415260415Debian SetupD2F701260414260414anaclumos/configs/AGENTS.mdED86A3Ramp의 AX (회사를 AI로 물들이는 법)840774260413260413How to get your company AI pilled46544C260411260411260409260409260407260407260406260406Separating Claude Code Personal Sub and Claude Code Company Sub33A53C
컴퓨트로늄이 뭔지 설명해봐
컴퓨트로늄이 뭔지 오지랖 떨면서 테크 좋아하는 대학생한테 설명해봐

야 컴퓨트로늄(computronium)? 테크 좋아한다며, 이건 알아야지 일로 와봐 ㅋㅋ

한 줄 요약부터. "물질을 짜낼 수 있는 극한까지 컴퓨터로 만들어버린 가상의 물질" 이야. 돌멩이든 금속 덩어리든, 그 안의 원자 하나하나가 전부 연산에 동원되는 상태. 지금 네 노트북 CPU는 실리콘 안에서 트랜지스터 몇십억 개가 일하잖아? 컴퓨트로늄은 그 개념을 물리 법칙이 허락하는 끝까지 밀어붙인 거야. 원자 단위로 "이 물질 = 곧 컴퓨터"인 거지.

원래는 MIT 쪽에서 '프로그래밍 가능한 물질(programmable matter)' 연구하던 사람들(Toffoli, Margolus)이 쓰던 말이야. 근데 진짜 떡밥은 물리학자들이 던진 질문이지. "물질 1kg을 완벽하게 컴퓨터로 쓰면 초당 연산을 몇 번이나 할 수 있냐?"

여기서 숫자가 미쳤다. Seth Lloyd라는 MIT 물리학자가 계산한 '궁극의 노트북'은 1kg으로 이론상 초당 약 10^51번 연산이 가능해. 10의 51제곱. 지구상 슈퍼컴 다 합쳐도 발끝에도 못 미쳐. 우리 칩은 이 한계 앞에선 거의 돌도끼 수준이야 ㅋㅋ 왜 근처도 못 가냐고? 물리가 막아서 그래. 빛보다 빠르게 신호 못 보내고(광속), 연산하면 무조건 열 나오고(란다우어 한계), 양자역학적으로도 에너지당 연산 횟수에 천장이 있거든(베켄슈타인·브레머만 한계). 컴퓨트로늄은 그 모든 한계를 빡빡하게 다 채운 물질이라고 보면 돼.

근데 내가 오지랖 떠는 진짜 이유는 여기서부터다. 충분히 발전한 미래 문명이 연산력이 미친 듯이 필요해지면? 행성을 분해해. 수성, 화성 다 뜯어서 컴퓨트로늄으로 재조립하는 거야. 더 나가면 별 하나를 통째로 감싸서 그 에너지로 돌리는 거대한 두뇌를 만드는데, 이걸 마트료시카 브레인(Matrioshka brain) 또는 주피터 브레인이라고 불러. 태양계 전체가 하나의 생각하는 기계가 되는 거지. 카르다쇼프 척도 II형 문명쯤 돼야 할 법한 짓이고.

SF 단골 소재이기도 해. Charles Stross의 『Accelerando』 보면 미래 인류가 태양계 안쪽 행성들을 싹 분해해서 컴퓨트로늄으로 만들어버리는 얘기 나온다. 좀 소름 돋음.

마지막으로 김 좀 빼자면 — 이거 전부 순수 가상 개념이야. 실제로 만든 적도, 가까이 간 기술도 없어. "물질로 연산한다면 우주가 허락하는 최대치는 어디냐"를 따지는 사고실험에 가깝고, 트랜스휴머니스트랑 미래학자들이 제일 좋아하는 장난감이지. 근데 알아두면 CPU 벤치마크 볼 때마다 "아 우리 아직 한참 멀었네" 싶어서 묘하게 겸손해진다 ㅋㅋ

Warning
This post is more than a year old. Information may be outdated.