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We consider the case where one correct process ρ mistakenly triggers a <crash, π> event for some correct process π. In the case where id(π) = 0, id(ρ) = 1, π proposes A, ρ proposes B: ● π broadcasts its proposal A and immediately decides for A. ● ρ believes that π crashed, and broadcasts and decides B. This breaks the ... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/4. consensus_sol.pdf | 2 | 4. consensus_sol | 0 |
Solution 1 - Algorithm II We consider the case where one faulty process ρ mistakenly triggers a <crash, π> event for some correct process π. In the case where N = 2, id(π) = 0, id(ρ) = 1, π proposes A, ρ proposes B: ● π broadcasts its proposal A. ● ρ believes that π crashed, and broadcasts its proposal B. 4 | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/4. consensus_sol.pdf | 3 | 4. consensus_sol | 0 |
Solution 1 - Algorithm II ● ρ delivers its own proposal B. Since N = 2, ρ decides B. ● ρ crashes before π gets its proposal. π correctly triggers <crash, ρ> and, since N = 2, decides A. This breaks the uniform agreement property of uniform consensus. 5 | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/4. consensus_sol.pdf | 4 | 4. consensus_sol | 0 |
Exercise 2 (Consensus & Eventually perfect failure detector) Explain why any fail-noisy consensus algorithm (one that uses an eventually perfect failure detector ◇P) actually solves uniform consensus (and not only the non-uniform variant). 6 | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/4. consensus_sol.pdf | 5 | 4. consensus_sol | 0 |
● We consider an algorithm that uses an eventually perfect failure detector. By contradiction, we assume that the algorithm satisfies agreement, but not uniform agreement. We consider two executions A and B of the algorithm. ● In A, two processes π and ρ decide differently, and π crashes. Let t denote the time when ρ d... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/4. consensus_sol.pdf | 6 | 4. consensus_sol | 0 |
● It is easy to see that ρ receives exactly the same messages and indications from the failure detector in A and B, and thus decides differently from π also in B. ● However, in B, π never failed. Therefore, if the algorithm violates uniform agreement, it also violates agreement. Solution 2 8 | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/4. consensus_sol.pdf | 7 | 4. consensus_sol | 0 |
Exercise 3 (Consensus & Correct majority) Explain why any fail-noisy consensus algorithm (one that uses an eventually perfect failure detector ◇P) requires a majority of the processes to be correct. More precisely, provide a “bad run” in the case where the majority of processes is faulty. 9 | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/4. consensus_sol.pdf | 8 | 4. consensus_sol | 0 |
Consider a system with an even number N of processes. Let A, B denote two distinct subsets of N/2 processes that propose values A and B respectively. By contradiction, let us assume that an algorithm exists that achieves consensus when N/2 processes fail. The two following executions are valid: ● Execution 1. All proce... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/4. consensus_sol.pdf | 9 | 4. consensus_sol | 0 |
Let us now consider the following Execution 3. All the processes in A are suspected by each process in B, and vice versa (at the same time as Executions 1 and 2, respectively). No message between a process in A and a process in B is delivered before max(TA, TB). No process restores any process in the other group before... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/4. consensus_sol.pdf | 10 | 4. consensus_sol | 0 |
Sequential Objects A sequential object is a tuple T = (Q, q0, O, R, ∆), where: ● Q is a set of states. ● q0∈ Q is an initial state. ● O is a set of operations. ● R is a set of responses. ● ∆ ⊆ (Q × Π × O) × (Q × R) is a relation that associates a state, a process, and an operation to a set of possible new states and re... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/4. consensus_sol.pdf | 11 | 4. consensus_sol | 0 |
Guided Exercise 4 (Asset Transfer Object) Define a sequential object representing Asset Transfer, i.e., an object that allows processes to exchange units of currency. 13 | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/4. consensus_sol.pdf | 12 | 4. consensus_sol | 0 |
Solution 4 (Asset Transfer Object) 14 | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/4. consensus_sol.pdf | 13 | 4. consensus_sol | 0 |
Bonus Exercise 5 (Total Order & Asset Transfer) Use Total Order Broadcast to implement an Asset Transfer sequential object. 15 | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/4. consensus_sol.pdf | 14 | 4. consensus_sol | 0 |
Solution 5 ● Every process initializes a balance[] array with the initial, agreed upon balances of each process. ● Upon requesting a payment, a process TO-broadcasts a message [PAY, source, recipient, amount]. ● Upon TO-delivering a message [PAY, source, recipient, amount], a process verifies if balance[source] is at l... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/4. consensus_sol.pdf | 15 | 4. consensus_sol | 0 |
SIGNAL PROCESSING FOR AUDIO AND ACOUSTICS Dr. Christof Faller Autumn Semester 2008 Teaching assistant: Fritz Menzer, Francisco Pinto Final Exam - Solution Tuesday Jan. 13th, 2009 Solution 1. Various Topics (6 points) 1. By delaying the stereo signals, the precedence effect results in that listeners will perceive the in... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/2008_final_solution.pdf | 0 | 2008_final_solution | 0 |
Solution 2. Noise Cancelation System (7 points) 1. For the noise to be canceled at the meeting table, i.e., at rs = Rs and rw = Rw, the total pressure generated by the noise source plus the line source must be zero. Therefore, PS(Rs, ω) = −PW (Rw, ω). 2. Following the result in (1), PS(Rs, ω) = −PW (Rw, ω) r2π jk S(ω)e... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/2008_final_solution.pdf | 1 | 2008_final_solution | 0 |
Solution 3. Interaural coherence (7 points) 1. sL(t) = 1 + cos(φ0 −θ) 2 n1(t) + 1 + cos(φ0 + θ) 2 n2(t) sR(t) = 1 + cos(φ0 + θ) 2 n1(t) + 1 + cos(φ0 −θ) 2 n2(t) 2. IC1 = max d E(sL(t)sR(t + d)) p E(s2 L(t)E(s2 R(t + d)) Let sL(t) = an1(t) + bn2(t) sR(t) = bn1(t) + an2(t) , with a = 1 + cos(φ0 −θ) 2 and b = 1 + cos(φ0 +... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/2008_final_solution.pdf | 2 | 2008_final_solution | 0 |
(t −τ + d)) + E(αβn2(t −τ)n2(t + d)) . Therefore max d E(eL(t)eR(t + d)) = αβσ2 (note that there are two maxima at d = τ and d = −τ). Consequently, IC2 = αβ α2 + β2 4. IC2 = 0.3/(1.09) ≈0.28 IC1,60circ ≈ 0.83 IC1,90circ = 0.6 IC1,120circ ≈ 0.26 IC1,150circ = 0 Obviously φ0 = 120◦gives an IC1 close to IC2. 5. If you hav... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/2008_final_solution.pdf | 2 | 2008_final_solution | 0 |
Solution 4. Listener envelopment (5 points) 1. Given the equation for the lateral fraction (course notes eq. 7.15), one should use an omni- directional microphone for recording the impulse response at the denominator, and a dipole microphone for the impulse response at the numerator. By pointing the dipole microphone p... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/2008_final_solution.pdf | 3 | 2008_final_solution | 0 |
Solution 5. STFT and Audio Coding (7 points) 1. To perform a STFT of 3 blocks, we define a DFT matrix Ψ of size L 3 × L 3 . The STFT transform matrix ̃Ψ is then ̃Ψ = <unk> <unk> Ψ Ψ Ψ <unk> <unk>, and has size L × L. 2. The transform matrix is given by ̃Ψ = <unk> <unk> Ψ0 Ψ1 Ψ0 Ψ1 Ψ0 Ψ1 <unk> <unk>, where Ψ0 and Ψ1 are... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/2008_final_solution.pdf | 4 | 2008_final_solution | 0 |
4. The first STFT block is mostly composed of harmonic components, where each component generates a masking effect over its neighboring frequencies. The resulting masked threshold curve is approximately the sum of the masking effect generated by all harmonics. In the second case, the dominant spectral component is a na... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/2008_final_solution.pdf | 5 | 2008_final_solution | 0 |
Solution 6. Wavefield Synthesis (7 points) 1. While the plane array system has a larger listening area where the soundfield is reconstructed correctly, the line array system is only correct in one listening point and approximately correct on a listener line. 2. Further limitations arise from the fact that the loudspeak... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/2008_final_solution.pdf | 6 | 2008_final_solution | 0 |
Advanced Algorithms April 14, 2022 Lecture 15: Polynomial Identity Testing and Matchings Notes by Ola Svensson1 These notes are based on the lecture notes of Lecture 7 in Shayan Oveis Gharan’s course “CSE 521: Design and Analysis of Algorithms I” available here: http://courses.cs.washington.edu/courses/cse521/17wi/ In ... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/Lecture15.pdf | 0 | Lecture15 | 0 |
= [Ai,j]n×n is a polynomial of n2 variables with degree n: det(A) = X σ:[n]→[n] sgn(σ) n Y i=1 Ai,σ(i), where σ is a permutation defined on [n] = {1, . . . , n} and sgn(σ) is either +1 or −1 depending on the nature of the permutation σ. One naive way to test the identity of p and q is to simply make the list of all mon... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/Lecture15.pdf | 0 | Lecture15 | 0 |
Definition 3 (Polynomial Identity Testing) Given polynomials p and q defined over a common set of variables x, we’d like to determine whether p(x) −q(x) = 0 is true for all values of x. We are only given oracle access: no individual term of p or q is known, but we may evaluate p and q at any specific input x. Example 2... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/Lecture15.pdf | 1 | Lecture15 | 0 |
The Schwartz-Zippel Lemma) Let p(x1, . . . , xn) be a nonzero polynomial of n variables with degree d. Let S be a finite subset of R, with at least d elements in it. If we assign x1, . . . , xn values from S independently and uniformly at random, then P[p(x1, . . . , xn) = 0] ≤d |S|. This is an amazing result — all it ... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/Lecture15.pdf | 1 | Lecture15 | 0 |
We test whether ABx = Cx; if ABx = Cx, then we conclude AB = C. This procedure costs at most O(n2), involving three matrix-vector multiplications. The cost is even lower when the matrices are sparse. Now how likely is the false positive under this regime? That is, if AB <unk>= C, how likely is the outcome ABx = Cx? We ... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/Lecture15.pdf | 2 | Lecture15 | 0 |
technique known as the principle of deferred decision: random choices are made only when they become relevant to the algorithm at hand. Since ai <unk>= ci, there exists a coordinate j such that ai,j <unk>= ci,j. Now, set x1, . . . xn except xj arbitrarily. Since all xi’s are chosen independently of one another, the ran... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/Lecture15.pdf | 2 | Lecture15 | 0 |
3 Proof of Schwartz-Zippel Lemma Proof [Proof of Lemma 4] We proceed by strong induction. Base case: n = 1. The problem is reduced to the univariate case presented in Example 2. Inductive step. Suppose that lemma holds for any polynomial with less than n variables; let’s show that it would also hold if we have n variab... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/Lecture15.pdf | 3 | Lecture15 | 0 |
write p(x) as follows: p(x) = d(x)q(x) + r(x) where the quotient q(x) has degree at most (d −k) and the remainder r(x) has degree at most k −1. The polynomial d(x) is the divisor. Let k be the largest degree xn in all monomials of p. So p can be “divided” by xk n as follows: p(x1, . . . , xn) = xk nq(x1, . . . , xn−1) ... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/Lecture15.pdf | 3 | Lecture15 | 0 |
] ≤k |S|. (5) We can now finish the proof using equations (4) and (5). by Bayes rule, P[p = 0] = P [p = 0|q = 0] · P [q = 0] + P [p = 0|q <unk>= 0] P [q <unk>= 0] ≤ P [q = 0] + P [p = 0|q <unk>= 0] ≤ d −k |S| + k |S| = d |S|. 4 | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/Lecture15.pdf | 3 | Lecture15 | 0 |
4 Bipartite Graph Matching Polynomial identity testing can be use to determine the existence of a perfect matching within a given bipartite graph G = (X, Y, E). Definition 6 A bipartite graph G = (X, Y, E) is a graph where every edge in E connects a vertex in X to a vertex in Y . Definition 7 A matching of graph G is a... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/Lecture15.pdf | 4 | Lecture15 | 0 |
X will be mapped to the same vertex in Y . Since the matching is perfect, no vertex in Y will be left out.) Therefore, we can see f as a permutation on the set of integers [n] = {1, 2, . . . , n}. It follows that n Y i=1 Ai,f(i) = n Y i=1 xi,f(i) is a nonzero monomial of the polynomial det(A), recall the formula for th... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/Lecture15.pdf | 4 | Lecture15 | 0 |
In particular, when σ = f in the above polynomial we get a monomial with a nonzero coefficient. This monomial is different from all other monomials of det(A), i.e., there is no cancellations. This means that det(A) is not a zero polynomial. [⇐] Now, suppose det(A) is not identical to zero. That is, for some values of x... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/Lecture15.pdf | 5 | Lecture15 | 0 |
number]), where w(M) is the sum of the weight of edges of the minimum weight prefect matching. Having this in hand, all we need to do is to test for every edge of G if that edge is a part of the minimum weight perfect matching. Note that w(M) is uniquely defined in the above, given det(A); in particular, w(M) is the th... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/Lecture15.pdf | 5 | Lecture15 | 0 |
Homework I, Advanced Algorithms 2022 Solutions to many homework problems, including problems on this set, are available on the Internet, either in exactly the same formulation or with some minor perturbation. It is not acceptable to copy such solutions. It is hard to make strict rules on what information from the Inter... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/Homework_I__Advanced_Algorithms_2022_solutions.pdf | 0 | Homework_I__Advanced_Algorithms_2022_solutions | 0 |
where x(uk−1,uk) = 1. Given these claims we will first show what the question asks, and then we’ll give the individual proofs of the claims which are rather mechanical and follows the standard technique of changing x values in the support by + or −ε. Theorem. Every extreme point solution to the linear program in the pr... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/Homework_I__Advanced_Algorithms_2022_solutions.pdf | 0 | Homework_I__Advanced_Algorithms_2022_solutions | 0 |
v3 v1 v2 v4 Figure 1: Cycles sharing more than one vertex. v1, v2, v3, v4 is an even cycle. (1a) You land on an already visited vertex. (1b) The vertex you are visiting doesn’t have any other edge you can traverse. (2a) You land on an already visited vertex during either of the walks. (2b) The vertex you are visiting d... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/Homework_I__Advanced_Algorithms_2022_solutions.pdf | 1 | Homework_I__Advanced_Algorithms_2022_solutions | 0 |
value at least 1. By Claim 2, we know that this path is of length at most 2. Conclusion : We have shown that the only acceptable structure in an extreme point are odd length cycles that are their own connected components, or paths of length at most 2. But the connected components of paths of length less than 2 are exac... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/Homework_I__Advanced_Algorithms_2022_solutions.pdf | 1 | Homework_I__Advanced_Algorithms_2022_solutions | 0 |
Proving the claims Proof of Non-Existence of Structure 1. Let x be a feasible solution of the linear problem, and let C = {e1, . . . , e2n} be a cycle of even length supporting x. Let ε = 1 2 mine∈[2n] xe. We define the following two feasible solutions y and z : yek = <unk> <unk> <unk> xek + ε if k ≤2n and k is even, x... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/Homework_I__Advanced_Algorithms_2022_solutions.pdf | 2 | Homework_I__Advanced_Algorithms_2022_solutions | 0 |
= <unk> <unk> <unk> xe −ε if e = (v2j−1, v2j), e /∈{(v1, v2), (vn−1, vn), xe + ε if e = (v2j, v2j+1), e /∈{(v1, v2), (vn−1, vn), xe otherwise. Because the two end edges have value at least1, the end vertices are not affected by the change of value, and still verify the constraints. On the other hand, the vertices that ... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/Homework_I__Advanced_Algorithms_2022_solutions.pdf | 2 | Homework_I__Advanced_Algorithms_2022_solutions | 0 |
j+1), xe + 2ε if e = (u2j−1, u2j), xe −2ε if e = (u2j, u2j+1), xe + ε if e = (w2j−1, w2j) or e = (w1, w2m+1), xe −ε if e = (w2j, w2j+1), xe otherwise. Refer to Figure 2. One can check that y is indeed a feasible solution. With similar ideas to the Page 3 (of 6) Advanced Algorithms • Spring 2022 Michael Kapralov, Ola Sv... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/Homework_I__Advanced_Algorithms_2022_solutions.pdf | 2 | Homework_I__Advanced_Algorithms_2022_solutions | 0 |
v3 v1 u2 w1 v2 w3 w2 +ε −2ε +2ε −ε +ε −ε −ε +ε Figure 2: Odd cycles with even path in between previous proofs, one can prove that the vertices that are in the middle of the path or in the cycles still verify the constraints of the LP. The only new vertices to consider are v1 and w1 (which are the vertices where the cyc... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/Homework_I__Advanced_Algorithms_2022_solutions.pdf | 3 | Homework_I__Advanced_Algorithms_2022_solutions | 0 |
or e = (v1, v2n+1), xe + ε if e = (v2j, v2j+1), xe + 2ε if e = (u2j−1, u2j), xe −2ε if e = (u2j, u2j+1), xe −ε if e = (w2j−1, w2j) or e = (w1, w2m+1), xe + ε if e = (w2j, w2j+1), xe otherwise. Proof of Non-Existence of Structure 4. The proof of this claim is completely analogous to the proof of claim 3, with a path of ... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/Homework_I__Advanced_Algorithms_2022_solutions.pdf | 3 | Homework_I__Advanced_Algorithms_2022_solutions | 0 |
v3 v1 w1 v2 w3 w2 +ε −2ε −ε +ε +ε +ε −ε Figure 3: Odd cycles with odd path in between ye = <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> xe + ε if e = (v2j−1, v2j) or e = (v1, v2n+1), xe −ε if e = (v2j, v2j+1), xe −2ε if e = (u2j−1, u2j), and e <unk>= (uk−1, uk) xe + 2ε if e = (u2j, u2j+1), and e <u... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/Homework_I__Advanced_Algorithms_2022_solutions.pdf | 4 | Homework_I__Advanced_Algorithms_2022_solutions | 0 |
]. L denotes the columns of the matrix and R denotes the partitions. An edge (i.j) is picked in the solution if column i is assigned to partition j. Thus on a high level we want the following constraints to be met: 1. Each column is assigned to exactly one such partition 2. Each partition has linearly independent colum... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/Homework_I__Advanced_Algorithms_2022_solutions.pdf | 4 | Homework_I__Advanced_Algorithms_2022_solutions | 0 |
all columns assigned to a vertex v ∈R are linearly independent. However a vertex v ∈L might have multiple edges incident to it but that’s fine. We need to prove that M2 is indeed a matroid. 1. Property 1 - Let Y ∈I2 and suppose X ⊆Y . Let v ∈R. Thus rank(ANY (v) = |NY (v)|). Thus the columns indexed by NY (v) are linea... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/Homework_I__Advanced_Algorithms_2022_solutions.pdf | 5 | Homework_I__Advanced_Algorithms_2022_solutions | 0 |
matroid intersection algorithm to check if there exists a base of weight nk. Clearly a base of weight nk exists if and only if there is a partition of the nk columns into k bases of Rn. Page 6 (of 6) Advanced Algorithms • Spring 2022 Michael Kapralov, Ola Svensson | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/Homework_I__Advanced_Algorithms_2022_solutions.pdf | 5 | Homework_I__Advanced_Algorithms_2022_solutions | 0 |
Digital 3D Geometry Processing Digital Fabrica<unk>on Prof. Dr. Mark Pauly Julian Paneza EPFL | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 0 | 08_Digital_Fabrication | 0 |
1 2 Digital fabrica<unk>on: bringing digital geometry to the physical world Fabrica<unk>on technologies Subtrac<unk>ve Laser cu<unk>ng CNC (computer numerical control) turning/milling Addi<unk>ve 3D prin<unk>ng How to design parts for 3d prin<unk>ng | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 1 | 08_Digital_Fabrication | 0 |
3 Laser cu<unk>ng Cut/engrave flat sheets with a laser Supports many materials (wood, metal, paper, acrylic, ...) | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 2 | 08_Digital_Fabrication | 0 |
4 Laser cu<unk>ng 0:00 / 1:15 | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 3 | 08_Digital_Fabrication | 0 |
5 Laser cu<unk>ng: how it works Mirrors direct laser beam to precise (x, y) loca<unk>on Laser intensity/exposure <unk>me tuned for the material and desired effect (cu<unk>ng vs engraving) | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 4 | 08_Digital_Fabrication | 0 |
6 Laser cu<unk>ng Machine specs Work area: 610 x 457mm Laser power: 60W Cuts paper, cardboard, wood, PMMA (Plexiglas) Universal Laser Systems, Model: VLS4.60 | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 5 | 08_Digital_Fabrication | 0 |
Laser cu<unk>ng: from file to object Design is just an image; mode chosen by color: vector cu<unk>ng, vector engraving, raster engraving | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 6 | 08_Digital_Fabrication | 0 |
78 Laser cu<unk>ng: tutorial Open a vector graphics program (e.g., Illustrator or Inkscape) Create a new 610x457mm document. Note: units must be mm, color mode RGB. Import/draw your design Apply proper colors and line sizes Cu<unk>ng lines red (255, 0, 0) Engraving lines blue (0, 0, 255) Raster images grayscale Line th... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 7 | 08_Digital_Fabrication | 0 |
9 Laser cu<unk>ng Great for fabrica<unk>ng 2D or 2.5D designs Can also produce 3D shapes: | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 8 | 08_Digital_Fabrication | 0 |
10 Laser cu<unk>ng: designing sliced parts Rhino/Grasshopper | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 9 | 08_Digital_Fabrication | 0 |
11 Laser cu<unk>ng: 3D shapes by folding Pepakura Designer | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 10 | 08_Digital_Fabrication | 0 |
12 Laser cu<unk>ng: curved surfaces | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 11 | 08_Digital_Fabrication | 0 |
Sheet metal/paper only bend into developable surfaces (K = 0) Inser<unk>ng cuts enables stretching, permi<unk>ng nonzero Gaussian curvature, K Beyond Developable: Konakovic et al. 2016 0:00 / 0:20 | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 12 | 08_Digital_Fabrication | 0 |
13 CNC turning (lathes) Rapidly spin material, insert cu<unk>ng tool Great for surfaces of revolu<unk>on, screw threads, etc. Glacern Machine Tools - YouTube 0:00 / 0:54 | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 13 | 08_Digital_Fabrication | 0 |
14 CNC milling Rota<unk>ng cuzer, complex inser<unk>on paths (5- axis) Produce much more general 3D shapes 0:00 / 1:59 | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 14 | 08_Digital_Fabrication | 0 |
15 CNC milling: advantages Great accuracy Large parts Great mechanical proper<unk>es: part is solid material, not built from slices | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 15 | 08_Digital_Fabrication | 0 |
CNC milling: limita<unk>ons Wasteful: material milled away is lost Geometry limita<unk>ons: cu<unk>ng tool must reach all surfaces Planning tool paths/selec<unk>ng cu<unk>ng tools can be challenging Higher fixed-costs Slower turnaround <unk>me Usually large, expensive and dangerous Infill Op<unk>miza<unk>on: Wu et al. ... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 16 | 08_Digital_Fabrication | 0 |
16 Desktop CNC milling Pocket NC Version 2 - YouTube 0:00 / 0:28 | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 17 | 08_Digital_Fabrication | 0 |
17 18 3D prin<unk>ng Slice 3D object into many layers Fabricate layers one at a <unk>me, from bozom to top | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 18 | 08_Digital_Fabrication | 0 |
19 3D prin<unk>ng: advantages Pay only for the material you use Material isn’t wasted like in CNC milling Cost and fabrica<unk>on <unk>me independent of complexity Objects with highly complex topology can be fabricated “Just click print” Less complicated planning process than CNC Shorter turn-around <unk>me Infill Op<u... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 19 | 08_Digital_Fabrication | 0 |
Applica<unk>ons: industrial design Rensselaer Polytechnic Ins<unk>tute | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 20 | 08_Digital_Fabrication | 0 |
20 21 Applica<unk>ons: customized products Makerbot Makerbot | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 21 | 08_Digital_Fabrication | 0 |
22 Applica<unk>ons: aerospace Boeing 787 - Wikipedia Titanium Part for 787 - Norsk Titanium Boeing is saving $3 million on each 787 jet using 3D prin<unk>ng (by avoiding material waste). | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 22 | 08_Digital_Fabrication | 0 |
23 Applica<unk>ons: medical/dental Robohand Amos Dudley | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 23 | 08_Digital_Fabrication | 0 |
24 Applica<unk>ons: new materials James Zhou 0:00 / 0:04 Create new materials by prin<unk>ng microstructure | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 24 | 08_Digital_Fabrication | 0 |
25 Applica<unk>ons: shoes Adidas and Carbon3D Spa<unk>ally varying proper<unk>es for op<unk>mal sports performance | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 25 | 08_Digital_Fabrication | 0 |
26 Applica<unk>ons: mechanical clock! Christoph Laimer - YouTube 0:00 / 0:45 | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 26 | 08_Digital_Fabrication | 0 |
27 3D prin<unk>ng: technologies Fused Deposi<unk>on Modeling (FDM) Stereolithography (SLA) Selec<unk>ve Laser Sintering/Mel<unk>ng (SLS/SLM) Material Je<unk>ng (MJP) ... | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 27 | 08_Digital_Fabrication | 0 |
28 FDM technology 3D Hubs Extrude melted plas<unk>c through a nozzle to form each layer | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 28 | 08_Digital_Fabrication | 0 |
29 FDM limita<unk>ons Lower resolu<unk>on; limited by nozzle diameter spa<unk>al precision of extruder head flow of melted plas<unk>c Poor bonding between layers (anisotropic proper<unk>es) 3D Hubs | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 29 | 08_Digital_Fabrication | 0 |
3D Hubs 30 FDM free-form use: wireprint 0:00 / 0:16 | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 30 | 08_Digital_Fabrication | 0 |
31 FDM free-form use: weaving / flowers Vessel - David Lobser drooloop flowers - Mark Peeters 0:00 / 0:05 | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 31 | 08_Digital_Fabrication | 0 |
32 SLA Technology 3D Hubs Cure liquid resin into a solid by shining UV light | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 32 | 08_Digital_Fabrication | 0 |
33 3D prin<unk>ng: Laser SLA (Form 2) formlabs form 2 | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 33 | 08_Digital_Fabrication | 0 |
34 3D prin<unk>ng: Laser SLA (Form 2) form 2 in ac<unk>on 0:00 / 0:14 | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 34 | 08_Digital_Fabrication | 0 |
35 3D prin<unk>ng: DLP SLA (B9 Creator) Project and cure en<unk>re layer geometry simultaneously | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 35 | 08_Digital_Fabrication | 0 |
36 SLA pros and cons Advantages Excellent resolu<unk>on (25um layer height) Layers almost perfectly bind; isotropic proper<unk>es Wide range of resins to choose from Disadvantages Requires cleaning and curing postprocess (place in UV oven) Color and mechanical proper<unk>es shiW over <unk>me as curing con<unk>nues | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 36 | 08_Digital_Fabrication | 0 |
37 SLS/SLM Technology 3D Hubs Laser shines into a bed of plas<unk>c or metal powder, fusing or mel<unk>ng the par<unk>cles into a solid part. | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 37 | 08_Digital_Fabrication | 0 |
38 Material Je<unk>ng Technology 3D Hubs | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 38 | 08_Digital_Fabrication | 0 |
39 3D prin<unk>ng: what can go wrong? | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 39 | 08_Digital_Fabrication | 0 |
40 3D prin<unk>ng: support structure The defect on the overhanging feature is caused by insufficient support; support structure required. SLA can print overhangs, but not local minima (e.g., bear hand). | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 40 | 08_Digital_Fabrication | 0 |
41 3D prin<unk>ng: support structure removal Removing support can be difficult or completely intractable. | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 41 | 08_Digital_Fabrication | 0 |
42 3D prin<unk>ng: support structure for SLS 3D Hubs SLS doesn’t need support structure! | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 42 | 08_Digital_Fabrication | 0 |
43 3D prin<unk>ng: support structure for Mul<unk>jet 3D Hubs Mul<unk>jet printers can print with dissolvable support! | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 43 | 08_Digital_Fabrication | 0 |
44 3D prin<unk>ng: stringing and gaps (FDM) Stringing can be fixed by lowering temperature/speed or asking printer to retract filament. Gaps can be fixed by increasing the extrusion rate. | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 44 | 08_Digital_Fabrication | 0 |
45 3D prin<unk>ng: warping (FDM and SLA) First few layers can warp/peel from build plaworm. Consider prin<unk>ng on “raW.” | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 45 | 08_Digital_Fabrication | 0 |
46 3D prin<unk>ng: design rules Minimum thickness constraints | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 46 | 08_Digital_Fabrication | 0 |
47 3D prin<unk>ng: design rules Minimum thickness constraints Support requirements | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 47 | 08_Digital_Fabrication | 0 |
48 3D prin<unk>ng: design rules Minimum thickness constraints Support requirements Clearance requirements | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 48 | 08_Digital_Fabrication | 0 |
49 3D prin<unk>ng: design rules Minimum thickness constraints Support requirements Clearance requirements No enclosed voids (SLA/SLS) | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 49 | 08_Digital_Fabrication | 0 |
50 Op<unk>mizing for 3D prin<unk>ng What exactly can we op<unk>mize? Prin<unk>ng parameters Support structure The design itself | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 50 | 08_Digital_Fabrication | 0 |
51 Op<unk>mizing prin<unk>ng parameters Adjust orienta<unk>on/slicing to minimize print error keep prints fast: use adap<unk>ve slicing! | /home/ricoiban/GEMMA/mnlp_chatsplaining/RAG/08_Digital_Fabrication.pdf | 51 | 08_Digital_Fabrication | 0 |
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