405 Matching Annotations
  1. Sep 2023

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  2. May 2023
    1. 4.7-3 Uniform Boundedness Theorem.

      A sequence of linear operator such that, pointwise the value of the linear operators are bounded is enough for the norm of the operator to be bounded for the limit of the sequence of linear operators.

    1. Clear from (5.1).

      Sequence \(\Vert x_n - c\Vert\) bounded, monotone, and for all \(c \in C\). The real sequence is bounded above and monotone decreasing, limit exists in norm for all \(c\in C\).

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    1. 8.3.1. I NTEGRAL C ONVERGENCE T HEOREM

      Integral of the limit of uniform converging function is the same as the integral of the lmit. Pay attention to conditions: 1. Closed and bounded interval \([a, b]\). 2. Sequence of CONTINUOUS functions. 3. Of course, converges uniformly.

  3. Apr 2023
    1. 11.16
      1. Convert the tree structure to a flow.
      2. compute the potential and the reduced costs.
      3. Select an arc to pivot, keeping the strong feasibility of the tree when changing the arcs on the tree!

      Strong feasibility of the tree is at pg 421

    2. Theorem 9.4 (Complementary Slackness Optimality Conditions)

      Some flow is optimal, if and only if, there is some potential, where, the reduced costs for the potential satisfies the complementary slackness conditions for the duality of the linear program of the min cos flow problem.

    1. The definition willinvolve a general field K, but in functional analysis, K will be R or C.The elements of K are called scalars; hence in our case they will ber.eal or complex numbers.

      This is probably very important due to the fact that, both field is complete, and they are equipped with norm. And of course, the reals and the complex are technically also a vector field with dim = 1.

      The total orderness property is probably important for vector fields in functional analysis.

    2. 2.7-10 Corollary (Continuity, null space)

      For a bounded linear operator \(T\), we have: 1. a sequence converge in the domain of T will converge in the image of T after applying the linear transformation 2. The null space of the operator is a closed linear subspace.

    3. Proof.

      The proof make a special kind of construction. It takes a point in \(Z\) but not \(Y\) , and then it find the cloest point in \(Y\) to approxiamte such a vector. Then, the unit vector of, the closest approximation point to the point being approximated can generate a unitary vector that satisfies the conditions we talked about.

    4. 2.4-1 Lemma (Linear combinations)

      Norm of the linear combinations of vectors are, larger than the sum of absolute value of all the scalar weight, by a strictly positive constant. Only for finite dimensional spaces.

      \( \Vert \alpha_1 + \cdots + \alpha_n\Vert \ge c(|\alpha_1| + \cdots + |\alpha_n|) \)

    5. 2.7-2 Lemma (Norm).

      operator norm is actually a type of norm. Linear operators as homomorphism between the vector spaces are indeed a vector space itself. See here.

      The properties of the operator norm basically interits the properties from the normed vector space. This is direct from the definition. However, these results are only for the bounded linear operators.

    6. 2.6-10 Theorem (Inverse operator)

      (1) Inverse exists is equivalen to having only \(\vec 0\) for the null space of the operator. (2) When the inverse exists, then it's a linear operator. (3) if dim of T is finite dimensional, then dim of range and domain is the same for T.

    7. 2.4-4 Definition (Equivalent norms).

      Take note that, if we treat the norm as a type of metric, then the conditions for equivalent norm is strictly stronger than the conditions needed for metric, which is stated in convergence of sequences.

    8. 1.

      A finite dimensional spaces/subspaces have equivalence between compactness and Closed and Bounded. In this case, we infinite dimensions. Then there is some closed and bounded spaces that is not compact.

      That was my first thought that is all.

    9. subspaces of a normed spaceX (of any dimension)

      I just discovered that the subspaces in vector spaces are very different compare to metric spaces.

      1. A subspace of a metric space just have to be a metric space.
      2. A subspace of a vector space will still have to retain the vector space structure. But if it's viewed as a metric space, this doesn't have to be the case.

      Also take note that this is talking about any spaces of dimensions.

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  4. Mar 2023
    1. Thus the maximum capacity augmenting path has residual capacity atleast (v* - v)/m. Now consider a sequence of 2m consecutive maximum capacityaugmentations starting with the flow x. If each of these augmentations augments atleast (v* - v)/2m units of flow, then within 2m or fewer iterations we will establisha maximum flow.

      Why is this \(2m\) what the fuck dude.

      Ok, check: here for more information.

    2. This argument shows that within 2m consecutive iterations, thealgorithm either establishes a maximum flow or reduces the residual capacity of themaximum capacity augmenting path by a factor of at least 2.

      This is an unstated claim where, the proof comes before this paragraph.

    3. 6.18

      gradute based, proof.

      Observe that: A minimum flow will require some lowerbound, or else this is trivial because zero flow will always be an optimal solution.

      We need to convert the problem to an max flow, with/without the lower bound. 2 Applications of the maxflow probably means that we need a lower bound yes.

    4. break ties infavor of a node with the smallest node number

      I think that, the set of active nodes can never contains the same \(j\) such that \(j'\) is also an active node.

    5. 7.2

      Undergraduate based, algorithm.

      Show your residual Δ network after each augmentation. Verify you have the maximum flow by exhibiting an appropriate cut.

      Why is there a double arc for the graph 7.21(b)? Is it a lower bound for the flow on the network?

    6. An augmentation on arc (i, j) might, however,create an additional arc (j, i) with rji > 0 and therefore also create an additionalinequality d(j) ~ d(i) + 1 that the distanceJabels must satisfy.

      This is just using the inductive hypothesis. The indecutive hypothesis hold for all the neighbours of the curren node \(i\), after the delection of the arcs, the neigboring vertices decreased, therefore, valid distance label condition still holds.

    7. Lemma 7.5

      The shortest augmenting path matains a correct distance label for all the vertices during the iterations. By valid, we mean that there exists some way such that \(d(j) = d(i) + 1\), where \((i, j)\in A\), for all \(i\in N\).

    8. We have seen earlier thatsome feasible circulation Xl in G(XO) satisfies the property that x = XO + Xl.

      Applying the flow decomposition theorem from before, we decompose any flows into a sum of cycles flow and paths flows.

    9. Figure 6.22 Examples for Exercises 6.12 and 6.13.

      Undergraduate based:

      Using the shortest augmenting path algorithm, solve the maximum flow problem give in Figure 6.22(b). Describe the augmenting paths at each iteration as well as any advances and/or retreats and the distance labels at each iteration. Verify you have the maximum flow by exhibiting an appropriate cut.

      In brief, run the Dinic's algorithm for this graph.

    10. Consider a node j that is connected to the source node by a shortest path s =io - i] - i2 - ... - ik - ik +] = j consisting of k + 1 arcs, but has no shortestpath containing fewer than k + 1 arcs.

      shortest path complexity proof is actually around here.

    11. If we apply themodified label-correcting algorithm to a problem with nonnegative arc lengths andwe always examine a node from LIST with the minimum distance label, the resultingalgorithm is the same as Dijkstra's algorithm discussed in Section 4.5

      Dijkstra's interpretation from the generic labeling algorithm.

    12. Consequently, we can store all nodeswhose distance labels change during a pass, and consider (or examine) only thosenodes in the next pass.

      this is the key that helps to improve the labeling algorithm and develope the idea of FIFO.

    13. Property 3.6. A circulation x can be represented as cycle flow along at mostm directed cycles.

      Circulation property. Why is it not along a \(m\) many closed walks?

      This is only true from the view of the flow decomposition algorithm and not the LP.

    1. Proof.

      It's about the maximum of partially order sets:

      It states that a partially ordered set containing upper bounds for every chain (that is, every totally ordered subset) necessarily contains at least one maximal element.

    2. (xn) converges, so it is Cauchy

      This is not directly from the definition of complete metric, it's from the fact that a convergence sequence is Cauchy for the same metric!

    3. Example 24

      WTF, absolute convergence, the convergene on the norm of some sequences of elements from the vector spaces doesn't imply that the sequence itself is going to a limit in the space.

      Relevant discussion: here, Please think and read carefully about them yes.

    4. Proof

      Why Cauchy when we already know that the sequence of the function is just the heaviside function at \(x = 1/2\)?????

      Because convering to some limit in the spaces never meant that the metric of the sequence can be Cauchy. The metric. To show that it's incomplete, we need to start with the \(d(x, y)\) metric first, and then show that the convergence of the sequence of piecwise linear function converges in metric and then show that the limit is not in the space. Just showing thta it's not in the space is not entirely valid. But we sure can make use of the fact that a converging sequence in metric should also be Cauchy yes.

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  5. Feb 2023
    1. 5.41.

      Graduate base, proof. Description:

      Gives a proof if the answer is yes or a no.

      We would need an arc that goes over $O(n^2)$ many pairs of paths for all the possible path, and we need to firstly show that the previous algorithm 5.40 is not going to work anymore and, if would have no choice to look over almost all the possible path to figure out the optimal path for the pertrubed costs!

    2. 4.20

      Graduate based, proof. Description:

      The state requires that \(s\neq t\) to be true.

      By length, the question probably mean the total cost of the directed path.

      (a) I think this is just a direct application of the characterization of shortest path conditions, \(d(i, j) = d(i, k) + d(k, j) \; \forall i, j \in N\). (b) I have a feeling that this is a similar proof compare to the Floyd Warshall algorithm as discussed before.

    3. 5.4

      Undergraduate homework question. Description from the assignment sheet:

      Order the nodes 1,...,7, and order the arcs with tail node i, A(i), increasing by their head nodes. See next page for the algorithm.

      The graph is at the same page. The FIFO Label-Correcting algorithm is exactly the same as Moore Bellman Ford Algorithm. Labels at the end should be: [0, -5, 15, -5, -10]

    4. (a) (b)Figure 4.15 Some shortest path networks.

      Undergraduate, description:

      Solve the all-pairs shortest path problem using the Floyd-Warshall algorithm to the network given in Fig. 4.15(a). Show the distance matrix for each iteration. To make the matrix more readable, relabel the nodes a,b, ..., f for 1,2,... ,6, and z for node 0.

      What the heck is this "z for node 0" under the homework discription, no such node exists in the graph, nor is thre additional nodes for the FW algorithm.

    5. Assumption 6.4.

      \((i, j)\in A\iff (j, i)\in A\). We allow for arcs with zero capacity. For all single direction arc, the opposite direction arc on the graph is having a capacity of zero.

      I think this is essentially a residual graph on the zero flow, which is trivially feasible. Residual graph on the network is covered in the later part of the chapter (pg177).

    1. Proof.

      It's proving by showing the existence of a basis that contains infinitely many elements in it.

      Recall that function is an element from an vector space and the inner product of the space is the integrals of stuff.

    2. This representation is unique for every nonzero x ∈ X

      I think it's unique also for zero vector since the set of vector is linear independent?

      Also observe that the Hamel basis seems to be finite.

    3. e1, e2, ... is not a Hamel basis of l2, but is a Schauder basis of l2

      I am guessing that orthogonality is the key. Shcuader Basis is an orthogonal basis and Hamel basis is not...? Also observe that Hamel basis seems to be finite.

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    1. not proven here

      Probably by strong convexity and the fact that the function is differentiable on its entire.

      Max Monotone Operator related, related to the Surjectivity of the operator or something.

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