Difference between pages "Chapter 12" and "7.17"

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(Created page with "1) We can determine that leafs should never be included into the cover. Therefore all leaves should be unmarked, which means that all of their parents should be marked. Now we...")
 
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=Dealing with Hard Problems=
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1) We can determine that leafs should never be included into the cover. Therefore all leaves should be unmarked, which means that all of their parents should be marked. Now we remove all leaves from the tree and all of their parents, together with all of their edges. We then repeat this process on the modified tree.
  
===Special Cases of Hard Problems===
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2) We consider again the leafs, each is degree 1. For each leaf, if we consider its parent, its degree is the number of children + 1. It means that we have to choose between removing the n children of degree one or the parent of degree n+1. We remove the parent, mark the leafs as included, and recurse as in 1) above.
  
:[[12.1]]. Dominos are tiles represented by integer pairs <math>(x_i, y_i)</math>, where each of the values <math>x_i</math> and <math>y_i</math> are integers between 1 and <math>n</math>. Let <math>S</math> be a sequence of m integer pairs <math>[(x_1, y_1),(x_2, y_2), ...,(x_m, y_m)]</math>. The goal of the game is to create long chains <math>[(x_{i1}, y_{i1}),(x_{i2}, y_{i2}), ...,(x_{it}, y_{it})]</math> such that <math>y_{ij} = x_{i(j+1)}</math>. Dominos can be flipped, so <math>(x_i, y_i)</math> equivalent to <math>(y_i, x_i)</math>. For <math>S = [(1, 3),(4, 2),(3, 5),(2, 3),(3, 8)]</math>, the longest domino sequences include <math>[(4, 2),(2, 3),(3, 8)]</math> and <math>[(1, 3),(3, 2),(2, 4)]</math>.
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3) We know we will be able to remove at most one every other node, so we use a two-coloring technique (Red/Black) and perform a post-order traversal. Let's assume we will remove all the Black node.  When we process a node, we also store with each node the sum over its immediate children of the respective Red and Black weight for the subtree.
::(a) Prove that finding the longest domino chain is NP-complete.
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If not all of the children are Red, we need to mark the current node as Red. But we also have the option to reverse the coloring of all the Red-children's subtree. So we look at the sum over the red-children for
::(b) Give an efficient algorithm to find the longest domino chain where the numbers increase along the chain. For S above, the longest such chains are <math>[(1, 3),(3, 5)]</math> and <math>[(2, 3),(3, 5)]</math>.
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Red and Black, and compare the difference of these sum to the current node's weight. If the current node's weight is above, we swap the coloring for these subtree.
[[12.1|Solution]]
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The current node will record the Black and Red sum of its children's subtree, and add its own weight to its color.
  
  
:12.2. Let <math>G = (V, E)</math> be a graph and <math>x</math> and <math>y</math> be two distinct vertices of <math>G</math>. Each vertex <math>v</math> contains a given number of tokens <math>t(v)</math> that you can collect if you visit <math>v</math>.
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Back to [[Chapter 7]]
::(a) Prove that it is NP-complete to find the path from <math>x</math> to <math>y</math> where you can collect the greatest possible number of tokens.
 
::(b) Give an efficient algorithm if <math>G</math> is a directed acyclic graph (DAG).
 
 
 
 
 
:[[12.3]]. The ''Hamiltonian completion problem'' takes a given graph <math>G</math> and seeks an algorithm to add the smallest number of edges to <math>G</math> so that it contains a Hamiltonian cycle. This problem is NP-complete for general graphs; however, it has an efficient algorithm if <math>G</math> is a tree. Give an efficient and provably correct algorithm to add the minimum number of possible edges to tree <math>T</math> so that <math>T</math> plus these edges is Hamiltonian.
 
[[12.3|Solution]]
 
 
 
===Approximation Algorithms===
 
 
 
:12.4. In the ''maximum satisfiability problem'', we seek a truth assignment that satisfies as many clauses as possible. Give an heuristic that always satisfies at least half as many clauses as the optimal solution.
 
 
 
 
 
:[[12.5]]. Consider the following heuristic for vertex cover. Construct a DFS tree of the graph, and delete all the leaves from this tree. What remains must be a vertex cover of the graph. Prove that the size of this cover is at most twice as large as optimal.
 
[[12.5|Solution]]
 
 
 
 
 
:12.6. The ''maximum cut problem'' for a graph <math>G = (V, E)</math> seeks to partition the vertices <math>V</math> into disjoint sets <math>A</math> and <math>B</math> so as to maximize the number of edges <math>(a, b) \in E</math> such that <math>a \in A</math> and <math>b \in B</math>. Consider the following heuristic for maximum cut. First assign <math>v_1</math> to <math>A</math> and <math>v_2</math> to <math>B</math>. For each remaining vertex, assign it to the side that adds the most edges to the cut. Prove that this cut is at least half as large as the optimal cut.
 
 
 
 
 
:[[12.7]]. [5] In the ''bin-packing problem'', we are given n objects with weights <math>w_1, w_2, ..., w_n</math>, respectively. Our goal is to find the smallest number of bins that will hold the <math>n</math> objects, where each bin has a capacity of at most one kilogram.
 
:The ''first-fit heuristic'' considers the objects in the order in which they are given. For each object, place it into the first bin that has room for it. If no such bin exists, start a new bin. Prove that this heuristic uses at most twice as many bins as the optimal solution.
 
[[12.7|Solution]]
 
 
 
 
 
:12.8. For the first-fit heuristic described just above, give an example where the packing it finds uses at least 5/3 times as many bins as optimal.
 
 
 
 
 
:[[12.9]]. Given an undirected graph <math>G = (V, E)</math> in which each node has degree ≤ d, show how to efficiently find an independent set whose size is at least <math>1/(d + 1)</math> times that of the largest independent set.
 
[[12.9|Solution]]
 
 
 
 
 
:12.10. A vertex coloring of graph <math>G = (V, E)</math> is an assignment of colors to vertices of <math>V</math> such that each edge <math>(x, y)</math> implies that vertices <math>x</math> and <math>y</math> are assigned different colors. Give an algorithm for vertex coloring <math>G</math> using at most <math>\Delta + 1</math> colors, where <math>\Delta</math> is the maximum vertex degree of <math>G</math>.
 
 
 
 
 
:[[12.11]]. Show that you can solve any given Sudoku puzzle by finding the minimum vertex coloring of a specific, appropriately constructed (9×9)+9 vertex graph.
 
[[12.11|Solution]]
 
 
 
===Combinatorial Optimization===
 
For each of the problems below, design and implement a simulated annealing heuristic to get reasonable solutions. How well does your program perform in practice?
 
 
 
 
 
:12.12. Design and implement a heuristic for the bandwidth minimization problem discussed in Section 16.2 (page 470).
 
 
 
 
 
:[[12.13]]. Design and implement a heuristic for the maximum satisfiability problem discussed in Section 17.10 (page 537).
 
[[12.13|Solution]]
 
 
 
 
 
:12.14. Design and implement a heuristic for the maximum clique problem discussed in Section 19.1 (page 586).
 
 
 
 
 
:[[12.15]]. Design and implement a heuristic for the minimum vertex coloring problem discussed in Section 19.7 (page 604).
 
[[12.15|Solution]]
 
 
 
 
 
:12.16. Design and implement a heuristic for the minimum edge coloring problem discussed in Section 19.8 (page 608).
 
 
 
 
 
:[[12.17]]. Design and implement a heuristic for the minimum feedback vertex set problem discussed in Section 19.11 (page 618).
 
[[12.17|Solution]]
 
 
 
 
 
:12.18. Design and implement a heuristic for the set cover problem discussed in Section 21.1 (page 678).
 
 
 
==="Quantum" Computing===
 
 
 
:[[12.19]]. Consider an <math>n</math> qubit “quantum” system <math>Q</math>, where each of the <math>N = 2^n</math> states start out with equal probability <math>p(i) = 1/2^n</math>. Say the ''Jack''<math>(Q, 0^n)</math> operation doubles the probability of the state where all qubits are zero. How many calls to this ''Jack'' operation are necessary until the probability of sampling this null state becomes ≥ 1/2?
 
[[12.19|Solution]]
 
 
 
 
 
:12.20. For the satisfiability problem, construct (a) an instance on <math>n</math> variables that has exactly one solution, and (b) an instance on <math>n</math> variables that has exactly <math>2^n</math> different solutions.
 
 
 
 
 
:[[12.21]]. Consider the first ten multiples of 11, namely 11, 22, . . . 110. Pick two of them (<math>x</math> and <math>y</math>) at random. What is the probability that ''gcd''<math>(x, y) = 11</math>?
 
[[12.21|Solution]]
 
 
 
 
 
:12.22. IBM quantum computing (https://www.ibm.com/quantum-computing/) offers the opportunity to program a quantum computing simulator. Take a look at an example quantum computing program and run it to see what happens.
 
 
 
 
 
Back to [[Chapter List]]
 

Latest revision as of 01:04, 21 September 2020

1) We can determine that leafs should never be included into the cover. Therefore all leaves should be unmarked, which means that all of their parents should be marked. Now we remove all leaves from the tree and all of their parents, together with all of their edges. We then repeat this process on the modified tree.

2) We consider again the leafs, each is degree 1. For each leaf, if we consider its parent, its degree is the number of children + 1. It means that we have to choose between removing the n children of degree one or the parent of degree n+1. We remove the parent, mark the leafs as included, and recurse as in 1) above.

3) We know we will be able to remove at most one every other node, so we use a two-coloring technique (Red/Black) and perform a post-order traversal. Let's assume we will remove all the Black node. When we process a node, we also store with each node the sum over its immediate children of the respective Red and Black weight for the subtree. If not all of the children are Red, we need to mark the current node as Red. But we also have the option to reverse the coloring of all the Red-children's subtree. So we look at the sum over the red-children for Red and Black, and compare the difference of these sum to the current node's weight. If the current node's weight is above, we swap the coloring for these subtree. The current node will record the Black and Red sum of its children's subtree, and add its own weight to its color.


Back to Chapter 7