 6.3.1: Draw the decision tree for sequential search on a list of three ele...
 6.3.2: Draw the decision tree for sequential search on a list of six eleme...
 6.3.3: Draw the decision tree for binary search on a sorted list of seven ...
 6.3.4: Draw the decision tree for binary search on a sorted list of four e...
 6.3.5: Consider a search algorithm that compares an item with the last ele...
 6.3.6: Consider a search algorithm that compares an item with an element o...
 6.3.7: a. Given the data 9, 5, 6, 2, 4, 7 construct the binary search tree...
 6.3.8: a. Given the data g, d, r, s, b, q, c, m construct the binary searc...
 6.3.9: a. For a set of six data items, what is the minimum worstcase numb...
 6.3.10: a. For a set of nine data items, what is the minimum worstcase num...
 6.3.11: An inorder tree traversal of a binary search tree produces a listin...
 6.3.12: Construct a binary search tree for In the high and faroff times th...
 6.3.13: Use the theorem on the lower bound for sorting to find lower bounds...
 6.3.14: Contrast the number of comparisons required for selection sort and ...
 6.3.15: One of five coinsis counterfeit and islighter than the other four. ...
 6.3.16: One of five coins is counterfeit and is either too heavy or too lig...
 6.3.17: One of four coins is counterfeit and is either too heavy or too lig...
 6.3.18: One of four coins is counterfeit and is either too heavy or too lig...
 6.3.19: Devise an algorithm to solve the problem of Exercise 18 using three...
 6.3.20: One of eight coins is counterfeit and is either too heavy or too li...
 6.3.21: In the decision tree for the binary search algorithm (and the binar...
 6.3.22: Our existing binary search algorithm (Chapter 3, Example 13) contai...
 6.3.23: To prove that log n! = (n log n), we can use the definition of orde...
Solutions for Chapter 6.3: Decision Trees
Full solutions for Mathematical Structures for Computer Science  7th Edition
ISBN: 9781429215107
Solutions for Chapter 6.3: Decision Trees
Get Full SolutionsMathematical Structures for Computer Science was written by Patricia and is associated to the ISBN: 9781429215107. Chapter 6.3: Decision Trees includes 23 full stepbystep solutions. Since 23 problems in chapter 6.3: Decision Trees have been answered, more than 4433 students have viewed full stepbystep solutions from this chapter. This expansive textbook survival guide covers the following chapters and their solutions. This textbook survival guide was created for the textbook: Mathematical Structures for Computer Science, edition: 7.

Back substitution.
Upper triangular systems are solved in reverse order Xn to Xl.

Column space C (A) =
space of all combinations of the columns of A.

Diagonalizable matrix A.
Must have n independent eigenvectors (in the columns of S; automatic with n different eigenvalues). Then SI AS = A = eigenvalue matrix.

Echelon matrix U.
The first nonzero entry (the pivot) in each row comes in a later column than the pivot in the previous row. All zero rows come last.

Free variable Xi.
Column i has no pivot in elimination. We can give the n  r free variables any values, then Ax = b determines the r pivot variables (if solvable!).

Full column rank r = n.
Independent columns, N(A) = {O}, no free variables.

Hessenberg matrix H.
Triangular matrix with one extra nonzero adjacent diagonal.

Linear combination cv + d w or L C jV j.
Vector addition and scalar multiplication.

Minimal polynomial of A.
The lowest degree polynomial with meA) = zero matrix. This is peA) = det(A  AI) if no eigenvalues are repeated; always meA) divides peA).

Nullspace matrix N.
The columns of N are the n  r special solutions to As = O.

Outer product uv T
= column times row = rank one matrix.

Partial pivoting.
In each column, choose the largest available pivot to control roundoff; all multipliers have leij I < 1. See condition number.

Particular solution x p.
Any solution to Ax = b; often x p has free variables = o.

Projection p = a(aTblaTa) onto the line through a.
P = aaT laTa has rank l.

Pseudoinverse A+ (MoorePenrose inverse).
The n by m matrix that "inverts" A from column space back to row space, with N(A+) = N(AT). A+ A and AA+ are the projection matrices onto the row space and column space. Rank(A +) = rank(A).

Row picture of Ax = b.
Each equation gives a plane in Rn; the planes intersect at x.

Row space C (AT) = all combinations of rows of A.
Column vectors by convention.

Singular matrix A.
A square matrix that has no inverse: det(A) = o.

Solvable system Ax = b.
The right side b is in the column space of A.

Spanning set.
Combinations of VI, ... ,Vm fill the space. The columns of A span C (A)!
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