 6.4.1: Is the following code a prefix code? Why or why not? Character m b ...
 6.4.2: Using the code of Exercise 1, decode the string 01101.
 6.4.3: Given the codes Character a e i o u Encoding scheme 00 01 10 110 11...
 6.4.4: Given the codes Character b h q w % Encoding scheme 1000 1001 0 11 ...
 6.4.5: Given the codes Character a p w ( ) Encoding scheme 001 1010 110 11...
 6.4.6: Given the nonprefix codes Character 1 3 5 7 9 Encoding scheme 1 111...
 6.4.7: Write the Huffman codes for a, b, c, and d in the binary tree shown
 6.4.8: Write the Huffman codes for r, s, t, u in the binary tree shown.
 6.4.9: a. Construct the Huffman tree for the following characters and freq...
 6.4.10: a. Construct the Huffman tree for the following characters and freq...
 6.4.11: a. Construct the Huffman tree for the following characters and freq...
 6.4.12: a. Construct the Huffman tree for the following characters and freq...
 6.4.13: Construct the Huffman tree and find the Huffman codes for the follo...
 6.4.14: Construct the Huffman tree and find the Huffman codes for the follo...
 6.4.15: JPEG can achieve various compression levels; the higher the compres...
 6.4.16: Explain why JPEG encoding results in less compression for grayscal...
 6.4.17: Someone does a global substitution on the text file of Exercise 11,...
 6.4.18: Consider the following paragraph. However, in my thoughts I could n...
 6.4.19: Recall the problem posed at the beginning of this chapter. You work...
 6.4.20: In the justification that the Huffman algorithm produces an optimal...
Solutions for Chapter 6.4: Huffman Codes
Full solutions for Mathematical Structures for Computer Science  7th Edition
ISBN: 9781429215107
Solutions for Chapter 6.4: Huffman Codes
Get Full SolutionsChapter 6.4: Huffman Codes includes 20 full stepbystep solutions. Since 20 problems in chapter 6.4: Huffman Codes have been answered, more than 4000 students have viewed full stepbystep solutions from this chapter. Mathematical Structures for Computer Science was written by Patricia and is associated to the ISBN: 9781429215107. This textbook survival guide was created for the textbook: Mathematical Structures for Computer Science, edition: 7. This expansive textbook survival guide covers the following chapters and their solutions.

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

CayleyHamilton Theorem.
peA) = det(A  AI) has peA) = zero matrix.

Diagonalization
A = S1 AS. A = eigenvalue matrix and S = eigenvector matrix of A. A must have n independent eigenvectors to make S invertible. All Ak = SA k SI.

Eigenvalue A and eigenvector x.
Ax = AX with x#O so det(A  AI) = o.

Fibonacci numbers
0,1,1,2,3,5, ... satisfy Fn = Fnl + Fn 2 = (A7 A~)I()q A2). Growth rate Al = (1 + .J5) 12 is the largest eigenvalue of the Fibonacci matrix [ } A].

Four Fundamental Subspaces C (A), N (A), C (AT), N (AT).
Use AT for complex A.

GaussJordan method.
Invert A by row operations on [A I] to reach [I AI].

Iterative method.
A sequence of steps intended to approach the desired solution.

lAII = l/lAI and IATI = IAI.
The big formula for det(A) has a sum of n! terms, the cofactor formula uses determinants of size n  1, volume of box = I det( A) I.

Length II x II.
Square root of x T x (Pythagoras in n dimensions).

Lucas numbers
Ln = 2,J, 3, 4, ... satisfy Ln = L n l +Ln 2 = A1 +A~, with AI, A2 = (1 ± /5)/2 from the Fibonacci matrix U~]' Compare Lo = 2 with Fo = O.

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 N (A)
= All solutions to Ax = O. Dimension n  r = (# columns)  rank.

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).

Rayleigh quotient q (x) = X T Ax I x T x for symmetric A: Amin < q (x) < Amax.
Those extremes are reached at the eigenvectors x for Amin(A) and Amax(A).

Right inverse A+.
If A has full row rank m, then A+ = AT(AAT)l has AA+ = 1m.

Semidefinite matrix A.
(Positive) semidefinite: all x T Ax > 0, all A > 0; A = any RT R.

Singular Value Decomposition
(SVD) A = U:E VT = (orthogonal) ( diag)( orthogonal) First r columns of U and V are orthonormal bases of C (A) and C (AT), AVi = O'iUi with singular value O'i > O. Last columns are orthonormal bases of nullspaces.

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

Vector v in Rn.
Sequence of n real numbers v = (VI, ... , Vn) = point in Rn.
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