- 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 gray-scal...
- 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
Upper triangular systems are solved in reverse order Xn to Xl.
peA) = det(A - AI) has peA) = zero matrix.
A = S-1 AS. A = eigenvalue matrix and S = eigenvector matrix of A. A must have n independent eigenvectors to make S invertible. All Ak = SA k S-I.
Eigenvalue A and eigenvector x.
Ax = AX with x#-O so det(A - AI) = o.
0,1,1,2,3,5, ... satisfy Fn = Fn-l + 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.
Invert A by row operations on [A I] to reach [I A-I].
A sequence of steps intended to approach the desired solution.
lA-II = 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).
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+ (Moore-Penrose 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.
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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