- 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
Change of basis matrix M.
The old basis vectors v j are combinations L mij Wi of the new basis vectors. The coordinates of CI VI + ... + cnvn = dl wI + ... + dn Wn are related by d = M c. (For n = 2 set VI = mll WI +m21 W2, V2 = m12WI +m22w2.)
When random variables Xi have mean = average value = 0, their covariances "'£ ij are the averages of XiX j. With means Xi, the matrix :E = mean of (x - x) (x - x) T is positive (semi)definite; :E is diagonal if the Xi are independent.
Diagonal matrix D.
dij = 0 if i #- j. Block-diagonal: zero outside square blocks Du.
Eigenvalue A and eigenvector x.
Ax = AX with x#-O so det(A - AI) = o.
A sequence of row operations that reduces A to an upper triangular U or to the reduced form R = rref(A). Then A = LU with multipliers eO in L, or P A = L U with row exchanges in P, or E A = R with an invertible E.
Fast Fourier Transform (FFT).
A factorization of the Fourier matrix Fn into e = log2 n matrices Si times a permutation. Each Si needs only nl2 multiplications, so Fnx and Fn-1c can be computed with ne/2 multiplications. Revolutionary.
Free columns of A.
Columns without pivots; these are combinations of earlier columns.
Invert A by row operations on [A I] to reach [I A-I].
Hermitian matrix A H = AT = A.
Complex analog a j i = aU of a symmetric matrix.
Kronecker product (tensor product) A ® B.
Blocks aij B, eigenvalues Ap(A)Aq(B).
Markov matrix M.
All mij > 0 and each column sum is 1. Largest eigenvalue A = 1. If mij > 0, the columns of Mk approach the steady state eigenvector M s = s > O.
Plane (or hyperplane) in Rn.
Vectors x with aT x = O. Plane is perpendicular to a =1= O.
Projection matrix P onto subspace S.
Projection p = P b is the closest point to b in S, error e = b - Pb is perpendicularto S. p 2 = P = pT, eigenvalues are 1 or 0, eigenvectors are in S or S...L. If columns of A = basis for S then P = A (AT A) -1 AT.
Projection p = a(aTblaTa) onto the line through a.
P = aaT laTa has rank l.
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).
Saddle point of I(x}, ... ,xn ).
A point where the first derivatives of I are zero and the second derivative matrix (a2 II aXi ax j = Hessian matrix) is indefinite.
Semidefinite matrix A.
(Positive) semidefinite: all x T Ax > 0, all A > 0; A = any RT R.
Simplex method for linear programming.
The minimum cost vector x * is found by moving from comer to lower cost comer along the edges of the feasible set (where the constraints Ax = b and x > 0 are satisfied). Minimum cost at a comer!
Combinations of VI, ... ,Vm fill the space. The columns of A span C (A)!
Special solutions to As = O.
One free variable is Si = 1, other free variables = o.