Solutions to review questions for Prelim 2
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1. allInterval
function B = allInterval(x,a,b)
% x is a length-n vector.
% B is assigned the value of 1 (true) if every x-value is in the
% interval [a,b].
% B is assigned the value of 0 (false) if at least one of the x-values
% is not in the interval [a,b].
% Your implementation should make effective use of a while-loop.
n = length(x);
B = (a <= x(1)) && (x(1) <= b);
k = 1;
% B is true if x(1),...,x(k) are in the interval...
while k<n && B
k = k+1;
% B "turns false" if x(k) not in [a,b]...
B = (a <= x(k)) && (x(k) <= b));
end
2. TotalValue
% Refer to the Cost-Inventory application in Lecture 14.
% Complete the following function...
function T = totalValue(Inv,Cost)
% T is the total value of all the inventory in all the factories
[nFact,nProd] = size(Inv);
T = 0;
for i=1:nFact
% Compute the value of factory i's inventory
S = 0;
for j=1:nProd
S = S + Inv(i,j)*Cost(i,j);
end
T = T + S;
end
3. Censoring text
function D = censor(str, A)
% Replace all occurrences of string str in character matrix A, regardless
% of case, with 'X's.
% A is a matrix of characters.
% str is a string. Assume that str is never split across two lines in A.
% D is A with 'X's replacing the censored string str.
%
% Example: A = ['Use MATLAB '; ...
% 'in that lab.']
%
% and str = 'lab'
%
% Then D = ['Use MATXXX '; ...
% 'in that XXX.']
D= A;
B= lower(A);
s= lower(str);
ns= length(str);
[nr,nc]= size(A);
% Traverse the matrix to censor string str
for r= 1:nr
for c= 1:nc-ns+1
if strcmp(s,B(r,c:c+ns-1))==1
for k= 1:ns
D(r,c+k-1)= 'X';
end
end
end
end
4. TwoClicks
% Refer to Lectures 14 and 15 and the function RandomLinks(n).
A = RandomLinks(1000);
% Note that if A(i,j) is one, then there is a link on webpage j
% to webpage i. Write a fragment that prints "yes" if it is
% possible to go from web page #100 to webpage #200 in one or two clicks.
% Thus, if A(101,100) = 1 and A(200,101) = 1 then "yes".
if A(200,100)==1
% One click away...
disp('yes')
else
% See if it is possible to reach in two clicks
% Idea: Check the "1-click" pages to see if they link
% to page 200. Quit as soon as success.
Found = 0;
k =0;
while k<1000 && ~Found
k = k+1;
% Is page k a 1-click page?
if A(k,100)==1
% Is there a link to page k
if A(200,k) == 1
% There is a link on page k to page 200
disp('yes')
Found = 1;
end
end
end
end
5. Checking for numbers in a vector/matrix
function B = vecHasNegAndPos(x)
% x is a length-n vector and n>=2
% B is assigned the value of 1 (true) if x has at least one component
% that is strictly negative and at least one component that is
% strictly positive.
% Otherise B should be assigned the value of 0.
% Your implementation should make effective use of a while-loop.
nPos = 0; % Number of positive components found so far.
nNeg = 0; % Number of negative components found so far;
k = 0;
while k<length(x) && (nPos==0 || nNeg==0)
k = k+1;
if x(k) > 0
nPos = nPos + 1;
end
if x(k) < 0
nNeg = nNeg + 1;
end
end
B = (nPos>=1 && nNeg>=1);
function B = matHasNegAndPos(A)
% A is an m-by-n real array with M>=2 and n>=2
% B is assigned the value of 1 (true) if A has at least one component
% that has a strictly negative value and at least one component that has a
% strictly positive value.
% Otherise B should be assigned the value of 0.
% Your implemention should make effective use of VecHasNegAndPos
[m,n] = size(A);
% Place all the values in A into a 1-dimensional array...
aVec = []
for i =1:m
aVec = [aVec A(i,:)];
end
B = VecHasNegAndPos(aVec);
6. Reduce
function B = Reduce(A)
% A is an n-by-n array with n odd and at least 3 in value.
% B is obtained by deleting all the even-indexed rows and columns.
% Thus if
% A = [ 1 2 3 4 5 ;...
% 6 7 8 9 10 ;...
% 11 12 13 14 15 ;...
% 16 17 18 19 20 ;...
% 21 22 23 24 25 ]
% then
% B = [ 1 3 5;...
% 11 13 15;...
% 21 23 25]
[n,n] = size(A);
% Assemble all the odd rows
D = [];
for i=1:m
if rem(i,2)==1
D = [D ; A(i,:)];
end
end
% Now assemble all the odd columns of D...
B = [];
for j=1:n
if rem(j,2)==1
B = [B D(:,j)];
end
end
7. Longest
function [len, ind] = longest(C)
% Find the longest string(s) in cell array C.
% C is an n-by-1 cell array of strings, n>=1.
% len is the length of the longest string in C.
% ind is a vector, possibly of length one, containing the index number(s)
% of the string(s) with length len
% len records the length of the longest string from C{1} to C{k}
len = -1;
% ind records indicies of strings whose lengths are len
ind = [];
for k=1:length(C)
l = length(C{k});
if l == len
ind = [ind k];
elseif l > len
ind = k;
len = l;
end
end
8. Insight questions
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% Sec 7.1 #3
function [p,q] = maxEntry(A)
% A is m-by-n matrix
% p and q are indices with the property that |A(p,q)|>=|A(i,j)| for
% all i and j that satisfy 1<=i<=m, 1<=j<=n
p = 1;
q = 1;
[m,n] = size(A);
for i =1:m
for j=1:n
if (A(i,j) > A(p,q))
p = i;
q = j;
end
end
end
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% Sec 9.1 #4
function s = Compress(t)
% t is a string
% s is obtained by deleting all the blanks in t.
% Thus, if t = 'ab cd efg ', the s should be assigned 'abcdef'
s= '';
k= 0;
for i= 1:length(t)
if t(i)~=' '
k= k+1;
s(k)= t(i);
end
end
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% Sec 9.1 #7
function k = FindSubstring(S1, S2)
%Find the first occurrence of string S1 in string S2.
%If S1 is a substring of S2 then k is the position of the first
% matching character of the first match in S2.
%If s1 is not a substring of S2, then k is zero.
len1 = length(S1);
len2 = length(S2);
lastPos2Check= len2-len1+1;
k = 1;
while k <= lastPos2Check && strcmp(S1, S2(k:k+len1-1))==0
k= k+1;
end
if k>lastPos2Check
k= 0;
end
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% Sec 9.2 #5
function a = compare1(s1,s2)
% s1 and s2 are strings
% If s1 comes before or is equal to s2 in the ASCII dictionary order then a
% is 1. otherwise, a is 0. Case is ignored.
if (length(s1) < length(s2))
for i = 1:length(s2)-length(s1)
s1 = [s1 ' '];
end
end
if (length(s2) < length(s1))
for i = 1:length(s1)-length(s2)
s2 = [s2 ' '];
end
end
k = 1;
while (k < length(s1)) && (s1(k) == s2(k))
k = k + 1;
end
if s1(k) <= s2(k)
a = 1;
else
a = 0;
end