Solutions to Homework 6
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Problem neighbor
function w = neighbor(v)
w = [];
if min(size(v)) == 1
% must be a vector
for ii = 1:length(v)-1
% if length is less than 2, loop won't do any
thing
w(ii) = abs(v(ii+1) - v(ii));
end
end
end
Problem neighbor (alternative solution)
no explicit loop
function w = neighbor(v)
if length(v) < 2 || min(size(v)) ~= 1
% must be a vector of at least two elements
w = [];
else
w = abs(v(1:end-1)-v(2:end));
end
% take the difference of two subvectors
% of length (n-1)
end
Problem replace_me
builds up the output one element at a time
function w = replace_me(v,a,b,c)
if nargin < 3
b = 0;
end
if nargin < 4
c = b;
end
w = [];
for k = 1:length(v);
if v(k) == a
% if a is found,
w = [w,b,c];
else
w = [w,v(k)];
% we insert b and c at the end of the current w
% otherwise,
% we insert the original element of v
end
end
end
Problem replace_me (alternative solution)
only changes the output vector when an instance of a is found
function w = replace_me(v,a,b,c)
if nargin < 3
b = 0;
end
if nargin < 4
c = b;
end
w = v;
% make w the same as v
wi = 1;
% wi is used to index into w
for vi = 1:length(v)
if v(vi) == a
w = [w(1:wi-1) b c w(wi+1:end)];
% insert b and c at position wi
wi = wi + 1;
% increment wi
end
wi = wi + 1;
% wi is incremented in either case
end
end
Problem halfsum
using nested loops
function s = halfsum(A)
[row col] = size(A);
s = 0;
for ii = 1:row
for jj = ii:col
% the column index only starts at the current row index
s = s + A(ii,jj);
end
end
end
Problem halfsum (alternative solution)
using a single loop and sum
function s = halfsum(A)
[nr,~] = size(A);
s = 0;
for r = 1:nr
% for each row
s = s + sum(A(r,r:end));
% sum adds up the elements right of the diagonal (inc
lusive)
end
% in the current row
end
Problem large_elements
function found = large_element(A)
[row col] = size(A);
found = [];
for ii = 1:row
for jj = 1:col
if A(ii,jj) > ii + jj
% if the element is larger than the sum of its in
dexes
found = [found; ii jj]; % add a new row to the output matrix
end
end
end
end
problem one_per_n
using while-loop
function n = one_per_n(x)
n = 0;
sum = 0;
while sum < x && n <= 10000
n = n + 1;
sum = sum + 1/n;
end
if n > 10000
n = -1;
end
end
problem one_per_n (alternative solution)
using for-loop
function n = one_per_n(x)
s = 0;
for n = 1:1e4
s = s + 1/n;
if s >= x
return;
end
end
n = -1;
end
Problem approximate_pi
function [a,k] = approximate_pi(delta)
k = 0;
f = sqrt(12);
% compute sqrt(12) only once
a = f;
% the value of a for k == 0
while abs(pi-a) > delta
% while we are further away than delta
k = k + 1;
% increment k
a = a + f*(-3)^(-k)/(2*k+1);
% add increment to current value of a
end
end
Problem separate_by_two
using division and rounding
function [even,odd] = separate_by_two(A)
even = A(fix(A/2) == A/2)';
% if A is even, rounding does not do anything to A/2
odd
% if A is odd, it gets rid of the .5 part, so they won't
= A(fix(A/2) ~= A/2)';
be equal
end
% note that this will put non-integers into odd
Problem separate_by_two (alternative solution)
using mod (or rem)
function [even, odd] = separate_by_two(A)
even = A(mod(A,2) == 0)';
% mod gives 0 if even
odd
% mod gives 1 if odd
= A(mod(A,2) == 1)';
end
% note that this one will not put non-integers in any of the outputs
Problem separate_by_two (alternative
solution)
using mod (or rem)
function [even,odd] = separate_by_two(A)
mod2 = logical(mod(A,2));
even = A(~mod2)';
% modulo 2 is zero for even numbers (logical false), so we ne
ed to negate it
odd
= A(mod2)';
% modulo 2 is non-zero for odd numbers, that is, logical true
end
% note that this will put non-integers into odd
Problem divvy
function A = divvy (A,k)
L = (mod(A,k) ~= 0);
% creates a logical matrix based on divisibility by k
A(L) = k * A(L);
% changes only the non-divisible elements of A by multiplying
them by k
end
% uses A as both input and output, so we only need to modify some elements of A
Problem divvy (alternative solution)
single line solution
function I = divvy(I,k)
I(mod(I,k) ~= 0) = I(mod(I,k) ~= 0) * k;
end
% same solution as above, but it repeats the modulo computation
Problem square_wave
using a for-loop
function sq = square_wave(n)
t = 0 : 4*pi/1000 : 4*pi;
% setup vector according to the specs
sq = zeros(1,length(t));
% initialize output to 0
for ii = 1:2:2*n
% run for first n odd numbers (2k-1)
sq = sq + cos(ii*t-pi/2)/ii;
end
end
% add the next cosine term
Problem square_wave (alternative solution)
tricky code with no explicit loops
function s = square_wave(n)
t = 0 : 4*pi/1000 : 4*pi;
% setup vector according to the specs
idx = (2*(1:n)' - 1);
% make column vector of fist n odd numbers (2k-1)
% idx*t makes a matrix; each row is (2k-1)*t, for a given k
% idx*ones(size(t)) also makes a matrix; each element of row k is just (2k-1)
% sum down the columns
s = sum(sin(idx*t) ./ (idx*ones(size(t))),1);
end
% the second argument to sum is needed in case n is 1
% remember that sum(x) sums x along columns unless x is a row vector!
Problem my_prime
using a for-loop
function a = myprime(n)
a = false;
if n > 1
% 1 is by definition not prime
for ii = 2:sqrt(n)
% see explanation below
if ~mod(n,ii)
return;
end
end
a = true;
end
end
% x is prime if it is NOT divisible by all integers from 2 to sqrt(x)
% because factors have to come in pairs -- one bigger than sqrt(x) and
% one smaller (or both equal)
Problem my_prime (alternative solution)
with no explicit loops
function prim = myprime(p)
v = 2:sqrt(p);
end
v = v(rem(p,v) == 0);
% if p is prime, none of the remainders can be 0
prim = ~length(v) && (p ~= 1);
% so if v has any elements, p is not prime
% 1 is handled by the (p ~= 1) condition
Published with MATLAB R2014a
Created Tue 2 Jun 2015 12:21 PM PET
Last Modified Tue 9 Jun 2015 9:59 AM PET
