1. Swapping Values (Octave)
In this problem your task is to write an octave function
swap that takes a matrix \(M\), numbers \(a\) and \(b\), and
replaces all of \(a\)’s in M with \(b\)’s, and vice-versa.
For full marks Your function should pass the tests below, and
be fully vectorized, i.e. not use loops, cellfun or
arrayfun.
%!test %!shared M1, M2 %! M1 = [0,0;0,1; %! 1,0;1,1]; %! M2 = [1,1;1,0; %! 0,1;0,0]; %! assert(swap(M1,1,0) == M2); %!assert (swap(M1,0,1) == M2); %!assert (swap(M1,1,1) == M1); %!test %! M3=[1,2; %! 3,4]; %! assert(swap(M3,1,4) == [4,2; %! 3,1])
1.1. Sample Solution
2. Go Territory (Octave)
In this problem, loosely based on the game of Go, your task
is to find captured spaces: \(0\)’s in a matrix that have \(1\)’s
above, below, left, and right. Write an octave function
territory that takes \(0/1\)-valued matrix as input and returns \(1\)
in the position of each captured space. For full marks your function should
pass the tests below, and be fully vectorized, i.e. not use loops, cellfun or
arrayfun
%!test %! board = [1,0,1,0; %! 0,1,0,1; %! 1,0,1,0; %! 0,1,0,1]; %! assert(territory(board) == [0,0,0,0; %! 0,0,1,0; %! 0,1,0,0; %! 0,0,0,0]); %!test %! board = [1,1,1,0; %! 0,1,0,1; %! 1,0,1,0; %! 0,1,0,1]; %! assert(territory(board) == [0,0,0,0; %! 0,0,1,0; %! 0,1,0,0; %! 0,0,0,0]); %!test %! board = [1,0,1,0, 1,0,1,0; %! 0,1,0,1, 0,1,0,1]; %! assert(territory(board) == zeros(2,8)) %!test %! board = [1,0,1,0,1,0,1,0; %! 0,1,0,1,0,1,0,1; %! 1,0,1,0,1,0,1,0; %! 1,0,1,0,1,1,1,0]; %! assert(territory(board) == [0,0,0,0,0,0,0,0; %! 0,0,1,0,1,0,1,0; %! 0,0,0,0,0,1,0,0; %! 0,0,0,0,0,0,0,0]);
3. Check Diet
Hint: Matrix times vector multiplication helps.
- Write an octave function
checkdietthat checks if a proposed diet meets the minimum daily requirements. Your function should pass the given tests. For full marks your function should be fully vectorized (i.e. no loops). - Identify two weaknesses of the given test suite for
checkdietand add two new tests.
## usage: passes = checkdiet(TABLE, MINS, DIET) ## ## Check if DIET passes the min daily requirements ## TABLE(i,j) is the amount of nutrient i in food j ## MINS(i) = minimum amount of nutrient i required ## DIET(j) = amount of food j in proposed diet function yesno = checkdiet(table, mins, diet) end %!test %! assert(checkdiet(eye(3),ones(3,1),ones(3,1)) == 1) %!test %! assert(checkdiet(eye(3),zeros(3,1),[1;0;0]) == 1) %!test %! assert(checkdiet(eye(3),[1;0;0],zeros(3,1)) == 0) %!test %! assert(checkdiet(eye(3),[1;0;0],0.5*ones(3,1)) == 0) %!test %! assert(checkdiet(ones(3,3),[1;0;0],0.5*ones(3,1)) == 1)
4. Mincol
Write the octave function mincol that finds the position of the
smallest element in each row of a matrix. You function should pass
the given test. For full marks your function should be fully
vectorized (i.e. no loops, arrayfun or cellfun).
## usage: given a matrix of integers, for each row ## return the column containing the minimum value function out = mincol(data) endfunction %!test %! A = [1,2,3; %! 3,2,4; %! 1,2,0; %! 5,3,4; %! 3,2,1; %! -10,0,0]; %! assert (mincol(A) == [1;2;3;2;3;1]);
4.1. Sample Solution
5. Neighbour Count
Complete the Octave function nbrcount that counts
for each element of a zero/one matrix, how many of the
neighbours are one. For full marks your solution should be vectorized.
function out = nbrcount(img) endfunction %!test %! A= [1,0,0; %! 0,0,0; %! 0,0,1; %! 1,0,0]; %! counts = [0,1,0; 1,2,1; 1,2,0; 0,2,1] %! assert(nbrcount(A) == counts)
5.1. Sample Solution
6. Isolated Pixels
Use the nbrcount function from
the previous question to define a function
isolated that finds all the isolated ones (those ones
whose neighbours are all zero). For full marks your solution should be vectorized.
function out = isolated(img) endfunction %!test %! A= [1,0,0; 0,0,0; 0,0,1; 1,0,0]; %! assert(isolated(A) == A) %!test %! A=[1,0,0; %! 0,0,0; %! 0,1,1; %! 1,0,1]; %! assert(isolated(A) == [1,0,0; 0,0,0; 0,0,0;0,0,0])
6.1. Sample Solution
7. Normalize
Complete the vectorized Octave function normalize according to the
given comment and tests. You do not need to copy the usage comment.
##usage: matrix = percent(raw, maxes) ## ## raw - raw scores, one row per student. ## maxes - maximum possible for that column ## ## Output is a matrix one row per student, with ratios function out=normalize(raw, maxes) endfunction %!test %! # journal,assgn, midterm,final %! mxs=[260, 60, 20, 60]; %! nrm = [0.9, 0.9, 0.9, 0.9; 0.75, 0.75, 0.75, 0.75; 1, 0, 1, 0]; %! raw=[234, 54, 18, 54; 195, 45, 15, 45; 260, 0, 20, 0]; %! assert(normalize(raw, mxs), nrm, eps);
8. Percent
Use the function normalize from the
previous question to complete the following vectorized Octave
function to calculate final percentages for students in a class. You
do not need to copy the usage comment.
##usage: scores = percent(raw, maxes, weights) ## ## raw - raw scores, one row per student. ## maxes - maximum possible for that column ## weights - weight for that column in percent ## ## Output is a column vector of a percent for each student. function out=percent(raw, maxes, weights) end %!test %! # journal,assgn, midterm,final %! mxs=[260, 60, 20, 60]; %! wgt=[20, 30, 20, 30]; %! raw=[234, 54, 18, 54; %! 195, 45, 15, 45; %! 260, 0, 20, 0; %! 0, 60, 0, 60; %! 200, 40, 17, 33]; %! assert(percent(raw, mxs, wgt), [90;75;40;60;68.88], .01);
9. Digits2num
Write a function to convert the digits of a positive base 10 number to the corresponding number. Your function should pass the following tests.
%!test "0" %! assert(digits2num([0]) == 0); %! assert(digits2num([0,0]) == 0;) %!test "1" %! assert(digits2num([1]) == 1); %! assert(digits2num([0,1]) == 1;) %!test "place value" %! assert(digits2num([2]) == 2) %! assert(digits2num([2,0]) == 20) %! assert(digits2num([2,3,0]) == 230)
9.1. Sample Solution