Basic exercises. Array exercises. 1. x = 32:2: x = [ ] a = x + 16 b = x(1:2:end) + 3 c = sqrt(x) or c = x.^(0.5) d = x.^2 or d = x.

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1 Basic exercises 1. x = 32:2:75 2. x = [ ] a = x + 16 b = x(1:2:) + 3 c = sqrt(x) or c = x.^(0.5) d = x.^2 or d = x.*x 3. x = [ ]', y = [ ]' a = y + sum(x) b = x.^y c = x./y z = x.*y w = sum(z) x'*y - w (same thing) 5. The function "rats" just displays the contents of a variable a = 2:2:20 (or whatever max #) b = 10:-2:-4 c1 = 1:5, c2 = 1./c1, rats(c2) d1 = 0:4, d2 = 1:5, d3 = d1./d2, rats(d3) 6. n = 1:100; x = ( (-1).^(n+1) )./ (2*n - 1); y = sum(x) 8. t = 1:0.2:2 a = log(2 + t + t.^2) b = exp(t).*(1 + cos(3*t)) c = cos(t).^2 + sin(t).^2 (all ones!) d = atan(t) e = cot(t) f = sec(t).^2 + cot(t) t = 1790:2000; term = 1 + exp( *(t )); P = /term; plot(t,p) xlabel('year'), ylabel('population') P2020 = /(1 + exp( *( ))) Array exercises 2. A = [ ; ; 3 5 9] x1 = A(1,:) y = A(-1:,:) c = sum(a) d = sum(a,2) or d = sum(a')' N = size(a,1), e = std(a)/sqrt(n) 5. A = [ ; ; ] B = A(:,2:2:) C = A(1:2:,:) c = reshape(a,4,3) or c = A' (they are different but are both 4x3) d = 1./A, rats(d) e = sqrt(a) 6. randn('seed', )

2 F = randn(5,10); N = size(f,1) avg = mean(f) s = std(f) tscore = (avg - 0)./(s/sqrt(N)) None were different at 90% LOC (all < 2.132). Relational/Logical exercises 4. x = [ ] a = x, idxa = x > 0, a(idxa) = 0 b = x, idxb = ~rem(x,3), b(idxb) = 3 c = x, idxc = ~rem(x,2), c(idxc) = 5*c(idxc) 5. x = 1:35; y = zeros(size(x)); idx1 = x < 6; idx2 = (x >= 6) & (x < 20); idx3 = (x >= 20) & (x <= 35); y(idx1) = 2; y(idx2) = x(idx2) - 4; y(idx3) = 36 - x(idx3); disp([x(:) idx1(:) idx2(:) idx3(:) y(:)]) plot(x,y,'o') Loop constructs: The answers here provide one version of the solutions. Alternatives are possible and encouraged, especially where time and efficiency of the code is important. 1. x = [ ] a. total = 0; for j = 1:length(x) total = total + x(j); b. runningtotal = zeros(size(x)); runningtotal(1) = x(1); for j = 2:length(x) runningtotal(j) = runningtotal(j-1) + x(j); c. s = zeros(size(x)); for j = 1:length(x) s(j) = sin(x(j)); 2. A = rand(4,7); [M,N] = size(a); for j = 1:M for k = 1:N if A(j,k) < 0.2 A(j,k) = 0; else A(j,k) = 1; 3. x = [4 1 6], y = [6 2 7]

3 N = length(x); for j = 1:N c(j) = x(j)*y(j); for k = 1:N a(j,k) = x(j)*y(k); b(j,k) = x(j)/y(k); d(j,k) = x(j)/(2 + x(j) + y(k)); e(j,k) = 1/min(x(j),y(k)); c = sum(c); % or use 1.a. loop 4. These code snippets do the job but their repeated use is much more interesting. An example is given for the first exercise. a. total = 0; % initialize current sum (the test variable) count = 0; % initialize the counter (output of the program) while total < 20 % loop until 20 is exceeded count = count + 1; % another loop repeat => another number added x = rand(1,1); total = total + x; % modify the test variable! disp(['it took ',int2str(count),' numbers this time.']) To do this many times, place the above code in a for-loop. Some simple (though perhaps subtle) changes are needed wth respect to retaining the counts for each repeat. Also, the summary has been changed from a single text message to a histogram. Nrep = 1000; % collect 1000 repeats of the above code count = zeros(nrep,1); for j = 1:Nrep total = 0; % reset the test variable each repeat!!! while total < 20 count(j) = count(j) + 1; % use a vector to capture each result total = total + rand(1,1); hist(count,min(count):max(count)) xlabel('number of random numbers from U(0,1) added to make 20') ylabel('count') title(['histogram of results for ',int2str(nrep),' repeats']) b. count = 0; while 1 % infinite loop use count = count + 1; x = rand(1,1); % get a number if (x < 0.85) & (x > 0.8) % check the value break % bail out if in selected range disp(['it took ',int2str(count),' numbers this time.']) c. count = 0; avg = 0; % test variable while abs(avg - 0.5) > 0.01 count = count + 1;

4 % The following line is one way to update the average. % (count-1)*avg is the sum of the first count-1 numbers % and rand just adds another number. Dividing by count % then gives the new average. avg = ((count-1)*avg + rand(1,1))/count; % modify the test var. % There are other ways to do this and you are encouraged % to come up with them disp(['it took ',int2str(count),' numbers this time.']) 5. while 1 % use of an infinite loop TinF = input('temperature in F: '); % get input if isempty(tinf) % how to get out break TinC = 5*(TinF - 32)/9; % conversion disp(' ') disp([' ==> Temperature in C = ',num2str(tinc)]) disp(' ') IF-block exercises 1. n = 7 gives m = 8 n = 9 gives m = -1 n = -10 gives m = z = 1 gives w = 2 z = 9 gives w = 0 z = 60 gives w = sqrt(60) z = 200 gives w = T = 50 gives h = 0 T = 15 gives h = 31 T = 0 gives h = 1 4. x = -1 gives y = -4 x = 5 gives y = 20 x = 30 gives y = 120 x = 100 gives y = 400

5 Arithmetic: >> 7+9 >> 3*2 >> 2(3+4) >> 2+3*4 >> (2+3)*4 >> 2^3 >> 21/7 >> ans*2 >> 3,14 >> 3.14 >> pi >> 2/3 >> format long >> 2/3 >> pi >> round(ans) >> ceil(3.1) >> floor(3.1) >> help >> help ops >> helpwin >> helpdesk Exercises: 1. Compute the sum of the product of 3 and 4 and the difference between 9 and Convert the binary numbers and to the decimal form. Hint: use 2^j.

6 3. Use helpwin or helpdesk to find a matlab command that does the conversion in previous exercise. Hint: search for binary. 4. Express pi rounded to 3 decimals. Hint: round 1000*pi. ================= Variables: >> a=3 >> b=5; (What is the effect of ; here?) >> b >> c=a+2*b; >> c=c+1 >> c=a*d; >> base=3 >> height=5 >> area=base*height/2 >> a_minus_b=a-b >> who >> whos >> help who >> help whos >> clear >> who ================== Vectors/lists: >> v=[ ] >> v(3) >> v(3)=8

7 >> v(5) >> w=[ ] >> v+w >> v-w >> v+1 >> 2*w >> w/4 >> sum(w) >> v*w >> v.*w >> w.^2 >> [v w] >> ans' >> u=1:4 >> u=0:2:10 >> u=0:3:9 >> u=0:3:10 >> u=0:.2:.6 >> u(2:3) >> help linspace >> linspace(0,1,6) Exercises: 1. Define a price list for magazines, movie, cd, and a corresponding list of how many of these items you consume during a year. Then use these lists to compute a list of the total cost for each of the three items and the total cost. 2. Test whether N ~= N^2/2 and 1+2^2+3^2+..+N^2 ~= N^3/3. Hint: first define x=1:n and use sum. What does ~= mean?

8 ====================== Matrices: >> m=[1 2 3; 4 5 6] >> m(2,1) >> m(1,3) >> m(1,3)=7 >> v >> w >> [v;w] >> v' >> ans(3) >> w' >> [v' w'] >> ans' >> [v';w'] >> who >> help save >> save junk >> clear >> who >> load junk >> who

9 False (=0) / True (=1): >> 2<3 >> 2>3 >> -3<2 >> 1+1 = = 2 >> 2 = = 3 >> 2 ~= 3 ================== Conditional statements: >> i=1; >> if i = = 1 disp('i is one'); >> i=2 >> if i = = 1 disp('i is one'); >> if i = = 1 disp('i is one'); else disp('i is not one'); >> help if ====================== Script files: >> edit (type disp('hi') and save file as hello.m) >> hello Replace disp('hi') by code that sorts a list/vector v=[a b], that is, gives v=[a b] if a < b and v=[b a] if b < a, for example if v(1) < v(2) v=v; elseif v(2) < v(1) v=[v(2) v(1)]; v or more elegantly

10 if v(2) < v(1) v=[v(2) v(1)]; v and save as MyFirstSort.m. Test the script file with >> v=[2 1] >> MyFirstSort >> MyFirstSort What happens in the two code versions if v(1)=v(2)? =================== Loops, for and while: Write code that makes the computer count to a given number N, for example. i=0; while i < N i=i+1; disp(i) Save as CountToN.m. Test with >> N=10; >> CountToN Alternative code construction for i=1:n disp(i) Save as CountToN2.m and test with >> CountToN2 Exercises: 1. a) Get the computer to count only the odd numbers up to N. Hint: elegant to use the for loop with i=1:2:n. b) Get the computer to count backwards from N down to 1, first with a "for loop" (hint: elegant to just replace i=1:n by i=n:-1:1), then with a "while loop". 2. Write code that for a given number N tests whether N = N^2/2 by computing N and N^2/2, and display their difference and ratio. What is the ratio for N=1000? Explain why two numbers with a considerable difference can have a ratio very close to one!

11 3. Given natural numbers m and n, have matlab find p and r with 0<=r < n such that m=pn+r. Try the following code p=0; r=m; while r>=n p=p+1; r=m-p*n; disp(['m=' num2str(p) 'n+' num2str(r)]) 4. Check if a turtle starting at 0 ever reaches 1. Consider and test the following code and explain the outcome! pos=0; while pos<1 step=(1-pos)/2; pos=pos+step; disp(pos) 5. How would the following code work? while 1<2 disp('this will go on forever') Try it! If you run into problem you can always press Ctrl-C 6. Try and then seek to understand the following code? while 1 disp('what is 3+4?') answer=input(''); if answer == 7 disp('correct') return else disp('wrong, but you get another chance.') >> help input ============================ Computer arithmetic: >> eps (a very small positive number in the computer. The smallest?) >> eps/2 >> eps/

12 >> help eps >> 1+eps = = 1 >> 1+eps/2 = = 1 >> a=10 >> a=a^2 (repeat, first until matlab changes to number representation of the form d.dddde+ddd, and make sure you can interprete this, then until matlab gets exhausted..) >> Inf=Inf+1 >> Inf*2 >> Inf/2 >> 1/eps >> 1/0 >> -1/0 (seems as if 0 is somewhat positive after all, strange..) >> 0/0 (Not a Number, without any reasonable interpretation..) >> Inf-Inf >> Inf/Inf >> 0*Inf Exercises: 1. Seek the largest number less than Inf, according to matlab. ======================== Composite conditions: >> 0<=1 >> 1<=1 >> 1<2 1>2

13 >> 1<2 & 1>2 >> ~(1<2)

14 Strings: >> s=1+2 >> s='1+2' >> eval(s) >> f='2*x' >> eval(f) >> x=7 >> eval(f) >> f(2) >> f(3) >> ['text1' 'text2'] >> ['text' 7] >> ['text' num2str(7)] >> help num2str >> disp(['f(x)=' num2str(eval(f))]) >> help disp >> whos ============================== Elementary graphics: >> x=0:.2:1 >> y=x.^2 >> plot(x,y) >> plot(x,1-x/2) >> plot(x,sin(10*x)) >> x=0:.01:1; >> plot(x,sin(10*x))

15 >> xlabel('x') >> ylabel('y') >> xlabel('x_1'); ylabel('x_2') >> title('my first plot') >> hold on >> plot(x,x.^2) >> grid on >> clf >> subplot(211); plot(x,sin(10*x)) >> subplot(212); plot(x,x.^2) >> get(gcf) ============================ Search priorities: >> max(2,1) >> max=[1 2 3;4 5 6] >> max(2,1) >> clear max >> max(2,1) >> hello ( I assume that your hello.m file is intact with the text disp('hi') ) >> hello=5; >> hello >> who >> clear hello >> hello

16 >> path >> help path >> dir >> ls >> help ls >> help dir >> help mkdir >> mkdir m-files >> help cd >> cd m-files >> dir >> edit ( type disp('hi from hello in m-files') and save as hello.m ) >> hello >> cd.. >> hello (the idea is that you will now get a 'hi' from your old hello.m file) ======================== Documentation: >> help max >> edit