One thing that new users of MATLAB mistake, is they want an if statement to apply to every element of a matrix or vector. For example:
M = magic(7)
M =
30 39 48 1 10 19 28
38 47 7 9 18 27 29
46 6 8 17 26 35 37
5 14 16 25 34 36 45
13 15 24 33 42 44 4
21 23 32 41 43 3 12
22 31 40 49 2 11 20
So M is a matrix. Now users want to have an if statement that will do something different for every element of the matrix, depending on some test, and MATLAB does not have such an ability, at least not as they want. But there are some things you can do. For example, suppose I wanted to divide all of the even numbers by 2, but for the odd numbers, I want to multiply by 3, and then add 1? (You may recognize what I am doing here, if not, then you could do some reading about the Collatz conjecture, sometimes just called the 3n+1 problem.)
Solution 1: Just use a loop.
M
M =
15 118 24 4 5 58 14
19 142 22 28 9 82 88
23 3 4 52 13 106 112
16 7 8 76 17 18 136
40 46 12 100 21 22 2
64 70 16 124 130 10 6
11 94 20 148 1 34 10
So it did exactly as we need it to do, operating on each element of M separately. Note that the loop uses a linear index, treating the matrix as if it were a vector of elements. This works in MATLAB, so I did not need to create a double loop on the rows AND the columns of M.
But the MATLAB way is to use matrices. That is how MATLAB is really written to be efficient. So we might have used this next idea. (I'll just apply it directly to the last result of M.)
Solution 2:
M(evenloc) = M(evenloc)/2;
M(oddloc) = 3*M(oddloc) + 1;
M
M =
46 59 12 2 16 29 7
58 71 11 14 28 41 44
70 10 2 26 40 53 56
8 22 4 38 52 9 68
20 23 6 50 64 11 1
32 35 8 62 65 5 3
34 47 10 74 4 17 5
As you can see, this also worked. No explicit loops were used, although internally inside the guts of MATLAB, it will be performing a loop in a lower level language, so an implicit loop. It employed what is called a boolean index. I could also have used find in a very similar way.
Solution 3:
evenloc = find(rem(M,2) == 0);
oddloc = find(rem(M,2) == 1);
M(evenloc) = M(evenloc)/2;
M(oddloc) = 3*M(oddloc) + 1;
M
M =
23 178 6 1 8 88 22
29 214 34 7 14 124 22
35 5 1 13 20 160 28
4 11 2 19 26 28 34
10 70 3 25 32 34 4
16 106 4 31 196 16 10
17 142 5 37 2 52 16
These are the common solutions you may see. Certainly there are others ways to do it too. I might have tried to be tricky, perhaps like this:
Solution 4:
M = mfac(rem(M,2) + 1) .* M + afac(rem(M,2) + 1)
M =
70 89 3 4 4 44 11
88 107 17 22 7 62 11
106 16 4 40 10 80 14
2 34 1 58 13 14 17
5 35 10 76 16 17 2
8 53 2 94 98 8 5
52 71 16 112 1 26 8
You may wish to look carefully at this last one. Why did it work? (Because it does work.) We could even use the last idea to usein a simple iteration scheme to play with the classic Collatz problem.
Miter = @(M) mfac(rem(M,2) + 1) .* M + afac(rem(M,2) + 1);
while any(~ismember(M(:),[1 2 4]))
M
M =
4 1 2 4 4 2 4
4 2 1 1 2 4 4
1 2 4 4 1 1 4
2 2 1 2 1 4 1
4 2 1 2 2 1 2
1 4 2 1 2 1 4
4 1 2 4 1 2 1
And there are cetainly other creative ways to solve the problem. In any such scheme I should have probably worried if the number was too large to be exactly representable as an integer in floating point arithematic in MATLAB, thus as a double. The limit for that is calling flintmax in MATLAB, but now I am digressing too far.
