converting Matlab program to Mathematica code

Question

I have the following Matlab program;

k=1:1:10;
a=[3,5,2];
for j=1:length(k)
for h =1:1:length(a)
    x=a(h);
    y1(h)=x^j-2;
end
minfind(j)=min(y1);
end
plot(k, minfind)

I tried to write Mathematica code that perform the same calculation. So far I wrote the following Mathematica code

a = ( { {3, 5, 2} } );
For[j = 1, j <= 10, j = j + 1,
 For[h = 1, h <= 3, h = h + 1,
  x = a [[1, h]];
  y22 = x^j-2;
  y1 = Append[y1, y22];
  ];
 ymin1 = Min[y1];
 minfind = Append[minfind, ymin1];
 ]

But the above code doesn't work.

Can anyone help in pointing what is wrong with the above code and what is the correct code?

Answer 1

It's hard to know quite where to start with this, but I'd start with the answers to this question for some initial guidance.

As a general guide, nested For loops are almost never necessary and using list-based operations is much more efficient, as well as readable and less prone to error.

Let's take the inner loop first.

 For[h = 1, h <= 3, h = h + 1,
  x = a [[1, h]];
  y22 = x^j-2;
  y1 = Append[y1, y22]; ];

All you are really doing is constructing a vector of each element of a (respecifying a as a vector {3, 5, 2}) taken to the power j and then subtracting 2. By the way, is that what you want? or did you mean x^(j-2)?

So eliminate this loop by using the Listable property of arithmetic operations and writing

a^j - 2

You can eliminate most of the outer loop by changing this to

a^# - 2 & /@ Range[10]

Where Range[10] is what I think your definition of k in your Matlab code does. Evaluate that and check.

The result of the last line above is:

 {{1, 3, 0}, {7, 23, 2}, {25, 123, 6}, {79, 623, 14}, {241, 3123, 
  30}, {727, 15623, 62}, {2185, 78123, 126}, {6559, 390623, 
  254}, {19681, 1953123, 510}, {59047, 9765623, 1022}}

(Incidentally Outer[#2^#1 - 2 &, Range[10], a] gives the same output. You might want to experiment with some other list-based functions.)

You want the minimum of each row of that, so just use Map (shortcut notation /@) like this

Min /@ ( a^# - 2 & /@ Range[10])

And your result should be

{0, 2, 6, 14, 30, 62, 126, 254, 510, 1022}

Yes, all you need to replace that convoluted nested For loop is:

a =  {3, 5, 2};
Min /@ ( a^# - 2 & /@ Range[10])