Anne Marsan: Software for Engineers

Sum the primes below 2,000,000.

This is the sort of problem that Matlab handles very well. The code looks pretty, and I got an answer right quick. Now if I type fast, I'll get my code posted before my battery drains and Baseball Tonight ends.

Now I'm really challenging myself - can I get a solution to a problem without the fan kicking on? This one took a few tries, but I finally found a good algorithm.

We're asked to find the pythagorean triplet (a, b, c) that sums to 1000. I made an educated guess as to a good range for c, to limit the search space.

This problem reminded me of Professor Rida Farouki, who teaches his students the value of conserving computations in CAD systems, and who has written several papers on Pythagorean hodograph curves.

This problem asks us to find the greatest product of five consecutive digits in a given 1000 digit number. First the 1000 digit number has to be stored as a string. I wanted to make sure that each digit of the number was converted from a string to an int only once, which complicates the code a bit. Second, it occurred to me that if I'm looking for the greatest product, I'm also looking for the greatest sum. This allows me to save some further compute time.

Is it cheating to use the Matlab built in isprime function? It saves me from going to Google to find a prime number algorithm.

Here's my code:

This completed in a reasonable amount of time on my slow Mac, but later I read about only testing odd numbers, and I wish I had thought of that.

I'm skipping around a bit, but this one shows off Matlab a little better.

function euler6

nat_nums = 1:1:100
sum(nat_nums,2)^2 - sum(nat_nums .* nat_nums, 2) 
 

.* is the array multiplcation operator (element-wise)

Arrays in Matlab are really 1xn matrices, so in the sum functions, the second argument indicates that I should sum across columns.

Matlab rocks. 

I thought I'd join the Euler problem bandwagon. In true mechanical engineering fashion, I'm using Matlab to solve them. Actually, just for an added twist, I'm using Octave, an open source math interpreter that is mostly compatible with Matlab, on my dog-slow Mac notebook, forcing me to not use brute-force approaches if they can be avoided.

I had already seen solutions to problems 1-3, so I started with 4.  So here goes:

function euler4

max_result = 0;
for ii = 999:-1:100
  for jj = 999:-1:100

    % Don't bother to check if this number is smaller than the max found
    result = ii*jj;
    if (result < max_result)
      continue;
    end
      
    % Check for palindrome
    result_string = int2str(result);
    success = 1;
    rlength = length(result_string);
    for kk = 1:rlength/2
      if result_string(kk) ~= result_string(rlength-kk+1)
        success = 0;
        break;
      end
    end
    if success && result > max_result
      max_result = result;
    end
    
  end
end
max_result 

Not pretty, but it gets the job done.