## C vs Java vs Python vs EMT

I recently tested Python to see if it is good as a first programming language. The answer is a limited „yes“. Python has certainly a lot to offer. It is a mainstream language and thus there is a lot of support in the net, and there are tons of libraries for all sorts of applications. A modern version like Anaconda even installs all these tools effortlessly.

However, I keep asking myself if it isn’t more important to learn the basics before flying high. Or to put it another way: Should we really start to teach the usage of something like a machine learning toolbox before the student has understood data types and simple loops? This question is important because most high-level libraries take away the tedious work. If you know the right commands you can start problem-solving without knowing much of the underlying programming language.

I cannot decide for all teachers. It depends on the type of students and the type of class. But I can give you an idea of how much a high-level language like Python is holding you back if you want to implement some algorithm yourself. You have to rely on a well-implemented solution being available, or you will be lost.

So here is some test code for the following simple problem. We want to count how many pairs (i,j) of numbers exist such that i^2+j^2 is prime. For the count, we restrict i and j to 1000.

#include#include int isprime (int n) { if (n == 1 || n == 2) return 1; if (n % 2 == 0) return 0; int i = 3; while (i * i <= n) { if (n % i == 0) return 0; i = i + 2; } return 1; } int count(int n) { int c = 0; for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) if (isprime(i * i + j * j)) c = c + 1; return c; } int main() { clock_t t = clock(); printf("%d\n",count(1000)); printf("%g", (clock() - t) / 1000.0); } /* Result was: 98023 0.132 */

So this takes 0.132 seconds. That is okay for one million prime-number checks.

Below is a direct translation in Java. Surprisingly, this is even faster than C, taking 0.127 seconds. The reason is not clear. But I might not have switched on every optimization of C.

public class Test { static boolean isprime (int n) { if (n == 1 || n == 2) return true; if (n % 2 == 0) return false; int i = 3; while (i * i <= n) { if (n % i == 0) return false; i = i + 2; } return true; } static int count(int n) { int c = 0; for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) if (isprime(i * i + j * j)) c = c + 1; return c; } static double clock () { return System.currentTimeMillis(); } public static void main (String args[]) { double t = clock(); System.out.println(count(1000)); System.out.println((clock() - t) / 1000.0); } /* Result was: 98023 0.127 */ }

Below is the same code in Python. It is more than 30 times slower. This is unexpected even to me. I have to admit that I have no idea if some compilation trick can speed that up. But in any case, it is the behavior that an innocent student will see. Python should be seen as an interactive interpreter language, not a basic language to implement fast algorithms.

def isprime (n): if n==2 or n==1: return True if n%2==0: return False i=3 while i*i<=n: if n%i==0: return False i=i+2 return True def count (n): c=0 for k in range(1,n+1): for l in range(1,n+1): if isprime(k*k+l*l): ## print(k*k+l*l) c=c+1 return c import time sec=time.time() print(count(1000)) print(time.time()-sec) ## Result was ## 98023 ## 4.791218519210815

I have a version in Euler Math Toolbox too. It uses matrix tricks and is thus very short, using built-in loops in C. But it still cannot compete with the versions above. It is about 3 times slower than Python and 100 times slower than C. The reason is that the isprime() function is not implemented in C but in the EMT language.

>n=1000; k=1:n; l=k'; tic; totalsum(isprime(k^2+l^2)), toc; 98023 Used 13.352 seconds

Below is a version where I use TinyC to check for primes. The function isprimc() can only handle real numbers, no vectors. Thus I vectorize it with another function isprimecv(). This takes a bit of performance.

>function tinyc isprimec (x) ... $ ARG_DOUBLE(x); $ int n = (int)x; $ int res = 1; $ if (n > 2) $ { $ if (n%2 == 0) res=0; $ int i=3; $ while (i*i<=n) $ { $ if (n%i==0) { res=0; break; } $ i = i+2; $ } $ } $ new_real(res); $ endfunction >function map isprimecv (x) := isprimec(x); >n=1000; k=1:n; l=k'; tic; totalsum(isprimecv(k^2+l^2)), toc; 98023 Used 0.849 seconds