calling subfunctions multiple times

Question

Todd Jones 2020-5-27

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/536003-calling-subfunctions-multiple-times

评论： Todd Jones 2020-5-27

I have a main function which runs 2 different subfunctions. Both subfunctions do the exact same thing but in different ways (Find the LCM of of all numbers below input n). I want to compare the efficiencies of the subfunctions across many different values and plot them. Here is what I have for my main function with the subfunctions written below it.

function problem5(n)
for i = 0:20
    tic; method1(n+i); t1(i+1) = toc;
    tic; method2(n+i); t2(i+1) = toc;
end
figure;
plot(0:20,t1,'ro',0:20,t2,'bs'); legend('Method 1','Method2');
figure
semilogy(0:20,t1,'ro',0:20,t2,'bs'); legend('Method 1','Method2');
%Method 1
function[N] = method1(n)
for N = n:n:1e15
    for i = n:-1:1
        if mod(N,i) ~= 0
            break
        end
    end
    if i == 1
        N;
        break
    end
end
%Method 2
function[N] = method2(n)
N = 1;
p = primes(n);
limit = sqrt(n);
a = 1;
i = 1;
while p(i) <= n
    if p(i) <= limit
        a(i) = floor( log(n) / log(p(i)) );
        N = N*(p(i)^a(i));
    else
        N = N*p(i);
    end
    i = i+1;
    if i == length(p) + 1
        break
    end
end

Both subfunctions work individually but when I run the script as shown above it doesn't work. Any tips?

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Answer 1

David Hill 2020-5-27

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/536003-calling-subfunctions-multiple-times#answer_440823

在 MATLAB Online 中打开

function problem5(n)
k=0:20;%method1 gets very slow causing the function to take a few minutes for low values of n=3
for i=k
    tic;
    method1(n+i);
    t1(i+1)=toc;
    tic;
    method2(n+i);
    t2(i+1)=toc;
end
figure;
plot(k,t1,k,t2); legend('Method 1','Method2');
figure
semilogy(k,t1,k,t2); legend('Method 1','Method2');
end
%Method 1
function[N] = method1(n)
for N = n:n:1e15
    for i = n:-1:1
        if mod(N,i) ~= 0
            break
        end
    end
    if i == 1
        N;
        break
    end
end
end
%Method 2
function[N] = method2(n)
N = 1;
p = primes(n);
limit = sqrt(n);
a = 1;
i = 1;
while p(i) <= n
    if p(i) <= limit
        a(i) = floor( log(n) / log(p(i)) );
        N = N*(p(i)^a(i));
    else
        N = N*p(i);
    end
    i = i+1;
    if i == length(p) + 1
        break
    end
end
end

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Todd Jones 2020-5-27

Thanks! This makes a lot of sense. My program ends up crashing though. I guess I just have to test for a smaller interval of values so I can still observe some data at least.

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