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Find the power of each factor in a given number using matlab function

Problem

Write a function in matlab to find the exponent of each prime factor of a given number. For example if the input is 28 then the output should be



Power of 2 in the number 28 is 2
Power of 3 in the number 28 is 0
Power of 5 in the number 28 is 0
Power of 7 in the number 28 is 1






Solution





The solution is simple and one has to know the function definition, factor function, for loop, nnz function, isprime function, and the if statement to understand the program. Don't worry if you don't know the function definition. You can even program this code without a function. The difference will be that there will not be a function definition line, and the value of n is directly defined instead. But it would be useful if you know them. If you want to learn any of the above then head on to the links below.













Find the explanation for the program below the program



Program








function ret = countfacpowers(n)
%this function will calculate the exponent of each prime factor and return
%the result as a printed statement

vec = factor(n);
highest = vec(length(vec));

for i = 1:highest
    if isprime(i) == 1
        fprintf('Power of %d in the number %d is %d\n',i,n,nnz(vec == i));
    end
end





To get the matlab for the above code click here: countfacpowers.m



Explanation



During the starting of the function a vector is created to store the prime factors of the given numbers using the factor function. Now digging deep a highest variable is created in order to store the largest prime factor of the number. Here you can also use the max function but not to complicate this program more for the novice programmers, I just have used the last number of the vector which is the largest prime factor. Now let us understand the for loop iteration by iteration. I am considering the input number as 28 for this example



First iteration



During the first iteration the value of i is 1 and this doesn't qualify the if statement and doesn't execute. Thus first iteration is over and the second iteration starts



Second iteration



During the second iteration the value of i is changed from 1 to 2 and the value of 2 qualifies the if statement and the fprintf statement gets executed. Here we will understand what is happening in the nnz function. The nnz function in general gives the number of non zero elements in the vector. But here we have given the input to the nnz function as the vector whose values are 1 only where the value becomes equal to i. Let us simplify this:



First the value of the vec is [2 2 7]



Now we will call vec == 2 and this will give the output as [1 1 0] , 1 will only be created at the places where there is 2(Here in this case 2 and where there is i in general). Finally nnz([1 1 0]) will give 2 i.e. 1+1 = 2.



and the second iteration comes to halt







Similarly all the other iterations continue until the i value changes to 7.







I have tried to explain this program as easy as possible, but if you don't understand then please do contact me so that many others will get benefited 







The above code was highlighted using hilite.me







To read the above post in pdf format then you can do so by downloading from here: countfacpowers.pdf















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