Im doing a homework problem to make a function sumOdd to computer the sum of the first n odd integers, but i cant seem to find any sort of elseif type statement to do so. What im trying to do is below but of course doesnt work:
fun sumOdd n = if n=0 then 0 elseif (n mod 2)=0 then sumOdd(n-1) elseif n + sumOdd(n-1);
Your function didn't compile because elseif is not a keyword in SML. Changing the last elseif to else and other elseif to else if should fix the error.
Furthermore, the function is more readable in the below format:
fun sumOdd n = if n = 0 then 0
else if n mod 2 = 0 then sumOdd(n-1)
else n + sumOdd(n-1)
You could also remove the need for the else if expression by separating the base case from the general case:
fun sumOdd 0 = 0
| sumOdd n = if n mod 2 = 0 then sumOdd(n-1)
else n + sumOdd(n-1)
You should also note that this solution does (and your own) does not actually sum the first N odd numbers. It computes the sum of all odd numbers less than N.
sumOdd(5) gives 9(5+3+1) when it should give 25(1+3+5+7+9).
Related
Hello I am learning the OCaml language and working on an assignment.
infinite precision natural numbers can be represented as lists of ints between 0 and 9
Write a function that takes an integer and represents it with a list of integers between 0 and 9 where the head
of the list holds the least significant digit and the very last element of the list represents the most significant digit.
If the input is negative return None. We provide you with some use cases:
For example:
toDec 1234 = Some [4; 3; 2; 1]
toDec 0 = Some []
toDec -1234 = None
I have written below code for it.
let rec toDec i =
(
if i < 10 then i::[]
else toDec ((i mod 10)::acc) (i/10) in toDec [] i;
);;
I am getting syntax error on line 4. Since I am new to this language, not able to get what's wrong. Can somebody please help on this.
The in keyword must go with a let. You could use a local function aux, as follows:
let toDec i =
let rec aux acc i =
if i < 10 then i::[]
else aux ((i mod 10)::acc) (i/10)
in
aux [] i
This doesn't do what you want but syntax and types are valid and I'm sure you can fix the rest.
Vicky, you forgot to define acc and also forgot to put else if statement.
Update your code as below,
let rec toDec ?acc:(acc=[]) i =
if i < 0 then None
else if i = 0 then Some acc
else toDec ~acc:((i mod 10)::acc) (i / 10)
This question already has answers here:
Does Fortran preserve the value of internal variables through function and subroutine calls?
(3 answers)
Closed 2 years ago.
I'm recently studying Fortran, and trying to make a program to check the prime number. The function works fine without any loop. It can give 1 when the given number is a prime number and 0 otherwise. However, it doesn't work properly when it is used in do while loop. In the range of 2 ~ 10, it is supposed to give 1 (for 2), 1(for 2), 0(for 4), 1(for 5), 0(for 6), etc. But, it keeps showing only 0. I'm pretty new to programming, so I'm not sure what I'm missing. I know there are many answers related to prime numbers, but I don't see any issue like this.
** Function checking prime numbers **
module prime_function
contains
integer function isPrime(inp_num)
implicit none
integer :: inp_num
integer :: i = 1
integer :: temp1 = 0
do while (i < inp_num)
i = i + 1
if(mod(inp_num, i) == 0) then
exit
end if
end do
if(inp_num == i) then
temp1 = 1
else
temp1 = 0
end if
isPrime = temp1
end function
end module
program fortran_q
use prime_function
implicit none
integer :: ii, a
a = isPrime(10)
print *, "10 is prime number, so the return : ", a
a = isPrime(11)
print *, "11 is prime number, so the return : ", a
ii = 1
do while (ii < 10)
ii = ii + 1
print *, isPrime(ii)
end do
end program
** Results **
10 is prime number, so the return : 0
11 is prime number, so the return : 1
0
0
0
0
0
0
0
0
0
You have a classic issue for people new to Fortran. The initialization of i and temp0 implies the SAVE attribute. When you call isPrime for the first time the values are set to 1 and 0. On the next invocation, the values of i and temp0 are set to whatever their previous values were when isPrime was last executed. The belong program fixes the issue.
module prime_function
implicit none
private
public isprime
contains
function isPrime(inp_num) result(res)
integer res
integer, intent(in) :: inp_num
integer i, temp1
i = 1
temp1 = 0
do while (i < inp_num)
i = i + 1
if (mod(inp_num, i) == 0) exit
end do
res = 0
if (inp_num == i) res = 1
end function
end module
program fortran_q
use prime_function
implicit none
integer :: ii, a
a = isPrime(10)
print *, "10 is prime number, so the return : ", a
a = isPrime(11)
print *, "11 is prime number, so the return : ", a
ii = 1
do while (ii < 10)
ii = ii + 1
print *, isPrime(ii)
end do
end program
I'm trying to write some Fortran 90 code to sum up the first 1234 multiples of 3 and 5 (including multiples of both). Here is my code so far:
program sum
implicit none
integer :: x
integer :: y = 5
integer :: z = 3
integer :: n
if (mod(x,y) == 0 .or. mod(x,z) ==0) then
print *, x
n = x
n = x + x
end if
end program sum
However, this code does not print anything to the terminal.
Your code tests the value of x in the if condition:
if (mod(x,y) == 0 .or. mod(x,z) ==0
but the value of x is not set at all. Therefore the result of the program is completely undefined. You need to create some kind of loop. Better two loops.
The most naive approach is to loop from 1 and test all numbers with the above if condition and stop when you have found the desired number of multiples.
This is a homework. OCaml seems to be made by a psychopath.
let prime : int -> bool
= fun n ->
if n > 2 then
let a = n - 1 in
let rec divisor n a =
if a > 1 && n mod a = 0 then false
else if a = 2 && n mod a <> 0 then true
else divisor n a-1 ;;
else if n = 2 then true
else if n = 1 then false
I am not good at coding and I know that my isPrime algorithm is wrong.
But I wonder where in my code is the mistake that produces the syntax error.
Also is there any way to define the isPrime function in a recursive form?
Example:
let rec prime n = ~
You'll get better responses from experts if you don't gratuitously insult their language :-) But I'm an easygoing guy, so I'll take a stab at your syntax error.
There are quite a few problems in this code. Here are 3 that I see right off:
The symbol ;; is used to tell the interpreter that you've entered a full expression that you want it to evaluate. It's definitely out of place in the middle of a function declaration.
Your second let doesn't have an associated in. Every let must have an in after it. The only exception is for defining values at the top level of a module (like your prime function).
The expression divisor n a-1 is parsed as (divisor n a) - 1. You want parentheses like this: divisor a (n - 1).
Assume the availability of a function is_prime. Assume a variable n has been associated with a positive integer. Write the statements needed to compute the sum of the first n prime numbers. The sum should be associated with the variable total.
Note: is_prime takes an integer as a parameter and returns True if and only if that integer is prime.
Well, I wrote is_prime function like this:
def is_prime(n):
n = abs(n)
i = 2
while i < n:
if n % i == 0:
return False
i += 1
return True
but it works except for n==0. How can I fix it to make it work for every integer?
I'm trying to find out answers for both how to write function to get the sum of first n prime numbers and how to modify my is_prime function, which should work for all possible input, not only positive numbers.
Your assignment is as follows.
Assume the availability of a function is_prime. Assume a variable n has been associated with a positive integer. Write the statements needed to compute the sum of the first n prime numbers. The sum should be associated with the variable total.
As NVRAM rightly points out in the comments (and nobody else appears to have picked up on), the question states "assume the availability of a function is_prime".
You don't have to write that function. What you do have to do is "write the statements needed to compute the sum of the first n prime numbers".
The pseudocode for that would be something like:
primes_left = n
curr_num = 2
curr_sum = 0
while primes_left > 0:
if is_prime(curr_num):
curr_sum = curr_sum + curr_num
primes_left = primes_left - 1
curr_num = curr_num + 1
print "Sum of first " + n + " primes is " + curr_sum
I think you'll find that, if you just implement that pseudocode in your language of choice, that'll be all you have to do.
If you are looking for an implementation of is_prime to test your assignment with, it doesn't really matter how efficient it is, since you'll only be testing a few small values anyway. You also don't have to worry about numbers less than two, given the constraints of the code that will be using it. Something like this is perfectly acceptable:
def is_prime(num):
if num < 2:
return false
if num == 2:
return true
divisor = 2
while divisor * divisor <= num:
if num % divisor == 0:
return false
divisor = divisor + 1
return true
In your problem statement it says that n is a positive integer. So assert(n>0) and ensure that your program outer-loop will never is_prime() with a negative value nor zero.
Your algorithm - trial division of every successive odd number (the 'odd' would be a major speed-up for you) - works, but is going to be very slow. Look at the prime sieve for inspiration.
Well, what happens when n is 0 or 1?
You have
i = 2
while i < n: #is 2 less than 0 (or 1?)
...
return True
If you want n of 0 or 1 to return False, then doesn't this suggest that you need to modify your conditional (or function itself) to account for these cases?
Why not just hardcode an answer for i = 0 or 1?
n = abs(n)
i = 2
if(n == 0 || n == 1)
return true //Or whatever you feel 0 or 1 should return.
while i < n:
if n % i == 0:
return False
i += 1
return True
And you could further improve the speed of your algorithm by omitting some numbers. This script only checks up to the square root of n as no composite number has factors greater than its square root if a number has one or more factors, one will be encountered before the square root of that number. When testing large numbers, this makes a pretty big difference.
n = abs(n)
i = 2
if(n == 0 || n == 1)
return true //Or whatever you feel 0 or 1 should return.
while i <= sqrt(n):
if n % i == 0:
return False
i += 1
return True
try this:
if(n==0)
return true
else
n = abs(n)
i = 2
while i < n:
if n % i == 0:
return False
i += 1
return True