Ackermann's Function is a recursive mathematical algorithm that can be used to test how well a computer performs recursion. Design a function ackermann(m,n), which solves Ackermann's Function. Use the following logic in your function:
If m = 0, then return n + 1
If n = 0 then return ackermann(m-1, 1)
Otherwise, return ackermann(m-1, ackermann(m, n-1))
The program stops after it hits 13. Can anyone tell me what I have done wrong?
declare main()
declare ackermann(m:int, n:int)
openconsole
main()
print:print "Press any key to quit...",
do:until inkey$<>""
closeconsole
end
sub main()
def counter, m, n:int
counter = 0
for counter = 1 to 100
print ackermann(counter)
next counter
return
sub ackermann(m, n)
if m = 0
return = n + 1
else
if n = 0
return = ackermann(m - 1, 1)
else
return = ackermann(m - 1, ackermann(m, n - 1))
endif
endif
return
Your print statement is
print ackermann(counter)
but your function is
sub ackermann(m, n)
You need to send a second parameter in the first call.
Do note that Ackermann's function grows incredibly quickly - values above m,n > (3,4) are going to have a LOT of digits and if you go more than even about (4,4) you'll quickly find numbers that could quite possibly fill up your memory entirely with digits...
Refer to Wikipedia to see the magnitude of the numbers you are trying to compute...
declare main()
declare ackermann(m:int, n:int)
openconsole
main()
print:print "Press any key to quit...",
do:until inkey$<>""
closeconsole
end
sub main()
print Ackermann(3,5)
return
sub ackermann(m, n)
if m = 0
return = n + 1
else
if n = 0
return = ackermann(m - 1, 1)
else
return = ackermann(m - 1, ackermann(m, n - 1))
endif
endif
return
Related
For the function
function function_maker(N)
if N == 1
x = 1.0
else
x = 1
end
f(y) = x+y
end
I want the output of this to not be type-stable, but I want it to generate an f that is type-stable, i.e. uses the type of x determined by the value of N to generate a function dependent on N. Basically, I want the functions that come out of this to be performant, but the function_maker itself doesn't need to be performant because it's only used in the global scope or above a function barrier.
f = function_maker(1)
#code_warntype f(1)
Variables:
#self#::#f#9
y::Int64
Body:
begin
return ((Core.getfield)((Core.getfield)(#self#::#f#9, :x)::ANY, :contents)::ANY + y::Int64)::ANY
end::ANY
This doesn't happen by default. I tried f(y) = x::typeof(x)+y but that didn't work either. Is there a simple way to do this?
There's:
julia> function function_maker2(N)
if N == 1
let x = 1.0
return f(y) = x + y
end
else
let x = 1
return f(y) = x + y
end
end
end
function_maker2 (generic function with 1 method)
julia> f2 = function_maker2(1)
(::f) (generic function with 1 method)
julia> #code_warntype f2(1)
Variables:
#self#::#f#5{Float64}
y::Int64
Body:
begin
return (Base.add_float)((Core.getfield)(#self#::#f#5{Float64}, :x)::Float64, (Base.sitofp)(Float64, y::Int64)::Float64)::Float64
end::Float64
This version separates the x in each branch inside let blocks. Otherwise the compiler seems to get confused.
I need to make a function that given natural number n, calculates the product
of the numbers below n that are not divisible by
2 or by 3 im confused on how to use nested functions in order to solve this problem (also new to sml ) here is my code so far
fun countdown(x : int) =
if x=0
then []
else x :: countdown(x-1)
fun check(countdown : int list) =
if null countdown
then 0
else
It is not clear from the question itself (part of an exercise in some class?) how we are supposed to use nested functions since there are ways to write the function without nesting, for example like
fun p1 n =
if n = 1 then 1 else
let val m = n - 1
in (if m mod 2 = 0 orelse m mod 3 = 0 then 1 else m) * p1 m
end
and there are also many ways to write it with nested functions, like
fun p2 n =
if n = 1 then 1 else
let val m = n - 1
fun check m = (m mod 2 = 0 orelse m mod 3 = 0)
in (if check m then 1 else m) * p2 m
end
or
fun p3 n =
let fun check m = (m mod 2 = 0 orelse m mod 3 = 0)
fun loop m =
if m = n then 1 else
(if check m then 1 else m) * loop (m + 1)
in loop 1
end
or like the previous answer by #coder, just to give a few examples. Of these, p3 is somewhat special in that the inner function loop has a "free variable" n, which refers to a parameter of the outer p3.
Using the standard library, a function that produces the numbers [1; n-1],
fun below n = List.tabulate (n-1, fn i => i+1);
a function that removes numbers divisible by 2 or 3,
val filter23 = List.filter (fn i => i mod 2 <> 0 andalso i mod 3 <> 0)
a function that calculates the product of its input,
val product = List.foldl op* 1
and sticking them all together,
val f = product o filter23 o below
This generates a list, filters it and collapses it. This wastes more memory than necessary. It would be more efficient to do what #FPstudent and #coder do and generate the numbers and immediately either make them a part of the end product, or throw them away if they're divisible by 2 or 3. Two things you could do in addition to this is,
Make the function tail-recursive, so it uses less stack space.
Generalise the iteration / folding into a common pattern.
For example,
fun folditer f e i j =
if i < j
then folditer f (f (i, e)) (i+1) j
else e
fun accept i = i mod 2 <> 0 andalso i mod 3 <> 0
val f = folditer (fn (i, acc) => if accept i then i*acc else acc) 1 1
This is similar to Python's xrange.
I am new to this, and I am looking for help. I currently am stuck in a program I'm trying to complete. Here it is:
def printStock(stockList, stockPrice, p):
for i in range(len(stockPrice)):
if stockPrice[i] > p:
p = stockList[i]
print("The stocks with a higher value are:", p)
def searchStock(stockList, stockPrice, s):
for i in range(len(stockList)):
if s == stockList[i]:
s = stockPrice[i]
elif s != stockList[i]:
s = -1
return s
def mainFun():
stockList= []
stockPrice = []
l = 1
while l > 0:
stocks = str(input("Enter the name of the stock:"))
stockList += [stocks]
if stocks == "done"or stocks == 'done':
l = l * -1
stockList.remove("done")
else:
price = int(input("Enter the price of the stock:"))
stockPrice += [price]
l = l + 1
print(stockList)
print(stockPrice)
s = input("Enter the name of the stock you're looking for:")
searchStock(stockList, stockPrice, s)
p = s
printStock(stockList, stockPrice, p)
Every time I run the program to the end, it never returns the variable s for some reason. If i replace return with print, it always prints -1 instead of the stockPrice if its on the list. I also get an error saying "unorderable types int() > str()" regarding line 3. Basically the function printStock takes the three parameters and once the function is done it should print the names of the stocks higher than the value 'p'. The value of 'p' is the same as the value I get after calling searchStock function, but I cant seem to get it to work. Can someone please help me?
s is being returned from the function, but the return value is not being assigned to any variable on the outer scope.
Just replace
searchStock(stockList, stockPrice, s)
with
s=searchStock(stockList, stockPrice, s)
And everything should work as expected
Here is the Main Program:
PROGRAM integration
EXTERNAL funct
DOUBLE PRECISION funct, a , b, sum, h
INTEGER n, i
REAL s
PARAMETER (a = 0, b = 10, n = 200)
h = (b-a)/n
sum = 0.0
DO i = 1, n
sum = sum+funct(i*h+a)
END DO
sum = h*(sum-0.5*(funct(a)+funct(b)))
PRINT *,sum
CONTAINS
END
And below is the Function funct(x)
DOUBLE PRECISION FUNCTION funct(x)
IMPLICIT NONE
DOUBLE PRECISION x
INTEGER K
Do k = 1,10
funct = x ** 2 * k
End Do
PRINT *, 'Value of funct is', funct
RETURN
END
I would like the 'Sum' in the Main Program to print 10 different sums over 10 different values of k in Function funct(x).
I have tried the above program but it just compiles the last value of Funct() instead of 10 different values in sum.
Array results require an explicit interface. You would also need to adjust funct and sum to actually be arrays using the dimension statement. Using an explicit interface requires Fortran 90+ (thanks for the hints by #francescalus and #VladimirF) and is quite tedious:
PROGRAM integration
INTERFACE funct
FUNCTION funct(x) result(r)
IMPLICIT NONE
DOUBLE PRECISION r
DIMENSION r( 10 )
DOUBLE PRECISION x
END FUNCTION
END INTERFACE
DOUBLE PRECISION a , b, sum, h
DIMENSION sum( 10)
INTEGER n, i
PARAMETER (a = 0, b = 10, n = 200)
h = (b-a)/n
sum = 0.0
DO i = 1, n
sum = sum+funct(i*h+a)
END DO
sum = h*(sum-0.5*(funct(a)+funct(b)))
PRINT *,sum
END
FUNCTION funct(x)
IMPLICIT NONE
DOUBLE PRECISION funct
DIMENSION funct( 10)
DOUBLE PRECISION x
INTEGER K
Do k = 1,10
funct(k) = x ** 2 * k
End Do
PRINT *, 'Value of funct is', funct
RETURN
END
If you can, you should switch to a more modern Standard such as Fortran 90+, and use modules. These provide interfaces automatically, which makes the code much simpler.
Alternatively, you could take the loop over k out of the function, and perform the sum element-wise. This would be valid FORTRAN 77:
PROGRAM integration
c ...
DIMENSION sum( 10)
c ...
INTEGER K
c ...
DO i = 1, n
Do k = 1,10
sum(k)= sum(k)+funct(i*h+a, k)
End Do
END DO
c ...
Notice that I pass k to the function. It needs to be adjusted accordingly:
DOUBLE PRECISION FUNCTION funct(x,k)
IMPLICIT NONE
DOUBLE PRECISION x
INTEGER K
funct = x ** 2 * k
PRINT *, 'Value of funct is', funct
RETURN
END
This version just returns a scalar and fills the array in the main program.
Apart from that I'm not sure it is wise to use a variable called sum. There is an intrinsic function with the same name. This could lead to some confusion...
I am trying to implement Miller-Rabin for the first time. My code is giving correct answer for all the testcases, i tried but still on SPOJ it is giving wrong answer.
Problem Statement: I am supposed to print "YES" if entered number is prime otherwise "NO"
Please help:
Problem Link: http://www.spoj.com/problems/PON/
CODE:
#include<stdio.h>
#include<stdlib.h>
#include<time.h>
#define LL long long
LL expo(LL a,LL b,LL c)
{
LL x=1,y=a;
if(b==0)
return 1;
while(b)
{
if(b%2==1)
x=(x*y)%c;
y=(y*y)%c;
b=b/2;
}
return x;
}
int main()
{
LL t,s,x,a,n,prime,temp;
scanf("%lld",&t);
srand(time(NULL));
while(t--)
{
scanf("%lld",&n);
if(n<2)
puts("NO");
else if(n==2)
puts("YES");
else if(n%2==0)
puts("NO");
else
{
s=n-1;
prime=1;
while(s%2==0)
s=s/2;
for(int i=0;i<20;i++)
{
a=rand()%(n-1)+1;
x=expo(a,s,n);
temp=s;
while((temp!=n-1)&&(x!=1)&&(x!=n-1))
{
x=(x*x)%n;
temp*=2;
}
if((x!=n-1)&&(temp%2==0))
{
prime=0;
break;
}
}
if(prime==0)
puts("NO");
else
puts("YES");
}
}
return 0;
}
Keep in mind that puts appends a newline character '\n' to the string that you're giving. You can try with printf instead.
I think your calculation of s and d is incorrect:
function isStrongPseudoprime(n, a)
d := n - 1; s := 0
while d % 2 == 0
d := d / 2; s := s + 1
t := powerMod(a, d, n)
if t == 1 return ProbablyPrime
while s > 0
if t == n - 1 return ProbablyPrime
t := (t * t) % n
s := s - 1
return Composite
I discuss the Miller-Rabin method in an essay at my blog.
You are getting wrong answer because of integer overflow as you are multiplying 2 long number which can't be holded in a single long long type.
Here is a solution in python to overcome the issue
import random
_mrpt_num_trials = 25 # number of bases to test
def is_probable_prime(n):
assert n >= 2
# special case 2
if n == 2:
return True
# ensure n is odd
if n % 2 == 0:
return False
# write n-1 as 2**s * d
# repeatedly try to divide n-1 by 2
s = 0
d = n - 1
while True:
quotient, remainder = divmod(d, 2)
if remainder == 1:
break
s += 1
d = quotient
assert(2 ** s * d == n - 1)
def try_composite(a):
if pow(a, d, n) == 1:
return False
for i in range(s):
if pow(a, 2 ** i * d, n) == n - 1:
return False
return True
for _ in range(_mrpt_num_trials):
a = random.randrange(2, n)
if try_composite(a):
return False
return True
for i in range(int(input())):
a = int(input())
if is_probable_prime(a):
print("YES")
else:
print("NO")