I want to convert the following c code to haskell code, without using lists. It returns the number of occurrences of two numbers for a given n , where n satisfies n=(a*a)*(b*b*b).
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
int main(void) {
int n = 46656;
int i,j,counter=0,res=1;
int tr = sqrt(n);
for(i=1; i<=tr; i++) {
for(j=1; j<=tr; j++) {
res = (i*i) * (j*j*j) ;
if(res==n) {
counter=counter+1;
}
printf("%d\n",res);
}
}
printf("%d\n",counter);
}
I've managed to do something similar in haskell in regarding to loops, but only for finding the overall sum. I find difficult implementing the if part and counter part(see on c code) in haskell also. Any help much appreciated! Heres my haskell code also:
sumF :: (Int->Int)->Int->Int
sumF f 0 = 0
sumF f n = sumF f (n-1) + f n
sumF1n1n :: (Int->Int->Int)->Int->Int
sumF1n1n f 0 = 0
sumF1n1n f n = sumF1n1n f (n-1)
+sumF (\i -> f i n) (n-1)
+sumF (\j -> f n j) (n-1)
+f n n
func :: Int->Int->Int
func 0 0 = 0
func a b = res
where
res = (a^2 * b^3)
call :: Int->Int
call n = sumF1n1n func n
I guess an idiomatic translation would look like this:
n = 46656
tr = sqrt n
counter = length
[ ()
| i <- [1..tr]
, j <- [1..tr]
, i*i*j*j*j == n
]
Not that it isn't possible, but definitely not the best looking:
counter n = go (sqrt n) (sqrt n)
where
go 0 _ = 0
go i tr = (go2 tr 0 i) + (go (i - 1) tr)
go2 0 c i = c
go2 j c i = go2 (j - 1) (if i^2 * j^3 == n then c + 1 else c) i
A general and relatively straightforward way to translate imperative code is to replace each basic block with a function, and give it a parameter for every piece of state it uses. If it’s a loop, it will repeatedly tail-call itself with different values of those parameters. If you don’t care about printing the intermediate results, this translates straightforwardly:
The main program prints the result of the outer loop, which begins with i = 1 and counter = 0.
main = print (outer 1 0)
where
These are constants, so we can just bind them outside the loops:
n = 46656
tr = floor (sqrt n)
The outer loop tail-calls itself with increasing i, and counter updated by the inner loop, until i > tr, then it returns the final counter.
outer i counter
| i <= tr = outer (i + 1) (inner 1 counter)
| otherwise = counter
where
The inner loop tail-calls itself with increasing j, and its counter (counter') incremented when i^2 * j^3 == n, until j > tr, then it returns the updated counter back to outer. Note that this is inside the where clause of outer because it uses i to calculate res—you could alternatively make i an additional parameter.
inner j counter'
| j <= tr = inner (j + 1) $ let
res = i ^ 2 * j ^ 3
in if res == n then counter' + 1 else counter'
| otherwise = counter'
This is a question from one of the old exams from algorithms and data structure that I recently came upon. I'm having a hard time understanding the solution.
I need to find big-O, big-ϴ and big-Ω bounds of a function:
void recursion(int n) {
int i;
if (n == 0) {
return;
}
for (i = 0; i < n; i++) {
recursion(i);
}
}
The solution is 2^n for all three and I can't understand why. I've tried writing things down and I can't even get close to the solution. I would appreciate if anyone would explain where the 2^n comes from here.
Let's look at a simpler recursion which is known to be O(2^n)
void fib(int n) {
if (n < 3) {
return 1;
} else {
return fib(n - 1) + fib(n - 2);
}
}
Here you can see, for the non-trivial case of n > 2, this will result in 2^(n-2) calls to itself. For example, if n = 5:
n = 5
n = 4
n = 3
n = 2
n = 1
n = 2
n = 3
n = 2
n = 1
There are 8 (2^3) recursive calls, because each call with n > 2 spawns two more recursive calls, so fib(n+1) has twice as many recursive calls as fib(n).
So for your example:
n = 3
n = 2
n = 1
n = 0
n = 0
n = 1
n = 0
n = 0
so we get 7 recursive calls when n = 3
for n = 4
n = 4
n = 3
n = 2
n = 1
n = 0
n = 0
n = 1
n = 0
n = 0
n = 2
n = 1
n = 0
n = 0
n = 1
n = 0
n = 0
Here, we have 15 calls. Looking at the execution tree above, you can see that recusrsion(4) is basically recursion(3) + recursion(3) + 1
n = 4
n = 3 // + 1
n = 2 //
n = 1 //
n = 0 // recursion(3)
n = 0 //
n = 1 //
n = 0 //
n = 0 //
n = 2 //
n = 1 //
n = 0 // recursion(3)
n = 0 //
n = 1 //
n = 0 //
n = 0 //
So in general, recursion(n + 1) will have one more recursive calls than 2 * recursion(n)....which is basically doubling for every +1 to n....which is O(2^n)
Let's denote the total runtime as f(n). Due to the loop in the function the f(n) is actually a sum of f(i) for i between 0 and n-1. That's a sum of n items. Let's try to simplify the expression. A standard trick in such situations is to find a complimentary equation. Let's see what is the value of f(n-1). Similary to the previous case, it's a sum of f(i) for i between 0 and n-2. So now we have 2 equations:
f(n)=f(1)+...+f(n-1)
f(n-1)=f(1)+...+f(n-2)
Let's subtract second from the first:
f(n)-f(n-1)=f(n-1)
--> f(n)=2f(n-1)
Now this is a homogeneous linear recurrence relation with constant coefficients.
The solution is immediate (see the link for more details):
f(n)=f(1)*2n=2n
Since this smells like a homework question, this answer is incomplete by design.
The usual trick behind these kind of problems is to create a recurrence equation. That is, the time complexity of recursion(k+1) is somehow related to the complexity of recursion(k). Just writing down the recurrence itself is not sufficient to prove the complexity, you have to demonstrate why the recurrence is true. But, for 2n, this suggests that recursion(k+1) takes twice as long as recursion(k).
Let T(k) denote the time complexity of recursion(k). Since recursion(0) returns immediately, let T(0) = 1. For k > 0, given the iterative implementation of recursion Thus You can inductively prove that T(k) = 2k.
r(n) = r(n-1)+r(n-2)+...+r(0) // n calls.
r(n-1) = r(n-2)+r(n-3)+...+r(0) // n-1 calls.
r(n-2) = r(n-3)+r(n-4)+...+r(0) // n-2 calls.
.
.
.
r(1) = r(0) // 1 call.
r(0) = return; // 0 call.
So,
r(n) = r(n-1)+r(n-2)+...+r(0) // n calls.
= 2 * (r(n-2)+...+r(0)) // 2 * (n - 1) calls.
= 2 * ( 2 * (r(n-3)+...+r(0)) ) // 2 * 2 * (n - 2) calls.
.
.
.
This follows that =>
2^(n-1) * (n - (n-1))
And that would be
2^n calls...
This is the code that I had tried to find the consecutive zero which are in the order of 5 or more.
a=[0,0,0,0,0,0,0,0,9,8,5,6,0,0,0,0,0,0,3,4,6,8,0,0,9,8,4,0,0,7,8,9,5,0,0,0,0,0,8,9,0,5,8,7,0,0,0,0,0];
[x,y]=size(a);
for i=0:y
i+1;
k=1;
l=0;
n=i;
count=0;
while (a==0)
count+1;
break;
n+1;
end
if(count>=5)
v([]);
for l=k:l<n
v(m)=l+1;
m+1;
end
end
count=1;
i=n;
end
for i = o : i<m
i+1;
fprintf('index of continous zero more than 5 or equal=%d',v(i));
end
If you want to find the starting indices of runs of n or more zeros:
v = find(conv(double(a==0),ones(1,n),'valid')==n); %// find n zeros
v = v([true diff(v)>n]); %// remove similar indices, indicating n+1, n+2... zeros
In your example, this gives
v =
1 13 34 45
One-liner strfind approach to find the starting indices of 5 consecutive zeros -
out = strfind(['0' num2str(a==0,'%1d')],'011111')
Output -
out =
1 13 34 45
The above code could be generalised like this -
n = 5 %// number of consecutive matches
match = 0 %// match to be used
out = strfind(['0' num2str(a==match,'%1d')],['0' repmat('1',1,n)]) %// starting indices of n consecutive matches
If you are looking to find all the indices where the n consecutive matches were found, you can add this code -
outb = strfind([num2str(a==match,'%1d'),'0'],[repmat('1',1,n) '0'])+n-1
allind = find(any(bsxfun(#ge,1:numel(a),out') & bsxfun(#le,1:numel(a),outb')))
If you want to find the general case of a "run of n or more values x in vector V", you could do the following:
% your particular case:
n = 5;
x = 0;
V = [0,0,0,0,0,0,0,0,9,8,5,6,0,0,0,0, ...
0,0,3,4,6,8,0,0,9,8,4,0,0,7,8,9, ...
5,0,0,0,0,0,8,9,0,5,8,7,0,0,0,0,0];
b = (V == x); % create boolean array: ones and zeros
d = diff( [0 b 0] ); % turn the start and end of a run into +1 and -1
startRun = find( d==1 );
endRun = find( d==-1 );
runlength = endRun - startRun;
answer = find(runlength > n);
runs = runlength(answer);
disp([answer(:) runs(:)]);
This will display the start of the run, and its length, for all runs > n of value x.
I am writing program to count Bell numbers,
it is my first big program in OCaml.
I want to use loop While in the loop While, but there is syntax error.
Please correct it. Thanks.
I'm using site http://try.ocamlpro.com/
let rec factorial n =
if n < 2
then 1
else
n * factorial(n-1)
let rec newton n k =
factorial n / (factorial k * factorial (n-k))
let bell = [|1;1;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0|]
let i = ref 2
let k = ref 0
let x = ref 0
let z = ref 0
let s = ref 0
here you need to choose number u want to calc e.g. 4
let n = ref 4
if !n != 0 || !n != 1 then
while !i <= !n do
while !k <= (!i-1) do
x := newton (!i-1) !k;
s := !s + (!x * bell.(!k));
k := !k + 1 ;
z := !k + 1
done
s:=0;
i:= !i + 1;
done
else
bell.(!n)<-1
should use Num to calc Bell numbers, but I first I I would like to make the program work on int
You can try to add ; after 1st done.
For a given b and N and a range of a say (0...n),
I need to find ans(0...n-1)
where,
ans[i] = no of a's for which pow(a, b)modN == i
What I am searching here is a possible repetition in pow(a,b)modN for a range of a, to reduce computation time.
Example:-
if b = 2 N = 3 and n = 5
for a in (0...4):
A[pow(a,b)modN]++;
so that would be
pow(0,2)mod3 = 0
pow(1,2)mod3 = 1
pow(2,2)mod3 = 1
pow(3,2)mod3 = 0
pow(4,2)mod3 = 1
so the final results would be:
ans[0] = 2 // no of times we have found 0 as answer .
ans[1] = 3
...
Your algorithm have a complexity of O(n).
Meaning it take a lot of time when n gets bigger.
You could have the same result with an algorithm O(N).
As N << n it will reduce your computation time.
Firts, two math facts :
pow(a,b) modulo N == pow (a modulo N,b) modulo N
and
if (i < n modulo N)
ans[i] = (n div N) + 1
else if (i < N)
ans[i] = (n div N)
else
ans[i] = 0
So a solution to your problem is to fill your result array with the following loop :
int nModN = n % N;
int nDivN = n / N;
for (int i = 0; i < N; i++)
{
if (i < nModN)
ans[pow(i,b) % N] += nDivN + 1;
else
ans[pow(i,b) % N] += nDivN;
}
You could calculate pow for primes only, and use pow(a*b,n) == pow(a,n)*pow(b,n).
So if pow(2,2) mod 3 == 1 and pow(3,2) mod 3 == 2, then pow(6,2) mod 3 == 2.