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I want to compute sequence of numbers like this:
n*(n-1)+n*(n-1)*(n-2)+n*(n-1)*(n-2)*(n-3)+n*(n-1)*(n-2)*(n-3)*(n-4)+...+n(n-1)...(n-n)
For example n=5 and sum equals 320.
I have a function, which compute one element:
int fac(int n, int s)
{
if (n > s)
return n*fac(n - 1, s);
return 1;
}
Recomputing the factorial for each summand is quite wasteful. Instead, I'd suggest to use memoization. If you reorder
n*(n-1) + n*(n-1)*(n-2) + n*(n-1)*(n-2)*(n-3) + n*(n-1)*(n-2)*(n-3)*...*1
you get
n*(n-1)*(n-2)*(n-3)*...*1 + n*(n-1)*(n-2)*(n-3) + n*(n-1)*(n-2) + n*(n-1)
Notice how you start with the product of 1..n, then you add the product of 1..n divided by 1, then you add the product divided by 1*2 etc.
I think a much more efficient definition of your function is (in Python):
def f(n):
p = product(range(1, n+1))
sum_ = p
for i in range(1, n-1):
p /= i
sum_ += p
return sum_
A recursive version of this definition is:
def f(n):
def go(sum_, i):
if i >= n-1:
return sum_
return sum_ + go(sum_ / i, i+1)
return go(product(range(1, n+1)), 1)
Last but not least, you can also define the function without any explicit recursion by using reduce to generate the list of summands (this is a more 'functional' -- as in functional programming -- style):
def f(n):
summands, _ = reduce(lambda (lst, p), i: (lst + [p], p / i),
range(1, n),
([], product(range(1, n+1))))
return sum(summands)
This style is very concise in functional programming languages such as Haskell; Haskell has a function call scanl which simplifies generating the summands so that the definition is just:
f n = sum $ scanl (/) (product [1..n]) [1..(n-2)]
Something like this?
function fac(int n, int s)
{
if (n >= s)
return n * fac(n - 1, s);
return 1;
}
int sum = 0;
int s = 4;
n = 5;
while(s > 0)
{
sum += fac(n, s);
s--;
}
print sum; //320
Loop-free version:
int fac(int n, int s)
{
if (n >= s)
return n * fac(n - 1, s);
return 1;
}
int compute(int n, int s, int sum = 0)
{
if(s > 0)
return compute(n, s - 1, sum + fac(n, s));
return sum;
}
print compute(5, 4); //320
Ok ther is not mutch to write. I would suggest 2 methodes if you want to solve this recursiv. (Becaus of the recrusiv faculty the complexity is a mess and runtime will increase drasticaly with big numbers!)
int func(int n){
return func(n, 2);
}
int func(int n, int i){
if (i < n){
return n*(fac(n-1,n-i)+func(n, i + 1));
}else return 0;
}
int fac(int i,int a){
if(i>a){
return i*fac(i-1, a);
}else return 1;
}
Recently i attended an interview where i was asked to write a recursive java code for (x^y)^z.
function power(x,y){
if(y==0){
return 1;
}else{
x*=power(x,y-1);
}
}
I could manage doing it for x^y but was not getting a solution for including the z also in the recursive call.
On asking for a hint, they told me instead of having 2 parameters in call u can have a array with 2 values. But even then i dint get the solution. can u suggest a solution both ways.
This is the solution I would use in python, but you could easily have done it in javascipt or any other language too:
def power(x, y):
if y == 0:
return 1
if y == 1:
return x
return x * power(x, y - 1)
def power2(x, y, z):
return power(power(x, y), z)
You can then use power2 to return your result. In another language you could probably overload the same function but I do not think this is possible in Python for this scenario.
For your javascript code, all you really needed to add to your solution was a second function along the lines of:
function power2(x,y,z)
{
return power(power(x, y), z);
}
As you can see, the solution itself is also recursive despite defining a new function (or overloading your previous one).
Michael's solution in Java Language
public void testPower()
{
int val = power(2, 3, 2);
System.out.println(val);
}
private int power(int x, int y, int z)
{
return power(power(x, y), z);
}
private int power(int x, int y)
{
if (y == 0)
{
return 1;
}
if (y == 1)
{
return x;
}
return x * power(x, y - 1);
}
output is 64
As we all know, the simplest algorithm to generate Fibonacci sequence is as follows:
if(n<=0) return 0;
else if(n==1) return 1;
f(n) = f(n-1) + f(n-2);
But this algorithm has some repetitive calculation. For example, if you calculate f(5), it will calculate f(4) and f(3). When you calculate f(4), it will again calculate both f(3) and f(2). Could someone give me a more time-efficient recursive algorithm?
I have read about some of the methods for calculating Fibonacci with efficient time complexity following are some of them -
Method 1 - Dynamic Programming
Now here the substructure is commonly known hence I'll straightly Jump to the solution -
static int fib(int n)
{
int f[] = new int[n+2]; // 1 extra to handle case, n = 0
int i;
f[0] = 0;
f[1] = 1;
for (i = 2; i <= n; i++)
{
f[i] = f[i-1] + f[i-2];
}
return f[n];
}
A space-optimized version of above can be done as follows -
static int fib(int n)
{
int a = 0, b = 1, c;
if (n == 0)
return a;
for (int i = 2; i <= n; i++)
{
c = a + b;
a = b;
b = c;
}
return b;
}
Method 2- ( Using power of the matrix {{1,1},{1,0}} )
This an O(n) which relies on the fact that if we n times multiply the matrix M = {{1,1},{1,0}} to itself (in other words calculate power(M, n )), then we get the (n+1)th Fibonacci number as the element at row and column (0, 0) in the resultant matrix. This solution would have O(n) time.
The matrix representation gives the following closed expression for the Fibonacci numbers:
fibonaccimatrix
static int fib(int n)
{
int F[][] = new int[][]{{1,1},{1,0}};
if (n == 0)
return 0;
power(F, n-1);
return F[0][0];
}
/*multiplies 2 matrices F and M of size 2*2, and
puts the multiplication result back to F[][] */
static void multiply(int F[][], int M[][])
{
int x = F[0][0]*M[0][0] + F[0][1]*M[1][0];
int y = F[0][0]*M[0][1] + F[0][1]*M[1][1];
int z = F[1][0]*M[0][0] + F[1][1]*M[1][0];
int w = F[1][0]*M[0][1] + F[1][1]*M[1][1];
F[0][0] = x;
F[0][1] = y;
F[1][0] = z;
F[1][1] = w;
}
/*function that calculates F[][] raise to the power n and puts the
result in F[][]*/
static void power(int F[][], int n)
{
int i;
int M[][] = new int[][]{{1,1},{1,0}};
// n - 1 times multiply the matrix to {{1,0},{0,1}}
for (i = 2; i <= n; i++)
multiply(F, M);
}
This can be optimized to work in O(Logn) time complexity. We can do recursive multiplication to get power(M, n) in the previous method.
static int fib(int n)
{
int F[][] = new int[][]{{1,1},{1,0}};
if (n == 0)
return 0;
power(F, n-1);
return F[0][0];
}
static void multiply(int F[][], int M[][])
{
int x = F[0][0]*M[0][0] + F[0][1]*M[1][0];
int y = F[0][0]*M[0][1] + F[0][1]*M[1][1];
int z = F[1][0]*M[0][0] + F[1][1]*M[1][0];
int w = F[1][0]*M[0][1] + F[1][1]*M[1][1];
F[0][0] = x;
F[0][1] = y;
F[1][0] = z;
F[1][1] = w;
}
static void power(int F[][], int n)
{
if( n == 0 || n == 1)
return;
int M[][] = new int[][]{{1,1},{1,0}};
power(F, n/2);
multiply(F, F);
if (n%2 != 0)
multiply(F, M);
}
Method 3 (O(log n) Time)
Below is one more interesting recurrence formula that can be used to find nth Fibonacci Number in O(log n) time.
If n is even then k = n/2:
F(n) = [2*F(k-1) + F(k)]*F(k)
If n is odd then k = (n + 1)/2
F(n) = F(k)*F(k) + F(k-1)*F(k-1)
How does this formula work?
The formula can be derived from the above matrix equation.
fibonaccimatrix
Taking determinant on both sides, we get
(-1)n = Fn+1Fn-1 – Fn2
Moreover, since AnAm = An+m for any square matrix A, the following identities can be derived (they are obtained from two different coefficients of the matrix product)
FmFn + Fm-1Fn-1 = Fm+n-1
By putting n = n+1,
FmFn+1 + Fm-1Fn = Fm+n
Putting m = n
F2n-1 = Fn2 + Fn-12
F2n = (Fn-1 + Fn+1)Fn = (2Fn-1 + Fn)Fn (Source: Wiki)
To get the formula to be proved, we simply need to do the following
If n is even, we can put k = n/2
If n is odd, we can put k = (n+1)/2
public static int fib(int n)
{
if (n == 0)
return 0;
if (n == 1 || n == 2)
return (f[n] = 1);
// If fib(n) is already computed
if (f[n] != 0)
return f[n];
int k = (n & 1) == 1? (n + 1) / 2
: n / 2;
// Applyting above formula [See value
// n&1 is 1 if n is odd, else 0.
f[n] = (n & 1) == 1? (fib(k) * fib(k) +
fib(k - 1) * fib(k - 1))
: (2 * fib(k - 1) + fib(k))
* fib(k);
return f[n];
}
Method 4 - Using a formula
In this method, we directly implement the formula for the nth term in the Fibonacci series. Time O(1) Space O(1)
Fn = {[(√5 + 1)/2] ^ n} / √5
static int fib(int n) {
double phi = (1 + Math.sqrt(5)) / 2;
return (int) Math.round(Math.pow(phi, n)
/ Math.sqrt(5));
}
Reference: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibFormula.html
Look here for implementation in Erlang which uses formula
. It shows nice linear resulting behavior because in O(M(n) log n) part M(n) is exponential for big numbers. It calculates fib of one million in 2s where result has 208988 digits. The trick is that you can compute exponentiation in O(log n) multiplications using (tail) recursive formula (tail means with O(1) space when used proper compiler or rewrite to cycle):
% compute X^N
power(X, N) when is_integer(N), N >= 0 ->
power(N, X, 1).
power(0, _, Acc) ->
Acc;
power(N, X, Acc) ->
if N rem 2 =:= 1 ->
power(N - 1, X, Acc * X);
true ->
power(N div 2, X * X, Acc)
end.
where X and Acc you substitute with matrices. X will be initiated with and Acc with identity I equals to .
One simple way is to calculate it iteratively instead of recursively. This will calculate F(n) in linear time.
def fib(n):
a,b = 0,1
for i in range(n):
a,b = a+b,a
return a
Hint: One way you achieve faster results is by using Binet's formula:
Here is a way of doing it in Python:
from decimal import *
def fib(n):
return int((Decimal(1.6180339)**Decimal(n)-Decimal(-0.6180339)**Decimal(n))/Decimal(2.236067977))
you can save your results and use them :
public static long[] fibs;
public long fib(int n) {
fibs = new long[n];
return internalFib(n);
}
public long internalFib(int n) {
if (n<=2) return 1;
fibs[n-1] = fibs[n-1]==0 ? internalFib(n-1) : fibs[n-1];
fibs[n-2] = fibs[n-2]==0 ? internalFib(n-2) : fibs[n-2];
return fibs[n-1]+fibs[n-2];
}
F(n) = (φ^n)/√5 and round to nearest integer, where φ is the golden ratio....
φ^n can be calculated in O(lg(n)) time hence F(n) can be calculated in O(lg(n)) time.
// D Programming Language
void vFibonacci ( const ulong X, const ulong Y, const int Limit ) {
// Equivalent : if ( Limit != 10 ). Former ( Limit ^ 0xA ) is More Efficient However.
if ( Limit ^ 0xA ) {
write ( Y, " " ) ;
vFibonacci ( Y, Y + X, Limit + 1 ) ;
} ;
} ;
// Call As
// By Default the Limit is 10 Numbers
vFibonacci ( 0, 1, 0 ) ;
EDIT: I actually think Hynek Vychodil's answer is superior to mine, but I'm leaving this here just in case someone is looking for an alternate method.
I think the other methods are all valid, but not optimal. Using Binet's formula should give you the right answer in principle, but rounding to the closest integer will give some problems for large values of n. The other solutions will unnecessarily recalculate the values upto n every time you call the function, and so the function is not optimized for repeated calling.
In my opinion the best thing to do is to define a global array and then to add new values to the array IF needed. In Python:
import numpy
fibo=numpy.array([1,1])
last_index=fibo.size
def fib(n):
global fibo,last_index
if (n>0):
if(n>last_index):
for i in range(last_index+1,n+1):
fibo=numpy.concatenate((fibo,numpy.array([fibo[i-2]+fibo[i-3]])))
last_index=fibo.size
return fibo[n-1]
else:
print "fib called for index less than 1"
quit()
Naturally, if you need to call fib for n>80 (approximately) then you will need to implement arbitrary precision integers, which is easy to do in python.
This will execute faster, O(n)
def fibo(n):
a, b = 0, 1
for i in range(n):
if i == 0:
print(i)
elif i == 1:
print(i)
else:
temp = a
a = b
b += temp
print(b)
n = int(input())
fibo(n)
What is the most efficient way to calculate the least common multiple of two integers?
I just came up with this, but it definitely leaves something to be desired.
int n=7, m=4, n1=n, m1=m;
while( m1 != n1 ){
if( m1 > n1 )
n1 += n;
else
m1 += m;
}
System.out.println( "lcm is " + m1 );
The least common multiple (lcm) of a and b is their product divided by their greatest common divisor (gcd) ( i.e. lcm(a, b) = ab/gcd(a,b)).
So, the question becomes, how to find the gcd? The Euclidean algorithm is generally how the gcd is computed. The direct implementation of the classic algorithm is efficient, but there are variations that take advantage of binary arithmetic to do a little better. See Knuth's "The Art of Computer Programming" Volume 2, "Seminumerical Algorithms" § 4.5.2.
Remember
The least common multiple is the least whole number that is a multiple of each of two or more numbers.
If you are trying to figure out the LCM of three integers, follow these steps:
**Find the LCM of 19, 21, and 42.**
Write the prime factorization for each number. 19 is a prime number. You do not need to factor 19.
21 = 3 × 7
42 = 2 × 3 × 7
19
Repeat each prime factor the greatest number of times it appears in any of the prime factorizations above.
2 × 3 × 7 × 19 = 798
The least common multiple of 21, 42, and 19 is 798.
I think that the approach of "reduction by the greatest common divider" should be faster. Start by calculating the GCD (e.g. using Euclid's algorithm), then divide the product of the two numbers by the GCD.
Best solution in C++ below without overflowing
#include <iostream>
using namespace std;
long long gcd(long long int a, long long int b){
if(b==0)
return a;
return gcd(b,a%b);
}
long long lcm(long long a,long long b){
if(a>b)
return (a/gcd(a,b))*b;
else
return (b/gcd(a,b))*a;
}
int main()
{
long long int a ,b ;
cin>>a>>b;
cout<<lcm(a,b)<<endl;
return 0;
}
First of all, you have to find the greatest common divisor
for(int i=1; i<=a && i<=b; i++) {
if (i % a == 0 && i % b == 0)
{
gcd = i;
}
}
After that, using the GCD you can easily find the least common multiple like this
lcm = a / gcd * b;
I don't know whether it is optimized or not, but probably the easiest one:
public void lcm(int a, int b)
{
if (a > b)
{
min = b;
max = a;
}
else
{
min = a;
max = b;
}
for (i = 1; i < max; i++)
{
if ((min*i)%max == 0)
{
res = min*i;
break;
}
}
Console.Write("{0}", res);
}
Here is a highly efficient approach to find the LCM of two numbers in python.
def gcd(a, b):
if min(a, b) == 0:
return max(a, b)
a_1 = max(a, b) % min(a, b)
return gcd(a_1, min(a, b))
def lcm(a, b):
return (a * b) // gcd(a, b)
Using Euclidean algorithm to find gcd and then calculating the lcm dividing a by the product of gcd and b worked for me.
int euclidgcd(int a, int b){
if(b==0)
return a;
int a_rem = a % b;
return euclidgcd(b, a_rem);
}
long long lcm(int a, int b) {
int gcd=euclidgcd(a, b);
return (a/gcd*b);
}
int main() {
int a, b;
std::cin >> a >> b;
std::cout << lcm(a, b) << std::endl;
return 0;
}
Take successive multiples of the larger of the two numbers until the result is a multiple of the smaller.
this might work..
public int LCM(int x, int y)
{
int larger = x>y? x: y,
smaller = x>y? y: x,
candidate = larger ;
while (candidate % smaller != 0) candidate += larger ;
return candidate;
}
C++ template. Compile time
#include <iostream>
const int lhs = 8, rhs = 12;
template<int n, int mod_lhs=n % lhs, int mod_rhs=n % rhs> struct calc {
calc() { }
};
template<int n> struct calc<n, 0, 0> {
calc() { std::cout << n << std::endl; }
};
template<int n, int mod_rhs> struct calc<n, 0, mod_rhs> {
calc() { }
};
template<int n, int mod_lhs> struct calc <n, mod_lhs, 0> {
calc() { }
};
template<int n> struct lcm {
lcm() {
lcm<n-1>();
calc<n>();
}
};
template<> struct lcm<0> {
lcm() {}
};
int main() {
lcm<lhs * rhs>();
}
Euclidean GCD code snippet
int findGCD(int a, int b) {
if(a < 0 || b < 0)
return -1;
if (a == 0)
return b;
else if (b == 0)
return a;
else
return findGCD(b, a % b);
}
Product of 2 numbers is equal to LCM * GCD or HCF. So best way to find LCM is to find GCD and divide the product with GCD. That is, LCM(a,b) = (a*b)/GCD(a,b).
There is no way more efficient than using a built-in function!
As of Python 3.8 lcm() function has been added in math library. And can be called with folowing signature:
math.lcm(*integers)
Returns the least common multiple of the specified integer arguments. If all arguments are nonzero, then the returned value is the smallest positive integer that is a multiple of all arguments. If any of the arguments is zero, then the returned value is 0. lcm() without arguments returns 1.
Extending #John D. Cook answer that is also marked answer for this question. ( https://stackoverflow.com/a/3154503/13272795), I am sharing algorithm to find LCM of n numbers, it maybe LCM of 2 numbers or any numbers. Source for this code is this
int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// Returns LCM of array elements
ll findlcm(int arr[], int n)
{
// Initialize result
ll ans = arr[0];
// ans contains LCM of arr[0], ..arr[i]
// after i'th iteration,
for (int i = 1; i < n; i++)
ans = arr[i] * ans/gcd(arr[i], ans);
return ans;
}
Since we know the mathematic property which states that "product of LCM and HCF of any two numbers is equal to the product of the two numbers".
lets say X and Y are two integers,
then
X * Y = HCF(X, Y) * LCM(X, Y)
Now we can find LCM by knowing the HCF, which we can find through Euclidean Algorithm.
LCM(X, Y) = (X * Y) / HCF(X, Y)
Hope this will be efficient.
import java.util.*;
public class Hello {
public static int HCF(int X, int Y){
if(X == 0)return Y;
return HCF(Y%X, X);
}
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int X = scanner.nextInt(), Y = scanner.nextInt();
System.out.print((X * Y) / HCF(X, Y));
}
}
Yes, there are numerous way to calculate LCM such as using GCD (HCF).
You can apply prime decomposition such as (optimized/naive) Sieve Eratosthenes or find factor of prime number to compute GCD, which is way more faster than calculate LCM directly. Then as all said above, LCM(X, Y) = (X * Y) / GCD(X, Y)
I googled the same question, and found this Stackoverflow page,
however I come up with another simple solution using python
def find_lcm(numbers):
h = max(numbers)
lcm = h
def check(l, numbers):
remainders = [ l%n==0 for n in numbers]
return all(remainders)
while (check(lcm, numbers) == False):
lcm = lcm + h
return lcm
for
numbers = [120,150,135,225]
it will return 5400
numbers = [120,150,135,225]
print(find_lcm(numbers)) # will print 5400
If I know the number number y and know that 2^x=y, how do I compute x?
Base 2 logarithm function:
log2(y)
which is equivalent to:
log(y) / log(2)
for arbitrary base.
And in case you don't have a log function handy, you can always see how many times you must divide y by 2 before it becomes 1. (This assumes x is positive and an integer.)
If you are sure that it is a power of 2, then you can write a loop and right shift the number until you get a 1. The number of times the loop ran will be the value of x.
Example code:
int power(int num)
{
if(0 == num)
{
return 0;
}
int count = 0;
do
{
++count;
num = num >> 1;
}while(! (num & 1) && num > 0);
return count;
}
If x is a positive integer, then, following code will be more efficient..
unsigned int y; // You know the number y for which you require x..
unsigned int x = 0;
while (y >>= 1)
{
x++;
}
x is the answer!