maths

Author: jeffin143

Math/collatz_conjecture/C++/collatz_conjecture.cpp

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+#include <iostream>
+#include <vector>
+#include <string>
+#include <algorithm>
+#include <utility>
+#include <iomanip>
+
+using namespace std;
+
+typedef unsigned long long int ulli;
+
+vector<ulli> collatz(ulli number){
+
+        vector<ulli> seq;
+
+        while(number > 1){
+
+                if( number%2 == 1 ){
+                      number = (3*number) + 1;
+                      seq.push_back(number);
+                }
+                else{
+                      number = number/2;
+                      seq.push_back(number);
+                }
+
+        }
+
+        return seq;
+}
+
+
+void printsequence(vector<ulli> &seq){
+
+      for(int i = 0;i < seq.size();i++){
+            cout<<seq[i]<<" ";
+      }
+      cout<<endl;
+
+}
+int main(){
+
+        ios_base::sync_with_stdio(false);
+
+        ulli number;
+        cout<<"Enter the number for which collatz sequence is to be displayed : ";
+        cin>>number;
+
+        vector<ulli> sequence;
+        sequence = collatz(number);
+
+        printsequence(sequence);
+
+        return 0;
+}

Math/collatz_conjecture/Python/collatz_conjecture.py

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+def collatz(n):
+    seq = [n]
+    while n > 1:
+        if n%2 == 1:
+            n = n*3 + 1
+            seq.append(n)
+        else:
+            n = n/2
+            seq.append(n)
+    return seq
+
+if __name__ == "__main__" :
+    number = input("Enter the number for which collatz sequence is to be displayed : ")
+    print collatz(number)

Math/collatz_conjecture/README.md

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+# Collatz Conjecture #
+In the field of mathematics the collatz conjecture was posed by L. Collatz in 1937 which states that given a number n, one can  always
+'get to one' by applying the following function recursively on the number.
+
+## Function for Collatz Conjecture ##
+
+1. **If number is even ** :<br />
+    number = number / 2
+
+2. **If number is odd ** :<br />
+    number = (3 * number) + 1

Math/convuxhull/grahamscan.cpp

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+// A simple C program to calculate Euler's Totient Function
+#include <stdio.h>
+ 
+// Function to return gcd of a and b
+int gcd(int a, int b)
+{
+    if (a == 0)
+        return b;
+    return gcd(b%a, a);
+}
+ 
+// A simple method to evaluate Euler Totient Function
+int phi(unsigned int n)
+{
+    unsigned int result = 1;
+    for (int i=2; i < n; i++)
+        if (gcd(i, n) == 1)
+            result++;
+    return result;
+}
+ 
+// Driver program to test above function
+int main()
+{
+    int n;
+    for (n=1; n<=10; n++)
+      printf("phi(%d) = %d\n", n, phi(n));
+    return 0;
+}

Math/eulers_totient_function/C/eulers_totient_func.c

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+def gcd(a, b):
+	if a == 0:
+		return b
+
+	return gcd(b % a, a)
+
+def phi(n):
+	res = 1
+
+	for i in range(2, n):
+		if gcd(i, n) == 1:
+			res += 1
+
+	return res

Math/eulers_totient_function/python/eulers_totient.py

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+from eulers_totient import phi
+
+if __name__ == "__main__":
+	for i in range(1, 11):
+		print ("phi({}) == {}".format(i, phi(i)))

Math/eulers_totient_function/python/test.py

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+# Euler's Totient Function
+
+Euler’s Totient function Φ(n) for an input n is count of numbers in {1, 2, 3, …, n} that are relatively prime to n, i.e., the numbers whose GCD (Greatest Common Divisor) with n is 1.
+
+```
+Φ(1) = 1
+gcd(1, 1) is 1
+
+Φ(2) = 1
+gcd(1, 2) is 1, but gcd(2, 2) is 2.
+
+Φ(3) = 2
+gcd(1, 3) is 1 and gcd(2, 3) is 1
+
+Φ(4) = 2
+gcd(1, 4) is 1 and gcd(3, 4) is 1
+
+Φ(5) = 4
+gcd(1, 5) is 1, gcd(2, 5) is 1, 
+gcd(3, 5) is 1 and gcd(4, 5) is 1
+
+Φ(6) = 2
+gcd(1, 6) is 1 and gcd(5, 6) is 1
+
+```
+
+### How to compute Φ(n) for an input n?
+
+A simple solution is to iterate through all numbers from 1 to n-1 and count numbers with gcd with n as 1. 
+
+