create the Math folder add some math algorithms.

src/main/python/algorithms/math/GCD.py

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+'''
+ * @author (original JAVA) William Fiset, [email protected]
+ *         (conversion to Python) Hiruna Vishwamith, [email protected]
+ * @date   11 Nov 2022
+
+
+    An implementation of finding the GCD of two numbers. Time Complexity: Time Complexity ~O(log(a + b))
+
+'''
+
+def gcd(a,b):
+    if a<0:
+        a=-a
+    if b==0:
+        return a
+    else:
+        return gcd(b,a%b)
+
+
+print(gcd(12, 18))   # 6
+print(gcd(-12, 18))  # 6
+print(gcd(12, -18))  # 6
+print(gcd(-12, -18)) # 6
+print(gcd(5, 0))     # 5
+print(gcd(0, 5))     # 5
+print(gcd(-5, 0))    # 5
+print(gcd(0, -5))    # 5
+print(gcd(0, 0))     # 0

src/main/python/algorithms/math/IsPrime.py

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+'''
+ * @author (original JAVA) William Fiset, [email protected]
+ *         (conversion to Python) Hiruna Vishwamith, [email protected]
+ * @date   11 Nov 2022
+
+    Tests whether a number is a prime number or not Time Complexity: O(sqrt(n))
+
+'''
+
+
+
+import math
+
+def isPrime(n):
+    if n<2:
+        return False
+    if (n==2)or(n==3):
+        return True
+    if n%2==0 or n%3==0:
+        return False
+
+    limit=int(math.sqrt(n))
+    for i in range(5,limit+1,6):
+        if n%i==0 or (n%(i+2)==0):
+            return False
+
+    return True
+
+print(isPrime(5))
+print(isPrime(31))
+print(isPrime(1433))
+print(isPrime(8763857775536878331))
+
+
+

src/main/python/algorithms/math/LCM.py

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+'''
+ * @author (original JAVA) William Fiset, [email protected]
+ *         (conversion to Python) Hiruna Vishwamith, [email protected]
+ * @date   11 Nov 2022
+
+
+    An implementation of finding the LCM of two numbers. Time Complexity ~O(log(a + b))
+
+'''
+
+
+def gcd(a, b):
+    if a < 0:
+        a = -a
+    if b == 0:
+        return a
+    else:
+        return gcd(b, a % b)
+
+
+def lcm(a, b):
+    lcm = (a//gcd(a, b))*b
+    if lcm > 0:
+        return lcm
+    else:
+        return -lcm
+
+
+print(lcm(12, 18))  # 36
+print(lcm(-12, 18))  # 36
+print(lcm(12, -18))  # 36
+print(lcm(-12, -18))  # 36

src/main/python/algorithms/math/ModInv.py

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+'''
+ * @author (original JAVA) William Fiset, [email protected]
+ *         (conversion to Python) Hiruna Vishwamith, [email protected]
+ * @date   11 Nov 2022
+
+
+    An implementation of the modPow(a, n, mod) operation.
+    Time Complexity O(lg(n))
+
+'''
+
+import math
+
+
+# This function performs the extended euclidean algorithm on two numbers a and b.
+# The function returns the gcd(a,b) as well as the numbers x and y such
+# that ax + by = gcd(a,b). This calculation is important in number theory
+# and can be used for several things such as finding modular inverses and
+# solutions to linear Diophantine equations.
+
+
+def egcd(a, b):
+    if b == 0:
+        return (a, 1, 0)
+    else:
+        g, y, x = egcd(b, a % b)
+        return (g, x, y - (a // b) * x)
+
+
+#  Returns the modular inverse of 'a' mod 'm'
+#  Make sure m > 0 and 'a' & 'm' are relatively prime.
+def modInv(a, m):
+
+    if m < 0:
+        raise Exception(' mod must be > 0')
+    a = ((a % m)+m) % m
+    g, x, y = egcd(a, m)
+    if g != 1:
+        return None
+    return ((x % m)+m) % m
+
+
+print(modInv(2, 5))  # 2*3 mod 5=1
+print(modInv(4, 18))  # 4*x mod 18=1

src/main/python/algorithms/math/ModPow.py

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+'''
+ * @author (original JAVA) William Fiset, [email protected]
+ *         (conversion to Python) Hiruna Vishwamith, [email protected]
+ * @date   11 Nov 2022
+
+
+    An implementation of the modPow(a, n, mod) operation.
+    Time Complexity O(lg(n))
+
+'''
+
+import math
+
+
+# Computes the Greatest Common Divisor (GCD) of a & b
+def gcd(a, b):
+    if a < 0:
+        a = -a
+    if b == 0:
+        return a
+    else:
+        return gcd(b, a % b)
+
+# This function performs the extended euclidean algorithm on two numbers a and b.
+# The function returns the gcd(a,b) as well as the numbers x and y such
+# that ax + by = gcd(a,b). This calculation is important in number theory
+# and can be used for several things such as finding modular inverses and
+# solutions to linear Diophantine equations.
+
+
+def egcd(a, b):
+    if a < 0:
+        a = -a
+    if b == 0:
+        return (b, 1, 0)
+    else:
+        g, y, x = egcd(b, a % b)
+        return (g, x, y - (a // b) * x)
+
+
+#  Returns the modular inverse of 'a' mod 'm'
+#  Make sure m > 0 and 'a' & 'm' are relatively prime.
+def modInv(a, m):
+
+    a = ((a % m)+m) % m
+    g, x, y = egcd(a, m)
+    return ((x % m)+m) % m
+
+# Computes a^n modulo mod very efficiently in O(lg(n)) time.
+# This function supports negative exponent values and a negative
+# base, however the modulus must be positive.
+
+
+def modPow(a, n, mod):
+    if mod < 0:
+        raise Exception('mod must be positive')
+
+    # To handle negative exponents we can use the modular
+    # inverse of a to our advantage since: a^-n mod m = (a^-1)^n mod m
+
+    if n < 0:
+        if gcd(a, mod) != 1:
+            raise Exception("If n < 0 then must have gcd(a, mod) = 1")
+        return modPow(modInv(a, mod), -n, mod)
+    if n == 0:
+        return 1
+    p = a
+    r = 1
+    i = 0
+    while n != 0:
+        mask = 1 << i
+        if n & mask == mask:
+            r = (((r * p) % mod) + mod) % mod
+            n -= mask
+        p = ((p * p) % mod + mod) % mod
+        i+=1
+    return ((r % mod) + mod) % mod
+
+
+# 3 ^ 4 mod 1000000
+
+r1 = modPow(3, 4, 10**6) #81
+r2 = modPow(-45, 12345, 987654321) #323182557
+r3 = modPow(6, -66, 101) #84
+r4 = modPow(-5, -7, 1009) #675
+
+print(r1,r2,r3,r4)