Diophantine Equation

data_structures/hashing/number_theory/chinese_remainder_theorem.py

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+# Chinese Remainder Theorem:
+
+# If GCD(a,b) = 1, then for any remainder ra modulo a and any remainder rb modulo b there exists integer n,
+# such that n = ra (mod a) and n = ra(mod b).  If n1 and n2 are two such integers, then n1=n2(mod ab)
+
+# Algorithm :
+
+# 1. Use extended euclid algorithm to find x,y such that a*x + b*y = 1
+# 2. Take n = ra*by + rb*ax
+
+
+# Extended Euclid
+def ExtendedEuclid(a, b):
+    if b == 0:
+        return (1, 0)
+    (x, y) = ExtendedEuclid(b, a % b)
+    k = a // b
+    return (y, x - k * y)
+
+
+# Uses ExtendedEuclid to find inverses
+def ChineseRemainderTheorem(n1, r1, n2, r2):
+    (x, y) = ExtendedEuclid(n1, n2)
+    m = n1 * n2
+    n = r2 * x * n1 + r1 * y * n2
+    return ((n % m + m) % m)
+
+
+# ----------SAME SOLUTION USING InvertModulo instead ExtendedEuclid----------------
+
+# This function find the inverses of a i.e., a^(-1)
+def InvertModulo(a, n):
+    (b, x) = ExtendedEuclid(a, n)
+    if b < 0:
+        b = (b % n + n) % n
+    return b
+
+
+# Same a above using InvertingModulo
+def ChineseRemainderTheorem2(n1, r1, n2, r2):
+    x, y = InvertModulo(n1, n2), InvertModulo(n2, n1)
+    m = n1 * n2
+    n = r2 * x * n1 + r1 * y * n2
+    return (n % m + m) % m

data_structures/hashing/number_theory/diophantine_equation.py

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+# Diophantine Equation : Given integers a,b,c ( at least one of a and b != 0), the diophantine equation
+# a*x + b*y = c has a solution (where x and y are integers) iff gcd(a,b) divides c.
+
+
+def diophantine(a, b, c):
+    assert c % gcd(a, b) == 0  # gcd(a,b) function implemented below
+
+    (d, x, y) = extended_gcd(a, b)  # extended_gcd(a,b) function implemented below
+    r = c / d
+    return (r * x, r * y)
+
+
+# Lemma : if n|ab and gcd(a,n) = 1, then n|b.
+
+# Finding All solutions of Diophantine Equations:
+
+# Theorem : Let gcd(a,b) = d, a = d*p, b = d*q. If (x0,y0) is a solution of Diophantine Equation a*x + b*y = c.
+# a*x0 + b*y0 = c, then all the solutions have the form a(x0 + t*q) + b(y0 - t*p) = c, where t is an arbitrary integer.
+
+# n is the number of solution you want, n = 2 by default
+
+def diophantine_all_soln(a, b, c, n=2):
+    (x0, y0) = diophantine(a, b, c)
+    d = gcd(a, b)
+    p = a // d
+    q = b // d
+
+    for i in range(n):
+        x = x0 + i * q
+        y = y0 - i * p
+        print(x, y)
+
+
+# Euclid's Lemma :  d divides a and b, if and only if d divides a-b and b
+
+# Euclid's Algorithm
+
+def gcd(a, b):
+    if a < b:
+        a, b = b, a
+
+    while a % b != 0:
+        a, b = b, a % b
+
+    return b
+
+
+# Extended Euclid's Algorithm : If d divides a and b and d = a*x + b*y for integers x and y, then d = gcd(a,b)
+
+
+def extended_gcd(a, b):
+    assert a >= 0 and b >= 0
+
+    if b == 0:
+        d, x, y = a, 1, 0
+    else:
+        (d, p, q) = extended_gcd(b, a % b)
+        x = q
+        y = p - q * (a // b)
+
+    assert a % d == 0 and b % d == 0
+    assert d == a * x + b * y
+
+    return (d, x, y)