From 43cd9c4100c5e7aa4d5429ce660cc16a880087bc Mon Sep 17 00:00:00 2001
From: Lex Berezhny <lex@damoti.com>
Date: Thu, 14 Jun 2018 16:27:17 -0400
Subject: [PATCH] removed unused msqr module

---
 torba/msqr.py | 96 ---------------------------------------------------
 1 file changed, 96 deletions(-)
 delete mode 100644 torba/msqr.py

diff --git a/torba/msqr.py b/torba/msqr.py
deleted file mode 100644
index beb5bed7f..000000000
--- a/torba/msqr.py
+++ /dev/null
@@ -1,96 +0,0 @@
-# from http://eli.thegreenplace.net/2009/03/07/computing-modular-square-roots-in-python/
-
-
-def modular_sqrt(a, p):
-    """ Find a quadratic residue (mod p) of 'a'. p
-    must be an odd prime.
-
-    Solve the congruence of the form:
-    x^2 = a (mod p)
-    And returns x. Note that p - x is also a root.
-
-    0 is returned is no square root exists for
-    these a and p.
-
-    The Tonelli-Shanks algorithm is used (except
-    for some simple cases in which the solution
-    is known from an identity). This algorithm
-    runs in polynomial time (unless the
-    generalized Riemann hypothesis is false).
-    """
-    # Simple cases
-    #
-    if legendre_symbol(a, p) != 1:
-        return 0
-    elif a == 0:
-        return 0
-    elif p == 2:
-        return p
-    elif p % 4 == 3:
-        return pow(a, (p + 1) / 4, p)
-
-    # Partition p-1 to s * 2^e for an odd s (i.e.
-    # reduce all the powers of 2 from p-1)
-    #
-    s = p - 1
-    e = 0
-    while s % 2 == 0:
-        s /= 2
-        e += 1
-
-    # Find some 'n' with a legendre symbol n|p = -1.
-    # Shouldn't take long.
-    #
-    n = 2
-    while legendre_symbol(n, p) != -1:
-        n += 1
-
-    # Here be dragons!
-    # Read the paper "Square roots from 1; 24, 51,
-    # 10 to Dan Shanks" by Ezra Brown for more
-    # information
-    #
-
-    # x is a guess of the square root that gets better
-    # with each iteration.
-    # b is the "fudge factor" - by how much we're off
-    # with the guess. The invariant x^2 = ab (mod p)
-    # is maintained throughout the loop.
-    # g is used for successive powers of n to update
-    # both a and b
-    # r is the exponent - decreases with each update
-    #
-    x = pow(a, (s + 1) / 2, p)
-    b = pow(a, s, p)
-    g = pow(n, s, p)
-    r = e
-
-    while True:
-        t = b
-        m = 0
-        for m in xrange(r):
-            if t == 1:
-                break
-            t = pow(t, 2, p)
-
-        if m == 0:
-            return x
-
-        gs = pow(g, 2 ** (r - m - 1), p)
-        g = (gs * gs) % p
-        x = (x * gs) % p
-        b = (b * g) % p
-        r = m
-
-
-def legendre_symbol(a, p):
-    """ Compute the Legendre symbol a|p using
-    Euler's criterion. p is a prime, a is
-    relatively prime to p (if p divides
-    a, then a|p = 0)
-
-    Returns 1 if a has a square root modulo
-    p, -1 otherwise.
-    """
-    ls = pow(a, (p - 1) / 2, p)
-    return -1 if ls == p - 1 else ls