[pypy-commit] pypy default: Test and -er- fix, by commenting out the buggy code.

arigo noreply at buildbot.pypy.org
Sun Jul 24 18:47:09 CEST 2011


Author: Armin Rigo <arigo at tunes.org>
Branch: 
Changeset: r45944:70d00af8294e
Date: 2011-07-24 18:46 +0200
http://bitbucket.org/pypy/pypy/changeset/70d00af8294e/

Log:	Test and -er- fix, by commenting out the buggy code.

diff --git a/pypy/rlib/rbigint.py b/pypy/rlib/rbigint.py
--- a/pypy/rlib/rbigint.py
+++ b/pypy/rlib/rbigint.py
@@ -40,7 +40,7 @@
 # In that case, do 5 bits at a time.  The potential drawback is that
 # a table of 2**5 intermediate results is computed.
 
-FIVEARY_CUTOFF = 8
+## FIVEARY_CUTOFF = 8   disabled for now
 
 
 def _mask_digit(x):
@@ -456,7 +456,7 @@
 
         # python adaptation: moved macros REDUCE(X) and MULT(X, Y, result)
         # into helper function result = _help_mult(x, y, c)
-        if b.numdigits() <= FIVEARY_CUTOFF:
+        if 1:   ## b.numdigits() <= FIVEARY_CUTOFF:
             # Left-to-right binary exponentiation (HAC Algorithm 14.79)
             # http://www.cacr.math.uwaterloo.ca/hac/about/chap14.pdf
             i = b.numdigits() - 1
@@ -469,26 +469,30 @@
                         z = _help_mult(z, a, c)
                     j >>= 1
                 i -= 1
-        else:
-            # Left-to-right 5-ary exponentiation (HAC Algorithm 14.82)
-            # This is only useful in the case where c != None.
-            # z still holds 1L
-            table = [z] * 32
-            table[0] = z
-            for i in range(1, 32):
-                table[i] = _help_mult(table[i-1], a, c)
-            i = b.numdigits() - 1
-            while i >= 0:
-                bi = b.digit(i)
-                j = SHIFT - 5
-                while j >= 0:
-                    index = (bi >> j) & 0x1f
-                    for k in range(5):
-                        z = _help_mult(z, z, c)
-                    if index:
-                        z = _help_mult(z, table[index], c)
-                    j -= 5
-                i -= 1
+##        else:
+##            This code is disabled for now, because it assumes that
+##            SHIFT is a multiple of 5.  It could be fixed but it looks
+##            like it's more troubles than benefits...
+##
+##            # Left-to-right 5-ary exponentiation (HAC Algorithm 14.82)
+##            # This is only useful in the case where c != None.
+##            # z still holds 1L
+##            table = [z] * 32
+##            table[0] = z
+##            for i in range(1, 32):
+##                table[i] = _help_mult(table[i-1], a, c)
+##            i = b.numdigits() - 1
+##            while i >= 0:
+##                bi = b.digit(i)
+##                j = SHIFT - 5
+##                while j >= 0:
+##                    index = (bi >> j) & 0x1f
+##                    for k in range(5):
+##                        z = _help_mult(z, z, c)
+##                    if index:
+##                        z = _help_mult(z, table[index], c)
+##                    j -= 5
+##                i -= 1
 
         if negativeOutput and z.sign != 0:
             z = z.sub(c)
diff --git a/pypy/rlib/test/test_rbigint.py b/pypy/rlib/test/test_rbigint.py
--- a/pypy/rlib/test/test_rbigint.py
+++ b/pypy/rlib/test/test_rbigint.py
@@ -373,6 +373,13 @@
         print '--->', v
         assert v.tolong() == pow(x, y, z)
 
+    def test_pow_lll_bug(self):
+        two = rbigint.fromint(2)
+        t = rbigint.fromlong(2655689964083835493447941032762343136647965588635159615997220691002017799304)
+        for n, expected in [(37, 9), (1291, 931), (67889, 39464)]:
+            v = two.pow(t, rbigint.fromint(n))
+            assert v.toint() == expected
+
     def test_pow_lln(self):
         x = 10L
         y = 2L


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