mirrors/cpython
mirrors/cpython
0
0
Fork 0
mirror of https://github.com/python/cpython.git synced 2024-11-21 12:59:38 +01:00
Code Issues Releases Activity
main
cpython/Lib/_pylong.py
Xie Yanbo 253c6a0b2f
Fix typos in comments and test code (#122846)
2024-08-11 21:16:41 -07:00

730 lines
28 KiB
Python
Raw Permalink Blame History

"""Python implementations of some algorithms for use by longobject.c.
The goal is to provide asymptotically faster algorithms that can be
used for operations on integers with many digits. In those cases, the
performance overhead of the Python implementation is not significant
since the asymptotic behavior is what dominates runtime. Functions
provided by this module should be considered private and not part of any
public API.
Note: for ease of maintainability, please prefer clear code and avoid
"micro-optimizations". This module will only be imported and used for
integers with a huge number of digits. Saving a few microseconds with
tricky or non-obvious code is not worth it. For people looking for
maximum performance, they should use something like gmpy2."""
import re
import decimal
try:
import _decimal
except ImportError:
_decimal = None
# A number of functions have this form, where `w` is a desired number of
# digits in base `base`:
#
# def inner(...w...):
# if w <= LIMIT:
# return something
# lo = w >> 1
# hi = w - lo
# something involving base**lo, inner(...lo...), j, and inner(...hi...)
# figure out largest w needed
# result = inner(w)
#
# They all had some on-the-fly scheme to cache `base**lo` results for reuse.
# Power is costly.
#
# This routine aims to compute all amd only the needed powers in advance, as
# efficiently as reasonably possible. This isn't trivial, and all the
# on-the-fly methods did needless work in many cases. The driving code above
# changes to:
#
# figure out largest w needed
# mycache = compute_powers(w, base, LIMIT)
# result = inner(w)
#
# and `mycache[lo]` replaces `base**lo` in the inner function.
#
# If an algorithm wants the powers of ceiling(w/2) instead of the floor,
# pass keyword argument `need_hi=True`.
#
# While this does give minor speedups (a few percent at best), the
# primary intent is to simplify the functions using this, by eliminating
# the need for them to craft their own ad-hoc caching schemes.
#
# See code near end of file for a block of code that can be enabled to
# run millions of tests.
def compute_powers(w, base, more_than, *, need_hi=False, show=False):
seen = set()
need = set()
ws = {w}
while ws:
w = ws.pop() # any element is fine to use next
if w in seen or w <= more_than:
continue
seen.add(w)
lo = w >> 1
hi = w - lo
# only _need_ one here; the other may, or may not, be needed
which = hi if need_hi else lo
need.add(which)
ws.add(which)
if lo != hi:
ws.add(w - which)
# `need` is the set of exponents needed. To compute them all
# efficiently, possibly add other exponents to `extra`. The goal is
# to ensure that each exponent can be gotten from a smaller one via
# multiplying by the base, squaring it, or squaring and then
# multiplying by the base.
#
# If need_hi is False, this is already the case (w can always be
# gotten from w >> 1 via one of the squaring strategies). But we do
# the work anyway, just in case ;-)
#
# Note that speed is irrelevant. These loops are working on little
# ints (exponents) and go around O(log w) times. The total cost is
# insignificant compared to just one of the bigint multiplies.
cands = need.copy()
extra = set()
while cands:
w = max(cands)
cands.remove(w)
lo = w >> 1
if lo > more_than and w-1 not in cands and lo not in cands:
extra.add(lo)
cands.add(lo)
assert need_hi or not extra
d = {}
for n in sorted(need | extra):
lo = n >> 1
hi = n - lo
if n-1 in d:
if show:
print("* base", end="")
result = d[n-1] * base # cheap!
elif lo in d:
# Multiplying a bigint by itself is about twice as fast
# in CPython provided it's the same object.
if show:
print("square", end="")
result = d[lo] * d[lo] # same object
if hi != lo:
if show:
print(" * base", end="")
assert 2 * lo + 1 == n
result *= base
else: # rare
if show:
print("pow", end='')
result = base ** n
if show:
print(" at", n, "needed" if n in need else "extra")
d[n] = result
assert need <= d.keys()
if excess := d.keys() - need:
assert need_hi
for n in excess:
del d[n]
return d
_unbounded_dec_context = decimal.getcontext().copy()
_unbounded_dec_context.prec = decimal.MAX_PREC
_unbounded_dec_context.Emax = decimal.MAX_EMAX
_unbounded_dec_context.Emin = decimal.MIN_EMIN
_unbounded_dec_context.traps[decimal.Inexact] = 1 # sanity check
def int_to_decimal(n):
"""Asymptotically fast conversion of an 'int' to Decimal."""
# Function due to Tim Peters. See GH issue #90716 for details.
# https://github.com/python/cpython/issues/90716
#
# The implementation in longobject.c of base conversion algorithms
# between power-of-2 and non-power-of-2 bases are quadratic time.
# This function implements a divide-and-conquer algorithm that is
# faster for large numbers. Builds an equal decimal.Decimal in a
# "clever" recursive way. If we want a string representation, we
# apply str to _that_.
from decimal import Decimal as D
BITLIM = 200
# Don't bother caching the "lo" mask in this; the time to compute it is
# tiny compared to the multiply.
def inner(n, w):
if w <= BITLIM:
return D(n)
w2 = w >> 1
hi = n >> w2
lo = n & ((1 << w2) - 1)
return inner(lo, w2) + inner(hi, w - w2) * w2pow[w2]
with decimal.localcontext(_unbounded_dec_context):
nbits = n.bit_length()
w2pow = compute_powers(nbits, D(2), BITLIM)
if n < 0:
negate = True
n = -n
else:
negate = False
result = inner(n, nbits)
if negate:
result = -result
return result
def int_to_decimal_string(n):
"""Asymptotically fast conversion of an 'int' to a decimal string."""
w = n.bit_length()
if w > 450_000 and _decimal is not None:
# It is only usable with the C decimal implementation.
# _pydecimal.py calls str() on very large integers, which in its
# turn calls int_to_decimal_string(), causing very deep recursion.
return str(int_to_decimal(n))
# Fallback algorithm for the case when the C decimal module isn't
# available. This algorithm is asymptotically worse than the algorithm
# using the decimal module, but better than the quadratic time
# implementation in longobject.c.
DIGLIM = 1000
def inner(n, w):
if w <= DIGLIM:
return str(n)
w2 = w >> 1
hi, lo = divmod(n, pow10[w2])
return inner(hi, w - w2) + inner(lo, w2).zfill(w2)
# The estimation of the number of decimal digits.
# There is no harm in small error. If we guess too large, there may
# be leading 0's that need to be stripped. If we guess too small, we
# may need to call str() recursively for the remaining highest digits,
# which can still potentially be a large integer. This is manifested
# only if the number has way more than 10**15 digits, that exceeds
# the 52-bit physical address limit in both Intel64 and AMD64.
w = int(w * 0.3010299956639812 + 1) # log10(2)
pow10 = compute_powers(w, 5, DIGLIM)
for k, v in pow10.items():
pow10[k] = v << k # 5**k << k == 5**k * 2**k == 10**k
if n < 0:
n = -n
sign = '-'
else:
sign = ''
s = inner(n, w)
if s[0] == '0' and n:
# If our guess of w is too large, there may be leading 0's that
# need to be stripped.
s = s.lstrip('0')
return sign + s
def _str_to_int_inner(s):
"""Asymptotically fast conversion of a 'str' to an 'int'."""
# Function due to Bjorn Martinsson. See GH issue #90716 for details.
# https://github.com/python/cpython/issues/90716
#
# The implementation in longobject.c of base conversion algorithms
# between power-of-2 and non-power-of-2 bases are quadratic time.
# This function implements a divide-and-conquer algorithm making use
# of Python's built in big int multiplication. Since Python uses the
# Karatsuba algorithm for multiplication, the time complexity
# of this function is O(len(s)**1.58).
DIGLIM = 2048
def inner(a, b):
if b - a <= DIGLIM:
return int(s[a:b])
mid = (a + b + 1) >> 1
return (inner(mid, b)
+ ((inner(a, mid) * w5pow[b - mid])
<< (b - mid)))
w5pow = compute_powers(len(s), 5, DIGLIM)
return inner(0, len(s))
# Asymptotically faster version, using the C decimal module. See
# comments at the end of the file. This uses decimal arithmetic to
# convert from base 10 to base 256. The latter is just a string of
# bytes, which CPython can convert very efficiently to a Python int.
# log of 10 to base 256 with best-possible 53-bit precision. Obtained
# via:
# from mpmath import mp
# mp.prec = 1000
# print(float(mp.log(10, 256)).hex())
_LOG_10_BASE_256 = float.fromhex('0x1.a934f0979a371p-2') # about 0.415
# _spread is for internal testing. It maps a key to the number of times
# that condition obtained in _dec_str_to_int_inner:
# key 0 - quotient guess was right
# key 1 - quotient had to be boosted by 1, one time
# key 999 - one adjustment wasn't enough, so fell back to divmod
from collections import defaultdict
_spread = defaultdict(int)
del defaultdict
def _dec_str_to_int_inner(s, *, GUARD=8):
# Yes, BYTELIM is "large". Large enough that CPython will usually
# use the Karatsuba _str_to_int_inner to convert the string. This
# allowed reducing the cutoff for calling _this_ function from 3.5M
# to 2M digits. We could almost certainly do even better by
# fine-tuning this and/or using a larger output base than 256.
BYTELIM = 100_000
D = decimal.Decimal
result = bytearray()
# See notes at end of file for discussion of GUARD.
assert GUARD > 0 # if 0, `decimal` can blow up - .prec 0 not allowed
def inner(n, w):
#assert n < D256 ** w # required, but too expensive to check
if w <= BYTELIM:
# XXX Stefan Pochmann discovered that, for 1024-bit ints,
# `int(Decimal)` took 2.5x longer than `int(str(Decimal))`.
# Worse, `int(Decimal) is still quadratic-time for much
# larger ints. So unless/until all that is repaired, the
# seemingly redundant `str(Decimal)` is crucial to speed.
result.extend(int(str(n)).to_bytes(w)) # big-endian default
return
w1 = w >> 1
w2 = w - w1
if 0:
# This is maximally clear, but "too slow". `decimal`
# division is asymptotically fast, but we have no way to
# tell it to reuse the high-precision reciprocal it computes
# for pow256[w2], so it has to recompute it over & over &
# over again :-(
hi, lo = divmod(n, pow256[w2][0])
else:
p256, recip = pow256[w2]
# The integer part will have a number of digits about equal
# to the difference between the log10s of `n` and `pow256`
# (which, since these are integers, is roughly approximated
# by `.adjusted()`). That's the working precision we need,
ctx.prec = max(n.adjusted() - p256.adjusted(), 0) + GUARD
hi = +n * +recip # unary `+` chops back to ctx.prec digits
ctx.prec = decimal.MAX_PREC
hi = hi.to_integral_value() # lose the fractional digits
lo = n - hi * p256
# Because we've been uniformly rounding down, `hi` is a
# lower bound on the correct quotient.
assert lo >= 0
# Adjust quotient up if needed. It usually isn't. In random
# testing on inputs through 5 billion digit strings, the
# test triggered once in about 200 thousand tries.
count = 0
if lo >= p256:
count = 1
lo -= p256
hi += 1
if lo >= p256:
# Complete correction via an exact computation. I
# believe it's not possible to get here provided
# GUARD >= 3. It's tested by reducing GUARD below
# that.
count = 999
hi2, lo = divmod(lo, p256)
hi += hi2
_spread[count] += 1
# The assert should always succeed, but way too slow to keep
# enabled.
#assert hi, lo == divmod(n, pow256[w2][0])
inner(hi, w1)
del hi # at top levels, can free a lot of RAM "early"
inner(lo, w2)
# How many base 256 digits are needed?. Mathematically, exactly
# floor(log256(int(s))) + 1. There is no cheap way to compute this.
# But we can get an upper bound, and that's necessary for our error
# analysis to make sense. int(s) < 10**len(s), so the log needed is
# < log256(10**len(s)) = len(s) * log256(10). However, using
# finite-precision floating point for this, it's possible that the
# computed value is a little less than the true value. If the true
# value is at - or a little higher than - an integer, we can get an
# off-by-1 error too low. So we add 2 instead of 1 if chopping lost
# a fraction > 0.9.
# The "WASI" test platform can complain about `len(s)` if it's too
# large to fit in its idea of "an index-sized integer".
lenS = s.__len__()
log_ub = lenS * _LOG_10_BASE_256
log_ub_as_int = int(log_ub)
w = log_ub_as_int + 1 + (log_ub - log_ub_as_int > 0.9)
# And what if we've plain exhausted the limits of HW floats? We
# could compute the log to any desired precision using `decimal`,
# but it's not plausible that anyone will pass a string requiring
# trillions of bytes (unless they're just trying to "break things").
if w.bit_length() >= 46:
# "Only" had < 53 - 46 = 7 bits to spare in IEEE-754 double.
raise ValueError(f"cannot convert string of len {lenS} to int")
with decimal.localcontext(_unbounded_dec_context) as ctx:
D256 = D(256)
pow256 = compute_powers(w, D256, BYTELIM, need_hi=True)
rpow256 = compute_powers(w, 1 / D256, BYTELIM, need_hi=True)
# We're going to do inexact, chopped arithmetic, multiplying by
# an approximation to the reciprocal of 256**i. We chop to get a
# lower bound on the true integer quotient. Our approximation is
# a lower bound, the multiplication is chopped too, and
# to_integral_value() is also chopped.
ctx.traps[decimal.Inexact] = 0
ctx.rounding = decimal.ROUND_DOWN
for k, v in pow256.items():
# No need to save much more precision in the reciprocal than
# the power of 256 has, plus some guard digits to absorb
# most relevant rounding errors. This is highly significant:
# 1/2**i has the same number of significant decimal digits
# as 5**i, generally over twice the number in 2**i,
ctx.prec = v.adjusted() + GUARD + 1
# The unary "+" chops the reciprocal back to that precision.
pow256[k] = v, +rpow256[k]
del rpow256 # exact reciprocals no longer needed
ctx.prec = decimal.MAX_PREC
inner(D(s), w)
return int.from_bytes(result)
def int_from_string(s):
"""Asymptotically fast version of PyLong_FromString(), conversion
of a string of decimal digits into an 'int'."""
# PyLong_FromString() has already removed leading +/-, checked for invalid
# use of underscore characters, checked that string consists of only digits
# and underscores, and stripped leading whitespace. The input can still
# contain underscores and have trailing whitespace.
s = s.rstrip().replace('_', '')
func = _str_to_int_inner
if len(s) >= 2_000_000 and _decimal is not None:
func = _dec_str_to_int_inner
return func(s)
def str_to_int(s):
"""Asymptotically fast version of decimal string to 'int' conversion."""
# FIXME: this doesn't support the full syntax that int() supports.
m = re.match(r'\s*([+-]?)([0-9_]+)\s*', s)
if not m:
raise ValueError('invalid literal for int() with base 10')
v = int_from_string(m.group(2))
if m.group(1) == '-':
v = -v
return v
# Fast integer division, based on code from Mark Dickinson, fast_div.py
# GH-47701. Additional refinements and optimizations by Bjorn Martinsson. The
# algorithm is due to Burnikel and Ziegler, in their paper "Fast Recursive
# Division".
_DIV_LIMIT = 4000
def _div2n1n(a, b, n):
"""Divide a 2n-bit nonnegative integer a by an n-bit positive integer
b, using a recursive divide-and-conquer algorithm.
Inputs:
n is a positive integer
b is a positive integer with exactly n bits
a is a nonnegative integer such that a < 2**n * b
Output:
(q, r) such that a = b*q+r and 0 <= r < b.
"""
if a.bit_length() - n <= _DIV_LIMIT:
return divmod(a, b)
pad = n & 1
if pad:
a <<= 1
b <<= 1
n += 1
half_n = n >> 1
mask = (1 << half_n) - 1
b1, b2 = b >> half_n, b & mask
q1, r = _div3n2n(a >> n, (a >> half_n) & mask, b, b1, b2, half_n)
q2, r = _div3n2n(r, a & mask, b, b1, b2, half_n)
if pad:
r >>= 1
return q1 << half_n | q2, r
def _div3n2n(a12, a3, b, b1, b2, n):
"""Helper function for _div2n1n; not intended to be called directly."""
if a12 >> n == b1:
q, r = (1 << n) - 1, a12 - (b1 << n) + b1
else:
q, r = _div2n1n(a12, b1, n)
r = (r << n | a3) - q * b2
while r < 0:
q -= 1
r += b
return q, r
def _int2digits(a, n):
"""Decompose non-negative int a into base 2**n
Input:
a is a non-negative integer
Output:
List of the digits of a in base 2**n in little-endian order,
meaning the most significant digit is last. The most
significant digit is guaranteed to be non-zero.
If a is 0 then the output is an empty list.
"""
a_digits = [0] * ((a.bit_length() + n - 1) // n)
def inner(x, L, R):
if L + 1 == R:
a_digits[L] = x
return
mid = (L + R) >> 1
shift = (mid - L) * n
upper = x >> shift
lower = x ^ (upper << shift)
inner(lower, L, mid)
inner(upper, mid, R)
if a:
inner(a, 0, len(a_digits))
return a_digits
def _digits2int(digits, n):
"""Combine base-2**n digits into an int. This function is the
inverse of `_int2digits`. For more details, see _int2digits.
"""
def inner(L, R):
if L + 1 == R:
return digits[L]
mid = (L + R) >> 1
shift = (mid - L) * n
return (inner(mid, R) << shift) + inner(L, mid)
return inner(0, len(digits)) if digits else 0
def _divmod_pos(a, b):
"""Divide a non-negative integer a by a positive integer b, giving
quotient and remainder."""
# Use grade-school algorithm in base 2**n, n = nbits(b)
n = b.bit_length()
a_digits = _int2digits(a, n)
r = 0
q_digits = []
for a_digit in reversed(a_digits):
q_digit, r = _div2n1n((r << n) + a_digit, b, n)
q_digits.append(q_digit)
q_digits.reverse()
q = _digits2int(q_digits, n)
return q, r
def int_divmod(a, b):
"""Asymptotically fast replacement for divmod, for 'int'.
Its time complexity is O(n**1.58), where n = #bits(a) + #bits(b).
"""
if b == 0:
raise ZeroDivisionError('division by zero')
elif b < 0:
q, r = int_divmod(-a, -b)
return q, -r
elif a < 0:
q, r = int_divmod(~a, b)
return ~q, b + ~r
else:
return _divmod_pos(a, b)
# Notes on _dec_str_to_int_inner:
#
# Stefan Pochmann worked up a str->int function that used the decimal
# module to, in effect, convert from base 10 to base 256. This is
# "unnatural", in that it requires multiplying and dividing by large
# powers of 2, which `decimal` isn't naturally suited to. But
# `decimal`'s `*` and `/` are asymptotically superior to CPython's, so
# at _some_ point it could be expected to win.
#
# Alas, the crossover point was too high to be of much real interest. I
# (Tim) then worked on ways to replace its division with multiplication
# by a cached reciprocal approximation instead, fixing up errors
# afterwards. This reduced the crossover point significantly,
#
# I revisited the code, and found ways to improve and simplify it. The
# crossover point is at about 3.4 million digits now.
#
# About .adjusted()
# -----------------
# Restrict to Decimal values x > 0. We don't use negative numbers in the
# code, and I don't want to have to keep typing, e.g., "absolute value".
#
# For convenience, I'll use `x.a` to mean `x.adjusted()`. x.a doesn't
# look at the digits of x, but instead returns an integer giving x's
# order of magnitude. These are equivalent:
#
# - x.a is the power-of-10 exponent of x's most significant digit.
# - x.a = the infinitely precise floor(log10(x))
# - x can be written in this form, where f is a real with 1 <= f < 10:
# x = f * 10**x.a
#
# Observation; if x is an integer, len(str(x)) = x.a + 1.
#
# Lemma 1: (x * y).a = x.a + y.a, or one larger
#
# Proof: Write x = f * 10**x.a and y = g * 10**y.a, where f and g are in
# [1, 10). Then x*y = f*g * 10**(x.a + y.a), where 1 <= f*g < 100. If
# f*g < 10, (x*y).a is x.a+y.a. Else divide f*g by 10 to bring it back
# into [1, 10], and add 1 to the exponent to compensate. Then (x*y).a is
# x.a+y.a+1.
#
# Lemma 2: ceiling(log10(x/y)) <= x.a - y.a + 1
#
# Proof: Express x and y as in Lemma 1. Then x/y = f/g * 10**(x.a -
# y.a), where 1/10 < f/g < 10. If 1 <= f/g, (x/y).a is x.a-y.a. Else
# multiply f/g by 10 to bring it back into [1, 10], and subtract 1 from
# the exponent to compensate. Then (x/y).a is x.a-y.a-1. So the largest
# (x/y).a can be is x.a-y.a. Since that's the floor of log10(x/y). the
# ceiling is at most 1 larger (with equality iff f/g = 1 exactly).
#
# GUARD digits
# ------------
# We only want the integer part of divisions, so don't need to build
# the full multiplication tree. But using _just_ the number of
# digits expected in the integer part ignores too much. What's left
# out can have a very significant effect on the quotient. So we use
# GUARD additional digits.
#
# The default 8 is more than enough so no more than 1 correction step
# was ever needed for all inputs tried through 2.5 billion digits. In
# fact, I believe 3 guard digits are always enough - but the proof is
# very involved, so better safe than sorry.
#
# Short course:
#
# If prec is the decimal precision in effect, and we're rounding down,
# the result of an operation is exactly equal to the infinitely precise
# result times 1-e for some real e with 0 <= e < 10**(1-prec). In
#
# ctx.prec = max(n.adjusted() - p256.adjusted(), 0) + GUARD
# hi = +n * +recip # unary `+` chops to ctx.prec digits
#
# we have 3 visible chopped operations, but there's also a 4th:
# precomputing a truncated `recip` as part of setup.
#
# So the computed product is exactly equal to the true product times
# (1-e1)*(1-e2)*(1-e3)*(1-e4); since the e's are all very small, an
# excellent approximation to the second factor is 1-(e1+e2+e3+e4) (the
# 2nd and higher order terms in the expanded product are too tiny to
# matter). If they're all as large as possible, that's
#
# 1 - 4*10**(1-prec). This, BTW, is all bog-standard FP error analysis.
#
# That implies the computed product is within 1 of the true product
# provided prec >= log10(true_product) + 1.602.
#
# Here are telegraphic details, rephrasing the initial condition in
# equivalent ways, step by step:
#
# prod - prod * (1 - 4*10**(1-prec)) <= 1
# prod - prod + prod * 4*10**(1-prec)) <= 1
# prod * 4*10**(1-prec)) <= 1
# 10**(log10(prod)) * 4*10**(1-prec)) <= 1
# 4*10**(1-prec+log10(prod))) <= 1
# 10**(1-prec+log10(prod))) <= 1/4
# 1-prec+log10(prod) <= log10(1/4) = -0.602
# -prec <= -1.602 - log10(prod)
# prec >= log10(prod) + 1.602
#
# The true product is the same as the true ratio n/p256. By Lemma 2
# above, n.a - p256.a + 1 is an upper bound on the ceiling of
# log10(prod). Then 2 is the ceiling of 1.602. so n.a - p256.a + 3 is an
# upper bound on the right hand side of the inequality. Any prec >= that
# will work.
#
# But since this is just a sketch of a proof ;-), the code uses the
# empirically tested 8 instead of 3. 5 digits more or less makes no
# practical difference to speed - these ints are huge. And while
# increasing GUARD above 3 may not be necessary, every increase cuts the
# percentage of cases that need a correction at all.
#
# On Computing Reciprocals
# ------------------------
# In general, the exact reciprocals we compute have over twice as many
# significant digits as needed. 1/256**i has the same number of
# significant decimal digits as 5**i. It's a significant waste of RAM
# to store all those unneeded digits.
#
# So we cut exact reciprocals back to the least precision that can
# be needed so that the error analysis above is valid,
#
# [Note: turns out it's very significantly faster to do it this way than
# to compute 1 / 256**i directly to the desired precision, because the
# power method doesn't require division. It's also faster than computing
# (1/256)**i directly to the desired precision - no material division
# there, but `compute_powers()` is much smarter about _how_ to compute
# all the powers needed than repeated applications of `**` - that
# function invokes `**` for at most the few smallest powers needed.]
#
# The hard part is that chopping back to a shorter width occurs
# _outside_ of `inner`. We can't know then what `prec` `inner()` will
# need. We have to pick, for each value of `w2`, the largest possible
# value `prec` can become when `inner()` is working on `w2`.
#
# This is the `prec` inner() uses:
# max(n.a - p256.a, 0) + GUARD
# and what setup uses (renaming its `v` to `p256` - same thing):
# p256.a + GUARD + 1
#
# We need that the second is always at least as large as the first,
# which is the same as requiring
#
# n.a - 2 * p256.a <= 1
#
# What's the largest n can be? n < 255**w = 256**(w2 + (w - w2)). The
# worst case in this context is when w ix even. and then w = 2*w2, so
# n < 256**(2*w2) = (256**w2)**2 = p256**2. By Lemma 1, then, n.a
# is at most p256.a + p256.a + 1.
#
# So the most n.a - 2 * p256.a can be is
# p256.a + p256.a + 1 - 2 * p256.a = 1. QED
#
# Note: an earlier version of the code split on floor(e/2) instead of on
# the ceiling. The worst case then is odd `w`, and a more involved proof
# was needed to show that adding 4 (instead of 1) may be necessary.
# Basically because, in that case, n may be up to 256 times larger than
# p256**2. Curiously enough, by splitting on the ceiling instead,
# nothing in any proof here actually depends on the output base (256).
# Enable for brute-force testing of compute_powers(). This takes about a
# minute, because it tries millions of cases.
if 0:
def consumer(w, limit, need_hi):
seen = set()
need = set()
def inner(w):
if w <= limit:
return
if w in seen:
return
seen.add(w)
lo = w >> 1
hi = w - lo
need.add(hi if need_hi else lo)
inner(lo)
inner(hi)
inner(w)
exp = compute_powers(w, 1, limit, need_hi=need_hi)
assert exp.keys() == need
from itertools import chain
for need_hi in (False, True):
for limit in (0, 1, 10, 100, 1_000, 10_000, 100_000):
for w in chain(range(1, 100_000),
(10**i for i in range(5, 30))):
consumer(w, limit, need_hi)
View Git Blame Copy Permalink