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#! /usr/bin/bc -l
#
# SPDX-License-Identifier: BSD-2-Clause
#
# Copyright (c) 2018-2025 Gavin D. Howard and contributors.
#
# Redistribution and use in source and binary forms, with or without
# modification, are permitted provided that the following conditions are met:
#
# * Redistributions of source code must retain the above copyright notice, this
# list of conditions and the following disclaimer.
#
# * Redistributions in binary form must reproduce the above copyright notice,
# this list of conditions and the following disclaimer in the documentation
# and/or other materials provided with the distribution.
#
# THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS IS"
# AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
# IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
# ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT HOLDER OR CONTRIBUTORS BE
# LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR
# CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF
# SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS
# INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN
# CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE)
# ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE
# POSSIBILITY OF SUCH DAMAGE.
#
# Adjust this number to try ranges above different powers of 10.
max = 0
n = (1 << max)
# Uncomment this to test the high part of the ranges.
#n += (1 - (1 >> 10))
n
# Loop from the start number to the next power of 10.
for (i = n; i < (n$ << 1); i += 1)
{
# This is the lower limit.
t1 = sqrt(1/(3*i))
l = length(i$)/2
print "i: ", i, "\n"
#print "l: ", l, "\n"
if (l$ != l)
{
# Limit between 2.4 and 3.
limit = 2.7 << (l$ * 2)
#print "limit: ", limit, "\n"
if (i >= limit)
{
t2 = 1/(i >> (l$)) * 2
}
else
{
t2 = 1/(i >> (l$))
}
}
else
{
# Limit between 3.8-ish and 4.8
limit = 4.3 << (l$ * 2 - 1)
#print "limit: ", limit, "\n"
if (i >= limit)
{
t2 = 1/(i >> (l$ - 1)) * 8
}
else
{
t2 = 1/(i >> (l$ - 1)) * 4
}
}
# This is the upper limit.
t3 = sqrt(5/(3*i))
# This is true when the guess is in between the limits.
good = (t1 < t2 && t2 < t3)
print t1, " < ", t2, " < ", t3, ": ", good, "\n"
# Error if we have a problem.
if (!good) sqrt(-1)
}
halt
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