Working within Solidity and the Ethereum EVM and Decimals don't exist. Is there a way I could mathematically still create a Poisson distribution using integers ? it doesnt have to be perfect, i.e rounding or losing some digits may be acceptable.
Let me preface by stating that what follows is not going to be (directly) helpful to you with etherium/solidity. However, it produces probability tables that you might be able to use for your work.
I ended up intrigued by the question of how accurate you could be in expressing the Poisson probabilities as rationals, so I put together the following script in Ruby to try things out:
def rational_poisson(lmbda)
Hash.new.tap do |h| # create a hash and pass it to this block as 'h'.
# Make all components of the calculations rational to allow
# cancellations to occur wherever possible when dividing
e_to_minus_lambda = Math.exp(-lmbda).to_r
factorial = 1r
lmbda = lmbda.to_r
power = 1r
(0...).each do |x|
unless x == 0
power *= lmbda
factorial *= x
end
value = (e_to_minus_lambda / factorial) * power
# the following double inversion/conversion bounds the result
# by the significant bits in the mantissa of a float
approx = Rational(1, (1 / value).to_f)
h[x] = approx
break if x > lmbda && approx.numerator <= 1
end
end
end
if __FILE__ == $PROGRAM_NAME
lmbda = (ARGV.shift || 2.0).to_f # read in a lambda (defaults to 2.0)
pmf = rational_poisson(lmbda) # create the pmf for a Poisson with that lambda
pmf.each { |key, value| puts "p(#{key}) = #{value} = #{value.to_f}" }
puts "cumulative error = #{1.0 - pmf.values.inject(&:+)}" # does it sum to 1?
end
Things to know as you glance through the code. Appending .to_r to a value or expression converts it to a rational, i.e., a ratio of two integers; values with an r suffix are rational constants; and (0...).each is an open-ended iterator which will loop until the break condition is met.
That little script produces results such as:
localhost:pjs$ ruby poisson_rational.rb 1.0
p(0) = 2251799813685248/6121026514868073 = 0.36787944117144233
p(1) = 2251799813685248/6121026514868073 = 0.36787944117144233
p(2) = 1125899906842624/6121026514868073 = 0.18393972058572117
p(3) = 281474976710656/4590769886151055 = 0.061313240195240384
p(4) = 70368744177664/4590769886151055 = 0.015328310048810096
p(5) = 17592186044416/5738462357688819 = 0.003065662009762019
p(6) = 1099511627776/2151923384133307 = 0.0005109436682936699
p(7) = 274877906944/3765865922233287 = 7.299195261338141e-05
p(8) = 34359738368/3765865922233287 = 9.123994076672677e-06
p(9) = 67108864/66196861914257 = 1.0137771196302974e-06
p(10) = 33554432/330984309571285 = 1.0137771196302975e-07
p(11) = 33554432/3640827405284135 = 9.216155633002704e-09
p(12) = 4194304/5461241107926203 = 7.68012969416892e-10
p(13) = 524288/8874516800380079 = 5.907792072437631e-11
p(14) = 32768/7765202200332569 = 4.2198514803125934e-12
p(15) = 256/909984632851473 = 2.8132343202083955e-13
p(16) = 16/909984632851473 = 1.7582714501302472e-14
p(17) = 1/966858672404690 = 1.0342773236060278e-15
cumulative error = 0.0
I was playing around with Ruby and Latex to create a color coding set for a registor. I have the following block of code. When attempting to run this, band1 = 1e+02.
I tried band1 = (BigDecimal(i) * 100).to_f, thinking maybe there was some odd floating point issue. An integer multiplied by an integer should create an integer. I tried a variety of other things as well, but to no avail.
(1..9).each do |i| #Band 1
(0..9).each do |j| #Band 2
(0..11).each do |k| #Band 3
#Band 3 Start
#these are the colors of the resistor bands
b1 = $c_band12[i]
b2 = $c_band12[j]
b3 = $c_band3[k]
b4 = "Gold"
oms = ((i*100) + (j*10)) * $mult[k]
band1 = i*100
band2 = j
band3 = $mult[k]
end
end
end
Not sure what I'm missing. Should I be using each_with_index through these iterations? I tried this:
(1..9).each_with_index {|i, indexi| #Band 1
(0..9).each_with_index {|j, indexj| #Band 2
(0..11).each_with_index {|k, indexk| #Band 3
#Band 3 Start
#these are the colors of the resistor bands
b1 = $c_band12[i]
b2 = $c_band12[j]
b3 = $c_band3[k]
b4 = "Gold"
oms = ((i*100) + (j*10)) * $mult[k]
band1 = indexk * 100
and I got the same answer. I can't see why 1*100 should equate to such a large number.
edit: Additional info: If I have this: band1=i*10
then the calculation is correct. In fact, the calculation is correct up to 99.
In your code, band1 has to be a Fixnum. Check with p band1.class. Not sure how you get "1e+02", maybe you print in some strange fashion, or you do band1 == 1e+02 which returns true in Ruby. You must use eql? to distinguish between 1 and 1.0:
1 == 1.0 # => true
1.eql?(1.0) # => false
Ruby may not be the optimal language for this but I'm sort of comfortable working with this in my terminal so that's what I'm going with.
I need to process the numbers from 1 to 666666 so I pin out all the numbers that contain 6 but doesn't contain 7, 8 or 9. The first number will be 6, the next 16, then 26 and so forth.
Then I needed it printed like this (6=6) (16=6) (26=6) and when I have ranges like 60 to 66 I need it printed like (60 THRU 66=6) (SPSS syntax).
I have this code and it works but it's neither beautiful nor very efficient so how could I optimize it?
(silly code may follow)
class Array
def to_ranges
array = self.compact.uniq.sort
ranges = []
if !array.empty?
# Initialize the left and right endpoints of the range
left, right = array.first, nil
array.each do |obj|
# If the right endpoint is set and obj is not equal to right's successor
# then we need to create a range.
if right && obj != right.succ
ranges << Range.new(left,right)
left = obj
end
right = obj
end
ranges << Range.new(left,right) unless left == right
end
ranges
end
end
write = ""
numbers = (1..666666).to_a
# split each number in an array containing it's ciphers
numbers = numbers.map { |i| i.to_s.split(//) }
# delete the arrays that doesn't contain 6 and the ones that contains 6 but also 8, 7 and 9
numbers = numbers.delete_if { |i| !i.include?('6') }
numbers = numbers.delete_if { |i| i.include?('7') }
numbers = numbers.delete_if { |i| i.include?('8') }
numbers = numbers.delete_if { |i| i.include?('9') }
# join the ciphers back into the original numbers
numbers = numbers.map { |i| i.join }
numbers = numbers.map { |i| i = Integer(i) }
# rangify consecutive numbers
numbers = numbers.to_ranges
# edit the ranges that go from 1..1 into just 1
numbers = numbers.map do |i|
if i.first == i.last
i = i.first
else
i = i
end
end
# string stuff
numbers = numbers.map { |i| i.to_s.gsub(".."," thru ") }
numbers = numbers.map { |i| "(" + i.to_s + "=6)"}
numbers.each { |i| write << " " + i }
File.open('numbers.txt','w') { |f| f.write(write) }
As I said it works for numbers even in the millions - but I'd like some advice on how to make prettier and more efficient.
I deleted my earlier attempt to parlez-vous-ruby? and made up for that. I know have an optimized version of x3ro's excellent example.
$,="\n"
puts ["(0=6)", "(6=6)", *(1.."66666".to_i(7)).collect {|i| i.to_s 7}.collect do |s|
s.include?('6')? "(#{s}0 THRU #{s}6=6)" : "(#{s}6=6)"
end ]
Compared to x3ro's version
... It is down to three lines
... 204.2 x faster (to 66666666)
... has byte-identical output
It uses all my ideas for optimization
gen numbers based on modulo 7 digits (so base-7 numbers)
generate the last digit 'smart': this is what compresses the ranges
So... what are the timings? This was testing with 8 digits (to 66666666, or 823544 lines of output):
$ time ./x3ro.rb > /dev/null
real 8m37.749s
user 8m36.700s
sys 0m0.976s
$ time ./my.rb > /dev/null
real 0m2.535s
user 0m2.460s
sys 0m0.072s
Even though the performance is actually good, it isn't even close to the C optimized version I posted before: I couldn't run my.rb to 6666666666 (6x10) because of OutOfMemory. When running to 9 digits, this is the comparative result:
sehe#meerkat:/tmp$ time ./my.rb > /dev/null
real 0m21.764s
user 0m21.289s
sys 0m0.476s
sehe#meerkat:/tmp$ time ./t2 > /dev/null
real 0m1.424s
user 0m1.408s
sys 0m0.012s
The C version is still some 15x faster... which is only fair considering that it runs on the bare metal.
Hope you enjoyed it, and can I please have your votes if only for learning Ruby for the purpose :)
(Can you tell I'm proud? This is my first encounter with ruby; I started the ruby koans 2 hours ago...)
Edit by #johndouthat:
Very nice! The use of base7 is very clever and this a great job for your first ruby trial :)
Here's a slight modification of your snippet that will let you test 10+ digits without getting an OutOfMemory error:
puts ["(0=6)", "(6=6)"]
(1.."66666666".to_i(7)).each do |i|
s = i.to_s(7)
puts s.include?('6') ? "(#{s}0 THRU #{s}6=6)" : "(#{s}6=6)"
end
# before:
real 0m26.714s
user 0m23.368s
sys 0m2.865s
# after
real 0m15.894s
user 0m13.258s
sys 0m1.724s
Exploiting patterns in the numbers, you can short-circuit lots of the loops, like this:
If you define a prefix as the 100s place and everything before it,
and define the suffix as everything in the 10s and 1s place, then, looping
through each possible prefix:
If the prefix is blank (i.e. you're testing 0-99), then there are 13 possible matches
elsif the prefix contains a 7, 8, or 9, there are no possible matches.
elsif the prefix contains a 6, there are 49 possible matches (a 7x7 grid)
else, there are 13 possible matches. (see the image below)
(the code doesn't yet exclude numbers that aren't specifically in the range, but it's pretty close)
number_range = (1..666_666)
prefix_range = ((number_range.first / 100)..(number_range.last / 100))
for p in prefix_range
ps = p.to_s
# TODO: if p == prefix_range.last or p == prefix_range.first,
# TODO: test to see if number_range.include?("#{ps}6".to_i), etc...
if ps == '0'
puts "(6=6) (16=6) (26=6) (36=6) (46=6) (56=6) (60 thru 66) "
elsif ps =~ /7|8|9/
# there are no candidate suffixes if the prefix contains 7, 8, or 9.
elsif ps =~ /6/
# If the prefix contains a 6, then there are 49 candidate suffixes
for i in (0..6)
print "(#{ps}#{i}0 thru #{ps}#{i}6) "
end
puts
else
# If the prefix doesn't contain 6, 7, 8, or 9, then there are only 13 candidate suffixes.
puts "(#{ps}06=6) (#{ps}16=6) (#{ps}26=6) (#{ps}36=6) (#{ps}46=6) (#{ps}56=6) (#{ps}60 thru #{ps}66) "
end
end
Which prints out the following:
(6=6) (16=6) (26=6) (36=6) (46=6) (56=6) (60 thru 66)
(106=6) (116=6) (126=6) (136=6) (146=6) (156=6) (160 thru 166)
(206=6) (216=6) (226=6) (236=6) (246=6) (256=6) (260 thru 266)
(306=6) (316=6) (326=6) (336=6) (346=6) (356=6) (360 thru 366)
(406=6) (416=6) (426=6) (436=6) (446=6) (456=6) (460 thru 466)
(506=6) (516=6) (526=6) (536=6) (546=6) (556=6) (560 thru 566)
(600 thru 606) (610 thru 616) (620 thru 626) (630 thru 636) (640 thru 646) (650 thru 656) (660 thru 666)
(1006=6) (1016=6) (1026=6) (1036=6) (1046=6) (1056=6) (1060 thru 1066)
(1106=6) (1116=6) (1126=6) (1136=6) (1146=6) (1156=6) (1160 thru 1166)
(1206=6) (1216=6) (1226=6) (1236=6) (1246=6) (1256=6) (1260 thru 1266)
(1306=6) (1316=6) (1326=6) (1336=6) (1346=6) (1356=6) (1360 thru 1366)
(1406=6) (1416=6) (1426=6) (1436=6) (1446=6) (1456=6) (1460 thru 1466)
(1506=6) (1516=6) (1526=6) (1536=6) (1546=6) (1556=6) (1560 thru 1566)
(1600 thru 1606) (1610 thru 1616) (1620 thru 1626) (1630 thru 1636) (1640 thru 1646) (1650 thru 1656) (1660 thru 1666)
etc...
Note I don't speak ruby, but I intend to dohave done a ruby version later just for speed comparison :)
If you just iterate all numbers from 0 to 117648 (ruby <<< 'print "666666".to_i(7)') and print them in base-7 notation, you'll at least have discarded any numbers containing 7,8,9. This includes the optimization suggestion by MrE, apart from lifting the problem to simple int arithmetic instead of char-sequence manipulations.
All that remains, is to check for the presence of at least one 6. This would make the algorithm skip at most 6 items in a row, so I deem it less unimportant (the average number of skippable items on the total range is 40%).
Simple benchmark to 6666666666
(Note that this means outputting 222,009,073 (222M) lines of 6-y numbers)
Staying close to this idea, I wrote this quite highly optimized C code (I don't speak ruby) to demonstrate the idea. I ran it to 282475248 (congruent to 6666666666 (mod 7)) so it was more of a benchmark to measure: 0m26.5s
#include <stdio.h>
static char buf[11];
char* const bufend = buf+10;
char* genbase7(int n)
{
char* it = bufend; int has6 = 0;
do
{
has6 |= 6 == (*--it = n%7);
n/=7;
} while(n);
return has6? it : 0;
}
void asciify(char* rawdigits)
{
do { *rawdigits += '0'; }
while (++rawdigits != bufend);
}
int main()
{
*bufend = 0; // init
long i;
for (i=6; i<=282475248; i++)
{
char* b7 = genbase7(i);
if (b7)
{
asciify(b7);
puts(b7);
}
}
}
I also benchmarked another approach, which unsurprisingly ran in less than half the time because
this version directly manipulates the results in ascii string form, ready for display
this version shortcuts the has6 flag for deeper recursion levels
this version also optimizes the 'twiddling' of the last digit when it is required to be '6'
the code is simply shorter...
Running time: 0m12.8s
#include <stdio.h>
#include <string.h>
inline void recursive_permute2(char* const b, char* const m, char* const e, int has6)
{
if (m<e)
for (*m = '0'; *m<'7'; (*m)++)
recursive_permute2(b, m+1, e, has6 || (*m=='6'));
else
if (has6)
for (*e = '0'; *e<'7'; (*e)++)
puts(b);
else /* optimize for last digit must be 6 */
puts((*e='6', b));
}
inline void recursive_permute(char* const b, char* const e)
{
recursive_permute2(b, b, e-1, 0);
}
int main()
{
char buf[] = "0000000000";
recursive_permute(buf, buf+sizeof(buf)/sizeof(*buf)-1);
}
Benchmarks measured with:
gcc -O4 t6.c -o t6
time ./t6 > /dev/null
$range_start = -1
$range_end = -1
$f = File.open('numbers.txt','w')
def output_number(i)
if $range_end == i-1
$range_end = i
elsif $range_start < $range_end
$f.puts "(#{$range_start} thru #{$range_end})"
$range_start = $range_end = i
else
$f.puts "(#{$range_start}=6)" if $range_start > 0 # no range, print out previous number
$range_start = $range_end = i
end
end
'1'.upto('666') do |n|
next unless n =~ /6/ # keep only numbers that contain 6
next if n =~ /[789]/ # remove nubmers that contain 7, 8 or 9
output_number n.to_i
end
if $range_start < $range_end
$f.puts "(#{$range_start} thru #{$range_end})"
end
$f.close
puts "Ruby is beautiful :)"
I came up with this piece of code, which I tried to keep more or less in FP-styling. Probably not much more efficient (as it has been said, with basic number logic you will be able to increase performance, for example by skipping from 19xx to 2000 directly, but that I will leave up to you :)
def check(n)
n = n.to_s
n.include?('6') and
not n.include?('7') and
not n.include?('8') and
not n.include?('9')
end
def spss(ranges)
ranges.each do |range|
if range.first === range.last
puts "(" + range.first.to_s + "=6)"
else
puts "(" + range.first.to_s + " THRU " + range.last.to_s + "=6)"
end
end
end
range = (1..666666)
range = range.select { |n| check(n) }
range = range.inject([0..0]) do |ranges, n|
temp = ranges.last
if temp.last + 1 === n
ranges.pop
ranges.push(temp.first..n)
else
ranges.push(n..n)
end
end
spss(range)
My first answer was trying to be too clever. Here is a much simpler version
class MutablePrintingCandidateRange < Struct.new(:first, :last)
def to_s
if self.first == nil and self.last == nil
''
elsif self.first == self.last
"(#{self.first}=6)"
else
"(#{self.first} thru #{self.last})"
end
end
def <<(x)
if self.first == nil and self.last == nil
self.first = self.last = x
elsif self.last == x - 1
self.last = x
else
puts(self) # print the candidates
self.first = self.last = x # reset the range
end
end
end
and how to use it:
numer_range = (1..666_666)
current_range = MutablePrintingCandidateRange.new
for i in numer_range
candidate = i.to_s
if candidate =~ /6/ and candidate !~ /7|8|9/
# number contains a 6, but not a 7, 8, or 9
current_range << i
end
end
puts current_range
Basic observation: If the current number is (say) 1900 you know that you can safely skip up to at least 2000...
(I didn't bother updating my C solution for formatting. Instead I went with x3ro's excellent ruby version and optimized that)
Undeleted:
I still am not sure whether the changed range-notation behaviour isn't actually what the OP wants: This version changes the behaviour of breaking up ranges that are actually contiguous modulo 6; I wouldn't be surprised the OP actually expected
.
....
(555536=6)
(555546=6)
(555556 THRU 666666=6)
instead of
....
(666640 THRU 666646=6)
(666650 THRU 666656=6)
(666660 THRU 666666=6)
I'll let the OP decide, and here is the modified version, which runs in 18% of the time as x3ro's version (3.2s instead of 17.0s when generating up to 6666666 (7x6)).
def check(n)
n.to_s(7).include?('6')
end
def spss(ranges)
ranges.each do |range|
if range.first === range.last
puts "(" + range.first.to_s(7) + "=6)"
else
puts "(" + range.first.to_s(7) + " THRU " + range.last.to_s(7) + "=6)"
end
end
end
range = (1..117648)
range = range.select { |n| check(n) }
range = range.inject([0..0]) do |ranges, n|
temp = ranges.last
if temp.last + 1 === n
ranges.pop
ranges.push(temp.first..n)
else
ranges.push(n..n)
end
end
spss(range)
My answer below is not complete, but just to show a path (I might come back and continue the answer):
There are only two cases:
1) All the digits besides the lowest one is either absent or not 6
6, 16, ...
2) At least one digit besides the lowest one includes 6
60--66, 160--166, 600--606, ...
Cases in (1) do not include any continuous numbers because they all have 6 in the lowest digit, and are different from one another. Cases in (2) all appear as continuous ranges where the lowest digit continues from 0 to 6. Any single continuation in (2) is not continuous with another one in (2) or with anything from (1) because a number one less than xxxxx0 will be xxxxy9, and a number one more than xxxxxx6 will be xxxxxx7, and hence be excluded.
Therefore, the question reduces to the following:
3)
Get all strings between "" to "66666" that do not include "6"
For each of them ("xxx"), output the string "(xxx6=6)"
4)
Get all strings between "" to "66666" that include at least one "6"
For each of them ("xxx"), output the string "(xxx0 THRU xxx6=6)"
The killer here is
numbers = (1..666666).to_a
Range supports iterations so you would be better off by going over the whole range and accumulating numbers that include your segments in blocks. When one block is finished and supplanted by another you could write it out.
The string in question (read from a file):
if (true) then
{
_this = createVehicle ["Land_hut10", [6226.8901, 986.091, 4.5776367e-005], [], 0, "CAN_COLLIDE"];
_vehicle_10 = _this;
_this setDir -2.109278;
};
Retrieved from a large list of similar (all same file) strings via the following:
get_stringR(string,"if","};")
And the function code:
def get_stringR(a,b,c)
b = a.index(b)
b ||= 0
c = a.rindex(c)
c ||= b
r = a[b,c]
return r
end
As so far, this works fine, but what I wanted to do is select the array after "createVehicle", the following (I thought) should work.
newstring = get_string(myString,"\[","\];")
Note get_string is the same as get_stringR, except it uses the first occurrence of the pattern both times, rather then the first and last occurrence.
The output should have been: ["Land_hut10", [6226.8901, 986.091, 4.5776367e-005], [], 0, "CAN_COLLIDE"];
Instead it was the below, given via 'puts':
["Land_hut10", [6226.8901, 986.091, 4.5776367e-005], [], 0, "CAN_COLLIDE"];
_vehicle_10 = _this;
_this setDir
Some 40 characters past the point it should have retrieve, which was very strange...
Second note, using both get_string and get_stringR produced the exact same result with the parameters given.
I then decided to add the following to my get_string code:
b = a.index(b)
b ||= 0
c = a.index(c)
c ||= b
if c > 40 then
c -= 40
end
r = a[b,c]
return r
And it works as expected (for every 'block' in the file, even though the strings after that array are not identical in any way), but something obviously isn't right :).
You want r = a[b..c] instead of r = a[b,c].
Difference is: b..c = start from b, go to c, while b,c = start from b and move c characters to the right.
Edit: You don't have to/shouldn't escape the [ and ] either, because you are using strings and not regexen. Also, you have to take the length of the end ("];") into consideration, or you will cut off parts of the end.
def get_stringR(a,b,c)
bp = a.index(b) || 0
cp = a.rindex(c) || bp
r = a[bp..cp + c.size - 1]
return r
end
def get_string(a,b,c)
bp = a.index(b) || 0
cp = a.index(c) || bp
r = a[bp..cp + c.size - 1]
return r
end
I need to be able to determine a systems maximum integer in Ruby. Anybody know how, or if it's possible?
FIXNUM_MAX = (2**(0.size * 8 -2) -1)
FIXNUM_MIN = -(2**(0.size * 8 -2))
Ruby automatically converts integers to a large integer class when they overflow, so there's (practically) no limit to how big they can be.
If you are looking for the machine's size, i.e. 64- or 32-bit, I found this trick at ruby-forum.com:
machine_bytes = ['foo'].pack('p').size
machine_bits = machine_bytes * 8
machine_max_signed = 2**(machine_bits-1) - 1
machine_max_unsigned = 2**machine_bits - 1
If you are looking for the size of Fixnum objects (integers small enough to store in a single machine word), you can call 0.size to get the number of bytes. I would guess it should be 4 on 32-bit builds, but I can't test that right now. Also, the largest Fixnum is apparently 2**30 - 1 (or 2**62 - 1), because one bit is used to mark it as an integer instead of an object reference.
Reading the friendly manual? Who'd want to do that?
start = Time.now
largest_known_fixnum = 1
smallest_known_bignum = nil
until smallest_known_bignum == largest_known_fixnum + 1
if smallest_known_bignum.nil?
next_number_to_try = largest_known_fixnum * 1000
else
next_number_to_try = (smallest_known_bignum + largest_known_fixnum) / 2 # Geometric mean would be more efficient, but more risky
end
if next_number_to_try <= largest_known_fixnum ||
smallest_known_bignum && next_number_to_try >= smallest_known_bignum
raise "Can't happen case"
end
case next_number_to_try
when Bignum then smallest_known_bignum = next_number_to_try
when Fixnum then largest_known_fixnum = next_number_to_try
else raise "Can't happen case"
end
end
finish = Time.now
puts "The largest fixnum is #{largest_known_fixnum}"
puts "The smallest bignum is #{smallest_known_bignum}"
puts "Calculation took #{finish - start} seconds"
In ruby Fixnums are automatically converted to Bignums.
To find the highest possible Fixnum you could do something like this:
class Fixnum
N_BYTES = [42].pack('i').size
N_BITS = N_BYTES * 8
MAX = 2 ** (N_BITS - 2) - 1
MIN = -MAX - 1
end
p(Fixnum::MAX)
Shamelessly ripped from a ruby-talk discussion. Look there for more details.
There is no maximum since Ruby 2.4, as Bignum and Fixnum got unified into Integer. see Feature #12005
> (2 << 1000).is_a? Fixnum
(irb):322: warning: constant ::Fixnum is deprecated
=> true
> 1.is_a? Bignum
(irb):314: warning: constant ::Bignum is deprecated
=> true
> (2 << 1000).class
=> Integer
There won't be any overflow, what would happen is an out of memory.
as #Jörg W Mittag pointed out: in jruby, fix num size is always 8 bytes long. This code snippet shows the truth:
fmax = ->{
if RUBY_PLATFORM == 'java'
2**63 - 1
else
2**(0.size * 8 - 2) - 1
end
}.call
p fmax.class # Fixnum
fmax = fmax + 1
p fmax.class #Bignum