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Posted by

David Haley USA (3,881 posts) Bio

Date

Reply #15 on Mon 03 Jul 2006 05:12 PM (UTC)

Message

%s is for strings, %d is for numbers. num_punct must be a function that converts doubles to strings. What is it that you want to do here? I guess I don't understand your question. :)

David Haley aka Ksilyan
Head Programmer,
Legends of the Darkstone

http://david.the-haleys.org

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Posted by

Metsuro USA (389 posts) Bio

Date

Reply #16 on Mon 03 Jul 2006 05:14 PM (UTC)

Message

oh... wow I didn't even notice that...

Everything turns around in the end

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Posted by Nick Gammon Australia (23,163 posts) Bio Forum Administrator

Date Reply #17 on Tue 04 Jul 2006 11:55 PM (UTC)

Message

Quote:

But for your case, I suspect it might be constant propagation ...

I can believe that for the case:



print (((1 / 3) * 3) == 1) --> true

However if I do these as separate commands in the command window, Lua can hardly know to optimize away a divide when it hasn't seen the multiply yet:



/a = 1/3 

/print (a) --> 0.33333333333333

/print (a * 3)  --> 1

/print (a * 3 == 1)  --> true

- Nick Gammon

www.gammon.com.au, www.mushclient.com

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Posted by

David Haley USA (3,881 posts) Bio

Date

Reply #18 on Wed 05 Jul 2006 04:30 AM (UTC)

Message

Yeah, you're right, it is pretty surprising. I did some further experiments of my own, trying to confuse it, and in order for it to do constant folding it would have to go back across the previous assignments of all variables which I truly doubt Lua does.

Perhaps doubles have some way of noting 'infinity', so that 1/3 is represented as 0.3bar (where 'bar' means 'repeat forever'). Mathematically 0.9bar is defined as 1 (I don't like that definition, but it's the "rule"), so 0.3bar times three would be 0.9bar which would be one.

If I had the time I'd look into how all this actually works. Maybe I will, in fact, sometime soon... this is quite interesting. :)

David Haley aka Ksilyan
Head Programmer,
Legends of the Darkstone

http://david.the-haleys.org

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Posted by

Nick Gammon Australia (23,163 posts) Bio Forum Administrator

Date

Reply #19 on Wed 05 Jul 2006 09:19 AM (UTC)

Message

I think you are making an assumption here, and that assumption is that numbers are stored internally to the base 10.

Hypothetically, if doubles were stored to the base 3, then 1/3 could be represented exactly.

More realistically, the scenario of 1/3 * 3, might result in a number that is so close to 1, that the hardware rounds it up to 1.

- Nick Gammon

www.gammon.com.au, www.mushclient.com

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Posted by

David Haley USA (3,881 posts) Bio

Date

Reply #20 on Wed 05 Jul 2006 01:41 PM (UTC)

Message

Well, numbers are stored to the base 2 sort of a fortiori, no? If I remember correctly, floating point numbers have two components to them: the digits and the exponent, and those are both basically binary numbers and the hardware knows how to convert them to the proper form.

Besides, this also works for 1/7 * 7, and the base 3 assumption wouldn't really help there.

It seems to me that rounding is more likely what's going on here. I'm sure you could find something such that playing around with it would in fact break it.

David Haley aka Ksilyan
Head Programmer,
Legends of the Darkstone

http://david.the-haleys.org

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Posted by Nick Gammon Australia (23,163 posts) Bio Forum Administrator

Date Reply #21 on Wed 05 Jul 2006 09:37 PM (UTC)

Message

Quote:

Well, numbers are stored to the base 2 sort of a fortiori, no?

Well, probably they are today, yes. However I am pointing out the assumption there. Years ago I used to work on Burroughs Medium Systems (B 4700 etc.) and they stored memory addressable to 4 bits, not 8 bits as is commonly done these days. The 4-bit unit was called a "nibble" (half a byte), and I think also a "digit". Since 4 bits could represent 0-9, plus A-F, numbers were stored decimally. Thus the number 12345 would be stored in memory as (in hex): 12345.

Quote:

I'm sure you could find something such that playing around with it would in fact break it.

A script seems called for:



for i = 1, 1000 do
  j = 1 / i
  k = j * i
  assert (k == 1, "Failed at " .. i)
end -- for

When run I get: Failed at 49

This confirms it:



print (string.format ("%2.20f", ((1 / 49) * 49)))  --> 0.99999999999999989000

If you juggle around the script and work with a starting number other than 1 you can make it fail earlier or later:


base = 1000

for i = 1, 100000 do
  j = base / i
  k = j * i
  assert (k == base, "Failed at " .. i)
end -- for

This shows: Failed at 15

I think the point of the Lua developers was that doubles will work fine for whole numbers up to a fairly large number.

If you are in fact going to be doing division on your currency, then long long will probably be worse anyway, as it simply will throw away information much faster than a double will.

For example:



long long i = 1;

long long j = i / 2;

long long k = j * 2; // k is now zero, not 1

- Nick Gammon

www.gammon.com.au, www.mushclient.com

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