I've read the manual (bible) on ncb's, and have 2 questions for the SET part:

If I put the following down:

R (b1 !b2 !b2)

Will it have a 2/3rds chance of clearing bit 2 and a 1/3rd chance of setting bit one?

If the Player's Ship's name was the following:

Babylonian

and I did T128 where string 128 was the following:

**S.S. ***

Would the ship's new name be S.S. Babylonian ?

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1. No. The R( ) operator can have a max of two arguements. You can, however, nest the statements.

2. Yes.

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Quote

OrigR> **1. No. The R( ) operator can have a max of two arguements. You can, however, nest the statements.

**

Would this accomplish the desired result?

R(!b2 R(!b2 b1))

with it having a 2/3-chance of clearing bit 2 and 1/3-chance of setting bit one?

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See I'm the king of the swingers, the Jungle V.I.P.,
But I've reached the top and had to stop,
And that's what's
botherin' me.

(This message has been edited by Limax (edited 08-30-2002).)

(This message has been edited by Limax (edited 08-30-2002).)

(This message has been edited by Limax (edited 08-30-2002).)

Quote

Originally posted by Limax:
R(!b2 R(!b2 b1))

Would give you a 75% chance (50% in the first clause + 50% of 50% = 25% in the second) of clearing b2 and a 25% (50% of 50%)chance of setting b1.

Kidglove

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Compilers - the ultimate god games.

How would one acomplish a 1/3 probability of three things (or close enough)?

R( b1 R( b2 b3 ) ) would not do this...

R( R( b1 b2 ) R( b3 R( R( b1 b2 ) R( b3 R(...) would though in theory, but is impossible to do in a closed system (aka, not infinity). Therefore, how do we accomplish the R( R( b1 b2 ) R( b3 R( R( b1 b2 ) R( b3 R(...) equivalent in way which is possible to represent in the Nova engine?

Doing a R( R( b1 b2 ) R( b3 R( R( b1 b2 ) R( b3 R(...) for a while (but not to infinity, cut off with a slight bias towards either b1, b2, or b3), would produce a "good-enough" result. But can it get closer? A crön may get us there.

Using the iterative crön entry/exit (set Flags 0x0001), it may be possible to have the crön continue to do something similar to or in the form of R( R( b1 b2 ) R( b3 b4) ), where getting the value of b4 makes it go back and try again until it has either b1, b2, or b3 set. It essentually "finishes" with b4 never being set, and a 1/3 chance of the others being set. Two cröns may be needed to pull off the effect. I'll have to do more experimentation to see if I can get it to happen without any infinite loops (and actually meet the probabilities required).

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Some call me strange. Some call me weird. Some just call me Normal Guy.

couldn't you just do this

(b2|!b1)|b2

2/3 of b2 and 1/3 of !b1.. i think.

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No, no. That is for test expressions. They're talking about set expressions.

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Quote

Originally posted by Normal Guy:
**(idea with cröns and nested r-operators)
**

After trying it for a while, I came to the conclusion that the R operator doesn't work when you try to nest it. The expression I used "!b4 r(r(b1 b2) r(b3 b4))" in an iterative exit clause of a crön set all the bits b1, b2 and b3.

It didn't go into an endless loop, though, so I'm guessing it evaluated the r-operators before the !b4...

Kidglove

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Compilers - the ultimate god games.

Quote

Originally posted by Normal Guy:
**How would one acomplish a 1/3 probability of three things (or close enough)?

R( b1 R( b2 b3 ) ) would not do this...

R( R( b1 b2 ) R( b3 R( R( b1 b2 ) R( b3 R(...) would though in theory, but is impossible to do in a closed system (aka, not infinity). Therefore, how do we accomplish the R( R( b1 b2 ) R( b3 R( R( b1 b2 ) R( b3 R(...) equivalent in way which is possible to represent in the Nova engine?

Doing a R( R( b1 b2 ) R( b3 R( R( b1 b2 ) R( b3 R(...) for a while (but not to infinity, cut off with a slight bias towards either b1, b2, or b3), would produce a "good-enough" result. But can it get closer? A crön may get us there.

Using the iterative crön entry/exit (set Flags 0x0001), it may be possible to have the crön continue to do something similar to or in the form of R( R( b1 b2 ) R( b3 b4) ), where getting the value of b4 makes it go back and try again until it has either b1, b2, or b3 set. It essentually "finishes" with b4 never being set, and a 1/3 chance of the others being set. Two cröns may be needed to pull off the effect. I'll have to do more experimentation to see if I can get it to happen without any infinite loops (and actually meet the probabilities required).

**

Urrrgh... now I know why I haven't dared to do much Nova mission experimentation.

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Well, 6 tries so far and I haven't been able to get it to work. I need to learn more about the crön resource before I try again. Although it is possible that the crön I'm using is actually working as planned, the mďsns I've set up to test this don't seem to like me much. Oh well. I'll keep you all posted.

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Some call me strange. Some call me weird. Some just call me Normal Guy.

If you want to figure out how many times the cron iterates, you could have it give the player an outfit (IFF decoder?).

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Ta Da!
I have done it!

Using two bits (4 possible combos) and non-nested R() things (which aparently don't work anyway), it can be done.

crön:
EnableOn: !b1 & !b2
OnStart: R(b1 !b1) R(b2 !b2)
Use Iterative Entry
Random: 100
Duration: 0

mďsn 1 (or any other resource you wish to be effected by the 1/3 chance):
AvailBit: b1 & !b2

mďsn 2:
AvailBit !b1 & b2

mďsn 3:
AvailBit: b1 & b2

I've done enough trials to satisfy myself that these all have equal chance of occuring (1/3 for each!!!). If !b1 & !b2 is the case, the crön will reactivate and has a 75% chance of changing to useful values in b1 and b2. If that 25% comes up, it will retry until it has a useful value.

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Some call me strange. Some call me weird. Some just call me Normal Guy.

Quote

Originally posted by Sheer_falacy:
**If you want to figure out how many times the cron iterates, you could have it give the player an outfit (IFF decoder?).

**

For the most part, only one iteration per trail. Theoretically, you should have more than one iteration only about 25% of the time, and extremely rarely any more than that. In a quick trial of ten, only twice was there 2 (20%), and none more than that. Fractions of iterations are not possible of course, or you would expect 2.5 instances.

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Some call me strange. Some call me weird. Some just call me Normal Guy.

Excellent. Do you mind emailing me the file you made?

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