logo Sign In

DanielB

This user has been banned.

User Group
Banned Members
Join date
15-Jul-2004
Last activity
5-Oct-2005
Posts
594

Post History

Post
#88906
Topic
<strong>The &quot;EditDroid&quot; Trilogy DVD Info and Feedback Thread</strong> (Released)
Link
https://originaltrilogy.com/post/id/88906/action/topic#88906
Time
Originally posted by: zion
I think the problem comes in where people think I'm going insane and flame is shooting out of my eyesockets because I used a cussword. Which isn't so. I guarantee almost everyone here, in the course of their normal day, cusses at least 10 times.
Well I don't do that. No, really, I don't. But the difference is I know that's how most people are and I'm used to it. The bottom line is that Jay doesn't have a problem with it for the most part and I'm not going to do anything about it except in extreme cases.
Neither do I, and even if that is normal, it doesn't make it okay. You can't excuse bad behaviour with "but everyone else does it".
Post
#88905
Topic
<strong>The &quot;EditDroid&quot; Trilogy DVD Info and Feedback Thread</strong> (Released)
Link
https://originaltrilogy.com/post/id/88905/action/topic#88905
Time
Originally posted by: Rikter
REALLY - MVerta does not have one HOW-TO do anything I've seen HE ONLY shows WHAT he has DONE -

Laserman is HELPING people and so (is) MeBeJedi...

- ALL Mike Verta is doing is Showing off a set and the steps he's taken and he's made it clear that NO ONE will see this set all he cares about is selling T-Shirts that are not being delievered in a timely manner if at all.

Why are you so angry, all I asked is why not share?
I concur Rik. However, my shirt was delivered in a timely-manner, so unless you are talking from personal experience it seems a bit daunting to use it against him. Certainly though, he is only interested in making money from them - but they are good shirts.
Post
#88901
Topic
<strong>The &quot;EditDroid&quot; Trilogy DVD Info and Feedback Thread</strong> (Released)
Link
https://originaltrilogy.com/post/id/88901/action/topic#88901
Time
And DanielB, go fuck yourself. I'm surprised your head hasn't imploded due to the black-hole level vaccuum currently sucking at the inside of your skull. Stop talking. Having someone as ass-backwards as you attempt to moralize to ANYONE about ANYTHING is the height of absurdity, as far as I'm concerned. Either that or your weird as hell justification for your asinine ramblings. "Lay off me, I made a girl kill herself." Next thing you know, you'll be telling us about how you caused a grandmother to jump from a building and impale her grandchildren on railroad spikes, but you had to brush it off because "That's just how you are." or something. Save it. I don't need your empty-headed speechifying. So shut the fuck up.
Oh how very mature. Firstly the girl has not killed herself, secondly I was the one who dragged her out of constant-hatred in the first place, so it can't be so surprising she returned to it. Thirdly the entire situation hurts me enough as it is, without someone as clueless as you trying to somehow blame me for it.

Yes, I probably could have done more. I am only human, and it isn't my responsibility. And I really don't appreciate you attacking me like that at all, I mean I know I'm not perfect, but I really did not have anything to do with her going back to a hatful attitude. And you do not have any right to claim otherwise. I believe I have a responsibility to do the best by people I value. But it isn't my responsibility to parent her. I know it sounds terrible, I can't seem to express it exactly the way I feel.

Bizzle, please talk civilly on these forums. What right do you have to claim that I shouldn't post here if you don't post maturely yourself? Is this another "do as I say not as I do" situations? I'm not going to post negativity back to you Bizzle – I understand you have feelings and I don't want to hurt them. So you can flame the forum all you want, but at the end of the day it will be you disrupting the forum with hate and flame, not me. I reckon almost everyone here would agree to conduct their posts here maturely (with no hate or flame). Oh, and Rik is right. Children view these forums.

Now for the sake of the forums I will ask you, and anyone else who wishes to insult me to at least have the decency to take it off the forums and into PM's.

One final thing, Rik I generally agree with you. When it comes to someone doing a personal project and making it publicly known, I don't take issue with that. That's fine by me and certainly welcome. I don't believe this is the place for it however, since this forum is for original trilogy preservation, and since you are not helping the preservation efforts it should be taken off these forums.

However, when someone decides to distribute their preservation effort, I do not believe they have the right to dictate how it is distributed. This is because the creator owns his own project's DVD as much as I do. Neither of us own it, Lucas is the true owner. I know that might turn some stomaches, however if I did the same thing (and I mean the same thing in theory, it doesn't have to be a SW DVD) then I would not expect to dictate how it is distributed. If I ever author a SW preservation DVD, you would expect to find the caption:

Let us not loose sight of the purpose of preservation.

While I created this set, from many hours of hard-labour, they contain the property of George Lucas. I do not claim ownership to any part herein and I feel that the true owner is Lucas. This set is designed to preserve the trilogy for home use until such a time as it is made legally available by George Lucas. When that happens, please show Lucas your support by purchasing the official product –DanielB.
Post
#88925
Topic
Riddles
Link
https://originaltrilogy.com/post/id/88925/action/topic#88925
Time
I may as well explain how phase 2 works, however firstly let's look at our averages going into it:

~21 people with no token.
~75 people with 1 token.
~3 people with arbitary tokens (2-20).
1 counter

In phase 2 what we aim to do is pair-up all the single tokens, and let the counter collect as well. So after the first cycle you want:

~64 people with no token.
~32 people with 2 tokens.
~3 people with arbitary tokens.
1 counter

Then you want to pair-up all the people with 2 tokens, so half have their tokens removed, and half are now on 4 tokens. You want your algorithim to remember the people with arbitary conts, to convert them to standard counts - and to have the probability to leave no one on on the previous standard count.

That is, when transfering the standard count from 1 to 2, you want no one left with 1 token at the end of it. Simerly when converting 2 to 4 you want no one left with 2 at the end of it. Any left-over people will have to be compleated by phase 3, but we want it so that phase 2 should finish of the counting in most cases.

You also want the arbitary counts to be standardized. That is, while converting the standard count from 1 to 2 you want to convert all odd counts to even. Simerly you want all arbitary counts to be divisible by 4 when conveting the standard count from 2 to 4 and so on...

To do this you need to to impliment stratigic rules. Concider the first cycle:

Rules:

For non-counters:
1. If the light is off and you have an odd number of tokens, then remove one token and turn the light on.
2. If the light is on and you have an odd number of tokens, then add one token and turn the light off.

For the counter:
1. If the light is on then switch it off and add one token to the count.

The last day of the cycle (you don't want this to happen):
1. If the light is on and you have an odd number of tokens, then remove one token and leave the light on.
2. If the light is on, then add one token and turn the light off.

I estimate by the end you should expect to have:

~3 or 4 people on 16 tokens.
~96/95 people on zero tokens.
1 counter. Then let the counter have time to count the max value (16).

Then enter Phase 3 which decreases the ammount the counter counts until it reaches 1 and then stands and runs until the end. Usually it should complete with never entering phase 3.
Post
#88910
Topic
Riddles
Link
https://originaltrilogy.com/post/id/88910/action/topic#88910
Time
I might add the place I got this riddle from had been able (after years and years and many many people trying) to get the average run-time down to 3,500 days. Well, I think - I didn't check the forum for it all the way through, so it is possible someone did better. Even so, I believe I will beat that. And that is without using (or even knowing) their formulas. I also believe my formula is the best possible method.
Post
#88909
Topic
Riddles
Link
https://originaltrilogy.com/post/id/88909/action/topic#88909
Time
It's all part of the challenge. Ironic I know, me being the one to post it... and then solve. I'm about half-way through phase 2, I completely programmed it, but it isn't optimal yet, I'm going to make a few fundamental changes to phase. The idea is that you can normally expect (in, say 99% of cases) the counter to have collected all 100 tokens by the end of phase 2. It should be less than 3000 days for this. Of course, this isn't, and can never be guaranteed - so after that you enter phase 3 which will basically last until the count is completed (so it'll have no set time-limit). It could still have recurring cycles, if it'd help, but I don't think it would. At the moment with my pseudo-phase two I am achieving more than 90 people on zero tokens (taking less than 2,000 days and with the counter on anything from 10 to 50 tokens) by the end of it. The way it's set up at the moment you can't expect the counter to have all the tokens by the end of it. I'm going to re-program it with more optimised rules.
Post
#88766
Topic
<strong>The &quot;EditDroid&quot; Trilogy DVD Info and Feedback Thread</strong> (Released)
Link
https://originaltrilogy.com/post/id/88766/action/topic#88766
Time
Originally posted by: Rikter
No disrespect to those that have labor hard to create these DVD's (myself included) but screw the elitist attitude of "don't share these" - DON'T CREATE THE DARN THING in the first place if you don't want it shared
What would this forum be like without your triple-posts Rik? I concur.
Post
#88755
Topic
<strong>The &quot;EditDroid&quot; Trilogy DVD Info and Feedback Thread</strong> (Released)
Link
https://originaltrilogy.com/post/id/88755/action/topic#88755
Time
In danger of breaking my break from this forum (sorry couldn't help myself entering this thread, I haven't gone into any others excepting to read my own posts), I have a couple of things to say. The Bizzle and Dr_Gonzo were instructed not to make copies of the set by the creators of the set.

I find such things hard to swallow. For instance, a friend of mine committed an act of vandalism last year, where he removed and destroyed private property. He then told me not to tell anyone. Of what authority does he have to say that to me? I mean I should have told the police outright. And it wasn't because he's my friend that I didn't. Heck I mean he shouldn't be the one to dictate who should know the truth, I really should have told the police. Next time I will, that time I was in the wrong too (for not telling the police).

I know distributing Star Wars bootlegs is not the same thing, however people should realize at least that it is the property of Lucas/LucasFilm. Therefore whatever creators of sets have to say about their distribution shouldn't matter. What Lucas says should matter most, and then what other people say should matter less. Not because Lucas deserves to dictate, but because he owns the copyright. If someone else breaches his copyright - as we all have done - then they/we do not have any authority to claim others shouldn't do it also. Because that is saying "do as I say, not as I do".

I know I've said that before, but I just wanted to give it a fresh run. I would like to hear what others think about this too.
Post
#88752
Topic
To Jay...
Link
https://originaltrilogy.com/post/id/88752/action/topic#88752
Time
You still have not responded to my concern (PM), it is important to me. It is not the largest concern in the world, yes I realize that - but it still warrants my request (and is fairly urgent). Please review my PM and get back to me. (Yes to everyone else I'm still on a break from this forum, it's just that Jay didn't seem to read this post in the Off Topic forum, so here's hoping).
Post
#88748
Topic
Riddles
Link
https://originaltrilogy.com/post/id/88748/action/topic#88748
Time
Phase 1:

As you know, the average after 1 cycle is 11 people on zero. The average after 2 cycles is 14 people on zero. The average after 3 cycles is 17 people on zero. The average after 4 cycles is 19 people on zero. Now if after phase 1 you have the counter on only 1 or 2 (there's a total 1% chance of this) then it means you might want a fifth cycle.

Optimise the day numbers for each cycle.

There is a 0.545% chance that after the first cycle you reach 30+ people on zero. So, allocate 30 days for it.

If you reach 11 people on zero after the first cycle then there is a 0.509% chance that after the second cycle you have 31+ people on zero. So allocate 20 days for it.

If you reach 14 people on zero after the second cycle then there is a 0.518% chance after the third cycle that you have 32+ people on zero. So allocate 18 days for it.

If you reach 17 people on zero after the third cycle then there is a 0.576% chance after the fourth cycle you have 34+ people on zero. So allocate 16 days for it.

If you reach 19 people on zero after the fourth cycle then there is a 0.564% chance after the fifth cycle you have 34+ people on zero. So allocate 15 days for it.

Complete. Allocate 5 cycles for phase 1 as above. Phase 2 coming next....
Post
#88741
Topic
Riddles
Link
https://originaltrilogy.com/post/id/88741/action/topic#88741
Time
Well I'm getting there. Currently I've only programmed in Phase1 (what I've discussed here) with the latter Phases to be coming (however, the program does take the cycles from the Phase1 settings textbox). You can see how the tokens are transferred:



Also, you'll notice the light is on at the bottom. All this means (in this instance) is that one token is currently being held by the light - so that when you add up the tokens they should tally. 21 (counter) + 1x9 + 1x6 + 1x2 + 61x1 + 1 (light bulb) = 100. I should note the example portrayed in the screenshot is not an average count at the end of those 4 cycles, it's just a random case. The get averages button does nothing yet.
Post
#88523
Topic
Riddles
Link
https://originaltrilogy.com/post/id/88523/action/topic#88523
Time
I'll give you my full answer shortly, when I have built a VB program to run the entire algorithim with different settings. It is 100% accurate, and I believe the best possible algorithim (you would have to find the most optimal settings yourself, however). Everyone else has to be on 0 tokens for the counter to get to 100 tokens.
Post
#88422
Topic
Riddles
Link
https://originaltrilogy.com/post/id/88422/action/topic#88422
Time
Just so you know I'm cruncing the numbers. There is a 1.097% chance that the counter counts 30+ people initially with that starting stratergy. So it seems it is a wise idea to allocate, say 35 days to the initial counting period (since there is a .11% chance you'll get to day 35 with the light off). the rest of the first 100 days would be wasted.
Post
#88410
Topic
Macrovision
Link
https://originaltrilogy.com/post/id/88410/action/topic#88410
Time
There was once a time when consumers could illegally copy their vhs tapes by wiring up two VCR's. To combat this, a company called Macrovision bought in a system to exploit the TV signal. In fact, it more than exploits the specifications - it breaks them. Therefore Macrovision's signal is an illegal, non-standard signal. Despite the fact the signal is illegal, studios and distributors all signed-up for it.

Macrovision:

Inconvenienced many people legitimately watching paid-for content on TVs.
Reduced the visible picture quality for everyone (even today, esp on 100hz displays).
Prevented the use of projectors.
Oh, and stoped some home users creating illegal copies.

So, everyone was treated like a criminal. Everyone suffered from the signal.

When DVD players were introduced, the signal changed. It is no longer a signal hard-encoded into the picture information, rather it is added "on demand" by a macrovision "chip". That's to say, an otherwise legal picture is fed in and out comes an illegal one. Artificial means of viewing control. One of the things that has disappointed me, is that while DVD-Region control was seen to be consumer-invasive, Macrovision's signal has not been!! I mean, it's not even a legal signal, the fact that it's there means someone has been tampering with it deliberately.

And it hasn't stoped the bigger fish from pirating. I wish Australian law would not require macrovision circuits in DVD Players (dishonour that part of the patent, like the do with RPC). All they do is ruin an otherwise legitimate signal.
Post
#88399
Topic
Riddles
Link
https://originaltrilogy.com/post/id/88399/action/topic#88399
Time
You're confusing yourself. Imagine there are only 5 prisoners. If they agree:

that the first time they enter the room while the light is off to remove their token and leave the light off,
the second time someone enters the room while the light is off to turn it on and deisgnate himself the role of counter.

then for the rest of that period the light remains on, to signal to the others that a counter has been chosen, and to ensure there is no error. Therefore on the last day - say day five, there are two posabilities. Either the light is on because the counter turned it on some time before, or it is off because there is no counter.

If it is off and it is your first time in the room, then you know everyone else was in the room. If it is off and it is your second time in the room you know one person is left to enter the room. if it is on you don't know how many people entered the room, but you know the counter counted all the ones to enter while the light was still off.
Post
#88045
Topic
Riddles
Link
https://originaltrilogy.com/post/id/88045/action/topic#88045
Time
Quote
Originally posted by: ricarleite
Transfers?? Well, like I said, and I'm PRETTY sure of myself: if the prisoners absolutely cannot comunicate with each other, with no "token transfer" or anything like that, then it's IMPOSSIBLE to solve this one.
They're using the light bulb to comunicate and transfer the tokens.
Post
#88037
Topic
Riddles
Link
https://originaltrilogy.com/post/id/88037/action/topic#88037
Time
Quote
Originally posted by: Warbler
I think I understand it a little the "tokens" are imaginary right?
Well yes, for the most part. They could be recorded on paper or something in each prisoner's cell, if he has memory problems - I'll talk about the transfers later.
Post
#87999
Topic
Riddles
Link
https://originaltrilogy.com/post/id/87999/action/topic#87999
Time
No, it's not impossible ricarleite. You just aren't thinking of all the possibilities. The lamp on its own can only hold one bit of information, however the lamp when used in conjunction with the day number can hold all sorts of information.

For instance, say days 1-99 are treated thus (all prisoners initially hold one conceptual token):

If the light is off:

1. And it is your first time in the room, remove your token and leave the light off.
2. And it is your second time in the room, designate yourself the counter, and count Days-1 as your initial count. Turn the light on.

If the light is on do nothing.

On day 100:

If the light is off:

1. And it is your first time in the room, then everyone has been in the room. Demand freedom.
2. And it is your second time in the room, designate yourself the counter, and count Days-1 (=99) as your initial count. Leave the light off in this instance since it is day 100.

If the light is on:

Turn it off.

That method gives you a head-start. On average the initial count is 12, however it is not an optimal head-start. For instance there is no reason to dedicate the first 100 days to it, in fact the first 20 days would be more than enough. It is so improbably that your initial count will be more than 20 you may as well save the other 60 days for better use.

Like I said, that concept I didn't think of initially myself. However my concept is that the more people you have on zero tokens the faster you will be able to transfer all the tokens to the counter. If you were to designate the first 20 days to this step you would instead have:

For instance, say days 1-19 are treated thus (all prisoners initially hold one conceptual token):

If the light is off:

1. And it is your first time in the room, remove your token and leave the light off.
2. And it is your second time in the room, designate yourself the counter, and count Days-1 as your initial count. Turn the light on.

If the light is on do nothing.

On day 20:

If the light is off:

1. And it is your first time in the room, then designate yourself the counter and start the count at 20. Leave the light off.
2. And it is your second time in the room, designate yourself the counter, and start the count at 19. Leave the light off.

That works much better, you can also improve your chances by having a special case for days 2 and 3 - that is, there is a 1% chance that the same person will be chosen twice in a row, if this happens you could insert the rules for days 2 and 3:

Day 2:

If it is your second time in the room, switch the light on but do not designate yourself the role of counter.

Day 3:

If the light is on, and it is your third time in the room (0.1% chance) then designate yourself the role of counter and start the count at 1.
If the light is on, and it is your second time in the room then designate yourself the role of counter and start the count at 2.

You get away with this because according to the rules for this period if on the third day the light is on, you know with certainty only one person had been in the room before you. On the forum which discussed this riddle over 3 years, no one ever mentioned that (at least I don't think so, I didn't go through all of it). What no one figured out, on the forum that discussed this riddle (in over 3 years), is that you can give yourself a further head-start by repeating this step! I couldn't believe it either. I know the average for repeating it goes down and down, but you only need to designate a certain number of days to it.

I would designate 20 more days for repeating it (possibly 1 or 2 days less). Thus:

The rules for days 21-39:

If the light is off:

And you are not the counter, and you have one token then remove your token and leave the light off.
And you are not the counter, and you have no tokens, then turn the light on and your new token count = days - 21.
And you are the counter then switch the light on and amend your count days - 21. Turn the light on.

I do not yet know the optimal number of times to repeat this, but I imagine it might be 3 or 4 times. The whole idea is to get as many people onto zero tokens as quickly as possible. Yes, you will have people with counts of any possible number, but they can be transferred to the counter in larger clumps at later times.

People on zero tokens are passive, that means when they enter the room they don't have to get rid of any tokens by switching the light on, and they don't have to collect any tokens by turning the light off. This is a huge advantage. by the end of 80 or 100 days, I imagine you could have average it to ~20 people on zero tokens (that is just an estimate at this stage). Compare that to running it once with 100 days allocated to it - which has an average of just 11 people on zero tokens. You've got a huge head-start.

Anyhow, that's not as far as I've got, but that's as far as hints from me come. I'm sure you can easily take this in the right direction to find the optimal plan now (something that apparantly no one from the original site could find).
Post
#87466
Topic
Riddles
Link
https://originaltrilogy.com/post/id/87466/action/topic#87466
Time
I know it's a particularly nasty riddle. I'll give you the same thing I got that gave me a head-start (I've worked off this idea).... the prisoners agree that:

#1 they are all assigned one token, and a counter will collect all these tokens as to be able to assert that they have all entered the central living area.

#2 During the first 100 days:

If it is their first time in the living room and the light is off to leave it off. They then remove their token.
If it is their second time in the living room and the light is off to turn it on and assign themselves the role of the counter. The count (number of tokens) is equal to the day number minus one (since he's been in the room twice).
If the light is on, do nothing regardless until day 100, in which case switch it off.

This gives a bit of a head-start, on average the initial count is 12. Meaning there are now on average 11 people with no token, 88 people with one token each and the counter with 12 tokens.

I will also give you my main concept, which I came up with myself and is a huge advantage if you couldn't think of it yourself.

The concept is if we were to assert that a prisoner releases his token, wherever possible for the counter to collect by turning the light on, then we expect the counter and everyone else to enter the room once every time a token is released. Meaning there is conceptually 100 room entries per token release. Reducing this ratio should be our main concern.

I believe I've worked out the perfect strategy, but I have to do some pretty heavy number-crunching until I can make it optimal.

Afterwards I believe I could then produce an optimal strategy with the rather nasty alteration:

The procedure ends not when a prisoner asserts that all prisoners have entered the room, rather the prisoner must assert that all other prisoners now know and have asserted that all prisoners have entered the central living room.

Ooooh that's nasty, I shouldn't have thought of it. But it can be done.
Post
#87281
Topic
Riddles
Link
https://originaltrilogy.com/post/id/87281/action/topic#87281
Time
Well to stick with the light-bulb riddles, here's one:

100 prisoners are imprisoned in solitary cells. Each cell is windowless and soundproof. There's a central living room with one light bulb; the bulb is initially off. No prisoner can see the light bulb from his or her own cell. Each day, the warden picks a prisoner equally at random, and that prisoner visits the central living room; at the end of the day the prisoner is returned to his cell. While in the living room, the prisoner can toggle the bulb if he or she wishes. Also, the prisoner has the option of asserting the claim that all 100 prisoners have been to the living room. If this assertion is false (that is, some prisoners still haven't been to the living room), all 100 prisoners will be shot for their stupidity. However, if it is indeed true, all prisoners are set free since the world can always use more smart people.

Before this whole procedure begins, the prisoners are allowed to get together in the courtyard to discuss a plan. What is the optimal plan they can agree on, so that eventually, someone will make a correct assertion with 100% accuracy?

...

I am currently working on the most optimal plan, I do believe that I've got it. Yes, I know if they wait 3 years then there's a 99.9% chance they've all been there, but we want to be 100% certain. 99.9999999999999999999999999999999999999999999999999% is not good enough. Good luck.