 | | Re: Geek Hierarchy
|
|
(...) Hey, where are the toy-collecting geeks on this thing? I feel so underrepresented! :) Hilarious. -jeremiah- (22 years ago, 27-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Geek Hierarchy
|
|
Posted without further comment... (URL) (22 years ago, 27-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Retina scanner?
|
|
You do realise that that 'retina' scanner on CLSOTW is actually just photographing the cornea, not the retina, don't you? Jason Railton (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: On a scale of 0 to 10?
|
|
(...) aHA! Much oblige! I now get (approx): s = 5.39 * e^(2.625q) - 0.39 Whew. Of course, now here's a totally different question. In order to get that point, I cheated. I couldn't solve: e^(B/4) + e^(-B) = 2 using algebra, but using other means, I (...) (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: On a scale of 0 to 10?
|
|
Hi all. The reason the equation is unsolvable is because B is an unneeded constant. Why? A*e^(q+B)=A*(e^q)*(e^B) It is impossible to distinquish between A & e^B. What you need to solve is an equation of the form s=A*exp(B*q) + C This is the form of (...) (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: On a scale of 0 to 10?
|
|
(...) Oh yeah! (...) Oh... yeah. (...) Hm. Double checked the math-- it appears solid [1], which would mean that there's no viable solution for: s = Ae^(q + B) + C for coordinates (-1,0), (0,5), (0.25,10). Darn. Hm. I guess that also means that any (...) (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: On a scale of 0 to 10?
|
|
(...) *snip* (...) summed exponents like that can be re-written in this way: e^B * e^.25 + e^B * e^-1 = 2e^B (e^.25 + e^-1) * e^B = 2e^B Uh oh. Divide both sides by e^B and we've got (e^.25 + e^-1) = 2 1.652 = 2 Which ain't right. I think your math (...) (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: On a scale of 0 to 10?
|
|
(...) Thought about that... If necessary I guess, but I'm pretty sure there's a way to do it. I think what I need is: s = Ae^(q + B) + C And at the moment, I've solved: C = -Ae^(B - 1) A = 10 / (e^(B + .25) - e^(B - 1)) And I'm scratching my head (...) (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: On a scale of 0 to 10?
|
|
(...) How about a double-linear solution? from -1 to 0, translate linearly from 0 to 5. From 0 to .25, translate linearly from 5 to 10. You won't have a single equation, but that seems to accomplish what you want, right? Adrian (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
|
| |
| | On a scale of 0 to 10?
|
|
I've got a function that represents a quality rating, based on 2 parameters, which are on a scale of 0 to 1: (d-k)*k => quality (q) Notice that this yields a sort of wacky result, insofar as the quality can vary between -1 and 0.25, not -1 and 1 or (...) (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: Website move
|
|
(...) The domain name is the fully qualified host name. If you need DNS to look it up, it's a domain name - the "host" in the RFC. That's what it's a "Domain Name Service" :) Anything after the / is part of the path, as defined in the HTTP RFC (...) (22 years ago, 26-Mar-03, to lugnet.space, lugnet.publish, lugnet.off-topic.geek)
|
| |
| | Re: Backwards Compatibility (Was Calling all Meta-commands)
|
|
(...) ROFL. I'm sorry, I can't reply in kind. My brain is too fogged. Steve (22 years ago, 22-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: Backwards Compatibility (Was Calling all Meta-commands)
|
|
(...) you mean $puctuation++ > ! $puctuation ? $goodness++ : $goodness-- ; ? (22 years ago, 22-Mar-03, to lugnet.cad.dev, lugnet.off-topic.geek)
|
| |
| | Re: connect four
|
|
(...) This interested me for a while-- I wrote a recursive program once that would look X ply into the game, though because I never stored state information, it took ridiculously exponentially longer with each value of X. Something like 6 ply was (...) (22 years ago, 19-Mar-03, to lugnet.org.ca.rtltoronto, lugnet.off-topic.geek)
|
| |
| | Re: Access Help
|
|
James & All, (...) Yes, I have a total query, where it is running a sum of all the data from a date range (say 1/1/02 to 12/31/02 or whatever). The value I am trying to evaluate is a calculated sum of various percentages of our recovery numbers, in (...) (22 years ago, 7-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: Access Help
|
|
Scott, Can you post and example of the type of queries you have been trying? I know you said they where giving you errors, but at least we'd have an idea of what you are trying to do. jt (...) (22 years ago, 7-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Access Help
|
|
To All, I have been working on a project at work for over a year now, converting an Excel Production Recap program into an Access program. I have been successful converting it, outside of some small issues, which I have finally resolved in the last (...) (22 years ago, 7-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: Now the stakes are higher....
|
|
(...) Effectively, he's right but he's not giving things the probability they're due. In his solution, the (r1 r2) choice is GUARANTEED that no matter which one you reveal (r1 or r2), the one revealed will be red. However, with, say, (r1 b), there's (...) (22 years ago, 6-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: Now the stakes are higher....
|
|
In lugnet.off-topic.geek, David Eaton writes: <snip> (...) That's perfect! I owe you a lunch! Now whetehr or not my co-worker succumbs to 'outside sources' as valid enuf to change his position remains to be seen I'm trying to explain his fallacy and (...) (22 years ago, 6-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: Now the stakes are higher....
|
|
(...) HA! Congrats on winning 100 bucks :) (...) Proof number one: (pure stats) #!/usr/bin/perl for(1..1000) { @marbles = ("r","r","b","w"); $pick = ''; for(1..2) { $n = int rand(@marbles); $pick .= $marbles[$n]; splice @marbles,$n,1; } $tot++; (...) (22 years ago, 6-Mar-03, to lugnet.off-topic.geek)
|
