|
You can factor the lefthand side b.a2-b=1 into b and something,
then you have the form x.y=z which just requires a division.
Luc Pattyn [Forum Guidelines] [My Articles]
This month's tips:
- before you ask a question here, search CodeProject, then Google;
- the quality and detail of your question reflects on the effectiveness of the help you are likely to get;
- use PRE tags to preserve formatting when showing multi-line code snippets.
|
|
|
|
|
Luc Pattyn wrote: You can factor the lefthand side b.a2-b=1 into b and something
Waheeeyyy!!! I just called him down and asked him to factorise
b.a2-b
b(a<sup>2</sup>-1) which gives
b(a<sup>2</sup>-1) = 1
b = 1 / (a<sup>2</sup> - 1) It's all about applying what he knows. He didn't think to factorise ( I didn't know you could!). Thanks a million Luc. He only has another 12 to do!! I'm going to have to really get into this stuff - it's actually quite cool.
so you answer don't be scared of failure
The only failure is never to try
Things You've Never Done - Passenger -2008
|
|
|
|
|
You're welcome.
Malcolm Smart wrote: It's all about applying what he knows
Of course. Don't apply what you don't know, but don't forget to apply what you do know; what is
the purpose of knowing something, if you don't apply it...
Malcolm Smart wrote: I'm going to have to really get into this stuff
Knowledge is hereditary, it will find its way up or down. 
Luc Pattyn [Forum Guidelines] [My Articles]
This month's tips:
- before you ask a question here, search CodeProject, then Google;
- the quality and detail of your question reflects on the effectiveness of the help you are likely to get;
- use PRE tags to preserve formatting when showing multi-line code snippets.
|
|
|
|
|
Of course this is probably way too late to help but...
The key here is that you need to isolate b
starting from your step:
(b+1) / b = a^2
this is equal to:
(b+1) * (1/b) = a^2
multiply out left side:
(b/b) + (1/b) = a^2
reduce (b/b):
1 + (1/b) = a^2
subtract 1 from both sides:
1/b = a^2 - 1
multiply both sides by b:
b/b = (a^2 - 1) * b
multiply both sides by 1/(a^2 -1):
(1/(a^2 - 1))*(b/b) = (a^2 - 1) * (1/(a^2 - 1)) * b
reduce both sides:
1/(a^2 - 1) = b
|
|
|
|
|
Simpler way
(b+1)/b=a^2
(b/b) + (1/b)=a^2
1 + (1/b) =a^2
(1/b)= a^2 - 1
=> b= 1/(a^2-1)
|
|
|
|
|
Thats the same thing as I posted. I just included all the steps that you left out which would be required by a math teacher who is actually teaching this stuff
Although I could have sworn that the entire other thread didn't exist from 2 weeks before walking them through it when I posted this.... 
|
|
|
|
|
Hello
Does anyone have a good idea of a decent line segment detection algorithm to be used on binary images (black and white)? The images are pages from newspapers, so they contain a lot of text.
The algorithm should detect almost straight vertical lines (maximum deviation of 5 degrees).
It should also be able to detect dotted lines, and some imperfect lines (with noise pixels, either white or black).
I've tried using Hough transform, but didn't get reasonable results.
Also tried some straight-forward methods, but they don't seem to give the same accuracy on different images.
I'm not so knowledgeable in this domain, and if there's someone to point me in the right direction, I would greatly appreciate it.
Thank you.
|
|
|
|
|
The Hough transform is probably the right approach, since it handles noise and imperfections. The problem is that the last step involves interpreting the transform results, which isn't straightforward.
It may help to do some noise removal before processing to get clearer results.
|
|
|
|
|
|
It looks very impressive. What does it do?
|
|
|
|
|
Found out about SheetCalculator[^] today.
Very nice!
There are II kinds of people in the world, those who understand binary and those who understand Roman numerals. Web - Blog - RSS - Math
|
|
|
|
|
Looks very interesting
"Real programmers just throw a bunch of 1s and 0s at the computer to see what sticks" - Pete O'Hanlon
|
|
|
|
|
Yeah, and the perfect excuse to learn Java.
There are II kinds of people in the world, those who understand binary and those who understand Roman numerals. Web - Blog - RSS - Math
|
|
|
|
|
Hello,
I'm an Italian student and I'm new on this site.
I have to implement in Java the Map abstract data type with separate chaining collison handlin without using java.util classes.
Is there anyone who could help me please?Because I've some doubts.
Thanks in advance,
Giorgio Vezzaro 
|
|
|
|
|
You might want to try some Java specific forum.
"I guess it's what separates the professionals from the drag and drop, girly wirly, namby pamby, wishy washy, can't code for crap types." - Pete O'Hanlon
|
|
|
|
|
This is a C#-forum - so nobody will help you with java-code.
So some simple questions: are the algorithms clear?
(You will want to implement a hash-function mapping integer-values (the hash-values) to some memory-address containing - in it's simplest form - the head of a linked list containing all the stored objects).
|
|
|
|
|
CKnig wrote: his is a C#-forum - so nobody will help you with java-code.
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke
|
|
|
|
|
What's the matter?
He was asking for help in java writing a hashtable with collision-managment.
So I tried to express my opinion that he won't get any java-code from here (I myself am hardly competent to give "working" java code 
|
|
|
|
|
Well, using Java is just one of his requirements. On the other hand this forum is focused on algorithms (hence is, to some level, language-independent), specifically C# language should have no privilege here.
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke
|
|
|
|
|
You are right on this - but I think this is a .net-"centric" forum and you will see mostly C#-code (or VB but never java - right?).
And I tried hinting at the basic points of a hashtable - but this concept is so simple that the hardest part will be implementing it!
|
|
|
|
|
OMG.
All these years I was under the impression that this was an VC++ only forum.
Glad that this is now sorted out
codito ergo sum
|
|
|
|
|
Well I think this is getting OT but one last remark:
of course you will find a lot of VC++ code here but a major part of articles and posts deals with .net - so I'm sorry if I hurt your feelings (yeah I know C++ is the only real thing - and all the nasty GC-noobs can kneel in the light of your superior coding-skills ) ....
|
|
|
|
|
CKnig wrote: but a major part of articles and posts deals with .net
perhaps because this site attracts a lot of inexperienced programmers... c# being a lot easier to learn, but difficult to do the master concepts, like writing a c# compiler or an os.... those we leave to those daring few c++ programmers who kept a few brain cells... 
_________________________
Asu no koto o ieba, tenjo de nezumi ga warau.
Talk about things of tomorrow and the mice in the ceiling laugh. (Japanese Proverb)
|
|
|
|
|
CKnig wrote: but I think this is a .net-"centric" forum and you will see mostly C#-code (or VB but never java - right?).
Probably pseudo-code (or no code at all) is more appropriate for this forum and no, I don't think it is .net centric (did you forget C/C++?).
CKnig wrote: And I tried hinting at the basic points of a hashtable - but this concept is so simple that the hardest part will be implementing it!
I think that is not a difficult task to implement a simple algorithm with a powerful language like Java.Do you really think that hashing is such a simple matter?
Have a look at [^].
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke
|
|
|
|
|
CPallini wrote: I think that is not a difficult task to implement a simple algorithm with a powerful language like Java.Do you really think that hashing is such a simple matter?
Have a look at [^].
The only difficult part of a hash-table is creating a "good" hash-function.
The OP wanted to write a hashtable with collision-handling so I assumed that the "difficulty" in his task would be to link the "lists" with the "table" - and this is a simple task!
And bye the way: he suggest some college course or whatever and on this level it's the most basic things (just wait for the LR(1)-grammar questions )
|
|
|
|
General 
News 
Suggestion 
Question 
Bug 
Answer 
Joke 
Praise 
Rant 
Admin 
Submit an article or tip
