|
Maybe someone should do a Wiki entry for "tall order" algorithm. A greedy algorithm that asks for alot to be done in little or no time. Generally has a time/space order complexity in the same neighborhood of any typical NP-Complete problem.
"Real programmers just throw a bunch of 1s and 0s at the computer to see what sticks" - Pete O'Hanlon
|
|
|
|
|
 Yeah, that sounds about right!
|
|
|
|
|
Isn't that what most of the posters in the C# / ASP.NET and web dev forums are asking for :- A lot, urgently, without them doing anything themselves?
"More functions should disregard input values and just return 12. It would make life easier." - comment posted on WTF
|
|
|
|
|
There does seem to be an unfortunate rash of laziness around some of these forums lately 
"Real programmers just throw a bunch of 1s and 0s at the computer to see what sticks" - Pete O'Hanlon
|
|
|
|
|
You could consider using regular expressions.
Steve
|
|
|
|
|
Yep. Not sure what kind of regular expression he could use, but maybe Expresso[^] could be useful 
"Real programmers just throw a bunch of 1s and 0s at the computer to see what sticks" - Pete O'Hanlon
|
|
|
|
|
Hi,
this could be very easy, depends on your definition of pattern.
if your example 236321236 contains two patterns for you, that is 23 (twice),
then the following will do. If the fact that 236 appears twice is a pattern too,
then it is harder.
Here is my algorithm in pseudo-code (i.e. approx C#); some details may
be incorrect, but you will get the general idea; it is quadratic which
is good enough for phone numbers I guess.
string s="my text to analyze";
int sLength=s.Length;
for (int i=0; i<sLength-1; i++) {
char c1=s[i];
char c2=s[i+1];
int diff=c2-c1;
if (diff==0) log("double char");
if (diff*diff==1) log("two char series");
for (int len=3; len<sLength-i; len++) {
c2+=diff;
char c=s[i+len-1];
if (c2==c) log("series extends to "+len+" chars");
else break;
}
}
As you can see, nothing fancy, no tricks, no regex...
|
|
|
|
|
Hi
I am not sure if this is true or not, but my initial tests gave a positive result
To check for primality of P
if 4 does not perfectly divide the perimeter of all the rectangles with area (P-1) then P is prime.
Regards
The Best Religion is Science.
Once you understand it, you will know God.
|
|
|
|
|
Do you have a link to this, so it can be investigated/verified?
"Real programmers just throw a bunch of 1s and 0s at the computer to see what sticks" - Pete O'Hanlon
|
|
|
|
|
Not true, fails for 15, 27, 35, ..
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."
|
|
|
|
|
cp9876 wrote: Not true, fails for 15, 27, 35
Observant and quick, I must say
"Real programmers just throw a bunch of 1s and 0s at the computer to see what sticks" - Pete O'Hanlon
|
|
|
|
|
Hi
| Not true, fails for 15, 27, 35
ture...
i was trying with big numbers and found this working...didnt notice that it failed for small numbers, infact it fails for large numbers also.
Regards
The Best Religion is Science.
Once you understand it, you will know God.
|
|
|
|
|
You ought to direct your reply to cp9876...
"Real programmers just throw a bunch of 1s and 0s at the computer to see what sticks" - Pete O'Hanlon
|
|
|
|
|
It is pretty easy to see that it identifies any number of the form 4k+3 as prime.
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."
|
|
|
|
|
Hey Gui you are giving some what like silly idea. Don't you know 15,27, 35 are not prime.
gyan
|
|
|
|
|
This idea is not clear cut for all primes. For example 13 is a prime number but 13-1=12 is divisible by 4. I mean that what happens if 4 perfectly divides p-1???
Gyan
|
|
|
|
|
Hi
Just found some wonderful properties of number 4.
1. 4 is the only perfect square whose reciprocal is also a perfect square.
2. Any perfect square when devided by 4 gives a reminder of either 1 or 0.
Is this something already known?
The Best Religion is Science.
Once you understand it, you will know God.
|
|
|
|
|
Zainu wrote: Is this something already known?
probably for a few hundred years.
|
|
|
|
|
Zainu wrote: Is this something already known?
Technically, yes, I'm fairly sure that it is known by at least one person. I'm sure others will claim knowledge of it, too. Whether they do or not, you'll never know for sure.
"A good athlete is the result of a good and worthy opponent." - David Crow
"To have a respect for ourselves guides our morals; to have deference for others governs our manners." - Laurence Sterne
|
|
|
|
|
Zainu wrote: 4 is the only perfect square whose reciprocal is also a perfect square.
hence 1/4 is a perfect square while, for instance, 1/9 not?
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.
|
|
|
|
|
Sounds like you are just discovering the magic of Number Theory, you can find lots of interesting resources on the web.
I'm not sure what you are implying with (1), but (2) is very well known. 0 and 1 are called the quadratic residues mod 4. Some other examples, the same thing is true mod 3, but in mod 8 the residues are 0,1,4.
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."
|
|
|
|
|
|
Mathworld is a great website, alot of good stuff there
"Real programmers just throw a bunch of 1s and 0s at the computer to see what sticks" - Pete O'Hanlon
|
|
|
|
|
Hiya guys.. Using C++ does anyone know how i can tell if a line is linear or not? I want the user to enter 5 readings which i can plot on a chart. However i would like a function to find out if the plotted line is linear.
Any help would be greatly appreciated!
|
|
|
|
|
EDIT: Not sure if this is needed or not: if the slope between x_1,y_1 and x_5,y_5 is greater than 1, swap the x and y values for each coordinate pair. This swap is needed to keep the delta value in the next step smaller than 1.
Create a function for the line y = f(x) using points 1 and 5. Then of each of the 3 points in the middle calculate f(x_n), if ABS(f(x_n) - y_n) < 1, then x_n,y_n is on on the line (give or take the rounding error from forcing integral values).
--
You have to explain to them [VB coders] what you mean by "typed". their first response is likely to be something like, "Of course my code is typed. Do you think i magically project it onto the screen with the power of my mind?" --- John Simmons / outlaw programmer
|
|
|
|
General 
News 
Suggestion 
Question 
Bug 
Answer 
Joke 
Praise 
Rant 
Admin 
Submit an article or tip
