Re: determinant of a matrix - Algorithms Discussion Boards

Kastellanos Nikos10-Oct-06 23:42

10-Oct-06 23:42

~~sounds like a programming question!~~
~~Is this a homework?~~
(..I don't think i am in The Lounge anymore...")

Anyway, Are there negatives values in that array?

-- modified at 6:10 Wednesday 11th October, 2006

Re: Search Algorithm

Kastellanos Nikos11-Oct-06 0:02

Kastellanos Nikos

11-Oct-06 0:02

That's what pop into my mind:

<br />
sumx(A[],x)<br />
{<br />
  Array S[x/2];<br />
  for(i=0,i<S.lenght;i++) S[i]=0;  //set all elements to 0 <br />
  for(i=0,i<A.lenght;i++)<br />
  { <br />
    a=A[i];<br />
    if(a<0 || a>x/2) continue;<br />
    if(a<x/2) S[a]   |= 1;<br />
    if(a>x/2) S[x-a] |= 2;<br />
  }<br />
  return S;<br />
}<br />

You get an array S.
For a number k<(x/2) you have:
if S[k] is 0 then neither number k, nor (x-k) is in the input array.
if S[k] is 1 then only number k was in the input array.
if S[k] is 2 then only number (x-k) was in the input array.
if S[k] is 3 then k and (x-k) was in the input array. The sum of those two is k+(x-k)=k+x-k=x. bingo! Rose | [Rose]

For a number j>(x/2) you have k=(x-j) and continue as above.

The array S can be more comlex, like storing the index of two number in the input array A, if you like. Dealling with negative numbers as well might need a more complex aproach.

Re: Search Algorithm

Anand Vivek Srivastava11-Oct-06 4:31

Anand Vivek Srivastava

11-Oct-06 4:31

Small bug:

Kastellanos Nikos wrote:
if(a<0 || a>x/2) continue;

if(a<0 || a>x) continue;

To put the algorithm in simple words, do a counting sort for numbers <x , so O(n).
lookup is O(1) for a hash table; will have to do x/2 lookups, so O(x).
Overall O(x+n).
The complexity of algorithm is O(x+n); If x is reasonably small, then the algorithm is good enough.
If x>n*log n; a quick sort followed by a binary search is better, but will be managable only in O(n*log n)

determinant of a matrix

cagdasanilan4-Oct-06 4:15

cagdasanilan

4-Oct-06 4:15

i have written a code to find the determinant of a matrix, i would like to ask is it fishy or can it be devoloped? thanks to you all for replies... my code is written in vb.. and one more thing... i wonder is it the right place to post? thanks anyway...
here is my code...
Sub deneme()
Dim a(1 To 10, 1 To 10) As Double

Dim t As Integer
a1 = 1
t = 0
GoTo git
For i = 1 To 3
For j = 1 To 3
t = t + 1
a(i, j) = t
Next j
Next i

git:

a(1, 1) = -1
a(1, 2) = 0
a(1, 3) = -50
a(2, 1) = 1
a(2, 2) = 4.92
a(2, 3) = -2
a(3, 1) = -1.4
a(3, 2) = 3
a(3, 3) = 1

For i = 1 To 3
For j = 1 To 3
If i = j Then
If a(i, i) = 0 Then GoTo devam
For g = i + 1 To 3
tut = a(g, i)
For b = 1 To 3
a(g, b) = a(g, b) - (a(i, b) * (tut / a(i, i)))
Next b
Next g
End If
Next j
Next i

devam:

For i = 1 To 3
For j = 1 To 3
Cells(i, j) = a(i, j)
Next j
Next i
For i = 1 To 3
a1 = a1 * a(i, i)
Next i

Cells(1, 10) = a1

End Sub

Re: determinant of a matrix

User 66588-Oct-06 13:28

User 6658

8-Oct-06 13:28

Well I don't know what your formula does, but why not use the Laplace Expansion?

regards

modified 12-Sep-18 21:01pm.

Re: determinant of a matrix

cagdasanilan8-Oct-06 22:58

cagdasanilan

8-Oct-06 22:58

first of all thanks for replying... i've examined the laplace expansion. But in high order matrices finding the minor matrices or the cofactors is hard to code for me and i think, i can turn the square matrice to the triangle square matrice. so i just multiply the a[j,j] s then i find the determinant of it.. by the way.. my code is just turn the square matrice to a triangle square matrice. thanks for comments...

Re: determinant of a matrix

ejuanpp18-Oct-06 11:48

ejuanpp

18-Oct-06 11:48

Hi,

I do not know if your code works or not, but if I were you I would use the definition of determinant to compute it.
The hardest part is to code a function to get permutations.

Regards

Re: determinant of a matrix

cagdasanilan21-Oct-06 8:07

cagdasanilan

21-Oct-06 8:07

thanks for replying my message. my code is working for me for all inputs and it works only square matrices you know... and yeah you are very right... and by the way if you want to try it you can use excel microsoft vb.. thanks for advance...

Re: determinant of a matrix

ejuanpp22-Oct-06 22:47

ejuanpp

22-Oct-06 22:47

But ... determinants are only defined for a square matrix... isn't then the range what you want to compute ?

Regards

Re: determinant of a matrix

cagdasanilan26-Oct-06 4:06

cagdasanilan

26-Oct-06 4:06

sure i am... i need only square matrices... n unkowns n linear equations. best regards....

Port To Application Mapping (aka PAM)...

Shy Agam29-Sep-06 11:06

Shy Agam

29-Sep-06 11:06

Hey all,

I'm not sure this is the right forum, as I'm not sure what I'm looking for is an algorithm. Confused | :confused:

I have an IP address along with a port.
My goal is to determine which running application is using this port.
Googling taught me that this process is called Port-To-Application Mapping (aka PAM),
but I couldn't find any code which does that...
The best thing I found was a program which knows how to do it.

Anyone knows any PAM algorithms?

Thanks in advance,
Shy.

image convolution

Prinz Ryoiji27-Sep-06 22:02

Prinz Ryoiji

27-Sep-06 22:02

anyone know step by step mathematical processes for image convolution using laplacian? plz explain...

Re: image convolution

Dominik Reichl29-Sep-06 7:43

Dominik Reichl

29-Sep-06 7:43

See http://en.wikipedia.org/wiki/Discrete_Laplace_operator[^]
or http://mathworld.wolfram.com/Laplacian.html[^]

Best regards
Dominik

_outp(0x64, 0xAD);
and
__asm mov al, 0xAD __asm out 0x64, al
do the same... but what do they do??
(doesn't work on NT)

Re: image convolution [modified]

ejuanpp2-Oct-06 1:46

ejuanpp

2-Oct-06 1:46

Hi,

If I remember well ...

If your image M is considered as a x*y matrix (ie gray scaled) of pixels and a point P is then described as P(i,j) where 0 <= i < n and 0 <= j < n, and the laplacian is considered as a 3*3 matrix L defined as [[a,b,a],[b,c,b],[a,b,a]], the discrete convolution gives you another n*m matrix R where R(x, y) = Sum(i=0, i < 3, Sum(j=0, j < 3, L(i,j)*P(x + i - 1, y + j -1))) (0 outbounds)

Regards

looking for a source code...

good_buttercup27-Sep-06 19:04

good_buttercup

27-Sep-06 19:04

anyone can help find out any source code for prediction using bayesian belief network algorithm pls...A.S.A.P

respect,relax,respon

Re: looking for a source code...

User 58385227-Sep-06 20:45

User 583852

27-Sep-06 20:45

bad_bad_buttercup

Re: looking for a source code...

good_buttercup27-Sep-06 21:22

good_buttercup

27-Sep-06 21:22

i am looking for bayes network free source code for my preliminary analysis research...

respect,relax,respon

bayesian belief network algorithm

good_buttercup27-Sep-06 18:29

good_buttercup

27-Sep-06 18:29

anyone know about this technique...amswering me pls???

respect,relax,respon

English Word Tense

chinajuanbob22-Sep-06 23:15

chinajuanbob

22-Sep-06 23:15

is there an algorithm(c++/c#) or existing database, which can be used to turn English words from tense or plural form to their original form?

thx~

http://chinajuanbob.spaces.live.com/

Re: English Word Tense

Dan Neely25-Sep-06 2:21

Dan Neely

25-Sep-06 2:21

acording to MS, you shouldn't try to do this because while it's (barely) possible to do it for a single language, internationalizing will require writing a set of special rules for every language supported. Instead write your text in a plurality nuetral fashion. Instead of "1 cat, 2 dogs" use "cats: 1, dogs: 2".

Extended Midy's Theorem

Bassam Abdul-Baki21-Sep-06 8:52

Bassam Abdul-Baki

21-Sep-06 8:52

Coolness! See references below.[^]

Re: Extended Midy's Theorem

Link20061-Oct-06 23:48

Link2006

1-Oct-06 23:48

You wrote that? That's cool. Cool | :cool:

Re: Extended Midy's Theorem

Shawn Poulson13-Oct-06 7:40

Shawn Poulson

13-Oct-06 7:40

Holy crap. My head hurts. I give you a 5 for being smarter than I. Smile | :)

---
Shawn Poulson
spoulson@explodingcoder.com

Simple path finding algorithm

f*** YOU20-Sep-06 19:13

f*** YOU

20-Sep-06 19:13

Im having a problem with a A* path finding algorithm I made. Its able to find the path but it will rarely make diagonal moves when it should.

Strart
X
X
X X X End

I think I got the math right, its the algorithm that isnt working right. Iv googled around and found some good articals but most fail to explain how to impliment it. Some come with source code but is too complex and bloated for me to grasp. Im close to giving up on using the internet to help solve my problem.

static int Sqrt(int x) { if (x<0) throw new ArgumentOutOfRangeException(); int temp, y=0, b=0x8000, bshft=15, v=x; do { if (v>=(temp=(y<<1)+b<<bshft--)) {="" y+="b;" v-="temp;" }="" while="" ((b="">>=1)>0); return y; OMG | :OMG:

Re: Simple path finding algorithm

Cedric Moonen21-Sep-06 2:40

Cedric Moonen

21-Sep-06 2:40

Henize wrote:
it will rarely make diagonal moves when it should

Are diagonal cells considered as neighbours in your algorithm ? If no, then that is your problem Wink | ;)

Cédric Moonen
Software developer

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