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Jan 28 2013 04:44am
Quote (Azrad @ 28 Jan 2013 09:17)
i totally understand why you said that, but in python that is not how it works...
here is a snippet with its output:
http://python.codepad.org/3pGNDUJ8


link doesn't load
but had a quick look at some other sides dealing with python
and yes, my comment is wrong but so is your flow control mechanism for the problem stated in the op
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Jan 28 2013 05:18am
Quote (brmv @ Jan 28 2013 03:44am)
and yes, my comment is wrong but so is your flow control mechanism for the problem stated in the op

how exactly is it wrong? (I'm not trying to start a pissing contest, I just want to know what is wrong with it, cuz it seems right to me).

This post was edited by Azrad on Jan 28 2013 05:19am
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Jan 28 2013 04:45pm
Quote (Azrad @ 28 Jan 2013 11:18)
how exactly is it wrong? (I'm not trying to start a pissing contest, I just want to know what is wrong with it, cuz it seems right to me).


from what i have seen you have to run through a "range" in python which has to have the three values 1,2,3 for the problem in the op
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Jan 28 2013 08:57pm
εijkεpqk = δipδjq - δiqδjp

epsilon (ε) will only have values 1, -1, or 0.

It is 1 if i, j, k have even permutations, i.e 1, 2, 3 or 2, 3, 1 or 3, 1, 2

It is -1 if i, j, k have odd permutations, i.e 3, 2, 1 or 2, 1, 3 or 1, 3, 2

It is 0, if it is none of the above, i.e has repeats of any subscribts, i.e 1, 1, 1 or 1, 2, 2, or 3, 3, 1 etc...

delta (δ) will only have values, 1 or 0

It is 1 if i = j

It is 0 if i does not equal to j

If you need anymore info, let me know
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Jan 28 2013 09:38pm
so the two functions would be:

INTEGER FUNCTION EPSILON(I,J,K)
C
C this function calculates the epsilon value
C
IMPLICIT INTEGER (A-Z)
C
C only using integers & assuming that I,J,K are all in the range 1 to 3
C
VALUE=0
TEST=100*I+10*J+K
C
C creating TEST for easy comparison & setting VALUE as default to improper permutation
C
IF(TEST.EQ.123) VALUE=1.
IF(TEST.EQ.231) VALUE=1
IF(TEST.EQ.312) VALUE=1
IF(TEST.EQ.321) VALUE=-1
IF(TEST.EQ.213) VALUE=-1
IF(TEST.EQ.132) VALUE=-1
C
C now have to return VALUE as EPSILON
C
EPSILON=VALUE
END

and

INTEGER FUNCTION DELTA(I,J)
C
INTEGER RESULT
C
RESULT=0
IF(I.EQ.J) RESULT=1
C
DELTA=RESULT
END

done those functions the easy way, EPSILON could be made more elegant using preset values but not worth the effort for such a small sample

got to do something, back a little later

PS: all statements exept the comment (C) ones have to start in column 6 - d2jsp swallows the spaces

This post was edited by brmv on Jan 28 2013 09:40pm
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Jan 28 2013 09:49pm
I compiled it and proceeded to see if it works, but it's flawed. Where's your declaration statements? and sub routines?

Is this in fortran? I don't recognize some of these process.

This post was edited by TempoONE on Jan 28 2013 09:58pm
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Jan 28 2013 10:11pm
Quote (TempoONE @ 29 Jan 2013 03:49)
I compiled it and proceeded to see if it works, but it's flawed.  Where's your declaration statements?  and sub routines?


those above are the SUBROUTINES, the main program follows (putting "." in the first 6 columns instead of space because of space suppresion):

......IMPLICIT INTEGER (A-Z)
......EXTERNAL FUNCTION DELTA,EPSILON
C
C only using integers in this program
C
......DO 1050 K=1,3
......DO 1040 I=1,3
......DO 1030 J=1,3
......DO 1020 P=1,3
......DO 1010 Q=1,3
......RIGHT=DELTA(I,P)*DELTA(J,Q)-DELTA(I,Q)*DELTA(J,P)
......LEFT=EPSILON(I,J,K)*EPSILON(P,Q,K)
......IF(LEFT.EQ.RIGHT) GOTO 1010
C
C we found a combination the formula does not work
C
......WRITE(6,1000) I,J,P,Q,K
.1000.FORMAT(' the identity is not true for I=',I1,' J=',I1,' P=',I1,' Q=',I1,' and K=',I1)
......GOTO 9000
C
C continue to check the other values
C
.1010.CONTINUE
.1020.CONTINUE
.1030.CONTINUE
.1040.CONTINUE
.1050.CONTINUE
C
C we tested all combinations and they worked
C
......WRITE(6,2000)
.2000.FORMAT(' the identity xxx is correct')
C
.9000.STOP
......END

remember that this is FORTRAN V and you have to sort out the compilation/binding depending on your operating system

PS: not column 6 but column 7 the statements start, column 6 is for statement continuation - sorry about saying 6 in previous post

This post was edited by brmv on Jan 28 2013 10:17pm
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Jan 28 2013 11:10pm
not the cleanest but here goes:

I don't want to paste it here as white space is important in python and it is hard to get correct here, even with the code tags:

I put it on codepad where you can execute on their site:

http://python.codepad.org/ZyFMF46S

This post was edited by Azrad on Jan 28 2013 11:30pm
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