Is there a nice way to write a Fortran ASSOCIATE statement to turn this
FORALL (i = 2:n-2)
v(:,i) = v(:,i) + MATMUL(A, &
c(2)*u(:,i-2) + c(1)*u(:,i-1) + c(0)*u(:,i) + c(1)*u(:,i+1) + c(2)*u(:,i+2))
END FORALL
into something like the following
ASSOCIATE ( U => ..., V => ...)
FORALL (i = 2:n-2)
V(i) = V(i) + MATMUL(A, &
c(2)*U(i-2) + c(1)*U(i-1) + c(0)*U(i) + c(1)*U(i+1) + c(2)*U(i+2))
END FORALL
END ASSOCIATE
I'm staring at Adams et al's The Fortran 2003 Handbook section 8.2, but I can't see how to write the associate-name => selector construct to allow for indexed access into the associate-name.
Obviously what I'm going for is overkill for a couple of lines. I've got a bunch of 'em that I'd like to condense.
Unless I'm misreading things, I don't think this is possible. The specification says (section 8.1.4.3):
Within a SELECT TYPE or ASSOCIATE construct, each associating entity has the same rank as its
associated selector.
and as far as I can see, you want a rank 1 associating entity (V) and will need a rank 2
associated selector (to hold v).
If the aim is to make the code shorter/nicer, I think the best way to achieve that does not need an ASSOCIATE construct:
forall(i=2:n-2)
v(:,i) = v(:,i) + MATMUL(A,MATMUL(u(:,i-2:i+2),c([2,1,0,1,2]))
end forall
which, as long as the 5-sized coefficient array does not change, could be preinitialized as a
real :: c5(5)
c5 = c([2,1,0,1,2])
And then run in a single line as
forall(i=2:n-2) v(:,i) = v(:,i) + MATMUL(A,MATMUL(u(:,i-2:i+2),c5))
note that it must be lbound(u,2)<=0.
Related
I am fairly new to fortran and am using it for some molecular dynamics work
I have a text file that contains xyz coordinates of multiple atoms. Edit: The data is repetitive
12
xyz
OH2 2.056771 0.152501 -3.407425
H1 2.086389 -0.658114 -2.899234
H2 1.325328 0.643692 -3.033321
OH2 -1.620865 1.026821 -4.353753
H1 -1.045534 1.344086 -5.049863
H2 -1.107708 1.130454 -3.552402
OH2 -2.064113 1.377066 -1.093998
H1 -1.228430 1.344786 -1.559641
H2 -2.692285 1.681116 -1.749120
OH2 1.451636 1.645941 -0.456822
H1 0.841741 1.630468 -1.194400
H2 1.251076 0.850951 0.037141
Where OH2 and H1/H2 are atoms.
I would like to be able subtract the xyz coor of the H2(hydrogen) following the preceding OH2(oxygen) for every OH2 that is recognized.
So far my code looks as follows:
program placepoint
implicit none
real(kind(0.0d0)) :: xCoor(1:12), yCoor(1:12), zCoor(1:12)
real(kind(0.0d0)) :: v1_x, v1_y, v1_z, dOH, nx, ny, nz
real(kind(0.0d0)) :: ip_x, ip_y, ip_z, norm
integer :: j, n
character*20 :: dumch(1:12)
open(unit = 10, file = 'hoh.xyz')
open(unit = 13, file = 'ipcoor.txt')
do j= 1, 12
read(10,*) dumch(j) , xCoor(j), yCoor(j), zCoor(j)
!Calculate vector 1 along OXR-HX bond that can be used to place a point ip (R1)
if (dumch(j) .eq. "OH2") then
v1_x = xCoor(j+2) - xCoor(j)
v1_y = yCoor(j+2) - yCoor(j)
v1_z = zCoor(j+2) - zCoor(j)
dOH = sqrt((v1_x)**2 + (v1_y)**2 + (v1_z)**2)
! Normalize vector
nx = v1_x/dOH
ny = v1_y/dOH
nz = v1_z/dOH
!Place ip at 0.7Å along the OH bond (this is not exactly at the correct HX-OXR-OR angle but these are for dummy atoms and shake should take care of this during dyn runs)
ip_x = xCoor(1) + 0.7*nx
ip_y = yCoor(1) + 0.7*ny
ip_z = zCoor(1) + 0.7*nz
end if
end do
write(13,*) 'ip' , ip_x, ip_y, ip_z, dOH
end program place point
I realize I am being to naive using the j+2 as a identifier to find the value of H2 immediately following the OH2, but unfortunately I have not found any other better way. I also get the fortran error "End of file" after proper compilation. So I am expecting that I am perhaps making multiple mistakes here. Any help with this will be so appreciated!
It is difficult for me to describe the problem in the title, so excuse any misleading description.
The easiest way to describe what I need is with an example. I have a table like:
A B C
1 x
2 x x
3 x x
Now what I want is the formula in a cell for every single column and row with each of the column or row name for every x that is placed. In the example like:
A B C
1,2 2,3 3
1 A x
2 A, B x x
3 B, C x x
The column and row names are not equivalent to the excel designation. It works with an easy WHEN statement for single cells (=WHEN(C3="x";C1)), but not for a bunch of them (=WHEN(C3:E3="x";C1:E1)). How should/can such a formula look like?
So I found the answer to my problem. Excel provides the normal CONCATENATE function. What is needed is something like a CONCATENATEIF (in German = verkettenwenn) function. By adding a module in VBA based on a thread from ransi from 2011 on the ms-office-forum.net the function verkettenwenn can be used. The code for the German module looks like:
Option Explicit
Public Function verkettenwenn(Bereich_Kriterium, Kriterium, Bereich_Verketten)
Dim mydic As Object
Dim L As Long
Set mydic = CreateObject("Scripting.Dictionary")
For L = 1 To Bereich_Kriterium.Count
If Bereich_Kriterium(L) = Kriterium Then
mydic(L) = Bereich_Verketten(L)
End If
Next
verkettenwenn = Join(mydic.items, ", ")
End Function
With that module in place one of the formula for the mentioned example looks like: =verkettenwenn(C3:E3;"x";$C$1:$K$1)
The English code for a CONCATENATEIF function should probably be:
Option Explicit
Public Function CONCATENATEIF(Criteria_Area, Criterion, Concate_Area)
Dim mydic As Object
Dim L As Long
Set mydic = CreateObject("Scripting.Dictionary")
For L = 1 To Criteria_Area.Count
If Criteria_Area(L) = Criterion Then
mydic(L) = Concate_Area(L)
End If
Next
CONCATENATEIF = Join(mydic.items, ", ")
End Function
"if then statement in VBA" I'm writing a program that puts a number in a cell in Excell if a variable reaches a certain value. I understand how to declare variables but I don't know how to tell excel to write x if A1 =34. Thanks
Add a listener to your worksheet to capture a Range. You can make the range [A1] if you are only watching a specific column/row, or you can add a range like I have below.
Private Sub Worksheet_Change(ByVal Target As Range)
Dim KeyCells As Range
Set KeyCells = Range("A:A")
If Not Application.Intersect(KeyCells, Range(Target.Address)) _
Is Nothing Then
If Target.Value = "34" Then
Cells(Target.Row, 2) = "X"
Else
Cells(Target.Row, 2) = ""
End If
End If
End Sub
Change "x" to if you want variable x and not literal x.
If your goal is to change the value of the cell to "X" (Literal X), and you are not having macros run constantly or with each cell change, you can use the following function (or similar) in each cell in which you have a conditional.
See the Microsoft support on this topic https://support.microsoft.com/en-us/kb/213612
It's not clear what you wish to do, but let's say you want to write the current value of your variable, x, into cell B2... if cell A1 is 34.
In the above case, you would do this:
If [a1] = 34 then [b2] = x
Private Sub CommandButton1_Click()
Dim lr As Long
lr = Worksheets("New").Range("A" & Rows.Count).End(xlUp).Row
Worksheets("New").Range("T2").Formula = "=LEFT(B2,2)"
Worksheets("New").Range("T2").AutoFill Destination:=Worksheets("New").Range("T2:T" & lr)
Worksheets("New").Range("U2").Formula = "=(T2&0&0)"
Worksheets("New").Range("U2").AutoFill Destination:=Worksheets("New").Range("U2:U" & lr)
Worksheets("New").Range("V2").Formula = "=IF(AND(a2=A1,U2=U1),"",A2")" (HOW TO AUTO FILL THIS FORMULA IN A CELL)
Worksheets("New").Range("V2").AutoFill Destination:=Worksheets("New").Range("V2:V" & lr)
End Sub
I need to write a regular expression that can detect a string that contains only the characters x,y, and z, but where the characters are different from their neighbors.
Here is an example
xyzxzyz = Pass
xyxyxyx = Pass
xxyzxz = Fail (repeated x)
zzzxxzz = Fail (adjacent characters are repeated)
I thought that this would work ((x|y|z)?)*, but it does not seem to work. Any suggestions?
EDIT
Please note, I am looking for an answer that does not allow for look ahead or look behind operations. The only operations allowed are alternation, concatenation, grouping, and closure
Usually for this type of question, if the regex is not simple enough to be derived directly, you can start from drawing a DFA and derive a regex from there.
You should be able to derive the following DFA. q1, q2, q3, q4 are end states, with q1 also being the start state. q5 is the failed/trap state.
There are several methods to find Regular Expression for a DFA. I am going to use Brzozowski Algebraic Method as explained in section 5 of this paper:
For each state qi, the equation Ri is a union of terms: for a transition a from qi to qj, the term is aRj. Basically, you will look at all the outgoing edges from a state. If Ri is a final state, λ is also one of the terms.
Let me quote the identities from the definition section of the paper, since they will come in handy later (λ is the empty string and ∅ is the empty set):
(ab)c = a(bc) = abc
λx = xλ = x
∅x = x∅ = ∅
∅ + x = x
λ + x* = x*
(λ + x)* = x*
Since q5 is a trap state, the formula will end up an infinite recursion, so you can drop it in the equations. It will end up as empty set and disappear if you include it in the equation anyway (explained in the appendix).
You will come up with:
R1 = xR2 + yR3 + zR4 + λ
R2 = + yR3 + zR4 + λ
R3 = xR2 + + zR4 + λ
R4 = xR2 + yR3 + λ
Solve the equation above with substitution and Arden's theorem, which states:
Given an equation of the form X = AX + B where λ ∉ A, the equation has the solution X = A*B.
You will get to the answer.
I don't have time and confidence to derive the whole thing, but I will show the first few steps of derivation.
Remove R4 by substitution, note that zλ becomes z due to the identity:
R1 = xR2 + yR3 + (zxR2 + zyR3 + z) + λ
R2 = + yR3 + (zxR2 + zyR3 + z) + λ
R3 = xR2 + + (zxR2 + zyR3 + z) + λ
Regroup them:
R1 = (x + zx)R2 + (y + zy)R3 + z + λ
R2 = zxR2 + (y + zy)R3 + z + λ
R3 = (x + zx)R2 + zyR3 + z + λ
Apply Arden's theorem to R3:
R3 = (zy)*((x + zx)R2 + z + λ)
= (zy)*(x + zx)R2 + (zy)*z + (zy)*
You can substitute R3 back to R2 and R1 and remove R3. I leave the rest as exercise. Continue ahead and you should reach the answer.
Appendix
We will explain why trap states can be discarded from the equations, since they will just disappear anyway. Let us use the state q5 in the DFA as an example here.
R5 = (x + y + z)R5
Use identity ∅ + x = x:
R5 = (x + y + z)R5 + ∅
Apply Arden's theorem to R5:
R5 = (x + y + z)*∅
Use identity ∅x = x∅ = ∅:
R5 = ∅
The identity ∅x = x∅ = ∅ will also take effect when R5 is substituted into other equations, causing the term with R5 to disappear.
This should do what you want:
^(?!.*(.)\1)[xyz]*$
(Obviously, only on engines with lookahead)
The content itself is handled by the second part: [xyz]* (any number of x, y, or z characters). The anchors ^...$ are here to say that it has to be the entirety of the string. And the special condition (no adjacent pairs) is handled by a negative lookahead (?!.*(.)\1), which says that there must not be a character followed by the same character anywhere in the string.
I've had an idea while I was walking today and put it on regex and I have yet to find a pattern that it doesn't match correctly. So here is the regex :
^((y|z)|((yz)*y?|(zy)*z?))?(xy|xz|(xyz(yz|yx|yxz)*y?)|(xzy(zy|zx|zxy)*z?))*x?$
Here is a fiddle to go with it!
If you find a pattern mismatch tell me I'll try to modify it! I know it's a bit late but I was really bothered by the fact that I couldn't solve it.
I understand this is quite an old question and has an approved solution as well. But then I am posting 1 more possible and quick solution for the same case, where you want to check your regular expression that contains consecutive characters.
Use below regular expression:
String regex = "\\b\\w*(\\w)\\1\\1\\w*";
Listing possible cases that above expression returning the result.
Case 1: abcdddd or 123444
Result: Matched
Case 2: abcd or 1234
Result: Unmatched
Case 3: &*%$$$ (Special characters)
Result: Unmatched
Hope this will be helpful...
Thanks:)
This is not homework, but an old exam question. I am curious to see the answer.
We are given an alphabet S={0,1,2,3,4,5,6,7,8,9,+}. Define the language L as the set of strings w from this alphabet such that w is in L if:
a) w is a number such as 42 or w is the (finite) sum of numbers such as 34 + 16 or 34 + 2 + 10
and
b) The number represented by w is divisible by 3.
Write a regular expression (and a DFA) for L.
This should work:
^(?:0|(?:(?:[369]|[147](?:0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[147]0*(?:\+?(?:0\
+)*[369]0*)*\+?(?:0\+)*[258])*(?:0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[258]|0*(?:
\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[147]0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[147])|[
258](?:0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[258]0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0
\+)*[147])*(?:0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[147]|0*(?:\+?(?:0\+)*[369]0*)
*\+?(?:0\+)*[258]0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[258]))0*)+)(?:\+(?:0|(?:(?
:[369]|[147](?:0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[147]0*(?:\+?(?:0\+)*[369]0*)
*\+?(?:0\+)*[258])*(?:0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[258]|0*(?:\+?(?:0\+)*
[369]0*)*\+?(?:0\+)*[147]0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[147])|[258](?:0*(?
:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[258]0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[147])*
(?:0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[147]|0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)
*[258]0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[258]))0*)+))*$
It works by having three states representing the sum of the digits so far modulo 3. It disallows leading zeros on numbers, and plus signs at the start and end of the string, as well as two consecutive plus signs.
Generation of regular expression and test bed:
a = r'0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*'
b = r'a[147]'
c = r'a[258]'
r1 = '[369]|[147](?:bc)*(?:c|bb)|[258](?:cb)*(?:b|cc)'
r2 = '(?:0|(?:(?:' + r1 + ')0*)+)'
r3 = '^' + r2 + r'(?:\+' + r2 + ')*$'
r = r3.replace('b', b).replace('c', c).replace('a', a)
print r
# Test on 10000 examples.
import random, re
random.seed(1)
r = re.compile(r)
for _ in range(10000):
x = ''.join(random.choice('0123456789+') for j in range(random.randint(1,50)))
if re.search(r'(?:\+|^)(?:\+|0[0-9])|\+$', x):
valid = False
else:
valid = eval(x) % 3 == 0
result = re.match(r, x) is not None
if result != valid:
print 'Failed for ' + x
Note that my memory of DFA syntax is woefully out of date, so my answer is undoubtedly a little broken. Hopefully this gives you a general idea. I've chosen to ignore + completely. As AmirW states, abc+def and abcdef are the same for divisibility purposes.
Accept state is C.
A=1,4,7,BB,AC,CA
B=2,5,8,AA,BC,CB
C=0,3,6,9,AB,BA,CC
Notice that the above language uses all 9 possible ABC pairings. It will always end at either A,B,or C, and the fact that every variable use is paired means that each iteration of processing will shorten the string of variables.
Example:
1490 = AACC = BCC = BC = B (Fail)
1491 = AACA = BCA = BA = C (Success)
Not a full solution, just an idea:
(B) alone: The "plus" signs don't matter here. abc + def is the same as abcdef for the sake of divisibility by 3. For the latter case, there is a regexp here: http://blog.vkistudios.com/index.cfm/2008/12/30/Regular-Expression-to-determine-if-a-base-10-number-is-divisible-by-3
to combine this with requirement (A), we can take the solution of (B) and modify it:
First read character must be in 0..9 (not a plus)
Input must not end with a plus, so: Duplicate each state (will use S for the original state and S' for the duplicate to distinguish between them). If we're in state S and we read a plus we'll move to S'.
When reading a number we'll go to the new state as if we were in S. S' states cannot accept (another) plus.
Also, S' is not "accept state" even if S is. (because input must not end with a plus).