I need to check if the imaginary part is very small and set it to zero if it is in order to eliminate some floating point errors that result in very small non-zero imaginary parts when it should be zero.
My code is as follows:
kz2 = SQRT((n2*(2.0*PI*eta))**2 - kxarray(p)**2)
kz1 = SQRT((n1*(2.0*PI*eta))**2 - kxarray(p)**2)
if (aimag(kz2) < 0.0005) then
kz2 = (REAL(kz2),0.0)
end if
if (aimag(kz1) < 0.0005) then
kz1 = (REAL(kz1), 0.0)
end if
Unfortunately the compiler just returns:
gaussian1.f90:122.18:
kz2 = (REAL(kz2),0.0)
1
Error: Expected a right parenthesis in expression at (1)
gaussian1.f90:126.18:
kz1 = (REAL(kz1), 0.0)
1
Error: Expected a right parenthesis in expression at (1)
Any advice would be greatly appreciated - am I even going about this problem the right way?
UPDATE: I managed to avoid the problem by using:
if (aimag(kz2) < 0.0005) then
kz2 = real(kz2)
end if
if (aimag(kz1) < 0.0005) then
kz1 = real(kz1)
end if
But what would I do if I wanted to set the imaginary part to a non-zero amount?
In Fortran 2008 there are even more possibilities. You can access real and imaginary parts as derived type components, e.g.
a = c%re
b%im = 5
So, to set imaginary part of z to zero in new compilers you can try z%im = 0 .
I think you are looking for the CMPLX function, which converts real or integer arguments to a complex number. So it you example you should be able to do something like this:
kz1 = cmplx(real(kz1), 0.)
The (1.0,1.0) style parenthesis notation you have tried is only valid for constant values, not forming a complex number from the values held in variables.
Related
I have an array of registers/buses and a single result bus defined as follows.
wire [BW-1:0] bus_array[NUM-1:0];
reg [BW-1:0] and_result;
where
parameter BW = 4;
parameter NUM = 8;
I wish to perform a BW-bit AND operation on the elements of the array and assign the result to the register and_result.
I try doing this as follows.
integer l;
generate
genvar m;
for (m=0; m<BW; m=m+1)
begin : BW_LOOP
always # (*)
begin
and_result[m] = 1'b1;
for (l=0; l<NUM; l=l+1)
and_result[m] = and_result[m] & bus_array[l][m];
end
end
endgenerate
However when I simulate this in Modelsim 10.1e, I get the following error.
Error: (vsim-3601) Iteration limit reached at time 2 ns
If I do not use the generate loop, and instead have BW instances of the always # (*) block, the simulation works okay.
I can infer from the error message, that there is a problem with the generate for loop, but I am not able to resolve the problem.
Most likely a bug with ModelSim. Reproducible on EDAplayground with ModelSim10.1d. Works fine with Riviera2014 (After localizing l inside the generate loop). I'm guessing that and_result[m] is somehow in the #(*) sensitivity list, which it shouldn't be.
l needs to be localized or it will be accessed in parallel with the generated always blocks; creating a potential raise condition.
One workaround is to use SystemVerilog and use always_comb instead of always #(*).
A backwarnd compatable solution is to change and_result[m] = and_result[m] & bus_array[l][m]; to if (bus_array[l][m]==1'b0) and_result[m] = 1'b0; which is equivalent code. This keeps and_result[m] only on as a left hand expression so it cannot be in the sensitivity list.
genvar m;
for (m=0; m<BW; m=m+1)
begin : BW_LOOP
integer l; // <== 'l' is local to this generate loop
always # (*)
begin
and_result[m] = 1'b1;
for (l=0; l<NUM; l=l+1) begin
if (bus_array[l][m]==1'b0) begin
and_result[m] = 1'b0;
end
end
end
Working code here
I have to calculate the value of arctan(x) . I have calculated the value of this by evaluating the following series :
Arctan (x) = x – x^3/3 + x^5/5 – x^7/7 + x^9/9 - …
But the following code can not calculate the actual value. For example, calculate_angle(1) returns 38.34 . Why?
const double DEGREES_PER_RADIAN = 57.296;
double calculate_angle(double x)
{
int power=5,count=3;
double term,result,prev_res;
prev_res = x;
result= x-pow(x,3)/3;
while(abs(result-prev_res)<1e-10)
{
term = pow(x,power)/power;
if(count%2==0)
term = term*(-1);
prev_res=result;
result=result+term;
++count;
power+=2;
// if(count=99)
// break;
}
return result*DEGREES_PER_RADIAN;
}
EDIT: I found the culprit. You forgot to include stdlib.h, where the function abs resides. You must have ignored the warning about abs being implicitly declared. I checked that removing the include yields the result 38.19 and including it yields the result ~45.
The compiler is not required to stop compilation when an undeclared function is being used (in this case abs). Instead, it is allowed to make assumptions on how the function is declared (in this case, wrong one.)
Besides, like other posters already stated, your use of abs is inappropriate as it returns an int, not a double or float. The condition in the while should be >1e-100 not <1e-100. The 1e-100 is also too small.
--
You forgot to increase count and power after calculating the first two summands:
prev_res = x;
result= x-pow(x,3)/3;
count = 4; <<<<<<
power = 5; <<<<<<
while(abs(result-prev_res)<1e-100)
{
term = pow(x,power)/power;
if(count%2==1)
term = term*(-1);
Also I consider your use of the count variable counterintuitive: it is intialized with 3 like if it denotes the last used power; but then, loop iterations increase it by 1 instead of 2 and you decide the sign by count%2 == 1 as opposed to power%4 == 3
The series converges to tan^{-1} x, but not very fast. Consider the series when x=1:
1 - 1/3 + 1/5 - 1/7 + 1/9 - ...
What is the error when truncating at the 1/9 term? It's around 1/9. To get 10^{-100} accuracy, you would need to have 10^{100} terms. The universe would end before you'd get that. And, catastrophic round-off error and truncation error would make the answer utterly unreliable. You only have 14 digits to play with for doubles.
Look at reference works like Abramowitz and Stegun [AMS 55] or the new NIST Digital Library of Mathematical Functions at http://dlmf.nist.gov to see how these are done in practice. Often, one uses Padé approximants instead of Taylor series. Even when you stick with Taylor series, you often use Chebyshev approximation to cut down on the total error.
I also recommend Numerical Methods that [Usually] Work, by Forman Acton. Or the Numerical Recipes in ... series.
Your sign is the wrong way around after the first two terms. It should be:
if(count%2==0)
term = term*(-1);
Your comparison is the wrong way around in the while condition. Also, you're expecting an unrealistically high level of precision. I would suggest something more like this:
while(fabs(result-prev_res)>1e-8)
Finally, you'll get a more accurate result with a better value for DEGREES_PER_RADIAN. Why not something like this:
const double DEGREES_PER_RADIAN = 180/M_PI;
What would be the most efficient algorithm to solve a linear equation in one variable given as a string input to a function? For example, for input string:
"x + 9 – 2 - 4 + x = – x + 5 – 1 + 3 – x"
The output should be 1.
I am considering using a stack and pushing each string token onto it as I encounter spaces in the string. If the input was in polish notation then it would have been easier to pop numbers off the stack to get to a result, but I am not sure what approach to take here.
It is an interview question.
Solving the linear equation is (I hope) extremely easy for you once you've worked out the coefficients a and b in the equation a * x + b = 0.
So, the difficult part of the problem is parsing the expression and "evaluating" it to find the coefficients. Your example expression is extremely simple, it uses only the operators unary -, binary -, binary +. And =, which you could handle specially.
It is not clear from the question whether the solution should also handle expressions involving binary * and /, or parentheses. I'm wondering whether the interview question is intended:
to make you write some simple code, or
to make you ask what the real scope of the problem is before you write anything.
Both are important skills :-)
It could even be that the question is intended:
to separate those with lots of experience writing parsers (who will solve it as fast as they can write/type) from those with none (who might struggle to solve it at all within a few minutes, at least without some hints).
Anyway, to allow for future more complicated requirements, there are two common approaches to parsing arithmetic expressions: recursive descent or Dijkstra's shunting-yard algorithm. You can look these up, and if you only need the simple expressions in version 1.0 then you can use a simplified form of Dijkstra's algorithm. Then once you've parsed the expression, you need to evaluate it: use values that are linear expressions in x and interpret = as an operator with lowest possible precedence that means "subtract". The result is a linear expression in x that is equal to 0.
If you don't need complicated expressions then you can evaluate that simple example pretty much directly from left-to-right once you've tokenised it[*]:
x
x + 9
// set the "we've found minus sign" bit to negate the first thing that follows
x + 7 // and clear the negative bit
x + 3
2 * x + 3
// set the "we've found the equals sign" bit to negate everything that follows
3 * x + 3
3 * x - 2
3 * x - 1
3 * x - 4
4 * x - 4
Finally, solve a * x + b = 0 as x = - b/a.
[*] example tokenisation code, in Python:
acc = None
for idx, ch in enumerate(input):
if ch in '1234567890':
if acc is None: acc = 0
acc = 10 * acc + int(ch)
continue
if acc != None:
yield acc
acc = None
if ch in '+-=x':
yield ch
elif ch == ' ':
pass
else:
raise ValueError('illegal character "%s" at %d' % (ch, idx))
Alternative example tokenisation code, also in Python, assuming there will always be spaces between tokens as in the example. This leaves token validation to the parser:
return input.split()
ok some simple psuedo code that you could use to solve this problem
function(stinrgToParse){
arrayoftokens = stringToParse.match(RegexMatching);
foreach(arrayoftokens as token)
{
//now step through the tokens and determine what they are
//and store the neccesary information.
}
//Use the above information to do the arithmetic.
//count the number of times a variable appears positive and negative
//do the arithmetic.
//add up the numbers both positive and negative.
//return the result.
}
The first thing is to parse the string, to identify the various tokens (numbers, variables and operators), so that an expression tree can be formed by giving operator proper precedences.
Regular expressions can help, but that's not the only method (grammar parsers like boost::spirit are good too, and you can even run your own: its all a "find and recourse").
The tree can then be manipulated reducing the nodes executing those operation that deals with constants and by grouping variables related operations, executing them accordingly.
This goes on recursively until you remain with a variable related node and a constant node.
At the point the solution is calculated trivially.
They are basically the same principles that leads to the production of an interpreter or a compiler.
Consider:
from operator import add, sub
def ab(expr):
a, b, op = 0, 0, add
for t in expr.split():
if t == '+': op = add
elif t == '-': op = sub
elif t == 'x': a = op(a, 1)
else : b = op(b, int(t))
return a, b
Given an expression like 1 + x - 2 - x... this converts it to a canonical form ax+b and returns a pair of coefficients (a,b).
Now, let's obtain the coefficients from both parts of the equation:
le, ri = equation.split('=')
a1, b1 = ab(le)
a2, b2 = ab(ri)
and finally solve the trivial equation a1*x + b1 = a2*x + b2:
x = (b2 - b1) / (a1 - a2)
Of course, this only solves this particular example, without operator precedence or parentheses. To support the latter you'll need a parser, presumable a recursive descent one, which would be simper to code by hand.
I have a whole series of assignments which I have put on the same ike using a ";" to seperate the statemnts but I get this error:
1.0; lb(1,9)
1
Error: Unclassifiable statement at (1)
In file LJ.F90:223
I do not understand where there is coming from, when I have the code working if each statement is on its own line. The code is really simple...
What am I stupidly doing wrong.. below code is all on one line.
lb(1,1) = 1.0; lb(1,2) = 1.0; lb(1,3) = 1.0; lb(1,4) = 1.0; lb(1,5) = 1.0; lb(1,6) = 1.0; lb(1,7) = 1.0; lb(1,8) = 1.0; lb(1,9) = 1.0
Your line of code is 134 characters long and, even with Fortran 90-style free format code, most compilers impose a maximum line length. For example, with Sun Studio the default limit is 132 characters.
You can usually increase this character limit using compiler flags, but I suggest splitting that code so that you have one statement per line. It is more legible to human readers and compile- and run-time error messages may be more easily diagnosed.
Adding to the comments of #Deditos, in this case you could use Fortran array notation to reduce the number of lines since all of the elements are being set to the same value:
lb (1, 1:9) = 1.0
Are all elements of the array being initialized to 1.0? Then simply:
lb = 1.0
Hey, I wrote this (fortran) with the aim of finding the minimum spanning tree of a bunch of points (syscount of them). I know for a fact that this approach works, since i wrote it in javascript earlier today. js is slow though, and i wanted to see how much faster fortran would be!!
only problem is it's not working, i'm getting an annoying error;
prims.f95:72.43:
if((check == 1) .and. (path(nodesin(j))(k) < minpath)) then
1
Error: Expected a right parenthesis in expression at (1)
What the hell is that about?! the 43rd character on the line is the "h" of "path"
nodesin(1) = 1
do i = 1,syscount-1
pathstart = -1
pathend = -1
minpath = 2000
do j = 1,i
do k = 1, syscount
check = 1
do l = 1, i
if(nodesin(l) == k) then
check = 0
end if
end do
if((check == 1) .and. (path(nodesin(j))(k) < minpath)) then
minpath = path(nodesin(j))(k)
pathstart = nodesin(j)
pathend = k
end if
end do
end do
nodesin(i+1) = pathend
minpaths(i)(1) = pathstart
minpaths(i)(2) = pathend
end do
Also, i'm fairly new to fortran, so i have a few other questions;
can i use && instead of .and. ?
is there a versions of the for(object in list){} loop found in many other languages?
is there a verion of the php function in_array ? i.e. bool in_array(needle,haystack), and if there is, is there a better way of doing it than:
check = false
Asize = size(array)
do i = 1, Asize
if(array(i) == needle) then
check = true
end if
end do
then to using the check variable to see if it's there?
(I haven't posted anything on stackoverflow before. please don't get angry if i've broken loads of etiquette things!)
It looks like you have defined path and minpaths as two-dimensional arrays. Multi-dimensional arrays are accessed differently in Fortran when compared to C-like languages. In Fortran you separate the indices by commas within one set of parentheses.
I'm guessing by the use of these variables they are integer arrays. Here is how you access elements of those arrays (since you didn't share your variable declarations I am making up the shape of these arrays):
integer :: path(n1, n2)
integer :: minpaths(n3, 2)
your if statement should be:
if((check == 1) .and. (path(nodesin(j), k) < minpath)) then
your access to minpaths should be:
minpaths(i, 1) = pathstart
minpaths(i, 2) = pathend
Also, if you are not using IMPLICIT NONE I recommend you consider it. Not using it is dangerous, and you are using variable names that are close to each other (minpath and minpaths). You could save hours of hair pulling debugging by using IMPLICIT NONE.
While .EQ. can be replaced with ==, there is still only .AND.
For your code block to check whether a "variable is there", you can use "where" and have much shorter code!
In Fortran >= 90 statements and functions can operate on arrays so that explicit loops don't have to be used as frequently.
There is no for (object in list), but using the where statement can do something very similar.
Many of the intrinsic functions that act on arrays also take masks as optional arguments to selectively operate.
I suggest reading a book to learn about these features. I like the one by Metcalf, Reid and Cohen. In the meantime, the second Wikipedia article may help: http://en.wikipedia.org/wiki/Fortran_95_language_features