I'm stuck in a process where I need to compute the values of a function f[x,y,z] on a grid. Here I put how I wrote the program, only evaluating on a one-dimensional grid.
I wrote the program:
program CHISQUARE_MINIMIZATION_VELOCITY_PROFILES
use distribution
IMPLICIT none
integer, parameter :: kp=1001 ! Parameter which states the number of points on the grid.
integer, parameter :: ndata=13 ! Parameter which states the number of elements of the data file.
integer, parameter :: nconst=3 ! Fixed integer parameter.
integer i, j, n
real*8 rc0, rcf, V00, V0f, d00, d0f, rc, V0, d, z
real*8 rcr(kp), V0r(kp), d0r(kp), chisq(kp)
!Scaling radius range
rc0=0.0d-5 ! kpc
rcf=1.0d2 ! kpc
call linspace(rc0,rcf,kp,rcr)
!**************If I call like this, it works normal*****************
!CHISQUARED(1.3d0, 130.2d0, 0.12d0, 1.0d0, 1.0d0, 2.0d0, 0.0d0, 0.0d0, 1, !ndata, nconst)
! **1.27000000000000 0.745818846396887**
! Press any key to continue
!**************If I call like this, it works normal*****************
!******* Here is where my problem is****************
do j=1, kp
rc=rcr(j)
write(*,*) rc, CHISQUARED(rc, 130.2d0, 0.12d0, 1.0d0, 1.0d0, 2.0d0, 0.0d0, 0.0d0, 1, ndata, nconst)
enddo
!******* Here is where my problem is****************
end program CHISQUARE_MINIMIZATION_VELOCITY_PROFILES
I use the module where I compute the chi^2 distribution, coming from a theoretical model...
MODULE distribution
IMPLICIT NONE
CONTAINS
! I define here the chi^2 function****
real*8 function CHISQUARED(rc, V0, d, alpha, gamma, chi, a, b, n, ndata, nconst)
integer i, n, ndata, nconst
real*8 rc, V0, d
real*8 alpha, gamma, chi, a, b, s
real*8, DIMENSION(ndata,3) :: X
open(unit=1, file="data.txt")
s=0.0d0
do i=1, ndata
Read(1,*) X(i,:)
s=s+((X(i,2)-VELOCITYPROFILE(X(i,1), rc, V0, d, alpha, gamma, chi, a, b, n))/(X(i,3)))**2.0d0
end do
CHISQUARED=s/(ndata-nconst)
end function CHISQUARED
!****Here I define the model function
real*8 function VELOCITYPROFILE(r, rc, V0, d, alpha, gamma, chi, a, b, n)
integer i, n
real*8 r, rc, V0, d, alpha, gamma, chi, a, b, z
if (rc < 0.0d0 .OR. d < 0.0d0 .OR. a <0.0d0 .OR. b <0.0d0 .OR. alpha < 0.0d0 .OR. gamma <0.0d0 .OR. chi < 0.0d0 .OR. n<1 ) then
VELOCITYPROFILE=0.0d0
return
else
z=0.0d0
do i=0,n
z=z+((V0*((r/rc)**(1.5d0))*(1+a+r/rc)**(-gamma*(2*n+0.5d0)))/((a+(r/rc)**alpha)**(chi/2.0d0)))*(((b+r/rc)**gamma)/d)**i
end do
VELOCITYPROFILE=z
end if
end function VELOCITYPROFILE
END MODULE distribution
!*****************END OF THE MODULE******************************
the data.txt file is of the form
0.24 37.31 6.15
0.28 37.92 5.5
0.46 47.12 3.9
0.64 53.48 2.8
0.73 55.14 3.3
0.82 58.47 2.5
1.08 66.15 3.3
1.22 69.39 2.75
1.45 74.55 5.
1.71 77.94 2.93
1.87 81.66 2.5
2.2 86.81 3.02
2.28 90.08 2.1
2.69 94.38 3.92
2.7 95.36 1.8
In order to get several values of the function CHISQUARED, I use the subroutine linspace to generate the partition of the 1-dimensional grid
subroutine linspace(xi,xf,jmax,y)
integer jmax,j
real*8 xi,xf,y(jmax)
y=(/(xi+dble(j-1)*(xf-xi)/(dble(jmax)-1.0d0), j=1, jmax)/)
end subroutine linspace
What happens is that if in the main program, I call the function CHISQUARED like this:
CHISQUARED(1.3d0, 130.2d0, 0.12d0, 1.0d0, 1.0d0, 2.0d0, 0.0d0, 0.0d0, 1, ndata, nconst)
**1.27000000000000 0.745818846396887**
Press any key to continue
I get some finite value, like, I don't know, 0.7 or something like this. (I restricted the data file so the result won't be the one written, I just put 0.7 as an example). However, when I put it inside a loop as it is in the program written above, to get the values on the one dimensional grid, it gives me the error
**0.000000000000000E+000 NaN**
forrtl: severe (24): end-of-file during read, unit 1, file C:\Users\Ernesto Lopez Fune\Desktop\Minimize\newone\chisquarerotationcurve\data.txt
Image PC Routine Line Source
chisquarerotation 0040B889 Unknown Unknown Unknown
Press any key to continue
Can anyone recommend me what to do in this case? How to overcome this barrier?
According to your error, you reach the end of your file.
When you call your subroutine once, it's OK but in a loop, your file is read multiple times. After the first iteration, your file is read until the EOF control but for the next iteration, the program can't read anymore because it has already reached the end of the file.
You need to use the REWIND(1) statement before end function CHISQUARED. With this, the cursor will be re-positioned at the beginning of the file. Besides, I think it would be better to OPEN your file in the main program and not in a function or subroutine to avoid multiple OPEN/CLOSE.
Don't forget to CLOSE your file when you are done dealing with it.
Related
I actually want to solve an n variable hamilton's sets of equations. In fortran,to define a function, we, generally do the following.
function H(x,p) result(s)
real::x,p,s
s=x**2+p**2
end function H
Now, if I wish to solve an n variable hamilton's equation, I need to define an n variable H(x(i),p(i)) where i runs from 1 to n. Suppose p(i) are the variables and H is p(i)^2, summed over i from 1 to n.
What are the possible ways of defining a function with an array as input? It is not possible to write H(x1,x2....x100...) manuaaly each time.
module aa
implicit none
public :: H
contains
function H(x,p) result(s)
real, dimension(:), intent(in) :: x,p
real, dimension(:), allocatable :: s
integer :: i, n
n = size(x, 1)
allocate(s(n))
do i=1, n
s(i) = x(i)**2 + p(i)**2
enddo
end function H
end module aa
program test
use aa
real, dimension(10) :: x, p
real, dimension(:), allocatable :: s
integer :: n
x(:) = 1.
p(:) = 1.
n = size(x, 1)
allocate(s(n))
s(:) = 0.
s = H(x,p)
print*, s
end program test
Compiled and tested with GNU Fortran (Ubuntu 7.4.0-1ubuntu1~18.04.1) 7.4.0
This is not really a real case program example because if you know the dimensions of x and p, you also know the dimensions of s, so you could have just defined it instead of allocating. But it can be used to generalize modules once there is no reference to any dimension in the module.
For this to work, you will notice that s must be allocatable and must have been allocated before calling the function.
I have a problem with a Fortran project and figured out maybe you could help me.
I'm using codeblocks as IDE and there you can make projects, so I created a project with two files in it: a main program and a fuction (I don't know what else to use, I could use something different from a fuction maybe).
So I have my function that reads values from a .txt and saves them as real numbers and everything is working good. What I want to do is, from the file main, call this function and save in main the data I collected with my function so that the main remains cleaner.
How would I do that? I can post the whole script if you want, but I don't think it would add that much more.
EDIT: As you asked, here it is (uncut):
program main
! Variables
real :: d1, r1, r2, a, teta, freq, Dt, mu, g0, r_t, height, r, omega, H, lx, ly, lz, m_c0, &
Jx, Jy, Jz, gmax, I_s, K, Jx0, Jy0, Jz0, Vmin, tsp_0, Fmax, Isp, c1, n, F, DV, tfin, cont, Tmax
real :: data_input
! Call the funcion
data_input=data_module(d1, r1, r2, a, teta, freq, Dt, mu, g0, r_t, height, r, omega, H, lx, ly, lz, m_c0, &
Jx, Jy, Jz, gmax, I_s, K, Jx0, Jy0, Jz0, Vmin, tsp_0, Fmax, Isp, c1, n, F, DV, tfin, cont, Tmax)
! Error
if (data_input/=1) then
print*, 'ERROR: data_module did not work'
end if
!Just to show it
print*,'After'
print*, d1, r1, r2, a, teta, freq, Dt, mu, g0, r_t, height, r, omega, H, lx, ly, lz, m_c0, &
Jx, Jy, Jz, gmax, I_s, K, Jx0, Jy0, Jz0, Vmin, tsp_0, Fmax, Isp, c1, n, F, DV, tfin, cont, Tmax
end program main
real function data_module ()
! Variables
implicit none
integer :: flag_read=0, w_int, d_int
real:: coefficient, d1, r1, r2, a, teta, freq, Dt, mu, g0, r_t, height, r, omega, H, lx, ly, lz, m_c0, &
Jx, Jy, Jz, gmax, I_s, K, Jx0, Jy0, Jz0, Vmin, tsp_0, Fmax, Isp, c1, n, F, DV, tfin, cont, Tmax
character (LEN=35) :: starting_string, name*15, coefficient_string*20, w_string, d_string, number_format
character :: w*2, d
! Open file
open (11, file = 'Data.txt', status = 'old', access = 'sequential', form = 'formatted')
! Read a new line for every iteration
sentence_reader: do while (flag_read==0)
read (11, fmt='(A)', iostat = flag_read) starting_string
! Error
if (flag_read>0)then
print*, 'ERROR: could not read data'
stop
end if
! Skip useless lines
if (starting_string(1:1)=='%' .OR. starting_string(1:1)==' ') then
cycle
end if
! Exit when you're done
if (flag_read<0)then
exit sentence_reader
end if
! Just stuff to prepare it
name=trim(starting_string(1:index(starting_string, '=')-1))
coefficient_string=trim(adjustl(starting_string(index(starting_string, '=')+1:index(starting_string,';')-1)))
if (scan(coefficient_string,'E')/=0) then
w_string=coefficient_string
w_int=len_trim(w_string)
write(w, '(BN,I2)') w_int
d_string=coefficient_string(index(coefficient_string, '.')+1:index(coefficient_string, 'E')-1)
d_int=len_trim(d_string)
write(d, '(BN,I1)') d_int
!All togheter
number_format='(BN,F' // trim(w) // '.' // d // ')'
else
w_string=coefficient_string
w_int=len_trim(w_string)
write(w, '(BN,I1)') w_int
d_string=coefficient_string(index(coefficient_string, '.')+1:len_trim(coefficient_string))
d_int=len_trim(d_string)
write(d, '(BN,I1)') d_int
number_format='(BN,F' // trim(w) // '.' // d // ')'
end if
! Read the number
read(coefficient_string,number_format) coefficient
! Save where it's needed (is there an easier way to do it?)
select case (name)
case ('d1')
d1=coefficient
case ('r1')
r1=coefficient
case ('r2')
r2=coefficient
case ('a')
exit
case ('teta')
exit
case ('freq')
freq=coefficient
case ('Dt')
exit
case ('mu')
mu=coefficient
case ('g0')
g0=coefficient
case ('r_t')
r_t=coefficient
case ('height')
height=coefficient
case ('lx')
lx=coefficient
case ('ly')
ly=coefficient
case ('lz')
lz=coefficient
case ('m_c0')
m_c0=coefficient
case ('Jx')
Jx=coefficient
case ('Jy')
Jy=coefficient
case ('Jz')
Jz=coefficient
case ('gmax')
gmax=coefficient
case ('I_s')
I_s=coefficient
case ('K')
K=coefficient
case ('Vmin')
Vmin=coefficient
case ('tsp_0')
tsp_0=coefficient
case ('Fmax')
Fmax=coefficient
case ('Isp')
Isp=coefficient
case ('n')
n=coefficient
case ('tfin')
tfin=coefficient
case ('cont')
cont=coefficient
case ('Tmax')
Tmax=coefficient
case default
print*, 'Variable ', name, ' is not recognized'
end select
end do sentence_reader
! Other stuff I need
teta=atan((r1 - r2)/d1)
a=sqrt(d1**2 + (r1 - r2)**2)
Dt=1/freq
r=r_t + height
omega=(mu/(r**3))**0.5
H=(r*mu)**0.5
Jx0=Jx - I_s
Jy0=Jy - I_s
Jz0=Jz - I_s
c1=Isp*g0
F=n*Fmax
DV=(F/m_c0)*tsp_0
! Shows that the function is correctly executed
data_module=1
print*,'Before'
print*, d1, r1, r2, a, teta, freq, Dt, mu, g0, r_t, height, r, omega, H, lx, ly, lz, m_c0, &
Jx, Jy, Jz, gmax, I_s, K, Jx0, Jy0, Jz0, Vmin, tsp_0, Fmax, Isp, c1, n, F, DV, tfin, cont, Tmax
end function data_module
PS. I know modules, but with open and all the other stuff I couldn't get them to work. Would love to.
What I want to do is to pass the data d1, r1, ecc that I collected in data_module to main and save them in main, but doing it this way it doesn't save them (if you run it, when you print them "before" everything is fine, when you print them "after" you got all zeros.
Okay, there are a few things I notice.
Your function is of type real, but you set it only to 1 (an integer), to, as you put it in the comment "Show that the function is correctly executed".
It's not uncommon to make a procedure return a value to show whether it executed correctly or not, but it's usually an error code, with zero meaning that no error occurred and everything went fine.
Also, you might want to declare the function as integer instead of real, as integers are better for that kind of thing. (More reliable to compare.)
As to your actual question: If you want to pass more than a single value back to the calling routine, you would want to declare intent(out) dummy variables. See this example:
integer function test_output(outdata)
integer, intent(out) :: outdata(10)
integer :: i
outdata = (/(i, i=1, 10)/)
! All worked well
test_output = 0
return
end function test_output
Modules are the way to go. Here is a very limited example on how to incorporate the function above into a module, and using that module in a program:
module mod_test
implicit none
! Here you can place variables that should be available
! to any procedure using this module
contains
! Here you can place all the procedures (functions and
! subroutines)
integer function test_output(outdata)
integer, intent(out) :: outdata(10)
integer :: i
outdata = (/(i, i=1, 10)/)
! All worked well
test_output = 0
return
end function test_output
end module mod_test
program test
! The 'USE' statement is the only thing that needs to be
! *ahead* of the 'implicit none'
use mod_test
implicit none
integer :: mydata(10) ! The variable that will contain the data
! from the function
integer :: status ! The variable that will contain the error
! code.
status = test_output(mydata)
if (status == 0) then
print*, mydata
end if
end program test
If the module is in a different source file, you need to compile them this way (assuming that you use gfortran):
$ gfortran -c -o mod_test.o mod_test.f90
$ gfortran -c -o test.o test.f90
$ gfortran -o test test.o mod_test.o
I have I simple problem but I cannot find a solution anywhere.
I have to integrate a function (for example using a Simpson's rule subroutine) but I am obliged to pass to my function more than one argument: one is the variable that I want to integrate later and another one is just a value coming from a different calculation which I cannot perform inside the function.
The problem is that the Simpson subroutine only accept f(x) to perform the integral and not f(x,y).
After Vladimir suggestions I modified the code.
Below the example:
Program main2
!------------------------------------------------------------------
! Integration of a function using Simpson rule
! with doubling number of intervals
!------------------------------------------------------------------
! to compile:
! gfortran main2.f90 -o simp2
implicit none
double precision r, rb, rmin, rmax, rstep, integral, eps
double precision F_int
integer nint, i, rbins
double precision t
rbins = 4
rmin = 0.0
rmax = 4.0
rstep = (rmax-rmin)/rbins
rb = rmin
eps = 1.0e-8
func = 0.0
t=2.0
do i=1,rbins
call func(rb,t,res)
write(*,*)'r, f(rb) (in main) = ', rb, res
!test = F_int(rb)
!write(*,*)'test F_int (in loop) = ', test
call simpson2(F_int(rb),rmin,rb,eps,integral,nint)
write(*,*)'r, integral = ', rb, integral
rb = rb+rstep
end do
end program main2
subroutine func(x,y,res)
!----------------------------------------
! Real Function
!----------------------------------------
implicit none
double precision res
double precision, intent(in) :: x
double precision y
res = 2.0*x + y
write(*,*)'f(x,y) (in func) = ',res
return
end subroutine func
function F_int(x)
!Function to integrate
implicit none
double precision F_int, res
double precision, intent(in) :: x
double precision y
call func(x,y,res)
F_int = res
end function F_int
Subroutine simpson2(f,a,b,eps,integral,nint)
!==========================================================
! Integration of f(x) on [a,b]
! Method: Simpson rule with doubling number of intervals
! till error = coeff*|I_n - I_2n| < eps
! written by: Alex Godunov (October 2009)
!----------------------------------------------------------
! IN:
! f - Function to integrate (supplied by a user)
! a - Lower limit of integration
! b - Upper limit of integration
! eps - tolerance
! OUT:
! integral - Result of integration
! nint - number of intervals to achieve accuracy
!==========================================================
implicit none
double precision f, a, b, eps, integral
double precision sn, s2n, h, x
integer nint
double precision, parameter :: coeff = 1.0/15.0 ! error estimate coeff
integer, parameter :: nmax=1048576 ! max number of intervals
integer n, i
! evaluate integral for 2 intervals (three points)
h = (b-a)/2.0
sn = (1.0/3.0)*h*(f(a)+4.0*f(a+h)+f(b))
write(*,*)'a, b, h, sn (in simp) = ', a, b, h, sn
! loop over number of intervals (starting from 4 intervals)
n=4
do while (n <= nmax)
s2n = 0.0
h = (b-a)/dfloat(n)
do i=2, n-2, 2
x = a+dfloat(i)*h
s2n = s2n + 2.0*f(x) + 4.0*f(x+h)
end do
s2n = (s2n + f(a) + f(b) + 4.0*f(a+h))*h/3.0
if(coeff*abs(s2n-sn) <= eps) then
integral = s2n + coeff*(s2n-sn)
nint = n
exit
end if
sn = s2n
n = n*2
end do
return
end subroutine simpson2
I think I'm pretty close to the solution but I cannot figure it out...
If I call simpson2(F_int, ..) without putting the argument in F_int I receive this message:
call simpson2(F_int,rmin,rb,eps,integral,nint)
1
Warning: Expected a procedure for argument 'f' at (1)
Any help?
Thanks in advance!
Now you have a code we can work with, good job!
You need to tell the compiler, that F_int is a function. That can be done by
external F_int
but it is much better to learn Fortran 90 and use modules or at least interface blocks.
module my_functions
implicit none
contains
subroutine func(x,y,res)
!----------------------------------------
! Real Function
!----------------------------------------
implicit none
double precision res
double precision, intent(in) :: x
double precision y
res = 2.0*x + y
write(*,*)'f(x,y) (in func) = ',res
return
end subroutine func
function F_int(x)
!Function to integrate
implicit none
double precision F_int, res
double precision, intent(in) :: x
double precision y
call func(x,y,res)
F_int = res
end function F_int
end module
Now you can easily use the module and integrate the function
use my_functions
call simpson2(F_int,rmin,rb,eps,integral,nint)
But you will find that F_int still does not know what y is! It has it's own y with undefined value! You should put y into the module instead so that everyone can see it.
module my_functions
implicit none
double precision :: y
contains
Don't forget to remove all other declarations of y! Both in function F_int and in the main program. Probably it is also better to call it differently.
Don't forget to set the value of y somewhere inside your main loop!
Locked. This question and its answers are locked because the question is off-topic but has historical significance. It is not currently accepting new answers or interactions.
Code Golf: Rotating Maze
Make a program that takes in a file consisting of a maze. The maze has walls given by #. The maze must include a single ball, given by a o and any number of holes given by a #. The maze file can either be entered via command line or read in as a line through standard input. Please specify which in your solution.
Your program then does the following:
1: If the ball is not directly above a wall, drop it down to the nearest wall.
2: If the ball passes through a hole during step 1, remove the ball.
3: Display the maze in the standard output (followed by a newline).
Extraneous whitespace should not be displayed.
Extraneous whitespace is defined to be whitespace outside of a rectangle
that fits snugly around the maze.
4: If there is no ball in the maze, exit.
5: Read a line from the standard input.
Given a 1, rotate the maze counterclockwise.
Given a 2, rotate the maze clockwise.
Rotations are done by 90 degrees.
It is up to you to decide if extraneous whitespace is allowed.
If the user enters other inputs, repeat this step.
6: Goto step 1.
You may assume all input mazes are closed. Note: a hole effectively acts as a wall in this regard.
You may assume all input mazes have no extraneous whitespace.
The shortest source code by character count wins.
Example written in javascript:
http://trinithis.awardspace.com/rotatingMaze/maze.html
Example mazes:
######
#o ##
######
###########
#o #
# ####### #
#### #
#########
###########################
# #
# # # #
# # # ##
# # ####o####
# # #
# #
# #########
# #
######################
Perl, 143 (128) char
172 152 146 144 143 chars,
sub L{my$o;$o.=$/while s/.$/$o.=$&,""/meg;$_=$o}$_.=<>until/
/;{L;1while s/o / o/;s/o#/ #/;L;L;L;print;if(/o/){1-($z=<>)||L;$z-2||L&L&L;redo}}
Newlines are significant.
Uses standard input and expects input to contain the maze, followed by a blank line, followed by the instructions (1 or 2), one instruction per line.
Explanation:
sub L{my$o;$o.="\n"while s/.$/$o.=$&,""/meg;$_=$o}
L is a function that uses regular expressions to rotate the multi-line expression $_ counterclockwise by 90 degrees. The regular expression was used famously by hobbs in my favorite code golf solution of all time.
$_.=<>until/\n\n/;
Slurps the input up to the first pair of consecutive newlines (that is, the maze) into $_.
L;1 while s/o / o/;s/o#/ */;
L;L;L;print
To drop the ball, we need to move the o character down one line is there is a space under it. This is kind of hard to do with a single scalar expression, so what we'll do instead is rotate the maze counterclockwise, move the ball to the "right". If a hole ever appears to the "right" of the ball, then the ball is going to fall in the hole (it's not in the spec, but we can change the # to an * to show which hole the ball fell into). Then before we print, we need to rotate the board clockwise 90 degrees (or counterclockwise 3 times) so that down is "down" again.
if(/o/) { ... }
Continue if there is still a ball in the maze. Otherwise the block will end and the program will exit.
1-($z=<>)||L;$z-2||L+L+L;redo
Read an instruction into $z. Rotate the board counterclockwise once for instruction "1" and three times for instruction "2".
If we used 3 more characters and said +s/o[#*]/ */ instead of ;s/o#/ */, then we could support multiple balls.
A simpler version of this program, where the instructions are "2" for rotating the maze clockwise and any other instruction for rotating counterclockwise, can be done in 128 chars.
sub L{my$o;$o.=$/while s/.$/$o.=$&,""/meg;$_=$o}$_.=<>until/
/;L;{1while s/o / o/+s/o#/ #/;L,L,L;print;if(/o/){2-<>&&L,L;redo}}
GolfScript - 97 chars
n/['']/~{;(#"zip-1%":|3*~{{." o"/"o "*"#o"/"# "*.#>}do}%|~.n*."o"/,(}{;\~(2*)|*~\}/\[n*]+n.+*])\;
This isn't done as well as I hoped (maybe later).
(These are my notes and not an explanation)
n/['']/~ #[M I]
{
;(# #[I c M]
"zip-1%":|3*~ #rotate
{{." o"/"o "*"#o"/"# "*.#>}do}% #drop
|~ #rotate back
.n* #"display" -> [I c M d]
."o"/,( #any ball? -> [I c M d ?]
}{ #d is collected into an array -> [I c M]
;\~(2*)|*~ #rotate
\ #stack order
}/
\[n*]+n.+*])\; #output
Rebmu: 298 Characters
I'm tinkering with with my own experiment in Code Golf language design! I haven't thrown matrix tricks into the standard bag yet, and copying GolfScript's ideas will probably help. But right now I'm working on refining the basic gimmick.
Anyway, here's my first try. The four internal spaces are required in the code as it is, but the line breaks are not necessary:
.fFS.sSC L{#o#}W|[l?fM]H|[l?m]Z|[Tre[wH]iOD?j[rvT]t]
Ca|[st[xY]a KrePC[[yBKx][ntSBhXbkY][ntSBhYsbWx][xSBwY]]ntJskPCmFkSk]
Ga|[rtYsZ[rtXfZ[TaRE[xY]iTbr]iTbr]t]B|[gA|[ieSlFcA[rnA]]]
MeFI?a[rlA]aFV[NbIbl?n[ut[++n/2 TfCnIEfLtBRchCbSPieTHlTbrCHcNsLe?sNsZ]]
gA|[TfCaEEfZfA[prT][pnT]nn]ulBbr JmoADjPC[3 1]rK4]
It may look like a cat was on my keyboard. But once you get past a little space-saving trick (literally saving spaces) called "mushing" it's not so bad. The idea is that Rebmu is not case sensitive, so alternation of capitalization runs is used to compress the symbols. Instead of doing FooBazBar => foo baz bar I apply distinct meanings. FOObazBAR => foo: baz bar (where the first token is an assignment target) vs fooBAZbar => foo baz bar (all ordinary tokens).
When the unmush is run, you get something more readable, but expanded to 488 characters:
. f fs . s sc l: {#o#} w: | [l? f m] h: | [l? m] z: | [t: re [w h] i od?
j [rv t] t] c: a| [st [x y] a k: re pc [[y bk x] [nt sb h x bk y] [nt sb
h y sb w x] [x sb w y]] nt j sk pc m f k s k] g: a| [rt y s z [rt x f z
[t: a re [x y] i t br] i t br] rn t] b: | [g a| [ie s l f c a [rn a]]]
m: e fi? a [rl a] a fv [n: b i bl? n [ut [++ n/2 t: f c n ie f l t br
ch c b sp ie th l t br ch c n s l e? s n s z]] g a| [t: f c a ee f z f
a [pr t] [pn t] nn] ul b br j: mo ad j pc [3 1] r k 4]
Rebmu can run it expanded also. There are also verbose keywords as well (first instead of fs) and you can mix and match. Here's the function definitions with some comments:
; shortcuts f and s extracting the first and second series elements
. f fs
. s sc
; character constants are like #"a", this way we can do fL for #"#" etc
L: {#o#}
; width and height of the input data
W: | [l? f m]
H: | [l? m]
; dimensions adjusted for rotation (we don't rotate the array)
Z: | [t: re [w h] i od? j [rv t] t]
; cell extractor, gives series position (like an iterator) for coordinate
C: a| [
st [x y] a
k: re pc [[y bk x] [nt sb h x bk y] [nt sb h y sb w x] [x sb w y]] nt j
sk pc m f k s k
]
; grid enumerator, pass in function to run on each cell
G: a| [rt y s z [rt x f z [t: a re [x y] i t br] i t br] t]
; ball position function
B: | [g a| [ie sc l f c a [rn a]]]
W is the width function and H is the height of the original array data. The data is never rotated...but there is a variable j which tells us how many 90 degree right turns we should apply.
A function Z gives us the adjusted size for when rotation is taken into account, and a function C takes a coordinate pair parameter and returns a series position (kind of like a pointer or iterator) into the data for that coordinate pair.
There's an array iterator G which you pass a function to and it will call that function for each cell in the grid. If the function you supply ever returns a value it will stop the iteration and the iteration function will return that value. The function B scans the maze for a ball and returns coordinates if found, or none.
Here's the main loop with some commenting:
; if the command line argument is a filename, load it, otherwise use string
m: e fi? a [rl a] a
; forever (until break, anyway...)
fv [
; save ball position in n
n: B
; if n is a block type then enter a loop
i bl? n [
; until (i.e. repeat until)
ut [
; increment second element of n (the y coordinate)
++ n/2
; t = first(C(n))
t: f C n
; if-equal(first(L), t) then break
ie f l t br
; change(C(B), space)
ch C B sp
; if-equal(third(L),t) then break
ie th L t br
; change(C(n), second(L))
ch C n s L
; terminate loop if "equals(second(n), second(z))"
e? s n s z
]
]
; iterate over array and print each line
g a| [t: f c a ee f z f a [pr t] [pn t] nn]
; unless the ball is not none, we'll be breaking the loop here...
ul b br
; rotate according to input
j: mo ad j pc [3 1] r k 4
]
There's not all that much particularly clever about this program. Which is part of my idea, which is to see what kind of compression one could get on simple, boring approaches that don't rely on any tricks. I think it demonstrates some of Rebmu's novel potential.
It will be interesting to see how a better standard library could affect the brevity of solutions!
Latest up-to-date commented source available on GitHub: rotating-maze.rebmu
Ruby 1.9.1 p243
355 353 characters
I'm pretty new to Ruby, so I'm sure this could be a lot shorter - theres probably some nuances i'm missing.
When executed, the path to the map file is the first line it reads. I tried to make it part of the execution arguments (would have saved 3 characters), but had issues :)
The short version:
def b m;m.each_index{|r|i=m[r].index(?o);return r,i if i}end;def d m;x,y=b m;z=x;
while z=z+1;c=m[z][y];return if c==?#;m[z-1][y]=" "; return 1 if c==?#;m[z][y]=?o;end;end;
def r m;m.transpose.reverse;end;m=File.readlines(gets.chomp).map{|x|x.chomp.split(//)};
while a=0;w=d m;puts m.map(&:join);break if w;a=gets.to_i until 0<a&&a<3;
m=r a==1?m:r(r(m));end
The verbose version:
(I've changed a bit in the compressed version, but you get the idea)
def display_maze m
puts m.map(&:join)
end
def ball_pos m
m.each_index{ |r|
i = m[r].index("o")
return [r,i] if i
}
end
def drop_ball m
x,y = ball_pos m
z=x
while z=z+1 do
c=m[z][y]
return if c=="#"
m[z-1][y]=" "
return 1 if c=="#"
m[z][y]="o"
end
end
def rot m
m.transpose.reverse
end
maze = File.readlines(gets.chomp).map{|x|x.chomp.split(//)}
while a=0
win = drop_ball maze
display_maze maze
break if win
a=gets.to_i until (0 < a && a < 3)
maze=rot maze
maze=rot rot maze if a==1
end
Possible improvement areas:
Reading the maze into a clean 2D array (currently 55 chars)
Finding and returning (x,y) co-ordinates of the ball (currently 61 chars)
Any suggestions to improve are welcome.
Haskell: 577 509 527 244 230 228 chars
Massive new approach: Keep the maze as a single string!
import Data.List
d('o':' ':x)=' ':(d$'o':x)
d('o':'#':x)=" *"++x
d(a:x)=a:d x
d e=e
l=unlines.reverse.transpose.lines
z%1=z;z%2=l.l$z
t=putStr.l.l.l
a z|elem 'o' z=t z>>readLn>>=a.d.l.(z%)|0<1=t z
main=getLine>>=readFile>>=a.d.l
Nods to #mobrule's Perl solution for the idea of dropping the ball sideways!
Python 2.6: ~ 284 ~ characters
There is possibly still room for improvement (although I already got it down a lot since the first revisions).
All comments or suggestions more then welcome!
Supply the map file on the command line as the first argument:
python rotating_maze.py input.txt
import sys
t=[list(r)[:-1]for r in open(sys.argv[1])]
while t:
x=['o'in e for e in t].index(1);y=t[x].index('o')
while t[x+1][y]!="#":t[x][y],t[x+1][y]=" "+"o#"[t[x+1][y]>" "];x+=1
for l in t:print''.join(l)
t=t[x][y]=='o'and map(list,(t,zip(*t[::-1]),zip(*t)[::-1])[input()])or 0
C# 3.0 - 650 638 characters
(not sure how newlines being counted)
(leading whitespace for reading, not counted)
using System.Linq;
using S=System.String;
using C=System.Console;
namespace R
{
class R
{
static void Main(S[]a)
{
S m=S.Join("\n",a);
bool u;
do
{
m=L(m);
int b=m.IndexOf('o');
int h=m.IndexOf('#',b);
b=m.IndexOf('#',b);
m=m.Replace('o',' ');
u=(b!=-1&b<h|h==-1);
if (u)
m=m.Insert(b-1,"o").Remove(b,1);
m=L(L(L(m)));
C.WriteLine(m);
if (!u) return;
do
{
int.TryParse(C.ReadLine(),out b);
u=b==1|b==2;
m=b==1?L(L(L(m))):u?L(m):m;
}while(!u);
}while(u);
}
static S L(S s)
{
return S.Join("\n",
s.Split('\n')
.SelectMany(z => z.Select((c,i)=>new{c,i}))
.GroupBy(x =>x.i,x=>x.c)
.Select(g => new S(g.Reverse().ToArray()))
.ToArray());
}
}
}
Reads from commandline, here's the test line I used:
"###########" "#o #" "# ####### #" "#### #" " #########"
Relied heavily on mobrule's Perl answer for algorithm.
My Rotation method (L) can probably be improved.
Handles wall-less case.
Locked. This question and its answers are locked because the question is off-topic but has historical significance. It is not currently accepting new answers or interactions.
In Google Code Jam 2009, Round 1B, there is a problem called Decision Tree that lent itself to rather creative solutions.
Post your shortest solution; I'll update the Accepted Answer to the current shortest entry on a semi-frequent basis, assuming you didn't just create a new language just to solve this problem. :-P
Current rankings:
107 Perl
121 PostScript (binary)
132 Ruby
154 Arc
160 PostScript (ASCII85)
170 PostScript
192 Python
196 JavaScript
199 Common Lisp
212 LilyPond
273 Scheme
280 R
281 sed w/ bc
312 Haskell
314 PHP
339 m4 w/ bc
346 C
381 Fortran
462 Java
718 OCaml
759 F#
1554 sed
C++ not qualified for now
sed in 1554 chars (pure) / 281 (with bc)
Yes, seriously.
Usage: sed -r -f thisfile.sed < input.in > output.out
(works on GNU sed)
1d
/ /!{x
s/^$/Case #Y:/
:i
s/9Y/Y0/
ti
s/#Y/#0Y/
s/:/:0123456789/
s/(.)Y(.*):[0-9]*\1(.).*/\3\2Y:/
x
G
s/.*\n|Y//gp
z
:p
N
/[()]/s/ |\n//g
y/()/JK/
tp
H
d}
G
s/\n[^J]*/ %/
s/[^JK]*$//
:c
s/J1?([.-9]+)(.*)K/\2#\1/
/%#/by
:b
/J/s/T//
s/J([^JK]*)K/TC\1B/
tb
/ (.+) .*%\1C/{s/%[^C]*/%/
s/T.*B//
by}
s/%.*T/%/
:y
y/CB/JK/
tc
s/.\.0*\b//g
:r
/#.*#/{s/\w*#\w*$/C&B/
s/C(\w)(.*B)/\1C\2~/
s/"[^"]*/&0/g
:t
s/(\w)(C.*)(\w)B(.*~)/\1\2B\3\4\1\3/
T
s/~(10|2[01]|3[0-2]|4[0-3]|5[0-4]|6[0-5]|7[0-6]|8[0-7]|9.)/&Q/
s/(.)(.)Q/\2\1/
s/~0\w/`00/
s/~1\B/`0/
s/~22/`04/
s/~23/`06/
s/~24/`08/
s/~33/`09/
s/~25/`10/
s/~26|~34/`12/
s/~27/`14/
s/~28|~44/`16/
s/~29|~36/`18/
s/~35/`15/
s/~45/`20/
s/~37/`21/
s/~38|~46/`24/
s/~55/`25/
s/~39/`27/
s/~47/`28/
s/~56/`30/
s/~48/`32/
s/~57/`35/
s/~49|~66/`36/
s/~58/`40/
s/~67/`42/
s/~59/`45/
s/~68/`48/
s/~77/`49/
s/~69/`54/
s/~78/`56/
s/~79/`63/
s/~88/`64/
s/~89/`72/
s/~99/`81/
s/`(.)(.)/~\1'\2/
bt
:
s/(~.)'/\1/
s/..'/K&/
/K/bk
:v
s/=(,?.)'/\1/
s/,/1'/
t
s/B(.*)~/\1B"/
tr
s/"(\w*)0/A\1/g
/A.*A/{s/A[^A]*$/J&K/
:k
s/([^A])(J.*)([^A])K/\2K\1\3/
s/K(10|2[01]|3[0-2]|4[0-3]|5[0-4]|6[0-5]|7[0-6]|8[^9]|9.)/&Q/
s/(.)(.)Q/\2\1/
s/K0/=/
s/K11/=2/
s/K12/=3/
s/K13|K22/=4/
s/K14|K23/=5/
s/K15|K24|K33/=6/
s/K16|K25|K34/=7/
s/K(17|26|35|44)/=8/
s/K(18|27|36|45)/=9/
s/K(19|28|37|46|55)/W0/
s/K(29|38|47|56)/W1/
s/K(39|48|57|66)/W2/
s/K49|K58|K67/W3/
s/K59|K68|K77/W4/
s/K69|K78/W5/
s/K79|K88/W6/
s/K89/W7/
s/K99/W8/
s/W/=,/
/'/bv
s/\b=/K:/
tk
s/[:JK]A?//g
s/,/,0123456789GF/
s/(.),.*\1(.).*F/\2/
s/G/,0/
tk}
/A.*A/bv}
s/\w*C.*A//
tr
s/.*#/./
This solution omits the leading zero in front of the decimal point, and does not handle cases where the answer is 1.00. Luckily, the GCJ judge accepts the lack of a zero, and does not have any cases where the answer is 1.00.
To include the leading zero, change the last line to s/.*#/0./; and to handle a 1.00 case, append the line s/^$/1/.
Here's a solution that outsources the multiplication to bc:
1d
/ /!{x
s/\n.*//
s/.*/echo 0&+1|bc/e
x
g
s/.*/Case #&:/p
:p
N
/[()]/s/ |\n//g
y/()/JK/
tp
H
d}
G
s/\n[^J]*/ %/
s/[^JK]*$//
:c
s/J([.-9]+)(.*)K/\2*\1/
/%\*/s/.*%.(.*)/echo \1|bc -l/e
:b
/J/s/T//
s/J([^JK]*)K/TC\1B/
tb
/ (.+) .*%\1C/{s/%[^C]*/%/
s/T.*B//
b}
s/%.*T/%/
:
y/CB/JK/
tc
Perl in 107 characters
say("Case #$_:"),
$_=eval"''".'.<>'x<>,
s:[a-z]+:*(/ $&\\s/?:g,s/\)\s*\(/):/g,
eval"\$_=<>;say$_;"x<>for 1..<>
Newlines for legibility; none of them is necessary or counted in.
It uses features found only in the latest versions of Perl, so run with perl -M5.010 or later.
I used to be a Perl noob too, so this works almost the same as the ruby one. Original version 126 chars, optimizations by peutri.
Backlinks:
Word Aligned - Power Programming
LilyPond: 212 characters
Craziness! Utter ridiculousness!! LilyPond, with its built-in Scheme interpreter, manages to outdo Scheme by more than FIFTY BYTES! Holy acrobatic flying mooses in tights!!
x=#lambda
w=#read
#(letrec((v(x(a)(map a(iota(w)1))))(c(x(f q)(*(car q)(if(any list? q)(c
f((if(memq(cadr q)f)caddr cadddr)q))1)))))(v(x(i)(w)(set! #(w))(format
#t"Case #~a:
~{~y~}"i(v(x i(w)(c(v(x i(w)))#)))))))
Usage: lilypond thisfile.ly <input.in >output.out 2>/dev/null
Credit goes to cky for writing the Scheme solution this was based on, though this version is now substantially different. Seriously, though, the Scheme could be golfed a bit further...
PostScript: 170 (regular) / 160 (ASCII85) / 121 (binary)
My shortest (regular) PostScript solution so far, provided that you rename the input file to "r" (170 characters, including newlines); uses a GhostScript-specific procedure (=only):
1[/:{repeat}/!{exch token{\ exch known{/<>}if]pop]]3 index mul
!}if}(]){token pop}/?(r)(r)file([){?]}>>begin
1[{(Case #)2{=only}:(:)=[/|[def[{[/\<<[{[/}:>>def |]! =}:}for
Usage: cp input.in r; gs -q -dNOPROMPT -dNODISPLAY -dBATCH thisfile.ps > output.out
Here's a binary version of this in 121 bytes (backslashes and unprintable characters escaped):
1[/!{\x92>\x92\xab{\\\x92>\x92`\x92p{]\x92u}if]]3\x92X\x92l!}if}(]){\x92\xab\x92u}/r(r)\x928\x92A([){r]}>>\x92\r1[{(Case #)\x92v=only[/:\x928[\x923=[{[/\\<<[{[/}\x92\x83>>\x923:]! =}\x92\x83}\x92H
If characters outside the ASCII printable range are disallowed, PS has built-in ASCII85 encoding of binary sources. We therefore have the following 160-byte solution in all ASCII printable characters:
1[([){r]}/r(r)<~OuSUj0-P\*5*Dsn>`q:6#$5JU?'9>YBkCXV1Qkk'Ca"4#Apl(5.=75YP')1:5*?#0>C.bc#<6!&,:Se!4`>4SH!;p_OuQ[/1Herh>;'5D4Bm/:07B"95!G,c3aEmO4aiKGI?I,~>cvx exec
Ruby in 132
Improved by leonid. Newlines are essential.
def j
'1
'..gets
end
j.map{|c|s=j.map{gets}*''
puts"Case #%d:"%c,j.map{gets;eval s.gsub(/[a-z]+/,'*(/ \&\b/?').gsub /\)\s*\(/,'):'}}
Ruby in 136
def j;1..gets.to_i;end;j.map{|c|m=j.map{gets}*"";puts"Case ##{c}:";j.map{gets;p eval m.gsub(/[a-z]+/,'*(/ \0\s/?').gsub /\)\s*\(/,'):'}}
I just learned about *"" being equivalent to .join"". Also realised that map could be used in a few places
Ruby in 150
1.upto(gets.to_i){|c|m=eval("gets+"*gets.to_i+"''");puts"Case ##{c}:";1.upto(gets.to_i){gets;p eval m.gsub(/[a-z]+/,'*(/ \0\s/?').gsub /\)\s*\(/,'):'}}
I am just a noob to ruby, so there is probably still a lot of room for improvement
Python in 192
import re;S=re.sub;R=raw_input;I=input;c=0;exec r"c+=1;L=S('\) *\(',')or ',S('([a-z]+)','*(\' \\1 \'in a and',eval(('+R()'*I('Case #%s:\n'%c))[1:])));exec'a=R()+\' \';print eval(L);'*I();"*I()
Common Lisp, 199 bytes
Wrapped every 80 characters:
(defun r()(read))(dotimes(i(r))(format t"~&Case #~D:"(1+ i))(r)(set'z(r))(dotime
s(a(r))(r)(print(do((g(mapcar'read(make-list(r))))(p 1(*(pop c)p))(c z(if(find(p
op c)g)(car c)(cadr c))))((not c)p)))))
Spaced and indented:
(defun r () (read))
(dotimes (i (r))
(format t "~&Case #~D:" (1+ i))
(r)
(set 'z (r))
(dotimes (a (r))
(r)
(print
(do ((g (mapcar 'read (make-list (r))))
(p 1 (* (pop c) p))
(c z (if (find (pop c) g)
(car c)
(cadr c))))
((not c) p)))))
C - 346 bytes
Compile with gcc -w
#define N{int n=atoi(gets(A));for(;n--;)
T[999];F[99];char*t,*f,*a,A[99];float p(){float
d,m=1;for(;*t++^40;);sscanf(t,"%f %[^ (]",&d,A);if(*A^41){for(f=F;m**f;){for(;*f&&*f++^32;);for(a=A;*a&&*f==*a;f++,a++);m=*a||*f&64;}d*=!m*p()+m*p();}return
d;}main(I)N{printf("Case #%d:\n",I++);t=T;N
for(gets(t);*++t;);}N gets(F),t=T,printf("%f\n",p());}}}
Arc, 143 154 characters
Very similar to the CL one, but Arc sure has terse identifiers. Wrapped every 40 chars:
(for i 1((= r read))(prn"Case #"i":")(r)
(= z(r))(repeat(r)(r)(loop(= g(n-of(r)(r
))c z p 1)c(= p(*(pop c)p)c(if(pos(pop c
)g)c.0 cadr.c)))prn.p))
Indented:
(for i 1 ((= r read))
(prn "Case #" i ":")
(r)
(= z (r))
(repeat (r)
(r)
(loop (= g (n-of (r) (r))
c z
p 1)
c
(= p (* (pop c) p)
c (if (pos (pop c) g)
(c 0)
(cadr c))))
(prn p)))
Backlink: Word Aligned - Power Programming
JavaScript in 196 bytes
r='replace'
q=readline
for(n=0,t=q();t-n++;){for(print('Case #'+n+':'),d='',x=q();x--;d+=q());for(x=q();x--;)print(eval(d[r](/([a-z]+)/g,'*({'+q()[r](/ /g,':1,z')+':1}.z$1?')[r](/\) *\(/g,'):')))}
Usage: $ smjs thisfile.js <input.in
With contributions by Hyperlisk.
PHP in 314
<?php function q(){return trim(fgets(STDIN));}for($n=q($x=0);$x++<$n;){for($s=q($t='');$s--;$t.=q());echo"Case #$x:\n";for($z=q();$z--;){$l=explode(' ',q());$l[0]=0;printf("%f\n",eval('return'.preg_replace(array('/\(/','/(\w+),/','/(\d\)*),\((\d)/','/^./'),array(',(','*(in_array("$1",$l,1)?','$1:$2'),$t).';'));}}
FORTRAN - 381
Save as a.F95
Compile with f95 a.F95
#define _ ENDDO
#define A READ(t(k:l-1),*),a
#define Q j=1,n;READ"(A)",s
#define R READ*,n;DO
#define S k+SCAN(t(k:),'()')
CHARACTER(999)s,t,u;R i=1,n;t="";PRINT"('Case #'i0':')",i
R Q;t=TRIM(t)//s;_;R Q;d=1;k=1;DO;k=S;l=S-1
IF(t(l:l)>"(")EXIT;A,u;d=d*a;k=l;m=0
IF(INDEX(s," "//TRIM(u)//" ")>0)CYCLE;DO;IF(')'>t(k:k))m=m+2;m=m-1;k=k+1
IF(1>m)EXIT;k=S-1;_;_;A;d=d*a;PRINT*,d;_;_;END
By using the default format, each of the results starts with 2 spaces, but the google judge permits it. Thanks google judge!
EXPANDED VERSION
CHARACTER(999)s,t,u
READ*,n
DO i=1,n
t=""
PRINT"('Case #'I0':')",i
READ*,n
DO j=1,n
READ"(A)",s
t=TRIM(t)//s
ENDDO
READ*,n
DO j=1,n
READ"(A)",s
d=1
k=1
DO
k=k+SCAN(t(k:),'()')
l=k+SCAN(t(k:),'()')-1
IF(t(l:l)>"(")THEN
READ(t(k:l-1),*),a
d=d*a
PRINT*,d
EXIT
ELSE
READ(t(k:l-1),*),a,u
d=d*a
k=l
m=0
IF(INDEX(s," "//TRIM(u)//" ")>0)CYCLE
DO
IF(')'>t(k:k))m=m+2
m=m-1
k=k+1
IF(1>m)EXIT
k=k+SCAN(t(k:),'()')-1
ENDDO
ENDIF
ENDDO
ENDDO
ENDDO
END
Haskell, 312 characters
Here's another aproach to Haskell. I left the dirty work to the Prelude's lex. The wrapping around it is Text.ParserCombinators.ReadP. Importing it cost 36 characters on its own—ugh!
The parser is a Features -> SExp -> Cuteness function, which spares me most of the type declarations in quibble's/yairchu's solution.
import Text.ParserCombinators.ReadP
main=f(\t->do putStrLn$"Case #"++show t++":";s<-r g;r$print.fst.head.($id=<<s).readP_to_S.d.tail.words=<<g)
d x=do"("<-e;w<-e;c<-do{f<-e;y<-d x;n<-d x;u$if elem f x then y else n}<++u 1.0;e;u$c*read w
f x=do n<-g;mapM x[1..read n]
e=readS_to_P lex
r=f.const
g=getLine
u=return
It used to use Control.Monad's join, forM_ and replicateM, but it turns out it takes less space to redefine them approximately than to import.
I also abandoned the Prelude's readParen in favor of just calling lex before and after. In the current version, there is no need to verify the closing parenthesis: on a valid input it will always be there. On the other hand, it is vital to check the opening one: since the number is only converted after the whole subexpression has been read, a lot of backtracking would be needed to align to the correct parse.
On a theoretical machine with infinite memory and time to spare, the "("<- part might be dropped (4 characters' gain, 308 in total). Unless the call to read just aborts. On mine, the stack just overflows pretty fast.
Java in 467 bytes
This uses the javascript interpreter contained in java 6.
import java.util.*;class D{static{Scanner c=new
Scanner(System.in);int n=c.nextInt(),i=0,l;while(i++<n){l=c.nextInt();String
s="(";while(l-->=0)s+=c.nextLine();System.out.println("Case #"+i+":");l=c.nextInt();while(l-->0)try{c.next();System.out.println(new
javax.script.ScriptEngineManager().getEngineByName("js").eval(s.replace(")","))").replaceAll("\\) *\\(",":(").replaceAll("[a-z]+","*(/ $0 /.test('"+c.nextLine()+" ')?")));}catch(Exception
x){}}System.exit(0);}}
Thanks Varan, Chris and pfn (indirectly) for helping me shorten it.
Please see my other (even shorter!) java answer.
m4 with echo and bc, 339 bytes
This solution is a complete and utter hack, and it gives me a headache. It contains, among other things, escaped double quotes, unescaped double quotes, unescapable backquote and single quote pairs (including a nested pair seven quotes deep), unquoted regular expressions, outsourcing decimal multiplication to bc, and the use of craZy caSE to circumvent macro expansion. But it had to be done, I guess. :p
This adds an "ultimate macroizing" solution to the previous kinds of solutions (iterated loops, recursion w/ lambda mapping, labels and branches, regexp and eval, etc.)
I think a good term for this is "macroni code" :D
(wrapped every 60 characters, for clarity)
define(T,`translit($#)')define(Q,`patsubst($#)')define(I,0)Q
(T(T(T(Q(Q(Q(Q(Q(Q(T(include(A),(),<>),>\s*>,>>),>\s*<,>;),\
([a-z]+\)\s*<,`*ifElsE<rEgExp<P;``````` \1 ''''''';0>;0;<'),
^<,`defiNe<````I';iNcr<I>>\\"Case `#'I:\\"defiNe<`A'''';'),^
[0-9]*),.+ [0-9]+.*,`dEfiNE<```P';`\& '''>A'),<>;N,`(),n'),E
,e),()),.*,`syscmd(`echo "\&"|bc -l')')
Usage: $ cp input.in A; m4 thisfile.m4 > output.out
I'm an m4 n00b, though, having learned it only an hour before writing this. So there's probably room for improvement.
C++ in 698 bytes
Compile with 'g++ -o test source.cpp -include iostream -include vector -include sstream'
#define R(x,f,t) for(int x=f;x<t;x++){
#define S(x) x.size()
#define H string
#define U while
#define I if
#define D cin>>
#define X t.substr(p,S(t))
using namespace std;
int main(){int h,l,n,a,p,Y,W;D h;for(int q=1;q<=h;q++){D l;H s;char c;D c;R(i,0,l)H L;getline(cin,L);R(j,0,S(L))I (L[j]==41||L[j]==40)s+=32;s+=L[j];I(L[j]==40)s+=32;}}D a;printf("Case #%d:\n",q);R(i,0,a)H N;D N;D n;vector<H>f;R(j,0,n)D N;f.push_back(N);}H t=s;float P=1;p=0;U(p<S(t)-1){p=0;U(t[p]!=48&&t[p]!=49)p++;t=X;stringstream T(t);float V;T>>V;H F;T>>F;P*=V;I(F[0]==41)break;Y=0;R(j,0,S(f))if(F==f[j])Y=1;}p=t.find(40)+1;t=X;p=0;I(Y==0){W=1;U (W>0){I(t[p]==40)W++;I(t[p]==41)W--;p++;}t=X;p=0;}}cout<<P<<endl;}}return 0;}
EDIT: I'm sorry; I thought it was ok for the includes (eg, C works even w/o including basic libraries), while I'm sure it would be if I decleared the defines this way.
I'm not home now, and I won't be for some time: I won't be able to modify it. Just ignore my submission.
OCaml in 718 bytes
I'm an OCaml n00b, so this is probably much longer than it needs to be.
Usage: ocaml thisfile.ml <input.in >output.out
#load"str.cma";;open List;;open String;;open Str;;let x=length and
y=Printf.printf and e=global_replace and h=float_of_string and b=regexp and
k=index and r=read_line and a=read_int and w s m c=sub s(c+1)(m-c-1);;for i=1to
a()do y"Case #%d:\n"i;let t=let n=a()in let rec g d j=if j>n then d else
g(d^(r()))(j+1)in e(b" ")""(e(b"\\b")"^"(g""1))and n=a()in let rec z j=if j>n
then()else let q=tl(split(b" ")(r()))in let rec g l j s p=let o=k s '('and c=k
s ')'in if j then let f=w s c o in if contains f '('then let m=k s '^'in let
c=index_from s(m+1)'^'in g 0(mem(w s c m)q)(w s(x s)c)(h(w s m o)*.p)else h f*.p
else if o<c then g(l+1)j(w s(x s)o)p else g(l-1)(l=1)(w s(x s)c)p in y"%f\n"(g
0(0=0)t 1.);z(j+1)in z 1done
Scheme (Guile 1.8)
Here's my version at 278 bytes (with improvements from KirarinSnow to bring it down to 273), after stripping off all the newlines (except ones in string literals, of course). It only works on Guile 1.8 (since in standard Scheme, define is a syntax, not an object, but Guile represents it as an object anyway).
(define ! define)
(!(c f p w . r)(if(null? r)(* p w)(apply c f(* p w)((if(memq(car r)f)cadr caddr)r))))
(!(d . l)(map display l))
(!(r . x)(read))
(! n(r))
(do((i 1(1+ i)))((> i n))(r)(let((t(r)))(d"Case #"i":
")(do((a(r)(1- a)))((= a 0))(r)(d(apply c(map r(iota(r)))1 t)"
"))))
Pure java in 440 bytes
A shorter java solution that doesn't use any eval trick. Can be reduced to 425 by removing System.exit(0) if stderr output is ignored.
import java.util.*;enum A{_;Scanner c,d;float p(String a){return
d.nextFloat()*(d.hasNext("\\D+")?a.contains(' '+d.next()+' ')?p(a)+0*p(a):0*p(a)+p(a):1);}{c=new
Scanner(System.in);for(int n=c.nextInt(),i=0,l;i++<n;){String
s="";for(l=c.nextInt();l-->=0;)s+=c.nextLine();System.out.println("Case #"+i+":");for(l=c.nextInt();l-->0;){c.next();d=new
Scanner(s.replaceAll("[()]"," "));System.out.println(p(c.nextLine()+' '));}}System.exit(0);}}
Haskell, 514 bytes (I suck?).
Based on quibble's solution:
import Control.Monad
import Text.ParserCombinators.Parsec
data F=N|F String(Float,F)(Float,F)
r=return
f=many1 letter>>= \i->w>>d>>= \t->d>>=r.F i t
d=char '('>>w>>many1(oneOf".0123456789")>>= \g->w>>(f<|>r N)>>= \p->char ')'>>w>>r(read g,p)
w=many$oneOf" \n"
g=getLine
l=readLn
m=replicateM
main=l>>= \n->forM_[1..n]$ \t->putStrLn("Case #"++show t++":")>>l>>=(`m`g)>>=(\(Right q)->l>>=(`m`p q)).parse d"".join
z(p,f)=(p*).y f
y N _=1
y(F n t f)x=z(if n`elem`x then t else f)x
p q=fmap(drop 2.words)g>>=print.z q
C in 489 bytes
Code wrapped at 80 chars, there are actually just 3 lines.
Save in a.c and compile with: gcc -w a.c -o a
#define S int I,N;scanf("%d\n",&N);for(I=-1;++I<N;)
#define M 1000
char B[M],Z[M],Q[M]={' '},*F[M],*V;float W[M],H;int J,C,L[M],R[M];t(){V=strtok(0
," \n()");}p(){int U=C++;F[U]=0;if(!V)t();sscanf(V,"%f",W+U);t();if(V&&*V>='a')s
trcpy(Q+1,V),V=0,F[U]=strdup(strcat(Q," ")),L[U]=p(),R[U]=p();return U;}main(){S
{printf("Case #%d:\n",I+1);*B=0;{S strcat(B,gets(Z));}V=strtok(B," \n(");C=0,p()
;{S{strcat(gets(B)," ");for(J=0,H=W[0];F[J];J=strstr(B,F[J])?L[J]:R[J],H*=W[J]);
printf("%f\n",H);};}}}
F#: 759 significant chars (Wow, I'm bad at this ;) )
Minimized version
open System.Text.RegularExpressions
type t=T of float*(string*t*t)option
let rec e=function x,T(w,Some(s,a,b))->e(x,if Set.contains s x then a else b)*w|x,T(w,_)->w
let rec h x=Regex.Matches(x, #"\(|\)|\d\.\d+|\S+")|>Seq.cast<Match>|>Seq.map (fun x -> x.Value)|> Seq.toList
let rec p=function ")"::y->p y|"("::w::x::y->match x with ")"->T(float w,None),y|n->let a,f=p y in let b,g=p f in T(float w,Some(n,a,b)),g
let solve input =
Regex.Matches(input,#"(\(((?<s>\()|[^()]|(?<-s>\)))*\)(?(s)(?!)))\s+\d+\s+((\S+\s\d(.+)?\s*)+)")
|>Seq.cast<Match>
|>Seq.map(fun m->fst(p(h(m.Groups.[1].Value))), [for a in m.Groups.[3].Value.Trim().Split([|'\n'|])->set(a.Split([|' '|]))])
|>Seq.iteri(fun i (r,c)->printfn"Case #%i"(i+1);c|>Seq.iter(fun x->printfn"%.7F"(e(x, r))))
Readable version
open System.Text.RegularExpressions
type decisionTree = T of float * (string * decisionTree * decisionTree) option
let rec eval = function
| x, T(w, Some(s, a, b)) -> eval(x, if Set.contains s x then a else b) * w
| x, T(w, _) -> w
// creates a token stream
let rec tokenize tree =
Regex.Matches(tree, #"\(|\)|\d\.\d+|\S+")
|> Seq.cast<Match>
|> Seq.map (fun x -> x.Value)
|> Seq.toList
// converts token stream into a decisionTree
let rec parse = function
| ")"::xs -> parse xs
| "("::weight::x::xs ->
match x with
| ")" -> T(float weight, None), xs
| name ->
let t1, xs' = parse xs
let t2, xs'' = parse xs'
T(float weight, Some(name, t1, t2)), xs''
// uses regex to transform input file into a Seq<decisionTree, list<set<string>>, which each item in our
// list will be tested against the decisionTree
let solve input =
Regex.Matches(input, #"(\(((?<s>\()|[^()]|(?<-s>\)))*\)(?(s)(?!)))\s+\d+\s+((\S+\s\d(.+)?\s*)+)")
|> Seq.cast<Match>
|> Seq.map (fun m -> fst(parse(tokenize(m.Groups.[1].Value))), [for a in m.Groups.[3].Value.Trim().Split([|'\n'|]) -> set(a.Split([|' '|])) ])
|> Seq.iteri (fun i (tree, testCases) ->
printfn "Case #%i" (i+1)
testCases |> Seq.iter (fun testCase -> printfn "%.7F" (eval (testCase, tree)))
)
R in 280 bytes
Note: On the standard distribution of R (as of v. 2.9.2), this program does not pass the large input and fails on just Case 28 (which is nested to 99 levels), generating a "contextstack overflow". To fix this, modify the line in src/main/gram.c that reads
#define CONTEXTSTACK_SIZE 50
and replace the 50 with something like 500. Then recompile. Et voilà!
n=0
g=gsub
eval(parse(text=g('[^
]* [0-9]+( [^
]*|
)','f=c(\\1)
cat(eval(d),"
")
',g('
\\(','
cat("Case #",n<-n+1,":
",sep="")
d=expression(',g('" "','","',g(')\\s*\\(',',',g(' *("[a-z]+")\\s*\\(','*ifelse(\\1%in%f,',g('([a-z]+)','"\\1"',paste(readLines('A'),collapse='
')))))))))
Usage (requires renaming input): cp input.in A; R -q --slave -f thisfile.R >output.out