I have tried adding parenthesis to the code but lindo gives an error stating that its a variable. According to lindo documentation "(" should be accepted as a parenthesis but this does not occur. Removing the parenthesis causes the code to terminate as soon as it encounters the division symbol. Does anybody know what I'm missing in my code? (error occurs on the 4th, 5th and 6th line of code).
min Y1 + Y2 + Y3 + Y4 + Y5 + Y6
ST
Y1 + Y2 + Y3 + Y4 + Y5 + Y6 >= 1
5X11 + 10X12 + 2X13 + 5X14 + 4X15 + 4X16 / Y1 <= 10
5X21 + 10X22 + 2X23 + 5X24 + 4X25 + 4X26 / Y2 <= 10
5X31 + 10X32 + 2X33 + 5X34 + 4X35 + 4X36 / Y3 <= 10
Related
I have following problem that I try to solve with LPSolve IDE:
min: x1;
r_1: 1.08 - k <= x1;
r_2: -1.08 + k <= x1;
c_1: y1 + y2 + y3 = k;
c_2: 2.29 a1 y1 + 2.28 a2 y1 + 2.27 a3 y1 = 1;
c_3: 1.88 b1 y2 + 1.89 b2 y2 + 1.9 b3 y2 = 1;
c_4: 8.98 c1 y3 + 8.99 c2 y3 + 9.0 c3 y3 = 1;
c_14: a1+a2+a3=1;
c_15: b1+b2+b3=1;
c_16: c1+c2+c3=1;
bin a1,a2,a3,b1,b2,b3,c1,c2,c3;
Not sure why I get output from LPSolve as INFEASIBLE when I can use following param values to solve this:
a1=0, a2=1, a3=0
b1=0, b2=1, b3=0
c1=0, c2=1, c3=0
0 + 2.28 0.438596491 + 0 = 1
0 + 1.89 0.529100529 + 0 = 1
0 + 8.99 0.111234705 + 0 = 1
0.438596491 + 0.529100529 + 0.111234705 = 1.0789 (this is k)
1.08 - 1.0789 == 0.0011 <= x1
-1.08 + 1.0789 == -0.0011 <= x1
x1 = 0.0011
Am I formulating the problem in a wrong way, or doing something else wrong? If I relax that =1 constraint to >=1 there are some results, but I need it to be 1 (as it is in my solution).
Lpsolve is for linear models only. You have products of variables in the model such as 2.29 a1 y1. Lpsolve can not solve such quadratic models.
Too bad you don't get a good error message. I guess they never expected this input.
It is noted that products of binary and continuous variables can be linearized resulting in so-called big-M constraints (see link).
This is really a duplicate of lpsolve - unfeasible solution, but I have example of 1. Embarrassingly, this was an earlier question from the same poster!
Objective function => x1 - 2x2
Subject to =>
x2 <= 5
x1 - x2 >= 2
x1 ,x2, x3 >= 0
Maximize?
convert to standard form :
Maximize -> -x1 + 2x2
Subject to ->
x2 <= 5
-x1 + x2 <= -2
convert to slack form :
Z = -x1 + 2x2
x3 = 5 - x2
x4 = -2 +x1 -x2
Basic solution (0,0,5,-2)
Can I found optimal solution in here? If not why?
I am working with GP and minimum polynomials as follows running on ASUS x75:
(19:25) gp > elt=Mod(a*x^3+b*x^2+c*x+d,('x^5-1)/('x-1))
%122 = Mod(a*x^3 + b*x^2 + c*x + d, x^4 + x^3 + x^2 + x + 1)
(19:25) gp > (poly=minpoly(elt,x='x))
%123 = x^4 + (a + (b + (c - 4*d)))*x^3 + (a^2 + (-3*b + (2*c - 3*d))*a + (b^2 + (2*c - 3*d)*b + (c^2 - 3*d*c + 6*d^2)))*x^2 + (a^3 + (-2*b + (3*c - 2*d))*a^2 + (-2*b^2 + (c + 6*d)*b + (-2*c^2 - 4*d*c + 3*d^2))*a + (b^3 + (-2*c - 2*d)*b^2 + (3*c^2 - 4*d*c + 3*d^2)*b + (c^3 - 2*d*c^2 + 3*d^2*c - 4*d^3)))*x + (a^4 + (-b + (-c - d))*a^3 + (b^2 + (2*c + 2*d)*b + (c^2 - 3*d*c + d^2))*a^2 + (-b^3 + (-3*c + 2*d)*b^2 + (2*c^2 - d*c - 3*d^2)*b + (-c^3 + 2*d*c^2 + 2*d^2*c - d^3))*a + (b^4 + (-c - d)*b^3 + (c^2 + 2*d*c + d^2)*b^2 + (-c^3 - 3*d*c^2 + 2*d^2*c - d^3)*b + (c^4 - d*c^3 + d^2*c^2 - d^3*c + d^4)))
The first command came out successfully, while the second one below did finish successfully and gave an allocatemem() error. How is it possible to get the second command to work, without overheating the computer or program exhaustion? And the WHOLE output to command below this is needed. Thanks for the help.
(19:23) gp > elt=Mod(a*x^5+b*x^4+c*x^3+d*x^2+e*x+f,('x^7-1)/('x-1))
%120 = Mod(a*x^5 + b*x^4 + c*x^3 + d*x^2 + e*x + f, x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)
(19:23) gp > (poly=minpoly(elt,x='x))
*** at top-level: poly=minpoly(elt,x='x)
*** ^-----------------
*** minpoly: the PARI stack overflows !
current stack size: 9000000 (8.583 Mbytes)
[hint] you can increase GP stack with allocatemem()
You can increase the PARI/GP's heap up to any limit you want at run-time following the example below (demonstrates how to set heap size to 120000000 bytes):
default(parisize, 120000000)
default(parisize, 10000000000) is more than 8 GB and in my case was
enough to make advanced calculations with matrices.
I wrote a simple Fortran program for the three body problem using Euler-Richardson algorithm. For some reason the output files give nothing but NaN. Would the issue be solved by using subroutines or functions?
PROGRAM Threebody
IMPLICIT none
!************** Variable declarations ***************
!real(8), parameter :: G = 6.6738D-11
INTEGER :: IOSTAT, dt, t_step, io_error, i
DOUBLE PRECISION:: st, g, m0, m1, m2, force0x, force0y, force1x, force1y, force2x, force2y, n0, n1, n2, x0,v0,x1,v1,x2,v2,y0,w0,y1
DOUBLE PRECISION:: w1,y2,w2
m0 = 1.d0 ! the masses of the three bodies
m1 = 1.d0
m2 = 1.d0
g = 9.80d0 ! m/s^2 s due to gravity
t_step = 1
y0 = 0.d0
x0 = 0.d0
v0 = 0.d0
w0 = 0.d0
y1 = 0.d0
x1 = 1.d0
v1 = 2.d0
w1 = 1.118d0
y2 = 0.d0
x2 = -1.d0
v2 = -1.118d0
w2 = 0.d0
dt = 1 ! time step
OPEN(unit=5, file='out.txt', status='replace',action='write', IOSTAT=io_error)
OPEN(unit=6, file='out1.txt', status='replace',action='write', IOSTAT=io_error)
OPEN(unit=7, file='out2.txt', status='replace',action='write', IOSTAT=io_error)
! particle no.1
DO i=0,1000,t_step ! 1000 is the total time
st=i!/10.d0
n0 = sqrt(((x2 - x1)**2 + (y2 - y1)**2)**3)
force0x = (- m1*(( x0 - x1 ) / n2)) - (m2*(( x0 - x2 ) / n1))
x0 = x0 + (w0 + 0.5*force0x*st)*st
w0 = w0 + force0x*st
force0y = (- m1 * (( y0 - y1 ) / n2)) - (m2 * (( y0 - y2 ) / n1))
y0 = y0 + (v0 + 0.5*force0y*st)*st
v0 = v0 + force0y*st
WRITE(5,*) i*t_step, x0, y0, n0
END DO
! particle no.2
DO i=0,1000,dt
st=i
n1 = sqrt(((x0 - x2 )**2 + (y0 - y2)**2)**3)
force1x = (- m2 * (( x1 - x0 ) / (n0))) - m0 * (( x1 - x2 ) / n2)
x1 = x1 + (w1 + 0.5*force1x*st)*st
w1 = w1 + force1x*st
force1y = (- m2 * ( y1 - y0 ) / n0) - (m0 * (( y1 - y2 ) / n2))
y1 = y1 + (v1 + 0.5*force1y*st)*st
v1 = v1 + force1y*st
WRITE(6,*) i*t_step, x1, y1
END DO
! particle no.3
DO i=0,1000,dt
st=i
n2 = sqrt ( ( ( x1 - x0 )**2 + ( y1 - y0 )**2 )**3 )
force2x = (- m0 * (( x2 - x0 ) / n1)) - (m1 * (( x2 - x1 ) / n0))
x2 = x2 + (w2 + 0.5*force2x*st)*st
w2 = w2 + force2x*st
force2y = (- m0 * (( y2 - y0 ) / n1)) - (m1 * (( y2 - y1 ) / n0))
y2 = y2 + (v2 + 0.5*force2y*st)*st
v2 = v2 + force2y*st
WRITE(7,*) i*t_step, x2, y2, n2
END DO
CLOSE(unit=5)
CLOSE(unit=6)
CLOSE(unit=7)
END PROGRAM Threebody
Compiling with -Wall in gfortran gives you a hint:
f.f90:56:0: warning: ‘n1’ may be used uninitialized in this function [-Wmaybe-uninitialized]
force0x = (- m1*(( x0 - x1 ) / n2)) - (m2*(( x0 - x2 ) / n1))
n1 and n2 are not initialized so they may contain anything. In my case, 6.95e-310, causing force0x to give NaN.
To debug this, you can print the variables on screen to check when they are becoming NaN. For that, please don't use units below 10, because they may be reserved. I think that Unit 5 is standard output so by reusing it you cannot print anything on screen.
I'm trying to implement this: from https://docs.google.com/viewer?url=http://www.tinaja.com/glib/bezdist.pdf&pli=1
The following BASIC program uses the method of finding distance. The
program also searches for the minimum squared distance between points and
a curve.
REM BEZIER.BAS JIM 20DEC92 12:37
DATA 2,3,5,8,8,14,11,17,14,17,16,15,18,11,-1
DATA 2,10,5,12,8,11,11,8,14,6,17,5,19,10,-1
DATA 2,5,5,7,8,8,12,12,13,14,12,17,10,18,8,17,7,14,8,12,12,8,15,7,18,5,-1
OPEN "BEZIER.OUT" FOR OUTPUT AS #1
OPEN "BEZ.ps" FOR OUTPUT AS #2
CLS
psscale = 20
FOR example% = 1 TO 3
REDIM rawdata(32)
FOR I% = 0 TO 32
READ rawdata(I%)
IF rawdata(I%) < 0! THEN EXIT FOR
NEXT I%
n% = I% - 1
PRINT "Example "; example%; (n% + 1) \ 2; " points"
PRINT #1, ""
PRINT #1, "Example "; example%; (n% + 1) \ 2; " points"
PRINT #1, " #
x
y"
J% = 0
FOR I% = 0 TO n% STEP 2
J% = J% + 1
PRINT #1, USING "### ####.### ####.###"; J%; rawdata(I%); rawdata(I% + 1)
LPRINT USING "####.### ####.### 3 0 360 arc fill"; rawdata(I%) * psscale; rawdata(I% + 1) * psscale
PRINT #2, USING "####.### ####.### 3 0 360 arc fill"; rawdata(I%) * psscale; rawdata(I% + 1) * psscale
NEXT I%
x0 = rawdata(0)
y0 = rawdata(1)
x1 = rawdata(2)
y1 = rawdata(3)
x2 = rawdata(n% - 3)
y2 = rawdata(n% - 2)
x3 = rawdata(n% - 1)
y3 = rawdata(n%)
IF example% = 3 THEN
’special guess for loop
x1 = 8 * x1 - 7 * x0
y1 = 8 * y1 - 7 * y0
x2 = 8 * x2 - 7 * x3
y2 = 8 * y2 - 7 * y3
ELSE
x1 = 2 * x1 - x0
y1 = 2 * y1 - y0
x2 = 2 * x2 - x3
y2 = 2 * y2 - y3
END IF
GOSUB distance
LPRINT ".1 setlinewidth"
PRINT #2, ".1 setlinewidth"
GOSUB curveto
e1 = totalerror
FOR Retry% = 1 TO 6
PRINT
PRINT "Retry "; Retry%
PRINT #1, "Retry "; Retry%
PRINT #1, " x1
y1
x2
y2
error"
e3 = .5
x1a = x1
DO
x1 = x1 + (x1 - x0) * e3
GOSUB distance
e2 = totalerror
IF e2 = e1 THEN
EXIT DO
ELSEIF e2 > e1 THEN
x1 = x1a
e3 = -e3 / 3
IF ABS(e3) < .001 THEN EXIT DO
ELSE
e1 = e2
x1a = x1
END IF
LOOP
e3 = .5
y1a = y1
DO
y1 = y1 + (y1 - y0) * e3
GOSUB distance
e2 = totalerror
IF e2 = e1 THEN
EXIT DO
ELSEIF e2 > e1 THEN
y1 = y1a
e3 = -e3 / 3
IF ABS(e3) < .01 THEN EXIT DO
ELSE
e1 = e2
y1a = y1
END IF
LOOP
e3 = .5
x2a = x2
DO
x2 = x2 + (x2 - x3) * e3
GOSUB distance
e2 = totalerror
IF e2 = e1 THEN
EXIT DO
ELSEIF e2 > e1 THEN
x2 = x2a
e3 = -e3 / 3
IF ABS(e3) < .01 THEN EXIT DO
ELSE
e1 = e2
x2a = x2
END IF
LOOP
e3 = .5
y2a = y2
DO
y2 = y2 + (y2 - y3) * e3
GOSUB distance
e2 = totalerror
IF e2 = e1 THEN
EXIT DO
ELSEIF e2 > e1 THEN
y2 = y2a
e3 = -e3 / 3
IF ABS(e3) < .01 THEN EXIT DO
ELSE
e1 = e2
y2a = y2
END IF
LOOP
IF Retry% = 6 THEN
LPRINT "1 setlinewidth"
PRINT #2, "1 setlinewidth"
END IF
GOSUB curveto
NEXT Retry%
LPRINT "100 200 translate"
PRINT #2, "100 200 translate"
NEXT example%
LPRINT "showpage"
PRINT #2, "showpage"
CLOSE #1
CLOSE #2
END
’
Bezier:
x = a0 + u * (a1 + u * (a2 + u * a3))
y = b0 + u * (b1 + u * (b2 + u * b3))
dx4 = x - x4: dy4 = y - y4
dx = a1 + u * (2 * a2 + u * 3 * a3)
dy = b1 + u * (2 * b2 + u * 3 * b3)
z = dx * dx4 + dy * dy4
s = dx4 * dx4 + dy4 * dy4
RETURN
’
distance:
totalerror = 0!
a3 = (x3 - x0 + 3 * (x1 - x2)) / 8
b3 = (y3 - y0 + 3 * (y1 - y2)) / 8
a2 = (x3 + x0 - x1 - x2) * 3 / 8
b2 = (y3 + y0 - y1 - y2) * 3 / 8
a1 = (x3 - x0) / 2 - a3
b1 = (y3 - y0) / 2 - b3
a0 = (x3 + x0) / 2 - a2
b0 = (y3 + y0) / 2 - b2
FOR I% = 2 TO n% - 2 STEP 2
x4 = rawdata(I%)
y4 = rawdata(I% + 1)
stepsize = 2 / (n% + 1)
FOR u = -1! TO 1.01 STEP stepsize
GOSUB Bezier
IF s = 0! THEN u1 = u: z1 = z: s1 = s: EXIT FOR
IF u = -1! THEN u1 = u: z1 = z: s1 = s
IF s < s1 THEN u1 = u: z1 = z: s1 = s
NEXT u
IF s1 <> 0! THEN
u = u1 + stepsize
IF u > 1! THEN u = 1! - stepsize
DO
GOSUB Bezier
IF s = 0! THEN EXIT DO
IF z = 0! THEN EXIT DO
u2 = u
z2 = z
temp = z2 - z1
IF temp <> 0! THEN
u = (z2 * u1 - z1 * u2) / temp
ELSE
u = (u1 + u2) / 2!
END IF
IF u > 1! THEN
u = 1!
ELSEIF u < -1! THEN
u = -1!
END IF
IF ABS(u - u2) < .0001 THEN EXIT DO
u1 = u2
z1 = z2
LOOP
END IF
totalerror = totalerror + s
NEXT I%
PRINT totalerror;
PRINT #1, USING "####.### ####.### ####.### ####.### ######.###"; x1; y1; x2; y2; totalerror
RETURN
’
curveto:
LPRINT USING "####.### ####.### moveto"; x0 * psscale; y0 * psscale
PRINT #2, USING "####.### ####.### moveto"; x0 * psscale; y0 * psscale
F$ = "####.### ####.### ####.### ####.### ####.### ####.### curveto stroke"
LPRINT USING F$; x1 * psscale; y1 * psscale; x2 * psscale; y2 * psscale; x3 * psscale; y3 * psscale
PRINT #2, USING F$; x1 * psscale; y1 * psscale; x2 * psscale; y2 * psscale; x3 * psscale; y3 * psscale
RETURN
I want to implement it in c++ because I'm trying to get my algorithm to best fit beziers from points.
What would the above look like in pseudo-code or c / c++?
thanks
The best approach here is to split the code bit by bit and do minor refactorings until it's in a usable state. Data can be changed into global variables at first.
Then start taking small chunks of the code and turning them into functions. At first they'll just use a bunch of global data. As you rewrite the pieces into C++ things will become more clear.
Once you have most of the code built out functionally, then you can start refactoring the variables. The goal would be to remove all the global non-const data and have all the working data be locals. const values can remain namespace level initialized data.
Finally once you have it procedure-based, you can decide if it's worth the effort to encapsulate the work into objects and methods. Depending on how long the program needs to be maintained grouping the data and methods may be a good long-term step.