I am using numerical recipes scheme to generate random numbers (ran3, page 7 in this PDF file). I didn't notice anything strange but this time, I got a negative numbers at the "warm up" stage which are larger than MBIG. The code look as if this shouldn't happen. I can easily fix this with changing the if statement to be a while statement at the line that says if(mk.lt.MZ)mk=mk+MBIG but I want to know what are the consequences.
Edit:here is the function
FUNCTION ran3a(idum)
INTEGER idum
INTEGER MBIG,MSEED,MZ
C REAL MBIG,MSEED,MZ
REAL ran3a,FAC
PARAMETER (MBIG=1000000000,MSEED=161803398,MZ=0,FAC=1./MBIG)
C PARAMETER (MBIG=4000000.,MSEED=1618033.,MZ=0.,FAC=1./MBIG)
INTEGER i,iff,ii,inext,inextp,k
INTEGER mj,mk,ma(55)
C REAL mj,mk,ma(55)
SAVE iff,inext,inextp,ma
DATA iff /0/
if(idum.lt.0.or.iff.eq.0)then
iff=1
mj=MSEED-iabs(idum)
mj=mod(mj,MBIG)
ma(55)=mj
mk=1
do 11 i=1,54
ii=mod(21*i,55)
ma(ii)=mk
mk=mj-mk
if(mk.lt.MZ)mk=mk+MBIG
mj=ma(ii)
11 continue
do 13 k=1,4
do 12 i=1,55
ma(i)=ma(i)-ma(1+mod(i+30,55))
if(ma(i).lt.MZ)ma(i)=ma(i)+MBIG
12 continue
13 continue
inext=0
inextp=31
idum=1
endif
inext=inext+1
if(inext.eq.56)inext=1
inextp=inextp+1
if(inextp.eq.56)inextp=1
mj=ma(inext)-ma(inextp)
if(mj.lt.MZ)mj=mj+MBIG
ma(inext)=mj
ran3a=mj*FAC
return
END
I was getting Seg Faults (using gfortran 4.8) because the function was trying to change the input value idum from the negative number to 1. There is no reason for that line (nor anything with iff), so I deleted it and printed out the array ma at several different places and found no negative numbers in the array.
One possibility, though, is if iabs(idum) is larger than MSEED, you might have a problem with the line mj=MSEED - iabs(idum). You should protect from this by using mj=abs(MSEED-abs(idum)) like the book has written.
Had a look at the pdf. What you need to do is
1) Seed it: value = ran3(-1)
2) Use it: value = ran3(0)
Related
I am trying to "skip forward" a few realizations by using the function Future.randjump(), but it doesn't seem to behave as I expect it to. The following code gives me the desired result, where jumping forward 1 steps gives the same result as if I had called rand(rng) twice, i.e. the two println display the same number:
using Random, Future
rng = MersenneTwister(123);
new_rng = Future.randjump(rng, 1)
rand(rng)
rand(rng)
println(rand(rng))
println(rand(new_rng))
However, if I add one extra call to rand(rng) before the call to randjump(), the two printed numbers are completely different:
using Random, Future
rng = MersenneTwister(123);
rand(rng) # Added line
new_rng = Future.randjump(rng, 1)
rand(rng)
rand(rng)
println(rand(rng))
println(rand(new_rng))
I expected that the two calls to println() would display the same thing even in the second case, how come they don't? Is there a way I can use randjump() in the second case to get the same realizations as if I had called rand(rng) several times? Thank you in advance.
One unit of randjump corresponds to generation of two floating point numbers.
Consider this example
julia> rng = MersenneTwister(123);
julia> rng2 = Future.randjump(rng, 1);
julia> rand(rng, 4)
4-element Vector{Float64}:
0.7684476751965699
0.940515000715187
0.6739586945680673
0.3954531123351086
julia> rand(rng2,2)
2-element Vector{Float64}:
0.6739586945680673
0.3954531123351086
Note that in the second call (that is rand(rng2,2)) the both numbers are identical to the two last numbers in the first call (taht is rand(rng,2)).
Another issue is that different distributions might "consume" Float64 numbers from the stream at a different speed - so you need to check with a particular distribution how fast it consumes floats for the stream (some might also use buffering etc...).
Looking at the source code of randn (#edit randn()) it consumes one float and hence you get the same results for those two calls:
julia> randn(MersenneTwister(123),6)[3:end]
4-element Vector{Float64}:
1.142650902867199
0.45941562040708034
-0.396679079295223
-0.6647125451916877
julia> randn(Future.randjump(MersenneTwister(123),1),4)
4-element Vector{Float64}:
1.142650902867199
0.45941562040708034
-0.396679079295223
-0.6647125451916877
EDIT
Regarding your comment the size of Mersenne Twister state is 19937 bits and half-unit jumps are not supported. Running rand is mutating this state but not half-the way - so you end up with different bits. Note that an RNG is a sequence of states and the actual values are calculated from that state.
The correct pattern to synchronize random numbers in your computations is the following:
master_rng = MersenneTwister(123);
rng1 = Future.randjump(master_rng, big(10)^20)
# do whatever you want
rng2 = Future.randjump(master_rng, 2*big(10)^20)
# do whatever you want
rng3 = Future.randjump(master_rng, 3*big(10)^20)
# do whatever you want
With this pattern you can correctly maintains synchronization between random number streams and have full control whether the should overlap or not.
I'm relatively new to Z3 and experimenting with it in python. I've coded a program which returns the order in which different actions is performed, represented with a number. Z3 returns an integer representing the second the action starts.
Now I want to look at the model and see if there is an instance of time where nothing happens. To do this I made a list with only 0's and I want to change the index at the times where each action is being executed, to 1. For instance, if an action start at the 5th second and takes 8 seconds to be executed, the index 5 to 12 would be set to 1. Doing this with all the actions and then look for 0's in the list would hopefully give me the instances where nothing happens.
The problem is: I would like to write something like this for coding the problem
list_for_check = [0]*total_time
m = s.model()
for action in actions:
for index in range(m.evaluate(action.number) , m.evaluate(action.number) + action.time_it_takes):
list_for_check[index] = 1
But I get the error:
'IntNumRef' object cannot be interpreted as an integer
I've understood that Z3 isn't returning normal ints or bools in their models, but writing
if m.evaluate(action.boolean):
works, so I'm assuming the if is overwritten in a way, but this doesn't seem to be the case with range. So my question is: Is there a way to use range with Z3 ints? Or is there another way to do this?
The problem might also be that action.time_it_takes is an integer and adding a Z3int with a "normal" int doesn't work. (Done in the second part of the range).
I've also tried using int(m.evaluate(action.number)), but it doesn't work.
Thanks in advance :)
When you call evaluate it returns an IntNumRef, which is an internal z3 representation of an integer number inside z3. You need to call as_long() method of it to convert it to a Python number. Here's an example:
from z3 import *
s = Solver()
a = Int('a')
s.add(a > 4);
s.add(a < 7);
if s.check() == sat:
m = s.model()
print("a is %s" % m.evaluate(a))
print("Iterating from a to a+5:")
av = m.evaluate(a).as_long()
for index in range(av, av + 5):
print(index)
When I run this, I get:
a is 5
Iterating from a to a+5:
5
6
7
8
9
which is exactly what you're trying to achieve.
The method as_long() is defined here. Note that there are similar conversion functions from bit-vectors and rationals as well. You can search the z3py api using the interface at: https://z3prover.github.io/api/html/namespacez3py.html
I am trying to use ran1 from Numerical Recipes in my FORTRAN 90 code. I think a common way is to compile the old subroutine separately, then use the object file. But here I want to know what change is necessary to use it directly in my code.
FUNCTION ran1(idum)
INTEGER idum,IA,IM,IQ,IR,NTAB,NDIV
REAL ran1,AM,EPS,RNMX
PARAMETER (IA=16807,IM=2147483647,AM=1./IM,IQ=127773,IR=2836,
! NTAB=32,NDIV=1+(IM-1)/NTAB,EPS=1.2e-7,RNMX=1.-EPS)
! “Minimal” random number generator of Park and Miller with Bays-Durham shuffle and
! added safeguards. Returns a uniform random deviate between 0.0 and 1.0 (exclusive of
! the endpoint values). Call with idum a negative integer to initialize; thereafter, do not
! alter idum between successive deviates in a sequence. RNMX should approximate the largest
! floating value that is less than 1.
INTEGER j,k,iv(NTAB),iy
SAVE iv,iy
DATA iv /NTAB*0/, iy /0/
iy = 0
if (idum.le.0.or.iy.eq.0) then !Initialize.
idum=max(-idum,1)
! Be sure to prevent idum = 0.
do 11 j=NTAB+8,1,-1
! Load the shuffle table (after 8 warm-ups).
k=idum/IQ
idum=IA*(idum-k*IQ)-IR*k
if (idum.lt.0) idum=idum+IM
if (j.le.NTAB) iv(j)=idum! Compute idum=mod(IA*idum,IM) without overflows by
enddo 11
iy=iv(1)
endif
k=idum/IQ
idum=IA*(idum-k*IQ)-IR*k
! Compute idum=mod(IA*idum,IM) without overflows by
if (idum.lt.0) idum=idum+IM ! Schrage’s method.
j=1+iy/NDIV
iy=iv(j) ! Output previously stored value and refill the shuffle table.
iv(j)=idum
ran1=min(AM*iy,RNMX) ! Because users don’t expect endpoint values.
return
END
Your code is malformed. It looks like you copied it manually from the book, but not exactly. The second problem is actually present even in the book.
Firstly, there should be a line continuation and not a comment in the parameter statement on the second and third line
PARAMETER (IA=16807,IM=2147483647,AM=1./IM,IQ=127773,IR=2836, &
NTAB=32,NDIV=1+(IM-1)/NTAB,EPS=1.2e-7,RNMX=1.-EPS)
(converted to free form, see the book for the original)
Secondly, the loop is a strange combination of a do loop with numeric label and a do loop with end do:
do 11 j=NTAB+8,1,-1
...
enddo 11
should be
do j=NTAB+8,1,-1
...
enddo
or
do 11 j=NTAB+8,1,-1
...
11 continue
There may be more problems present.
I am writing a simple code in matlab which has the purpose of creating the histogram of a grayscale image without using the function hist. I am stuck at the point in which mathlab displays the error "Subscript indices must either be real positive integers or logicals." Can you help me finding where is the wrong indices?
indirizzo='file.jpg';
immagine=imread(indirizzo);
immaginebn=rgb2gray(immagine);
n=zerps(0,255);
for x=0:255;
numeroennesimo=sum(sum(immaginebn==x));
n(x)=numeroennesimo;
end
plot(x,n)
you cant use 0 as index. Either make n(x+1) or for x = 1:256 and substract the 1 in your comparison. And there is a typo, I guess it means zeros instead of zerps, which also doesnt work with a 0. And one more, your plot will also not work as the x has only a size of 1 while n is an array of 266 and for a histogram I would use a barplot instead.
indirizzo='file.jpg';
immagine=imread(indirizzo);
immaginebn=rgb2gray(immagine);
n=zeros(1,256);
for x=0:255;
numeroennesimo=sum(sum(immaginebn==x-1));
n(x+1)=numeroennesimo;
end
bar(0:255,n)
or
indirizzo='file.jpg';
immagine=imread(indirizzo);
immaginebn=rgb2gray(immagine);
n=zeros(1,256);
xplot=zeros(1,256);
for x=1:256;
numeroennesimo=sum(sum(immaginebn==x-1));
n(x)=numeroennesimo;
xplot(x) = x-1;
end
plot(xplot,n)
I want to make a list with its elements representing the logic map given by
x_{n+1} = a*x_n(1-x_n)
I tried the following code (which adds stuff manually instead of a For loop):
x0 = Input["Enter x0"]
a = Input["a"]
M = {x0}
L[n_] := If[n < 1, x0, a*M[[n]]*(1 - M[[n]])]
Print[L[1]]
Append[M, L[1]]
Print[M]
Append[M, L[2]]
Print[M]
The output is as follows:
0.3
2
{0.3}
0.42
{0.3,0.42}
{0.3}
Part::partw: Part 2 of {0.3`} does not exist. >>
Part::partw: Part 2 of {0.3`} does not exist. >>
{0.3, 2 (1 - {0.3}[[2]]) {0.3}[[2]]}
{0.3}
It seems that, when the function definition is being called in Append[M,L[2]], L[2] is calling M[[2]] in the older definition of M, which clearly does not exist.
How can I make L use the newer, bigger version of M?
After doing this I could use a For loop to generate the entire list up to a certain index.
P.S. I apologise for the poor formatting but I could find out how to make Latex code work here.
Other minor question: What are the allowed names for functions and lists? Are underscores allowed in names?
It looks to me as if you are trying to compute the result of
FixedPointList[a*#*(1-#)&, x0]
Note:
Building lists element-by-element, whether you use a loop or some other construct, is almost always a bad idea in Mathematica. To use the system productively you need to learn some of the basic functional constructs, of which FixedPointList is one.
I'm not providing any explanation of the function I've used, nor of the interpretation of symbols such as # and &. This is all covered in the documentation which explains matters better than I can and with which you ought to become familiar.
Mathematica allows alphanumeric (only) names and they must start with a letter. Of course, Mathematic recognises many Unicode characters other than the 26 letters in the English alphabet as alphabetic. By convention (only) intrinsic names start with an upper-case letter and your own with a lower-case.
The underscore is most definitely not allowed in Mathematica names, it has a specific and widely-used interpretation as a short form of the Blank symbol.
Oh, LaTeX formatting doesn't work hereabouts, but Mathematica code is plenty readable enough.
It seems that, when the function definition is being called in
Append[M,L2], L2 is calling M[2] in the older definition of M,
which clearly does not exist.
How can I make L use the newer, bigger version of M?
M is never getting updated here. Append does not modify the parameters you pass to it; it returns the concatenated value of the arrays.
So, the following code:
A={1,2,3}
B=Append[A,5]
Will end up with B={1,2,3,5} and A={1,2,3}. A is not modfied.
To analyse your output,
0.3 // Output of x0 = Input["Enter x0"]. Note that the assignment operator returns the the assignment value.
2 // Output of a= Input["a"]
{0.3} // Output of M = {x0}
0.42 // Output of Print[L[1]]
{0.3,0.42} // Output of Append[M, L[1]]. This is the *return value*, not the new value of M
{0.3} // Output of Print[M]
Part::partw: Part 2 of {0.3`} does not exist. >> // M has only one element, so M[[2]] doesn't make sense
Part::partw: Part 2 of {0.3`} does not exist. >> // ditto
{0.3, 2 (1 - {0.3}[[2]]) {0.3}[[2]]} (* Output of Append[M, L[2]]. Again, *not* the new value of M *)
{0.3} // Output of Print[M]
The simple fix here is to use M=Append[M, L[1]].
To do it in a single for loop:
xn=x0;
For[i = 0, i < n, i++,
M = Append[M, xn];
xn = A*xn (1 - xn)
];
A faster method would be to use NestList[a*#*(1-#)&, x0,n] as a variation of the method mentioned by Mark above.
Here, the expression a*#*(1-#)& is basically an anonymous function (# is its parameter, the & is a shorthand for enclosing it in Function[]). The NestList method takes a function as one argument and recursively applies it starting with x0, for n iterations.
Other minor question: What are the allowed names for functions and lists? Are underscores allowed in names?
No underscores, they're used for pattern matching. Otherwise a variable can contain alphabets and special characters (like theta and all), but no characters that have a meaning in mathematica (parentheses/braces/brackets, the at symbol, the hash symbol, an ampersand, a period, arithmetic symbols, underscores, etc). They may contain a dollar sign but preferably not start with one (these are usually reserved for system variables and all, though you can define a variable starting with a dollar sign without breaking anything).