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Years ago in college,I tinkered with some prolog, but that's long forgotten, so I count as a complete beginnner again (humbling!)
Anyway, I was playing with some of Bruce Tate's code, and came up with what I thought was a sudoku solver for the full (9x9) game. But, when I run it, it generates some very odd output:
Solution = [_#3(2..3),_#24(2:7),_#45(2..3:5:7),_#66(2..3:8),_#87(2..3:5..6:8),4,_#121(2:5..6),1,9,6,8,_#194(2..5:7:9),_#215(1..3:9),_#236(2..3:5:9),_#257(1..2:5:9),_#278(2:4..5),_#299(4:7),_#320(5:7),_#341(1..2),_#362(2:4),_#383(2:4..5:9),_#404(1..2:9),_#425(2:5..6:9),7,3,_#472(4:6),8,4,1,_#532(2:8),_#553(2:8),7,3,9,5,6,7,5,_#689(6:8),_#710(4:8..9),_#731(4:6:8..9),_#752(6:8..9),1,2,3,_#828(2..3),9,_#862(2..3:6),5,1,_#909(2:6),7,8,4,8,_#990(2:4:7),1,6,_#1037(2..5:9),_#1058(2:5:9),_#1079(4..5),_#1100(3..4:7),_#1121(5:7),5,_#1163(4:6..7),_#1184(4:6..7),_#1205(1:3..4:8),_#1226(3..4:8),_#1247(1:8),_#1268(4:6:8),9,2,9,3,_#1341(2:4:6),7,_#1375(2:4..5:8),_#1396(1..2:5:8),_#1417(4..6:8),_#1438(4:6),_#1459(1:5)]
yes
I was expecting ... well, frankly I was half expecting total failure :) but I thought that only numbers could show up in this output. What's it trying to tell me with those #-tagged things, and stuff in parens that looks like ranges? Is it trying to say there are many possible solutions and it's telling me all at once (seems unlikely as it's very unhelpful if it is) or is this some kind of error state (in which case, why does it compile my code and say "yes" to this query?)
Any insight gratefully received!
I think it's the result of a set of constraints not sufficiently strong to determine a solution without search. For instance, _#3(2..3) could means that a variable named _#3 could assume values in range 2..3. You could try to label the variables, something like
..., labeling([], Solution).
Syntax details depend on your solver, of course...
I am currently building random forest.
I have made my dataset very small, just only binary variable and three numeric variables (double precision). There are around 400,000 samples.
Model_cforest <- cforest(result ~ score1 + score2 + score3
, data=trainData, controls = cforest_unbiased(ntree=100))
I really don't think it would consume too much memory, as the package randomForest can handle this easily if I disable approximation.
However, in party package, cforest function, it ate up all of my memory (16GB, at least 12GB free) and still want more. I have to do nothing but terminate the program. I believe there must be something wrong in my setting, but I can not figure it out.
Would you please help me out? Thanks
I just started working with Mathematica (5.0) for the first time, and while the manual has been helpful, I'm not entirely sure my technique has been correct using (Full)Simplify. I am using the program to check my work on a derived transform to change between reference frames, which consisted of multiplying a trio of relatively large square matrices.
A colleague and I each did the work by hand, separately, to make sure there were no mistakes. We hoped to get a third check from the program, which seemed that it would be simple enough to ask. The hand calculations took some time due to matrix size, but we came to the same conclusions. The fact that we had the same answer made me skeptical when the program produced different results.
I've checked and double checked my inputs.
I am definitely . (dot-multiplying) the matrices for correct multiplication.
FullSimplify made no difference.
Neither have combinations with TrigReduce / expanding algebraically before simplifying.
I've taken indices from the final matrix and tryed to simplify them while isolated, to no avail, so the problem isn't due to the use of matrices.
I've also tried to multiply the first two matrices, simplify, and then multiply that with the third matrix; however, this produced the same results as before.
I thought Simplify automatically crossed into all levels of Heads, so I didn't need to worry about mapping, but even where zeros would be expected as outputs in the matrix, there are terms, and where we would expect terms, there are close answers, plus a host of sin and cosine terms that do not reduce.
Does anyone frequent any type of technique with Simplify to get more preferable results, in contrast to solely using Simplify?
If there are assumptions on parameter ranges you will want to feed them to Simplify. The following simple examples will indicate why this might be useful.
In[218]:= Simplify[a*Sqrt[1 - x^2] - Sqrt[a^2 - a^2*x^2]]
Out[218]= a Sqrt[1 - x^2] - Sqrt[-a^2 (-1 + x^2)]
In[219]:= Simplify[a*Sqrt[1 - x^2] - Sqrt[a^2 - a^2*x^2],
Assumptions -> a > 0]
Out[219]= 0
Assuming this and other responses miss the mark, if you could provide an example that in some way shows the possibly bad behavior, that would be very helpful. Disguise it howsoever necessary in order to hide proprietary features: bleach out watermarks, file down registration numbers, maybe dress it in a moustache.
Daniel Lichtblau
Wolfram Research
As you didn't give much details to chew on I can only give you a few tips:
Mma5 is pretty old. The current version is 8. If you have access to someone with 8 you might ask him to try it to see whether that makes a difference. You could also try WolframAlpha online (http://www.wolframalpha.com/), which also understands some (all?) Mma syntax.
Have you tried comparing your own and Mma's result numerically? Generate a Table of differences for various parameter values or use Plot. If the differences are negligable (use Chop to cut off small residuals) the results are probably equivalent.
Cheers -- Sjoerd
As all developers do, we constantly deal with some kind of identifiers as part of our daily work. Most of the time, it's about bugs or support tickets. Our software, upon detecting a bug, creates a package that has a name formatted from a timestamp and a version number, which is a cheap way of creating reasonably unique identifiers to avoid mixing packages up. Example: "Bug Report 20101214 174856 6.4b2".
My brain just isn't that good at remembering numbers. What I would love to have is a simple way of generating alpha-numeric identifiers that are easy to remember.
It takes about 5 minutes to whip up an algorithm like the following in python, which produces halfway usable results:
import random
vowels = 'aeiuy' # 0 is confusing
consonants = 'bcdfghjklmnpqrstvwxz'
numbers = '0123456789'
random.seed()
for i in range(30):
chars = list()
chars.append(random.choice(consonants))
chars.append(random.choice(vowels))
chars.append(random.choice(consonants + numbers))
chars.append(random.choice(vowels))
chars.append(random.choice(vowels))
chars.append(random.choice(consonants))
print ''.join(chars)
The results look like this:
re1ean
meseux
le1ayl
kuteef
neluaq
tyliyd
ki5ias
This is already quite good, but I feel it is still easy to forget how they are spelled exactly, so that if you walk over to a colleagues desk and want to look one of those up, there's still potential for difficulty.
I know of algorithms that perform trigram analysis on text (say you feed them a whole book in German) and that can generate strings that look and feel like German words and are thus easier to handle generally. This requires lots of data, though, and makes it slightly less suitable for embedding in an application just for this purpose.
Do you know of any published algorithms that solve this problem?
Thanks!
Carl
As you said, your sample is quite good. But if you want random identifiers that can easily be remembered, then you should not mix alphanumeric and numeric characters. Instead, you could opt to postfix an alphanumeric string with a couple of digits.
Also, in your sample You wisely excluded 'o', but forgot about the 'l', which you can easily confuse with '1'. I suggest you remove the 'l' as wel. ;-)
I am not sure that this answers your question, but maybe think about how many unique bug report number you need.
Simply using a four letter uppercase alphanumeric key like "BX-3D", you can have 36^4 = 1.7 million bug reports.
Edit: I just saw your sample. Maybe the results could be considerably improved if you used syllables instead of consonants and vowels.
I am a beginner trying to do some engineering experiments using fortran 77. I am using Force 2.0 compiler and editor. I have the following queries:
How can I generate a random number between a specified range, e.g. if I need to generate a single random number between 3.0 and 10.0, how can I do that?
How can I use the data from a text file to be called in calculations in my program. e.g I have temperature, pressure and humidity values (hourly values for a day, so total 24 values in each text file).
Do I also need to define in the program how many values are there in the text file?
Knuth has released into the public domain sources in both C and FORTRAN for the pseudo-random number generator described in section 3.6 of The Art of Computer Programming.
2nd question:
If your file, for example, looks like:
hour temperature pressure humidity
00 15 101325 60
01 15 101325 60
... 24 of them, for each hour one
this simple program will read it:
implicit none
integer hour, temp, hum
real p
character(80) junkline
open(unit=1, file='name_of_file.dat', status='old')
rewind(1)
read(1,*)junkline
do 10 i=1,24
read(1,*)hour,temp,p,hum
C do something here ...
10 end
close(1)
end
(the indent is a little screwed up, but I don't know how to set it right in this weird environment)
My advice: read up on data types (INTEGER, REAL, CHARACTER), arrays (DIMENSION), input/output (READ, WRITE, OPEN, CLOSE, REWIND), and loops (DO, FOR), and you'll be doing useful stuff in no time.
I never did anything with random numbers, so I cannot help you there, but I think there are some intrinsic functions in fortran for that. I'll check it out, and report tomorrow. As for the 3rd question, I'm not sure what you ment (you don't know how many lines of data you'll be having in a file ? or ?)
You'll want to check your compiler manual for the specific random number generator function, but chances are it generates random numbers between 0 and 1. This is easy to handle - you just scale the interval to be the proper width, then shift it to match the proper starting point: i.e. to map r in [0, 1] to s in [a, b], use s = r*(b-a) + a, where r is the value you got from your random number generator and s is a random value in the range you want.
Idigas's answer covers your second question well - read in data using formatted input, then use them as you would any other variable.
For your third question, you will need to define how many lines there are in the text file only if you want to do something with all of them - if you're looking at reading the line, processing it, then moving on, you can get by without knowing the number of lines ahead of time. However, if you are looking to store all the values in the file (e.g. having arrays of temperature, humidity, and pressure so you can compute vapor pressure statistics), you'll need to set up storage somehow. Typically in FORTRAN 77, this is done by pre-allocating an array of a size larger than you think you'll need, but this can quickly become problematic. Is there any chance of switching to Fortran 90? The updated version has much better facilities for dealing with standardized dynamic memory allocation, not to mention many other advantages. I would strongly recommend using F90 if at all possible - you will make your life much easier.
Another option, depending on the type of processing you're doing, would be to investigate algorithms that use only single passes through data, so you won't need to store everything to compute things like means and standard deviations, for example.
This subroutine generate a random number in fortran 77 between 0 and ifin
where i is the seed; some great number such as 746397923
subroutine rnd001(xi,i,ifin)
integer*4 i,ifin
real*8 xi
i=i*54891
xi=i*2.328306e-10+0.5D00
xi=xi*ifin
return
end
You may modifies in order to take a certain range.