I am attempting to create a linear combination of two numbers to create their GCD. The code I have so far can find the expanded solution. I have done all of the (hard) math for it (i.e. find the GCD using Euclid's Algorithm, then work backward essentially) and it will result in something like this for example (the two starting numbers are 1215 and 960):
((960-(3*(1215-(1*960))))-(3*((1215-(1*960))-(1*(960-(3*(1215-(1*960))))))))
In my actual solution there is a space between every component (e.g. '( ( 960 - (3 * '...)
but I am trying to simplify this into the equation:
((-15*1215)+(19*960))
I feel like the best approach is to create an Expression Tree, but I don't know how to without actually just evaluating the answer.
It sounds like you are looking for a symbolic computation system. Here's one approach using Maxima (https://maxima.sourceforge.net). I'll enable stardisp to show * between terms of a product. I'll also input numbers like 960 as symbols in order to suppress arithmetic on them by writing them as \960 etc. Note that 1 and 3 are input as ordinary numbers so arithmetic is carried out on them.
(%i13) stardisp:true;
(%o13) true
(%i14) 2*3;
(%o14) 6
(%i15) \2*\3;
(%o15) 2*3
(%i16) ((\960-(3*(\1215-(1*\960))))-(3*((\1215-(1*\960))-(1*(\960-(3*(\1215-(1*\960))))))));
(%o16) 960 - 3*(1215 - 960) - 3*((- 2*960) + 3*(1215 - 960)
+ 1215)
(%i17) factor(%);
(%o17) 19*960 - 15*1215
Maybe you want to replace the numbers with symbols a, b, c, etc to get a more general solution.
There are many other symbol computation systems, a web search will find them. Good luck and have fun.
Sometimes a device has commands to be sent to another device. In simple cases, I often use strings of digits and letters like
"123X456YG"
with the meaning
set the value of parameter X to 123, 456 to Y, and then Go.
The interpretation algorithm is quite simple : process char by char, "push" the digit to build a number (n = 10*n + ch-'0'), execute action (and reset number) when it is a letter.
It very convenient when there are not too many different actions (and you can remember the letters), of course you may use SCPI for more complicated things.
Is there a name for this (rather obvious) way to to?
This is an example of postfix notation. It's a notation that fairly naturally falls out of anything that uses a stack.
Examples include:
Reverse Polish Notation -- common on programmable calculators from the late 1960s onwards.
... in which 4 3 - 5 + results in 6 being left at the head of the stack, because the stack builds up like this:
Stack Unconsumed input
1 [ ] 4 3 - 5 +
2 [ 4 ] 3 - 5 +
3 [ 4 3 ] - 5 +
4 [ 1 ] 5 +
5 [ 1 5 ] +
6 [ 6 ]
At steps 4 and 6, the operator is executed by popping elements off the stack, then pushing the result onto the stack.
Many stack-based programming languages -- including Forth and Postscript. Postfix syntax is also fairly common in the world of esoteric programming languages.
If you're curious, I thoroughly recommend having a go at hand-writing Postscript. It's a fully-fledged programming language; for example, you can write a recursive Postscript program in a few bytes, to draw a fractal curve. Test it in Ghostscript. PDF is closely related to Postscript.
The language you have created differs from these in your approach to tokenisation, but at its core -- putting the operand(s) before the operator -- it's a postfix language.
You are unlikely to get a patent granted. :)
In Maple,
restart; with(LinearAlgebra);
E := Matrix([[A, B]]);
E. Transpose(E);
yields
A^2 + B^2
However, I would like that Maple treat A and B as block matrices and yield
A.Transpose(A) + B.Transpose(B)
Is this possible?
You'll want to use the Maple assume() command for this (link). Scroll down that link, or ctrl-f and find the part where they show how to assume that a variable is a "SquareMatrix" type. Basically, Maple is treating your variables like they are real numbers, and you need to tell it not to do that. Once you get the assume statement right, it should print out the matrix-based solution.
If you get a lot of crufty extra symbols, this might be because Maple usually flags variables for which the assume() function was used (so the user remembers they are making an assumption about that variable). For example, it often replaces a with ~a if you issue an assume() regarding a. You can turn this off with the command interface(showassumed=0).
I have been playing with an implementation of lookandsay (OEIS A005150) in J. I have made two versions, both very simple, using while. type control structures. One recurs, the other loops. Because I am compulsive, I started running comparative timing on the versions.
look and say is the sequence 1 11 21 1211 111221 that s, one one, two ones, etc.
For early elements of the list (up to around 20) the looping version wins, but only by a tiny amount. Timings around 30 cause the recursive version to win, by a large enough amount that the recursive version might be preferred if the stack space were adequate to support it. I looked at why, and I believe that it has to do with handling intermediate results. The 30th number in the sequence has 5808 digits. (32nd number, 9898 digits, 34th, 16774.)
When you are doing the problem with recursion, you can hold the intermediate results in the recursive call, and the unstacking at the end builds the results so that there is minimal handling of the results.
In the list version, you need a variable to hold the result. Every loop iteration causes you to need to add two elements to the result.
The problem, as I see it, is that I can't find any way in J to modify an extant array without completely reassigning it. So I am saying
try. o =. o,e,(0&{y) catch. o =. e,(0&{y) end.
to put an element into o where o might not have a value when we start. That may be notably slower than
o =. i.0
.
.
.
o =. (,o),e,(0&{y)
The point is that the result gets the wrong shape without the ravels, or so it seems. It is inheriting a shape from i.0 somehow.
But even functions like } amend don't modify a list, they return a list that has a modification made to it, and if you want to save the list you need to assign it. As the size of the assigned list increases (as you walk the the number from the beginning to the end making the next number) the assignment seems to take more time and more time. This assignment is really the only thing I can see that would make element 32, 9898 digits, take less time in the recursive version while element 20 (408 digits) takes less time in the loopy version.
The recursive version builds the return with:
e,(0&{y),(,lookandsay e }. y)
The above line is both the return line from the function and the recursion, so the whole return vector gets built at once as the call gets to the end of the string and everything unstacks.
In APL I thought that one could say something on the order of:
a[1+rho a] <- new element
But when I try this in NARS2000 I find that it causes an index error. I don't have access to any other APL, I might be remembering this idiom from APL Plus, I doubt it worked this way in APL\360 or APL\1130. I might be misremembering it completely.
I can find no way to do that in J. It might be that there is no way to do that, but the next thought is to pre-allocate an array that could hold results, and to change individual entries. I see no way to do that either - that is, J does not seem to support the APL idiom:
a<- iota 5
a[3] <- -1
Is this one of those side effect things that is disallowed because of language purity?
Does the interpreter recognize a=. a,foo or some of its variants as a thing that it should fastpath to a[>:#a]=.foo internally?
This is the recursive version, just for the heck of it. I have tried a bunch of different versions and I believe that the longer the program, the slower, and generally, the more complex, the slower. Generally, the program can be chained so that if you want the nth number you can do lookandsay^: n ] y. I have tried a number of optimizations, but the problem I have is that I can't tell what environment I am sending my output into. If I could tell that I was sending it to the next iteration of the program I would send it as an array of digits rather than as a big number.
I also suspect that if I could figure out how to make a tacit version of the code, it would run faster, based on my finding that when I add something to the code that should make it shorter, it runs longer.
lookandsay=: 3 : 0
if. 0 = # ,y do. return. end. NB. return on empty argument
if. 1 ~: ##$ y do. NB. convert rank 0 argument to list of digits
y =. (10&#.^:_1) x: y
f =. 1
assert. 1 = ##$ y NB. the converted argument must be rank 1
else.
NB. yw =. y
f =. 0
end.
NB. e should be a count of the digits that match the leading digit.
e=.+/*./\y=0&{y
if. f do.
o=. e,(0&{y),(,lookandsay e }. y)
assert. e = 0&{ o
10&#. x: o
return.
else.
e,(0&{y),(,lookandsay e }. y)
return.
end.
)
I was interested in the characteristics of the numbers produced. I found that if you start with a 1, the numerals never get higher than 3. If you start with a numeral higher than 3, it will survive as a singleton, and you can also get a number into the generated numbers by starting with something like 888888888 which will generate a number with one 9 in it and a single 8 at the end of the number. But other than the singletons, no digit gets higher than 3.
Edit:
I did some more measuring. I had originally written the program to accept either a vector or a scalar, the idea being that internally I'd work with a vector. I had thought about passing a vector from one layer of code to the other, and I still might using a left argument to control code. With I pass the top level a vector the code runs enormously faster, so my guess is that most of the cpu is being eaten by converting very long numbers from vectors to digits. The recursive routine always passes down a vector when it recurs which might be why it is almost as fast as the loop.
That does not change my question.
I have an answer for this which I can't post for three hours. I will post it then, please don't do a ton of research to answer it.
assignments like
arr=. 'z' 15} arr
are executed in place. (See JWiki article for other supported in-place operations)
Interpreter determines that only small portion of arr is updated and does not create entire new list to reassign.
What happens in your case is not that array is being reassigned, but that it grows many times in small increments, causing memory allocation and reallocation.
If you preallocate (by assigning it some large chunk of data), then you can modify it with } without too much penalty.
After I asked this question, to be honest, I lost track of this web site.
Yes, the answer is that the language has no form that means "update in place, but if you use two forms
x =: x , most anything
or
x =: most anything } x
then the interpreter recognizes those as special and does update in place unless it can't. There are a number of other specials recognized by the interpreter, like:
199(1000&|#^)199
That combined operation is modular exponentiation. It never calculates the whole exponentiation, as
199(1000&|^)199
would - that just ends as _ without the #.
So it is worth reading the article on specials. I will mark someone else's answer up.
The link that sverre provided above ( http://www.jsoftware.com/jwiki/Essays/In-Place%20Operations ) shows the various operations that support modifying an existing array rather than creating a new one. They include:
myarray=: myarray,'blah'
If you are interested in a tacit version of the lookandsay sequence see this submission to RosettaCode:
las=: ,#((# , {.);.1~ 1 , 2 ~:/\ ])&.(10x&#.inv)#]^:(1+i.#[)
5 las 1
11 21 1211 111221 312211
So I was playing with the newly standardized unordered_map from the STL. The code I have is kinda like this, I just create an unordered_map, fill it up, and print it out:
unordered_map<int,string> m1;
m1[5]="lamb";
m1[2]="had";
m1[3]="a";
m1[1]="mary";
m1[4]="little";
m1[7]="fleece";
m1[6]="whose";
m1[10]="fleecey";
m1[8]="was";
m1[9]="all";
for(unordered_map<int,string>::const_iterator i = m1.begin(); i != m1.end(); ++i)
cout<<i->first<<" "<<i->second<<endl;
However, the output I get is ordered thusly:
1 mary
2 had
3 a
4 little
5 lamb
6 whose
7 fleece
8 was
9 all
10 fleecey
But I don't want to pay the price to have my map ordered! That is why I am using an unordered_map... What is going on here?
additional note: I am using gcc version 4.3.4 20090804 (release) 1 (GCC) and am compiling like this g++ -std=c++0X maptest.cpp
"Unordered" doesn't mean it will store the items randomly or maintain the order you put them in the map. It just means you can't rely on any particular ordering. You don't pay a price for ordering, quite the contrary - the implementation isn't explicitly ordering the items, it's a hashmap and stores its elements in whatever way it pleases, which usually is a pretty performant way. It just so happens that the the hashing algorithm and other internal workings of the map, when using exactly these keys and this number and order of operations on the map, end up storing the items in a order that looks ordered. Strings, for example, may lead to an apparently randomized layout.
On a side note, this is probably caused by the map using a hash that maps (at least some) integers to itself and using the lower bits (as many as the map size mandates) of the hash the to determine the index for the underlying array (for instance, CPython does this - with some very clever additions to handle collisions relatively simply and efficiently; for the same reason the hashes of CPython strings and tuples are very predictable).
For your amusement, here's the output from libc++, which also has an identity function for std::hash<int>.
9 all
8 was
10 fleecey
6 whose
7 fleece
4 little
1 mary
3 a
2 had
5 lamb
There are several ways to implement a hash container, each with its own tradeoffs.