I want define the digits 0-9 to say the word when used as arithmetic in scheme. for example. (+ 2 5) would be two + five. How can i do this, thanks!
Okay, I'm going to repeat my earlier suggestion, but this time as an answer :).
I think that you really need to formulate a set of concrete test cases. The point is partly to help us understand, but perhaps more importantly, to help you organize your thinking. A test case should include a call to the function, and the expected result. So, for instance, it might look like this:
(check-expect (gorzify '(+ 2 5)) "ouch! my toe")
... though I'm guessing that's not what your program is supposed to produce.
Related
I'm reading a note about the definition of algorithm, it has two requirements that I don't know what's the differences between them
Definiteness: Every instruction should be clear and unambiguous. (I found a source with exactly the same statement)
From the resource I have there are 5 requirements: Input, Output, Definiteness, Finiteness, Effectiveness. I can understand the other 4 except the Definiteness. Can anyone provide some better definition if the above is not precise?
From the above I only suspect that there are at least two subtleties should be considered...
For conclusion from answers below: definiteness = defined(clear) + only_one(unambiguous).
Algorithm should be clear and unambiguous. Each of its steps (or phases), and their inputs/outputs should be clear and must lead to only one meaning.
For example, if one step is to add two integers, we must define both “integers” as well as the “add” operation: we cannot for example use the same symbol to mean addition in one place and multiplication somewhere else.
If presented to an educated human, the text should allow him to simulate execution by hand in exactly the way you had in mind (same steps taken, same results obtained).
When you don't quite understand the definition of a term provided by some author, it's often helpful to look for other definitions of it. I especially like the one for "definite" from wiktionary.org:
Free from any doubt.
In this context, clear becomes understandable, and unambiguous becomes with a single meaning.
It just means that instructions in an algorithm should have one and only one interpretation. Moreover, the interpretation should be obvious.
A statement like "Repeat steps 1 to 4 a few times" does not fit the criteria as "few times" can mean different number of tries to different people.
On the other hand, a statement like "Repeat steps 1 to 4 until x is equal to y" where x and y are some parameters in the algorithm is indeed clear and unambiguous.
Is it possible to quote result of function call?
For example
(quoteresult (+ 1 1)) => '2
At first blush, your question doesn’t really make any sense. “Quoting” is a thing that can be performed on datums in a piece of source code. “Quoting” a runtime value is at best a no-op and at worst nonsensical.
The example in your question illustrates why it doesn’t make any sense. Your so-called quoteresult form would evaluate (+ 1 1) to produce '2, but '2 evaluates to 2, the same thing (+ 1 1) evaluates to. How would the result of quoteresult ever be different from ordinary evaluation?
If, however, you want to actually produce a quote expression to be handed off to some use of dynamic evaluation (with the usual disclaimer that that is probably a bad idea), then you need only generate a list of two elements: the symbol quote and your function’s result. If that’s the case, you can implement quoteresult quite simply:
(define (quoteresult x)
(list 'quote x))
This is, however, of limited usefulness for most programs.
For more information on what quoting is and how it works, see What is the difference between quote and list?.
//My question was so long So I reduced.
In scheme, user-made procedures consume more time than built-in procedures?
(If both's functions are same)
//This is my short version question.
//Below is long long version question.
EX 1.23 is problem(below), why the (next) procedure isn't twice faster than (+ 1)?
This is my guess.
reason 1 : (next) contains 'if' (special-form) and it consumes time.
reason 2 : function call consumes time.
http://community.schemewiki.org/?sicp-ex-1.23 says reason 1 is right.
And I want to know reason 2 is also right.
SO I rewrote the (next) procedure. I didn't use 'if' and checked the number divided by 2 just once before use (next)(so (next) procedure only do + 2). And I remeasured the time. It was more fast than before BUT still not 2. SO I rewrote again. I changed (next) to (+ num 2). Finally It became 2 or almost 2. And I thought why. This is why I guess the 'reason 2'. I want to know what is correct answer.
ps. I'm also curious about why some primes are being tested (very?) fast than others? It doesn't make sense because if a number n is prime, process should see from 2 to sqrt(n). But some numbers are tested faster. Do you know why some primes are tested faster?
Exercise 1.23. The smallest-divisor procedure shown at the start of this section does lots of needless testing: After it checks to see if the number is divisible by 2 there is no point in checking to see if it is divisible by any larger even numbers. This suggests that the values used for test-divisor should not be 2, 3, 4, 5, 6, ..., but rather 2, 3, 5, 7, 9, .... To implement this change, define a procedure next that returns 3 if its input is equal to 2 and otherwise returns its input plus 2. Modify the smallest-divisor procedure to use (next test-divisor) instead of (+ test-divisor 1). With timed-prime-test incorporating this modified version of smallest-divisor, run the test for each of the 12 primes found in exercise 1.22. Since this modification halves the number of test steps, you should expect it to run about twice as fast. Is this expectation confirmed? If not, what is the observed ratio of the speeds of the two algorithms, and how do you explain the fact that it is different from 2?
Where the book is :
http://mitpress.mit.edu/sicp/full-text/book/book-Z-H-11.html#%_sec_1.2.6
Your long ans short questions are actually addressing two different problems.
EX 1.23 is problem(below), why the (next) procedure isn't twice faster than (+ 1)?
You provide two possible reasons to explain the relative lack of speed up (and 30% speed gain is already a good achievement):
The apparent use of a function next instead of a simple arithmetic expression (as I understand it from your explanations),
the inclusion of a test in that very next function,
The first reason is an illusion: (+ 1) is a function incrementing its argument, so in both cases there is a function call (although the increment function is certainly a builtin one, a point which is addressed by your other question).
The second reason is indeed relevant: any test in a block of code will introduce a potential redirection in the code flow (a jump from the current executing instruction to some other address in the program), which may incur a delay. Note that this is analogous to a function call, which will also induce an address jump, so both reasons actually resolve to only one potential cause.
Regarding your short question, builtin functions are indeed usually faster, because the compiler is able to apply a special treatment to them in certain cases. This is due to the facts that:
knowing the semantics of the builtins, compiler designers are able to include special rules pertaining to the algebraic properties of these builtins, and for instance fuse successive incréments in a single call, or suppress a combination of increment and decrement called in sequence.
A builtin function call, when not optimised away, will be converted into a native machine code function call, which may not have to abide to all the calling convention rules. If your scheme compiler produce machine code from the source, then there might be only a marginal gain, but if it produce so called bytecode, the gain might be quite substantial, since user written functions will be translated to that bytecode format, and still require some form of interpretation. If you are only using an interpreter, then the gain is even more important.
I believe this is highly implementation and setting dependent. In many implementations there are different kinds of optimizations and in some there are none. To get the best performance you may need to compile your code or use settings to reduce debug information / stack traces. Getting the best performance in one implementation can worsen the performance in another.
Primitive procedures are usually compiled to be native and in some implementations, like ikarus, it's even inlined. (When you do (map car lst) ikarus changes it to (map (lambda (x) (car x)) lst) since car isn't a procedure) Lambdas are supposed to be cheap.. Remember many scheme implementations change your code to CPS and that is one procedure call for each expression in the body of a procedure call. It will never be as fast as machine code since it needs to do load closure variables.
To check which of the two options which are correct for your implementation make next do the same as it originally did. eg. no if but just increment the argument. The difference now is the extra call and nothing else. Then you can inline next by writing it's code directly in your procedure and substituting arguments for the operands. Is it still slower, then it's if. you need to run the tests several times, preferably with large enough number of primes to produce that it runs for a minute or so. Use time or similar in both the Scheme implementations to get the differences in ms. I use unix time command as well to see how the OS reflects on it.
You should also test to see if you get the same reason in some other implementation. It's not like it's not enough Scheme implementations out there so know yourself out! The differences between them might amaze you. I always use racket (raco exe source to make executable) and Ikarus. WHen doing a large test I include Chicken, Gambit and Chibi.
Let me establish that this is part of a class assignment, so I'm definitely not looking for a complete code answer. Essentially we need to write a converter in Scheme that takes a list representing a mathematical equation in infix format and then output a list with the equation in postfix format.
We've been provided with the algorithm to do so, simple enough. The issue is that there is a restriction against using any of the available imperative language features. I can't figure out how to do this in a purely functional manner. This is our fist introduction to functional programming in my program.
I know I'm going to be using recursion to iterate over the list of items in the infix expression like such.
(define (itp ifExpr)
(
; do some processing using cond statement
(itp (cdr ifExpr))
))
I have all of the processing implemented (at least as best I can without knowing how to do the rest) but the algorithm I'm using to implement this requires that operators be pushed onto a stack and used later. My question is how do I implement a stack in this function that is available to all of the recursive calls as well?
(Updated in response to the OP's comment; see the new section below the original answer.)
Use a list for the stack and make it one of the loop variables. E.g.
(let loop ((stack (list))
... ; other loop variables here,
; like e.g. what remains of the infix expression
)
... ; loop body
)
Then whenever you want to change what's on the stack at the next iteration, well, basically just do so.
(loop (cons 'foo stack) ...)
Also note that if you need to make a bunch of "updates" in sequence, you can often model that with a let* form. This doesn't really work with vectors in Scheme (though it does work with Clojure's persistent vectors, if you care to look into them), but it does with scalar values and lists, as well as SRFI 40/41 streams.
In response to your comment about loops being ruled out as an "imperative" feature:
(let loop ((foo foo-val)
(bar bar-val))
(do-stuff))
is syntactic sugar for
(letrec ((loop (lambda (foo bar) (do-stuff))))
(loop foo-val bar-val))
letrec then expands to a form of let which is likely to use something equivalent to a set! or local define internally, but is considered perfectly functional. You are free to use some other symbol in place of loop, by the way. Also, this kind of let is called 'named let' (or sometimes 'tagged').
You will likely remember that the basic form of let:
(let ((foo foo-val)
(bar bar-val))
(do-stuff))
is also syntactic sugar over a clever use of lambda:
((lambda (foo bar) (do-stuff)) foo-val bar-val)
so it all boils down to procedure application, as is usual in Scheme.
Named let makes self-recursion prettier, that's all; and as I'm sure you already know, (self-) recursion with tail calls is the way to go when modelling iterative computational processes in a functional way.
Clearly this particular "loopy" construct lends itself pretty well to imperative programming too -- just use set! or data structure mutators in the loop's body if that's what you want to do -- but if you stay away from destructive function calls, there's nothing inherently imperative about looping through recursion or the tagged let itself at all. In fact, looping through recursion is one of the most basic techniques in functional programming and the whole point of this kind of homework would have to be teaching precisely that... :-)
If you really feel uncertain about whether it's ok to use it (or whether it will be clear enough that you understand the pattern involved if you just use a named let), then you could just desugar it as explained above (possibly using a local define rather than letrec).
I'm not sure I understand this all correctly, but what's wrong with this simpler solution:
First:
You test if your argument is indeed a list:
If yes: Append the the MAP of the function over the tail (map postfixer (cdr lst)) to the a list containing only the head. The Map just applies the postfixer again to each sequential element of the tail.
If not, just return the argument unchanged.
Three lines of Scheme in my implementation, translates:
(postfixer '(= 7 (/ (+ 10 4) 2)))
To:
(7 ((10 4 +) 2 /) =)
The recursion via map needs no looping, not even tail looping, no mutation and shows the functional style by applying map. Unless I'm totally misunderstanding your point here, I don't see the need for all that complexity above.
Edit: Oh, now I read, infix, not prefix, to postfix. Well, the same general idea applies except taking the second element and not the first.
I am trying to put the following statement in Dr.Scheme:
{with {x {+ 5 5}} {+ x x}}
but I got an error:
expand: unbound identifier in module in: with
anyone could help me?Thanks.
You're taking some PLAI-based course, and you confuse the language that you're working in (Scheme) with the language that you're implementing (WAE, or one of the extensions). These two are very different things, and the book uses curly braces in the latter to avoid confusion.
I can tell you from experience of teaching this class a number of times that it's a dangerous confusion, and the sooner you clarify things the better. If you leave it behind things might get more confusing in the near future. So spend some time on the differences between the two languages, and make sure that you know which parts of the book talk about which language.
Are you trying to do this:
(let ([x (+ 5 5)] ) (+ x x ))
It would be really helpful if you could say what dialect of Scheme you are trying to use.