I try to place N rectangular blocsks with different sizes into a grid, by formulating it
as a CSP propblem.
The blocks should not overlap with each other, they can touch on the edges, and there can be
empty places.
For example place 4 rectangular blocks of size 2x2 into a 8x8 grid. (Vary the number of blocks, the sizes of the blocks, and the size of the grid.) I know the formula as
I try to write a program or script generates the formula but I am confused too much I cant write in an SMT syntax. İf anyone helps I aprreciate too much. Thank you.
You should be specific about what you tried and what didn't work. If your problem is with syntax, then here's something to get you started:
(set-option :produce-models true)
(declare-fun xi () Int)
(declare-fun wi () Int)
(declare-fun xj () Int)
(declare-fun wj () Int)
(assert (or (<= (+ xi wi) xj)
(<= (+ xj wj) xi)))
The above encodes the first two disjuncts in your formula. You can add the other variables and assert all the other conditions as required.
Related
I am trying to test some of the lambda calculus functions that I wrote using Racket but not having much luck with the testcases. For example given a definition
; successor function
(define my_succ (λ (one)
(λ (two)
(λ (three)
(two ((one two) three))))))
I am trying to apply it to 1 2 3, expecting the successor of 2 to be 3 by doing
(((my_succ 1) 2) 3)
logic being that since my_succ is a function that takes one arg and passes it to another function that takes one arg which passes it to the third function that takes one arg. But I get
application: not a procedure;
expected a procedure that can be applied to arguments
given: 1
arguments.:
I tried Googling and found a lot of code for the rules, but no examples of application of these rules. How should I call the above successor function in order to test it?
You are mixing two completely different things: lambda terms and functions in Racket.
In Racket you can have anonymous functions, that can be written in the λ notation (like (λ(x) (+ x 1)) that returns the successor of an integer, so that ((λ(x) (+ x 1)) 1) returns 2),
in Pure Lambda Calculus you have only lambda terms, that are written with in a similar notation, and that can be interpreted as functions.
In the second domain, you do not have natural numbers like 0, 1, 2, ..., but you have only lambda terms, and represent numbers as such. For instance, if you use the so-called Church numerals, you represent (in technical term encode) the number 0 with the lambda term λf.λx.x, 1 with λf.λx.f x, 2 with λf.λx.f (f x) and so on.
So, the function successor (for numbers represented with this encoding) correspond to a term which, in Racket notation, is the function that you have written, but you cannot apply it to numbers like 0, 1, etc., but only to other lambda expressions, that is you could write something like this:
(define zero (λ(f) (λ (x) x))) ; this correspond to λf.λx.x
(successor zero)
The result in Racket is a procedure (it will be printed as: #<procedure>), but if you try to test that your result is correct, comparing it with the functional encoding of 1, you will find something strange. In fact:
(equal? (successor zero) (λ(f) (λ(x) (f x))))
produces #f, since if you compare two procedures in Racket you obtain always false (e.g. (equal? (λ(x)x) (λ(x)x)) produces #f), unless you compare the “identical” (in the sense of “same memory cell”) value ((equal? zero zero) gives #t). This is due to the fact that, for comparing correctly two functions, you should compare infinite sets of couples (input, output)!
Another possibility would be representing lambda terms as some kind of structure in Racket, so you can represent Church numerals, as well as "normal" lambda terms, and define a function apply (or better reduce) the perform lambda-reduction.
You are trying to apply currying.
(define my_succ
(lambda(x)(
lambda(y)(
lambda(z)(
(f x y z)))))
(define (add x y z)
(+ x y z))
((( (my_succ add)1)2)3)
Implementation in DR Racket:
I'm converting my problem to SMT and I have noticed that the SMT solvers (MathSat5 and CVC4) are slow when solving sat instances. My suspension is that there is something in my translation that is making it go slow.
I'm attaching a sample cnf instance and the smt2 translation for reference and below I'm providing solver times (excluding the translation time) for a larger instance to compare MathSat5, CVC4 and MiniSat.
Solver Solver Time (s)
-------------------------------------
MiniSat 0.028062 s
MathSat5 2.629702 s
CVC4 7.488870 s
CVC4(QF_SAT) 1.253978 s
So does anyone have an idea why these times are drastically different?
PS. cvc4 says it spent 5.862 seconds in: theory uf symmetry_breaker
Sample cnf:
-------------------------------------
p cnf 20 91
4 -18 19 0
...
4 -16 -5 0
Sample smt2:
-------------------------------------
(set-logic QF_UF)
(set-info :smt-lib-version 2.0)
(set-option :produce-models true)
(declare-fun v1 () Bool)
...
(declare-fun x20 () Bool)
(assert (or v4 (not x18) x19))
...
(assert (or v4 (not v16) (not v5)))
(check-sat)
(get-value ( v1 ... x20))
(exit)
Thanks
SMT solvers have extra overhead because of theory solvers. In CVC4, you can avoid this by using the following commands:
(set-logic QF_UF)
(set-info :cvc4-logic QF_SAT)
instead of
(set-logic QF_UF)
Note that this is a CVC4 extension, not part of the SMT-LIB standard. But if you are truly using only Boolean reasoning, this should give you competitive performance.
I am reading a Gentle Introduction to Symbolic Computation and it asks this question. Basically, the previous content deals with making up bigger functions with small ones. (Like 2- will be made of two 1- (decrement operators for lisp))
So one of the questions is what are the two different ways to define a function HALF which returns one half of its input. I have been able to come up with the obvious one (dividing number by 2) but then get stuck. I was thinking of subtracting HALF of the number from itself to get half but then the first half also has to be calculated...(I don't think the author intended to introduce recursion so soon in the book, so I am most probably wrong).
So my question is what is the other way? And are there only two ways?
EDIT : Example HALF(5) gives 2.5
P.S - the book deals with teaching LISP of which I know nothing about but apparently has a specific bent towards using smaller blocks to build bigger ones, so please try to answer using that approach.
P.P.S - I found this so far, but it is on a completely different topic - How to define that float is half of the number?
Pdf of book available here - http://www.cs.cmu.edu/~dst/LispBook/book.pdf (ctrl+f "two different ways")
It's seems to be you are describing peano arithmetic. In practice it works the same way as doing computation with fluids using cups and buckets.
You add by taking cups from the source(s) to a target bucket until the source(s) is empty. Multiplication and division is just advanced adding and substraction. To halve you take from source to two buckets in alterations until the source is empty. Of course this will either do ceil or floor depending on what bucket you choose to use as answer.
(defun halve (x)
;; make an auxillary procedure to do the job
(labels ((loop (x even acc)
(if (zerop x)
(if even (+ acc 0.5) acc)
(loop (- x 1) (not even) (if even (+ acc 1) acc)))))
;; use the auxillary procedure
(loop x nil 0)))
Originally i provided a Scheme version (since you just tagged lisp)
(define (halve x)
(let loop ((x x) (even #f) (acc 0))
(if (zero? x)
(if even (+ acc 0.5) acc)
(loop (- x 1) (not even) (if even (+ acc 1) acc)))))
Edit: Okay, lets see if I can describe this step by step. I'll break the function into multiple lines also.
(defun half (n)
;Takes integer n, returns half of n
(+
(ash n -1) ;Line A
(if (= (mod n 2) 1) .5 0))) ;Line B
So this whole function is an addition problem. It is simply adding two numbers, but to calculate the values of those two numbers requires additional function calls within the "+" function.
Line A: This performs a bit-shift on n. The -1 tells the function to shift n to the right one bit. To explain this we'll have to look at bit strings.
Suppose we have the number 8, represented in binary. Then we shift it one to the right.
1000| --> 100|0
The vertical bar is the end of the number. When we shift one to the right, the rightmost bit pops off and is not part of the number, leaving us with 100. This is the binary for 4.
We get the same value, however if we perform the shift on nine:
1001| --> 100|1
Once, again we get the value 4. We can see from this example that bit-shifting truncates the value and we need some way to account for the lost .5 on odd numbers, which is where Line B comes in.
Line B: First this line tests to see if n is even or odd. It does this by using the modulus operation, which returns the remainder of a division problem. In our case, the function call is (mod n 2), which returns the remainder of n divided by 2. If n is even, this will return 0, if it is odd, it will return 1.
Something that might be tripping you up is the lisp "=" function. It takes a conditional as its first parameter. The next parameter is the value the "=" function returns if the conditional is true, and the final parameter is what to return if the conditional is false.
So, in this case, we test to see if (mod n 2) is equal to one, which means we are testing to see if n is odd. If it is odd, we add .5 to our value from Line A, if it is not odd, we add nothing (0) to our value from Line A.
I wrote an answer to what I thought was a quite interesting question, but unfortunately the question was deleted by its author before I could post. I'm reposting a summary of the question and my answer here in case it might be of use to anyone else.
Suppose I have a SAT solver that, given a Boolean formula in conjunctive normal form, returns either a solution (a variable assignment that satisfies the formula) or the information that the problem is unsatisfiable.
Can I use this solver to find all the solutions?
There is definitely a way to use the SAT solver you described to find all the solutions of a SAT problem, although it may not be the most efficient way.
Just use the solver to find a solution to your original problem, add a clause that does nothing except rule out the solution you just found, use the solver to find a solution to the new problem, and so forth. Keep going until you get a problem that's unsatisfiable.
For example, suppose you want to satisfy (X or Y) and (X or Z). There are five solutions:
Four with X true, Y and Z arbitrary.
One with X false, Y and Z true.
So you run your solver, and let's say it gives you the solution (X, Y, Z) = (T, F, F). You can rule out this solution---and only this solution---with the constraint
not (X and (not Y) and (not Z))
This constraint can be rewritten as the clause
(not X) or Y or Z
So now you can run your solver on the new problem
(X or Y) and (X or Z) and ((not X) or Y or Z)
and so forth.
Like I said, this is a way to do what you want, but it probably isn't the most efficient way. When your SAT solver is looking for a solution, it learns a lot about the problem, but it doesn't return all that information to you---it just gives you the solution it found. When you run the solver again, it has to re-learn all the information that was thrown away.
Sure it can. When MiniSat[1] finds a solution
s SATISFIABLE
v 1 2 -3 0
(solution 1=True, 2=True, 3=False) then you have to put into the original CNF[2] a clause that bans this solution:
-1 -2 3 0
(which means, either 1 or 2 must be False or 3 must be True). Then you solve again. You do this until the solver returns UNSAT i.e. that there are no more solutions to the problem. You will insert one clause for each iteration, and each clause will have the same format as the solution except that it's all inverted and has a 0 at the end
It's much faster to do this using the C++ interface of MiniSat, as it can then save intermediate data and the iterations will be faster.
[1] http://minisat.se/
[2] http://fairmut3x.wordpress.com/2011/07/29/cnf-conjunctive-normal-form-dimacs-format-explained/
I'm trying to write a function that can calculate GPA. Now I can do limited calculation(only 3 ),but I stuck on how to calculate more , without using loop or recursion (that's the requirement of subject) how to expend nth function? like: (nth n) ,if so ,is that mean i need to write a lambda expression? As an newbie, I maynot describe the question clearly, really need some help..
Glist is grade points Clist is credit hours.
GPA=( gradepoint *credithour + gradepoint *credithour) / ( the sum of credithour) like: (3*1+3*2+4*1)/(1+2+1)
here is my code:
(defun gpa (Glist Clist)
(format t "~3,2f~%"
(/
(+(nth 0 (mapcar #' * Glist Clist))
(nth 1 (mapcar #' * Glist Clist))
(nth 2 (mapcar #' * Glist Clist)))
(+ (nth 0 Clist)
(nth 1 Clist)
(nth 2 Clist))
);end "/"
);end "format"
(values) );end
EDIT
This seems like a good opportunity to emphasize some common (little c) Lisp ideas, so I fleshed out my answer to illustrate.
As mentioned in another answer, you could use a sum function that operates on lists (of numbers):
(defun sum (nums)
(reduce #'+ nums))
The dot product is the multiplicative sum of two (equal-length) vectors:
(defun dot-product (x y)
(sum (mapcar #'* x y)))
The function gpa is a simple combination of the two:
(defun gpa (grades credits)
(/ (dot-product grades credits) (sum credits)))
The example from the question results in the answer we expect (minus being formatted as a float):
(gpa '(3 3 4) '(1 2 1))
> 13/4
There are a few things worth mentioning from this example:
You should learn about map, reduce, and their variants and relatives. These functions are very important to Lisp and are very useful for operating on lists. map* functions generally map sequences to a sequence, and reduce usually transforms a sequence into to a single value (you can however use forms like (reduce #'cons '(1 2 3))).
This is a good example of the "bottom-up" approach to programming; by programming simple functions like sum that are often useful, you make it easy to write dot-product on top of it. Now the gpa function is a simple, readable function built on top of the other two. These are all one-liners, and all are easily readable to anyone who has a basic knowledge of CL. This is in contrast to the methodology usually applied to OOP.
There is no repetition of code. Sure, sum is used more than once, but only where it makes sense. You can do very little more to abstract the notion of a sum of the elements of a list. It's more natural in Scheme to write functions with functions, and that's a whole different topic. This is a simple example, but no two functions are doing the same thing.
If you're using nth to traverse a list, you're doing it wrong. In this case, you might want to write a summing function:
(defun sum (items)
(reduce #'+ items))