I have been trying to run the following boolean query in lucene but it seems to fail. Please help.
(a or b) and c -> works fine
(a AND b) or c -> gives result for a AND b OR c. So a becomes must and b and c becomes should, and the search result is wrong. Where it should work like a, b must be available or c may be available.
Another example:
If you search for "(a AND b)" it will return x results
If you search for "c" it will return y results
If you search for "(a and b) or c" the number of results cannot be less than the larger of x or y. But this is not happening. Please help how should I proceed to implement this?
(a AND b) or c gets converted to (+a +b) c. What you want is this without parentheses: +a +b c.
From a logical perspective, "A and B must occur, C might occur" is equivalent to "A and B must occur." (There is no way to say "might be true" in classical logic.) So you will have difficulty saying (+a +b) c in boolean terms. One way you could do it is "(a AND b AND c) or (a AND b)".
Related
I'm just starting playing with idris and theorem proving in general. I can follow most of the examples of proofs of basic facts on the internet, so I wanted to try something arbitrary by my own. So, I want to write a proof term for the following basic property of map:
map : (a -> b) -> List a -> List b
prf : map id = id
Intuitively, I can imagine how the proof should work: Take an arbitrary list l and analyze the possibilities for map id l. When l is empty, it's obvious; when
l is non-empty it's based on the concept that function application preserves equality.
So, I can do something like this:
prf' : (l : List a) -> map id l = id l
It's like a for all statement. How can I turn it into a proof of the equality of the functions involved?
You can't. Idris's type theory (like Coq's and Agda's) does not support general extensionality. Given two functions f and g that "act the same", you will never be able to prove Not (f = g), but you will only be able to prove f = g if f and g are defined the same, up to alpha and eta equivalence or so. Unfortunately, things only get worse when you consider higher-order functions; there's a theorem about such in the Coq standard library, but I can't seem to find or remember it right now.
I'm really loving Elm, up to the point where I encounter a function I've never seen before and want to understand its inputs and outputs.
Take the declaration of foldl for example:
foldl : (a -> b -> b) -> b -> List a -> b
I look at this and can't help feeling as if there's a set of parentheses that I'm missing, or some other subtlety about the associativity of this operator (for which I can't find any explicit documentation). Perhaps it's just a matter of using the language more until I just get a "feel" for it, but I'd like to think there's a way to "read" this definition in English.
Looking at the example from the docs…
foldl (::) [] [1,2,3] == [3,2,1]
I expect the function signature to read something like this:
Given a function that takes an a and a b and returns a b, an additional b, and a List, foldl returns a b.
Is that correct?
What advice can you give to someone like me who desperately wants inputs to be comma-delineated and inputs/outputs to be separated more clearly?
Short answer
The missing parentheses you're looking for are due to the fact that -> is right-associative: the type (a -> b -> b) -> b -> List a -> b is equivalent to (a -> b -> b) -> (b -> (List a -> b)). Informally, in a chain of ->s, read everything before the last -> as an argument and only the rightmost thing as a result.
Long answer
The key insight you may be missing is currying -- the idea that if you have a function that takes two arguments, you can represent it with a function that takes the first argument and returns a function that takes the second argument and then returns the result.
For instance, suppose you have a function add that takes two integers and adds them together. In Elm, you could write a function that takes both elements as a tuple and adds them:
add : (Int, Int) -> Int
add (x, y) = x+y
and you could call it as
add (1, 2) -- evaluates to 3
But suppose you didn't have tuples. You might think that there would be no way to write this function, but in fact using currying you could write it as:
add : Int -> (Int -> Int)
add x =
let addx : Int -> Int
addx y = x+y
in
addx
That is, you write a function that takes x and returns another function that takes y and adds it to the original x. You could call it with
((add 1) 2) -- evaluates to 3
You can now think of add in two ways: either as a function that takes an x and a y and adds them, or as a "factory" function that takes x values and produces new, specialized addx functions that take just one argument and add it to x.
The "factory" way of thinking about things comes in handy every once in a while. For instance, if you have a list of numbers called numbers and you want to add 3 to each number, you can just call List.map (add 3) numbers; if you'd written the tuple version instead you'd have to write something like List.map (\y -> add (3,y)) numbers which is a bit more awkward.
Elm comes from a tradition of programming languages that really like this way of thinking about functions and encourage it where possible, so Elm's syntax for functions is designed to make it easy. To that end, -> is right-associative: a -> b -> c is equivalent to a -> (b -> c). This means if you don't parenthesize, what you're defining is a function that takes an a and returns a b -> c, which again we can think of either as a function that takes an a and a b and returns a c, or equivalently a function that takes an a and returns a b -> c.
There's another syntactic nicety that helps call these functions: function application is left-associative. That way, the ugly ((add 1) 2) from above can be written as add 1 2. With that syntax tweak, you don't have to think about currying at all unless you want to partially apply a function -- just call it with all the arguments and the syntax will work out.
Question I saw on site that explains the issue of mutual-exclusion (http://www.faculty.idc.ac.il/gadi/PPTs/Chapter2-Mutex-BasicTopics.pptx - page 8). Unfortunately there is no answer. Also, the original question is only about C but I didn't understand If the order is changing how it affects the result, so I added D.
Let A and B be two algorithms designed to solve the mutual-exclusion problem. In other words, their structure consists of an entry-section, critical-section and exit-section (but you cannot assume they satisfy mutual-exclusion or deadlock-freedom unless written otherwise). Assume that algorithms A and B do not use the same variables. We construct a new mutual-excusion algorithm, C, as follows:
Algorithm C
entry code of A
entry code of B
Critical Section
exit code of B
exit code of A
For each of the following assertions, please prove or disprove its correctness.
If both A and B are deadlock-free, C is deadlock-free.
If both A and B are starvation-free, C is starvation-free.
If A or B satisfy mutual-exclusion, C satisfies mutual-exclusion.
If A is deadlock-free and B is starvation-free, C is starvation-free.
If A is starvation-free and B is deadlock-free, C is starvation-free.
Also that same questions, but this time on D instead of C, where D is:
Algorithm D
entry code of A
entry code of B
Critical Section
exit code of A
exit code of B
Thanks!
There's a certain over-verbosity that I have to engage in when writing certain Boolean expressions, at least with all the languages I've used, and I was wondering if there were any languages that let you write more concisely?
The way it goes is like this:
I want to find out if I have a Thing that can be either A, B, C, or D.
And I'd like to see if Thing is an A or a B.
The logical way for me to express this is
//1: true if Thing is an A or a B
Thing == (A || B)
Yet all the languages I know expect it to be written as
//2: true if Thing is an A or a B
Thing == A || Thing == B
Are there any languages that support 1? It doesn't seem problematic to me, unless Thing is a Boolean.
Yes. Icon does.
As a simple example, here is how to get the sum of all numbers less than 1000 that are divisble by three or five (the first problem of Project Euler).
procedure main ()
local result
local n
result := 0
every n := 1 to 999 do
if n % (3 | 5) == 0 then
result +:= n
write (result)
end
Note the n % (3 | 5) == 0 expression. I'm a bit fuzzy on the precise semantics, but in Icon, the concept of booleans is not like other languages. Every expression is a generator and it may pass (generating a value) or fail. When used in an if expression, a generator will continue to iterate until it passes or exhausts itself. In this case, n % (3 | 5) == 0 is a generator which uses another generator (3 | 5) to test if n is divisible by 3 or 5. (To be entirely technical, this isn't even syntactic sugar.)
Likewise, in Python (which was influenced by Icon) you can use the in statement to test for equality on multiple elements. It's a little weaker than Icon though (as in, you could not translate the modulo comparison above directly). In your case, you would write Thing in (A, B), which translates exactly to what you want.
There are other ways to express that condition without trying to add any magic to the conditional operators.
In Ruby, for example:
$> thing = "A"
=> "A"
$> ["A","B"].include? thing
=> true
I know you are looking for answers that have the functionality built into the language, but here are two other means that I find work better as they solve more problems and have been in use for many decades.
Have you considered using a preprocessor?
Also languages like Lisp have macros which is part of the language.
Given:
I have no idea what the accepted language is.
From looking at it you can get several end results:
1.) bb
2.) ab(a,b)
3.) bbab(a, b)
4.) bbaaa
How to write regular expression for a DFA
In any automata, the purpose of state is like memory element. A state stores some information in automate like ON-OFF fan switch.
A Deterministic-Finite-Automata(DFA) called finite automata because finite amount of memory present in the form of states. For any Regular Language(RL) a DFA is always possible.
Let's see what information stored in the DFA (refer my colorful figure).
(note: In my explanation any number means zero or more times and Λ is null symbol)
State-1: is START state and information stored in it is even number of a has been come. And ZERO b.
Regular Expression(RE) for this state is = (aa)*.
State-4: Odd number of a has been come. And ZERO b.
Regular Expression for this state is = (aa)*a.
Figure: a BLUE states = EVEN number of a, and RED states = ODD number of a has been come.
NOTICE: Once first b has been come, move can't back to state-1 and state-4.
State-5: comes after Yellow b. Yellow b means b after odd numbers of a.
Once you gets b after odd numbers of a(at state-5) every thing is acceptable because there is self a loop for (b,a) at state-5.
You can write for state-5 : Yellow-b followed-by any string of a, b that is = Yellow-b (a + b)*
State-6: Just to differentiate whether odd a or even.
State-2: comes after even a then b then any number of b. = (aa)* bb*
State-3: comes after state-2 then first a then there is a loop via state-6.
We can write for state-3 comes = state-2 a (aa)* = (aa)*bb* a (aa)*
Because in our DFA, we have three final states so language accepted by DFA is union (+ in RE) of three RL (or three RE).
So the language accepted by the DFA is corresponding to three accepting states-2,3,5, And we can write like:
State-2 + state-3 + state-5
(aa)*bb* + (aa)*bb* a (aa)* + Yellow-b (a + b)*
I forgot to explain how Yellow-b comes?
ANSWER: Yellow-b is a b after state-4 or state-3. And we can write like:
Yellow-b = ( state-4 + state-3 ) b = ( (aa)*a + (aa)*bb* a (aa)* ) b
[ANSWER]
(aa)*bb* + (aa)*bb* a (aa)* + ( (aa)*a + (aa)*bb* a (aa)* ) b (a + b)*
English Description of Language: DFA accepts union of three languages
EVEN NUMBERs OF a's, FOLLOWED BY ONE OR MORE b's,
EVEN NUMBERs OF a's, FOLLOWED BY ONE OR MORE b's, FOLLOWED BY ODD NUMBERs OF a's.
A PREFIX STRING OF a AND b WITH ODD NUMBER OF a's, FOLLOWED BY b, FOLLOWED BY ANY STRING OF a AND b AND Λ.
English Description is complex but this the only way to describe the language. You can improve it by first convert given DFA into minimized DFA then write RE and description.
Also, there is a Derivative Method to find RE from a given Transition Graph using Arden's Theorem. I have explained here how to write a regular expression for a DFA using Arden's theorem. The transition graph must first be converted into a standard form without the null-move and single start state. But I prefer to learn Theory of computation by analysis instead of using the Mathematical derivation approach.
I guess this question isn't relevant anymore :) and it's probably better to guide you through it then just stating the answer, but I think I got a basic expression that covers it (it's probably minimizable), so i'll just write it down for future searchers
(aa)*b(b)* // for stoping at 2
U
(aa)*b(b)*a(aa)* // for stoping at 3
U
(aa)*b(b)*a(aa)*b((a)*(b)*)* // for stoping at 5 via 3
U
a(aa)*b((a)*(b)*)* // for stoping at 5 via 4
The examples (1 - 4) that you give there are not the language accepted by the DFA. They are merely strings that belong to the language that the DFA accepts. Therefore, they all fall in the same language.
If you want to figure out the regular expression that defines that DFA, you will need to do something called k-path induction, and you can read up on it here.