Here is the question:
Show that:
L = {0m1n, m > 1, n > 1, n < m } , where m & n are superscripts
is not regular.
I am not sure what superscripts mean in this situation? Does it mean something like this:
0^5 = 00000 or 1^7 = 1111111
P.S. How can I show superscripts in SO??
Related
Let T (n) = 1, and let
T(n) = x^k T(n/x)+ Cn^k
where C and k are constants. Prove that
T (n) = O(n^k log(n))
I want to prove the time complexity using Induction and not sure where to start. Can someone please help me.
Thanks in advance...
I think this is pretty straight forward:
Base case, n = 1.
T(1) = 1 is a constant so it is O(n^klog(n))
Induction hypothesis:
T(m) = O(m^k log(m)) for all m < n.
Induction step:
T(n) = x^k T(n/x)+ Cn^k
by IH we have that this is
T(n) = x^k D(n/x)^k log n/x + Cn^k for some constant D
T(n) = D x^k n^k /x^k (log n - log x) + C*n^k , x^k now cancels out
T(n) = D n^k log n - D n^k log x + C*n^k
largest term here is D n^k log n, so in O-notation
T(n) = O( n^k log n)
I want to find a maximal interval in which an expression e is true for all x. A way to write such a formula should be: Exists d : ForAll x in (-d,d) . e and ForAll x not in (-d,d) . !e.
To get such a d, the formula f in Z3 (looking at the one above) could be the following:
from z3 import *
x = Real('x')
delta = Real('d')
s = Solver()
e = And(1/10000*x**2 > 0, 1/5000*x**3 + -1/5000*x**2 < 0)
f = ForAll(x,
And(Implies(And(delta > 0,
-delta < x, x < delta,
x != 0),
e),
Implies(And(delta > 0,
Or(x > delta, x < -delta),
x != 0),
Not(e))
)
)
s.add(Not(f))
s.check()
print s.model()
It prints: [d = 2]. This is surely not true (take x = 1). What's wrong?
Also: by specifying delta = RealVal('1'), a counterexample is x = 0, even when x = 0 should be avoided.
Your constants are getting coerced to integers. Instead of writing:
1/5000
You should write:
1.0/5000.0
You can see the generated expression by:
print s.sexpr()
which would have alerted you to the issue.
NB. Being explicit about types is always important when writing constants. See this answer for a variation on this theme that can lead to further problems: https://stackoverflow.com/a/46860633/936310
I try to remove 1 whitespace from this string:
m y r e a l n a m e i s d o n a l d d u c k
Expected result:
my real name is donald duck
My code are:
def solve_cipher(input)
input.split('').map { |c| (c.ord - 3).chr }.join(' ') # <- Here I tried everything
end
puts solve_cipher('p| uhdo qdph lv grqdog gxfn')
# => m y r e a l n a m e i s d o n a l d d u c k
I tried everything to solve my problem, example:
input.....join(' ').squeeze(" ").strip # => m y r e a l n a m e...
or
input.....join.gsub(' ','') # => myrealnameisdonaldduck
or
input.....join(' ').lstrip # => m y r e a l n a m e...
and so on...
Well, you could split the string into words first, then split each word into characters. Using the same method you used in your code, it could look like this.
def solve_cipher(input) input.split(' ').map{ |w| w.split('').map { |c| (c.ord - 3).chr}.join('')}.join(' ') end
When joining the characters in the same word, we put no space between them; when joining the words together we put one space between them.
As stated in the question, you are using Rails, so you can also try squish method:
def solve_cipher( input )
input.split(' ').map(&:squish).join(' ')
end
str = "m y r e a l n a m e i s d o n a l d d u c k"
str.gsub(/\s(?!\s)/,'')
#=> "my real name is donald duck"
The regex matches a whitespace character not followed by another whitespace character and replaces the matched characters with empty strings. (?!\s) is a negative lookahead that matches a whitespace.
If more than two spaces may be present between words, first replace three or more spaces with two spaces, as follows.
str = "m y r e a l n a m e i s d o n a l d d u c k"
str.gsub(/\s{3,}/, " ").gsub(/\s(?!\s)/,'')
#=> "my real name is donald duck"
I know that it is not a fancy way of doing it but you could just try to create a new string and have a function traversal(input) with a counter initiated at 0, that would return this new string.
It would go through your input (which is here your string) and if the counter is 0 and it sees a space it just ignores it, increments a counter and go to the next character of the string.
If the counter is different of 0 and it sees a space it just concatenates it to the new string.
And if the counter is different of 0 and it sees something different of a space, it concatenates it to the new string and counter equals 0 again.
The trick is to use a capture group
"m y r e a l n a m e i s d o n a l d d u c k".gsub(/(\s)(.)/, '\2')
=> "my real name is donald duck
I have two collections (they happen to be arrays, but it doesn't really matter, I think): L and R. They are both sorted and now I want to compare them. I want to end up with two collections: one for each input array containing the items which were not in the other.
I could just take the first item from L and then search R and, if there isn't a match, add it to my "unique" collection (Lu). But that's extremely inefficient, and I am expecting to have some very large collections to process in the near future.
I though about possibly "playing hopscotch":
Step 1: Take two lists, L and R, and compare the head of each list ( l :: L and r :: R):
Branch 1: if l < r, then add l to Lu and recurse, passing in L and r :: R
Branch 2: if l > r, then add r to Ru and recurse, passing in l :: L and R
Branch 3: if l = r, then recurse, passing in L and R
Step 2: return Lu and Ru
I can write this function, but before I put in the effort I was wondering if a function already exists which can do this for me. It seems like a not-to-uncommon scenario, and I'd always rather use an existing solution to rolling my own.
(Also, if there's a more recognizable name for this algorithm, I'd like to know what it's called.)
(I wrote the question above about 2 hours ago. Since then, I found the answer on my own. The following is what I discovered.)
In set theory, the "list" of items in L but not in R is known as "the relative complement of R in L", also known as "set-theoretic difference of L and R"
(See Wikipedia's Complement (set theory) article)
F#, being a mathematical language, has this concept baked right in to it's Core library. First, you need to build your collections as sets:
// example arrays:
let arr1 = [| 1; 2; 3 |]
let arr2 = [| 2; 3; 4 |]
// build the L and R sets
let L = set arr1
let R = set arr2
Now you can call the "difference" function and quickly get the relative complement for each array:
let Lu = Set.difference L R |> Set.toArray
let Ru = Set.difference R L |> Set.toArray
> val Lu : int [] = [|1|]
> val Ru : int [] = [|4|]
There's also a shorter syntax. The Set type has overloaded the minus operator. Set.difference just subtracts the second parameter from the first, so you can actually just use the following:
let Lu = L - R |> Set.toArray
let Ru = R - L |> Set.toArray
> val Lu : int [] = [|1|]
> val Ru : int [] = [|4|]
I used the Z3_ast fs = Z3_parse_smtlib2_file(ctx,arg[1],0,0,0,0,0,0) to read file.
Additionally to add into the solver utilized the expr F = to_expr(ctx,fs) and then s.add(F).
My question is how can I get the number of total constraints in each instance?
I also tried the F.num_args(), however, it is giving wrong size in some instances.
Are there any ways to compute the total constraints?
Using Goal.size() may do what you want, after you add F to some goal. Here's a link to the Python API description, I'm sure you can find the equivalent in the C/C++ API: http://research.microsoft.com/en-us/um/redmond/projects/z3/z3.html#Goal-size
An expr F represents an abstract syntax tree, so F.num_args() returns the number of (one-step) children that F has, which is probably why what you've been trying doesn't always work. For example, suppose F = a + b, then F.num_args() = 2. But also, if F = a + b*c, then F.num_args() = 2 as well, where the children would be a and b*c (assuming usual order of operations). Thus, to compute the number of constraints (in case your definition is different than what Goal.size() yields), you can use a recursive method that traverses the tree.
I've included an example below highlighting all of these (z3py link here: http://rise4fun.com/Z3Py/It5E ).
For instance, my definition of constraint (or rather the complexity of an expression in some sense) might be the number of leaves or the depth of the expression. You can get as detailed as you want with this, e.g., counting different types of operands to fit whatever your definition of constraint might be, since it's not totally clear from your question. For instance, you might define a constraint as the number of equalities and/or inequalities appearing in an expression. This would probably need to be modified to work for formulas with quantifiers, arrays, or uninterpreted functions. Also note that Z3 may simplify things automatically (e.g., 1 - 1 gets simplified to 0 in the example below).
a, b, c = Reals('a b c')
F = a + b
print F.num_args() # 2
F = a + b * c
print F.num_args() # 2
print F.children() # [a,b*c]
g = Goal()
g.add(F == 0)
print g.size() # number of constraints = 1
g.add(Or(F == 0, F == 1, F == 2, F == 3))
print g.size() # number of constraints = 2
print g
g.add(And(F == 0, F == 1, F == 2, F == 3))
print g.size() # number of constraints = 6
print g
def count_constraints(c,d,f):
print 'depth: ' + str(d) + ' expr: ' + str(f)
if f.num_args() == 0:
return c + 1
else:
d += 1
for a in f.children():
c += count_constraints(0, d, a)
return c
exp = a + b * c + a + c * c
print count_constraints(0,0,exp)
exp = And(a == b, b == c, a == 0, c == 0, b == 1 - 1)
print count_constraints(0,0,exp)
q, r, s = Bools('q r s')
exp = And(q, r, s)
print count_constraints(0,0,exp)