I just want a function that returns true if all the elements of a List[Integer] follow each other, i.e.
noGaps(List(3,4,5)) // true
noGaps(List(4,3,5)) // false
noGaps(List(3,4,6)) // false
I have something that works but it's a bit verbose - what's the most elegant solution?
How about this?
def noGaps(xs: Seq[Int]) =
xs.size < 2 || xs.sliding(2).forall { case Seq(x, y) => y == x + 1 }
def noGaps(xs: Seq[Int]) = xs.isEmpty||xs.tail == xs.map(_+1).init
You can make it explicit that you compare the List to a Range:
def noGaps(l: Seq[Int]): Boolean =
l.isEmpty || l.sameElements(l.head to l.last)
Note that, although elegant, this is slightly less efficient than the sliding solution, due to l.last which is O(n). If n is the size of the list and i the first element for which there is a gap (or n if no gap), then the sliding solution would be performed in i steps whereas this one is performed in n + i steps.
Related
I'm trying to use Z3 (with C++ API) to check if lots of variable configurations satisfy my constraints, but I'm having big performance issues.
I'm looking for advice about which logic or parameter setting I might be able to use to improve the runtime, or hints about how I could try and feed the problem to Z3 in a different way.
Short description of what I'm doing and how I'm doing it:
//_______________Pseudocode and example_______________
context ctx()
solver s(ctx)
// All my variables are finite domain, maybe some 20 values at most, but usually less.
// They can only be ints, bools, or enums.
// There are not that many variables, maybe 10 or 20 for now.
//
// Since I need to be able to solve constraints of the type (e == f), where
// e and f are two different enum variables, all my
// enum types are actually contained in only one enumeration_sort(), populated
// with all the different values.
sort enum_sort = {"green", "red", "yellow", "blue", "null"}
expr x = ctx.int_const("x")
expr y = ctx.int_const("y")
expr b = ctx.bool_const("b")
expr e = ctx.constant("e", enum_sort)
expr f = ctx.constant("f", enum_sort)
// now I assert the finite domains, for each variable
// enum_value(s) is a helper function, that returns the matching enum expression
//
// Let's say that these are the domains:
//
// int x is from {1, 3, 4, 7, 8}
// int y is from {1, 2, 3, 4}
// bool b is from {0, 1}
// enum e is from {"green", "red", "yellow"}
// enum f is from {"red", "blue", "null"}
s.add(x == 1 || x == 3 || x == 3 || x == 7 || x == 8)
s.add(y == 1 || y == 2 || y == 3 || y == 4)
s.add(b == 0 || b == 1)
s.add(e == enum_value("green") || e == enum_value("red") || enum_value("yellow"))
s.add(f == enum_value("red") || f == enum_value("blue") || enum_value("null"))
// now I add in my constraints. There are also about 10 or 20 of them,
// and each one is pretty short
s.add(b => (x + y >= 5))
s.add((x > 1) => (e != f))
s.add((y == 4 && x == 1) || b)
// setup of the solver is now done. Here I start to query different combinations
// of values, and ask the solver if they are "sat" or "unsat"
// some values are left empty, because I don't care about them
expr_vector vec1 = {x == 1, y == 3, b == 1, e == "red"}
print(s.check(vec1))
expr_vector vec2 = {x == 4, e == "green", f == "null"}
print(s.check(vec2))
....
// I want to answer many such queries.
Of course, in my case this isn't hardcoded, but I read and parse the constraints, variables and their domains from files, then feed the info to Z3.
But it's slow.
Even for something like ten thousand queries, my program is already running over 10s. All of this is inside s.check(). Is it possible to make it run faster?
Hopefully it is, because what I'm asking of the solver doesn't look like it's overly difficult.
No quantifiers, finite domain, no functions, everything is a whole number or an enum, domains are small, the values of the numbers are small, there's only simple arithmetic, constraints are short, etc.
If I try to use parameters for parallel processing, or set the logic to "QF_FD", the runtime doesn't change at all.
Thanks in advance for any advice.
Is it always slow? Or does it get progressively slower as you query for more and more configurations using the same solver?
If it's the former, then your problem is just too hard and this is the price to pay. I don't see anything obviously wrong in what you've shown; though you should never use booleans as integers. (Just looking at your b variable in there. Stick to booleans as booleans, and integers as integers, and unless you really have to, don't mix the two together. See this answer for some further elaboration on this point: Why is Z3 slow for tiny search space?)
If it's the latter, you might want to create a solver from scratch for each query to clean-up all the extra stuff the solver created. While additional lemmas always help, they could also hurt performance if the solver cannot make good use of them in subsequent queries. And if you follow this path, then you can simply "parallelize" the problem yourself in your C++ program; i.e., create many threads and call the solver separately for each problem, taking advantage of many-cores your computer no doubt has and OS-level multi-tasking.
Admittedly, this is very general advice and may not apply directly to your situation. But, without a particular "running" example that we can see and inspect, it's hard to be any more specific than this.
Some Ideas:
1. Replace x == 1 || x == 3 || x == 3 || x == 7 || x == 8 with (1 <= x && x <= 8) && (x <= 1 || (3 <= x) && (x <= 4 || 7 <= x). Similar change with y.
rationale: the solver for linear arithmetic now knows that x is always confined in the interval [1,8], this can be useful information for other linear equalities/inequalities; it may be useful to also learn the trivial mutual exclusion constraints not(x <= 1) || not(3 <= x) and not(x <= 4) || not(7 <= x); there are now exactly 3 boolean assignments that cover your original 5 cases, this makes the reasoning of the linear arithmetic solver more cost-efficient because each invocation deals with a larger chunk of the search space. (Furthermore, it is more likely that clauses learned from conflicts are going to be useful with subsequent calls to the solver)
(Your queries may also contain set of values rather than specific assignments of values; this may allow one to prune some unsatisfiable ranges of values with fewer queries)
2. Just like #alias mentioned, Boolean variables ought to be Booleans and not 0/1 Integer variables. The example you provided is a bit confusing, b is declared as a bool const but then you state b == 0 || b == 1
3. I am not familiar with the enum_sort of z3, meaning that I don't know how it is internally encoded and what solving techniques are applied to deal with it. Therefore, I am not sure whether the solver may try to generate trivially inconsistent truth-assignments in which e == enum_value("green") e e == enum_value("red") are both assigned to true at the same time. This might be worth a bit of investigation. For instance, another possibility could be to declare e and f as Int and give them an appropriate interval domain (as contiguous as possible) with the same approach shown in 1., that will be interpreted by your software as a list of enum values. This should remove a number of Boolean assignments from the search space, make conflict clauses more effective and possibly speed-up the search.
4. Given the small number of problem variables, values and constraints, I would suggest you to try to encode everything using just the Bit-Vector theory and nothing else (using small-but-big-enough domains). If you then configure the solver to encode Bit-Vectors eagerly, then everything is bit-blasted into SAT, and z3 should only use Boolean Constraint Propagation for satisfiability, which is the cheapest technique.
This might be an X Y problem, why are you performing thousands of queries, what are you trying to achieve? Are you trying to explore all possible combination of values? Are you trying to perform model counting?
Wrote this code in comp sci class and I cant get it to work, it always returns as false every time I run it. Its supposed to be a recursive binary search method... Any idea why it only returns false?
arr = [1,10,12,15,16,122,132,143,155]
def binarysearch(arr,num):
arr2 = []
if (len(arr) == 1):
if (arr[0] == num):
return 1
else:
return 0
for i in range (len(arr)/2):
arr2.append(0)
if (arr[len(arr)/2]>num):
for x in range (len(arr)/2,len(arr)):
arr2[x-(len(arr)/2)]=arr[x]
return binarysearch(arr2,num)
if(arr[len(arr)/2]<num):
for x in range(0, len(arr) / 2 ):
arr2[x] = arr[x]
return binarysearch(arr2, num)
num = raw_input("put number to check here please: ")
if(binarysearch(arr,num)==1):
print "true"
else:
print "false"
You're doing vastly more work than you need to on things that Python can handle for you, and the complexity is masking your problems.
After your base case, you have two if statements, that don't cover the full range—you've overlooked the possibility of equality. Use if/else and adjust the ranges being copied accordingly.
Don't create a new array and copy stuff, use Python's subranges.
Don't keep repeating the same division operation throughout your program, do it once and store the result.
If you want to print True/False, why not just return that rather than encoding the outcome as 0/1 and then decoding it to do the print?
Recall that raw_input returns a string, you'll need to convert it to int.
The end result of all those revisions would be:
def binary_search(arr,num):
if (len(arr) == 1):
return (arr[0] == num)
mid = len(arr) / 2
if (arr[mid] > num):
return binary_search(arr[:mid], num)
else:
return binary_search(arr[mid:], num)
num = int(raw_input("put number to check here please: "))
print binary_search(arr,num)
import math
def is_prime(num):
if num < 2:
return False
for i in range(2, int(math.sqrt(num))+ 1):
if num % i == 0:
return False
return True
Primes seems to be a popular topic but in the book in which I am learning Python, I am on chpt 6 out of 21 and in the iteration chapter which it teaches while loops. I have not learned for loops yet although I understand what they do. So, let's say I have not learned for loops yet and am given only if/elif/else statements and the while loops as my tools. How can I change the for line of code into something more simple using the above tools? While asking this question I quickly came up with this code:
def formula(num):
i = 2
while i >= 2:
return int(math.sqrt(num)+ 1)
def is_primetwo(num):
i = 2
if num < 2:
return False
formula(num)
if num % i == 0:
return False
return True
It works but would this be a simple version of the for loop or is there something even more simple where I do not have to wrap a function within a function?
Absolutely, you do not need a function to replace a for loop.
So you've got this
for i in range(2, int(math.sqrt(num))+ 1):
which is your for loop. Take a second to think what it's doing.
1.) It's taking the variable i, and it's starting it at a value of 2.
2.) It's checking whether to do the loop every time by checking if i is less than the (square root of num) plus 1
3.) Every time through the loop, it adds one to i.
We can do all of these things using a while loop.
Here's the original
for i in range(2, int(math.sqrt(num))+ 1):
if num % i == 0:
return False
let's rename the second and third lines loop contents just so we're focusing on the looping part, not what logic we're doing with the variables i and num.
for i in range(2, int(math.sqrt(num))+ 1):
loop contents
ok, now let's just rearrange it to be a while loop. We need to set i to 2 first.
i = 2
Now we want to check that i is in the range we want
i = 2
while i <= int(math.sqrt(num) + 1):
loop contents
Now we're almost set, we just need to make i actually change, instead of staying at a value of 2 forever.
i = 2
while i <= int(math.sqrt(num) + 1):
loop contents
i = i + 1
Your example seemed to do some of these elements, but this way is a simple way to do it, and no extra function is necessary. It could be the range() function that is confusing. Just remember, the for loop is doing three things; setting a variable to an initial value, checking a condition (one variable is less than another), and incrementing your variable to be one large than previously to run the loop again.
How about something like:
from math import sqrt
def is_prime(num):
if (num < 2):
return False
i = 2
limit = int(sqrt(num) + 1)
while (i <= limit):
if num % i == 0:
return False
i = i + 1
return True
Not sure if this is what you want, but the for loop:
for i in range(2, int(math.sqrt(num))+ 1):
if num % i == 0:
return False
return True
can be expressed as:
i = 2
while i < int(math.sqrt(num))+ 1):
if num % i == 0:
return False
i += 1
return True
Probably a good idea to determine int(math.sqrt(num))+ 1) once:
i = 2
n = int(math.sqrt(num))+ 1)
while i < n:
if num % i == 0:
return False
i += 1
return True
If I have a boolean and some code which maybe changes it, and then I want to set it to true, should I check if it's false?
For example:
bool b = false;
// Some code
// Here "b" can be true or false
if (cond) {
b = true;
}
vs
bool b = false;
// Some code
// Here `b` can be `true` or `false`
if (cond && !b){
b = true;
}
Which is faster?
Note:
I ask that because of the following implementation of Sieve of Eratosthenes: http://bloc.gerardfarras.com/wp-content/uploads/2011/12/erastotenes.txt
if (( i % divisor == 0 ) && ( numsprimers[i] == 0 )) {
numsprimers[i] = 1;
}
(If numsprimers[i]==1 it means that i isn't a prime number. And if it's 0 it can be prime or not)
It's being very very nitpicky, but generally speaking it's better to just change the value.
Checking and setting a value have roughly the same overhead anyway, so why would you want to have to do both in some cases?
Now if you're wondering if you should overwrite some custom type (lets say a list of 100000 words) or if you should check to see if it needs to be overwritten first (let's say by simply checking a boolean value or a timestamp) then you should check first, because the cost of checking a boolean or timestamp is much less than writing so many words to memory.
This is of course all dependent on various things such as whether or not the memory you are editing is in cache, how expensive the "check" is, how often you need to overwrite the value versus how often it does not need to be overwritten, and of course the size of the memory.
How about:
if ( b = !!cond ) {
}
Where you check the condition and apply the value to b, if it is necessary for b to have a value. If you want b to stay true then I say to use one of your other examples. It shouldn't make a difference.
Sorry for this newbie question, but I can't find on google what I need to know.
I understand return, but don't understand this... What does it mean this?
return (tail+1)%N == head%N;
Thanks a lot for patience.
It returns true or false, depending on whether the expression is true or not.
It's the same as:
if ( (tail+1)%N == head%N )
return true;
else
return false;
this
(tail+1)%N == head%N
returns a boolean value, either true or false. This statement means that after adding 1 to trail (trail + 1) and the remainder obtained after division with N is equal to remainder of head divided with N. % is used for division with remainder
(%). Modulo is the operation that gives the remainder of a division of two values.
Check this link for c++ operators : http://www.cplusplus.com/doc/tutorial/operators/
you're returning a boolean value. The value represents whether or not the remainder of (tail+1) divided by N is the same as that of head.
It evaluates the expression, and return the result. In this case it's two modulo operations that are compared, and the result is either true or false which will be returned.
Short Answer:
Because of the == operator your function will return a bool, meaning it can only be trueor false. An equivalent would be something like:
return 5 == 4;
which would return false since 5 is not equal to 4.
Long Answer:
Instead of writing this in a single line you could split it up into more lines of code. Let's just assume that tail, head and N are integer values, then you could write it like this:
int x, y;
x = (tail+1)%N;
y = head%N;
if ( x == y )
{
return true;
}
else
{
return false;
}
Now in this code there may be also that %confuses you a bit. The %is called the Modulus Operator and can give you the remainder of arithmetic operations. In a simple example this would mean:
10 % 3 = 1 because 10/3 is 3 with a remainder of 1. So to make it more clear let's just make another example with your specific problem:
Lets just assume that tail=10,head=6 and N=2. Then you would get something like this:
x = (10+1)%2
x = 11 % 2
x = 1
y = 6 % 2
y = 0
y != x
This would return false cause x and y are not equal. ( If you would run your code with the given example values )
To learn more about Modulus you can look here, or just on any other basic C++ Tutorial.
it returns true if remainder of the division for tail + 1 and head is the same
for example if tail is 2, head is 1 and N is 2
(tail + 1) % N is 1
head % N is 1 too
so whole expression returns true