I need to determine whether the following Hash Functions Set is universal or not:
Let U be the set of the keys - {000, 001, 002, 003, ... ,999} - all the numbers between 0 and 999 padded with 0 in the beginning where needed. Let n = 10 and 1 < a < 9 ,an integer between 1 and 9. We denote by ha(x) the rightmost digit of the number a*x.
For example, h2(123) = 6, because, 2 * 123 = 246.
We also denote H = {h1, h2, h3, ... ,h9} as our set of hash functions.
Is H is universal? prove.
I know I need to calculate the probability for collision of 2 different keys and check if it's smaller or equal to 1/n (which is 1/10), so I tried to separate into cases - if a is odd or even, because when a is even the last digit of a*x will be 0/2/4/6/8, else it could be anything. But it didn't help me so much as I'm stuck on it.
Would be very glad for some help here.
Could someone explain to me the logic behind this hashMap algorithm? I'm getting confused about how the algorithm receives the total sum. I'm starting to learn about algorithms, so it's a little confusing for me. I made comments in my code to pinpoint each line code, but I'm not sure I'm grasping logic correctly. I'm just looking for an easier way to understand how the algorithm works to avoid confusing myself.
//**calculate Two Number Sum
func twoNumberSum(_ array: [Int], _ targetSum: Int) -> [Int] {
//1) initilize our Array to hold Integer Value: Boolean value to store value into hashTable
var numbersHashMap = [Int:Bool]()
//2) create placeHolder called number that iterates through our Array.
for number in array {
//3) variable = y - x
let match = targetSum - number
//4) ??
if let exists = numbersHashMap[match], exists {
//5) match = y / number = x
return [match, number] //
} else {
//6) Store number in HashTable and repeats
numbersHashMap[number] = true
}
}
return []
}
twoNumberSum([3,5,-4, 8, 11, 1, -1, -6], 10)
// x = Number
// y = Unknown *Solve for Y*
Sure, I can walk you through it. So we have a list of numbers, are we are trying to find two numbers that add together to make the specified target. To do this, for each number x, we check if (target - x) is in the list. If it is not, then we add x to the list. If it is, then we return x and (target - x).
Step 4 in your code is the part where we check if (target - x) is in the list. To see why this makes sense, let's walk through an example.
Say we have [2, 3, -1] and our target is 1. In this case, we first consider x = 2 and check our hashmap for (target - x) = (1 - 2) = -1. Since -1 is not in the hashmap, we add 2 to the hashmap. We then consider x = 3 and check for (1 - 3) = -2. Again, -2 is not in the hashmap, so we add it. Now we check x - -1. In this case, when we check (target - x) = (1 - (-1)) = 2, 2 is in the hashmap. Intuitively, we have already "seen" 2, and know that 2 and -1 can be added to get our value.
This is what provides the speed optimization over checking every two numbers in the list.
I am trying to do an assignment in JES a student jython program. I need to convert our student number taken as a string input variable to pass through our function i.e.
def assignment(stringID) and convert it into integers. The exact instructions are:
Step 1
Define an array called id which will store your 7 digit number as integers (the numbers you set in the array does not matter, it will be over written with your student number in the next step).
Step 2 Your student number has been passed in to your function as a String. You must separate the digits and assign them to your array id. This can do this manually line by line or using a loop. You will need to type cast each character from stringID to an integer before storing it in id.
I have tried so many different ways using the int and float functions but I am really stuck.
Thanks in advance!
>>> a = "545.2222"
>>> float(a)
545.22220000000004
>>> int(float(a))
545
I had to do some jython scripting for a websphere server. It must be a really old version of python it didn't have the ** operator or the len() function. I had to use an exception to find the end of a string.
Anyways I hope this saves someone else some time
def pow(x, y):
total = 1;
if (y > 0):
rng = y
else:
rng = -1 * y
print ("range", rng)
for itt in range (rng):
total *= x
if (y < 0):
total = 1.0 / float(total)
return total
#This will return an int if the percision restricts it from parsing decimal places
def parseNum(string, percision):
decIndex = string.index(".")
total = 0
print("decIndex: ", decIndex)
index = 0
string = string[0:decIndex] + string[decIndex + 1:]
try:
while string[index]:
if (ord(string[index]) >= ord("0") and ord(string[index]) <= ord("9")):
times = pow(10, decIndex - index - 1)
val = ord(string[index]) - ord("0")
print(times, " X ", val)
if (times < percision):
break
total += times * val
index += 1
except:
print "broke out"
return total
Warning! - make sure the string is a number. The function will not fail but you will get strange and almost assuredly, useless output.
I am new in fortran90 (30 minutes ago...) and I have this program:
program example1
implicit none
real (kind=8) :: x,y,z
x = 3.d0
y = 2.d-1
z = x + y
print *, "y = ", y
print *, "x = ", x
print *, "z = ", z
end program example1
but when i run it with:
gfortran example1.f90
./a.out
the output is:
y = 0.20000000000000001
x = 3.0000000000000000
z = 3.2000000000000002
why is not 3.2000000000000000 ???
What I am doing wrong?
Why y has a 1 in last digit?? and why z has a 2 in the last digit??
Sorry if it is a dumb question, but I just dont understand what I am doing wrong...
Thanks!!
There's absolutely nothing wrong with your program. The issue has to do with real's inability to represent decimals precisely, without an error. The problem is that numbers not composed of negative powers of 2 must be represented approximately. That's why there is a small error in the 16-th decimal place. For more information about representation of reals take a look at the article on wikipedia. This is another great article on the same subject.
If you replace 0.2 with 0.25, the problem will go away, because 0.25 is 2 ^ -2.
So I've spent hours trying to work out exactly how this code produces prime numbers.
lazy val ps: Stream[Int] = 2 #:: Stream.from(3).filter(i =>
ps.takeWhile{j => j * j <= i}.forall{ k => i % k > 0});
I've used a number of printlns etc, but nothings making it clearer.
This is what I think the code does:
/**
* [2,3]
*
* takeWhile 2*2 <= 3
* takeWhile 2*2 <= 4 found match
* (4 % [2,3] > 1) return false.
* takeWhile 2*2 <= 5 found match
* (5 % [2,3] > 1) return true
* Add 5 to the list
* takeWhile 2*2 <= 6 found match
* (6 % [2,3,5] > 1) return false
* takeWhile 2*2 <= 7
* (7 % [2,3,5] > 1) return true
* Add 7 to the list
*/
But If I change j*j in the list to be 2*2 which I assumed would work exactly the same, it causes a stackoverflow error.
I'm obviously missing something fundamental here, and could really use someone explaining this to me like I was a five year old.
Any help would be greatly appreciated.
I'm not sure that seeking a procedural/imperative explanation is the best way to gain understanding here. Streams come from functional programming and they're best understood from that perspective. The key aspects of the definition you've given are:
It's lazy. Other than the first element in the stream, nothing is computed until you ask for it. If you never ask for the 5th prime, it will never be computed.
It's recursive. The list of prime numbers is defined in terms of itself.
It's infinite. Streams have the interesting property (because they're lazy) that they can represent a sequence with an infinite number of elements. Stream.from(3) is an example of this: it represents the list [3, 4, 5, ...].
Let's see if we can understand why your definition computes the sequence of prime numbers.
The definition starts out with 2 #:: .... This just says that the first number in the sequence is 2 - simple enough so far.
The next part defines the rest of the prime numbers. We can start with all the counting numbers starting at 3 (Stream.from(3)), but we obviously need to filter a bunch of these numbers out (i.e., all the composites). So let's consider each number i. If i is not a multiple of a lesser prime number, then i is prime. That is, i is prime if, for all primes k less than i, i % k > 0. In Scala, we could express this as
nums.filter(i => ps.takeWhile(k => k < i).forall(k => i % k > 0))
However, it isn't actually necessary to check all lesser prime numbers -- we really only need to check the prime numbers whose square is less than or equal to i (this is a fact from number theory*). So we could instead write
nums.filter(i => ps.takeWhile(k => k * k <= i).forall(k => i % k > 0))
So we've derived your definition.
Now, if you happened to try the first definition (with k < i), you would have found that it didn't work. Why not? It has to do with the fact that this is a recursive definition.
Suppose we're trying to decide what comes after 2 in the sequence. The definition tells us to first determine whether 3 belongs. To do so, we consider the list of primes up to the first one greater than or equal to 3 (takeWhile(k => k < i)). The first prime is 2, which is less than 3 -- so far so good. But we don't yet know the second prime, so we need to compute it. Fine, so we need to first see whether 3 belongs ... BOOM!
* It's pretty easy to see that if a number n is composite then the square of one of its factors must be less than or equal to n. If n is composite, then by definition n == a * b, where 1 < a <= b < n (we can guarantee a <= b just by labeling the two factors appropriately). From a <= b it follows that a^2 <= a * b, so it follows that a^2 <= n.
Your explanations are mostly correct, you made only two mistakes:
takeWhile doesn't include the last checked element:
scala> List(1,2,3).takeWhile(_<2)
res1: List[Int] = List(1)
You assume that ps always contains only a two and a three but because Stream is lazy it is possible to add new elements to it. In fact each time a new prime is found it is added to ps and in the next step takeWhile will consider this new added element. Here, it is important to remember that the tail of a Stream is computed only when it is needed, thus takeWhile can't see it before forall is evaluated to true.
Keep these two things in mind and you should came up with this:
ps = [2]
i = 3
takeWhile
2*2 <= 3 -> false
forall on []
-> true
ps = [2,3]
i = 4
takeWhile
2*2 <= 4 -> true
3*3 <= 4 -> false
forall on [2]
4%2 > 0 -> false
ps = [2,3]
i = 5
takeWhile
2*2 <= 5 -> true
3*3 <= 5 -> false
forall on [2]
5%2 > 0 -> true
ps = [2,3,5]
i = 6
...
While these steps describe the behavior of the code, it is not fully correct because not only adding elements to the Stream is lazy but every operation on it. This means that when you call xs.takeWhile(f) not all values until the point when f is false are computed at once - they are computed when forall wants to see them (because it is the only function here that needs to look at all elements before it definitely can result to true, for false it can abort earlier). Here the computation order when laziness is considered everywhere (example only looking at 9):
ps = [2,3,5,7]
i = 9
takeWhile on 2
2*2 <= 9 -> true
forall on 2
9%2 > 0 -> true
takeWhile on 3
3*3 <= 9 -> true
forall on 3
9%3 > 0 -> false
ps = [2,3,5,7]
i = 10
...
Because forall is aborted when it evaluates to false, takeWhile doesn't calculate the remaining possible elements.
That code is easier (for me, at least) to read with some variables renamed suggestively, as
lazy val ps: Stream[Int] = 2 #:: Stream.from(3).filter(i =>
ps.takeWhile{p => p * p <= i}.forall{ p => i % p > 0});
This reads left-to-right quite naturally, as
primes are 2, and those numbers i from 3 up, that all of the primes p whose square does not exceed the i, do not divide i evenly (i.e. without some non-zero remainder).
In a true recursive fashion, to understand this definition as defining the ever increasing stream of primes, we assume that it is so, and from that assumption we see that no contradiction arises, i.e. the truth of the definition holds.
The only potential problem after that, is the timing of accessing the stream ps as it is being defined. As the first step, imagine we just have another stream of primes provided to us from somewhere, magically. Then, after seeing the truth of the definition, check that the timing of the access is okay, i.e. we never try to access the areas of ps before they are defined; that would make the definition stuck, unproductive.
I remember reading somewhere (don't recall where) something like the following -- a conversation between a student and a wizard,
student: which numbers are prime?
wizard: well, do you know what number is the first prime?
s: yes, it's 2.
w: okay (quickly writes down 2 on a piece of paper). And what about the next one?
s: well, next candidate is 3. we need to check whether it is divided by any prime whose square does not exceed it, but I don't yet know what the primes are!
w: don't worry, I'l give them to you. It's a magic I know; I'm a wizard after all.
s: okay, so what is the first prime number?
w: (glances over the piece of paper) 2.
s: great, so its square is already greater than 3... HEY, you've cheated! .....
Here's a pseudocode1 translation of your code, read partially right-to-left, with some variables again renamed for clarity (using p for "prime"):
ps = 2 : filter (\i-> all (\p->rem i p > 0) (takeWhile (\p->p^2 <= i) ps)) [3..]
which is also
ps = 2 : [i | i <- [3..], and [rem i p > 0 | p <- takeWhile (\p->p^2 <= i) ps]]
which is a bit more visually apparent, using list comprehensions. and checks that all entries in a list of Booleans are True (read | as "for", <- as "drawn from", , as "such that" and (\p-> ...) as "lambda of p").
So you see, ps is a lazy list of 2, and then of numbers i drawn from a stream [3,4,5,...] such that for all p drawn from ps such that p^2 <= i, it is true that i % p > 0. Which is actually an optimal trial division algorithm. :)
There's a subtlety here of course: the list ps is open-ended. We use it as it is being "fleshed-out" (that of course, because it is lazy). When ps are taken from ps, it could potentially be a case that we run past its end, in which case we'd have a non-terminating calculation on our hands (a "black hole"). It just so happens :) (and needs to ⁄ can be proved mathematically) that this is impossible with the above definition. So 2 is put into ps unconditionally, so there's something in it to begin with.
But if we try to "simplify",
bad = 2 : [i | i <- [3..], and [rem i p > 0 | p <- takeWhile (\p->p < i) bad]]
it stops working after producing just one number, 2: when considering 3 as the candidate, takeWhile (\p->p < 3) bad demands the next number in bad after 2, but there aren't yet any more numbers there. It "jumps ahead of itself".
This is "fixed" with
bad = 2 : [i | i <- [3..], and [rem i p > 0 | p <- [2..(i-1)] ]]
but that is a much much slower trial division algorithm, very far from the optimal one.
--
1 (Haskell actually, it's just easier for me that way :) )