🔒 Locked
Difficult computer science problem
Advertisement
Original Post
so this algorithm can't read/receive the numbers from an array or a file? well then how are you sending the numbers to the program? manual input?
The term you are looking for is "constant space".
This problem is provably unsolvable in constant space. Assume for a moment that we have inserted all integers between 1 and (N-1) into the program. The last integer we insert can result in any number of integers between N/2 and (N-1) being lower than the average (we can insert, to this effect, any number between 0 and N²). Therefore, the program must store enough information to determine which of the N/2 possibilities is valid, which cannot be done for any N if we restrict ourselves to constant space.
This problem is provably unsolvable in constant space. Assume for a moment that we have inserted all integers between 1 and (N-1) into the program. The last integer we insert can result in any number of integers between N/2 and (N-1) being lower than the average (we can insert, to this effect, any number between 0 and N²). Therefore, the program must store enough information to determine which of the N/2 possibilities is valid, which cannot be done for any N if we restrict ourselves to constant space.
Quote:
Original post by Deyja
The obvious problem is that you need to know all the numbers to get the average.
It can be done if you iterated the sequence twice, once to find the average and again to count how many are higher. Without storing them, you'd have the input the whole sequence twice.
Ah, I misread it. I thought it was a given average as opposed to the average of the numbers inputted.
Quote:
Original post by Enigma
This sounds suspiciously like homework. All I'll say is that it can be solved within the constraints listed using a technique that teachers/professors love to set homework problems about.
EDIT: This does depend on the interpretation of the list of constraints.
Σnigma
what is "interpretation of the list of constraints"?
Quote:
Original post by Semion_N
what is "interpretation of the list of constraints"?
It is the interpretation of the ect (sic) in the OP's question. Does his question mean to rule out any memory storage (constant space solving, which is provably impossible)? Does his question mean to rule out only the standard imperative storage approaches (hinting at an easily expressed recursive solution)?
I've interpreted it as the first option, since a recursive solution technically uses a stack (unless it uses tail recursion, but then the problem becomes provably unsolvable again). If it's homework, then the recursive solution would be a good answer, though.
Quote:
Original post by ToohrVyk
So it's a constant-space constraint. There is no solution to your problem in constant space.
It's not impossible - it's trivial. Loop once over the input to find the average. Loop over the inout again to count how many inputs are bigger than the average.
Quote:
Original post by Bander
If I understand the problem specification then it is not trivial, the input is not given all at once for processing.
Quote:
recieves an unknown amount of numbers(it stops when it recieves -1)
That's actually pretty vague.
I assumed that the function would be something like void algorithm(int* input); and you'd iterate over the input array until you found a -1.
Quote:
Original post by ToohrVyk
...
Assume for a moment that we have inserted all integers between 1 and (N-1) into the program. The last integer we insert can result in any number of integers between N/2 and (N-1) being lower than the average ...
If negative numbers are allowed in the input then you could affect the status of ANY number of elements, not just [n/2, n-1].
Quote:
Original post by Nitage Quote:
Original post by ToohrVyk
So it's a constant-space constraint. There is no solution to your problem in constant space.
It's not impossible - it's trivial. Loop once over the input to find the average. Loop over the inout again to count how many inputs are bigger than the average.
But to store the input in order to loop over it again will require more than constant space.
Quote:
Original post by Nitage
It's not impossible - it's trivial. Loop once over the input to find the average. Loop over the inout again to count how many inputs are bigger than the average.
This requires the input being given to you again (you cannot store it, because it's not constant in size). If you will, you can liken the input to an oracle which gives you integers.
If you have access to the entire set of integers as a memory portion, then memory is pretty obivously not constant.
Topic Locked
This topic has been locked by a moderator. New replies are not allowed.
Sign in to reply to this topic.
Advertisement