Skip to main content
gamedev.net
🔒 Locked

Difficult computer science problem

Started by someboddy Oct 1, 2006 at 12:37 PM 58 replies 10.1k views
Original Post
someboddy
someboddy
Does anyone know an algorithm that recieves an unknown amount of numbers(it stops when it recieves -1) and calculates how many of them are above the average without using an array, list, vector, stack, file ect.? Is it even possible?
-----------------------------------------Everyboddy need someboddy!
Alpha_ProgDes
Alpha_ProgDes
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?
Beginner in Game Development?  Read here. And read here.  
ToohrVyk
ToohrVyk
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.
phi
phi
I'm assuming you can hold variables. Can't you have a variable that's initialised to zero. Then if the inputted number is greater than the average then increase the variable. If it's -1 then quit. Or have I missed an obvious problem?
Enigma
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
Deyja
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.
phi
phi
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.
krikkit
krikkit
Well..I'm curious. Maybe the OP can let us know when teh homework gets turned in so someone can tell me what this is all about :)
Semion_N
Semion_N
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"?
ToohrVyk
ToohrVyk
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.
someboddy
someboddy
A friend of mine gave me that question. I believe the memory usage must be O(1)
-----------------------------------------Everyboddy need someboddy!
ToohrVyk
ToohrVyk
So it's a constant-space constraint. There is no solution to your problem in constant space.
Ezbez
Ezbez
Mind telling us the answer, then?
ToohrVyk
ToohrVyk
Quote:
Original post by Ezbez
Mind telling us the answer, then?


I was under the impression that he came here asking for it, because he didn't know it.

Nitage
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.
Bander
Bander
If I understand the problem specification then it is not trivial, the input is not given all at once for processing.
M2tM
M2tM
PHP variable variables and a similar naming convention :D

${"name$num"}

Not an array, it's a series of regular variables, but it acts as one. ;)

Alright, alright, that was terrible, I apologize.
_____"You're using a screwdriver to nail some glue to a ming vase. " -ToohrVyk
Nitage
Nitage
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.

rileyriley
rileyriley
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.
--Riley
ToohrVyk
ToohrVyk
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.