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Why do we prefer not to specify the constant factor in Big-O notation?
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Does it mean anything at all to have a function with time complexity O(2)?
For example, how would one describe a function that must check two lookup tables rather than one. Is that not strictly describable in big-O, or is O(2) a real way to describe this? Or something else?
Thanks.
O(something) is a set of functions.
O(1) and O(2) are the same set.
A constant time function is a member of O(1). It's also a member of O(2) because O(1) and O(2) are exactly the same thing. Use whichever one you prefer. Normally you'd use O(1), but you be you.
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Does every code of Dynamic Programming have the same time complexity in a table method or memorized recursion method?
A Solution with an appropriate example would be appreciated.
Time complexity- Yes (if you ignore the function calls/returns in Memoization)
Space complexity- No. Tabulation can save space by overwriting previously calculated but no longer needed values.
As mentioned in the "Optimality" section of this answer- https://stackoverflow.com/a/6165124/7145074
Either approach may not be time-optimal if the order you happen (or try to) visit subproblems is not optimal, specifically if there is more than one way to calculate a subproblem (normally caching would resolve this, but it's theoretically possible that caching might not in some exotic cases). Memoization will usually add on your time-complexity to your space-complexity (e.g. with tabulation you have more liberty to throw away calculations, like using tabulation with Fib lets you use O(1) space, but memoization with Fib uses O(N) stack space).
Further reading- https://www.geeksforgeeks.org/tabulation-vs-memoization/
I do not know if the language I am using in the title is correct, but here is an example that illustrates what I am asking.
What would the time complexity for this non-optimal algorithm that removes character pairs from a string?
The function will loop through a string. When it finds two identical characters next to each other it will return a string without the found pair. It then recursively calls itself until no pair is found.
Example (each line is the return string from one recurisive function call):
iabccba
iabba
iaa
i
Would it be fair to describe the time complexity as O(|Characters| * |Pairs|)?
What about O(|Characters|^2) Can pairs be used to describe the time complexity even though the number of pairs is not knowable at the initial function call?
It was argued to me that this algorithm was O(n^2)because the number of pairs is not known.
You're right that this is strictly speaking O(|Characters| * |Pairs|)
However, in the worst case, number of pairs can be same as number of charachters (or same order of magnitude), for example in the string 'abcdeedcba'
So it also makes sense to describe it as O(n^2) worst-case.
I think this largely depends on the problem you mean to solve and and it's definition.
For graph algorithms for example, everyone is comfortable with a writing as complexity O(|V| + |E|), although in the worst case of a dense graph |E| = |V|^2. In other problems we just look at the worst possible case and write O(n^2), without breaking it into more specific variables.
I'd say that if there's no special convention, or no special data in the problem regarding number of pairs, O(...) implies worst-case performance and hence O(n^2) would be more appropriate.
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I've got a linear system of 17 equations, 506 variables that solve for a minimum summation of the total variables. This works fine, so far, but the solution is a result of a combination of 19 variables.
But in the end I want to limit the amount of chosen variables to 10, without knowing in advance which ones are the optimal ones (The solver figures that out for me, as well as their ratio).
I figured I can set a boolean = 1 if the value becomes larger than 0: (meaning the variable is picked), and 0 if the variable is not picked for an optimal solution.
And then have the sum of the booleans be 10 at most.
However this seems a bit elaborate, and I was wondering whether there was a built in option in the opensolver, for I think it is quite a common problem to solve a large set with a subset.
So does anyone have a suggestion on:
How my elaborate way drastically decreases performance? (*I have no intrinsic comprehension of the opensolver algorithms, yet.)
A suggestion to more easily/within the opensolver options account for my desire of max. 10 solution variables?
Based on the information provided below, I first scaled down the size of the problem:
I have three lists of data with 18 entries in columns:
W7:W23,AC7:AD23
which manually (with: W28 = 6000, AC28=600,W29 = 1,AC29 =1), in a linear combination,equal/exceed the target list:
EGM34:EG50
So what I did was put the descion variables in W28:W29, AC28:AD29
Where I added the constraint W28,AC28:AD28 = integer in the solver (both the original excel solver as in opensolver)
And I added the constraint W29,AC29:AD29 = Boolean in the solver (both the original excel solver as in opensolver)
Then I have a multiplication of the integer*boolean = the actual multiplication factor for the above lists in (W7:W23 etc)
In order to limit the nr of chosen variables I have also tried, in addition to the described constraints, to limit the cell with =sum(W29,AC29:AD29) to <= 10 (effectively reducing the amount of booleans set to true below 11, or so I thought, but the booleans aren't evaluated as booleans by the solver).
These new multiplied lists are placed in W34:W50,AC34:AD50, and the summation is situated in: EGY34:EGY50 Hence the final check is added as a constraint as:
EGY34:EGY50 =>EGM34:EGM50
And I had a question about how the linear solver evaluates these constraints, does it:
a. Think the sum of EGY34:EGY50 must be larger or equal than/to EGM34:EGM50
or
b. Does it think: "for every row x EGYx must be larger or equal than/to EGMx
So far I've noted b. but I would like to make sure.
But my main question concerns:
After using the Evolutionary algorithm as was kindly suggested in the comments below, how/why does it try values as 0.99994 for the desicion variables designated as booleans?
The introduction of binary variables is indeed the standard way to implement such constraints. Unfortunately, it transforms the problem from being a linear programming problem to being an integer programming problem (specifically a mixed integer linear programming problem). A standard approach to such problems is the branch and bound algorithm. This is what Excel's built-in solver seems to use, I'm not sure about the open solver that you are using. In the best case (where there is a lot of bounding) it will run fairly rapidly, even with problems of your size. In the worst case, for your problem it could be little better than what you would get by running the simplex algorithm C(506,10) = 2.8 x 10^20 times (once for each possible set of 10 decision variables). In other words, it might be infeasible. Integer programming is known to be NP-hard.
If an exact solution is infeasible, you could always use a heuristic algorithm such as an evolutionary approach.
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Fastest way to clamp a real (fixed/floating point) value?
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Closed 6 years ago.
What I am looking for is a math function that constrains a primitive variable between a minimum and a maximum value in only one single function call, if it exists, in the standard math libraries for Objective-C.
I currently use:
float constrainedValue = fminf( maxValue, fmaxf( minValue, inValue ) );
Since I know that both fminf and fmaxf could potentially have instruction jumps or branches, it seems likely there could be a simple routine that could conjoin both of these operations into one, optimized function.
This topic is thoroughly discussed here: Fastest way to clamp a real (fixed/floating point) value?
'clamp' is the keyword I was looking for.
Not sure what exactly to google for this question, so I'll post it directly to SO:
Variables in Haskell are immutable
Pure functions should result in same values for same arguments
From these two points it's possible to deduce that if you call somePureFunc somevar1 somevar2 in your code twice, it only makes sense to compute the value during the first call. The resulting value can be stored in some sort of a giant hash table (or something like that) and looked up during subsequent calls to the function. I have two questions:
Does GHC actually do this kind of optimization?
If it does, what is the behaviour in the case when it's actually cheaper to repeat the computation than to look up the results?
Thanks.
GHC doesn't do automatic memoization. See the GHC FAQ on Common Subexpression Elimination (not exactly the same thing, but my guess is that the reasoning is the same) and the answer to this question.
If you want to do memoization yourself, then have a look at Data.MemoCombinators.
Another way of looking at memoization is to use laziness to take advantage of memoization. For example, you can define a list in terms of itself. The definition below is an infinite list of all the Fibonacci numbers (taken from the Haskell Wiki)
fibs = 0 : 1 : zipWith (+) fibs (tail fibs)
Because the list is realized lazily it's similar to having precomputed (memoized) previous values. e.g. fibs !! 10 will create the first ten elements such that fibs 11 is much faster.
Saving every function call result (cf. hash consing) is valid but can be a giant space leak and in general also slows your program down a lot. It often costs more to check if you have something in the table than to actually compute it.