So in the tag prolog someone wanted to solve the "the giant cat army riddle" by Dan Finkel (see video / Link for description of the puzzle).
Since I want to improve in answer set programming I hereby challenge you to solve the puzzle more efficient than me. You will find my solution as answer. I'll accept the fastest running answer (except if it's using dirty hacks).
Rules:
hardcoding the length of the list (or something similar) counts as dirty hack.
The output has to be in the predicate r/2, where it's first argument is the index of the list and the second its entry.
Time measured is for the first valid answer.
num(0..59).
%valid operation pairs
op(N*N,N):- N=2..7.
% no need to add operations that start with 14
op(Ori,New):- num(Ori), New = Ori+7, num(New), Ori!=14.
op(Ori,New):- num(Ori), New = Ori+5, num(New), Ori!=14.
%iteratively create new numbers from old numbers
l(0,0).
{l(T+1,New) : op(Old,New)} = 1 :- l(T,Old), num(T+1), op(Old,_).
%no number twice
:- 2 #sum {1,T : l(T,Value)}, num(Value).
%2 before 10 before 14
%linear encoding
reached(T,10) :- l(T,10).
reached(T+1,10) :- reached(T,10), num(T+1).
:- reached(T,10), l(T,2).
:- l(T,14), l(T+1,_).
%looks nicer, but quadratic
%:- l(T2,2), l(T10,10), T10<T2.
%:- l(T14,14), l(T10,10), T14<T10.
%we must have these three numbers in the list somewhere
:- not l(_,2).
:- not l(_,10).
:- not l(_,14).
#show r(T,V) : l(T,V).
#show.
Having a slightly more ugly encoding improves grounding a lot (which was your main problem).
I restricted op/2 to not start with 14, as this should be the last element in the list
I do create the list iteratively, this may not be as nice, but at least for the start of the list it already removed impossible to reach values via grounding. So you will never have l(1,33) or l(2,45) etc...
Also list generation stops when reaching the value 14, as no more operation is possible/needed.
I also added a linear scaling version of the "before" section, although it is not really necessary for this short list (but a cool trick in general if you have long lists!) This is called "chaining".
Also note that your show statement is non-trivial and does create some constraints/variables.
I hope this helps, otherwise feel free to ask such questions also on our potassco mailing list ;)
My first attempt is to generate a permutation of numbers and force successor elements to be connected by one of the 3 operations (+5, +7 or sqrt). I predefine the operations to avoid choosing/counting problems. Testing for <60 is not necessary since the output of an operation has to be a number between 0 and 59. The generated List l/2 is forwarded to the output r/2 until the number 14 appears. I guess there is plenty of room to outrun my solution.
num(0..59).
%valid operation pairs
op(N*N,N):- N=2..7.
op(Ori,New):- num(Ori), New = Ori+7, num(New).
op(Ori,New):- num(Ori), New = Ori+5, num(New).
%for each position one number
l(0,0).
{l(T,N):num(N)}==1:-num(T).
{l(T,N):num(T)}==1:-num(N).
% following numbers are connected with an operation until 14
:- l(T,Ori), not op(Ori,New), l(T+1,New), l(End,14), T+1<=End.
% 2 before 10 before 14
:- l(T2,2), l(T10,10), T10<T2.
:- l(T14,14), l(T10,10), T14<T10.
% output
r(T,E):- l(T,E), l(End,14), T<=End.
#show r/2.
first Answer:
r(0,0) r(1,5) r(2,12) r(3,19) r(4,26) r(5,31) r(6,36) r(7,6)
r(8,11) r(9,16) r(10,4) r(11,2) r(12,9) r(13,3) r(14,10) r(15,15)
r(16,20) r(17,25) r(18,30) r(19,37) r(20,42) r(21,49) r(22,7) r(23,14)
There are multiple possible lists with different length.
This question already has answers here:
Given a set of intervals, how to find the maximum number of intersections among them,
(3 answers)
Closed 3 years ago.
On the input, we are given a number N stating how many presentations are to be given, which is followed by N rows of start and end times given in HHMM format
Example:
3
0800 0900
0830 1000
0900 1030
The code must calculate the maximum nuber of occupied rooms (one room can host only one presentation at a time, therefore expected output to the provided example is 2.
My first idea was to create a table of bools 1440 x N (number of minutes in a day by number of presentations) and fill each minute when a presentation is being held, and later go column by column and find the maximum value of presentations at a time. It works, but I'm sure it can be done faster and better. Can someone suggest how to do it in a better manner?
Pretty simple actually: we just simulate the process. First of all, it doesn't matter which presentations are going on right now. All we care about is the number of presentations happening. So we'll just have a counter which we update when a presentation starts or ends.
We could iterate over every minute for the simulation, but our counter only changes when a presentation starts or ends, so we can just make a big list of all the start and end events, sort the list by time, and iterate through the list adjusting our counter appropriately.
The way you propose needs 1440 x 3 = 4320 values. As you can tell, that's very inefficient. A better way is to store just the minutes that are actually required. To save even less values, divide the time into slots of 30 minutes.
Now use a dictionary (std::map) to keep a count of how many presentations are during the same time slot. For your example this gives:
std::map<std::string, int> slots =
{
{"0800", 1},
{"0830", 2},
{"0900", 2},
{"0930", 2},
{"1000", 1},
{"1030", 1},
}
I'll let you figure out how to implement this.
Okay, so this is going to be very complicated to explain through text but I will do my best to try.
I am making a universal calculator where one of the function of the calculator is to process a formula when given an unknown number of variables. I have seen some ways to do this but for how i'm trying to use this calculator, it wont work.
Example for sum of function:
while (cin >> input)
count++;
Normally this would work but the problem is that I can't have the user input the values over and over again for one formula like for this formula: Formula Example
(Sorry its easier for me to explain through a picture) In it there are multiple times where I have to use the same numbers over and over again. Here is the entire process if you need it to understand what I'm saying:
Entire problem
The problem is that normally I would add another float for every point graph but I don't know ahead of time number of floats the user is going to enter in. The ideal way to do this is for the program to ask the user for all the points on the table and for the user to input those points in a format like: "(1,2) (2,4) (3,6)..."
Thinking ahead, would I make a function where the program creates an integer and assigns the integer to a value on the fly? But then how would the actual math formula interact with the new integers if they haven't been created yet?
Talking about this makes my head hurt....
I actually want to say more:
One idea that I tried to make in my head was something like
string VariableName = A [or something]
Then you would reassign VariableName = "A" to VariableName = "B" by something like VariableName = "A"+ 1 (which would equal B).
Then you would repeat that step until the user inputs a invalid input. But obviously you can't do math with letters so I wouldn't know how to do it.
I think that you are overthinking this. It's pretty simple and it doesn't need to store the input values.
The main thing to note is that you need to compute (step 2) the sum of the values of X and Y, the sum of their product and the sum of X squared. To compute the sum of a lot of values you don't need all the values together, but just one at the time. Exactly as when a user provides them. So declare four variables: sx, sy, sxy, sxx. Initialize them to 0. At every couple of X and Y you get, add it to sx and sy, add their product to sxy and the product of X with itself to sxx.
Now you've got all you need to compute the final result for a and b.
Anyway a good C++ book would be useful.
I am writing a code to do some template matching using cv::matchTemplate but I have run into some problems with the 2-dimensional vector of vectors (vov) I created which I have called vvABC. At the moment, my vov has 10 elements which can change based on the values I pass while running the code.
My problem is moving from one column in my vov to the next so I can calculate the size. From my understanding of how vov works, if I have my elements stored in my vov as:
C_A C_B
0 0
1 1
2 2
3
4
5
6
To calculate the size of the first column, I should simply do something like:
vvABC[0].size() to get the size of the first column (which would give 3 in this case) and vvABC[1].size() to get the size of the second column (which would give 7). The problem I am now faced with is both of them give '3' in both cases which is obviously wrong.
Can someone please help me out on how I can get the correct size of the next column?
I stored my detections in my vvABC, now I want to match them one at a time.
It seems like you made a mistake here:
for (uint iCaTemplate = iCa + 1; iCaTemplate < vvABC[iCa].size(); ++iCaTemplate) {
iCa is an index on the 'first level' of vector (of size 2 in your example above), i.e. columns, and you use it to go through the elements of the 'second level' of vector, i.e. rows.
Thanks a lot guys, esp. JGab, after several debug outputs, I finally found that my vector of vectors wasn't being filled up the way I thought it was...thanks once more and my apologies for my belated response.
Given a linked list of T size , select first 2n nodes and delete first n nodes from them; Then do it for the next 2n nodes and so on...
For example-
Let's consider a linked list of size 7:
`1->2->3->4->5->6->7`
If n = 2, the desired output is :
`1->2->5->6->7`
I didn't understand what this problem is actually indicating.Could somebody help me to understand the problem ?
EDIT : Adding C and C++ tags so that this may reach to more eye balls, and of-course those are only two languages allowed in the interview itself.
That actually looks like it should say:
Given a linked list of T size , select first 2n nodes and delete last n nodes from them; Then do it for the next 2n nodes and so on...
or:
Given a linked list of T size , select first 2n nodes and keep first n nodes from them; Then do it for the next 2n nodes and so on...
That would mean select 1,2,3,4 then delete 3,4 (or keep 1,2 which is the same thing). Then select 5,6,7,8, not possible so stop.
I think it's even simpler than #paxdiablo indicates ...
do
take n
skip n
until you run out of elements to take or skip