I have a graph of multi-level dependecies like this, and I need to detect any circular reference in this graph.
A = B
B = C
C = [D, B]
D = [C, A]
Somebody have a problem like this?
Any solution???
Thanks and sorry by english.
========= updated ==========
I had another situation.
1
2 = 1
3 = 2
4 = [2, 3]
5 = 4
In this case, my recursive code iterate two times in "4" reference, but this references don't generate a infinite loop. My problem is to know when function iterate more than one time a reference and is not infinite loop and when is a infinite loop, to inform user.
1 = 4
2 = 1
3 = 2
4 = [2, 3]
5 = 4
This case is a bit diferent from 2th example. This generate a infinite loop. how can I know when cases generate a infinite loop or not?
Topological sorting. The description on Wikipedia is clear and works for all your examples.
Basically you start with a node that has no dependencies, put it in a list of sorted nodes, and then remove that dependency from every node. For you second example that means you start with 1. Once you remove all dependencies on 1 you're left with 2. You end up sorting them 1,2,3,4,5 and seeing that there's no cycle.
For your third example, every node has a dependency, so there's nowhere to start. Such a graph must contain at least one cycle.
Keep a list of uniquely identified nodes. Try to loop through the entire tree but keep checking nodes in the list till you get a node being referred as a child which is already there in the unique list - take it from there (handle the loop or simply ignore it depending on your requirement)
One way to detect circular dependency is to keep a record of the length of the dependency chains that your ordering algorithm detects. If a chain becomes longer than the total number of nodes (due to repetition over a loop) then there is a circular dependency. This should work both for an iterative and for a recursive algorithm.
Related
I am trying to put together an interactive meal tracker / planner in Google Sheets.
What I have is a table with food and info about kcal and their macro nutrition values.
I have already put together a logic for how much kcal does somebody requires and this gives a kcal and macro nutrition for each meal per day, e.g.
Breakfast: 518 kcal, carbohydrates 207, protein 62, fat 19
Now I want to randomly put together foods/meals from the table mentioned above until those numbers are hit automatically.
There a ways to randomly select things from an array, but I'm stuck how to combine this with a loop and a deadline.
Any idea?
Many thanks
I do not see any update so I will just go with my assumption on what you are stuck on and say that it's building the array and making sure you stop once you reach the kCal limit.
Basically you will be working with 3 variables, foodTable as your food array, mealTable which is the meal table you will generate and some kind of kCal tracker, let's just call it kCal.
In essence you want to use a do .. while loop and your while statement is basically kCal < target where target is the value you want to reach. If you want to use each value from the foodTable only once then you will need to use selectedFood = foodTable.splice(idx,1), if you want to be able to re-use the sime item then use slice() instead.
Your idx is a randomly generated number between 0 and mealTable.length (in case you are using splice to remove an item from the possible foods list this will ensure you do not generate an invalid number)
Then add the kCal value to your tracker with kCal += selectedFood[0][col] where col is the column number where you store the kCal value. Make sure that every time before you start the while loop you have kCal = 0 to reset, if you repeat this. And keep in mind that this method does not care how much over the target kCal you will go and you might want to add some kind of if statement to make sure you do not run out of items before you reach your target.
I have this code:
let sumfunc(n: int byref) =
let mutable s = 0
while n >= 1 do
s <- n + (n-1)
n <- n-1
printfn "%i" s
sumfunc 6
I get the error:
(8,10): error FS0001: This expression was expected to have type
'byref<int>'
but here has type
'int'
So from that I can tell what the problem is but I just dont know how to solve it. I guess I need to specify the number 6 to be a byref<int> somehow. I just dont know how. My main goal here is to make n or the function argument mutable so I can change and use its value inside the function.
Good for you for being upfront about this being a school assignment, and for doing the work yourself instead of just asking a question that boils down to "Please do my homework for me". Because you were honest about it, I'm going to give you a more detailed answer than I would have otherwise.
First, that seems to be a very strange assignment. Using a while loop and just a single local variable is leading you down the path of re-using the n parameter, which is a very bad idea. As a general rule, a function should never modify values outside of itself — and that's what you're trying to do by using a byref parameter. Once you're experienced enough to know why byref is a bad idea most of the time, you're experienced enough to know why it might — MIGHT — be necessary some of the time. But let me show you why it's a bad idea, by using the code that s952163 wrote:
let sumfunc2 (n: int byref) =
let mutable s = 0
while n >= 1 do
s <- n + (n - 1)
n <- n-1
printfn "%i" s
let t = ref 6
printfn "The value of t is %d" t.contents
sumfunc t
printfn "The value of t is %d" t.contents
This outputs:
The value of t is 7
13
11
9
7
5
3
1
The value of t is 0
Were you expecting that? Were you expecting the value of t to change just because you passed it to a function? You shouldn't. You really, REALLY shouldn't. Functions should, as far as possible, be "pure" -- a "pure" function, in programming terminology, is one that doesn't modify anything outside itself -- and therefore, if you run it twice with the same input, it should produce the same output every time.
I'll give you a way to solve this soon, but I'm going to post what I've written so far right now so that you see it.
UPDATE: Now, here's a better way to solve it. First, has your teacher covered recursion yet? If he hasn't, then here's a brief summary: functions can call themselves, and that's a very useful technique for solving all sorts of problems. If you're writing a recursive function, you need to add the rec keyword immediately after let, like so:
let rec sumExampleFromStackOverflow n =
if n <= 0 then
0
else
n + sumExampleFromStackOverflow (n-1)
let t = 7
printfn "The value of t is %d" t
printfn "The sum of 1 through t is %d" (sumExampleFromStackOverflow t)
printfn "The value of t is %d" t
Note how I didn't need to make t mutable this time. In fact, I could have just called sumExampleFromStackOverflow 7 and it would have worked.
Now, this doesn't use a while loop, so it might not be what your teacher is looking for. And I see that s952163 has just updated his answer with a different solution. But you should really get used to the idea of recursion as soon as you can, because breaking the problem down into individual steps using recursion is a really powerful technique for solving a lot of problems in F#. So even though this isn't the answer you're looking for right now, it is the answer you're going to be looking for soon.
P.S. If you use any of the help you've gotten here, tell your teacher that you've done so, and give him the URL of this question (http://stackoverflow.com/questions/39698430/f-how-to-call-a-function-with-argument-byref-int) so he can read what you asked and what other people told you. If he's a good teacher, he won't lower your grade for doing that; in fact, he might raise it for being honest and upfront about how you solved the problem. But if you got help with your homework and you don't tell your teacher, 1) that's dishonest, and 2) you'll only hurt yourself in the long run, because he'll think you understand a concept that you maybe haven't understood yet.
UPDATE 2: s952163 suggests that I show you how to use the fold and scan functions, and I thought "Why not?" Keep in mind that these are advanced techniques, so you probably won't get assignments where you need to use fold for a while. But fold is basically a way to take any list and do a calculation that turns the list into a single value, in a generic way. With fold, you specify three things: the list you want to work with, the starting value for your calculation, and a function of two parameters that will do one step of the calculation. For example, if you're trying to add up all the numbers from 1 to n, your "one step" function would be let add a b = a + b. (There's an even more advanced feature of F# that I'm skipping in this explanation, because you should learn just one thing at a time. By skipping it, it keeps the add function simple and easy to understand.)
The way you would use fold looks like this:
let sumWithFold n =
let upToN = [1..n] // This is the list [1; 2; 3; ...; n]
let add a b = a + b
List.fold add 0 upToN
Note that I wrote List.fold. If upToN was an array, then I would have written Array.fold instead. The arguments to fold, whether it's List.fold or Array.fold, are, in order:
The function to do one step of your calculation
The initial value for your calculation
The list (if using List.fold) or array (if using Array.fold) that you want to do the calculation with.
Let me step you through what List.fold does. We'll pretend you've called your function with 4 as the value of n.
First step: the list is [1;2;3;4], and an internal valueSoFar variable inside List.fold is set to the initial value, which in our case is 0.
Next: the calculation function (in our case, add) is called with valueSoFar as the first parameter, and the first item of the list as the second parameter. So we call add 0 1 and get the result 1. The internal valueSoFar variable is updated to 1, and the rest of the list is [2;3;4]. Since that is not yet empty, List.fold will continue to run.
Next: the calculation function (add) is called with valueSoFar as the first parameter, and the first item of the remainder of the list as the second parameter. So we call add 1 2 and get the result 3. The internal valueSoFar variable is updated to 3, and the rest of the list is [3;4]. Since that is not yet empty, List.fold will continue to run.
Next: the calculation function (add) is called with valueSoFar as the first parameter, and the first item of the remainder of the list as the second parameter. So we call add 3 3 and get the result 6. The internal valueSoFar variable is updated to 6, and the rest of the list is [4] (that's a list with one item, the number 4). Since that is not yet empty, List.fold will continue to run.
Next: the calculation function (add) is called with valueSoFar as the first parameter, and the first item of the remainder of the list as the second parameter. So we call add 6 4 and get the result 10. The internal valueSoFar variable is updated to 10, and the rest of the list is [] (that's an empty list). Since the remainder of the list is now empty, List.fold will stop, and return the current value of valueSoFar as its final result.
So calling List.fold add 0 [1;2;3;4] will essentially return 0+1+2+3+4, or 10.
Now we'll talk about scan. The scan function is just like the fold function, except that instead of returning just the final value, it returns a list of the values produced at all the steps (including the initial value). (Or if you called Array.scan, it returns an array of the values produced at all the steps). In other words, if you call List.scan add 0 [1;2;3;4], it goes through the same steps as List.fold add 0 [1;2;3;4], but it builds up a result list as it does each step of the calculation, and returns [0;1;3;6;10]. (The initial value is the first item of the list, then each step of the calculation).
As I said, these are advanced functions, that your teacher won't be covering just yet. But I figured I'd whet your appetite for what F# can do. By using List.fold, you don't have to write a while loop, or a for loop, or even use recursion: all that is done for you! All you have to do is write a function that does one step of a calculation, and F# will do all the rest.
This is such a bad idea:
let mutable n = 7
let sumfunc2 (n: int byref) =
let mutable s = 0
while n >= 1 do
s <- n + (n - 1)
n <- n-1
printfn "%i" s
sumfunc2 (&n)
Totally agree with munn's comments, here's another way to implode:
let sumfunc3 (n: int) =
let mutable s = n
while s >= 1 do
let n = s + (s - 1)
s <- (s-1)
printfn "%i" n
sumfunc3 7
I am looking for ideas on the data model for the following problem (and the proper CS terminology):
A (horizontal) "timeline" with several rows (A,B,C) contains "events" (1,2,3) width different durations (width) at different times (absolute x position or by delay "." after previous event):
A 1111....222222
B 33333
------------------
T 0123456789ABCDEF
(The rows are only interesting for graphical representation of overlapping/parallel "events", so they probably are not essential to the data model.)
Event duration may vary, affecting the whole timing:
A 11....222222
B 33333+3
------------------
T 0123456789ABCDEF
But let event 2 require events 1 and 3 to be finished, so the timing should look like this:
A 11.... 222222
B 33333+3
------------------
T 0123456789ABCDEF
(let's ignore that the original delay at T=7 is now missing.)
Originally I thought I'd have to have some "elastic" synchronization elements, one for each row:
A 11....####222222
B 33333+3#
------------------
T 0123456789ABCDEF
Thus the original problem of how to model and sync the sync elements in the two different "rows". But, as established above, this is only a matter of graphical/parallel representation.
Rather, the sync is a condition that could be "attached" to event 2, modifiying or determining its beginning.
If an event "has" a condition, it will not have an absolute or relative start time. Its start can only be determined at the ends of the "linked" events (1 and 3).
So, given (a list of) some events with variable duration and either an absolute start time or a delay relative to another event's end, how could the condition "events 1 and 3 ended" be modelled to determine the start of "event 2"?
(I will prototype this in JavaScript and eventually implement in C/C++, so any sample code provided should not use high-level data types or libraries.)
What you need is an object that I would call a TimeFrame. The object would have the attributes duration, link and type, where link can be a precise time or a link to another TimeFrame and type accounts for the kind of link. For instance, a given TimeFrame that starts at a known time would have that time as its link attribute and the type would be TIME. A TimeFrame that is linked to the end of another would have that other TimeFrame as its link attribute and START-END as its type and so on.
Using the combination between link and type you could also support other types of links such as START-START, END-START or END-END.
UPDATE
Also, in order to allow some time interval between say, the end of a TimeFrame and the start of the next, one can add the attribute lag, which represents any delay between events. So, for instance if tf1 and tf2 are TimeFrames such that tf2 must start 5 time units after the end of tf1 the attributes of tf2 would be link = tf1, type = START-END, duration = <something> and lag = 5. Note also that the lag could be negative, which would extend the expressiveness of the model to a broad range of relationships.
While #Leandro Caniglia nicely rephrased my question into an Object and Attributes, essentially, I see two options:
the whole list of "events" needs to be evaluated at "condition" (start/end) to check dependent "events".
adding a "link" to a "parent" also creates a link to the "child" (no need to evaluate all pending event's links).
Also:
The "link" property needs to be a List or Array to be able hold several references (e.g. 2:[1,3]).
Analogous to the link property start_me_on_condition a stop_me_on_condition association may be desirable (see Leandro's suggestion of type, it would need to be extended to support multiple links+type)
An independet delay "event" might be more practical than a lag property.
This is not a homework question. I'm merely trying to understand the process for my own edification. As a computer science student, I have attended several lectures where the concept of recursion was discussed. However, the lecturer was slightly vague, in my opinion, regarding the concept of a stack frame and how the call stack is traversed in order to calculate the final value. The manner in which I currently envision the process is analogous to building a tree from the top down (pushing items onto the call stack - a last in, first out data structure) then climbing the newly constructed tree where upon the final value is obtained upon reaching the top. Perhaps the canonical example:
def fact(n):
if n == 0:
ans = 1
else:
ans = n * fact(n-1)
return ans
value = fact(5)
print (value)
As indicated above, I think the call stack eventually resembles the following (crudely) drawn diagram:
+----------+
| 5 |
| 4 |
| 3 |
| 2 |
| 1 |
+----------+
Each number would be "enclosed" within a stack frame and control now proceeds from the bottom (with the value of 1) to the 2 then 3, etc. I'm not entirely certain where the operator resides in the process though. Would I be mistaken in assuming an abstract syntax tree (AST) in involved at some point or is a second stack present that contains the operator(s)?
Thanks for the help.
~Caitlin
Edit: Removed the 'recursion' tag and added 'function' and 'stackframe' tags.
The call stack frame stores arguments, return address and local variables. The code (not only the operator) itself is stored elsewhere. The same code is executed on different stack frames.
You can find some more information and visualization here: http://www.programmerinterview.com/index.php/recursion/explanation-of-recursion/
This question is more about how function calls work, rather than about recursion. When a function is called a frame is created and pushed on the stack. The frame includes a pointer to the calling code, so that the program knows where to return after the function call. The operator resides in the executable code, after the call point.
I need to use a for-loop in a function in order to find spring constants of all possible combinations of springs in series and parallel. I have 5 springs with data therefore I found the spring constant (K) of each in a new matrix by using polyfit to find the slope (using F=Kx).
I have created a function that does so, however it returns data not in a matrix, but as individual outputs. So instead of KP (Parallel)= [1 2 3 4 5] it says KP=1, KP=2, KP=3, etc. Because of this, only the final output is stored in my workspace. Here is the code I have for the function. Keep in mind that the reason I need to use the +2 in the for loop for b is because my original matrix K with all spring constants is ten columns, with every odd number being a 0. Ex: K=[1 0 2 0 3 0 4 0 5] --- This is because my original dataset to find K (slope) was ten columns wide.
function[KP,KS]=function_name1(K)
L=length(K);
c=1;
for a=1:2:L
for b=a+2:2:L
KP=K(a)+K(b)
KS=1/((1/K(a))+(1/K(b)))
end
end
c=c+1;
and then a program calling that function
[KP,KS]=function_name1(K);
What I tried: - Suppressing and unsuppressing lines of code (unsuccessful)
Any help would be greatly appreciated.
hmmm...
your code seems workable, but you aren't dealing with things in the most practical manner
I'd start be redimensioning K so that it makes sense, that is that it's 5 spaces wide instead of your current 10 - you'll see why in a minute.
Then I'd adjust KP and KS to the size that you want (I'm going to do a 5X5 as that will give all the permutations - right now it looks like you are doing some triangular thing, I wouldn't worry too much about space unless you were to do this for say 50,000 spring constants or so)
So my code would look like this
function[KP,KS]=function_name1(K)
L=length(K);
KP = zeros(L);
KS = zeros(l);
c=1;
for a=1:L
for b=1:L
KP(a,b)=K(a)+K(b)
KS(a,b)=1/((1/K(a))+(1/K(b)))
end
end
c=c+1;
then when you want the parallel combination of springs 1 and 4 KP(1,4) or KP(4,1) will do the trick