sorry for this blatent and not very conceptual question, but I am trying to parse a spreadsheet of attributes to find the Pareto set using Kung's algorithm. I'm not sure if I'm misunderstanding the algorithm and I know my julia code is horrible at this point, but I'm not sure how I can improve it so that this actually works. I might be looping infinitely, I overflow on very small cases as well.
The introduction to Kung's algorithm I'm referring to can be found here https://engineering.purdue.edu/~sudhoff/ee630/Lecture09.pdf
I'd really appreciate some help with making my code use less memory (it always fails in the "solve" function) and perhaps function better/more correctly. Kung's algorithm is completely recursive by nature, but I'm sure there are better ways than I'm doing it now. Here's what I have!
using PyPlot
function read_graph_data(fname)
f=open(fname)
arr=readdlm(f, ',', Float64, header=true)
close(f)
return arr
end
function solve(data)
if size(data, 1) == 1
return Set(data)
else
T= solve(data[1:convert(Int64, floor(size(data,1)/2))])
B= solve(data[convert(Int64, ceil(size(data,1)/2)): size(data,1)])
for (b in B)
for (t in T)
if (b.>t == all_true)
setdiff!(T, b)
continue
end
if b.>t == all_false
continue
end
end
push!(T, b)
end
return T
end
end
num_obj = 2
all_true = (trues(num_obj))
all_false = transpose(falses(num_obj))
data = first(read_graph_data("Lake-Problem-Reference-Set.csv"))
data[:, 1] *= -1
data = data[:, 1:num_obj]
solve(data)
Related
I've been studying a few searching algorithms and my last problem comes down to binary searching. I watched a few youtube videos to understand the concept and then tried to solve the problem, but keep getting an endless loop error. I've looked through stack overflow, and reddit, and wherever Google would lead me, but I can't quite find a solution that fits my method of coding. Also, please excuse the term 'monkey patching', it's been brought to my attention that the technical term is called 'extending' so the fault lies on my instructors for teaching it to us as 'monkey patching'.
Here's my code:
class Array
def my_bsearch(target)
return nil if self.empty?
middle_idx = self.length/2
left = self.take(middle_idx)
right = self.drop(middle_idx + 1)
return middle_idx if self[middle_idx] == target
until self[middle_idx] == target || self.nil? == nil
if self[middle_idx] < target
right.my_bsearch(target)
elsif self[middle_idx] > target
left.my_bsearch(target)
end
end
end
end
I have a solution, but I don't want to just memorize it-- and I'm having trouble understanding it; as I'm trying to translate it, learn from it, and implement what I'm missing into my own code.
class Array
def my_bsearch(target)
return nil if size == 0
mid = size/2
case self[mid] <=> target
when 0
return mid
when 1
return self.take(mid).my_bsearch(target)
else
search_res = self.drop(mid+1).my_bsearch(target)
search_res.nil? ? nil : mid + 1 + search_res
end
end
end
I guess I understand case/when despite not use to using it. I've tried following it with debugger, but I think I'm hung up on what's going on in the ELSE section. The syntactic sugar, while making this obviously more concise than my logic, isn't straight-forward/clean to someone of my ruby literacy level. So, yeah, my ignorance is most of the problem I guess.
Is there someone who is a little more literate, and patient, able to help me break this down into something I can understand a bit better so I can learn from this?
First, take and drop have sufficiently similar interfaces that you don't actually want your + 1 for drop. It will disregard one element in the array if you do.
Next, self.nil? will always be false (and never nil) for instances of this class. In fact, .nil? is a method exactly to avoid having to ever compare against nil with ==.
You want self.empty?. Furthermore, with the exception of setters, in Ruby messages are sent to self by default. In other words, the only time self. is a useful prefix is when the message ends in = and operates as an lvalue, as in self.instance_var = 'a constant', since without the self., the tokens instance_var = would be interpreted as a local variable rather than an instance variable setting. That's not the case here, so empty? will suffice just as well as self.empty?
So I figured it out, and I decided to answer my own post in hopes to help someone else out if they run into this issue.
So, if I have an Array and the target is the middle_element, then it will report middle_element_idx. That's fine. What if the target is less than middle_element? It recursively searches the left-side of the original Array. When it finds it, it reports the left_side_idx. There's no problem with that because elements in an array are sequentially counted left to right. So, it starts at 0 and goes up.
But what if the target is on the right side of the middle element?
Well, searching for the right side is easy. Relatively the same logic as searching left. Done recursively. And it will return a target_idx if it's found on that right side --however that's the target's idx as it was found in the right-side array! So, you need to take that returned target_idx and add 1 to it and the original middle_element_idx. See below:
def my_bsearch(target)
return nil if self.empty?
middle_idx = self.length/2
left = self.take(middle_idx)
right = self.drop(middle_idx + 1)
if self[middle_idx] == target
return middle_idx
elsif self[middle_idx] > target
return left.my_bsearch(target)
else
searched_right_side = 1 + right.my_bsearch(target)
return nil if searched_right_side.nil? == true
return searched_right_side + middle_idx
end
end
end
Notice how many more lines this solution is? The spaceship operator used in conjunction with case/when and a ternary method will reduce the number of lines significantly.
Based on suggestions/feedback from Tim, I updated it to:
def my_bsearch(target)
return nil if empty?
middle_idx = self.length/2
left = self.take(middle_idx)
right = self.drop(middle_idx)
if self[middle_idx] == target
return middle_idx
elsif self[middle_idx] > target
return left.my_bsearch(target)
else
searched_right_side = right.my_bsearch(target)
return nil if searched_right_side.nil?
return searched_right_side + middle_idx
end
end
end
After working on my code for a while, optimizing the most obvious things, I've resulted in this:
function FindPath(start, finish, path)
--Define a table to hold the paths
local paths = {}
--Make a default argument
path = path or {start}
--Loop through connected nodes
for i,v in ipairs(start:GetConnectedParts()) do
--Determine if backtracking
local loop = false
for i,vv in ipairs(path) do
if v == vv then
loop = true
end
end
if not loop then
--Make a path clone
local npath = {unpack(path)}
npath[#npath+1] = v
if v == finish then
--If we reach the end add the path
return npath
else
--Otherwise add the shortest part extending from this node
paths[#paths+1] = FindPath(v, finish, npath) --NOTED HERE
end
end
end
--Find and return the shortest path
if #paths > 0 then
local lengths = {}
for i,v in ipairs(paths) do
lengths[#lengths+1] = #v
end
local least = math.min(unpack(lengths))
for i,v in ipairs(paths) do
if #v == least then
return v
end
end
end
end
The problem being, the line noted gets some sort of game script timeout error (which I believe is a because of mass recursion with no yielding). I also feel like once that problem is fixed, it'll probably be rather slow even on the scale of a pacman board. Is there a way I can further optimize it, or perhaps a better method I can look into similar to this?
UPDATE: I finally decided to trash my algorithm due to inefficiency, and implemented a Dijkstra algorithm for pathfinding. For anybody interested in the source code it can be found here: http://pastebin.com/Xivf9mwv
You know that Roblox provides you with the PathfindingService? It uses C-side A* pathing to calculate quite quickly. I'd recommend using it
http://wiki.roblox.com/index.php?title=API:Class/PathfindingService
Try to remodel your algorithm to make use of tail calls. This is a great mechanism available in Lua.
A tail call is a type of recursion where your function returns a function call as the last thing it does. Lua has proper tail calls implementation and it will dress this recursion as a 'goto' under the scenes, so your stack will never blow.
Passing 'paths' as one of the arguments of FindPath might help with that.
I saw your edit about ditching the code, but just to help others stumbling on this question:
ipairs is slower than pairs, which is slower than a numeric for-loop.
If performance matters, never use ipairs, but use a for i=1,#tab loop
If you want to clone a table, use a for-loop. Sometimes, you have to use unpack (returning dynamic amount of trailing nils), but this is not such a case. Unpack is also a slow function.
Replacing ipairs with pairs or numeric for-loops and using loops instead of unpack will increase the performance a lot.
If you want to get the lowest value in a table, use this code snippet:
local lowestValue = values[1]
for k,v in pairs(values) do
if v < lowestValue then
lowestValue = k,v
end
end
This could be rewritten for your path example as so:
local least = #path[1]
for k,v in pairs(path) do
if #v < least then
least = v
end
end
I have to admit, you're very inventive. Not a lot of people would use math.min(unpack(tab)) (not counting the fact it's bad)
I'm banging my head against the wall trying to implement negamax for tic-tac-toe
def negamax(board_obj, mark, depth)
if board_obj.game_over?
return value(board_obj)
else
max = -1.0/0 # negative infinity
if mark == #mark
next_mark = #opponent_mark
else
next_mark = #mark
end
board_obj.empty_squares.each do |square|
board_obj[square] = mark
x = -negamax(board_obj, next_mark, depth + 1)
board_obj[square] = ' '
if x > max
max = x
#scores << x
#best_move = square if depth == 1
end
end
return max
end
end
# determines value of final board state
def value(board_obj)
if board_obj.mark_win?(#mark)
return 1
elsif board_obj.mark_win?(#opponent_mark)
return -1
else
return 0
end
end
the rest of the code is here: https://github.com/dave-maldonado/tic-tac-doh/blob/AI/tic-tac-doh.rb
It does produce a result but the AI is easily beat so I know something's wrong, any help
is appreciated!
The problem is that value needs to be relative to the mark in the current execution of negamax rather than always relative to the computer. If you pass in the mark argument to value from negamax with the following modified definition for value, you'll get the right results:
def value(board_obj, mark)
if board_obj.mark_win?(mark)
return 1
elsif board_obj.mark_win?(mark == 'X' ? 'O' : 'X')
return -1
else
return 0
end
end
That is, the first two lines of the negamax body need to be:
if board_obj.game_over?
return value(board_obj, mark)
That said, this overall program leaves an awful lot to be desired relative to Ruby, good design principles, etc (no offense intended). Now that you have it running, you might want to head over to the Code Review SE for some feedback. :-) And while it's too late to use TDD ;-), it would also be a good one to put "under test".
Also, please understand that per one of the other comments, this is not a kind of question that you'll typically get an answer to here at SO. I don't even know if this question will survive the review process without getting deleted. I worked on it for a variety of personal reasons.
Update: Looking at your reference implementation, you'll note that the negamax code includes the expression sign[color]*Analysis(b). It's that sign[color] that you were missing, effectively.
I have this program that I am working on that is supposed to find the sum of the first 1000 prime numbers. Currently all I am concerned with is making sure that the program is finding the first 1000 prime numbers, I will add the functionality for adding them later. Here is what I have:
#!/usr/bin/ruby
def prime(num)
is_prime = true
for i in 2..Math.sqrt(num)
if (num % i) == 0
is_prime = false
else
is_prime = true
end
end
return is_prime
end
i = 2
number_of_primes = 0
while number_of_primes < 1000
prime = prime(i)
if prime == true
number_of_primes++
end
i++
end
When i try to run the program I get the following feedback:
sumOfPrimes.rb:32: syntax error, unexpected keyword_end
sumOfPrimes.rb:34: syntax error, unexpected keyword_end
what gives? Any direction is appreciated.
Ruby doesn't have ++ operator, you need to do += 1
number_of_primes += 1
Unasked for, but a few pieces of advice if you're interested:
One of the cool things about Ruby is that question marks are legal in method names. As such you'll often find that 'predicate' methods (methods that test something and return true or false) end with a question mark, like this: odd?. Your prime method is a perfect candidate for this, so we can rename it prime?.
You use a local variable, is_prime, to hold whether you have found a factor of the number you're testing yet - this is the kind of thing you'd expect to do in an imperative language such as java or C - but Ruby has all sorts of cool features from functional programming that you will gain great power and expressiveness by learning. If you haven't come across them before, you may need to google what a block is and how the syntax works, but for this purpose you can just think of it as a way to get some code run on every item of a collection. It can be used with a variety of cool methods, and one of them is perfectly suited to your purpose: none?, which returns true if no items in the collection it is called on, when passed to the code block you give, return true. So your prime? method can be rewritten like this:
def prime? num
(2..Math.sqrt(num)).none? { |x| num % x == 0 }
end
Apart from being shorter, the advantage of not needing to use local variables like is_prime is that you give yourself fewer opportunities to introduce bugs - if for example you think the contents of is_prime is one thing but it's actually another. It's also, if you look carefully, a lot closer to the actual mathematical definition of a prime number. So by cutting out the unnecessary code you can get closer to exposing the 'meaning' of what you're writing.
As far as getting the first 1000 primes goes, infinite streams are a really cool way to do this but are probably a bit complex to explain here - definitely google if you're interested as they really are amazing! But just out of interest, here's a simple way you could do it using just recursion and no local variables (remember local variables are the devil!):
def first_n_primes(i = 2, primes = [], n)
if primes.count == n then primes
elsif prime? i then first_n_primes(i + 1, primes + [i], n)
else first_n_primes(i + 1, primes, n)
end
end
And as far as summing them up goes all I'll say is have a search for a ruby method called inject - also called reduce. It might be a bit brain-bending at first if you haven't come across the concept before but it's well worth learning! Very cool and very powerful.
Have fun!
Recursion makes backtracking easy as it guarantees that you won't go through the same path again. So all ramifications of your path are visited just once. I am trying to convert a backtracking tail-recursive (with accumulators) algorithm to iteration. I heard it is supposed to be easy to convert a perfectly tail-recursive algorithm to iteration. But I am stuck in the backtracking part.
Can anyone provide a example through code so myself and others can visualize how backtracking is done? I would think that a STACK is not needed here because I have a perfectly tail-recursive algorithm using accumulators, but I can be wrong here.
If the function is actually recursive, then the transformation is as follows (and this is what a compiler which understand TCO will do for you, so you shouldn't have to do it yourself):
Original:
function func(a1, a2, a3...)
... doesn't contain either return or call
return val
...
return func(x1, x2, x3...)
...
... etc.
Converted to:
function func(a1, a2, a3...)
func: // label for goto (yuk!)
...
return val // No change
...
a1 = x1; a2 = x2; a3 = x3...; goto func;
...
... etc.
In order to make this transformation work with mutually co-recursive functions, you need to combine them into a single function, each of which comes with a label. As above, simple return statements are not altered, and return foo(...) turn into assignment to parameter variables followed by goto foo.
Of course, when combining the functions, you may need to rename local variables to avoid conflicts. And you will also lose the ability to use more than one top-level function, unless you add something like a switch statement (with gotos) at the top entry point, before any label. (In fact, in a language in which allowed goto case foo, you could just use the case labels as labels.)
The use of goto is, of course, ugly. If you use a language which preferably guarantees tail-call optimization, or failing that, at least makes a reasonable attempt to do it and reports when it fails, then there is absolutely no motivation to replace the recursive solution, which (in my opinion) is almost always more readable.
In some cases, it's possible to replace the goto and label with something like while (1) { ... }or other such loops, but that involves replacing thegotos withcontinue` (or equivalent), and that won't work if they're nested inside of other loops. So you can actually waste quite a lot of time making the ugly transformation slightly less ugly, and still not end up with a program as readable as the original.
I'll stop proselytizing recursion now. :)
Edited (I couldn't help it, sorry)
Here's a back-tracking n-queens solution in Lua (which does do TCO), consisting of a tail-recursive solver and a tail-recursive verifier:
function solve(legal, n, first, ...)
if first == nil -- Failure
then return nil
elseif first >= n -- Back-track
then return solve(legal, n, ...)
elseif not legal(first + 1, ...) -- Continue search
then return solve(legal, n, first + 1, ...)
elseif n == 1 + select("#", ...) -- Success
then return first + 1, ...
else -- Forward
return solve(legal, n, 0, first + 1, ...)
end
end
function queens_helper(dist, first, second, ...)
if second == nil
then return true
elseif first == second or first - dist == second or first + dist == second
then return false
else
return queens_helper(dist + 1, first, ...)
end
end
function queens_legal(...) return queens_helper(1, ...) end
-- in case you want to try n-rooks, although the solution is trivial.
function rooks_legal(first, second, ...)
if second == nil then return true
elseif first == second then return false
else return rooks_legal(first, ...)
end
end
function queens(n) return solve(queens_legal, n, 0) end