I've hit a comprehension wall with respect to the following code. It is intended to search through a list of items that have both a personal utility and a cost, as if it is looking for an optimal solution in a binary search tree of possible items. Specifically, it contemplates the maximum utility that can be obtained by combining the items, subject to a weight constraint that is passed into the function.
The first recursive call under the else statement is where the trouble starts. Once the initial recursive call is made, and values are thrown into withVal and withToTake, does the code continue to recurse - piling up stack frames until a base case is hit - or does it continue to the bottom of the code and then consider the recursive calls? Print statements were of no help on this one. Any insight would be greatly appreciated!
def maxVal(toConsider, avail):
'''Assumes toConsider is a list of items (nodes higher up in the tree
corresponding to earlier calls in the recursive stack. That have not
yet been considered. avail a weight (amt of space still available).
Returns a tuple of the total value of a solution to the 0/1
knapsack problem and the items of that solution.
*Note we are NOT building the search tree. The local variable
result returns the best result found so far.To be used with
food class*
'''
#Base Cases: Nothing left to consider, or available weight is 0
if toConsider == [] or avail == 0:
result = (0, ())
#look at the first item's weight in toConsider & check against available space
#if the item's weight is > avail, recurse and slice off the first item
elif toConsider[0].getCost() > avail:
result = maxVal(toConsider[1:], avail)
else:
#look at left branch
nextItem = toConsider[0]
**withVal, withToTake = maxVal(toConsider[1:], avail - nextItem.getCost())**
withVal += nextItem.getValue()
#look at right branch
withoutVal, withoutToTake = maxVal(toConsider[1:], avail)
#tests whether left or right branch is better
if withVal > withoutVal:
result = (withVal, withToTake + (nextItem,))
else:
result = (withoutVal, withToTake)
return result
toCosider would take a list of Food objects with utility values & maxUnits is an arbitrary integer indicating the maximum amount of food one can carry.
Related
so i guess you are all fimilliar with a binary heap data structure if not.. Brilliant. org say
i.e. a binary tree which obeys the property that the root of any tree is greater than or equal to (or smaller than or equal to) all its children (heap property). The primary use of such a data structure is to implement a priority queue.
will one of the properties of a binary heap is that it must be filled from top to bottom (from root) and from right to left
I coded this algorithm to find the next available spot to insert the next number I add (I hard coded the first nodes so I can track more further down the tree
this search method is inspired by BFS(Breadth First Search) algorithm
note that in this code I only care about finding the next empty node without the need to keep the heap property
I tested the code but I don't think I tested it enough so if you spot problems, bugs or suggest any ideas, every comment is welcomed
def insert(self, data):
if self.root.data == None:
self.root.data = data
print('root', self.root.data)
else:
self.search()
def search(self):
print('search..L31')
queue = [self.root]
while queue:
curr = queue.pop(0)
print(curr.data)
if curr.right_child == None:
print('made it')
return
else:
queue.append(curr.left_child)
queue.append(curr.right_child)
h = Min_heap(10)
h.insert(2)
h.root.left_child = Node(3)
h.root.right_child = Node(5)
h.root.left_child.left_child = Node(8)
h.root.left_child.right_child = Node(7)
h.root.right_child.left_child = Node(9)
# The tree I am building...
# __2__
# / \
# 3 5
# / \ / \
# 8 7 9 ⨂
# ↑
# what am
# looking for
h.search()
there is another way to figuring this out which is basically translating the tree into an array/list using special formulas and then we just assume that the next data we want to insert is the last element in the previous array and then work back through the same formulas but I already know that algorithm and I thought why not trying to solve it as a graph soooo...
You should better implement a binary heap as a list (array). But if you want to do it with node objects that have left/right attributes, then the position for the next node can be derived from the size of the tree.
So if you enrich your heap class instances with a size attribute and maintain that attribute to reflect the current number of nodes in the tree, then the following method will tell you where the next insertion point is, in O(logn) time:
Take the binary representation of the current size plus 1. So if the tree currently has 4 nodes, take the binary representation of 5, i.e. 101. Then drop the leftmost (most significant) bit. The bits that then remain are an encoding of the path towards the new spot: 0 means "left", 1 means "right".
Here is an implementation of a method that will return the parent node of where the new insertion spot is, and whether it would become the "left" or the "right" child of it:
def next_spot(self):
if not self.root:
raise ValueError("empty tree")
node = self.root
path = self.size + 1
sides = bin(path)[3:-1] # skip "0b1" and final bit
for side in sides:
if side == "0":
node = node.left
else:
node = node.right
# use final bit for saying "left" or "right"
return node, ("left", "right")[path % 2]
If you want to guarantee balanced, just add to each node how many items are there or below. Maintain that with the heap. And when placing an element, always go to where there are the fewest things.
If you just want a simple way to place, just randomly place it. You don't have to be perfect. You will still on average be O(log(n)) levels, just with a worse constant.
(Of course your constants are better with the array approach, but you say you know that one and are deliberately not implementing it.)
I am solving a question where we are asked to return the sum of the Depths of all the nodes in a Binary Tree.
For example:
Usually I would use a debugger to find the error in my code, but I don't know how to set up trees/binary tree in my IDE.
My code below passes most of the tests, but fails some. It fails the above test, and produces an output of 20 instead of 16.
def nodeDepths(root):
queue = [root]
sumOfDepths = 0
currentDepth = 0
while len(queue):
for _ in range(len(queue)):
currentNode = queue.pop()
sumOfDepths += currentDepth
if currentNode.left:
queue.append(currentNode.left)
if currentNode.right:
queue.append(currentNode.right)
currentDepth += 1
return sumOfDepths
Any suggestions where the code fails/is doing something unexpected.
I believe the source of your error is in your current_node = queue.pop() statement. According to the docs "The argument passed to the method is optional. If not passed, the default index -1 is passed as an argument (index of the last item)." Because you are pulling the last entry from the queue everything works okay until you have entries in the queue from different depths. To fix this problem use current_node = queue.pop(0), this will always pull the oldest entry from the queue.
I know this might be trivial but i just want to make sure. I believe it's runtime will be at most O(n). My reasoning is that every node will return a height value once throughout this recursive method. Or in other words we will visit every node in the tree once.
def height(self):
if self.is_empty():
return 0
else:
left_max = self._left.height()
right_max = self._right.height()
return max(left_max, right_max) + 1
You are performing DFS traversal on tree all nodes will be visited only one time.
So obviously it will take time of O(N) only.
I have an xml file like the following:
<edge from="0/0" to="0/1" speed="10"/>
<edge from="0/0" to="1/0" speed="10"/>
<edge from="0/1" to="0/0" speed="10"/>
<edge from="0/1" to="0/2" speed="10"/>
...
Note, that there exist pairs of from-to and vice versa. (In the example above only the pair ("0/0","0/1") and ("0/1","0/0") is visible, however there is a partner for every entry.) Also, note that those pairs are not ordered.
The file describes edges within a SUMO network simulation. I want to assign new speeds randomly to the different streets. However, every <edge> entry only describes one direction(lane) of a street. Hence, I need to find its "partner".
The following code distributes the speed values lane-wise only:
import xml.dom.minidom as dom
import random
edgexml = dom.parse("plain.edg.xml")
MAX_SPEED_OPTIONS = ["8","9","10"]
for edge in edgexml.getElementsByTagName("edge"):
x = random.randint(0,2)
edge.setAttribute("speed", MAX_SPEED_OPTIONS[x])
Is there a simple (pythonic) way to maybe gather those pairs in tuples and then assign the same value to both?
If you know a better way to solve my problem using SUMO tools, I'd be happy too. However I'm still interested in how I can solve the given abstract list problem in python as it is not just a simple zip like in related questions.
Well, you can walk the list of edges and nest another iteration over all edges to search for possible partners. Since this is of quadratic complexity, we can even reduce calculation time by only walking over not yet visited edges in the nested run.
Solution
(for a detailed description, scroll down)
import xml.dom.minidom as dom
import random
edgexml = dom.parse('sampledata/tmp.xml')
MSO = [8, 9, 10]
edge_groups = []
passed = []
for idx, edge in enumerate(edgexml.getElementsByTagName('edge')):
if edge in passed:
continue
partners = []
for partner in edgexml.getElementsByTagName('edge')[idx:]:
if partner.getAttribute('from') == edge.getAttribute('to') \
and partner.getAttribute('to') == edge.getAttribute('from'):
partners.append(partner)
edge_groups.append([edge] + partners)
passed.extend([edge] + partners)
for e in edge_groups:
print('NEW EDGE GROUP')
x = random.choice(MSO)
for p in e:
p.setAttribute('speed', x)
print(' E from "%s" to "%s" at "%s"' % (p.getAttribute('from'), p.getAttribute('to'), x))
Yields the output:
NEW EDGE GROUP
E from "0/0" to "0/1" at "8"
E from "0/1" to "0/0" at "8"
NEW EDGE GROUP
E from "0/0" to "1/0" at "10"
NEW EDGE GROUP
E from "0/1" to "0/2" at "9"
Detailed description
edge_groups = []
passed = []
Initialize the result structure edge_groups, which will be a list of lists holding partnered edges in groups. The additional list passed will help us to avoid redundant edges in our result.
for idx, edge in enumerate(edgexml.getElementsByTagName('edge')):
Start iterating over the list of all edges. I use enumerate here to obtain the index at the same time, because our nested iteration will only iterate over a sub-list starting at the current index to reduce complexity.
if edge in passed:
continue
Stop, if we have visited this edge at any point in time before. This does only happen if the edge has been recognized as a partner of another list before (due to index-based sublisting). If it has been taken as the partner of another list, we can omit it with no doubt.
partners = []
for partner in edgexml.getElementsByTagName('edge')[idx:]:
if partner.getAttribute('from') == edge.getAttribute('to') \
and partner.getAttribute('to') == edge.getAttribute('from'):
partners.append(partner)
Initialize helper list to store identified partner edges. Then, walk through all edges in the remaining list starting from the current index. I.e. do not iterate over edges that have already been passed in the outer iteration. If the potential partner is an actual partner (from/to matches), then append it to our partners list.
edge_groups.append([edge] + partners)
passed.extend([edge] + partners)
The nested iteration has passed and partners holds all identified partners for the current edge. Push them into one list and append it to the result variable edge_groups. Since it is unneccessarily complex to check against the 2-level list edge_groups to see whether we have already traversed an edge in the next run, we will additionally keep a list of already used nodes and call it passed.
for e in edge_groups:
print('NEW EDGE GROUP')
x = random.choice(MSO)
for p in e:
p.setAttribute('speed', x)
print(' E from "%s" to "%s" at "%s"' % (p.getAttribute('from'), p.getAttribute('to'), x))
Finally, we walk over all groups of edges in our result edge_groups, randomly draw a speed from MSO (hint: use random.choice() to randomly choose from a list), and assign it to all edges in this group.
Im writing an algorithm in Python which plays this game.
The current state of the board of tiles in the game is a dictionary in the form of:
{
<tile_id>: {
'counters': <number of counters on tile or None>,
'player': <player id of the player who holds the tile or None>,
'neighbours': <list of ids of neighbouring tile>
},
...
}
I have another dictionary which stores all of my tiles which are 'full' (i.e. a tile which has one less counter than its number of neighbours and where the player is me) This dictionary, full_tiles, is in the same form as the board dictionary above.
I am now trying to create a list, chains, where each element in the list is a dictionary of my full tiles that are neighbouring at least one other full tile (i.e a chain of full tiles). So this will be a list of all my seperate chains on the board.
Here is my code so far:
for tile_id, tile in full_tiles.items(): #iterates through all full tiles
current_tile = {tile_id : tile} #temporarily stores current tile
if not chains: #if chains list is empty
chains.append(current_tile) #begin list
else: #if list is not empty
for index, chain in enumerate(chains): #iterate though list of chains
if not (current_tile in chain): #if current tile is not in current chain
for tile_id2, tile2 in chain.items(): #iterate through tiles in current chain
for neighbour in tile2["neighbours"]: #iterate through each neighbour of current tile
#neighbour is in the form of tile_id
if neighbour in chain: #if current tile's neighbour is in chain
chain[tile_id] = tile #add tile to chain
It is very difficult for me to test and debug my code and check if it is working correctly as the code can only be run in an application that simulates the game. As you can see, there is quite a lot going on in this block of code with all of the nested loops which are difficult to follow. I cannot seem to think straight at the minute and so I cannot determine if this mess, in all honesty, will function as I hope.
While I am writing this, I have just realised that - on line 7 of this code - I am only checking if the current tile is not in the current chain and so there will be intersecting chains which, of course, will be a mess. Instead of this, I need to first check if the current tile is in not in any of the chains, not just one.
Apart from this error, will my code achieve what I am attempting? Or can you recommend a simpler, neater way to do it? (There has to be!)
Also, let me know if I have not given enough information on how the code is run or if I need to explain anything further, such as the contents of the board dictionary.
Thank you for any help in advance.
EDIT: Unfortunately, I was under a time constraint to complete this project, and as it was my first time ever working with Python, I currently lack the knowledge in the language to optimise my solution using the sources given below. Here is my final extremely ugly and messy solution to this problem which, in the end, worked fine and wasn't terribly inefficient given the small data set that the code works on.
for x in range(0, len(my_hexplode_chains) - 1):
match_found = False
for y in range(x + 1, len(my_hexplode_chains)):
for tile_id_x, tile_x in my_hexplode_chains[x].items(): #compare each chain in list
for tile_id_y, tile_y in my_hexplode_chains[y].items(): #to every other chain
for neighbour in tile_x["neighbours"]: #if tiles in different lists
if neighbour == tile_id_y: #are neighbours
match_found = True
my_hexplode_chains[x].update(my_hexplode_chains[y]) #append one chain to the other
del my_hexplode_chains[y] #delete appended chain
if match_found: #continue loop at next chain
break #very ugly way to do this
if match_found:
break
if match_found:
break
if match_found:
break
How about this optimization?
def find_match (my_hexplode_chains):
x = 0
len_chain = len(my_hexplode_chains)
while x <= len_chain:
y = x + 1
for tile_id_x, tile_x in my_hexplode_chains[x].items():
for tile_id_y, tile_y in my_hexplode_chains[y].items():
if tile_id_y in tile_x["neighbours"]:
my_hexplode_chains[x].update(my_hexplode_chains[y])
del my_hexplode_chains[y]
return True
x += 1
return False
You could pass this function after each move in your game and trace the output.