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Lines of Code : 67License : Permissive (MIT License)
Copydef find_optimal_binary_search_tree(nodes):
"""
This function calculates and prints the optimal binary search tree.
The dynamic programming algorithm below runs in O(n^2) time.
Implemented from CLRS (Introduction to Algorithms) book.
Community Discussions
Trending Discussions on CLRS
QUESTION
For the code below I am trying to create a barplot. If the column name in column ACQUISITION_CHANNEL = 'Referral' then the bar should be red else grey.
...ANSWER
Answered 2021-May-31 at 17:05You can specify a dict that maps the values used for the hue parameter to matplotlib colours, see the second example under point plots.
QUESTION
I came upon the recursive relation for the max-heapify algorithm when going through CLRS. My teacher had justified, quite trivially in fact, that the time complexity of the max-heapify process was O(logn), simply because the worst case is when the root has to 'bubble/float down' from the top all the way to the last level. This means we travel layer by layer, and hence the number of steps equals the number of levels/height of the heap, which, as we know, is bounded by logn. Fair enough.
The same however was proven in CLRS in a more rigorous manner via a recurrence relation. The worst case was said to occur when the last level is half filled and this has already been explained here. So as far as I understand from that answer, they arrived at this conclusion mathematically: we want to maximise the size of the left subtree relative to the heap size n i.e to maximise the value of L/n. And to achieve this we have to have the last level half filled so that the number of nodes in L (left subtree) is maximized and L/n is maximized.
Adding any more nodes to the last level will increase the number of nodes but bring no change to the value of L. So L/n will decrease, as heap becomes more balanced. All is fine and good as long as it's mathematical.
Now this is where I get stuck: Let's say I add one more node in this half-filled level. Practically, I fail to see how this somehow reduces the number of steps/comparisons that occur and is no longer the worst case. Even though I have added one more node, all the comparisons occur only in the left subtree and have nothing to do with the right subtree. Can someone convince me/help me realise why and how exactly does it work out that L/n has to be maximized for the worst case? I would appreciate an example input and how adding more nodes no longer makes it the worst possible case?
...ANSWER
Answered 2021-May-22 at 14:50Let's say I add one more node in this half-filled level. Practically, I fail to see how this somehow reduces the number of steps/comparisons that occur and is no longer the worst case . Even though I have added one more node, all the comparisons occur only on the left subtree and has nothing to do with the right subtree.
It is correct that this does not reduce the number of steps. However, when we speak of time complexity, we look for a relation between the number of steps and 𝑛. If were to only look at the number of steps, we would only conclude that the worst case happens when the tree is infinitely large. Although that is a really bad case (the worst), that is not what the book means with "worst case" here. It is not just the number of steps that interests us, it is how that number relates to 𝑛.
We can argue about the terminology here, because usually "worst case" is not about something that depends on 𝑛, but about variations that can exist for a given 𝑛. For instance, when discussing worst case scenarios for a sorting algorithm, the worst and best cases are dependent on how the input data is organised (already sorted, reversed, ...etc). Here "worst case" is used for the shape of the (bottom layer of the) tree, which is directly inferred by the value of 𝑛. Once you have 𝑛, there is no variation possible there.
However, for the recurrence relation, we must find the formula -- in terms of 𝑛 -- that gives an upper limit for the number of children in the left subtree, with the constraint that we want this formula to use simple arithmetic (for example: no flooring).
Here is a graph where the blue bars represent the value of 𝑛, and the orange bars represent the number of nodes in the left subtree.
The recurrence relation is based on the idea that the greatest subtree of both subtrees is the left subtree, so it represents the worst case. That subtree has a number of nodes that is somewhere between (𝑛-1)/2 and 2𝑛/3. The ratio between the number of nodes in the left subtree and the total number of nodes is maximised when the left subtree is full, and the right subtree has a lesser height.
Here is the same data represented as a ratio:
You can see where these maxima occur: when 𝑛 is 2, 5, 11, 23, ... the ratio between the number of nodes in the left subtree and 𝑛 approaches 40%. This 40% represents the upper limit for the ratio, and is a safe "catch-all" for all values of 𝑛.
We need this ratio in the recurrence relation: that 40% can be rephrased: the number of nodes in the subtree has an upper bound of 2𝑛/3. And so the recurrence relation is
𝑇(𝑛) = 𝑇(2𝑛/3) + O(1)
QUESTION
I'm going through the Dynamic Programming chapter in the CLRS book. In the rod cutting problem, this recurrence relation is obtained when we don't use dynamic programming (with base case T(0) = 1). The solution is directly given as T(n) = 2^n.
I can verify that the solution is correct using induction. But I can't seem to figure out how to arrive at this solution step-by-step from the given recurrence using iteration (plug and chug) method. I would really appreciate some help on this matter.
...ANSWER
Answered 2021-May-19 at 10:56T(0) = 1
T(1) = 1 + T(0)
= 2
T(2) = 1 + T(0) + T(1)
\_,____/
= T(1) + T(1)
= 2*T(1)
= 4
T(3) = 1 + T(0) + T(1) + T(2)
\_,___________/
= T(2) + T(2)
= 2*T(2)
= 8
T(4) = 1 + T(0) + T(1) + T(2) + T(3)
\_,__________________/
= T(3) + T(3)
= 2*T(3)
= 16
:
T(n) = 2*T(n-1) = 2^n
QUESTION
I'm trying to answer clrs intro to alg edition2 exercises. in chapter 27.1-4 there is an exercise which says: "Prove that any sorting network on n inputs has depth at least lg n". so I think that we can use at most (n/2) comparators in each depth and if we assume that we have found a combination of comparators which can sort (n/2) of the numbers in depth1 then we need to sort the other (n/2) of the numbers. So if we keep doing the same thing we're dividing n by two in each depth so the depth of the sorting network would be lgn. Is this conclusion wrong? if it is what is the right way of proving the lower bound of a sorting networks depth.
...ANSWER
Answered 2021-May-07 at 19:21I can think of two.
The first is that you can view a sorting network for n elements as a comparison-based sorting algorithm, and the lower bound on the latter implies that the network does lg n! = n lg n − n + O(log n) comparisons, divided by n/2 comparisons per level is 2 lg n − 1 + O((log n)/n) ≥ lg n if n ≥ 2 (and you can verify n = 1 manually).
The other is that after r rounds, each input can have been shuffled to at most 2r different locations. This can be proved by induction. Each input must be able to reach each output, so 2r ≥ n, which implies r ≥ lg n.
(Better to ask this kind of question on cs.stackexchange.com in the future.)
QUESTION
Wrong output! I have tried each and every condition but failed to get the real result
...I tried to accomplish this from the clrs book pseudo-code but I failed. I am trying to write merge sort using iterators to implement myself pseudo-code in c language, but for some reason, this code is compiling but the outcome is not sorted. Can someone figure out what is wrong with it? it seems perfectly fine to my untrained eyes.
ANSWER
Answered 2021-Apr-19 at 19:46Please pay attention to array bounds and sizes:
Your parameter
ris not the size of the array, but the index of the rightmost element, so you should callmerge_sort(a, 0, size - 1);.When you want to use a large sentinel value, after the actual array, you must allocate space for it, so:
QUESTION
I would like to color my pie but the #numbers doesn't work. What am I doing wrong? The code gives the pie but with other colors. I am using the package "Palmerpenguins" in R.
...ANSWER
Answered 2021-Apr-15 at 08:30Include col = clrs :
QUESTION
I would like to apply the same background color to cells that have for each PEOPLE instance the name and the related name. I have tried to df.style.applymap, it does not return an error but it does not seem to work. Anyone has any ideas why? Thank you.
...ANSWER
Answered 2021-Mar-26 at 09:22Here is some more info on df.style. Here I'm using some simple example because I don't have your data available:
QUESTION
Often in CLRS, when proving recurrences via substitution, Ө(f(n)) is replaced with cf(n).
For example,on page 91, the recurrence
T(n) = 3T(⌊n/4⌋) + Ө(n^2)
is written like so in the proof
T(n) <= 3T(⌊n/4⌋) + cn^2
But can't Ө(n^2) stand for, let's say, cn^2 + n? Would that not make such a proof invalid? Further in the proof, the statement
T(n) <= (3/16)dn^2 + cn^2
<= dn^2
is reached. But if cn^2 +n was used instead, it would instead be the following
T(n)<= (3/16)dn^2 + cn^2 + n
Can it still be proven that T(n) <= dn^2 if this is so? Do such lower order terms not matter in proving recurrences via substitution?
...ANSWER
Answered 2021-Feb-09 at 18:28Yes, it does not matter.
T(n) <= (3/16)dn^2 + cn^2 + n still less than or equal to dn^2 if n is big enough. Because as n goes to infinity, two sides of the equation have the same increasing rate (which is n^2), so the lower-order term will never matter if there is a constant number of lower-order terms in the cost function. But if there is not a constant number of them, that is a different story.
Edit: as n goes to infinity, you will find suitable d and c for T(n) <= (3/16)dn^2 + cn^2 + n to be less than or equal to dn^2, for example d = 2 and c = 1
QUESTION
I'm trying to sort the graph with different colors in the same order as the dataframe, but when I sort the values, the colors don't change.
...ANSWER
Answered 2021-Feb-05 at 23:34You need to use inplace=True to have the sorting act on the dataframe itself. Otherwise, the function returns the sorted dataframe without changing the original.
Also, you need to give the column from the sorted dataframe as the list of colors, not the original unsorted color list.
(Note that in Python strings need either single or double quotes, and commands aren't ended with a semicolon.)
QUESTION
There are two key attributes that a problem must have in order for dynamic programming to be applicable: optimal substructure and overlapping subproblems [1]. For this question, we going to focus on the latter property only.
There are various definitions for overlapping subproblems, two of which are:
- A problem is said to have overlapping subproblems if the problem can be broken down into subproblems which are reused several times OR a recursive algorithm for the problem solves the same subproblem over and over rather than always generating new subproblems [2].
- A second ingredient that an optimization problem must have for dynamic programming to apply is that the space of subproblems must be "small" in the sense that a recursive algorithm for the problem solves the same subproblems over and over, rather than always generating new subproblems (Introduction to Algorithms by CLRS)
Both definitions (and lots of others on the internet) seem to boil down to a problem having overlapping subproblems if finding its solution involves solving the same subproblems multiple times. In other words, there are many small sub-problems which are computed many times during finding the solution to the original problem. A classic example is the Fibonacci algorithm that lots of examples use to make people understand this property.
Until a couple of days ago, life was great until I discovered Kadane's algorithm which made me question the overlapping subproblems definition. This was mostly due to the fact that people have different views on whether or NOT it is a DP algorithm:
- Dynamic programming aspect in Kadane's algorithm
- Is Kadane's algorithm consider DP or not? And how to implement it recursively?
- Is Kadane's Algorithm Greedy or Optimised DP?
- Dynamic Programming vs Memoization (see my comment)
The most compelling reason why someone wouldn't consider Kadane's algorithm a DP algorithm is that each subproblem would only appear and be computed once in a recursive implementation [3], hence it doesn't entail the overlapping subproblems property. However, lots of articles on the internet consider Kadane's algorithm to be a DP algorithm, which made me question my understanding of what overlapping subproblems means in the first place.
People seem to interpret the overlapping subproblems property differently. It's easy to see it with simple problems such as the Fibonacci algorithm but things become very unclear once you introduce Kadane's algorithm for instance. I would really appreciate it if someone could offer some further explanation.
...ANSWER
Answered 2020-Oct-23 at 13:23You've read so much about this already. The only thing I have to add is this:
The overlapping subproblems in Kadane's algorithm are here:
max_subarray = max( from i=1 to n [ max_subarray_to(i) ] )
max_subarray_to(i) = max(max_subarray_to(i-1) + array[i], array[i])
As you can see, max_subarray_to() is evaluated twice for each i. Kadane's algorithm memoizes these, turning it from O(n2) to O(n)
... But as @Stef says, it doesn't matter what you call it, as long as you understand it.
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