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Size of the universe for van Emde Boas Trees |
|algorithms|data-structures|binary-trees|trees| |
This assumption makes the analysis easier. If the universe $\mathcal{U}$ is of size $2^{2^k}$ then every element in $\mathcal{U}$ can be represented by $2^k$ bits. Roughly speaking, when executing one of the vEB-tree operations you halve the "relevant bits" in every level of the recursion. So when you start with $2^k$ ... |
In order to achieve the time complexity of $O(\log \log u)$ for van Emde Boas trees I read in [this lecture][1] that the the universe size $u$ is chosen as $u = 2^{2^k}$ for some integer $k$ for van Emde Boas trees. Why choose $u$ to be of this specific form ?
[1]:http://fileadmin.cs.lth.se/cs/Personal/Rolf_Karlss... |
|algorithms|data-structures|algorithm-analysis|binary-trees|trees| |
In my variant of the assignment problem I have a set $A$ of agents and a set (of possibly different cardinality) $T$ of tasks. Each agent needs to be assigned exactly $n$ or $n+1$ tasks, and each task needs to be assigned to exactly $m$ or $m+1$ agents.
It is guaranteed this this is possible: the segment $\left[ |A|... |
I was studying operating system concepts from [Silberschatz, Galvin and Gagne's book](http://www.amazon.com/Operating-System-Concepts-Abraham-Silberschatz/dp/0471417432) (sixth edition) and I have some questions about the flow of execution of a program. A figure explains the processing of the user program:
![program... |
Program compilation and execution flow |
|compilers|operating-systems|memory-management|virtual-memory| |
The dynamic programming solution computes a table that for each $i\le j$ has the optimal solution for $Ai$, ... $Aj$. The computation is done bottom-up, and (as one of the comments notes) many values are reused. Total complexity order $n^3$ as there is a quadratic number of $i,j$ each taking linear time to compute.
... |
|algorithms|data-structures|algorithm-analysis|search-trees|trees| |
From [Van Emde Boas trees lecture][1]:
> We will use the idea of superimposing a tree of degree ${u^{1/2}}$ on top of
> a bit vector, but **shrink the universe size recursively** by a square
> root at each tree level. The ${u^{1/2}}$ items on the first level each hold
> structures of ${u^{1/4}}$ items, which hold... |
From [Van Emde Boas trees lecture][1]:
> We will use the idea of superimposing a tree of degree ${u^{1/2}}$ on top of
> a bit vector, but **shrink the universe size recursively** by a square
> root at each tree level. The ${u^{1/2}}$ items on the first level each hold
> structures of ${u^{1/4}}$ items, which hold... |
Explanation of recursive structure of Van Emde Boas Tree |
Well the crucial idea about vEB trees is the following: You store all the elements as a 0/1 bitvector of size $s$. For example the vector $(0,1,1,0)$ denotes that in your universe of size $s=4$, the third and second elements are present. Now you subdivide the vectors in blocks of size $\sqrt{s}$ and for every block you... |
null |
There is a very simple undecidable problem for finite state automata. Break the alphabet into two halves $\Sigma \cup \bar\Sigma$, where the letters in $\bar\Sigma$ are "barred" copies. Now, given a finite state automaton $A$ over $\Sigma\cup\bar\Sigma$ decide whether it accepts a string such that the unbarred part equ... |
null |
Speeding up a new floating-point unit by 2 slows down data cache accesses by 2/3 (or a 1.5 slowdown for data caches). If old FP unit took 20% of program's execution time and data cache accesses took 10% of program's execution time, what is the overall speed up?
I solved this problem using amdahl's law:
FeFP = ... |
Using Amdahl's law how do you determine execution time after an improvement? |
|computer-architecture| |
This is a homework problem for my introduction to algorithms course.
>Recall the scheduling problem from Section 4.2 in which we sought to
minimize the maximum lateness. There are $n$ jobs, each with a deadline
$d_i$ and a required processing time $t_i$, and all jobs are available to be
scheduled starting at time... |
Maximum Schedulable Set Zero-Lateness Deadline Scheduling |
|dynamic-programming| |
(Partial answer/Too long for comment)
Note that if $|T| \neq |A|$, we can introduce dummy tasks or agents carrying zero profit, and make the cardinalities equal. Let $t_d,a_d$ be respectively the number of dummy tasks and agents. Suppose now that the tasks are $T_1,\ldots T_s$ and the agents are $A_1,\ldots,A_s$. Let ... |
Since the FP speedup <i> slows down </i> the cache by a factor of $2/3$, the cache is "sped up" by $3/2$. Hence the overall speedup is
$ S = \frac{1}{0.7 + 0.2/2 + 0.1 \times \frac {3}{2}} = 1.05 \equiv 5\%$
The execution times now have the ratio: $0.7:0.1:0.15$ for remaining operations:FP:cache. The percentage t... |
null |
I have been thinking about why the dynamic programming approach to finding the optimal matrix chain order is better than a brute force approach that finds the optimal order by exploring all nested orders. The more I think about it, the more I feel strongly that the dynamic programming solution considers all possible ch... |
I have been thinking about why the dynamic programming approach to finding the optimal matrix chain order is better than a brute force approach that finds the optimal order by exploring all nested orders. The more I think about it, the more I feel strongly that the dynamic programming solution considers all possible ch... |
Speeding up a new floating-point unit by 2 slows down data cache accesses by a factor of 2/3 (or a 1.5 slowdown for data caches). If old FP unit took 20% of program's execution time and data cache accesses took 10% of program's execution time, what is the overall speed up?
I solved this problem using amdahl's law:... |
$H(x)$ is the program size complexity of $x$ for some universal Turing machine $U$.
$H$ is not computable, however $H$ is *computable in the limit from above*.
From my notes:
>i.e the set $\{(x,n) | x \in B^*,n \geq 0, H(x) \leq n\}$ is computably enumerable.
Can someone tell me what this means? I understand ... |
What does 'computable in the limit from above' mean? |
|complexity-theory|space-complexity| |
$H(x)$ is the program size complexity of $x$ for some universal Turing machine $U$.
$H$ is not computable, however $H$ is *computable in the limit from above*.
From my notes:
>i.e the set $\{(x,n) \mid x \in B^*,n \geq 0, H(x) \leq n\}$ is computably enumerable.
Can someone tell me what this means? I understa... |
|complexity-theory| |
Here is my prove,but not sure it is correct or not:
We know:
$\begin{array}{l}
\forall {c_1},\exists {n_1},0 \le f\left( n \right) \le {c_1}g\left( n \right),\forall n \ge {n_1} \\
\forall {c_1},\exists {n_2},0 \le g\left( n \right) \le {c_1}h\left( n \right),\forall n \ge {n_2} \\
\end{array}$
Hope to... |
How to prove Transitivity in small-o of Asymptotic analysis? |
|asymptotics| |
I'm a none-computer-science-student and get some knowledge on AI by taking the CS188.1x Course (Artificial Intelligence) on www.edx.org .
Currently, I am working on the "Search in Pacman" Project; the sources can be found online at [Berkley CS188][1] . I have problems finding an good solution for "Finding All the Co... |
Heuristic for Finding Multiple Goals in Graph - e.g. using Kruskals Algorithm |
|algorithms|graphs|heuristics|search-problem| |
In my variant of the assignment problem I have a set $A$ of agents and a set (of possibly different cardinality) $T$ of tasks. Each agent needs to be assigned exactly $n$ or $n+1$ tasks, and each task needs to be assigned to exactly $m$ or $m+1$ agents.
It is guaranteed this this is possible: the segment $\left[ |A|... |
null |
In my variant of the assignment problem I have a set $A$ of agents and a set (of possibly different cardinality) $T$ of tasks. Each agent needs to be assigned exactly $n$ or $n+1$ tasks, and each task needs to be assigned to exactly $m$ or $m+1$ agents.
It is guaranteed this this is possible: the segment $\left[ |A|... |
I have a problem that can be reduced to an assignment problem.
(In a previous [question][1] i found out how to do that.)
Which means we have a set $A$ of agents and a set $T$ of tasks as well as a cost function $c(i,j)$. We need to find an assignment so that the total cost is minimal.
The [hungarian algorithm][2... |
Assignment problem for multiple days |
|algorithms|graphs|assignment-problem| |
null |
I have a problem that can be reduced to an assignment problem.
(In a previous [question][1] i found out how to do that.)
Which means we have a set $A$ of agents and a set $T$ of tasks as well as a cost function $c(i,j)$. We need to find an assignment so that the total cost is minimal.
The [hungarian algorithm][2... |
>This is a problem from Interview Street in Dynamic Programming section.
https://www.interviewstreet.com/challenges/dashboard/#problem/4f2c2e3780aeb
>Billboards(20 points)
>ADZEN is a very popular advertising firm in your city. In every road you can see their advertising billboards. Recently they are facing a se... |
How can I improve my Algorithm? |
|algorithms|data-structures|dynamic-programming| |
To show that $f(n) \in o(h(n))$, you need to show that $\lim_{n\rightarrow\infty}\frac{f(n)}{h(n)}=0$. Or, in other words, for any constant $\epsilon$, there exists $N>0$ such that
$$
\forall n \geq N\colon |f(n)| \leq \epsilon|h(n)|. \qquad \text{(A)}
$$
Since we have assumed $f(n) \in o(g(n))$, we know that for a... |
This can be formulated as an instance of min-cost (or in this case, max-profit) flow.
Set up a network as follows. There will be four layers.
The first layer is a single node we call the source. The next layer consists of a node for each agent. The next layer has a node for each task. The final layer is one node ... |
This can be formulated as an instance of min-cost (or in this case, max-profit) flow.
Set up a network as follows. There will be four layers.
The first layer is a single node we call the source. The next layer consists of a node for each agent. The next layer has a node for each task. The final layer is one node ... |
This can be formulated as an instance of min-cost (or in this case, max-profit) flow.
Set up a network as follows. There will be four layers.
The first layer is a single node we call the source. The next layer consists of a node for each agent. The next layer has a node for each task. The final layer is one node ... |
null |
null |
To quote [Wikipedia][1]/[Scholarpedia][2],
> A partial function $f$ from the rational numbers to the real numbers is
> **upper semicomputable** if it is defined by a rational-valued partial
> computable function $\phi(x,k)$ with $x$ a rational number and $k$ a
> nonnegative integer such that $\phi(x,k+1)≤ϕ(x,k)$... |
There's a way to this in polynomial time. I'll sketch the algorithm (in reverse order ... do step 2 first and step 1 second).
1. if we can find a set of $nk$ agent-task pairs $(i,j)$ such that each task is in exactly $k$ pairs, each agent is in exactly $k$ pairs, and no pair appears more than once, then we can fi... |
Least cost travel by intermixing different airline routes having
linear discount functions:
Lowest air fare route chosen by mixing different routes provided
by different airline having different discount functions(like
some airline can give 25 % discount if fare crosses $5k) so that
total cost of tra... |
Proving following problem NP Hard using known NP Hard partition problem |
|complexity-theory| |
One question we could ask is "Can we reduce this back to the subset sum problem?" In this case, the answer is **yes**: for each duplicate $z$ we replace it with two numbers $x$ and $y$ such that $x+y=z$.
$$[-1,-1,2,3]\to [-7,-1,2,3,6]$$
However, we need to be careful that we don't introduce additional solutions (t... |
Least cost travel by intermixing different airline routes having linear discount functions:
Lowest air fare route chosen by mixing different routes provided by different airline having different discount functions(like some airline can give 25 % discount if fare crosses $5k) so that total cost of travel is minimized... |
|complexity-theory|graphs| |
**No**. The easiest way to prove this is by counterexample. Expanding [matching pennies][1], we can create a game with two mixed strategies (players either mix between A and B or between C and D):
![Dual Matching Pennies][2]
What *might* be true about that statement is for a mix of particular pure strategies for ... |
null |
From [Wikipedia][1] about randomized algorithms
> One has to distinguish between **algorithms** that use the random
> input to reduce the expected running time or memory usage, but always
> terminate with a correct result in a bounded amount of time, and
> **probabilistic algorithms**, which, depending on the ran... |
Classfication of randomized algorithms |
|randomized-algorithms| |
From [Wikipedia][1] about randomized algorithms
> One has to distinguish between **algorithms** that use the random
> input to reduce the expected running time or memory usage, but always
> terminate with a correct result in a bounded amount of time, and
> **probabilistic algorithms**, which, depending on the ran... |
|algorithms|terminology|randomized-algorithms|nondeterminism|machine-models| |
1) An example of such an algorithm is randomized Quick Sort, where you randomly permute the list or randomly pick the pivot value, then use Quick Sort as normal. Quick Sort has a worst case running time of $O(n^{2})$, but on a random list has an expected running time of $O(n\log n)$, so it always terminates after $O(n^... |
null |
Given an adjacency matrix $A_G$ of an undirected graph $G$, it is easy and straightforward to compute the characteristic polynomial $\chi_G(\lambda)$. What about the other way around? The problem can be formulated as follows.
> **Problem** Given a polynomial $P$, decide whether there is a graph $G$ with the correspo... |
This is a homework problem for my introduction to algorithms course.
>Recall the scheduling problem from Section 4.2 in which we sought to
minimize the maximum lateness. There are $n$ jobs, each with a deadline
$d_i$ and a required processing time $t_i$, and all jobs are available to be
scheduled starting at time... |
Questions on space-efficient probabilistic data structures known as Bloom filters. In membership testing, false positives are possible but false negatives are not. |
null |
Questions on multigraphs, which are graphs with multiple edges allowed. |
null |
Questions on the Master theorem, a method for obtaining asymptotic bounds on recurrences of a specific form. |
null |
Least cost travel by intermixing different airline routes having linear discount functions:
Lowest air fare route chosen by mixing different routes provided by different airline having different discount functions (like some airline can give 25% discount if fare crosses $5k) so that total cost of travel is minimized... |
Assuming that an "airport" cannot be visited more than once, you can reduce HAMILTONIAN PATH to your problem: given a graph $G = (V,E)$, $|V|=n$; suppose that there are two airlines such that $E(1)=E(2)=E$ (i.e. both airlines cover all edges). Then set the price of each edge = $1$. One of the two companies gives no dis... |
I know the general concept of recursion . I came across the concept of tail recursion while studying up on Quick sort algorithm . But I am not clear in my mind yet yet as to what is meant by tail recursion . Can some one explain to me the concept with a proper example ? |
What is tail recursion? |
|algorithms|data-structures|recurrence-relation|recursion| |
From [Van Emde Boas trees lecture][1]:
> We will use the idea of superimposing a tree of degree ${u^{1/2}}$ on top of
> a bit vector, but **shrink the universe size recursively** by a square
> root at each tree level. The ${u^{1/2}}$ items on the first level each hold
> structures of ${u^{1/4}}$ items, which hold... |
I know the general concept of recursion. I came across the concept of **tail recursion** while studying the quicksort algorithm. I am not clear in my mind, what is meant by tail recursion. Can someone explain the concept with a proper example? |
|algorithms|reference-request|recursion| |
I know the general concept of recursion. I came across the concept of **tail recursion** while studying the quicksort algorithm.However I was through this [video of quick sort algorithm from MIT][1] and at 18:30 seconds the professor says that this is tail recursive algorithm. I am not clear in my mind, what is meant ... |
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