I have tried to measure the performance counters using perf for different sampling periods and for this I used the -I option. But this measures the sampling interval in milliseconds.
Is it possible to use with perf a sampling interval measured in cycles so that we get the information about performance counters every X number of cycles (i.e. collect information periodically)?
Related
I have read many explanations of amortized analysis and how it differs from average-case analysis. However, I have not found a single explanation that showed how, for a particular example for which both kinds of analysis are sensible, the two would give asymptotically different results.
The most wide-spread example of amortized running time analysis shows that appending an element to a dynamic array takes O(1) amortized time (where the running time of the operation is O(n) if the array's length is an exact power of 2, and O(1) otherwise). I believe that, if we consider all array lengths equally likely, then the average-case analysis will give the same O(1) answer.
So, could you please provide an example to show that amortized analysis and average-case analysis may give asymptotically different results?
Consider a dynamic array supporting push and pop from the end. In this example, the array capacity will double when push is called on a full array and halve when pop leaves the array size 1/2 of the capacity. pop on an empty array does nothing.
Note that this is not how dynamic arrays are "supposed" to work. To maintain O(1) amortized complexity, the array capacity should only halve when the size is alpha times the capacity, for alpha < 1/2.
In the bad dynamic array, when considering both operations, neither has O(1) amortized complexity, because alternating between them when the capacity is near 2x the size can produce Ω(n) time complexity for both operations repeatedly.
However, if you consider all sequences of push and pop to be equally likely, both operations have O(1) average time complexity, for two reasons:
First, since the sequences are random, I believe the size of the array will mostly be O(1). This is a random walk on the natural numbers.
Second, the array will be near size a power of 2 only rarely.
This shows an example where amortized complexity is strictly greater than average complexity.
They never have different asymptotically different results. average-case means that weird data might not trigger the average case and might be slower. asymptotic analysis means that even weird data will have the same performance. But on average they'll always have the same complexity.
Where they differ is the worst-case analysis. For algorithms where slowdowns come every few items regardless of their values, then the worst-case and the average-case are the same, and we often call this "asymptotic analysis". For algorithms that can have slowdowns based on the data itself, the worst-case and average-case are different, and we do not call either "asymptotic".
In "Pairing Heaps with Costless Meld", the author gives a priority queue with O(0) time per meld. Obviously, the average time per meld is greater than that.
Consider any data structure with worst-case and best-case inserts and removes taking I and R time. Now use the physicist's argument and give the structure a potential of nR, where n is the number of values in the structure. Each insert increases the potential by R, so the total amortized cost of an insert is I+R. However, each remove decreases the potential by R. Thus, each removal has an amortized cost of R-R=0!
The average cost is R; the amortized cost is 0; these are different.
Can someone explain what is the meaning of the time complexity in distributed networking algorithms? The definition given in DNA book by Panduranga is as follow :
"In the synchronous model, time is measured by the number of clock ticks called rounds, i.e., processors are said to compute in “lock step”. When running a distributed algorithm, different nodes might take a different number of rounds to finish. In that case, the maximum time needed over all nodes is taken as the time complexity"
Can you explain the above definition with a simple example
Let's say you want to compute the sum of a really large list (say, 1 billion numbers). To speed things up, you use 4 threads, each computing the sum of 250 million rows, which can then be added to get the total sum. If the time taken for each thread to run is:
thread1 takes 43 seconds
thread2 takes 39 seconds
thread3 takes 40 seconds
thread4 takes 41 seconds
Then you would say that the runtime of this operation is bounded by the thread that takes the longest, in this case 43 seconds. It doesn't matter if the other threads take 2 seconds, the longest task determines the runtime of your algorithm.
I am trying to explore genetic algorithms (GA) for the bin packing problem, and compare it to classical Any-Fit algorithms. However the time complexity for GA is never mentioned in any of the scholarly articles. Is this because the time complexity is very high? and that the main goal of a GA is to find the best solution without considering the time? What is the time complexity of a basic GA?
Assuming that termination condition is number of iterations and it's constant then in general it would look something like that:
O(p * Cp * O(Crossover) * Mp * O(Mutation) * O(Fitness))
p - population size
Cp - crossover probability
Mp - mutation probability
As you can see it not only depends on parameters like eg. population size but also on implementation of crossover, mutation operations and fitness function implementation. In practice there would be more parameters like for example chromosome size etc.
You don't see much about time complexity in publications because researchers most of the time compare GA using convergence time.
Edit Convergence Time
Every GA has some kind of a termination condition and usually it's convergence criteria. Let's assume that we want to find the minimum of a mathematical function so our convergence criteria will be the function's value. In short we reach convergence during optimization when it's no longer worth it to continue optimization because our best individual doesn't get better significantly. Take a look at this chart:
You can see that after around 10000 iterations fitness doesn't improve much and the line is getting flat. Best case scenario reaches convergence at around 9500 iterations, after that point we don't observe any improvement or it's insignificantly small. Assuming that each line shows different GA then Best case has the best convergence time becuase it reaches convergence criteria first.
I have an objective function having parameters of power consumption (p) and latency (d). I want to minimize the power consumption given a latency constraint (seconds). The optimization problem can be expressed in terms of Lagrange function as follows:
f(p,d) = p + L*d
Where L is Lagrange variable. Since power consumption and latency are inversely proportional to each other and decreasing the former results in increasing the later, the objective function can also be written in terms of relative weights as:
f(p,d) = L*p + (1-L)*d
The questions is, "given a latency constraint of d seconds, how do I find an appropriate value of L that can minimize the variable p?". I want to use reinforcement learning for this purpose, where at each state, the system takes a decision and assigns a cost to the previous action in next state in terms of the above function. Every action results in certain power consumption and latency in processing the requests. The goal is to minimize the power consumption given a latency constraint. Any suggestions/hints in this respect will be highly appreciated.
I'v got some problem to understand the difference between Logarithmic(Lcc) and Uniform(Ucc) cost criteria and also how to use it in calculations.
Could someone please explain the difference between the two and perhaps show how to calculate the complexity for a problem like A+B*C
(Yes this is part of an assignment =) )
Thx for any help!
/Marthin
Uniform Cost Criteria assigns a constant cost to every machine operation regardless of the number of bits involved WHILE Logarithm Cost Criteria assigns a cost to every machine operation proportional to the number of bits involved
Problem size influence complexity
Since complexity depends on the size of the
problem we define complexity to be a function
of problem size
Definition: Let T(n) denote the complexity for
an algorithm that is applied to a problem of
size n.
The size (n) of a problem instance (I) is the
number of (binary) bits used to represent the
instance. So problem size is the length of the
binary description of the instance.
This is called Logarithmic cost criteria
Unit Cost Criteria
If you assume that:
- every computer instruction takes one time
unit,
- every register is one storage unit
- and that a number always fits in a register
then you can use the number of inputs as
problem size since the length of input (in bits)
will be a constant times the number of inputs.
Uniform cost criteria assume that every instruction takes a single unit of time and that every register requires a single unit of space.
Logarithmic cost criteria assume that every instruction takes a logarithmic number of time units (with respect to the length of the operands) and that every register requires a logarithmic number of units of space.
In simpler terms, what this means is that uniform cost criteria count the number of operations, and logarithmic cost criteria count the number of bit operations.
For example, suppose we have an 8-bit adder.
If we're using uniform cost criteria to analyze the run-time of the adder, we would say that addition takes a single time unit; i.e., T(N)=1.
If we're using logarithmic cost criteria to analyze the run-time of the adder, we would say that addition takes lgn time units; i.e., T(N)=lgn, where n is the worst case number you would have to add in terms of time complexity (in this example, n would be 256). Thus, T(N)=8.
More specifically, say we're adding 256 to 32. To perform the addition, we have to add the binary bits together in the 1s column, the 2s column, the 4s column, etc (columns meaning the bit locations). The number 256 requires 8 bits. This is where logarithms come into our analysis. lg256=8. So to add the two numbers, we have to perform addition on 8 columns. Logarithmic cost criteria say that each of these 8 addition calculations takes a single unit of time. Uniform cost criteria say that the entire set of 8 addition calculations takes a single unit of time.
Similar analysis can be made in terms of space as well. Registers either take up a constant amount of space (under uniform cost criteria) or a logarithmic amount of space (under uniform cost criteria).
I think you should do some research on Big O notation... http://en.wikipedia.org/wiki/Big_O_notation#Orders_of_common_functions
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