Optimization – Can all types of computing problems be modeled as decision problems?

Can all types of computing problems (searching, counting, optimizing …) be modeled as (sets of) decision problems? Rewritten: Are there any set of decision problems for any type of computing problem where you get the result of the original problem if you solve those problems?

Colloquially: Can all kinds of questions you can ask a string be answered by (a series of) results of YES? NO questions … So the game is: you get a question about a string and can only ask the string YES NO questions and it answers the predictable. Can you compile all the answers you get to narrow down a SINGLE answer to the original question (no distribution of answers / this result, but with some probability)?

Decision problems can be modeled as word-to-speech membership. On the surface, this only models the functions that start from f: {0,1}.-> 1 not all other types of functions. Some f: {0,1}-> {0,1} for example. But I have the feeling that the answer could be yes …

Does anyone have a clue or evidence?

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How do I convert a decision table into a consistent decision tree?

I work from the school presentation, but I can't understand how they go from one step to another. This is the decision table that I have.

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First we have to select the root node from these nodes:

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But I don't understand why, for example, from Q1 the Y branch has a1, a2 and the N branch a1, a3.

This is the final solution for the tree:

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I'm not sure if this is the correct name in English. Maybe that was the reason why I couldn't find any literature online.

I would appreciate an explanation, or at least a link to a guide to converting decision tables to optimal decision trees that match the decision table.

What is important for forex trading before you make this decision – discussions and help

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machine learning – selection of the decision tree according to error rate (what does this mean?)

I'm asked to choose an attribute based on its error rate

Could someone please explain this to me?

For example, suppose I have this data for reasons

For an attribute

3 Yes, that is classified as +

1 Yes, classified as –

4 No, classified as –

2 No, classified as +

What is the error rate?

uk – Received passport from VFS without a decision email

I had applied for a British visitor visa with my wife. I received a text message from the VFS office today that my passport had been received, but I did not receive an email from UKVI Chennai that the decision had been made. I received the message about 10-12 days ago that your application has been received and is being processed.

Yes, I will find out what happened when I have my passport, but the tension kills me. Did someone have the same experience when they did not receive an email with the decision but received a passport directly?

Help with the decision between 2 Hetzner servers with old hard drives

Hi there,

I have been running a Hetzner server for several years. Technical data: Intel Xeon E3-1245V2, 2x hard disk 3.0 TB SATA Enterprise (in software RAID 0), 2x RAM 8192 MB DDR3 ECC and an Intel network (82574L).
I've had some serious operating system issues lately and decided to do a full reinstall. Since I didn't create separate partitions, I decided to upgrade to another auction server to transfer my data (~ 5 TB).

My new server has the same specifications as well as a hardware RAID card (LSI MegaRAID SAS 9260-4i).

The first thing I did was a quick, clever test. One of the drives has approximately 60,000 operating hours and the other approximately 35,000.
I have set up the server in hardware RAID 0. I have activated the write-back cache, but the write speed is actually somewhat slower on the new server. Although I have activated the write-back cache without BBU, "megacli -AdpAllInfo -aALL" still outputs the following: "Cache on BBU error: Deactivated.
I understand that I need to set this parameter when setting up RAID 0 for it to work.

Given these problems and the fact that I basically have no experience with hardware raid, I wonder if I should just reinstall the old server and transfer everything back. The drives on my other server are also very old, but I haven't noticed any problems with them.

Intelligent protocols:

New server: hdd1: pastebin.com/BDMuxgrg hdd2: pastebin.com/2M6hNw6B

Old server: hdd1: pastebin.com/Jq2khhy0 hdd2: pastebin.com/yPrCnykF

The invoice date on my old server is the 31st, so until then (or maybe the day before) I made the decision whether or not to cancel it without having to pay for both.

Thank you very much

nt.number Theory – Does finding positive integer solutions of $ zeta (a / b) = c $ correspond to the decision about the rationality of $ gamma $?

This question requires a little background explanation and is therefore a bit lengthy. Note: The question was originally posted in MSE but received no replies, so it was posted in MO.

For every integer $ n ge 2 $ there is a unique real $ 1 <x_n <$ 1.73 so the Riemann-Nzeta function $ zeta (x_n) = n $, example
$$ zeta (1.72865 ..) = 2, text {} zeta (1.41785 ..) = 3, text {} zeta (1.29396 ..) = 4, …
$$

It is likely that everyone $ x_n $ are irrational, but that's an open problem. I have developed a methodology to achieve some partial results in this direction.

If a reasonable one $ x_n> 1 $ then exists, WLOG, let $ x_n = 1 + dfrac {a_n} {b_n} = 1 + dfrac {1} {i_n + f_n} $ Where $ i_n $ is the integer part of the irreducible part $ b_n / a_n $ and $ f_n $ is its fraction. We get from the expansion of the Riemann zeta function in the Stieltjes series

$$
f_2 approx. 0.37241 le f_n le f_ {n + 1}
<1 -? about 0.422785
$$

Where $ gamma $ is the Euler-Mascheroni constant. $ f_n $ Strictly increasing it follows that $ zeta (x) $ decreases sharply in our region of interest $ x ge x_2 $,

So if $ zeta (1 + a_n / b_n) = n $ then we can write $ b_n = ka_n + r_n $ for some integers $ k $ and $ r_n $ so that $ gcd (r_n, a_n) = 1 $ and $ 0.37241 < dfrac {r_n} {a_n} <0.422789 $,

Now for everyone $ a_n ge 1 $ we can eliminate the possible values ​​of $ r_n $ through computer verification to increase the lower limit $ a_n $ As shown below.

Example: Prove $ a_n ge $ 6 to the $ n ge 2 $,

To the $ a_n = $ 1,2,3,4, there is no $ r_n $ satisfying $ 0.37241 < dfrac {r_n} {a_n} <0.422789 $ So these values ​​of $ a_n $ be eliminated. To the $ a_n = 5 $ there is a possible value $ r_n = 2 $ which satisfies both $ gcd (r_n, a_n) = 1 $ and $ 0.37241 < dfrac {r_n} {a_n} <0.422789 $, So we take $ r_n = 2 $ for further testing. We watch that

$$ f_3 approx.0.3932265 < frac {2} {5} <f_4 approx.0.4018059 $$

Since $ dfrac {2} {5} $ lies between two successive values ​​of $ f_3 $ and $ f_4 $ is getting bigger, we'll never have a situation where $ f_n = dfrac {2} {5} $ for each $ n $, Therefore $ r_n = 2 $ is excluded for $ a_n = 5 $, The whole numbers $ 3 $ and $ 4 $ are Witnesses for the elimination of $ dfrac {2} {5} $, With this we have exhausted all possibilities $ a_n = 5 $ and so we can conclude if $ zeta (x_n) in N $ then $ a_n ge $ 6, We can also eliminate them $ a_n = 6.7, ldots $ and so on.

Note: How $ a_n $ increases the number of possible $ r_n $ increases. Everyone $ r_n $ must be removed individually with a witness to completely eliminate $ a_n $, So after a point $ a_n $ will have several witnesses.

definition: A witness for a particular integer pair $ (a, r), a> r $ is an integer $ w_ {a, r} $ so that $ f_ {w_ {a, r}} < dfrac {r} {a} <f_ {1 + w_ {a, r}} $

The existence of a witness $ w_ {a, r} $ is therefore sufficient to prove this $ zeta Big (1 + dfrac {a} {x + r} Big) $ is not an integer for an integer $ x> a> r $, Through actual calculation, I found the witness for every allowed pair $ (a, r), a le 10 ^ 4 $ what that implies

If $ x_n> 1 $ is rational and $ zeta (x_n) in N $ then the difference between
the numerator and denominator of $ x_n $ is bigger than $ 10 ^ 4 $,

Size of the witness

I noticed that most integers have small witnesses, but some of them have big witnesses. I looked at the list of integers $ a $ whose greatest witness for some $ r $ is greater than all witnesses of all integers less than $ a $, Let us call them the maximum witnesses. The first values ​​of the maximum witnesses are given below:

a    Max of w_{a,r}
5    3
12   12
19   42
45   130
30   292
97   701
123  3621
518  16503
913  31689
1308 49858
1703 71984
2098 99521
2493 134724
2888 181317
3283 245888
3678 341249
4073 496221
4468 790559
4863 1598102
5258 11274164

Some of the values $ a = $ 5.19,123.5258 are the non-consecutive denominators in the further group expansion of $ gamma $ but generally each of these values ​​of $ a $ are the denominators in ascending order for the best rational approximation $ gamma $, In other words, the closer it is $ dfrac {r} {a} $ to $ 1- gamma $, the bigger his testimony $ w_ {a, r} $ that eliminates mathematically $ a $ more difficult.

I finally come to my question.

question: The problem is finding positive integer solutions $ zeta (p / q) = u $ synonymous with the decision about the rationality of $ gamma $,

Related question

Decision tree: How do you decide the next node?

record

I have to decide which value of "class"

How do I do it?

I know that I have to opt for maximum information gain

So first I calculated the entropy of "class"

That means E (class) = – (3/11 * log (3/11) + 3/11 * log (3/11) + 5/11 * log (5/11)) = 1.067

How do I assume?

I have to find the first decision node now, right?

And what do I do when I find a decision node?

Thank you very much