Migraine Problem is Common

Migraine: What your doctor doesn’t know

The increase in facilities and luxury often leads to inactivity of the body. The inactivity might lead to major problems in the body. One such Migraine Problem is Common. Headache can occur due to numerous reasons such as food poisoning, gas problems and other eating habit problems. However, it is very hard to reach the diagnosis of migraine because of the various reasons that causes severe headache. The usually is considered as Migraine Common Headache and it occurs in once in every five females and once in every 15 males. However, there is no specific treatment for migraine and only eating habits can help in curing the issue.

Migraine Problem is common

Migraine pain can be considered as common because many times it looks like regular headache. There are two type of migraines – one that occurs with aura and other is without aura. Both these migraines might show some different effects. The regular Symptoms of Migraine are nauseas, numbness, speaking difficulty, vomiting and sensitivity to light and sound. The main reason for migraine will remain unknown because it can occur due to family history also. Thus, the person will have to rely on the sources that their ancestor used to cure the problem. Healthy eating and healthy life will help in solving the problem but at some point you will need to take up medication.

Migraine in women

It has been observes by many researchers that migraine is more prominent in women than in men. However, in the younger years, boys suffer more than the girls but as the puberty hits the ratio inverses. The main reason for migraine in females is the estrogen level. The hormonal changes that occur in the body of the female after their first period remarks the beginning of migraine. The women often suffer from severe pain in the head which might require rest and soothing time to reduce the pain. However, with the beginning of 40s, the pain will subside. It has been observes that proper source of food and nutrients can reduce the pain of Migraine Problem is Common but the person will have to maintain an easy lifestyle.
SEMrush

Migraine awareness challenge

There is no particular test or signs that would show that the person is suffering from migraine. The usual course of action adopted by the doctors is the ruling out method. They will analyze the brain for all sorts of diseases including cancer and start ruling out the problems one by one. The ruling out method will help in understanding the main cause of the headaches. The headaches can be severe tension headaches or the ones that subsides with time or aura. These are the general symptoms for many diseases. So, it is important to create migraine awareness for the people. So, the basic Migraine Awareness will include the basics of the disease and how to identify the dominating symptoms in shorter time to reduce the chances of the disease.
Treatment for Migraine

There is no specific Treatment for Migraine as there is no specific test to identify migraine. There can be many reasons for migraine and it is important to find out the core reason of the disease. Usually, the general treatment of the disease will include pain killers that will reduce the headache. However, apart from that, the person needs to do some diet modifications that can impart instant relief from the pain. The basic modifications would include healthy food and getting proper sleep. However, therapy would help in reducing the stress which can also lead to reduction in pain. The treatment of migraine is purely based on trial and error. It would help, if the person observes the pattern through which they are suffering from maximum pain. If the source of the pain is known then they would be able to reduce the pain.

Thus, migraine is a disease that occurs purely due to theoretical reasons and there are no specific symptoms, tests or treatment. The person Suffering from Migraine has only two options – adopt the methods from their ancestors or observe their body to understand the patterns. Thus, this will help in reducing the problem of pain to a greater extent without any issues in the body.

 

linear algebra – Common basis property

Let F be a field and let $U_1, ldots, U_k$ be subspaces of $F^n$. Say that $U_1, ldots, U_k$ has a common basis if there is a basis $b_1,ldots,b_n$ of $F^n$ such that each $U_i$ is a span of some subset of ${b_1,ldots,b_n}$. Suppose $V_1,ldots,V_p$ and $W_1,ldots,W_q$ are subspaces such that:

  1. For all $i$, $V_1,ldots,V_p,W_i$ has a common basis.

  2. $W_1 < W_2< ldots < W_q$.

Is it true that $V_1,ldots,V_p, W_1,ldots,W_q$ has a common basis?

europe – Is it common to travel by giving private lessons at host family?

So, a while back I had this idea, to make travelling a bit less expensive, by giving private lessons and tutoring in target country.

In summary, I travel to a target country, stay with some family, they provide food and place to sleep and in exchange I give private lessons, probably to the kids, probably in some school subject.

My question is this.
Is such a thing even heard of, and if yes, how to begin?

Did You hear of someone who did something like this, how did it turn out?

Maybe there is some website that will help or people use Couchsurfing website to do it?

Note: I’m asking mostly about Central Europe, but feel free to say anything You know.

postgresql – Is there a common pattern for storing many boolean flags other than adding a ton of columns?

I have a bunch of booleans I need to associate with user accounts. They need to be toggled true and false, though not particularly frequently – they’re going to be read more often than they’re written. Most commonly they’ll just be set from False to True once, but that won’t always be the case.

Right now I’m thinking of appending columns to the user table like this:

UserID other_user_cols did_a did_b did_c
1 False True False
2 False False False
3 True False False

The thing is, there’s quite a lot of these boolean columns.

Let’s say you needed to add like, 100 columns. Is this really still the best way to do it?

Assume:

  • It’s not going to be very sparse – most users will toggle most booleans at some point
  • The booleans aren’t logically related in a way that could naturally be better organized
  • New ones may need to be added from time to time

I feel this must be a pretty common pattern, and yet this way of just jamming more boolean columns into my table seems rather messy. Is there a more typical way of doing it that’s a bit cleaner? Or is this actually the best way?

I’m using Postgres for this.

sharepoint online – Is it possible to have multiple rows in a list where some fields are duplicate to save user from having to complete a common field?

Sharepoint in Office 365

ISSUE
Form has 10 fields a user needs to complete before they can submit. 8 of the 10 form fields are unique whereas the remaining 2 are just their userID and dueDate (NOTE: userID is not the submitter’s userID). In some cases a user may have 20 records to submit. When the user has one record to enter, it’s not a big deal but when there are 20 records to enter, the time it takes waiting for the page to load as well as repeating the userID and dueDate is a long time to the user.

REQUEST
I’m looking for a more efficient way for a user to submit data such as implementing a “duplicate” button so the user only needs open the form once, enter the data for the 10 form fields, click duplicate and the row is duplicated but only the userID and the dueDate is copied from the source row, the other fields should be blank).

REQUIREMENTS
A user has 5 records they need to submit, all with the same userID and dueDate. The user should be able to enter the record for the first item, click a “duplicate” button. The button would duplicate the form fields (not the data, just the form field for the first 8 form fields but should include the userID and the dueDate of the last two form fields). Our form runs horizontally rather than vertically for reference so the concept of rows is accurate.

SCENARIO
There are real-world scenarios in which the user may have to enter 20+ items and this duplicate button would help the users A) not have to enter every field every time, B) accuracy for userID and dueDates, and B) would save the user time not having to wait for the page to load for every record they need to submit the data.

EXPERIENCE
I have zero experience with Sharepoint but I do have experience with JavaScript, jQuery, JSON, etc, and various other web development technologies which is why I was asked to assist with this endeavor.

Common approach to get a node’s old value when modifying or deleting node in a graph?

My application contains a graph of nodes. When user modifies a node, the change may propagate to other nodes. For each nodes that are changed directly or indrectly, an event handler is triggered. The code in the event handler needs to access the node’s old property value (the value before its changed). I wonder what’s the typical and robust way to implement it?

A straightforward approach may be determining the nodes to be changed (both directly and indirectly) and sending an event, say, “willChange” to them to notiffy them to save current value. But that requires the app to determine ahead which nodes will be changed, which is specific to application logic. I didn’t try this approach.

My current approach is saving the entire graph before making change, then accessing the counterpart node in the backed graph to get old value, as below:

-> backup the entire graph
-> make change to the active graph
   -> event handler is triggered on a node
      -> retrieve the node's old value from the backed up graph

This works, but has one limietaion. For example, the following doesn’t work because when we make the second change, the node value saved in the backed up graph is not up-to-date.

-> backup the entire graph
-> make a first change to the active graph
   -> event handler is triggered on a node
      -> retrieve the node's old value from the backed up graph
-> make a second change to the active graph
   -> event handler is triggered on a node
      -> retrieve the node's old value from the backed up graph

At first sight, the issue can be easily resolved by simply backing up the graph again before making the second change:

-> backup the entire graph
-> make a first change to the active graph
   -> event handler is triggered on a node
      -> retrieve the node's old value from the backed up graph
-> backup the entire graph
-> make a second change to the active graph
   -> event handler is triggered on a node
      -> retrieve the node's old value from the backed up graph

But it’s actually not that simple, because changes may be nested, like the following:

-> backup the entire graph
-> make a first change to the active graph
   -> event handler is triggered on a node
      -> retrieve the node's old value from the backed up graph
      -> make a second/nested change to the active graph
         -> event handler is triggered on a node
            -> retrieve the node's old value from the backed up graph

I resolve the issue in my current application with two approaches:

  1. Backup the entire graph again before the seond/nested change. This works. But backing up entire graph takes cpu time. So I suppose I should avoid doing this very often. Note that my application is event driven, so the event handler trigged in the second change may make a third change and this goes on.

  2. Design application code carefully to avoid the above nested call. This works, but it’s not a general solution and imposes restriction on how to design the code.

This is the first time I work on an application based on a graph of nodes. I didn’t have experience in this area before. I have made a long progress (the current application works fine) but I can’t help thinking if there is any better way to deal with the issue? I wonder what’s the common way to get a node’s old value when modifying or deleting node in a graph? Thanks.

common criteria – Can a software application / OS get past EAL4 if no source code is available

The EAL for Common Criteria are described briefly as:

EAL1: Functionally Tested. ...
EAL2: Structurally Tested. ...
EAL3: Methodically Tested and Checked. ...
EAL4: Methodically Designed, Tested, and Reviewed. ...
EAL5: Semi-Formally Designed and Tested. ...
EAL6: Semi-Formally Verified Design and Tested. ...
EAL7: Formally Verified Design and Tested.

Does “Formally Verified Design” refer to the source code undergoing static code analysis? Thus meaning that the source code would have to be disclosed to gain EAL > 4?

algorithms – Find non common numbers in two arrays

Given two arrays of integers, please write a function that returns all elements present in one of the two arrays but not both. E.g. f(( 1, 3, 5 ), ( 1, 2, 4, 5 )) -> ( 2, 3, 4 )

I know I can do this with two for loops that is O(n^2) but is there any better way ? Please help

How to list the common elements of two lists?

The elements of both lists are given as

A={9,5,9,3}, B={9,9,5,5}

In this case, how to list the list {9,5,9} of the common elements of the two lists?

(The order does not matter.)

search algorithms – Finding 2 nodes which sum equals twice their common ancestor in RBT in $Theta(nlg n)$

I have a red black tree, $T$, and I need to write an algorithm to find 2 nodes $x$ and $y$ so that $key(x) + key(y) = 2 cdot key(p(x, y))$, where $p(x, y)$ is the lowest common ancestor of $x$ and $y$, i.e $x$ is a node on $p(x, y)$‘s left subtree and $y$ a node on the right subtree (or the other way around)

My original solution:

FIND-XY(root)
 1. x <- TREE-MIN(root)
 2. y <- TREE-MAX(root)
 3. sum <- x + y
 4. while x != root and y != root and sum != 2 * key(root)
 5.   if sum < 2 * key(root(T))
 6.    x <- TREE-SUCCESSOR(x)
 7.   else
 8.    y <- TREE-PREDECESSOR(y)
 9.   sum <- x + y
 10. if sum = 2 * key(root)
 11.   return x, y
 12. return null

then for every node in the tree that isn’t a leaf I will run FIND-XY with himself as root. FIND-XY runs in $Theta(n)$ but there are $O(n)$ nodes that aren’t leaves so the total run time is $O(n ^ 2)$

I know that there are $O(n)$ nodes that can be $p(x, y)$ so if I can change FIND-XY to run in $O(lg n)$ then my original algorithm will run in $O(nlg n)$, but this seems unlikely as finding these $x$, $y$ for the entire tree requires traversing in-order on all $n$ nodes in the tree which is $O(n)$