## Ubuntu thinking directory size is larger than it is duting copy

I have just mounted a 1TB hdd (using Disks utility) at media/username/drivename. It contains a folder folder that is ~140GB in size as per

1. df command applied to the media/username/drivename/ which shows 138299660KB used and 773225836KB available.

2. right clicking the folder and seeing its properties which shows 791.8Gb free with folder being 140.7GB

3. using Disk Usage Analyzer utility which shows 841.8GB of 983.4GB available

Inspite of all this when I try to copy a file from the hdd to Desktop(in the boot drive ssd with ~170gb available) the system says that there isn’t enough memory in ssd

273.2 GB more space is required to copy to the destination.

This message is displayed when the “Preparing” progress bar(at the start of copy) is showing

Preparing to copy 2859633 files(438.5GB)

i.e its gauging folder‘s size to be ~711Gb.

Can anyone explain whats happening?

Env:

Ubuntu 18.04.2 LTS
256Gb sdd, available ~251GB Total(As per Disk Usage Analyzer)
1TB hdd, available ~983GB Total(As per Disk Usage Analyzer)

PS:
The ultimate goal is to make a 700Gb partition on the hdd to install CentOS and then copy folder back into the remaining space where it will keep getting used by the backup software running on ssd’s Ubuntu.

## Ubuntu 18.04.02, default (Gnome) spin, live session: am I right in thinking one can’t change theme, fonts?

I’m trying out the default (Gnome) spin of Ubuntu and I’m liking it a lot so far, except that it seems I can’t change the theme or the fonts. Now, Gnome lockdown hasn’t gone so far that I will not be able to change those things if I actually install Ubuntu – right?

## Here’s how I’m thinking about these DT models.

The Stanford d.school describes the Design Thinking method in (5) steps: Empathize, Define, Ideate, Prototype, and Test. IDEO describes it in (3) modes that are often but not always sequential: Inspiration, Ideation, and Implementation.

I’m assuming that these models describe the same set of activities. For example, user interviews in the initial research phase are part of the d.school’s Empathize step and IDEO’s “Inspiration” mode.

## So here’s my question.

What I’m wondering is where the lines are drawn (or if they’re inherently blurry). For example, does Prototype map onto Ideation, Implementation, or both? Or is my assumption am I wrong, and these models are orthogonal rather than overlapping? And if so, what is the relationship between them?

## Here’s what the models look like.

### The d.school’s model:

(source: kashifzaman.com)

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## Design Thinking: User testing approaches

I am in the process of starting to facilitate design thinking workshop sessions and from past experience I feel fairly comfortable with each step except for the user testing phase. I realize this phase would largely be dictated by the preceding phases (eg. empathy/research/define/ideate/prototype) but I’m struggling with the basic rules of engagement and logistics around the testing phase in Design thinking. For example:

1. Say we talk to users during the empathy phase in the morning, do we have them come back later in the day to test our prototypes? Maybe next day if its a 2 day session? That’s a lot of organizing/logistical effort.
2. If we’re testing a prototype, are we taking an MVP approach. Are we focusing on only a few key tasks? Testing the entire experience? This one in particular has me thinking about time constraints and getting as many users to test our prototype in a potentially short amount of time

I would love to hear how people have handled and approached Design thinking in general but specifically would love to hear about people’s approach to the testing phase

## Can a bard cast Vicious Mockery without passersby thinking it’s an attack?

So vicious mockery is, to my understanding, the bard insulting a target, and the target taking psychic damage, and gaining disadvantage.

The flavor, as I understand it, is that they’re so insulted it actually hurts.

Which lead me to wonder – could a bard get a vicious mockery off, and the target not know they’ve been attacked?

Of course, the target knows they’ve been viciously insulted, and assuming they fail their Wisdom check, are mad at the bard and would probably like to murder said bard. But, if they’re all in a bar, and the target wants to attack the bard (attempting to murder him in broad daylight), it seems reasonable that other people would want to stop said murder – after all, it is just a bard insulting someone, and bards do that all the time – it doesn’t quite rise to the level of trying to knock his lights out. (Although most of the patrons of said bar are probably fairly understanding of his position…)

To look at a slightly different situation, if a wizard cast fireball on a patron, the entire bar would be up in arms attempting to lynch said wizard. Would there be a similar effect on a bard mocking someone?

So I guess my question boils down to: Do other people recognize vicious mockery as an actual attack, or do they simply think it’s a devastating insult?

## (Calculus) Derivative Thinking Question

Recently, my Calculus and Vectors (Grade 12) teacher gave our class a thinking question/assignment to work on over the march break, and after working on for some time, I’ve become stuck on it.

The Question:

Consider f(x), a general quadratic function in standard form, and g(x) its reciprocal. For which values of x are the slopes of their respective tangent lines equal?

Consider two cases: one where it is true for exactly one value of x, and the other where it is true for exactly two values of x. In the latter case, you can assume that the steepness a does not equal 0 of f(x) is equal to both it’s y-intercept and also to the slope of its tangent at x = 1.

Find the required conditions on the parameters a,b,c in terms of a.

My Progress So Far:

So I know that f(x) = ax^2 +bx + c and g(x) = 1/f(x). After I solved for the derivative of each function, I set them both equal to each other and started solving for it. But then I came to a equation of (ax^2 + bx + c)^2 = -1 and that doesn’t work.

Next I tried something else. Since I know from the 2nd case that f(x) = c and f(x) = f'(1) when a cannot equal 0, I set c = f'(1) and got c = 2a + b. But after that, I don’t know where to go.

I’m not expecting a full solution, but if anyone could give me a hint for solving this question, I would appreciate it. I know that it says to write everything in terms of a, but I’m not sure how to approach that method.

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## Thinking of boolean variables as sets, and $\mathbf{P=NP=co-NP}$?

This is an algorithm I came up with, that seeks to solve the ‘Boolean Tautology Problem’ in polynomial time, using $$3-DNF$$ clauses. I am posting this algorithm here seeking advice, and correction, if any can be provided.

I treat boolean variables as sets. The set itself denotes the variable to be $$true$$, and complement of the set denotes the variable to be $$false$$. $$\bigcap$$ of sets denotes $$\land(AND)$$ of the variables, and $$\bigcup$$ of sets denotes $$\lor(OR)$$ of the variables.

The sets are arranged so that they perfectly intersect, and every section is a boolean expression. Here are some examples(using upto $$4$$ sets/variables, since drawing more clutters the picture) :

$$3$$ variable example :

$$4$$ variable example :

Now, each clause in $$3-DNF$$ has $$3$$ literals(using the exact $$3-DNF$$ version). So, there are only $$4$$ possible cases for the literals :
1) All $$3$$ literals are positive – $$(a\land b\land c)$$
2) $$1$$ literal negative, rest positive – $$(\lnot i\land j\land k)$$
3) $$2$$ literals negative, remaining positive – $$(\lnot p\land\lnot q\land r)$$
4) All $$3$$ literals negative – $$(\lnot x\land\lnot y\land\lnot z)$$

Any $$3-DNF$$ expression consists of $$\lor$$ of these $$4$$ kinds of clauses, and each clause itself is a subset of our total set of variables. Each clause highlights a certain part of the Venn Diagram of the set of variables. So, a tautological expression will have clauses that highlight the entirety of the Venn Diagram. The important part is that using $$3-DNF$$, the total set can be highlighted only using these specific sets, and not just any arbitrarily small subset.

The way the algorithm works is by highlighting all such clauses found within the expression, and their intersections with all such other subsets. The reasoning is that since only the $$4$$ clause cases can highlight their respective subsets in the Venn Diagram, upon highlighting one such clause, other clauses which may intersect with it also have portions that need to be highlighted. Since the possible highlighter subsets are limited to the $$4$$ kinds of clauses, we just need to check intersections between them, two at a time.

Implementation :
The implementation would be as follows(given a $$3-DNF$$ expression with $$n$$ variables) :
1) Create a graph, starting with all possible clauses as beginning nodes.(total number of clauses $$=8\times {n\choose 3}$$)
2) Create a set of children nodes for each parent node, by adding one more literal each to the parent clause.
3) For each such child, there is a dual child which has the negation of that literal.
4) Using these children as parents, follow procedure (2).
5) Continue this process until you have exhausted your variables, or have $$6$$ literals in your last created children nodes (intersection between two $$3-DNF$$ clauses).
6) A child can have multiple parents.

An example case of graph building is shown below. The directed arrow represents parent-to-child relation. Dual children are attached to the same relation for ease of view. The graph shown is partially built, so as to not clutter.

Solving :
Now, on to the solving the $$3-DNF$$ expression :
1) Scan through the expression and take up each clause.
2) For each clause, mark the corresponding node as true.
3) If a parent node is true, all child nodes are marked true.
4) If both dual children are true, the parent is marked true.
5) If every node is marked true, the expression is a tautology, otherwise not.

Remarks :
Step (3) of the solver ensures that if a particular clause is true, then all possible intersections of it, and any other clause also gets marked true. Since the graph stops building child nodes once those nodes have $$6$$ literals, we deal only with cases of intersection of two clauses. E.g.
$$(a\land b\land c)\land (\lnot d\land e\land\lnot f)=(a\land b\land c\land\lnot d\land e\land\lnot f)$$
It should be obvious that if common literals exist in these clauses then it results in less than $$6$$ literals in the child clause. We do not need to check for intersections of more than two clauses, since the intersection will be contained within the intersection of two clauses each, since each true assignment is in disjunction with others.

Time Complexity :
Since we stop at child nodes with $$6$$ literals each, no matter the value of $$n$$, the number of variables, it is obviously polynomially bounded.

Closing words :
Please read through the method, and do tell me if there are any errors, or logical leaps that I made, or if the algorithm does actually work, which would be great.

EDIT : Added some subtitles for paragraphs.

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