Is CloudFlare full SSL less secure than Cloudflace full (strict) SSL?

I have a Linux shared hosting on Hostgator, and am using a Full SSL from Cloudflare which Encrypts end-to-end, using a self-signed certificate on the server.

  1. How secure is it?

  2. What are different options to enable full security if this method has any drawback?

Cloudflare has a stricter SSL option that is Full (strict) which Encrypts end-to-end, but requires a trusted CA or Cloudflare Origin CA certificate on the server, I’m not using it as of now.

Linear programs with strict inequalities and supremum objectives

Linear programming can solve only problems with weak inequalities, such as “maximize $ c x$ such that $ A x \leq b$ “. This makes sense, since problems with strict inequality often do not have a solution. For example “maximize $ x$ such that $ x<5$ ” does not have a solution.

But suppose we are interested in finding supremum instead of maximum. In this case, the above program does have a solution – the supremum is $ 5$ .

Given a linear program with strict inequalities and a supremum or infimum objective, is it possible to solve it by reduction to a standard linear program?

Start another application and access ioio ports from a snap application with strict confinement

I built two applications using snapcraft(say snap1 and snap2) with confinement as devmode. snap1 is a simple daemon application which starts snap2(not a daemon) and both work normally. I’m developing for Dell Edge Gateway device with Ubuntu core-16 on it.

Now for security purpose I have to change the confinement of both these snaps to strict confinement and build again. But after this modification,

snap1 is unable to start snap2. snap2 is unable to open port or read data from serial-port-ioioix However I’m able to access network via the applications after the following changes. I have done following changes for snap1 and snap2 in their respective yaml files:

1. snap1.yaml

*name: snap1

version: ‘0.8’

summary: This is a snap1

description: |

Some description.

grade: stable #devel # must be ‘stable’ to release into candidate/stable channels

confinement: strict #devmode # use ‘strict’ once you have the right plugs and slots

apps:

snap1:

command: bin/snap1

plugs:

  • network
  • network-bind
  • home
  • removable-media
  • shutdown
  • snapd-control
  • daemon-notify
  • system-trace
  • timezone-control

    daemon: simple

parts:

snap1-app:

plugin: cmake

configflags:

  • -DAPPVERSION=0.8

    source: .*

2. snap1.yaml

*name: snap2

version: ‘1.0.7’

summary: This is a snap2

description: |

Some description.

grade: stable #devel # must be ‘stable’ to release into candidate/stable channels

confinement: strict #devmode # use ‘strict’ once you have the right plugs and slots

apps:

snap2:

command: bin/snap2

plugs:

  • network
  • network-bind
  • home
  • removable-media
  • serial-port
  • raw-usb
  • io-ports-control
  • gpio

parts:

snap2-app:

plugin: cmake

configflags:

  • -DBUILDFROMSNAP=TRUE

  • -DAPPVERSION=1.0.7

    source: .*

Please let me know if any other change is to be done.

– Thanks in advance… 🙂 :+1:

Strict WQOs and Strict WPOs

The Wikipedia article on WQOs does not mention a strict version. I came across a particular relation, which I could only describe as a strict WQO, but I am wondering if my reasoning is correct and if the notion of a “strict WQO” even makes sense (I think it does).

Consider the set $ X \times \mathbb{N}$ where $ X = \{1,\ldots,k\}$ for some integer $ k$ and $ \mathbb{N}$ denotes the set of natural numbers. We define the relation $ \prec$ as: $ $ (x_1,n_1) \prec (x_2,n_2) \quad \text{iff} \quad (x_1 \leq x_2) \wedge (n_1 < n_2) $ $

Strict Quasi Order: Clearly $ \prec$ is irreflexive and transitive, and therefore a strict quasi order.

No Infinite Antichains: It’s also easy to observe that there are no infinite antichains. Due to the Pigeonhole principle, any sequence $ (x_1,n_1), (x_2,n_2), \ldots$ of length $ \geq k$ must have at least one pair $ \langle (x_i,n_i),(x_j,n_j)\rangle$ such that $ x_i = x_j$ . Then, since either $ n_i < n_j$ or $ n_j < n_i$ must hold, we have at least one pair in the sequence on which $ \prec$ holds.

Well-Foundedness: Finally, we show that there are no infinite descending chains. For any element $ (i,j)$ , there are only $ |\{1,\ldots,i\} \times \{1,\ldots,(j-1)\}| = i(j-1)$ elements that can precede $ (i,j)$ .

Is this reasoning correct? Then is $ \prec$ a strict WQO on $ X \times \mathbb{N}$ ?

More questions: Since antisymmetry is implied by irreflexivity and transitivity, is any strict quasi order also a strict partial order? And so is any strict WQO also a strict WPO?

Thanks.

Does this overly strict reading of Two-Weapon Fighting Work?

Two-Weapon fighting states:

“When you take the Attack action and Attack with a light melee weapon that you’re holding in one hand, you can use a Bonus Action to Attack with a different light melee weapon that you’re holding in the other hand.” Emphasis mine

There have already been some questions regarding when you have to actually be wielding the light weapons required for Two-Weapon Fighting. Here and Here.

And to satisfies any of those, say the following scenario occurs:

It is Alice’s turn, and she starts her turn wielding two light melee weapons, say, two daggers.

She makes an opportunity attack against a creature using one of these weapons (this is possible, for example, if a creature had used the ready action to move).

She drops one of her daggers (a free action), then takes the attack action and as part of the attack she draws a non-light weapon, say, a quarterstaff, and finally she attacks using this quarterstaff.

Is there any reason, RAW, that she would not be able to use her bonus action to attack with the other dagger which she is still holding because she has technically fulfilled both requirements of Two-Weapon Fighting: “taking the attack action” and “attacking with a light melee weapon…”?

Prove that the upper bound in the Noiseless-coding theorem is strict

Given a probability distribution $ p$ across an alphabet, we define redundancy as:

Expected Length of codewords – entropy of p = $ \ E(S) – h(p)$

Prove that for each $ \epsilon$ with $ 0 \le \epsilon \lt 1$ there exists a $ p$ such that the optimal encoding has redundancy $ \epsilon$ .

Attempts

I have tried constructing a probability distribution like $ p_o = \epsilon, p_1 = 1 – \epsilon $ based on a previous answer, but I can’t get it to work.

Any help would be much appreciated.

Strict Convexity and Uniqueness of Dual norm

So, I have trouble proving the following, I’d be grateful if somebody helps me with this.

Let $ z$ be a given point in $ \mathbb{R}^m$ . Then, $ x\in \mathbb{R}^m$ is a dual vector of $ z$ with respect to $ \|.\|$ if it satisfies $ \|x\|=1$ and $ z^Tx=\|z\|’$ .
A norm $ \|.\|$ is said to be strictly convex if the unit sphere $ \{x:\|x\|=1\}$ contains no line segment.

Now, how does one prove that

The norm $ \|.\|$ is strictly convex if and only if each $ z\in \mathbb{R}^m$ has a unique dual vector.

Does this strict reading allow both Extra attack and Thirsthing blade to be used together?

The warlock invocation Thirsting Blade cannot be normally used with the class feature Extra attack since both feature explicitly say attack twice instead of once. But it seems to me that you could use both features when using two weapon fighting.

Assuming two light weapons, one of which is your pact weapon. You take the Attack action on your turn to attack twice with your non-pact weapon using you Extra Attack class feature.

Then, given the two fighting rules, you use your bonus action to attack once using your pact weapon this time.

Two weapon fighting rules (PHB p.195):

When you take the Attack action and attack with a light melee weapon that you’re holding in one hand, you can use a bonus action to attack with a different light melee weapon that you’re holding in the other hand.

But since your have the Thirsting Blade invocation, and you used the Attack action, you get to attack twice with your pact weapon.

Thirsting Blade (PHB p.111):

Vou can attack with your pact weapon twice, instead of once, whenever you take the Attack action on your turn.

Does this combo work or is there any interaction I’m not aware of ?