## Password seal through paper or password manager?

In my current company, the IT Risk Team has decided to provide root passwords for about 90 systems using the following conventional method:

1. The length of each password must be 16 characters and 8 characters must be entered while the IT infra team enters the remaining one.
2. Each team prints the part password on paper and seals it with an envelope. So there are two envelopes to combine the entire password for each account
3. The plenipotentiaries sign and seal the envelopes. Without the permission of the IT Security Officer, these envelopes can not be unsealed or disclosed to third parties.

My question is, is this way so manual and time consuming? I suggested using Password Manager, but no one shared the same idea.

## White Paper on AWS Disaster Recovery

The link is broken https://media.amazonwebservices.com/AWS_Disaster_Recovery.pdf.

I have a mistake:

``````
AccessDenied
B80C4C4115C5CEFE

JVaBsDqLa/6/gfpoqZyfe7lNu7ALgH/PCanmT11Bfn34UJTFaE9Rr/IMwBh6ARpNK21BdzrCzuo=

``````

Does anyone know where I can get it?

## Agal Algebraic Geometry – The explanation of a paper showing a morphism is etale

I read this paper (in French, which may be why I did not fully understand the idea). Theorem 2.6: There is a morphism between schemata (actually between closed sets in projective spaces over $$mathbb {C}$$) $$f: X to Y$$, We want to show that it will eventually be a fairy tale $$x$$, We also know the fiber $$y: = f (x)$$ contains only $$x$$, Then the author constructs the following Cartesian diagram, which in my opinion makes no sense:
$$require {AMScd} begin {CD} Z @ >>> X \ @V {a & A; VV @VV {a} V \ Spec ( mathscr {O} _ {Y, y}) @ >>> Y end {CD}$$
And he claims: $$a & # 39;$$ is finite morphism, suppose $$Z = Spec (B)$$ (locally?), $$B$$ is $$mathscr {O} _ {Y, y}$$-Algebra, which is finite and local. We will get from Nakayama $$B cong mathscr {O} _ {Y, y}$$? First $$Spec ( mathscr {O} _ {Y, y})$$ looks weird for me. What is $$Z$$, the stalk $$x$$? I do not understand why this is enough to say that it's a fairytale, because we have to show it's flat. Does the author show that?

## Paper Wallet – Handwritten Paper Wallet or Hardwarewallet for Saving Bitcoins?

Most of the people I talk to are convinced that a hardware wallet is the safest place to keep your bitcoins. But I'm not sure it's safer to take the keys offline and write them on a piece of paper on a non-compromised new system (two copies in two secure locations). Of course someone could steal the keys. When I look at the hardware wallets, I see the following risks:

1. Seed words must be written down – theft reserved
2. If I or someone else wants to use the stick in 10 or 15 years, the firmware is outdated, the solution may be hacked or I have the problem of not finding a secure computer with a USB port or accepting the stick. This reminds me when I want to get the movies from the 20 year old cameras.
3. The bar itself plus seeds is more difficult to hide than a seed / key alone. If a criminal finds the stick, he could force me to hand over the seed.
4. If not open source, I have to trust the manufacturer. For example, The selection of keys could not be random. If the manufacturer only chose from a fixed selection of 10 million private keys, I would not realize that after a few years, he could take over all the Bitcoin credits from the customers.

What do you think is the best strategy for keeping the coins safe?

## Paperwork – Is it necessary to present a paper version of your e-Visa when traveling to Myanmar with an e-Visa or is it accepted on a phone / tablet / laptop?

E-Visa for the state of Myanmar:

You must bring this visa approval letter as the Ministry of Labor, Immigration and Population requires you to present it for verification upon your arrival in Myanmar.

Is it necessary to provide a printed version of your e-Visa when traveling to Myanmar with an e-Visa, or will the E-Visa be accepted on a telephone / tablet / laptop?

## What combination of digital printer, ink and paper provides the highest DMAX currently available for adequate archival printing?

After reading recent allegations that color transfers would approach a DMAX of 3.0, I am curious what the inkjet art is in this regard

## Python Rock Paper Scissors on a class to handle the game

The original inspiration came from this Python beginner and caused me to rewrite some things with my flair and my Python experience: First try: Python Rock Paper Scissors

Okay, so I looked at the above post and was bored and took an hour to kill at work. So I killed an hour – I took her RPS game and turned it into a class and made it look less evil / ugly.

This is by no means a full-fledged program that I have created cleanly and really thoroughly tested, but this is something that I can at least ask for input.

Runs pretty clean and uses many strings that had the OP of the original inspiration post. But it also has a lot of docstrings. And the whole game is in a class calling over the class and so on.

Because this version uses f-strings, you must have Python 3.6 or later to use this program / code.

`rps.py`:

``````import random

class RockPaperScissors:
"""
Class to handle an instance of a Rock-Paper-Scissors game
with unlimited rounds.
"""

def __init__(self):
"""
Initialize the variables for the class
"""
self.wins = 0
self.losses = 0
self.ties = 0
self.options = {'rock': 0, 'paper': 1, 'scissors': 2}

def random_choice(self):
"""
Chooses a choice randomly from the keys in self.options.
:returns: String containing the choice of the computer.
"""

return random.choice(list(self.options.keys()))

def check_win(self, player, opponent):
"""
Check if the player wins or loses.
:param player: Numeric representation of player choice from self.options
:param opponent: Numeric representation of computer choice from self.options
:return: Nothing, but will print whether win or lose.
"""

result = (player - opponent) % 3
if result == 0:
self.ties += 1
print("The game is a tie! You are a most worthy opponent!")
elif result == 1:
self.wins += 1
print("You win! My honor demands a rematch!")
elif result == 2:
self.losses += 1
print("Haha, I am victorious! Dare you challenge me again?")

def print_score(self):
"""
Prints a string reflecting the current player score.
:return: Nothing, just prints current score.
"""
print(f"You have {self.wins} wins, {self.losses} losses, and "
f"{self.ties} ties.")

def run_game(self):
"""
Plays a round of Rock-Paper-Scissors with the computer.
:return: Nothing
"""
while True:
userchoice = input("Choices are 'rock', 'paper', or 'scissors'.n"
"Which do you choose? ").lower()
if userchoice not in self.options.keys():
print("Invalid input, try again!")
else:
break
opponent_choice = self.random_choice()
print(f"You've picked {userchoice}, and I picked {opponent_choice}.")
self.check_win(self.options(userchoice), self.options(opponent_choice))

if __name__ == "__main__":
game = RockPaperScissors()
while True:
game.run_game()
game.print_score()

while True:

continue_prompt = input('nDo you wish to play again? (y/n): ').lower()
if continue_prompt == 'n':
print("You are weak!")
exit()
elif continue_prompt == 'y':
break
else:
print("Invalid input!n")
continue
``````

Suggestions and suggestions are welcome as this is a rough attempt. 🙂

## Wrong address when transferring BTC to Paper Wallet

I think I screwed it up when I transferred my BTC from the exchange to the wallet.
I used the private bitcoin address you get when creating a paper wallet instead of getting the address generated in blockchain.info. The transaction was confirmed, but the BTC never received.
Where did the BTC go?

## fa.functional analysis – help in understanding a proof of an article of a paper

I am trying to reproduce the point (iv) of Lemma 4.4 from the work "traveling waves for non-zero Schrödinger non-linear equations in non-zero terms". by M. Mariç (found here https://arxiv.org/abs/0903.0354). The proof of (iv) from Lemma 4.4 is short, but I have many doubts. Can someone help me and explain step by step what the author has done, specifically (4.14) and in the next line?

Thank you, I am happy about every answer!