Install Simple Screen recorder on Fedora 30

Procedure below has been performed for Spice console video and sound fairly impressive demonstration. Repository rpmfusion is supposed to be installed first to provide `dnf install`  ability to resolve dependencies 

# dnf install https://download1.rpmfusion.org/free/fedora/rpmfusion-free-release-$(rpm -E %fedora).noarch.rpm https://download1.rpmfusion.org/nonfree/fedora/rpmfusion-nonfree-release-$(rpm -E %fedora).noarch.rpm

# dnf install simplescreenrecorder

# dnf info simplescreenrecorder

Last metadata expiration check: 0:37:25 ago on Fri 28 Jun 2019 11:09:36 AM MSK.Installed PackagesName         : simplescreenrecorderVersion      : 0.3.11Release      : 7.fc30Architecture : x86_64Size         : 3.7 MSource       : simplescreenrecorder-0.3.11-7.fc30.src.rpmRepository   : @SystemFrom repo    : rpmfusion-freeSummary      : Simple Screen Recorder is a screen recorder for LinuxURL          : http://www.maartenbaert.be/simplescreenrecorder/License      : GPLv3Description  : It is a screen recorder for Linux.             : Despite the name, this program is actually quite complex.             : It's 'simple' in the sense that it's easier to use than ffmpeg/avconv or VLC

   Fair Video recorded for Spice Console functionality on RH's systems, in particular Fedora 30

  https://youtu.be/-FjQDk3ZFb0

Dual booting Fedora 30 via traditional style system wide grub.cfg on Linux boxes

Turn off BLS inside Fedora and regenerate grub config in old format.
Set GRUB_ENABLE_BLSCFG=false in /etc/default/grub and then run as root `grub2-mkconfig -o /boot/grub2/grub.cfg`.
Could you to be able use BIOS boot menu to get into Fedora? That would be easiest. In particular, that shouldn't be a problem at all on UEFI boxes, presuming that fedora 30 was installed on second drive. Then mount partition containing /boot/grub2/grub.cfg managing the box boot up. Just cut and paste content of "Menu entry" for fedora 30 into system wide grub.cfg. Obviously it won't be an issue when f30 is installed on first drive responsible to manage systems wide booting up. Regarding last option I have to warn that f30 anaconda installer didn't work smoothly for me. A trick which helped me out several times. When f30 installer hangs some time latter it pops up the window "Quit Install " or "Wait". DON'T SELECT either one of suggested options just WAIT. In about 3-5 sec. installer closes the window and goes ahead performing smooth install. For now I have another box with f30 installed on system wide boot partition.
Keep in mind :- 
After each one `dnf update -y` you are supposed to rerun 
`grub2-mkconfig -o /boot/grub2/grub.cfg`.

   

I believe you understand that this hack was not issued by Fedora 30 official support. However, I don't see any other way to manage on Debian Linux system.Spice console for f30 kvm guest is running

   

   

   To create a screen-cast on Fedora 30 , use the key-combination Ctrl + Shift + Alt +R to start your recording, and a small orange dot will appear in the status icon area in the top right of your screen.

https://youtu.be/c9iIXT1XQiU

Install Simple Screen recorder on Fedora 30

[root@fedora30wks ~]# dnf install https://download1.rpmfusion.org/free/fedora/rpmfusion-free-release-$(rpm -E %fedora).noarch.rpm https://download1.rpmfusion.org/nonfree/fedora/rpmfusion-nonfree-release-$(rpm -E %fedora).noarch.rpm

[root@fedora30wks ~]# dnf install simplescreenrecorder

[root@fedora30wks ~]# dnf info simplescreenrecorder

Last metadata expiration check: 0:37:25 ago on Fri 28 Jun 2019 11:09:36 AM MSK.

Installed Packages

Name         : simplescreenrecorder

Version      : 0.3.11

Release      : 7.fc30

Architecture : x86_64

Size         : 3.7 M

Source       : simplescreenrecorder-0.3.11-7.fc30.src.rpm

Repository   : @System

From repo    : rpmfusion-free

Summary      : Simple Screen Recorder is a screen recorder for Linux

URL          : http://www.maartenbaert.be/simplescreenrecorder/

License      : GPLv3

Description  : It is a screen recorder for Linux.

             : Despite the name, this program is actually quite complex.

             : It's 'simple' in the sense that it's easier to use than ffmpeg/avconv or VLC

Fair Video recorded for Spice Console functionality on RH's systems, 

 in particular Fedora 30

https://youtu.be/KSQjBb51-FM

Solution of one system in Boolean variables via Classic Mapping Method

"Ignorance of the mapping method is still half the misfortune ..." B.D.  

Original system

((x1vy1)^z1)⊕((x2=>y2)=>z2)=1
((x2vy2)^z2)⊕((x3=>y3)=>z3)=1
(x1≡y2)=>(y2≡z3) =1

Solving system (1)
((x1vy1)^z1)⊕((x2=>y2)=>z2)=1
((x2vy2)^z2)⊕((x3=>y3)=>z3)=1
Result 140

Solving system (2)
((x1vy1)^z1)⊕((x2=>y2)=>z2)=1
((x2vy2)^z2)⊕((x3=>y3)=>z3)=1
(x1≡y2)=>(y2≡z3) = 0

Two additional passes with predefined variables which are based on the same fork diagram as for system (1). I am targeting to focus yours attention on details of technique had been published in 2013
in http://kpolyakov.spb.ru/download/mea-2013-10.pdf

x1=1 ; y2=1 ; z3=0  gives 28
x1=0 ; y2=0 ; z3=1  gives 16

                    Total 44  

Answer is 140-44 = 96 
Click on snapshot to switch to perfect slides show, either collapse
menu of current month blog archive

   Passing Polyakov's Control

A bit more complicated sample

((x1vy1)^z1)⊕((x2=>y2)=>z2)=1
((x2vy2)^z2)⊕((x3=>y3)=>z3)=1
((x3vy3)^z3)⊕((x4=>y4)=>z4)=1
((x1≡y2)=>(y2≡z3)) =>z4=1

Truth Table obviously stays the same  

Fork Mapping Method tables (total 6) for the following system

((x1vy1)^z1)⊕((x2=>y2)=>z2)=1
((x2vy2)^z2)⊕((x3=>y3)=>z3)=1
((x3vy3)^z3)⊕((x4=>y4)=>z4)=1
((x1≡y2)=>(y2≡z3)) =>z4=0

Total six passes have been performed due to this and
only this bit's chains matches equation
    ((x1≡y2)=>(y2≡z3)) =>z4=0

1) x1=1; y2=1; z3=1; z4=0  (1=>1) =1
2) x1=0; y2=0; z3=0; z4=0  (1=>1) =1
------------------------------------------
3) x1=0; y2=1; z3=0; z4=0  (0=>0) =1
4) x1=1; y2=0; z3=1; z4=0  (0=>0) =1  
------------------------------------------
5) x1=1; y2=0; z3=0; z4=0  (0=>1) =1
6) x1=0; y2=1; z3=1; z4=0  (0=>1) =1

Predefined zeros have been typed bold in
excel spreadsheets

Total for false of last implication is 60+40+20+16+20+32=168

Passing Polyakov's Control

Solution of one system of boolean equations via reverse pass to center per Helen Mironchick (II)

"Ignorance of the mapping method is still half the misfortune ..."
  B.D. 

***********************************************************************
  I do emphasize again that core idea of reverse pass to center belongs 
 to Helen Mironchick (unpublished manuscript) .
***********************************************************************  
The advantage of method being applied is its tolerance to number of equations in the system. Original system :-

(x1^x2=>x3)^x4=>x5=1
(y1^y2=>y3)^y4=>y5=1
(z1^z2=>z3)^z4=>z5=1
(x1=>y3)=>z4=1

#1 - x1
#2 - x1^x2
#3 - (x1^x2=>x3)
#4 - (x1^x2=>x3)^x4
#5 - (x1^x2=>x3)^x4=>x5)

   Passing Polyakov's Control

Solution of one system of boolean equations via complete reverse pass per Helen Mironchick

"Ignorance of the mapping method is still half the misfortune ..."
  B.D.

Down here we rely on technique proposed in 
https://mapping-metod.blogspot.com/2019/03/blog-post.html
Original system

((x1^x2=>x3)^x4=>x5)^x6=>x7=1
((y1^y2=>y3)^y4=>y5)^y6=>y7=1
x1=>y7=1

#1 - x1
#2 - x1^x2
#3 - (x1^x2=>x3)
#4 - (x1^x2=>x3)^x4
#5 - (x1^x2=>x3)^x4=>x5)
#6 - (x1^x2=>x3)^x4=>x5)^x6
#7 - ((x1^x2=>x3)^x4=>x5)^x6=>x7=1

   Passing Polyakov's Control

Метод Отображений - тестирование проход к центру и обратный проход

Оригинальна схема прохода к центру предложена 
Еленой А. Мирончик . Смотри 
https://vk.com/doc6125348_505417587?hash=44fb5048df995959ea&dl=210d2b189eb483566a

Одно из основных преимуществ метода - это толерантность к числу уравнений системы

Исходная система

((((x1=>x2)=>x3)=>x4)=>x5)=>x6 =1
((((y1=>y2)=>y3)=>y4)=>y5)=>y6 =1
((((z1=>z2)=>z3)=>z4)=>z5)=>z6 =1
(x3=>y4)=>z5 =1

                                  X3 Y4 Z5

                            0    23  19  27
                            1    20  24  16

 Контроль по Полякову

   Ссылки

  1. https://vk.com/doc6125348_505417587?hash=44fb5048df995959ea&dl=210d2b189eb483566a

Mapping method technique ( v. 08.2016 ) on real 23-rd problem of USE in Informatics 2019

In my very personal opinion pretty poor understanding the flexibility of 08.2016 technique affects  "USE in Informatics " on permanent basis

Original system

Building diagram per ГРАФЫ И СИСТЕМЫ ЛОГИЧЕСКИХ УРАВНЕНИЙ

  Passing Polyakov's control
  

Solution of one problem 23 from USE in Informatics 2019 or addressing one ugly 23-rd problem

UPDATE 22/06/2019 
  The most recent version of ege23.doc contains this problem as P-48.

 

https://www.youtube.com/watch?v=yb66JLfRTtI
END UPDATE

Original task

Down here we intend to demonstrate one more time well known axiom of well known USE's in Informatics developer ( sure not myself ), which states any system of boolean equations could be solved via Mapping Method. Would you expirience trouble in doing so that appears to be the limitation of yours skilset rather then the limitation of the method.

You might want to see the most impressive and up to date Mapping
Method solution of more or less complicated  system at
Helen Mironchick's Solution of one sysem of boolean equations from USE in Informatics forum (16.05.2019)

System below is equivalent to original one.
The previous system is supposed to make convertion itself
a real headaches generator

(x1=>x2)^(y2=>y1)^(x1 v y1)=1
(x2=>x3)^(y3=>y2)^(x2 v y2)=1
(x3=>x4)^(y4=>y3)^(x3 v y3)=1
(x4=>x5)^(y5=>y4)^(x4 v y4)=1
(x5=>x6)^(y6=>y5)^(x5 v y5)=1
x6 v y6=1

Proceed with building fork diagram with transition pair (x2,y2) &&  forking matrix

   Passing Polyakov's Control
  

Mapping Method and solution of system of boolean equations as of 12/06/2019

Original system

(x1⊕y1) => (x2⊕y2) =1
(x2⊕y2) => (x3⊕y3) =1
(x3⊕y3) => (x4⊕y4) =1
(x1⊕x4) => (y1⊕y4) =1

Solve first system 1

(x1⊕y1) => (x2⊕y2) =1
(x2⊕y2) => (x3⊕y3) =1
(x3⊕y3) => (x4⊕y4) =1

Solve second system 2

(x1⊕y1) => (x2⊕y2) =1
(x2⊕y2) => (x3⊕y3) =1
(x3⊕y3) => (x4⊕y4) =1
(x1⊕x4) => (y1⊕y4) =0

and deduct from number of solutions
system 1 number of solutions system 2

 
   Passing Polyakov's control
    

 

 

Mapping method power unleashed && solving binding condition in 08.2016 style

Original System

(x1=>x2)^(x2⊕x3)^(x3=>x4)^(x4=>x5)^(x5=>x6)=1
(y1=>y2)^(y2=>y3)^(y3⊕y4)^(y4=>y5)^(y5=>y6)=1
(z1=>z2)^(z2=>z3)^(z3=>z4)^(z4⊕z5)^(z5=>z6)=1
(x6+y6+z6) =1

    Solving final equation via splitting graph nodes
   

 

         Passing Polyakov's control
 

Consider another system

(x1=>x2)^(x2⊕x3)^(x3=>x4)^(x4=>x5)^(x5=>x6)=1
(y1=>y2)^(y2=>y3)^(y3⊕y4)^(y4=>y5)^(y5=>y6)=1
(z1=>z2)^(z2=>z3)^(z3=>z4)^(z4⊕z5)^(z5=>z6)=1
(x1+y2+z3) =1 

Here we intend to manage via pass to centre per Helen Mironchick

Passing Polyakov's Control

         

Mapping method && Solution of one system of boolean equations via pass to centre

The core idea of obtaining needed values for z1,x2,y3,w4 via pass to centre belongs to Helen Mironchick. Down here a fair sample 
is provided when bitmasks approach appears to be at no help at all

Original system looks as follows

(x1=>x2)^(x2=>x3)^(x3=>x4)^(x4=>x5)=1
(y1=>y2)^(y2=>y3)^(y3=>y4)^(y4=>y5)=1
(z1=>z2)^(z2=>z3)^(z3=>z4)^(z4=>z5)=1
(w1=>w2)^(w2=>w3)^(w3=>w4)^(w4=>w5)=1
(((z1=>x2)=>y3)=>w4) =1

First calculating diagrams  for x,y,w,z

for z we are doing complete reverse pass
for x reverse pass from x5 to x2
for y reverse pass from y5 to y3
for w reverse pass from w5 to w4

   Solving final implication

    

 
  Passing Polyakov's control

  

Mapping method as it works at nowdays or another lesson been taught from Helen Mironchick regarding technique of solving the systems of equations in boolean variables

System usually treated via bitmasks tables
concatenation. In particular case could hardly be
solved as it happens with fourth equation kind of
x6^y6^z6=0


(x1=>x2)^(x2=>x3)^(x3=>x4)^(x4=>x5)^(x5=>x6)=1
(y1=>y2)^(y2=>y3)^(y3=>y4)^(y4=>y5)^(y5=>y6)=1
(z1=>z2)^(z2=>z3)^(z3=>z4)^(z4=>z5)^(z5=>z6)=1
(x6=>y6)=>z6=1

(x1=>x2)^(x2=>x3)^(x3=>x4)^(x4=>x5)^(x5=>x6)=1
(y1=>y2)^(y2=>y3)^(y3=>y4)^(y4=>y5)^(y5=>y6)=1
(z1=>z2)^(z2=>z3)^(z3=>z4)^(z4=>z5)^(z5=>z6)=1
(x1=>y1)=>z1=1

Получаем x1,y1,z1 обратным проходом (6,1)

   Контроль по Полякову
 

Решение одной классической задачи ЕГЭ Информатика в стиле 08.2016 ( без сцепления таблиц битовых масок )

Система

Решение

***********************************
Первый вариант решения

***********************************
Множество решений (1) Х6 =0,У6,Z6 любые - решений 1*7*7=49 
Множество решений (2) Y6=0,X6=1,Z6 любое - решений 1*6*7=42
Множества (2) и (1) не имеют общих кортежей. 
В (1) Х6=0 в (2) Х6=1
Множество решений (3) Z6=0 Y6=1 X6=1 - решений 1*6*6 =36
Множества (3) и (1) не имеют общих кортежей . 
В (3) Х6=1 в (1) Х6=0
Множества (3) и (2) не имеют общих кортежей . 
В (3) У6=1 в (2) У6=0
Объединение множеств кортежей (1),(2),(3) даст полное 
множество кортежей - решений исходной системы.
Ответ : 49+42+36 = 127  
************************************************************
Второй вариант решения (Елена А. Мирончик)
************************************************************
Применяем метод отображений к уравнению x6^y6^z6 =0 
(раздвоение узлов будет иметь место в этой ситуации)

Cамое короткое решение