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Installing ns-3.37 and ns-3.35 in Ubuntu | Ubuntu 22.04 | NS3

Multiple Versions of ns3 in Ubuntu 22.04 In this post, we are going to install two versions of ns3 namely ns-3.35 and ns-3.37  My OS is : Ubuntu 22.04 LTS (Long Term Support) ns-3.35 uses waf and (./waf --run scratch/first)  ns-3.37 uses cmake  (./ns3 run scratch/first.cc) So we will install both the packages  Go through the video for complete instructions To start with  $ sudo apt update  $ sudo apt install build-essential autoconf automake libxmu-dev g++ python3 python3-dev pkg-config sqlite3 cmake python3-setuptools git qtbase5-dev qtchooser qt5-qmake qtbase5-dev-tools gir1.2-goocanvas-2.0 python3-gi python3-gi-cairo python3-pygraphviz gir1.2-gtk-3.0 ipython3 openmpi-bin openmpi-common openmpi-doc libopenmpi-dev autoconf cvs bzr unrar gsl-bin libgsl-dev libgslcblas0 wireshark tcpdump sqlite sqlite3 libsqlite3-dev  libxml2 libxml2-dev libc6-dev libc6-dev-i386 libclang-dev llvm-dev automake python3-pip libxml2 libxml2-dev libboost-all-dev  I have downloaded both the versions of ns3 fr

Radix Complement and Diminished Radix complement

Complements are used in digital computers for simplifying the subtraction and for logical manipulations. There are two types of complement

  • Radix complements ( r complement)
  • diminished radix complements (r-1 complement)

For base 2 or binary number system the r’s complement is 2’s complement and (r-1)’s complement is 1’s complement

For base 10 or decimal system the r’s complement is 10’s complement and (r-1)’s complement is 9’s complement.

10’s Complement

Let us assume the number 3567890, for finding the 10’s complement, there is a rule

3 5 6 7 8 9 0
Most significant Digit           Least Significant Digit

10’s complement can be obtained by

  • leaving all least significant 0’s unchanged
  • subtracting the first non zero least significant digit from 10
  • and subtracting all higher significant digits from 9

For the above example, The 10’s complement is

Given number 3 5 6 7 8 9 0
Process 9-3 9-5 9-6 9-7 9-8 10-9 unchanged
10’s Complement 6 4 3 2 1 1 0

so 10’s complement of 3567890 is 6432110

9’s Complement

9’s complement is a diminished radix complement and can be easily found out by subtracting all the given digits by 9.

For example, the 9’s complement of 3567890 is

9999999 – 3567890 = 6432109

given number 3 5 6 7 8 9 0
Process 9-3 9-5 9-6 9-7 9-8 9-9 9-0
9’s complement 6 4 3 2 1 0 9

In short,

10’s Complement = 9’s Complement + 1

2’s Complement

For binary numbers, there is 2’s complement and 1’s complement

2’s complement can be obtained by

  • leaving the least significant 0’s unchanged and the first 1 unchanged
  • replacing 1’s with 0’s and 0’s with 1’s in all other higher significant digits
given number 1 1 0 1 1 0 0
process 0 0 1 0 unchanged unchanged unchanged
2’s complement 0 0 1 0 1 0 0

ie.

2’s complement of 1101100 is 0010100

1’s complement

Finding 1’s complement is just to replace all 1’s by 0’s and all 0’s by 1’s

1’s complement of 1101100 is 0010011

in short

2’s complement = 1’s complement + 1

Comments

  1. very easy to understand and simlified form as well.

    ReplyDelete
  2. Thank you so much! I really appreciate this because it helps me a lot to understand directly about the types of complement namely the radix and diminished radix complement.In fact, this is so resourceful.

    ReplyDelete

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