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import tensorflow as tf

tf.reset_default_graph()
s1 = tf.Session()
g = tf.get_default_graph()

foo_var = tf.Variable(42, name='foo')
assign_14 = foo_var.assign(14, name="assign_14")
assign_17 = foo_var.assign(17, name="assign_17")

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Setup JanusGraph with HBASE backend

Install following

• hbase-1.2.9
• janusgraph-0.2.2-hadoop2

HBASE Setup and inspection

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IFACE=wlo1
VPN_MTU=1380

WL_MTU=$(cat /sys/class/net/$IFACE/mtu)

ip a | grep 'state UP'

echo "Checking: $IFACE MTU=$WL_MTU and VPN requires MTU=$VPN_MTU"
if [ "$WL_MTU" -gt "$VPN_MTU" ]
then

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Getting started with git-svn

git-svn is a git command that allows using git to interact with Subversion repositories.git-svn is part of git, meaning that is NOT a plugin but actually bundled with your git installation. SourceTree also happens to support this command so you can use it with your usual workflow.

Reference: http://git-scm.com/book/en/v1/Git-and-Other-Systems-Git-and-Subversion

Cloning the SVN repository

You need to create a new local copy of the repository with the command

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Mointoring NVIDIA GPUs

List NVIDIA GPUs available

$ nvidia-smi --query-gpu=gpu_name --format=csv,noheader
GeForce GTX 1050

Monitor GPU stats ( temperature, utlilization, memory )

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Git references

Git clone without deep history

Git clone without deep history make cloning faster when we only need the latest code.

With full history

First run

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$ python3 pivot-pandas.py
        date stock_code  price
0 2018-08-27        001   10.0
1 2018-08-27        002   11.0
2 2018-08-27        003   12.0
3 2018-08-27        004   13.0
4 2018-08-26        001   14.0
5 2018-08-26 002 15.0

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Groovy

//System.out.println("Hello World!")
//println("Hello World!")
//
//println "Hello World!"
//

class MyClass {

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Given a 6 sided dice, how many number of throws are required to guarantee a 6 as output?

This solution is basically finding expectation over 1 roll, 2 rolls, 3 rolls ... and so on until infinite rolls of dice. Basically SUM( k * P(rolls=k)) for k = 1 to infinity. Now probability we get a 6 in exactly kth roll is defined as P(rolls=k) = q^(k-1) * p, where p is probability we get a 6 in kth roll, and q^(k-1) is probability we do not get a 6 in all k-1 rolls. I guess it still cannot guarantee that we will always get a 6, rather that if we roll on average of 6 times in a row, we will get definitely get a 6. Quite interesting problem indeed.

References: