biobot

5 ms·

1.
A Service Mesh for Kubernetes, Part VI: Staging Microservices Without the Tears(blog.buoyant.io)
4 points by biobot 10y ago|0 comments
2.
Apache Mesos 0.23.0 Released(mesosphere.com)
1 points by biobot 11y ago|0 comments
3.
Web application with analytics using Docker and Marathon(mesosphere.com)
4 points by biobot 11y ago|0 comments
4.
Launching Thousands of Docker Containers with DCOS on Azure – Mesosphere(mesosphere.com)
5 points by biobot 11y ago|0 comments
5.
CoreOS Just Got Easier to Try with Panamax(coreos.com)
4 points by biobot 12y ago|0 comments
6.
Dynamic Docker links with an ambassador powered by etcd(coreos.com)
25 points by biobot 12y ago|3 comments
7.
by biobot 12y ago
Congratulation Beyang and Quinn. Looks really slick!
8.
by biobot 13y ago
Too bad it is GPL v2. https://github.com/facebook/flashcache/blob/master/LICENSE
9.
Linux Performance Analysis and Tools(joyent.com)
5 points by biobot 13y ago|0 comments
10.
by biobot 14y ago
Apache Software License, Version 2.0 https://oss.sonatype.org/content/repositories/snapshots/org/...
11.
by biobot 15y ago
You should've tried this: https://wiki.archlinux.org/index.php/Yaourt
12.
by biobot 15y ago
Interesting! But I think $70k salary for early employees is not realistic. They will probably take $10k-$20k less than market price for significant shares.
13.
by biobot 15y ago
NO (Samsung galaxy S II, T-Mobile)
14.
by biobot 15y ago
One of my friend came up with this solution: m is missing,d is duplicate. I is 1..N, A is input. Step 1: Compute m XOR d by XOR all values of A and I together. let say x = m XOR d. Then x must has at least one 1 bit (or else m == d). Call t
15.
by biobot 15y ago
OK. I think I find a way around. Assume d XOR m = a. Find k = position of the first bit (from MSB to LSB) in a. Then d + m can be in [a + 2^(k+1),a + 2^(k+2),...,a + 2^63]. We will have less than 63 cases. In each case, after we find d and
16.
by biobot 15y ago
Make sense! That fails my solution!
17.
by biobot 15y ago
There are at most 2 solutions: if m XOR d = a then either m+d = a or m+d = a+N. With the combination of m-d = b, you will find m and d. If both of them satisfied ( remember m and d are bounded by 1 and N) then we actually have two solutions
18.
by biobot 15y ago
This is the solution I posted there: Called the missing number m, duplicate d. The main idea is to find m+d and m-d. Then it’s trivial to find m and d. 1) Find n XOR d: X_array = 1; I_array = 1; for (i=2..N) X_array = X_a
19.
by biobot 15y ago
For every open position, they will bring in onsite about 10 people. And most of those 10 are pretty good too. So you had to make a very good impression with the interviewer to make the cut. And of course, a some luck won't hurt too. Again,
20.
by biobot 15y ago
I think you left the interviewers some bad impressions about your technical skills. Blame the interviewers throw easy questions at you and think that it's all they get is blatantly wrong. Most Microsoft's devs are better than that! Based on
21.
by biobot 15y ago
I have applied there for a fulltime position. I got a friend at Google refer and an HR called my immediately to set up a phone interview. It went well and they flew me in for onsite. After more than a week, they sent me an offer after the q
22.
by biobot 16y ago
One stark difference between programmers in the east and west is that in the east, people tend to stay in one company much longer. I think that effectively limit their development. Those who are in the west tend to be much more well rounded
23.
by biobot 16y ago
JVM is written in C/C++ so technically Java cannot be faster than C/C++ because everything in Java are (direct / indirect) products of C/C++. :)