Make Your Own Command-line Calculator with bc

I do not know about you, but I am bothered by the lack of a proper command line calculator for doing small single line calculations in Linux (or any other UNIX or Mac system). Sure bc is there but it does not support floating point by default, so one ends up writing things like:

echo "scale=3; 22/7" | bc

which is quite cumbersome for day to day work.

To subvert this trouble, I wrote the following two-liner and saved it as "calc" in a directory in PATH:

#! /bin/bash
echo "scale=3; $1" | bc; exit

Then do a "chmod +x" on the file to give it execute permission. This does the same thing as the above script, but is way simpler to use:

calc 22/7

The command "scale=3" was to instruct bc to print upto 3 decimal digits. You can change it as per your requirements.

The bc utility can do way more than this simple calculations. For example it can handle different bases and load functions from files. To know more about its capabilities try the good old "man bc" in terminal or read an online reference like this one on about.com.  You can read more about its advanced capabilities from an old but quite relevant article in LinuxJournal "bc: a handy utility".

Disclaimer : Although the script was by me, the idea is not new. In fact a lot of my colleagues use similar scripts for their own daily work.

Do-while loop in Ruby

As you know, I'm trying out Ruby. While solving one of the exercise problems I felt the need for a do-while loop. Unfortunately ruby has no documented do-while loop. Thankfully, somebody has already asked this same question in stackoverflow : Is there a do-while loop in ruby?

Here is a demonstration of a do-while code in ruby:


Remember that using undocumented features is always ill-advised !!

I'm trying out Ruby

I have always been interested in programming languages. After coding in a plethora of mainstream procedural languages C, C++ and Java, last year I tried my hands on Python. Apart from being an enjoyable experience, learning Python has paid off quite well while building prototypes in my work. This prompted me to try learning some other languages.

While searching for a cool language to try out, I came across the book, "Seven Languages in Seven Weeks" which promises to go beyond the simple tutorials and "introduction to" and introduce some non-trivial aspects of the language. I have started from page 1 and am exploring Ruby. So far the author has kept up his promise.

Ruby has got very friendly constructs and syntactic sugars like:

• the keyword "until", which is simply a negation of the popular "while"
• instead of simple "for" loops, constructs like "10.times"

which gives a warm and friendly feeling to the language. Besides the book mentioned above I am using "Programming Ruby 1.9" from the same publishers for a reference.

Demystifying bulk SMS sender name

With the high penetration of mobile phones, companies have started sending out automated sms. I have always wondered about the cryptic sender names. Careful observations about the sender name reveals:

• all letters are in capitals
• name format is **-******
• the first two letters seem random
• followed by an hyphen
• followed by a six letter
• last six letters seem to identify the sender name
• for example presently, "SBIINB" represents SBI bank
• sometimes the same sender can send some sms with different combinations of first two letters
• SBI sends sms as "BX-SBIINB", "TD-SBIINB" etc.

Some researching in internet points out that my observations about the last six letters are correct. The first two letters before the hyphen represents service provider and the region from which it is sent. For the examples above:

• "BX" = BSNL from Karnataka
• "TD" = Tata from Delhi

View the official list of bulk sms codes in the NDNC website.

GATE 2012 CS solution

I had appeared in GATE 2008 and 2009. After that I had not been interested much in GATE. However this time (GATE 2012) a close friend of mine appeared for the same, which got me interested again. Here is a solution set to the GATE 2012 Computer Science paper. I have solved question from Algorithms, Automata, C programming, Probability ... in short, the topics that I enjoyed enough to remember them after 2 years.

As I mentioned earlier, I have not appeared in for the examination. I got the question GATE 2012 CS and IT question set C from a fellow netizen. Please download the paper first as I will be using that as the reference for objective type answers.

2 years earlier I had posted solutions to GATE 2009 CS question paper. Comments and suggestions from that post made me realize the importance of explaining every answer. Feel free to ask questions if you need further clarifications.

GATE 2012 Computer Science and IT solutions to question set C

1. b
2. -
3. -
4. b
• Draw the sine graph for the interval. Only one minima at 3*pi/2 where sin(x) = -1
5. c
• fork() will spawn off a new child process every call. Three calls to fork() results in 8 processes, out of which one is original, other 7 are children.
6. b
• this is the hidden one concept
7. a
• Obviously. Look at f(X,Y) and compare it with X
8. c
• Balanced binary search trees require log (number of elements) time to search
• log (n*2^n) = log (n) + log(2^n) = O(n)
9. -
10. c
• Standard switch case behaviour called "fall through", when the break statement is not there
11. c
• Look into any DB book, I studied Navathe.
12. -
13. c
• Do the checking manually
14. c or d
• 1 is false, 3 and 4 are true
• not sue about 2
15. -
16. -
17. c
• follows from definition of CDF
18. -
19. d
• 4 bit multiplier requires 2x4 = 8 bit input, i.e. 2^8 set of inputs
• multiplication of two 4-bit numbers will result in at most 8-bit result, so 8 bits for every multiplication
• 2^8 * 8 = 2^11 bits = 2 Kbits
20. c
• Average case is at lease as good as worst case
• A(n) better than W(n) :: Quicksort
• A(n) same as W(n) :: Heapsort
21. c
• Draw it and find out. They key thing is "any embedding of G on the plane"
22. d
• T(n-1) for moving n-1 disk to aux from source
• 1 for moving the n-th disk to destination
• T(n-1) for moving n-1 disk back to source from aux
23. b
• S is correct
• regarding P or Q, there is some confusion. It seems that the SQL syntax allows it. You can test it out in an SQL validator.
24. -
25. c
26. -
27. --
28. -
29. b
• examine the cases:
• P(6 in first throw) = 1/6
• P(1 in first throw followed by 5 or 6 in second throw) = (1/6)*(2/6)
• P(2 in first throw followed by 4, 5 or 6 in second throw) = (1/6)*(3/6)
• P(3 in first throw followed by 3,4, 5 or 6 in second throw) = (1/6)*(4/6)
•  summing up = (1/6)*(1+2/6+3/6+4/6) = 5/12
30. -
31. -
32. b
• b'd' -- group the 4 corners
• b'c' -- group the a'b' columns to 2 cells with the top 2 cells in ab' column
33. a
• MST algorithm does not depend on the value of the weights, it depends only on the relative ordering of the edge weights.
• Ranking remains unchanged as all the weights are >= 1
34. b
• I did it by hand, don't know if there is a shortcut
35. b
• Observe the graphs
36. -
37. d
• observe that q is a dead state, so anything beginning with 000 should go to q -- A and B gets eliminated by this logic
• after 10 if there is a 0 the state should move to 00 state, so D is correct
38. a
• look into any Data Structure book
39. -
40. -
41. -
42. -
43. -
44. d
• Dijkstra's algorithm probes the best solution so far
• Thanks to Shivam for pointing this out
45. b
• merge-sort takes O(n log n) comparisons
• here to compare two strings you need O(n) time, so O(n^2 log n)
46. c
• represent a cycle by a permutation of 4 labeled vertices
• cycles are identical if one is a rotation of another
• same cycle of length 4 can have 4 different representations each starting with a unique letter
• (6 P 4) / 4 = 90
47. -
48. -
49. -
50. c
• line 2 has a static variable, which retains the old value when he function is called again
51. d
• "auto int" is same as normal "int"
• "register int" is same as normal "int" as far as value retention is considered
52. -
53. -
54. -
55. -
56. -
57. a
58. -
59. a
• maxima-minima problem
• maximize the profit where, profit = 50q-5q^2
60. b
61. b
• Given:
• P(X) = 0.6
• P(Y) = 0.4
• P(R | X) = 0.96
• P(R | Y) = 0.72
• P(R) = P(X) * P(R | X) + P(Y) * P(R | Y) = 0.864
  
• Applying Bayes Rule:
• P(Y | R) = [ P(R | Y) * P(Y) ] / P(R) = 0.333
62. c
•  follows from fundamental definitions of mean and standard deviation
63. -
64. -
65. b
•  another maxima-minima problem and the function is already given, just maximize y

Please leave a comment if you were helped by my effort. Feel free to ask questions if you need further clarifications or have found some error in my solution.