{Complexity} » Algorithm

Interview questions (2)

admin — Thu, 07 Jan 2010 05:51:49 +0000

1) There are integer arrays, A and B. what’s the best way to generate an array which contains all duplicated elements of A and B.
2) There are integer arrays, A and B. how to figure it out if those contain same elements.
3) There many of files in a folder and some of them contains phone numbers. how to find them.
4) Design restaurant reservation system. class design and data structures.

Calculating distances with geographic data

admin — Mon, 17 Aug 2009 20:28:05 +0000

The great circle distance formula:
d = acos(sin(x1)*sin(x2)+cos(x1)*cos(x2)*cos(y2-y1)) * r

CREATE TABLE ZCTA (
zcta CHAR(6) NOT NULL
, lat_degrees DECIMAL(9,6) NOT NULL
, long_degrees DECIMAL(9,6) NOT NULL
, PRIMARY KEY(zcta));

ALTER TABLE ZCTA
ADD COLUMN lat_radians DECIMAL(9,6) NOT NULL
, ADD COLUMN long_radians DECIMAL(9,6) NOT NULL;

UPDATE ZCTA
SET lat_radians = lat_degrees * (PI() / 180)
, long_radians = long_degrees * (PI() / 180);

/* calculating the distance between two points */
SELECT
ROUND(ACOS(SIN(x1.lat_radians) * SIN(x2.lat_radians)
+ COS(x1.lat_radians) * COS(x2.lat_radians)
* COS(x2.long_radians – x1.long_radians)) * 3956, 2) AS “Distance”
FROM ZCTA x1, ZCTA x2
WHERE x1.zcta = ’1001′
AND x2.zcta = ’21236′;

/* zip codes within a given radius */
SELECT
x2.zcta AS zip
, ROUND(ACOS(SIN(x1.lat_radians) * SIN(x2.lat_radians)
+ COS(x1.lat_radians) * COS(x2.lat_radians)
* COS(x2.long_radians – x1.long_radians)) * 3956, 2) AS “Distance”
FROM ZCTA x1, ZCTA x2
WHERE x1.zcta = ’21236′
AND ACOS(SIN(x1.lat_radians) * SIN(x2.lat_radians)
+ COS(x1.lat_radians) * COS(x2.lat_radians)
* COS(x2.long_radians – x1.long_radians)) * 3956 <= 5
ORDER BY Distance;

/* StoreLocation table definition */
CREATE TABLE StoreLocation (
store_id INT NOT NULL AUTO_INCREMENT
, address VARCHAR(100) NOT NULL
, city VARCHAR(35) NOT NULL
, state CHAR(2) NOT NULL
, zip VARCHAR(6) NOT NULL
, PRIMARY KEY (store_id)
, KEY (zip));

/* zip codes within a given radius with StoreLocation table*/
SELECT
address
, city
, state
, zip
FROM StoreLocation
WHERE zip IN (
SELECT x2.zcta
FROM ZCTA x1, ZCTA x2
WHERE x1.zcta = ’21236′
AND ACOS(SIN(x1.lat_radians) * SIN(x2.lat_radians)
+ COS(x1.lat_radians) * COS(x2.lat_radians)
* COS(x2.long_radians – x1.long_radians)) * 3956 <= 5
);

SELECT
address
, city
, state
, zip
FROM StoreLocation s1
INNER JOIN (
SELECT x2.zcta
FROM ZCTA x1, ZCTA x2
WHERE x1.zcta = ’21236′
AND ACOS(SIN(x1.lat_radians) * SIN(x2.lat_radians)
+ COS(x1.lat_radians) * COS(x2.lat_radians)
* COS(x2.long_radians – x1.long_radians)) * 3956 <= 5
) AS zips
ON s1.zip = zips.zcta;

SELECT
address
, city
, state
, zip
FROM ZCTA x1, ZCTA x2
INNER JOIN StoreLocation s1
ON x2.zcta = s1.zip
WHERE x1.zcta = ’21236′
AND ACOS(SIN(x1.lat_radians) * SIN(x2.lat_radians)
+ COS(x1.lat_radians) * COS(x2.lat_radians)
* COS(x2.long_radians – x1.long_radians)) * 3956 <= 5;

SELECT
address
, city
, state
, zip
, ROUND(ACOS(SIN(x1.lat_radians) * SIN(x2.lat_radians)
+ COS(x1.lat_radians) * COS(x2.lat_radians)
* COS(x2.long_radians – x1.long_radians)) * 3956, 2) AS “Distance”
FROM ZCTA x1, ZCTA x2
INNER JOIN StoreLocation s1
ON x2.zcta = s1.zip
WHERE x1.zcta = ’21236′
AND ACOS(SIN(x1.lat_radians) * SIN(x2.lat_radians)
+ COS(x1.lat_radians) * COS(x2.lat_radians)
* COS(x2.long_radians – x1.long_radians)) * 3956 <= 5
ORDER BY Distance;

Interview questions

admin — Wed, 06 Feb 2008 17:33:23 +0000

what’s difference C++ and java
explain about java static
is java use call by reference or call by value
get the algorithm to fine nth value from the last element in single linked list, and write code

all questions look pretty easy though,

anyway, the last algorithm question. because it’s just one direction single linked list, we should travel from the start node to the last node. then we figure it out how many elements in that linked list when we reached the last element. then we can simply travel again from the start node to (m-n)th element. complexity is O(n).
I was trying to bring up some other data structure like stack or hash table but, all those are not good in terms of space efficiency. we do not have to wast precious resource which doesn’t increase the performance that much.

Tomasulo's Algorithm

admin — Fri, 17 Nov 2006 01:02:23 +0000

An Efficient Algorithm for Exploiting Multiple Arithmetic Units, R. M. Tomasulo, IBM Journal, January 1967

instructions and it’s execution plan would be like this.

div Rg, Rb, Rc
add Rh, Rg, Rd (data hazard)
mul Rg, Re, Rf
add Ra, Rh, Rg (data hazard)

then, we should maintain RS(Reservation Station) for 2nd and 4th instructions because those instruction need to wait for previous result. and, RS may look like this. source operands of the 1st instruction are ready and the result would be tagged T1 for register Rg and then put this tag T1 and mark busy bit into register till the result comes out. but the first operand of 2nd instruction is not ready when this instruction decode. it need to wait for Rg which is tagged for T1 by previous instruction. by looking up the register, we can figure out. if busy bit is set, the register is not available. so, we should wait for this tagged result.

Ready	Tag	Contents	Ready	Tag	Contents	Register
Y		Rb	Y		Rc	T1
N	T1		Y		Rd	T2
Y		Re	Y		Rf	T3
N	T2		N	T3		T4

RS and register will watch the result bus to get tagged result. for this example, we will get T1 first then 2nd instruction will be ready to go. when we got 2nd tagged result(T3), left operand of 4th instruction will be ready and the busy bit of the register Rg will be marked free(N).

busy	Tab	register
Y	T4	Ra
Y	T2	Rh
N	T3	Rg

3D Rotation(and Align) about an arbitrary axis

admin — Fri, 17 Nov 2006 00:46:37 +0000

we need several steps to do this job.
1. translate to origin
2. rotate Z
3. rotate Y

Z-Axis Rotation
x’ = x*cos q – y*sin q
y’ = x*sin q + y*cos q
z’ = z

X-Axis Rotation
y’ = y*cos q – z*sin q
z’ = y*sin q + z*cos q
x’ = x

Y-Axis Rotation
z’ = z*cos q – x*sin q
x’ = z*sin q + x*cos q
y’ = y

Comparative Protein Modeling

admin — Fri, 17 Nov 2006 00:44:15 +0000

Once we figure out the protein sequence, then how we can figure out its 3D structure?
There are two kinds of methods. Physical and empirical methods. The physical prediction methods are based on interactions between atoms and include molecular dynamics and energy minimization, whereas the empirical methods depend on the protein structures that have been already determined by experiments.
A comparative protein modeling method designed to find the most probable structure for a sequence given its alignment with related structures. 3D model is obtained by optimally satisfying spatial restraints derived from the alignment and expressed as probability density functions for the features restrained.

Evolution and Physics in comparative protein structure modeling.pdf
Comparative_Protein_Modeling.pdf

Error Detection and Correction:

admin — Fri, 17 Nov 2006 00:33:53 +0000

Everyone knows that it is easy to make mistakes when lots of numbers are involved – and I’m not talking about all those slip-ups that occur in exams. It is so easy to get a few digits of a bank account number mixed up, or for fingers to slip on a keyboard and enter the wrong numbers. Imagine the consequences of these errors: things might go your way, you might get access to Bill Gates’s bank account, for example, but the chances of that are relatively slim. Alternatively, if a bar code gets printed wrongly, you might end up paying the price of a bottle of champagne for your carton of apple juice or if the number on your airline ticket is wrong, who knows where you or your luggage might end up? Luckily, there are schemes in place to detect, and in some cases even correct, such errors almost immediately.

Error Detection and Correction