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ACM Training Session

MACMACMACMACMACMACMAC BRAC UNIVERSITY

8/7/2012

Mahbub

This work is licensed under a Creative Commons Attribution-

NonCommercial-ShareAlike 3.0 Unported License.


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CHECK A NUMBER IS EVEN OR ODD MORE EFFICIENTLY

We all know if a number is divisible by two that is an even number else odd. But in computer
divide is not an easy task, it comes up with a cost.

We know, computer works with bits so if we can convert this odd, even check using bits that
will be faster than our common method. Look carefully the following numbers.

Decimal Binary
Representations Representations
0 0 0 0
1 0 0 1
2 0 1 0
3 0 1 1
4 1 0 0
5 1 0 1
6 1 1 0
7 1 1 1

Look how the last bit is changed, for every odd number the last bit is always one and for even
its zero. Using this pattern we can check if a number is odd or even. Here we will use bitwise
operator & (and), we know when both bit is one the answer is one else zero. So if we and
with one with an odd number the answer will be always one else zero if the number is even.

Java implementation of this method is,

// this method returns true if odd
public static boolean isOdd(int i) {

return (i & 1) == 1;

}

S. Mahbub – Uz – Zaman (09301004)

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Binary representation of 6 (even) is 110 and 1 is 001

6 & 1 = 110 & 001 = 000

Binary representation of 7 (odd) and 1 is 001

7 & 1 = 111 & 001 = 001


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HOW TO CHECK EFFICIENTLY IF A NUMBER IS PRIME OR NOT

A number is a prime number if that number has precisely two distinct divisors, one and itself.
First ten prime numbers are

2, 3, 5, 7, 11, 13, 17, 19, 23, 29

So, if we can find that N has two divisors than it’s a prime number else not, this is actually
brute force approach and the complexity is O (N). How we do that, starting from 1 to N we
have check if N is divisible by 1, 2, 3, ….., N each time it divide we increment our divisor
count by one and at the end we will check if the divisor count is 2 or not.

Can we do better, yes we can. Look carefully the only even prime is 2. If we add an if
condition that if the number is 2 return true else false if the number is even, because other
even numbers can’t not be a prime number. For even numbers the complexity becomes O (1).
So what about odd numbers? How can we improve that? We can reduce the complexity for
odd number O (N / 2). See we don’t need to divide by even numbers because the Number N
is an odd number, so it will never be divide by an even number. So we have to check if N is
divisible by 1, 3, 5, 7, 9, 11, 13, 15 …. N.

We never satisfied! We need more yes the ultimate complexity of an odd number to check
whether it’s prime or not is O (√ ).

For finding if the number has any divisors other then 1 and itself it will appear under the
square root of N, we don’t need to check up to N.


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public static boolean IsPrime(long num) {

if(num < 2)

return false;

if(num == 2)
return true;

if( (num & 1) == 0) // the number is even

return false;

long sqrt = (int) Math.sqrt(num);

for(int i = 3; i <= sqrt; i += 2) {

if(num % i == 0)

return false;

}

return true;

}


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Fibonacci number

Leonardo Pisano Bigollo introduced the Fibonacci sequence. He was an Italian
Mathematician.

0, 1, 1, 2, 3, 5, 8, 13, 21

Each number is the sum of the previous two numbers.

Mathematical representation of Fibonacci sequence is

Fn = Fn-1 + Fn-2, where F0 = 0 and F1 = 1

We can Generate Fibonacci numbers using Dynamic Programming in O (n) and also find a
particular Fibonacci number in O (log n).


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Linear Searching

In our daily life we often search thing like books, movie etc. Similarly in computer science
searching is an important area. There are several search technique some are easy to
understand today I will discuss an easy search technique called linear search which
complexity is O (n).

Suppose we have an array of String which contains name of 5 chocolates now we want to do
some search.

Index 0 1 2 3 4

Name of chocolates a b c d e

So in linear search we start form the first or last and do a check that the current element is our
searched item or not, if we found it then return true or else move to next element until we
have searched the whole array. That’s why the complexity is O (n) because in worst case the
searched item can be in the last position of the array or can be not present.

Case 1: Search a. Found it; a is in the index 0 position.

Case 2: Search g. Not found it. (Worst case)

Case 3: Search e. found it; e is in the index 4 position. (Worst case)

But in real life the input data is not too small always. For 1000000 data linear search will be
too slow, yes there are other search techniques which are faster than linear search.


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Perfect Square Number

Perfect square number is an integer that is the square of an integer. For example we can say
that 25 is a perfect square number, since it is a square of 5.

5 * 5 = 25

We can also say that the square root of a perfect square number is also a integer.

√ =6

We can check if a number is perfect square or not. First we get the square root of the number,
then we multiply the result by itself, if it is a squre number then it shuld macth to original
number.

√ =6

6 * 6 = 36 [36 is a perfect square number]

√ = 5.9160 (cast it into long) = 5

5 * 5 35 [whether 35 is not]


boolean isPSN(long num) {

if (num < 0)
return false;

long sqr = (long) Math.sqrt(num);
return sqr * sqr == num;
}


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