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Buacm 3


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Buacm 3

  2. 2. Page 1 of 7 CHECK A NUMBER IS EVEN OR ODD MORE EFFICIENTLYWe all know if a number is divisible by two that is an even number else odd. But in computerdivide 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 thatwill 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 1Look how the last bit is changed, for every odd number the last bit is always one and for evenits zero. Using this pattern we can check if a number is odd or even. Here we will use bitwiseoperator & (and), we know when both bit is one the answer is one else zero. So if we andwith 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 oddpublic static boolean isOdd(int i) { return (i & 1) == 1;}S. Mahbub – Uz – Zaman (09301004)
  3. 3. Page 2 of 7Binary representation of 6 (even) is 110 and 1 is 0016 & 1 = 110 & 001 = 000Binary representation of 7 (odd) and 1 is 0017 & 1 = 111 & 001 = 001S. Mahbub – Uz – Zaman (09301004)
  4. 4. Page 3 of 7 HOW TO CHECK EFFICIENTLY IF A NUMBER IS PRIME OR NOTA number is a prime number if that number has precisely two distinct divisors, one and itself.First ten prime numbers are2, 3, 5, 7, 11, 13, 17, 19, 23, 29So, if we can find that N has two divisors than it’s a prime number else not, this is actuallybrute force approach and the complexity is O (N). How we do that, starting from 1 to N wehave check if N is divisible by 1, 2, 3, ….., N each time it divide we increment our divisorcount 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 ifcondition that if the number is 2 return true else false if the number is even, because othereven 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 forodd number O (N / 2). See we don’t need to divide by even numbers because the Number Nis an odd number, so it will never be divide by an even number. So we have to check if N isdivisible 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 checkwhether it’s prime or not is O (√ ).For finding if the number has any divisors other then 1 and itself it will appear under thesquare root of N, we don’t need to check up to N.S. Mahbub – Uz – Zaman (09301004)
  5. 5. Page 4 of 7Java implementation of this method is,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;}S. Mahbub – Uz – Zaman (09301004)
  6. 6. Page 5 of 7 Fibonacci numberLeonardo Pisano Bigollo introduced the Fibonacci sequence. He was an ItalianMathematician.0, 1, 1, 2, 3, 5, 8, 13, 21Each number is the sum of the previous two numbers.Mathematical representation of Fibonacci sequence isFn = Fn-1 + Fn-2, where F0 = 0 and F1 = 1We can Generate Fibonacci numbers using Dynamic Programming in O (n) and also find aparticular Fibonacci number in O (log n).S. Mahbub – Uz – Zaman (09301004)
  7. 7. Page 6 of 7 Linear SearchingIn our daily life we often search thing like books, movie etc. Similarly in computer sciencesearching is an important area. There are several search technique some are easy tounderstand today I will discuss an easy search technique called linear search whichcomplexity is O (n).Suppose we have an array of String which contains name of 5 chocolates now we want to dosome search. Index 0 1 2 3 4 Name of chocolates a b c d eSo in linear search we start form the first or last and do a check that the current element is oursearched item or not, if we found it then return true or else move to next element until wehave searched the whole array. That’s why the complexity is O (n) because in worst case thesearched 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 betoo slow, yes there are other search techniques which are faster than linear search.S. Mahbub – Uz – Zaman (09301004)
  8. 8. Page 7 of 7 Perfect Square NumberPerfect square number is an integer that is the square of an integer. For example we can saythat 25 is a perfect square number, since it is a square of 5. 5 * 5 = 25We can also say that the square root of a perfect square number is also a integer. √ =6We 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 originalnumber. √ =6 6 * 6 = 36 [36 is a perfect square number] √ = 5.9160 (cast it into long) = 5 5 * 5 35 [whether 35 is not]Java implementation of this method is,boolean isPSN(long num) { if (num < 0) return false; long sqr = (long) Math.sqrt(num); return sqr * sqr == num;}S. Mahbub – Uz – Zaman (09301004)