Thursday, June 20, 2013

Problem 3- The Power of Cryptography

 Power of Cryptography 

Background

Current work in cryptography involves (among other things) large prime numbers and computing powers of numbers modulo functions of these primes. Work in this area has resulted in the practical use of results from number theory and other branches of mathematics once considered to be of only theoretical interest. This problem involves the efficient computation of integer roots of numbers.

The Problem

Given an integer tex2html_wrap_inline32 and an integer tex2html_wrap_inline34 you are to write a program that determines tex2html_wrap_inline36 , the positive tex2html_wrap_inline38 root of p. In this problem, given such integers n and p, p will always be of the form tex2html_wrap_inline48 for an integer k (this integer is what your program must find).

The Input

The input consists of a sequence of integer pairs n and p with each integer on a line by itself. For all such pairs tex2html_wrap_inline56 , tex2html_wrap_inline58 and there exists an integer k, tex2html_wrap_inline62 such that tex2html_wrap_inline64 .

The Output

For each integer pair n and p the value tex2html_wrap_inline36 should be printed, i.e., the number k such that tex2html_wrap_inline64 .

Sample Input

2
16
3
27
7
4357186184021382204544

Sample Output

4
3
1234
 
 
------------
scroll down for solution working till 10^18 but not more than that






























































//--------Main class
package
crypt;
import
java.io.InputStreamReader;
import
java.util.ArrayList;
import
java.util.Scanner;
public
class Main {
/**
*
@param args
*/
public static void main(String[] args) {
long n, p;
int k;
String more =
"n";
Scanner scan =
new Scanner(new InputStreamReader(System.in));
ArrayList<Long> nlist =
new ArrayList<Long>();
ArrayList<Long> plist =
new ArrayList<Long>();
ArrayList<Integer> klist =
new ArrayList<Integer>();
System.
out.println("Enter Numbers Less than 10^18");
do {
System.
out.println("Enter n ,p:");
n = scan.nextLong();
p = scan.nextLong();
nlist.add(n);
plist.add(p);
System.
out.println("MORE ?(y)");
more = scan.next();
}
while (more.equalsIgnoreCase("y"));
System.
out.println("USER INPUT");
for (int i = 0; i < nlist.size(); i++) {
System.
out.println(nlist.get(i) + " " + plist.get(i));
k = Divider.getPower(n, p);
klist.add(k);
}
System.
out.println("OUTPUT POWERS");
for (Integer integer : klist) {
System.
out.println(integer);
}
}
}
//------------------2nd class
package
crypt;
public
class Divider {
public static int getPower(long n, long p) {
int k = 0;
while (p > 1) {
p /= n;
k++;
}
return k;
}
}
//-----------EOP
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

Problem 2- Fibonacci word sequence

PROBLEM 2 The Fibonacci word sequence of bit strings is defined as:
F(n) = $\displaystyle \cases{0 & if $n = 0$ \cr 1 & if $n = 1$ \cr F (n - 1) + F (n - 2) & if $n \ge 2$ \cr}
$

Here + denotes concatenation of strings. The first few elements are:
n F(n)
0 0
1 1
2 10
3 101
4 10110
5 10110101
6 1011010110110
7 101101011011010110101
8 1011010110110101101011011010110110
9 1011010110110101101011011010110110101101011011010110101
Given a bit pattern p and a number n, how often does p occur in F(n)?

Input 

The first line of each test case contains the integer n (0$ \le$n$ \le$100). The second line contains the bit pattern p. The pattern p is nonempty and has a length of at most 100 000 characters.

Output 

For each test case, display its case number followed by the number of occurrences of the bit pattern p in F(n). Occurrences may overlap. The number of occurrences will be less than 263.

Sample Input 

6
10
7
10
6
01
6
101
96
10110101101101

Sample Output 

Case 1: 5
Case 2: 8
Case 3: 4
Case 4: 4
Case 5: 7540113804746346428

Scroll down for answer:




































//--------------main class
package
fib;
import
java.io.InputStreamReader;
import
java.util.Scanner;
public
class Fibonacci {
/**
*
@param args
*/
public static void main(String[] args) {
int n;
int occurrance;
String numString;
String pattern;
Scanner scan =
new Scanner(new InputStreamReader(System.in));
System.
out.println("Enter Number");
n = scan.nextInt();
numString = Creator.createString(n);
System.
out.println(n + "=" + numString);
System.
out.println("{EXTRAINFO: The representation requires " + numString.length()
+
" bits]");
// take aid of file for displaying non-visible strings on console
// Creator.writeTofile(numString);
System.
out.println("enter Pattern:");
pattern = scan.next();
// find number of occurrence of given pattern in this string
occurrance = Creator.matchPattern(numString, pattern);
System.
out.println("pattern " + pattern + " has occurred " + occurrance
+
" times");
}
}
//-------------------2nd class
package
fib;
import
java.io.File;
import
java.io.FileOutputStream;
import
java.io.IOException;
public
class Creator {
public
static String createString(int n) {
if
(n == 0)
return
"0";
else
if (n == 1)
return
"1";
else
return
"" + createString(n - 1) + createString(n - 2);
}
public
static void writeTofile(String s) {
File f =
new File("D:\\text.txt");
try
{
f.createNewFile();
FileOutputStream fos =
new FileOutputStream(f);
fos.write(s.getBytes());
}
catch (IOException e) {
//
TODO Auto-generated catch block
e.printStackTrace();
}
}
public
static int matchPattern(String numString, String pattern) {
int
count = 0;
int
pos = 0;
String replaceWith =
"";
int
patternLength = pattern.length();
for
(int i = 0; i < patternLength; i++) {
replaceWith +=
"x";
}
int
lastIndex = numString.lastIndexOf(pattern);
while
(pos < lastIndex) {
pos = numString.indexOf(pattern);
count++;
numString = numString.replaceFirst(pattern, replaceWith);
}
return
count;
}
}
-------------------------------------

Wednesday, June 19, 2013

Problem 1- 3n+1 problem

This is Problem1.
You can see the source code(solution) by scrolling little below the problem or try the problem first and see the solution later.
(This is my personal collection)
 The 3n + 1 problem 

Background

Problems in Computer Science are often classified as belonging to a certain class of problems (e.g., NP, Unsolvable, Recursive). In this problem you will be analyzing a property of an algorithm whose classification is not known for all possible inputs.

The Problem

Consider the following algorithm:
 
  1.    input n
  2.    print n
  3.    if n = 1 then STOP

  4.       if n is odd then  tex2html_wrap_inline44 

  5.       else  tex2html_wrap_inline46 

  6.    GOTO 2
Given the input 22, the following sequence of numbers will be printed 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1 It is conjectured that the algorithm above will terminate (when a 1 is printed) for any integral input value. Despite the simplicity of the algorithm, it is unknown whether this conjecture is true. It has been verified, however, for all integers n such that 0 < n < 1,000,000 (and, in fact, for many more numbers than this.) Given an input n, it is possible to determine the number of numbers printed (including the 1). For a given n this is called the cycle-length of n. In the example above, the cycle length of 22 is 16. For any two numbers i and j you are to determine the maximum cycle length over all numbers between i and j.

The Input

The input will consist of a series of pairs of integers i and j, one pair of integers per line. All integers will be less than 1,000,000 and greater than 0. You should process all pairs of integers and for each pair determine the maximum cycle length over all integers between and including i and j. You can assume that no operation overflows a 32-bit integer.

The Output

For each pair of input integers i and j you should output i, j, and the maximum cycle length for integers between and including i and j. These three numbers should be separated by at least one space with all three numbers on one line and with one line of output for each line of input. The integers i and j must appear in the output in the same order in which they appeared in the input and should be followed by the maximum cycle length (on the same line).

Sample Input

1 10
100 200
201 210
900 1000

Sample Output

1 10 20
100 200 125
201 210 89
900 1000 174



scroll Down for source -code written by me
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||---------------------------------------------------------------------
//Programming Language JAVA
//tool Eclipse IDE
package problem1;
public class Series {
 int lower;
 int upper;
 int maxCycleLength;

 public int getLower() {
  return lower;
 }
 public void setLower(int lower) {
  this.lower = lower;
 }
 public int getUpper() {
  return upper;
 }
 public void setUpper(int upper) {
  this.upper = upper;
 }
 public int getMaxCycleLength() {
  return maxCycleLength;
 }
 public void setMaxCycleLength(int maxCycleLength) {
  this.maxCycleLength = maxCycleLength;
 }

 public Series(){
  this.lower = 0;
  this.upper = 0;
  this.maxCycleLength = 0;
 }
 public Series(int i,int j){
  this.lower = i;
  this.upper = j;
  this.maxCycleLength = 0;
 }
}
//-----------------------------------------2nd Class---------------------------
//MAIN CLASS
package problem1;

import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Scanner;
public class CycleLength {
 /**
  * @param args
  */
 public static ArrayList<Series> list = new ArrayList<Series>();

 public static void main(String[] args) {

  int i,j;
  String moreSeries = "y";
  Scanner scanner = new Scanner(new InputStreamReader(System.in));


 do{
      System.out.println("Enter series [Less than 1,000,000 greater than 0]");
     
   i = scanner.nextInt(); j = scanner.nextInt();

   i = (i>0 && i<1000000 )? i :0;
   j = (j>0 && j<1000000 )? j :0;
   if(i ==0 || j ==0){
    System.out.println("Invalid Input.Exiting...");
    System.exit(1);
  
   }
 
   Series s = new Series(i,j);
   list.add(s);
   System.out.println("More inputs?");
   moreSeries = scanner.next();
 }while(moreSeries.equalsIgnoreCase("y"));
   processInput();
  
 }
 static void processInput(){
  System.out.println("Processing input:");
  Series s;
  for(int i=0;i<list.size();i++){
   s = (Series)list.remove(i);
   System.out.println(s.getLower()+"-"+s.getUpper());
   s = processSeries(s);
   list.add(i, s);
  }

  System.out.println("---------OUTPUT---------");
  for(int i=0;i<list.size();i++){
   s = (Series)list.get(i);
   System.out.println(s.getLower()+"-"+s.getUpper()+"-"+s.getMaxCycleLength());
   }

 }
 static Series processSeries(Series s){
  int i,j;
  int n,end,count=0,maxCount = 0;
  i=s.getLower();
  j=s.getUpper();

  if(i<=j){
   n = i;
   end = j;
  }
  else{
   n = j;
   end = i;
  }
 
  do{
  count = getCycleLength(n);
  maxCount = maxCount>count ? maxCount :count;
  n++;
  }while(n<=end);
  s.setMaxCycleLength(maxCount);
  return s;
 }
 static int getCycleLength(int n){
  int count=0;
  System.out.println("**CYCLE FOR "+n+"**");
  do{
   System.out.print(n+" ");
   n = n%2==0 ? n/2 : 3*n+1;
   count++;
  }while(n!=1);
  System.out.println("1");
  ++count;
  System.out.println("-------------->"+count);
  return count;

 }
}
//-------------END-OF-CODE----------------------