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I converted decimal to binary number however i dont know how to represent on label. I have a list of numbers 0 and 1,Now, how do I display the information on labels.In fact, i dont know how to represent on label.
private void btnRun_Click(object sender, EventArgs e)
{
var decimaltoBinary = fnDecimalToBinary(Convert.ToInt32(txtenterNumber.Text));
}
private List<int> fnDecimalToBinary(int number)
{
int[] decimalNumbers = new int[] { 1, 2, 4, 8, 16, 32, 64, 128, 256 };
List<int> binaryNumbers = new List<int>();
int locDecimalArray = 0;
int sumNumber = 0;
for (int i = 0; i < decimalNumbers.Length; i++)
{
if (number < decimalNumbers[i])
{
sumNumber = number;
locDecimalArray = i - 1;
for (int j = locDecimalArray; j >= 0; j--)
{
if (sumNumber == 0)
{
binaryNumbers.Add(0);
return binaryNumbers;
}
else if (sumNumber >= decimalNumbers[j])
{
sumNumber = sumNumber - decimalNumbers[j];
binaryNumbers.Add(1);
}
else if (sumNumber < decimalNumbers[j])
{
binaryNumbers.Add(0);
}
}
return binaryNumbers;
}
}
return binaryNumbers;
}
It seems that you've received a comment that explains how you can convert your List<int> to the string value you need for the Label control. However, it seems to me that for the purposes of this exercise, you might benefit from some help with the decimal-to-binary conversion itself. There are already a number of similar questions on Stack Overflow dealing with this scenario (as you can guess, converting to binary text is a fairly common programming exercise), but of course none will start with your specific code, so I think it's worth writing yet another answer. :)
Basing the conversion on a pre-computed list of numeric values is not a terrible way to go, especially for the purposes of learning. But your version has a bunch of extra code that's just not necessary:
Your outer loop doesn't accomplish anything except verify that the number passed is within the range permitted by your pre-computed values. But this can be done as part of the conversion itself.
Furthermore, I'm not convinced that returning an empty list is really the best way to deal with invalid input. Throwing an exception would be more appropriate, as this forces the caller to deal with errors, and allows you to provide a textual message for display to the user.
The value 0 is always less than any of the digit values you've pre-computed, so there's no need to check for that explicitly. You really only need the if and a single else inside the inner loop.
Since you are the one populating the array, and since for loops are generally more readable when they start at 0 and increment the index as opposed to starting at the end and decrement, it seems to me that you would be better off writing the pre-computed values in reverse.
Entering numbers by hand is a pain and it seems to me that the method could be more flexible (i.e. support larger binary numbers) if you allowed the caller to pass the number of digits to produce, and used that to compute the values at run-time (though, if for performance reasons that's less desirable, pre-computing the largest digits that would be used and storing that in a static field, and then just using whatever subset of that you need, would be yet another suitable approach).
With those changes, you would get something like this:
private List<int> DecimalToBinary(int number, int digitCount)
{
// The number can't itself have more than 32 digits, so there's
// no point in allowing the caller to ask for more than that.
if (digitCount < 1 || digitCount > 32)
{
throw new ArgumentOutOfRangeException("digitCount",
"digitCount must be between 1 and 32, inclusive");
}
long[] digitValues = Enumerable.Range(0, digitCount)
.Select(i => (long)Math.Pow(2, digitCount - i - 1)).ToArray();
List<int> binaryDigits = new List<int>(digitCount);
for (int i = 0; i < digitValues.Length; i++)
{
if (digitValues[i] <= number)
{
binaryDigits.Add(1);
number = (int)(number - digitValues[i]);
}
else
{
binaryDigits.Add(0);
}
}
if (number > 0)
{
throw new ArgumentOutOfRangeException("digitCount",
"digitCount was not large number to accommodate the number");
}
return binaryDigits;
}
And here's an example of how you might use it:
private void button1_Click(object sender, EventArgs e)
{
int number;
if (!int.TryParse(textBox1.Text, out number))
{
MessageBox.Show("Could not convert user input to an int value");
return;
}
try
{
List<int> binaryDigits = DecimalToBinary(number, 8);
label3.Text = string.Join("", binaryDigits);
}
catch (ArgumentOutOfRangeException e1)
{
MessageBox.Show("Exception: " + e1.Message, "Could not convert to binary");
}
}
Now, the above example fits the design you originally had, just cleaned it up a bit. But the fact is, the computer already knows binary. That's how it stores numbers, and even if it didn't, C# includes operators that treat the numbers as binary (so if the computer didn't use binary, the run-time would be required to translate for you anyway). Given that, it's actually a lot easier to convert just by looking at the individual bits. For example:
private List<int> DecimalToBinary2(int number, int digitCount)
{
if (digitCount < 1 || digitCount > 32)
{
throw new ArgumentOutOfRangeException("digitCount",
"digitCount must be between 1 and 32, inclusive");
}
if (number > Math.Pow(2, digitCount) - 1)
{
throw new ArgumentOutOfRangeException("digitCount",
"digitCount was not large number to accommodate the number");
}
List<int> binaryDigits = new List<int>(digitCount);
for (int i = digitCount - 1; i >= 0; i--)
{
binaryDigits.Add((number & (1 << i)) != 0 ? 1 : 0);
}
return binaryDigits;
}
The above simply starts at the highest possible binary digit (given the desired count of digits), and checks each individual digit in the provided number, using the "bit-shift" operator << and the logical bitwise "and" operator &. If you're not already familiar with binary arithmetic, shift operations, and these operators, this might seem like overkill. But it's actually a fundamental aspect of how computers work, worth knowing, and of course as shown above, can dramatically simplify code required to deal with binary data (to the point where parameter validation code takes up half the method :) ).
One last thing: this entire discussion ignores the fact that you're using a signed int value, rather than the unsigned uint type. Technically, this means your code really ought to be able to handle negative numbers as well. However, doing so is a bit trickier when you also want to deal with binary digit counts that are less than the natural width of the number in the numeric type (e.g. 32 bits for an int). Conversely, if you don't want to support negative numbers, you should really be using the uint type instead of int.
I figured that trying to address that particular complication would dramatically increase the complexity of this answer and take away from the more fundamental details that seemed worth conveying. So I've left that out. But I do encourage you to look more deeply into how computers represent numbers, and why negative numbers require more careful handling than the above code is doing.
I have recently come across this line of code and what it does is that it goes through an array and returns the value that is seen most often. For example 1,1,2,1,3 so it will return 1 because it appears more than 2 and 3. What I am trying to do is understand how it works so what I did was I went through it with visual studio step by step but it is not ringing any bells.
Can anyone help me understand what is going on here? It would be a total plus if someone can tell me what does c do and what is the logic behind the arguments in the if statements.
int[] arr = a;
int c = 1, maxcount = 1, maxvalue = 0;
int result = 0;
for (int i = 0; i < arr.Length; i++)
{
maxvalue = arr[i];
for (int j = 0; j <arr.Length; j++)
{
if (maxvalue == arr[j] && j != i)
{
c++;
if (c > maxcount)
{
maxcount = c;
result = arr[i];
}
}
else
{
c=1;
}
}
}
return result;
EDIT: On closer examination, the code snippet has a nested loop and is conventionally counting the maximum seen element by simply keeping track of the maximum seen times and the element that was seen and keeping them in sync.
That looks like an implementation of the Boyer-Moore majority vote counting algorithm. They have a nice illustration here.
The logic is simple, and is to compute the majority in a single pass, taking O(n) time. Note that majority here means that more than 50% of the array must be filled with that element. If there is no majority element, you get an "incorrect" result.
Verifying if the element is actually forming a majority is done in a separate pass usually.
It is not computing the most frequent element - what it is computing is the longest run of elements.
Also, it is not doing it very efficiently, the inner loop only needs to compute upto i-1, not upto arr.Length.
c is keeping track of the current run length. The first "if" is to check if this is a "continouous run". The second "if" (after reaching the last element in the run) will check if this run is longer than any run you have seen so far.
In the above input sample, you are getting 1 as answer because it is the longest run. Try with an input where the element with the longest run is not the same as the most frequent element. (e.g., 2,1,1,1,3,2,3,2,3,2,3,2 - here 2 is the most frequent element, but 1,1,1 is the longest run).
I've always loved reducing number of code lines by using simple but smart math approaches. This situation seems to be one of those that need this approach. So what I basically need is to sum up digits in the odd and even places separately with minimum code. So far this is the best way I have been able to think of:
string number = "123456789";
int sumOfDigitsInOddPlaces=0;
int sumOfDigitsInEvenPlaces=0;
for (int i=0;i<number.length;i++){
if(i%2==0)//Means odd ones
sumOfDigitsInOddPlaces+=number[i];
else
sumOfDigitsInEvenPlaces+=number[i];
{
//The rest is not important
Do you have a better idea? Something without needing to use if else
int* sum[2] = {&sumOfDigitsInOddPlaces,&sumOfDigitsInEvenPlaces};
for (int i=0;i<number.length;i++)
{
*(sum[i&1])+=number[i];
}
You could use two separate loops, one for the odd indexed digits and one for the even indexed digits.
Also your modulus conditional may be wrong, you're placing the even indexed digits (0,2,4...) in the odd accumulator. Could just be that you're considering the number to be 1-based indexing with the number array being 0-based (maybe what you intended), but for algorithms sake I will consider the number to be 0-based.
Here's my proposition
number = 123456789;
sumOfDigitsInOddPlaces=0;
sumOfDigitsInEvenPlaces=0;
//even digits
for (int i = 0; i < number.length; i = i + 2){
sumOfDigitsInEvenPlaces += number[i];
}
//odd digits, note the start at j = 1
for (int j = 1; i < number.length; i = i + 2){
sumOfDigitsInOddPlaces += number[j];
}
On the large scale this doesn't improve efficiency, still an O(N) algorithm, but it eliminates the branching
Since you added C# to the question:
var numString = "123456789";
var odds = numString.Split().Where((v, i) => i % 2 == 1);
var evens = numString.Split().Where((v, i) => i % 2 == 0);
var sumOfOdds = odds.Select(int.Parse).Sum();
var sumOfEvens = evens.Select(int.Parse).Sum();
Do you like Python?
num_string = "123456789"
odds = sum(map(int, num_string[::2]))
evens = sum(map(int, num_string[1::2]))
This Java solution requires no if/else, has no code duplication and is O(N):
number = "123456789";
int[] sums = new int[2]; //sums[0] == sum of even digits, sums[1] == sum of odd
for(int arrayIndex=0; arrayIndex < 2; ++arrayIndex)
{
for (int i=0; i < number.length()-arrayIndex; i += 2)
{
sums[arrayIndex] += Character.getNumericValue(number.charAt(i+arrayIndex));
}
}
Assuming number.length is even, it is quite simple. Then the corner case is to consider the last element if number is uneven.
int i=0;
while(i<number.length-1)
{
sumOfDigitsInEvenPlaces += number[ i++ ];
sumOfDigitsInOddPlaces += number[ i++ ];
}
if( i < number.length )
sumOfDigitsInEvenPlaces += number[ i ];
Because the loop goes over i 2 by 2, if number.length is even, removing 1 does nothing.
If number.length is uneven, it removes the last item.
If number.length is uneven, then the last value of i when exiting the loop is that of the not yet visited last element.
If number.length is uneven, by modulo 2 reasoning, you have to add the last item to sumOfDigitsInEvenPlaces.
This seems slightly more verbose, but also more readable, to me than Anonymous' (accepted) answer. However, benchmarks to come.
Well, the compiler seems to think my code more understandable as well, since he removes it all if I don't print the results (which explains why I kept getting a time of 0 all along...). The other code though is obfuscated enough for even the compiler.
In the end, even with huge arrays, it's pretty hard for clock_t to tell the difference between the two. You get about a third less instructions in the second case, but since everything's in cache (and your running sums even in registers) it doesn't matter much.
For the curious, I've put the disassembly of both versions (compiled from C) here : http://pastebin.com/2fciLEMw
I'm trying to calculate
If we calculated every possible combination of numbers from 0 to (c-1)
with a length of x
what set would occur at point i
For example:
c = 4
x = 4
i = 3
Would yield:
[0000]
[0001]
[0002]
[0003] <- i
[0010]
....
[3333]
This is very nearly the same problem as in the related question Logic to select a specific set from Cartesian set. However, because x and i are large enough to require the use of BigInteger objects, the code has to be changed to return a List, and take an int, instead of a string array:
int PossibleNumbers;
public List<int> Get(BigInteger Address)
{
List<int> values = new List<int>();
BigInteger sizes = new BigInteger(1);
for (int j = 0; j < PixelArrayLength; j++)
{
BigInteger index = BigInteger.Divide(Address, sizes);
index = (index % PossibleNumbers);
values.Add((int)index);
sizes *= PossibleNumbers;
}
return values;
}
This seems to behave as I'd expect, however, when I start using values like this:
c = 66000
x = 950000
i = (66000^950000)/2
So here, I'm looking for the ith value in the cartesian set of 0 to (c-1) of length 950000, or put another way, the halfway point.
At this point, I just get a list of zeroes returned. How can I solve this problem?
Notes: It's quite a specific problem, and I apologise for the wall-of-text, I do hope it's not too much, I was just hoping to properly explain what I meant. Thanks to you all!
Edit: Here are some more examples: http://pastebin.com/zmSDQEGC
Here is a generic base converter... it takes a decimal for the base10 value to convert into your newBase and returns an array of int's. If you need a BigInteger this method works perfectly well with just changing the base10Value to BigInteger.
EDIT: Converted method to BigInteger since that's what you need.
EDIT 2: Thanks phoog for pointing out BigInteger is base2 so changing the method signature.
public static int[] ConvertToBase(BigInteger value, int newBase, int length)
{
var result = new Stack<int>();
while (value > 0)
{
result.Push((int)(value % newBase));
if (value < newBase)
value = 0;
else
value = value / newBase;
}
for (var i = result.Count; i < length; i++)
result.Push(0);
return result.ToArray();
}
usage...
int[] a = ConvertToBase(13, 4, 4) = [0,0,3,1]
int[] b = ConvertToBase(0, 4, 4) = [0,0,3,1]
int[] c = ConvertToBase(1234, 12, 4) = [0,8,6,10]
However the probelm you specifically state is a bit large to test it on. :)
Just calculating 66000 ^ 950000 / 2 is a good bit of work as Phoog mentioned. Unless of course you meant ^ to be the XOR operator. In which case it's quite fast.
EDIT: From the comments... The largest base10 number that can be represented given a particular newBase and length is...
var largestBase10 = BigInteger.Pow(newBase, length)-1;
The first expression of the problem boils down to "write 3 as a 4-digit base-4 number". So, if the problem is "write i as an x-digit base-c number", or, in this case, "write (66000^950000)/2 as a 950000-digit base 66000 number", then does that make it easier?
If you're specifically looking for the halfway point of the cartesian product, it's not so hard. If you assume that c is even, then the most significant digit is c / 2, and the rest of the digits are zero. If your return value is all zeros, then you may have an off-by-one error, or the like, since actually only one digit is incorrect.
I'm a beginner in C#, I'm trying to write an application to get primes between two numbers entered by the user. The problem is: At large numbers (valid numbers are in the range from 1 to 1000000000) getting the primes takes long time and according to the problem I'm solving, the whole operation must be carried out in a small time interval. This is the problem link for more explanation:
SPOJ-Prime
And here's the part of my code that's responsible of getting primes:
public void GetPrime()
{
int L1 = int.Parse(Limits[0]);
int L2 = int.Parse(Limits[1]);
if (L1 == 1)
{
L1++;
}
for (int i = L1; i <= L2; i++)
{
for (int k = L1; k <= L2; k++)
{
if (i == k)
{
continue;
}
else if (i % k == 0)
{
flag = false;
break;
}
else
{
flag = true;
}
}
if (flag)
{
Console.WriteLine(i);
}
}
}
Is there any faster algorithm?
Thanks in advance.
I remember solving the problem like this:
Use the sieve of eratosthenes to generate all primes below sqrt(1000000000) = ~32 000 in an array primes.
For each number x between m and n only test if it's prime by testing for divisibility against numbers <= sqrt(x) from the array primes. So for x = 29 you will only test if it's divisibile by 2, 3 and 5.
There's no point in checking for divisibility against non-primes, since if x divisible by non-prime y, then there exists a prime p < y such that x divisible by p, since we can write y as a product of primes. For example, 12 is divisible by 6, but 6 = 2 * 3, which means that 12 is also divisible by 2 or 3. By generating all the needed primes in advance (there are very few in this case), you significantly reduce the time needed for the actual primality testing.
This will get accepted and doesn't require any optimization or modification to the sieve, and it's a pretty clean implementation.
You can do it faster by generalising the sieve to generate primes in an interval [left, right], not [2, right] like it's usually presented in tutorials and textbooks. This can get pretty ugly however, and it's not needed. But if anyone is interested, see:
http://pastie.org/9199654 and this linked answer.
You are doing a lot of extra divisions that are not needed - if you know a number is not divisible by 3, there is no point in checking if it is divisible by 9, 27, etc. You should try to divide only by the potential prime factors of the number. Cache the set of primes you are generating and only check division by the previous primes. Note that you do need to generate the initial set of primes below L1.
Remember that no number will have a prime factor that's greater than its own square root, so you can stop your divisions at that point. For instance, you can stop checking potential factors of the number 29 after 5.
You also do can increment by 2 so you can disregard checking if an even number is prime altogether (special casing the number 2, of course.)
I used to ask this question during interviews - as a test I compared an implementation similar to yours with the algorithm I described. With the optimized algorithm, I could generate hundreds of thousands of primes very fast - I never bothered waiting around for the slow, straightforward implementation.
You could try the Sieve of Eratosthenes. The basic difference would be that you start at L1 instead of starting at 2.
Let's change the question a bit: How quickly can you generate the primes between m and n and simply write them to memory? (Or, possibly, to a RAM disk.) On the other hand, remember the range of parameters as described on the problem page: m and n can be as high as a billion, while n-m is at most a million.
IVlad and Brian most of a competitive solution, even if it is true that a slower solution could be good enough. First generate or even precompute the prime numbers less than sqrt(billion); there aren't very many of them. Then do a truncated Sieve of Eratosthenes: Make an array of length n-m+1 to keep track of the status of every number in the range [m,n], with initially every such number marked as prime (1). Then for each precomputed prime p, do a loop that looks like this:
for(k=ceil(m/p)*p; k <= n; k += p) status[k-m] = 0;
This loop marks all of the numbers in the range m <= x <= n as composite (0) if they are multiple of p. If this is what IVlad meant by "pretty ugly", I don't agree; I don't think that it's so bad.
In fact, almost 40% of this work is just for the primes 2, 3, and 5. There is a trick to combine the sieve for a few primes with initialization of the status array. Namely, the pattern of divisibility by 2, 3, and 5 repeats mod 30. Instead of initializing the array to all 1s, you can initialize it to a repeating pattern of 010000010001010001010001000001. If you want to be even more cutting edge, you can advance k by 30*p instead of by p, and only mark off the multiples in the same pattern.
After this, realistic performance gains would involve steps like using a bit vector rather than a char array to keep the sieve data in on-chip cache. And initializing the bit vector word by word rather than bit by bit. This does get messy, and also hypothetical since you can get to the point of generating primes faster than you can use them. The basic sieve is already very fast and not very complicated.
One thing no one's mentioned is that it's rather quick to test a single number for primality. Thus, if the range involved is small but the numbers are large (ex. generate all primes between 1,000,000,000 and 1,000,100,000), it would be faster to just check every number for primality individually.
There are many algorithms finding prime numbers. Some are faster, others are easier.
You can start by making some easiest optimizations. For example,
why are you searching if every number is prime? In other words, are you sure that, given a range of 411 to 418, there is a need to search if numbers 412, 414, 416 and 418 are prime? Numbers which divide by 2 and 3 can be skipped with very simple code modifications.
if the number is not 5, but ends by a digit '5' (1405, 335), it is not prime bad idea: it will make the search slower.
what about caching the results? You can then divide by primes rather by every number. Moreover, only primes less than square root of the number you search are concerned.
If you need something really fast and optimized, taking an existing algorithm instead of reinventing the wheel can be an alternative. You can also try to find some scientific papers explaining how to do it fast, but it can be difficult to understand and to translate to code.
int ceilingNumber = 1000000;
int myPrimes = 0;
BitArray myNumbers = new BitArray(ceilingNumber, true);
for(int x = 2; x < ceilingNumber; x++)
if(myNumbers[x])
{
for(int y = x * 2; y < ceilingNumber; y += x)
myNumbers[y] = false;
}
for(int x = 2; x < ceilingNumber; x++)
if(myNumbers[x])
{
myPrimes++;
Console.Out.WriteLine(x);
}
Console.Out.WriteLine("======================================================");
Console.Out.WriteLine("There is/are {0} primes between 0 and {1} ",myPrimes,ceilingNumber);
Console.In.ReadLine();
I think i have a very fast and efficient(generate all prime even if using type BigInteger) algorithm to getting prime number,it much more faster and simpler than any other one and I use it to solve almost problem related to prime number in Project Euler with just a few seconds for complete solution(brute force)
Here is java code:
public boolean checkprime(int value){ //Using for loop if need to generate prime in a
int n, limit;
boolean isprime;
isprime = true;
limit = value / 2;
if(value == 1) isprime =false;
/*if(value >100)limit = value/10; // if 1 number is not prime it will generate
if(value >10000)limit = value/100; //at lest 2 factor (not 1 or itself)
if(value >90000)limit = value/300; // 1 greater than average 1 lower than average
if(value >1000000)limit = value/1000; //ex: 9997 =13*769 (average ~ sqrt(9997) is 100)
if(value >4000000)limit = value/2000; //so we just want to check divisor up to 100
if(value >9000000)limit = value/3000; // for prime ~10000
*/
limit = (int)Math.sqrt(value); //General case
for(n=2; n <= limit; n++){
if(value % n == 0 && value != 2){
isprime = false;
break;
}
}
return isprime;
}
import java.io.*;
import java.util.Scanner;
class Test{
public static void main(String args[]){
Test tt=new Test();
Scanner obj=new Scanner(System.in);
int m,n;
System.out.println(i);
m=obj.nextInt();
n=obj.nextInt();
tt.IsPrime(n,m);
}
public void IsPrime(int num,int k)
{
boolean[] isPrime = new boolean[num+1];
// initially assume all integers are prime
for (int i = 2; i <= num; i++) {
isPrime[i] = true;
}
// mark non-primes <= N using Sieve of Eratosthenes
for (int i = 2; i*i <= num; i++) {
// if i is prime, then mark multiples of i as nonprime
// suffices to consider mutiples i, i+1, ..., N/i
if (isPrime[i]) {
for (int j = i; i*j <=num; j++) {
isPrime[i*j] = false;
}
}
}
for (int i =k; i <= num; i++) {
if (isPrime[i])
{
System.out.println(i);
}
}
}
}
List<int> prime(int x, int y)
{
List<int> a = new List<int>();
int b = 0;
for (int m = x; m < y; m++)
{
for (int i = 2; i <= m / 2; i++)
{
b = 0;
if (m % i == 0)
{
b = 1;
break;
}
}
if (b == 0) a.Add(m)`
}
return a;
}