Getting extra empty user input while scanning - kotlin

I'm kinda of new to Kotlin programming.
I have been trying to scan the user input and calculate the fibonacci series for this input.
The problem that I'm facing: I have to enter either a Enter Button or any extra text before entering the real scanner value that should calculate the result.
here's the code snippet:
import java.util.Scanner;
fun main() {
val scanner = Scanner(System.`in`);
print("Geben Sie bitte eine Zahl ein :");
calculateFibonacciSequence(scanner.nextLong()).forEach { element->print("${element}\n") };
}
fun calculateFibonacciSequence(n : Long) : Array<Long>{
val tempArr = mutableListOf<Long>();
var term1 = 0;
var term2 = 1;
var count = 1;
while (count <= n){
tempArr.add(term1.toLong());
val s = term1+term2;
term1 = term2;
term2 = s;
count++;
}
val list: Array<Long> = tempArr.toTypedArray();
return list;
}

You can use readLine(). It worked for me:
val n = readLine()?.toLongOrNull() ?: error("Please enter a long number")
calculateFibonacciSequence(n)
toLongOrNull() will try to parse the String as Long, it will return null if the String does not represent any Long number.
?. are the safe calls
And ?: is the Elvis operator

Related

How to try every possible permutation in Kotlin

fun main () {
var integers = mutableListOf(0)
for (x in 1..9) {
integers.add(x)
}
//for or while could be used in this instance
var lowerCase = listOf("a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z")
var upperCase = listOf('A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z')
println(integers)
println(lowerCase)
println(upperCase)
//Note that for the actual program, it is also vital that I use potential punctuation
val passwordGeneratorKey1 = Math.random()*999
val passwordGeneratorKey2 = passwordGeneratorKey1.toInt()
var passwordGeneratorL1 = lowerCase[(Math.random()*lowerCase.size).toInt()]
var passwordGeneratorL2 = lowerCase[(Math.random()*lowerCase.size).toInt()]
var passwordGeneratorL3 = lowerCase[(Math.random()*lowerCase.size).toInt()]
var passwordGeneratorU1 = upperCase[(Math.random()*upperCase.size).toInt()]
var passwordGeneratorU2 = upperCase[(Math.random()*upperCase.size).toInt()]
var passwordGeneratorU3 = upperCase[(Math.random()*upperCase.size).toInt()]
val password = passwordGeneratorKey2.toString()+passwordGeneratorL1+passwordGeneratorL2+passwordGeneratorL3+passwordGeneratorU1+passwordGeneratorU2+passwordGeneratorU3
println(password)
//No, this isn't random, but it's pretty close to it
//How do I now run through every possible combination of the lists //lowerCase, integers, and upperCase?
}
How do I run through every possible permutation to eventually solve for the randomly generated password? This is in Kotlin.
I think you should append all the lists together and then draw from it by random index, this way you ensure that position of numbers, lower cases and uppercases is random too. Also you don't need to write all the characters, you can use Range which generates them for you.
fun main() {
val allChars = mutableListOf<Any>().apply {
addAll(0..9) // creates range from 0 to 9 and adds it to a list
addAll('a'..'z') // creates range from a to z and adds it to a list
addAll('A'..'Z') // creates range from A to Z and adds it to a list
}
val passwordLength = 9
val password = StringBuilder().apply {
for (i in 0 until passwordLength) {
val randomCharIndex =
Random.nextInt(allChars.lastIndex) // generate random index from 0 to lastIndex of list
val randomChar = allChars[randomCharIndex] // select character from list
append(randomChar) // append char to password string builder
}
}.toString()
println(password)
}
Even shorter solution can be achieved using list methods
fun main() {
val password = mutableListOf<Any>()
.apply {
addAll(0..9) // creates range from 0 to 9 and adds it to a list
addAll('a'..'z') // creates range from a to z and adds it to a list
addAll('A'..'Z') // creates range from A to Z and adds it to a list
}
.shuffled() // shuffle the list
.take(9) // take first 9 elements from list
.joinToString("") // join them to string
println(password)
}
As others pointed out there are less painful ways to generate the initial password in the format of: 1 to 3 digits followed by 3 lowercase characters followed by 3 uppercase characters.
To brute force this password, you will need to consider all 3-permutations of "a..z" and all 3-permitations of "A..Z". In both cases the number of such 3-permutations is 15600 = 26! / (26-3)!. In worst case you will have to examine 1000 * 15600 * 15600 combination, half of this on the average.
Probably doable in a few hours with the code below:
import kotlin.random.Random
import kotlin.system.exitProcess
val lowercaseList = ('a'..'z').toList()
val uppercaseList = ('A'..'Z').toList()
val lowercase = lowercaseList.joinToString(separator = "")
val uppercase = uppercaseList.joinToString(separator = "")
fun genPassword(): String {
val lowercase = lowercaseList.shuffled().take(3)
val uppercase = uppercaseList.shuffled().take(3)
return (listOf(Random.nextInt(0, 1000)) + lowercase + uppercase).joinToString(separator = "")
}
/**
* Generate all K-sized permutations of str of length N. The number of such permutations is:
* N! / (N-K)!
*
* For example: perm(2, "abc") = [ab, ac, ba, bc, ca, cb]
*/
fun perm(k: Int, str: String): List<String> {
val nk = str.length - k
fun perm(str: String, accumulate: String): List<String> {
return when (str.length == nk) {
true -> listOf(accumulate)
false -> {
str.flatMapIndexed { i, c ->
perm(str.removeRange(i, i + 1), accumulate + c)
}
}
}
}
return perm(str, "")
}
fun main() {
val password = genPassword().also { println(it) }
val all3LowercasePermutations = perm(3, lowercase).also { println(it) }.also { println(it.size) }
val all3UppercasePermutations = perm(3, uppercase).also { println(it) }.also { println(it.size) }
for (i in 0..999) {
println("trying $i")
for (l in all3LowercasePermutations) {
for (u in all3UppercasePermutations) {
if ("$i$l$u" == password) {
println("found: $i$l$u")
exitProcess(0)
}
}
}
}
}

How to solve PW Generator can´t handle specific Chars?

I´m new to Kotlin so first I start with a Password Generator. I do use IntelliJIDea
fun main() {
var helper = Int
var counter = 0
val counterOfPwCha = 10
var pwString = ""
val opportunityArray = arrayOf('A','C','a','c','B','b','D','d','E','e','F','f','#','+','*','$','%','&','§','G','g','H','h','I','i','J','j','k','K','1','2','3','4','5','6','7','8','9',
'l','L','M','m','N','n','O','o','P','p','Q','q','R','r','S','s','T','t','U','u','V','v','W','w','Y','y','X','x','Z','z','<','>','|','?','!',)
while (counter <= counterOfPwCha)
{
helper = (0..73).random()
pwString += opportunityArray[helper]
counter++
}
println(pwString)
}
it workes fine, but the output isn´t what I expected.
Example: +3�2L%II>J�
So it cant handle some of my Chars, I can debug it until I know what kind of Char he can´t handle with and delete them from the Array, but they are all some typical PW Chars, so my question is:
How can I let Kotlin know, how to Handle these ASCI2 chars?
Or is it because of the $?
Avoid hardcoded numbers like the "73" in your range, it's way too error prone!
If you still get funny characters, put in some println that shows each character's numeric value so that you can lookup it using "man ascii".
A more Kotlin-like code would be:
fun pwgen() {
val counterOfPwCha = 10
val charset = ('A'..'Z') + ('a'..'z') + ('0'..'9') + listOf('#', '+', '*', '$', '%', '&', '§', '<', '>', '|', '?', '!')
val pwString = (1..counterOfPwCha)
.map {
val randomArrayIndex = charset.indices.random()
charset[randomArrayIndex]
}
.onEach { c -> println("Character $c = " + c.toInt()) }
.joinToString("")
println(pwString)
}
fun main() {
var helper = Int
var counter = 0
val counterOfPwCha = 10
var pwString = ""
val opportunityArray = arrayOf('A','C','a','c','B','b','D','d','E','e','F','f','#','+','*','$','%','&','§','G','g','H','h','I','i','J','j','k','K','1','2','3','4','5','6','7','8','9',
'l','L','M','m','N','n','O','o','P','p','Q','q','R','r','S','s','T','t','U','u','V','v','W','w','Y','y','X','x','Z','z','<','>','|','?','!',)
while (counter <= counterOfPwCha)
{
helper = (0..73).random()
pwString += opportunityArray[helper]
counter++
}
println(pwString)
}

Kotlin - The caracter literal does not conform expect type Int

I'm struggling with types with my program, I've been asked to do it in JS first and it worked fine but now I can't achieve the result.
Do you think I should make another 'algorithm' ? In advance, thank you for your time.
fun main(){
// the idea is to put numbers in a box
// that cant be larger than 10
val data = "12493419133"
var result = data[0]
var currentBox = Character.getNumericValue(data[0])
var i = 1
while(i < data.length){
val currentArticle = Character.getNumericValue(data[i])
currentBox += currentArticle
println(currentBox)
if(currentBox <= 10){
result += Character.getNumericValue(currentArticle)
}else{
result += '/'
//var resultChar = result.toChar()
// result += '/'
currentBox = Character.getNumericValue(currentArticle)
result += currentArticle
}
i++
}
print(result) //should print 124/9/341/91/33
}
The result is actually of a Char type, and the overload operator function + only accepts Int to increment ASCII value to get new Char.
public operator fun plus(other: Int): Char
In idomatic Kotlin way, you can solve your problem:
fun main() {
val data = "12493419133"
var counter = 0
val result = data.asSequence()
.map(Character::getNumericValue)
.map { c ->
counter += c
if (counter <= 10) c.toString() else "/$c".also{ counter = c }
}
.joinToString("") // terminal operation, will trigger the map functions
println(result)
}
Edit: If the data is too large, you may want to use StringBuilder because it doesn't create string every single time the character is iterated, and instead of using a counter of yourself you can use list.fold()
fun main() {
val data = "12493419133"
val sb = StringBuilder()
data.fold(0) { acc, c ->
val num = Character.getNumericValue(c)
val count = num + acc
val ret = if (count > 10) num.also { sb.append('/') } else count
ret.also { sb.append(c) } // `ret` returned to ^fold, next time will be passed as acc
}
println(sb.toString())
}
If you want a result in List<Char> type:
val data = "12493419133"
val result = mutableListOf<Char>()
var sum = 0
data.asSequence().forEach {
val v = Character.getNumericValue(it)
sum += v
if (sum > 10) {
result.add('/')
sum = v
}
result.add(it)
}
println(result.joinToString(""))

Kotlin decomposing numbers into powers of 2

Hi I am writing an app in kotlin and need to decompose a number into powers of 2.
I have already done this in c#, PHP and swift but kotlin works differently somehow.
having researched this I believe it is something to do with the numbers in my code going negative somewhere and that the solution lies in declaring one or more of the variable as "Long" to prevent this from happening but i have not been able to figure out how to do this.
here is my code:
var salads = StringBuilder()
var value = 127
var j=0
while (j < 256) {
var mask = 1 shl j
if(value != 0 && mask != 0) {
salads.append(mask)
salads.append(",")
}
j += 1
}
// salads = (salads.dropLast()) // removes the final ","
println("Salads = $salads")
This shoud output the following:
1,2,4,8,16,32,64
What I actually get is:
1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536,131072,262144,524288,1048576,2097152,4194304,8388608,16777216,33554432,67108864,134217728,268435456,536870912,1073741824,-2147483648,1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536,131072,262144,524288,1048576,2097152,4194304,8388608,16777216,33554432,67108864,134217728,268435456,536870912,1073741824,-2147483648,1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536,131072,262144,524288,1048576,2097152,4194304,8388608,16777216,33554432,67108864,134217728,268435456,536870912,1073741824,-2147483648,1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536,131072,262144,524288,1048576,2097152,4194304,8388608,16777216,33554432,67108864,134217728,268435456,536870912,1073741824,-2147483648,1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536,131072,262144,524288,1048576,2097152,4194304,8388608,16777216,33554432,67108864,134217728,268435456,536870912,1073741824,-2147483648,1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536,131072,262144,524288,1048576,2097152,4194304,8388608,16777216,33554432,67108864,134217728,268435456,536870912,1073741824,-2147483648,1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536,131072,262144,524288,1048576,2097152,4194304,8388608,16777216,33554432,67108864,134217728,268435456,536870912,1073741824,-2147483648,1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536,131072,262144,524288,1048576,2097152,4194304,8388608,16777216,33554432,67108864,134217728,268435456,536870912,1073741824,-2147483648,
Any ideas?
This works for the one input that you specified, at the very least:
fun powersOfTwo(value :Long): String {
val result = ArrayList<String>()
var i = 0
var lastMask = 0
while (lastMask < value) {
val mask = 1 shl i
if (value != 0.toLong() && mask < value) {
result.add(mask.toString())
}
lastMask = mask
i += 1
}
return result.joinToString(",")
}
Ran it in a unit test:
#Test
fun addition_isCorrect() {
val result = powersOfTwo(127)
assertEquals("1,2,4,8,16,32,64", result)
}
Test passed.
You can get a list of all powers of two that fit in Int and test each of them for whether the value contains it with the infix function and:
val value = 126
val powersOfTwo = (0 until Int.SIZE_BITS).map { n -> 1 shl n }
println(powersOfTwo.filter { p -> value and p != 0}.joinToString(","))
// prints: 2,4,8,16,32,64
See the entire code in Kotlin playground: https://pl.kotl.in/f4CZtmCyI
Hi I finally managed to get this working properly:
fun decomposeByTwo(value :Int): String {
val result = ArrayList<String>()
var value = value
var j = 0
while (j < 256) {
var mask = 1 shl j
if ((value and mask) != 0) {
value -= mask
result.add(mask.toString())
}
j += 1
}
return result.toString()
}
I hope this helps someone trying to get a handle on bitwise options!
Somehow you want to do the "bitwise AND" of "value" and "mask" to determine if the j-th bit of "value" is set. I think you just forgot that test in your kotlin implementation.

How to get length of a character array in kotlin?

I am trying to find the length of two different character array but I am only being allowed to use array.size only once.
How to bypass this problem?
fun chararraytostr(inp1: CharArray): String{
var arlen: Int = inp1.size //here lies the problem
var out1: String = ""
for(j in 0..arlen-1){
var str = inp1[j].toString()
out1+=str
}
return out1
}
fun uppercase(input: String): String{
var temp1: CharArray = input.toCharArray()
var len = input.size //here lies the problem
var temp3: Char
for(i in 0..len-1){
var temp2: Char = temp1[i]
var ascii: Int = temp2.toInt()
if(ascii<=122 && ascii>=97){
ascii-=32
temp3 = ascii.toChar()
temp1[i] = temp3
}else{}
}
var output = chararraytostr(temp1)
return output
}
fun main(arg: Array<String>){
var toupper = "Hi my friend!"
println(uppercase(toupper))
}
It is always showing Unresolved reference: size. I don't know why. Please help.
You're calling size on the String variable instead of the CharArray variable. Use temp1.size instead of input.size.