The task is to concat the binary of 2 given numbers.
Example:
Given 5 (101) and 3 (011), the result is 46 (concat(101, 011) = 101011)
The code thus far:
public class Concat {
public static void main(String[] args) {
int t = 0;
int k = 5;
int x = 3;
int i = 0;
while (i < 3) {
t = x % 2;
x /= 2;
k <<= 1;
k |= t;
++i;
}
System.out.println(k);
}
}
But the problem is that the above code gives 101110, not 101011.
What's the problem?
Your problem is that you're feeding the bits of the second number in backwards. That's because x%2 is the low order bit:
+---+---+---+ <110
| 1 | 0 | 1 | <-----------------+^
+---+---+---+ |1
+---+---+---+ |1
| 0 | 1 | 1 | ----+0
+---+---+---+ 011>
Cringe at my awesome artistic abilities :-) However, if you already know that it's 3 bits wide, just use:
public class concat {
public static void main (String[] args) {
int t = 0;
int k = 5;
int x = 3;
k <<= 3;
k |= x;
// or, in one line: k = (k << 3) | x;
System.out.println(k);
}
}
In terms of how that looks graphically:
+---+---+---+
k:| 1 | 0 | 1 |
+---+---+---+
+---+---+---+
x:| 0 | 1 | 1 |
+---+---+---+
+---+---+---+---+---+---+
k<<=3:| 1 | 0 | 1 | 0 | 0 | 0 |
+---+---+---+---+---+---+
+---+---+---+
x:| 0 | 1 | 1 |
+---+---+---+
+---+---+---+---+---+---+
k|=3:| 1 | 0 | 1 | 0 | 1 | 1 |
+---+---+---+---+---+---+
^ ^ ^
+---+---+---+
x:| 0 | 1 | 1 |
+---+---+---+
There's no apparent reason for doing it one bit at a time.
You just shift one number and then or with the other number:
int a = 5;
int b = 3;
int c = (a << 3) | b;
I don't know what language you are using, it's almost Java, so I am going with that.
This returns the result you are asking for, though you haven't given rules for determining how you know that 3 should be 011 instead of 11.
I have made the assumption that you want to assume that both numbers have the same number of bits, so 3 is 011 because 5 requires 3 bits.
public class Concat {
public static void main(String[] args) {
System.out.println( joinNums(3,5) );
}
public static int numBits( int n ) {
return (int)Math.ceil( Math.log(n) / Math.log(2) );
}
public static int joinNums( int num1 , int num2 ) {
int bits = Math.max( numBits(num1) , numBits(num2) );
return (num1 << bits) + num2;
}
}
Related
I'm trying to write an iterative program that will help me palce 4 queens on a 4x4 board without them hitting each other. The problem is after looping through each position and backtracking a couple of times my main while loop that keeps looping until a solution is found gets terminated and the program ends even though the condition is not yet met.
I tried the following code:
static int[] solve(char[][] board){
int[] position = new int[4];
int row = 0;
int column = 0;
while(row < 4){
for(boolean check; column < board.length; column++){
System.out.println("["+row+","+column+"]");
check = true;
for(int queen= 0; queen < row; queen++){
if (position[queen] == column || queen- position[queen] == row - column || queen + position[queen] == row + column) {
check = false;
break;
}
}
if(check){
position[row] = column;
column = 0;
row++;
}
if(column > 2){
column = position[--row];
}
}
}
return position;
}
I'm currently getting the following output:
| Q | X | X | X |
| X | X | X | Q |
| X | Q | X | X |
| Q | X | X | X |
To check when exactly the while loop is getting terminated I printed the location (row and column)
System.out.println("["+row+","+column+"]"); and got the following:
[0,0][1,0][1,1][1,2][2,0][2,1][2,2][2,3][1,3][2,0][2,1][3,0][3,1][3,2][3,3][2,2][2,3]
After backtracking to [2,3] the while loop ends even though my row count is still less than 4.
I was expecting the following output:
| X | Q | X | X |
| X | X | X | Q |
| Q | X | X | X |
| X | X | Q | X |
I tried the code in a different compiler and still got the same wrong output. Is there a logical mistake that I missed out?
I'm new to programming so I'm still trying to get the hang of the fundamentals.
I have 2 variables A and B
if A =1 then B should B=2
and if A=2 then B should B=1
Like this, there are 3 pairs 1-2,3-4,5-6
What's the best way of making a code instead of just if-else
It is possible to use simple addition and subtraction to get the other element of the two (x, x + 1):
int a = 1; // the other is 2, sum is 3
int b = 3 - a; // if a = 2, b = 1
int c = 3; // the other is 4, sum is 7
int d = 7 - c; // if c = 4, d = 3
int m = 5; // the other is 6, sum is 11
int n = 11 - m;
Another approach could be using the following logic:
if (a % 2 == 1) b = a + 1;
else b = a - 1;
So, an array could be used to provide +/- 1:
static int[] signs = {-1, 1};
public static int nextWithArrPositive(int a) {
return a + signs [a % 2];
}
This expression fails to work for negative a as in this case a % 2 == -1 and more advanced logic would be required to calculate the value properly to take into account the negative remainder:
public static int nextWithArr(int a) {
int sign = (a & 0x80000000) >> 31; //-1 if a < 0, 0 otherwise
// a >= 0 : 0 - even, 1 - odd;
// a < 0 : 1 - even, 0 - odd
return a + signs[a % 2 - sign];
}
However, a simpler expression can be designed:
public static int nextWithMod(int a) {
return a + a % 2 - (a - 1) % 2;
}
Let's compare the results of the three implementations including xor solution b = ((a - 1) ^ 1) + 1 offered in the comments by user3386109:
public static int nextXor(int a) {
return ((a - 1) ^ 1) + 1;
}
Tests:
System.out.println("+-----+-----+-----+-----+");
System.out.println("| a | arr | mod | xor |");
System.out.println("+-----+-----+-----+-----+");
for (int i = -6; i < 7; i++) {
System.out.printf("| %2d | %2d | %2d | %2d |%n", i, nextWithArr(i), nextWithMod(i), nextXor(i));
}
System.out.println("+-----+-----+-----+-----+");
Output:
+-----+-----+-----+-----+
| a | arr | mod | xor |
+-----+-----+-----+-----+
| -6 | -5 | -5 | -7 |
| -5 | -6 | -6 | -4 |
| -4 | -3 | -3 | -5 |
| -3 | -4 | -4 | -2 |
| -2 | -1 | -1 | -3 |
| -1 | -2 | -2 | 0 |
| 0 | -1 | 1 | -1 |
| 1 | 2 | 2 | 2 |
| 2 | 1 | 1 | 1 |
| 3 | 4 | 4 | 4 |
| 4 | 3 | 3 | 3 |
| 5 | 6 | 6 | 6 |
| 6 | 5 | 5 | 5 |
+-----+-----+-----+-----+
One simple solution is a table lookup. In an array for each possible value of a I store the corresponding value of b:
private static final int[] B_PER_A = { -1, 2, 1, 4, 3, 6, 5 };
Since array indices always start at 0 in Java, we need to put a dummy value at index 0. This value is never used (or should never be, at least).
Let’s try it out:
for (int a = 1; a <= 6; a++) {
int b = B_PER_A[a];
System.out.format("a: %d; b: %d.%n", a, b);
}
Output:
a: 1; b: 2.
a: 2; b: 1.
a: 3; b: 4.
a: 4; b: 3.
a: 5; b: 6.
a: 6; b: 5.
Generalized to more than 3 pairs
If you need to handle a variable number of pairs, resort to math.
public static int calcB(int a) {
// 0-based index of pair (0 = 1-2, 1 = 3-4, etc.)
int pairNumber = (a - 1) / 2;
// a + b for given pair
int pairSum = 4 * pairNumber + 3;
int b = pairSum - a;
return b;
}
In each pair the sum is equivalent to 3 modulo 4. I am exploiting this fact in finding the sum for a given pair. When I subtract a from that sum, I get b. Let’s see that demonstrated too:
for (int a = 1; a <= 8; a++) {
int b = calcB(a);
System.out.format("a: %d; b: %d.%n", a, b);
}
a: 1; b: 2.
a: 2; b: 1.
a: 3; b: 4.
a: 4; b: 3.
a: 5; b: 6.
a: 6; b: 5.
a: 7; b: 8.
a: 8; b: 7.
The latter solution is more complicated and harder to read. So if you always have got three pairs, no more, no less, I recommend the simpler table lookup presented first.
I'm trying to do a Lotto game where I have to generate a random card and in the first column numbers go from 1-9, second 10-19 all the way to 90. Another rule of the card is that it can only have 5 numbers at random positions in each line and that's what I'm having trouble with.
I've started with this to put numbers in each column:
int[][] numberCard = new int[3][9];
for (int i = 0; i < numberCard.length; i++) {
numberCard[i][0] = randInt(1, 9);
}
for (int j = 0; j < numberCard.length; j++) {
numberCard[j][1] = randInt(10, 19);
}
And then do put 5 numbers in a number position in each line of the array i tried this:
//Five numbers first line
Random random0 = new Random();
int randomLocation = 0;
while (randomLocation < 5) {
int z0 = random0.nextInt(numberCard.length);
int b0 = random0.nextInt(numberCard[0].length);
if (numberCard[z0][b0] > 0) {
numberCard[z0][b0] = 0;
randomLocation++;
}
}
//Five numbers second line
Random random1 = new Random();
int randomLocation1 = 0;
while (randomLocation1 < 5) {
int z1 = random1.nextInt(numberCard.length);
int b1 = random1.nextInt(numberCard[1].length);
if (numberCard[z1][b1] > 0) {
numberCard[z1][b1] = 0;
randomLocation1++;
}
}
And then the same for the third one.
Output:
0 | 18 | 0 | 0 | 46 | 0 | 61 | 72 | 88 |
0 | 18 | 0 | 31 | 0 | 55 | 0 | 0 | 0 |
0 | 0 | 23 | 34 | 45 | 0 | 0 | 0 | 80 |
The problem is that sometimes I get 5 numbers, sometimes 6 and sometimes 4 when sometimes i get the perfect 5 numbers in each line. More frustrating that not working is only working sometimes...
I thought and I think it is because of the random numbers that sometimes repeat themselves so they go to the same index, but then sometimes I have more than the 5 numbers so that makes no sense.
Correct output expected:
0 | 18 | 23 | 30 | 48 | 0 | 0 | 76 | 0 |
0 | 12 | 24 | 0 | 0 | 58 | 61 | 78 | 0 |
1 | 17 | 0 | 0 | 42 | 54 | 0 | 0 | 86 |
Here is the pretty much the same code as before. The cards have become the rows.
public static void lottery() {
System.out.println(" 1 2 3 4 5 6 7 8 9");
System.out.println("---------------------------");
for (int card = 1; card<= 5; card++) {
List<Integer> numbers = new ArrayList<>();
for (int i = 0; i <= 8; i++) {
int s = i > 0 ? 0 : 1;
numbers.add(ThreadLocalRandom.current().nextInt(i*10 + s,(i+1)*10));
}
Collections.shuffle(numbers);
int[] row = new int[9];
for (int s = 0; s < 5; s++) {
int pick = numbers.get(s);
row[pick/10] = pick;
}
for (int i = 0; i < 9; i++) {
System.out.printf("%3d", row[i]);
}
System.out.println();
}
}
Prints
1 2 3 4 5 6 7 8 9
---------------------------
5 0 0 0 47 0 67 71 81
1 0 22 0 0 55 65 0 88
1 11 0 30 0 58 0 76 0
0 10 24 0 43 0 0 75 88
3 16 0 0 0 0 60 71 81
I think your approach is complicated than it should be. I would recomend to do it like below:
For each row of your 2D-Array, generate a random number between [1 - 90) and put it to the spot where it belongs by dividing the random number with 10 (will give you values [0 - 8]) while checking that that position doesn't have a random value aready assigned. Reapet this 5 times.
public static void main(String[] args) {
int[][] numberCard = new int[3][9];
Random rand = new Random();
for (int i = 0; i < numberCard.length; i++) {
for (int j = 0; j < 5; j++) {
int x = rand.nextInt(89) + 1;
while (numberCard[i][x / 10] != 0) {
x = rand.nextInt(89) + 1;
}
numberCard[i][x / 10] = x;
}
}
//print or do whatever with your numberCard array
for (int[] row : numberCard) {
System.out.println(Arrays.toString(row));
}
}
Sample output:
[0, 0, 24, 0, 40, 55, 0, 71, 86]
[0, 16, 0, 0, 42, 56, 0, 70, 84]
[5, 12, 0, 0, 49, 0, 69, 0, 82]
You can do it like this:
The card consists of 3 rows.
Each row consists of 9 random cells: 5 filled and 4 hollow.
Column ranges: 1-9 first, 80-90 last, x0-x9 others.
No duplicate numbers in every column.
List<int[]> card = new ArrayList<>();
IntStream.range(0, 3)
// prepare a list of cell indices of the row
.mapToObj(row -> IntStream.range(0, 9)
.boxed().collect(Collectors.toList()))
// random order of cell indices
.peek(Collections::shuffle)
// list of 5 random cell indices in the row
.map(list -> list.subList(0, 5))
// fill the cells with random numbers
.map(list -> IntStream.range(0, 9)
// if this cell is in the list, then fill
// it with a random number, otherwise 0
.map(i -> {
int number = 0;
if (list.contains(i)) {
// unique numbers in each column
boolean isPresent;
do {
// [1-9] first, [80-90] last, [x0-x9] others
number = (int) ((i > 0 ? i * 10 : 1)
+ Math.random()
* (i < 8 ? (i > 0 ? 10 : 9) : 11));
isPresent = false;
for (int[] row : card)
if (row[i] == number)
isPresent = true;
} while (isPresent);
}
return number;
}).toArray())
// add this row to the card
.forEach(card::add);
// output
card.stream()
// row of numbers to string
.map(row -> Arrays.stream(row)
.mapToObj(j -> j == 0 ? " " : String.format("%2d", j))
// concatenate cells into one row
.collect(Collectors.joining(" | ", "| ", " |")))
.forEach(System.out::println);
Output:
| 7 | 20 | | | 47 | 53 | | | 86 |
| | | 27 | 38 | | | 63 | 80 | 85 |
| 4 | | 26 | | | | 69 | 77 | 81 |
I'm working on an implementation of the "Merge Numbers" game, seen here on the Play store: https://play.google.com/store/apps/details?id=com.ft.game.puzzle.cubelevels
The idea is that if I have a 2D array of cells, some of which are empty and others of which have numbers, and I place a given number into an empty cell, then if that number matches at least 3 of its neighbors, the neighbors are made empty and the cell value is incremented. A simple example:
| 1 | 2 | 3 | 4 |
| 3 | | 2 | |
| 1 | 2 | 3 | |
| | 2 | 1 | 4 |
If the next number (generated randomly by the game) is 2, and I place it into cell (1, 1) (where (0, 0) is the upper left cell), the new game board should look like this:
| 1 | | 3 | 4 |
| 3 | 3 | | |
| 1 | | 3 | |
| | 2 | 1 | 4 |
The newly-placed 2 is incremented to 3, and the neighboring cells with value of 2 have been cleared.
I'm having trouble figuring out how to do this. Here is my code so far:
package com.company;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
int[][] w = {{1, 2, 3, 4}, {3, 0, 4, 0}, {1, 2, 3, 0}, {0, 2, 1, 4}};
Scanner input = new Scanner(System.in);
System.out.println("Enter space for putting 1: ");
int a = input.nextInt();
int b = input.nextInt();
if (a == 1 && b == 1) {
w[1][1] = 1;
for (a = 0; a < w.length; a++) {
for (b = 0; b < w[0].length; b++) {
System.out.print(w[a][b] + " ");
}
System.out.print('\n');
}
System.out.println("Enter space for putting 4: ");
a = input.nextInt();
b = input.nextInt();
if (a == 1 && b == 3) {
w[1][3] = 4;
for (a = 0; a < w.length; a++) {
for (b = 0; b < w[0].length; b++) {
System.out.print(w[a][b] + " ");
}
System.out.print('\n');
}
}
}
}
}
How can I check to see if the neighbors of the newly-added number have the same value?
How can I increment the newly-added number?
How do I find the complexity in tilde notation of the following algorithm:
for (int j = 0; j < N; j++) {
for (int k = j + 1; k < N; k++) {
array[k] = array[j];
}
array[j] = k
}
I've made a table with how many times the inner for-loop loops if N = 9:
| j | # of loops |
|:-----------|------------:|
| 0 | 8 |
| 1 | 7 |
| 2 | 6 |
| 3 | 5 |
| 4 | 4 |
| 5 | 3 |
| 6 | 2 |
| 7 | 1 |
| 8 | 0 |
As you evaluate, the number of inner iterations decreases linearly from 8 down to 0, i.e. it is 4 on average, for a total of 4.9=36.
More generally, the average is (N-1)/2 and the total N.(N-1)/2.
Consequently, I(N) ~ N²/2, in terms of the iteration count.
In terms of memory accesses (R+W), it's the double: A(N) ~ N². (The extra access in the outer loop adds a negligible N contribution.)