The problem is to reverse the bits of a 32 bit unsigned integer (since Java doesn't have unsigned integers we use long).
Here are two versions of my code. I have two concerns:
(1) why my 1st and 2nd solution don't return the same value (correct or not)
(2) where my 1st and 2nd solution went wrong in not getting the correct answer
//reverse(3) returns 0
public static long reverse(long a) {
long numBits = 32;
long finalResult = 0;
for(int i = 0; i < numBits; i++){
long ithBit = a & (1 << i);
finalResult = finalResult + ithBit * (1 << (numBits - i - 1));
}
return finalResult;
}
Second version:
//reverse(3) return 4294967296
public static long reverse(long a) {
long numBits = 32L;
long finalResult = 0L;
for(long i = 0L; i < numBits; i++){
long ithBit = a & (1L << i);
finalResult = finalResult + ithBit * (1L << (numBits - i - 1L));
}
return finalResult;
}
This code (the solution) returns the correct answer, however:
//reverse(3) returns 3221225472
public static long reverse(long A) {
long rev = 0;
for (int i = 0; i < 32; i++) {
rev <<= 1;
if ((A & (1 << i)) != 0)
rev |= 1;
}
return rev;
}
Thanks!
Let's have a look at your values as you iterate. For clarification, we'll have a look at the intermediate values, so we'll change code to:
int n = (1 << (numBits - i - 1));
long m = ithBit * n;
finalResult = finalResult + m;
Your starting value is 3:
a = 0000 0000 0000 0000 0000 0000 0000 0011
First loop iteration (i = 0):
ithBit = 00000000 00000000 00000000 00000001
n = 11111111 11111111 11111111 11111111 10000000 00000000 00000000 00000000
m = 11111111 11111111 11111111 11111111 10000000 00000000 00000000 00000000
finalResult = 11111111 11111111 11111111 11111111 10000000 00000000 00000000 00000000
Second loop iteration (i = 1):
ithBit = 00000000 00000000 00000000 00000010
n = 01000000 00000000 00000000 00000000
m = 10000000 00000000 00000000 00000000
finalResult = 00000000 00000000 00000000 00000000
As you can see, the first iterate sets n = 1 << 31, which is -2147483648. In your second version you do n = 1L << 31, which is 2147483648, and that's why your two versions give different results.
As you can also see, you definitely don't want to do the m = ithBit * n part.
Have a look at your number by printing them yourself, and you'll figure it out.
BTW, here's my version. If you have trouble understanding it, try printing the intermediate values to see what's going on.
public static long reverse4(long a) {
long rev = 0;
for (int i = 0; i < 32; i++, a >>= 1)
rev = (rev << 1) | (a & 1);
return rev;
}
since Java doesn't have unsigned integers we use long.
That's generally unecessary since all arithmetic operations except division and comparison result in identical bit patterns for unsigned and signed numbers in two's complement representation, which java uses. And for the latter two ops Integer.divideUnsigned(int, int) and Integer.compareUnsigned(int, int) are available.
The problem is to reverse the bits of a 32 bit unsigned integer
There's Integer.reverse(int)
Relevant docs, spending some time reading them is highly recommended.
Problem with your both examples is that the "i-th bit" isn't a 0 or 1 but rather masked off. In either case, the 31'th bit is 0x8000_0000. In the first case, this is an int, so it is negative, and when converted to a long it stays negative. In the second case it is already a long, so it stays positive. To fix it so it does what you intended, do:
ithBit = (a >>> i) & 1;
By the way, using long is silly; unsigned vs. signed makes no difference as long as you understand that there are two types of shifts in Java.
By the way, all three examples are terrible. If you are doing bit manipulation, you want speed, right? (Why else bother with bits?)
This is how to do it right (not mine, stolen from http://graphics.stanford.edu/~seander/bithacks.html#ReverseByteWith32Bits):
a = ((a >>> 1) & 0x55555555) | ((a & 0x55555555) << 1);
a = ((a >>> 2) & 0x33333333) | ((a & 0x33333333) << 2);
a = ((a >>> 4) & 0x0F0F0F0F) | ((a & 0x0F0F0F0F) << 4);
a = ((a >>> 8) & 0x00FF00FF) | ((a & 0x00FF00FF) << 8);
a = ( a >>> 16 ) | ( a << 16);
In both versions you have a logic problem here:
ithBit * (1 << (numBits - i - 1));
because the two numbers multiply to the same bit pattern of bit 31 (the 32nd) being set.
In version 1, the amount added is -2^31 if bit 0 is set because you're bit-shifting an int so the bit-shifted result is int which represents -2^31 as the high bit being set, or 2^31 for every other bit, which is possible due to the auto-cast to long of the result. You have two of each kind of bit (0 and non 0) so the result is zero.
Version 2 has the same problem, but without the negative int issue because you're bit shifting a long. Each bit that is 1 will add 2^31. The number 3 has 2 bits set, so your result is 2 * 2^31 (or 2^32) which is 4294967296.
To fix your logic, use version 2 but remove the multiply by ithBit.
Related
I need to switch the first half and the second half of a byte: Make 0011 0101 to 0101 0011 for example
I thought it might work this way:
For example, i have 1001 1100
i bitshift to the left 4 times and get 1111 1001(because if the first bit is a 1 the others become a one too)
i bitshift to the right 4 times and get 1100 0000(the second half of the byte gets filled with 0s)
i don't want 1111 1001 but 0000 1001 so i do 0x00001111 & 1111 1001 (which filters the frist 4 bits) to make 1111 1001 to 0000 1001
then i add everything up:
0000 1001 + 1100 0000 = 1100 1001
I got this:
bytes[i] = (byte) (( 0x00001111 & (bytes[i] >> 4)) + (bytes[i] << 4)
);
here is one output: 11111111 to 00000001
I do not really understand why this is happening, I know the binary System and I think I know how bitshifting works but I can't explain this one.
Sorry for bad english :)
Be careful with the >>> operation, which shifts the sign bits without sign extending so zero bits will fill in on the left. The problem is that it is an integer operation. The >> works the same way except it sign extends thru the int.
int i = -1;
i >>>= 30;
System.out.println(i); // prints 3 as expected.
byte b = -1;
b >>>= 6;
System.out.printnln(b); // prints -1 ???
The byte is still -1 because byte b = -1 was shifted as though it was an int then reassigned to a byte. So the byte remained negative. To get 3, you would need to do something that seems strange, like the following.
byte b = -1;
b >>>=30;
System.out.println(b); // prints 3
So to do your swap you need to do the following;
byte b = 0b10100110;
b = (byte)(((b>>>4)&0xF)|(b<<4));
The 0xF mask, masks off those lingering high order bits left over from the conversion from integer back to byte.
I'm not sure about the syntax for bit manipulation in Java, although here's how you can do it.
bitmask = 0x1111;
firstHalf = ((bytes[i] >> 4) & bitmask);
secondHalf = ((bytes[i] & bitmask) << 4)
result = firstHalf | secondHalf;
I don't want 1111 1001 but 0000 1001
If so, you need to use shift right zero fill operator(>>>) instead of preserving sign of the number.
I don't think the formula found works properly.
public byte reverseBitsByte(byte x) {
int intSize = 8;
byte y=0;
for(int position=intSize-1; position>0; position--){
y+=((x&1)<<position);
x >>= 1;
}
return y;
}
static int swapBits(int a) {
// Написать решение сюда ↓
int right = (a & 0b00001111);
right= (right<<4);
int left = (a & 0b11110000);
left = (left>>4);
return (right | left);
}
I don't understand what is this doCalculatePi means or does, in the following example:
public static double doCalculatePi(final int sliceNr) {
final int from = sliceNr * 10;
final int to = from + 10;
final int c = (to << 1) + 1;
double acc = 0;
for (int a = 4 - ((from & 1) << 3), b = (from << 1) + 1; b < c; a = -a, b += 2) {
acc += ((double) a) / b;
}
return acc;
}
public static void main(String args[]){
System.out.println(doCalculatePi(1));
System.out.println(doCalculatePi(2));
System.out.println(doCalculatePi(3));
System.out.println(doCalculatePi(4));
System.out.println(doCalculatePi(10));
System.out.println(doCalculatePi(100));
}
I have printed the values to understand what the results are but I still have no clue what this code calculates. The conditions inside the loop are not clear.
<< means left shift operation, which shifts the left-hand operand left by the number of bits specified by the right-hand operand (See oracle docs).
Say, you have a decimal value, 5 which binary representation is 101
Now for simplicity, consider,
byte a = (byte)0x05;
Hence, the bit representation of a will be,
a = 00000101 // 1 byte is 8 bit
Now if you left shift a by 2, then a will be
a << 2
a = 00010100 //shifted place filled with zero/s
So, you may now understand that, left shift a by 3 means
a << 3
a = 00101000
For better understanding you need to study Bitwise operation.
Note, you are using int instead of byte, and by default, the int data type is a 32-bit signed integer (reference here), so you have to consider,
int a = 5;
in binary
a << 3
a = 00000000 00000000 00000000 00101000 // total 32 bit
My guess is that it approximates PI with
PI = doCalculatePi(0)+doCalculatePi(1)+doCalculatePi(2)+...
Just a guess.
Trying this
double d = 0;
for(int k = 0; k<1000; k++) {
System.out.println(d += doCalculatePi(k));
}
gives me
3.0418396189294032
3.09162380666784
3.1082685666989476
[...]
3.1414924531892394
3.14149255348994
3.1414926535900394
<< is the Bitshift operator.
Basically, every number is represented as a series of binary digits (0's and 1's), and you're shifting each of those digits to the left by however many places you indicate. So for example, 15 is 00001111 and 15 << 1 is 00011110 (or 30), while 15 << 2 is (00111100) which is 60.
There's some special handling that comes into play when you get to the sign bit, but you should get the point.
I have this statement:
Assume the bit value of byte x is 00101011. what is the result of x>>2?
How can I program it and can someone explain me what is doing?
Firstly, you can not shift a byte in java, you can only shift an int or a long. So the byte will undergo promotion first, e.g.
00101011 -> 00000000000000000000000000101011
or
11010100 -> 11111111111111111111111111010100
Now, x >> N means (if you view it as a string of binary digits):
The rightmost N bits are discarded
The leftmost bit is replicated as many times as necessary to pad the result to the original size (32 or 64 bits), e.g.
00000000000000000000000000101011 >> 2 -> 00000000000000000000000000001010
11111111111111111111111111010100 >> 2 -> 11111111111111111111111111110101
The binary 32 bits for 00101011 is
00000000 00000000 00000000 00101011, and the result is:
00000000 00000000 00000000 00101011 >> 2(times)
\\ \\
00000000 00000000 00000000 00001010
Shifts the bits of 43 to right by distance 2; fills with highest(sign) bit on the left side.
Result is 00001010 with decimal value 10.
00001010
8+2 = 10
When you shift right 2 bits you drop the 2 least significant bits. So:
x = 00101011
x >> 2
// now (notice the 2 new 0's on the left of the byte)
x = 00001010
This is essentially the same thing as dividing an int by 2, 2 times.
In Java
byte b = (byte) 16;
b = b >> 2;
// prints 4
System.out.println(b);
These examples cover the three types of shifts applied to both a positive and a negative number:
// Signed left shift on 626348975
00100101010101010101001110101111 is 626348975
01001010101010101010011101011110 is 1252697950 after << 1
10010101010101010100111010111100 is -1789571396 after << 2
00101010101010101001110101111000 is 715824504 after << 3
// Signed left shift on -552270512
11011111000101010000010101010000 is -552270512
10111110001010100000101010100000 is -1104541024 after << 1
01111100010101000001010101000000 is 2085885248 after << 2
11111000101010000010101010000000 is -123196800 after << 3
// Signed right shift on 626348975
00100101010101010101001110101111 is 626348975
00010010101010101010100111010111 is 313174487 after >> 1
00001001010101010101010011101011 is 156587243 after >> 2
00000100101010101010101001110101 is 78293621 after >> 3
// Signed right shift on -552270512
11011111000101010000010101010000 is -552270512
11101111100010101000001010101000 is -276135256 after >> 1
11110111110001010100000101010100 is -138067628 after >> 2
11111011111000101010000010101010 is -69033814 after >> 3
// Unsigned right shift on 626348975
00100101010101010101001110101111 is 626348975
00010010101010101010100111010111 is 313174487 after >>> 1
00001001010101010101010011101011 is 156587243 after >>> 2
00000100101010101010101001110101 is 78293621 after >>> 3
// Unsigned right shift on -552270512
11011111000101010000010101010000 is -552270512
01101111100010101000001010101000 is 1871348392 after >>> 1
00110111110001010100000101010100 is 935674196 after >>> 2
00011011111000101010000010101010 is 467837098 after >>> 3
>> is the Arithmetic Right Shift operator. All of the bits in the first operand are shifted the number of places indicated by the second operand. The leftmost bits in the result are set to the same value as the leftmost bit in the original number. (This is so that negative numbers remain negative.)
Here's your specific case:
00101011
001010 <-- Shifted twice to the right (rightmost bits dropped)
00001010 <-- Leftmost bits filled with 0s (to match leftmost bit in original number)
public class Shift {
public static void main(String[] args) {
Byte b = Byte.parseByte("00101011",2);
System.out.println(b);
byte val = b.byteValue();
Byte shifted = new Byte((byte) (val >> 2));
System.out.println(shifted);
// often overloked are the methods of Integer
int i = Integer.parseInt("00101011",2);
System.out.println( Integer.toBinaryString(i));
i >>= 2;
System.out.println( Integer.toBinaryString(i));
}
}
Output:
43
10
101011
1010
byte x = 51; //00101011
byte y = (byte) (x >> 2); //00001010 aka Base(10) 10
You can't write binary literals like 00101011 in Java so you can write it in hexadecimal instead:
byte x = 0x2b;
To calculate the result of x >> 2 you can then just write exactly that and print the result.
System.out.println(x >> 2);
You can use e.g. this API if you would like to see bitString presentation of your numbers. Uncommons Math
Example (in jruby)
bitString = org.uncommons.maths.binary.BitString.new(java.math.BigInteger.new("12").toString(2))
bitString.setBit(1, true)
bitString.toNumber => 14
edit: Changed api link and add a little example
00101011 = 43 in decimal
class test {
public static void main(String[] args){
int a= 43;
String b= Integer.toBinaryString(a >> 2);
System.out.println(b);
}
}
Output:
101011 becomes 1010
I need int 32 in binary as 00100000 or int 127 in binary 0111 1111.
The variant Integer.toBinaryString returns results only from 1.
If I build the for loop this way:
for (int i= 32; i <= 127; i + +) {
System.out.println (i);
System.out.println (Integer.toBinaryString (i));
}
And from binary numbers I need the number of leading zeros (count leading zeros (clz) or number of leading zeros (nlz)) I really meant the exact number of 0, such ex: at 00100000 -> 2 and at 0111 1111 - > 1
How about
int lz = Integer.numberOfLeadingZeros(i & 0xFF) - 24;
int tz = Integer.numberOfLeadingZeros(i | 0x100); // max is 8.
Count the number of leading zeros as follows:
int lz = 8;
while (i)
{
lz--;
i >>>= 1;
}
Of course, this supposes the number doesn't exceed 255, otherwise, you would get negative results.
Efficient solution is int ans = 8-(log2(x)+1)
you can calculate log2(x)= logy (x) / logy (2)
public class UtilsInt {
int leadingZerosInt(int i) {
return leadingZeros(i,Integer.SIZE);
}
/**
* use recursion to find occurence of first set bit
* rotate right by one bit & adjust complement
* check if rotate value is not zero if so stop counting/recursion
* #param i - integer to check
* #param maxBitsCount - size of type (in this case int)
* if we want to check only for:
* positive values we can set this to Integer.SIZE / 2
* (as int is signed in java - positive values are in L16 bits)
*/
private synchronized int leadingZeros(int i, int maxBitsCount) {
try {
logger.debug("checking if bit: "+ maxBitsCount
+ " is set | " + UtilsInt.intToString(i,8));
return (i >>>= 1) != 0 ? leadingZeros(i, --maxBitsCount) : maxBitsCount;
} finally {
if(i==0) logger.debug("bits in this integer from: " + --maxBitsCount
+ " up to last are not set (i'm counting from msb->lsb)");
}
}
}
test statement:
int leadingZeros = new UtilsInt.leadingZerosInt(255); // 8
test output:
checking if bit: 32 is set |00000000 00000000 00000000 11111111
checking if bit: 31 is set |00000000 00000000 00000000 01111111
checking if bit: 30 is set |00000000 00000000 00000000 00111111
checking if bit: 29 is set |00000000 00000000 00000000 00011111
checking if bit: 28 is set |00000000 00000000 00000000 00001111
checking if bit: 27 is set |00000000 00000000 00000000 00000111
checking if bit: 26 is set |00000000 00000000 00000000 00000011
checking if bit: 25 is set |00000000 00000000 00000000 00000001
bits in this integer from: 24 up to last are not set (i'm counting from msb->lsb)
This question already has answers here:
Difference between >>> and >>
(9 answers)
Java >>> Operator [duplicate]
(3 answers)
Closed 9 years ago.
I found this code to find duplicates in SO post here.
but I dont understand what this line means int mid = (low + high) >>> 1;
private static int findDuplicate(int[] array) {
int low = 0;
int high = array.length - 1;
while (low <= high) {
int mid = (low + high) >>> 1;
System.out.println(mid);
int midVal = array[mid];
if (midVal == mid)
low = mid + 1;
else
high = mid - 1;
}
return high;
}
The >>> operator is the unsigned right bit-shift operator in Java. It effectively divides the operand by 2 to the power of the right operand, or just 2 here.
The difference between >> and >>> would only show up when shifting negative numbers. The >> operator shifts a 1 bit into the most significant bit if it was a 1, and the >>> shifts in a 0 regardless.
UPDATE:
Let's average 1 and 2147483647 (Integer.MAX_VALUE). We can do the math easily:
(1 + 2147483647) / 2 = 2147483648 / 2 = 1073741824
Now, with the code (low + high) / 2, these are the bits involved:
1: 00000000 00000000 00000000 00000001
+2147483647: 01111111 11111111 11111111 11111111
================================================
-2147483648: 10000000 00000000 00000000 00000000 // Overflow
/2
================================================
-1073741824: 11000000 00000000 00000000 00000000 // Signed divide, same as >> 1.
Let's make the "shift" to >>>:
1: 00000000 00000000 00000000 00000001
+2147483647: 01111111 11111111 11111111 11111111
================================================
-2147483648: 10000000 00000000 00000000 00000000 // Overflow
>>> 1
================================================
+1073741824: 01000000 00000000 00000000 00000000 // Unsigned shift right.
The significance of
int mid = (low + high) >>> 1;
is; by using the unsigned shift, it avoids overflows which result in a negative number. This is needed as Java doesn't support unsigned int values. (BTW char is unsigned)
The traditional way to write this was
int mid = (low + high) / 2; // don't do this
however this could overflow for larger sums and you get a negative number for the mid.
e.g.
int high = 2100000000;
int low = 2000000000;
System.out.println("mid using >>> 1 = " + ((low + high) >>> 1));
System.out.println("mid using / 2 = " + ((low + high) / 2));
prints
mid using >>> 1 = 2050000000
mid using / 2 = -97483648
clearly the second result is incorrect.
its a bitwise operator .. It works on the bit value .
Suppose if A holds 60 then A>>>2 will give 15 (bit value 0000 1111)
Its actual name is "Shift Zero right Operator"
here the left operands value is moved right by the number of bits specified (2 in this case) by the right operand and shifted values are filled up with zeros(0000).