Find first and last set bit of a long integer
Clarification: This task is asking for the position of two bits in the binary representation of a positive integer. Some parts of this task assume that this is the native representation in the language you are working in. Any part of this task which makes assumptions about native representation should be treated as a recommendation which is only relevant in some contexts. A bit is defined as the exponent in a binary polynomial -- an exponent corresponding to a power of 2 which has a non-zero multiplier in the summation sequence of powers of two which yields the desired positive integer, where the only allowed coefficients are 0 and 1.
Define routines (or operators) lwb and upb that find the first and last set bit in a binary value. Implement using a binary search to find the location of the particular upper/lower bit.
Also: Define the reverse routines (or operators) rlwb and rupb that find host's positive integers least- and most-significant set bit in a binary value expressed in LSB 0 bit numbering, i.e. indexed from the extreme right bit.
Use primarily bit operations, such as and, or, and bit shifting. Avoid additions, multiplications and especially avoid divisions.
Two implementations:
- For the host word size on the host platform, implement the routine "efficiently" in without looping or recursion.
- For the extended precision/long word implement the algorithm more generally - maybe as a template, and maybe with looping - so that any bits width for a binary type can be accommodated.
Test cases:
- For the host machine word size: Use the powers of 42 up to host's the "natural" word size to calculate the index of the first and last set bit.
- For the extended precision: Use the powers of 1302 up to the host's next "natural" long host word size to calculate the index of the first and last set bit.
- Output bit indexes in LSB 0 bit numbering.
Additionally:
In a particular language, there maybe (at least) two alternative approaches of calculating the required values:
- Using an external library.
- Using a built-in library.
If any of these approaches are available, then also note the library or built-in name.
See also:
- Find the log base 2 of an N-bit integer in O(lg(N)) operations
- 80386 Instruction Set - BSF -- Bit Scan Forward
ALGOL 68
File: Template.Find_first_and_last_set_bit.a68<lang algol68>INT lbits width = UPB []BOOL(LBITS(2r0));
OP LWB = (BITS in x)INT: bits width - RUPB in x;
OP RUPB = (BITS in x)INT:
### 32 bit LWB Find Lower Set Bit using an unrolled loop ###
- Note: BITS ELEM 1 is actually numerically the Most Significant Bit!! #
IF in x = 2r0 THEN -1 # EXIT # ELSE BITS x := in x, out := 2r0; IF(x AND NOT 2r1111111111111111)/=2r0 THEN x := x SHR 16; out := out OR 2r10000 FI; IF(x AND NOT 2r11111111) /=2r0 THEN x := x SHR 8; out := out OR 2r1000 FI; IF(x AND NOT 2r1111) /=2r0 THEN x := x SHR 4; out := out OR 2r100 FI; IF(x AND NOT 2r11) /=2r0 THEN x := x SHR 2; out := out OR 2r10 FI; IF(x AND NOT 2r1) /=2r0 THEN out := out OR 2r1 FI; ABS out # EXIT # FI;
OP LWB = (LBITS in x)INT: lbits width - RUPB in x;
OP RUPB = (LBITS in x)INT:
### Generalised Find Lower Set Bit using a loop ###
- Note: BITS ELEM 32 is actually numerically the Least Significant Bit!! #
IF in x = 2r0 THEN
-1 # EXIT #
ELSE
LBITS x := in x;
BITS
out bit := BIN 1 SHL (bits width - LWB BIN lbits width),
out := BIN 0;
WHILE
LBITS mask := NOT BIN (ABS (LONG 2r1 SHL ABS out bit) - 1);
IF(x AND mask) /= 2r0 THEN
x := x SHR ABS out bit;
out := out OR out bit FI;
out bit := out bit SHR 1;
# WHILE # out bit /= 2r0 DO SKIP OD;
ABS out # EXIT #
FI;
OP UPB = (BITS in x)INT: bits width - RLWB in x;
OP RLWB = (BITS in x)INT:
### 32 bit Find Upper Set Bit using an unrolled loop ###
- Note: BITS ELEM 1 is actually numerically the Most Significant Bit!! #
IF in x = 2r0 THEN 0 # EXIT # ELSE BITS x := in x, out := 2r0; IF(x AND 2r1111111111111111)=2r0 THEN x := x SHR 16; out := out OR 2r10000 FI; IF(x AND 2r11111111) =2r0 THEN x := x SHR 8; out := out OR 2r1000 FI; IF(x AND 2r1111) =2r0 THEN x := x SHR 4; out := out OR 2r100 FI; IF(x AND 2r11) =2r0 THEN x := x SHR 2; out := out OR 2r10 FI; IF(x AND 2r1) =2r0 THEN out := out OR 2r1 FI; ABS out # EXIT # FI;
OP UPB = (LBITS in x)INT: lbits width - RLWB in x;
OP RLWB = (LBITS in x)INT:
### Generalised Find Upper Set Bit using a loop ###
- Note: BITS ELEM 1 is actually numerically the Most Significant Bit!! #
IF in x = 2r0 THEN
0 # EXIT #
ELSE
LBITS x := in x;
BITS
out bit := BIN 1 SHL (bits width - LWB BIN lbits width),
out := BIN 0;
WHILE
LBITS mask := BIN (ABS (LONG 2r1 SHL ABS out bit) - 1);
IF(x AND mask) = 2r0 THEN
x := x SHR ABS out bit;
out := out OR out bit FI;
out bit := out bit SHR 1;
# WHILE # out bit /= 2r0 DO SKIP OD;
ABS out # EXIT #
FI;</lang>File: test.Find_first_and_last_set_bit.a68<lang algol68>#!/usr/local/bin/a68g --script #
MODE LBITS = LONG BITS; PR READ "Template.Find_first_and_last_set_bit.a68" PR
INT bits of prod; FORMAT header fmt = $g 36k"|RLWB|RUPB|Bits"l$; FORMAT row fmt0 = $g(-35)"|"2(g(-3)" |"),2rd l$; FORMAT row fmt = $g(-35)"|"2(g(-3)" |"),2rn(bits of prod+1)d l$;
test int:(
printf((header fmt, "INT: find first & last set bit")); INT prod := 0; # test case 0 # prod := 0; bits of prod := RUPB BIN prod; printf((row fmt0, prod, RLWB BIN prod, RUPB BIN prod, BIN prod)); prod := 1; # test case 1 etc ... # INT zoom := 2 * 3 * 7; WHILE bits of prod := RUPB BIN prod; printf((row fmt, prod, RLWB BIN prod, RUPB BIN prod, BIN prod));
- WHILE # prod <= max int / zoom DO
prod *:= zoom OD
);
test long int:(
printf(($l$,header fmt, "LONG INT:")); LONG INT prod := 0; # test case 0 # prod := 0; bits of prod := RUPB BIN prod; printf((row fmt0, prod, RLWB BIN prod, RUPB BIN prod, BIN prod)); prod := 1; # test case 1 etc ... # INT zoom := 2 * 3 * 7 * 31; WHILE bits of prod := RUPB BIN prod; printf((row fmt, prod, RLWB BIN prod, RUPB BIN prod, BIN prod));
- WHILE # prod <= long max int / zoom DO
prod *:= zoom OD
)</lang>Output:
INT: find first & last set bit |RLWB|RUPB|Bits
0| 0 | -1 |0
1| 0 | 0 |1
42| 1 | 5 |101010
1764| 2 | 10 |11011100100
74088| 3 | 16 |10010000101101000
3111696| 4 | 21 |1011110111101100010000
130691232| 5 | 26 |111110010100011000010100000
LONG INT: |RLWB|RUPB|Bits
0| 0 | -1 |0
1| 0 | 0 |1
1302| 1 | 10 |10100010110
1695204| 2 | 20 |110011101110111100100
2207155608| 3 | 31 |10000011100011101000010110011000
2873716601616| 4 | 41 |101001110100010110110110110111001100010000
3741579015304032| 5 | 51 |1101010010101111001001000000000110110011001101100000
4871535877925849664| 6 | 62 |100001110011011001011000001001000001010010101110100101001000000
6342739713059456262528| 7 | 72 |1010101111101011100110010001000111100000010010111111100111010000110000000
8258247106403412053811456| 8 | 82 |11011010100110000000111100100000001110101011000010011010001000101110110000100000000
10752237732537242494062515712| 9 | 93 |1000101011111000001010111001110110111101010011111100010111111101101100111001110101011000000000
13999413527763489727269395457024| 10 |103 |10110000101100101000101101110101000100000011010011101110001111100001001111100000100011110110010000000000
18227236413148063624904752885045248| 11 |113 |111000001010101100000100010100010101100000011011010011001110101111101110010001100000011001010001101001100000000000
C
<lang c>#include <stdio.h>
- include <stdint.h>
uint32_t msb32(uint32_t n) { uint32_t b = 1; if (!n) return 0;
- define step(x) if (n >= ((uint32_t)1) << x) b <<= x, n >>= x
step(16); step(8); step(4); step(2); step(1);
- undef step
return b; }
int msb32_idx(uint32_t n) { int b = 0; if (!n) return -1;
- define step(x) if (n >= ((uint32_t)1) << x) b += x, n >>= x
step(16); step(8); step(4); step(2); step(1);
- undef step
return b; }
- define lsb32(n) ( (uint32_t)(n) & -(int32_t)(n) )
/* finding the *position* of the least significant bit
rarely makes sense, so we don't put much effort in it*/
inline int lsb32_idx(uint32_t n) { return msb32_idx(lsb32(n)); }
int main() { int32_t n; int i;
for (i = 0, n = 1; ; i++, n *= 42) { printf("42**%d = %10d(x%08x): M x%08x(%2d) L x%03x(%2d)\n", i, n, n, msb32(n), msb32_idx(n), lsb32(n), lsb32_idx(n));
if (n >= INT32_MAX / 42) break; }
return 0; }</lang>output ("x###" are in base 16)
42**0 = 1(x00000001): M x00000001( 0) L x001( 0) 42**1 = 42(x0000002a): M x00000020( 5) L x002( 1) 42**2 = 1764(x000006e4): M x00000400(10) L x004( 2) 42**3 = 74088(x00012168): M x00010000(16) L x008( 3) 42**4 = 3111696(x002f7b10): M x00200000(21) L x010( 4) 42**5 = 130691232(x07ca30a0): M x04000000(26) L x020( 5)
GCC extension
<lang c>#include <stdio.h>
- include <limits.h>
int msb_int(unsigned int x) { int ret = sizeof(unsigned int) * CHAR_BIT - 1; return x ? ret - __builtin_clz(x) : ret; }
int msb_long(unsigned long x) { int ret = sizeof(unsigned long) * CHAR_BIT - 1; return x ? ret - __builtin_clzl(x) : ret; }
int msb_longlong(unsigned long long x) { int ret = sizeof(unsigned long long) * CHAR_BIT - 1; return x ? ret - __builtin_clzll(x) : ret; }
- define lsb_int(x) (__builtin_ffs(x) - 1)
- define lsb_long(x) (__builtin_ffsl(x) - 1)
- define lsb_longlong(x) (__builtin_ffsll(x) - 1)
int main() { int i;
printf("int:\n");
unsigned int n1; for (i = 0, n1 = 1; ; i++, n1 *= 42) { printf("42**%d = %10u(x%08x): M %2d L %2d\n", i, n1, n1, msb_int(n1), lsb_int(n1));
if (n1 >= UINT_MAX / 42) break; }
printf("long:\n");
unsigned long n2; for (i = 0, n2 = 1; ; i++, n2 *= 42) { printf("42**%02d = %20lu(x%016lx): M %2d L %2d\n", i, n2, n2, msb_long(n2), lsb_long(n2));
if (n2 >= ULONG_MAX / 42) break; }
return 0; }</lang>output ("x###" are in base 16)
int: 42**0 = 1(x00000001): M 0 L 0 42**1 = 42(x0000002a): M 5 L 1 42**2 = 1764(x000006e4): M 10 L 2 42**3 = 74088(x00012168): M 16 L 3 42**4 = 3111696(x002f7b10): M 21 L 4 42**5 = 130691232(x07ca30a0): M 26 L 5 long: 42**00 = 1(x0000000000000001): M 0 L 0 42**01 = 42(x000000000000002a): M 5 L 1 42**02 = 1764(x00000000000006e4): M 10 L 2 42**03 = 74088(x0000000000012168): M 16 L 3 42**04 = 3111696(x00000000002f7b10): M 21 L 4 42**05 = 130691232(x0000000007ca30a0): M 26 L 5 42**06 = 5489031744(x00000001472bfa40): M 32 L 6 42**07 = 230539333248(x00000035ad370e80): M 37 L 7 42**08 = 9682651996416(x000008ce6b086100): M 43 L 8 42**09 = 406671383849472(x000171dd8f5fea00): M 48 L 9 42**10 = 17080198121677824(x003cae5985bc6400): M 53 L 10 42**11 = 717368321110468608(x09f49aaff0e86800): M 59 L 11
Go
<lang go>package main
import (
"fmt" "math/big"
)
const (
mask0, bit0 = (1 << (1 << iota)) - 1, 1 << iota mask1, bit1 mask2, bit2 mask3, bit3 mask4, bit4 mask5, bit5
)
func rupb(x uint64) (out int) {
if x == 0 {
return -1
}
if x&^mask5 != 0 {
x >>= bit5
out |= bit5
}
if x&^mask4 != 0 {
x >>= bit4
out |= bit4
}
if x&^mask3 != 0 {
x >>= bit3
out |= bit3
}
if x&^mask2 != 0 {
x >>= bit2
out |= bit2
}
if x&^mask1 != 0 {
x >>= bit1
out |= bit1
}
if x&^mask0 != 0 {
out |= bit0
}
return
}
func rlwb(x uint64) (out int) {
if x == 0 {
return 0
}
if x&mask5 == 0 {
x >>= bit5
out |= bit5
}
if x&mask4 == 0 {
x >>= bit4
out |= bit4
}
if x&mask3 == 0 {
x >>= bit3
out |= bit3
}
if x&mask2 == 0 {
x >>= bit2
out |= bit2
}
if x&mask1 == 0 {
x >>= bit1
out |= bit1
}
if x&mask0 == 0 {
out |= bit0
}
return
}
// Big number versions of functions do not use the techniques of the ALGOL 68 // solution. The big number version of rupb is trivial given one of the // standard library functions, And for rlwb, I couldn't recommend shifting // the whole input number when working with smaller numbers would do. func rupbBig(x *big.Int) int {
return x.BitLen() - 1
}
// Binary search, for the spirit of the task, but without shifting the input // number x. (Actually though, I don't recommend this either. Linear search // would be much faster.) func rlwbBig(x *big.Int) int {
if x.BitLen() < 2 {
return 0
}
bit := uint(1)
mask := big.NewInt(1)
var ms []*big.Int
var y, z big.Int
for y.And(x, z.Lsh(mask, bit)).BitLen() == 0 {
ms = append(ms, mask)
mask = new(big.Int).Or(mask, &z)
bit <<= 1
}
out := bit
for i := len(ms) - 1; i >= 0; i-- {
bit >>= 1
if y.And(x, z.Lsh(ms[i], out)).BitLen() == 0 {
out |= bit
}
}
return int(out)
}
func main() {
show() showBig()
}
func show() {
fmt.Println("uint64:")
fmt.Println("power number rupb rlwb")
const base = 42
n := uint64(1)
for i := 0; i < 12; i++ {
fmt.Printf("%d^%02d %19d %5d %5d\n", base, i, n, rupb(n), rlwb(n))
n *= base
}
}
func showBig() {
fmt.Println("\nbig numbers:")
fmt.Println(" power number rupb rlwb")
base := big.NewInt(1302)
n := big.NewInt(1)
for i := 0; i < 12; i++ {
fmt.Printf("%d^%02d %36d %5d %5d\n", base, i, n, rupbBig(n), rlwbBig(n))
n.Mul(n, base)
}
}</lang>
- Output:
uint64: power number rupb rlwb 42^00 1 0 0 42^01 42 5 1 42^02 1764 10 2 42^03 74088 16 3 42^04 3111696 21 4 42^05 130691232 26 5 42^06 5489031744 32 6 42^07 230539333248 37 7 42^08 9682651996416 43 8 42^09 406671383849472 48 9 42^10 17080198121677824 53 10 42^11 717368321110468608 59 11 big numbers: power number rupb rlwb 1302^00 1 0 0 1302^01 1302 10 1 1302^02 1695204 20 2 1302^03 2207155608 31 3 1302^04 2873716601616 41 4 1302^05 3741579015304032 51 5 1302^06 4871535877925849664 62 6 1302^07 6342739713059456262528 72 7 1302^08 8258247106403412053811456 82 8 1302^09 10752237732537242494062515712 93 9 1302^10 13999413527763489727269395457024 103 10 1302^11 18227236413148063624904752885045248 113 11
Icon and Unicon
The task definition makes some assumptions that don't work in Icon/Unicon and are going to require some reinterpretation. In Icon/Unicon all integers appear to be implemented as a single common type. A specific implementation may or may not have large integers, but if it does they are essentially indistinguishable from regular integers. Given all of this, implementing "efficient" procedures for the platform word size without loops or recursion makes little sense.
Instead of this, to meet the spirit of the task, these lsb and msb routines are generalized to reduce the integer in blocks of bits and then zoom in on the desired bit by binary search (i.e. successively looking a blocks that are half the size again). The exponent for the initial power used to create the masks does not need to be itself a power of two. The xsb_initial procedure uses introspection to determine the word size of a basic integer type. This is used to build a mask that fits within the basic word size of the implementation. In this way we won't create unnecessary large integers through implicit type conversions.
<lang Icon>link printf,hexcvt
procedure main()
every B := [42,2^32-1] | [1302,2^64-1] do {
base := B[1]
lim := B[2]
fmt := sprintf("%%i^%%i = %%%is (x%%0%is) : MSB=%%s LSB=%%s\n",*lim,*hexstring(lim))
every e := seq(0) do {
if (i := base^e) > lim then break
printf(fmt,base,e,i,hexstring(i),msb(i)|"-",lsb(i)|"-")
}
}
end
procedure msb(i) #: return the most significant set bit index or fail static mask initial mask := xsb_initial()
if i > 0 then {
b := 0
every m := mask[j := 1 to *mask by 2] & r := mask[j+1] do {
repeat {
l := iand(i,m)
i := ishift(i,r)
if i = 0 then break
b -:= r
}
i := l
}
return b
}
end
procedure lsb(i) #: return the least significant set bit index or fail static mask initial mask := xsb_initial()
if i > 0 then {
b := 0
every m := mask[j := 1 to *mask by 2] & r := mask[j+1] do
until iand(i,m) > 0 do {
i := ishift(i,r)
b -:= r
}
return b
}
end
procedure xsb_initial() #: setup tables for lsb/msb static mask initial { # build
a := &allocated # bigint affects allocation
p := if 2^63 & a=&allocated then 63 else 31 # find wordsize-1
p *:= 2 # adjust pre-loop
mask := []
until (p := p / 2) = 0 do put(mask,2^p-1,-p) # list of masks and shifts
}
return mask # return pre-built data
end</lang>
printf.icn provides formatting hexcvt.icn provides hexstring
Output:
42^0 = 1 (x00000001) : MSB=0 LSB=0 42^1 = 42 (x0000002A) : MSB=5 LSB=1 42^2 = 1764 (x000006E4) : MSB=10 LSB=2 42^3 = 74088 (x00012168) : MSB=16 LSB=3 42^4 = 3111696 (x002F7B10) : MSB=21 LSB=4 42^5 = 130691232 (x07CA30A0) : MSB=26 LSB=5 1302^0 = 1 (x0000000000000001) : MSB=0 LSB=0 1302^1 = 1302 (x0000000000000516) : MSB=10 LSB=1 1302^2 = 1695204 (x000000000019DDE4) : MSB=20 LSB=2 1302^3 = 2207155608 (x00000000838E8598) : MSB=31 LSB=3 1302^4 = 2873716601616 (x0000029D16DB7310) : MSB=41 LSB=4 1302^5 = 3741579015304032 (x000D4AF2401B3360) : MSB=51 LSB=5 1302^6 = 4871535877925849664 (x439B2C120A574A40) : MSB=62 LSB=6
J
Implementation:
<lang j>lwb=: 0: upb=: (#: i: 1:)"0 rlwb=: #@#:"0 - 1: rupb=: rlwb - upb</lang>
Notes:
J's #: converts integers to bit lists.
This implementation is agnostic to numeric storage format.
Example use:
<lang j> (,.lwb,.upb,.rlwb,.rupb) <.i.@>.&.(42&^.) 2^64
1 0 0 0 0
42 0 4 5 1
1764 0 8 10 2
74088 0 13 16 3
3111696 0 17 21 4
130691232 0 21 26 5
5489031744 0 26 32 6
230539333248 0 30 37 7
9682651996416 0 35 43 8
406671383849472 0 39 48 9
17080198121677824 0 43 53 10
717368321110468608 0 48 59 11
(,.lwb,.upb,.rlwb,.rupb) i.@x:@>.&.(1302&^.) 2^128
1 0 0 0 0
1302 0 9 10 1
1695204 0 18 20 2
2207155608 0 28 31 3
2873716601616 0 37 41 4
3741579015304032 0 46 51 5
4871535877925849664 0 56 62 6
6342739713059456262528 0 65 72 7
8258247106403412053811456 0 74 82 8
10752237732537242494062515712 0 84 93 9
13999413527763489727269395457024 0 93 103 10
18227236413148063624904752885045248 0 102 113 11
23731861809918778839625988256328912896 0 112 124 12</lang>
Note, in the above sentences, the rightmost part of each sentence is about generating an arbitrary sequence of values. The phrase <.i.@>.&.(42&^.) 2^64 generates the sequence 1 42 1764 74088 3111696 130691232 ... and the phrase i.@x:@>.&.(1302&^.) 2^128 generates the sequence 1 1302 1695204 2207155608 ...
The left part of each sentence uses the words we defined here, organizing their results as columns in a table.
Java
Library
Notes:
- least significant bit is bit 0 (such that bit i always has value 2i, and the result is independent of integer type width)
- when the integer 0 is given, mssb() and lssb() both return no bits set; mssb_idx() and lssb_idx() will return -1 and the integer type width, respectively
<lang java>public class FirstLastBits {
// most significant set bit
public static int mssb(int x) {
return Integer.highestOneBit(x);
}
public static long mssb(long x) {
return Long.highestOneBit(x);
}
public static int mssb_idx(int x) {
return Integer.SIZE - 1 - Integer.numberOfLeadingZeros(x);
}
public static int mssb_idx(long x) {
return Long.SIZE - 1 - Long.numberOfLeadingZeros(x);
}
public static int mssb_idx(BigInteger x) {
return x.bitLength() - 1;
}
// least significant set bit
public static int lssb(int x) {
return Integer.lowestOneBit(x);
}
public static long lssb(long x) {
return Long.lowestOneBit(x);
}
public static int lssb_idx(int x) {
return Integer.numberOfTrailingZeros(x);
}
public static int lssb_idx(long x) {
return Long.numberOfTrailingZeros(x);
}
public static int lssb_idx(BigInteger x) {
return x.getLowestSetBit();
}
public static void main(String[] args) {
System.out.println("int:");
int n1 = 1;
for (int i = 0; ; i++, n1 *= 42) {
System.out.printf("42**%d = %10d(x%08x): M x%08x(%2d) L x%03x(%2d)\n",
i, n1, n1,
mssb(n1), mssb_idx(n1),
lssb(n1), lssb_idx(n1));
if (n1 >= Integer.MAX_VALUE / 42)
break;
}
System.out.println();
System.out.println("long:");
long n2 = 1;
for (int i = 0; ; i++, n2 *= 42) {
System.out.printf("42**%02d = %20d(x%016x): M x%016x(%2d) L x%06x(%2d)\n",
i, n2, n2,
mssb(n2), mssb_idx(n2),
lssb(n2), lssb_idx(n2));
if (n2 >= Long.MAX_VALUE / 42)
break;
}
System.out.println(); System.out.println("BigInteger:"); BigInteger n3 = BigInteger.ONE; BigInteger k = BigInteger.valueOf(1302); for (int i = 0; i < 10; i++, n3 = n3.multiply(k)) { System.out.printf("1302**%02d = %30d(x%28x): M %2d L %2d\n", i, n3, n3, mssb_idx(n3), lssb_idx(n3)); }
}
}</lang> output:
int: 42**0 = 1(x00000001): M x00000001( 0) L x001( 0) 42**1 = 42(x0000002a): M x00000020( 5) L x002( 1) 42**2 = 1764(x000006e4): M x00000400(10) L x004( 2) 42**3 = 74088(x00012168): M x00010000(16) L x008( 3) 42**4 = 3111696(x002f7b10): M x00200000(21) L x010( 4) 42**5 = 130691232(x07ca30a0): M x04000000(26) L x020( 5) long: 42**00 = 1(x0000000000000001): M x0000000000000001( 0) L x000001( 0) 42**01 = 42(x000000000000002a): M x0000000000000020( 5) L x000002( 1) 42**02 = 1764(x00000000000006e4): M x0000000000000400(10) L x000004( 2) 42**03 = 74088(x0000000000012168): M x0000000000010000(16) L x000008( 3) 42**04 = 3111696(x00000000002f7b10): M x0000000000200000(21) L x000010( 4) 42**05 = 130691232(x0000000007ca30a0): M x0000000004000000(26) L x000020( 5) 42**06 = 5489031744(x00000001472bfa40): M x0000000100000000(32) L x000040( 6) 42**07 = 230539333248(x00000035ad370e80): M x0000002000000000(37) L x000080( 7) 42**08 = 9682651996416(x000008ce6b086100): M x0000080000000000(43) L x000100( 8) 42**09 = 406671383849472(x000171dd8f5fea00): M x0001000000000000(48) L x000200( 9) 42**10 = 17080198121677824(x003cae5985bc6400): M x0020000000000000(53) L x000400(10) 42**11 = 717368321110468608(x09f49aaff0e86800): M x0800000000000000(59) L x000800(11) BigInteger: 1302**00 = 1(x 1): M 0 L 0 1302**01 = 1302(x 516): M 10 L 1 1302**02 = 1695204(x 19dde4): M 20 L 2 1302**03 = 2207155608(x 838e8598): M 31 L 3 1302**04 = 2873716601616(x 29d16db7310): M 41 L 4 1302**05 = 3741579015304032(x d4af2401b3360): M 51 L 5 1302**06 = 4871535877925849664(x 439b2c120a574a40): M 62 L 6 1302**07 = 6342739713059456262528(x 157d73223c097f3a180): M 72 L 7 1302**08 = 8258247106403412053811456(x 6d4c07901d584d1176100): M 82 L 8 1302**09 = 10752237732537242494062515712(x 22be0ae76f53f17f6ce75600): M 93 L 9
Mathematica
<lang Mathematica>MSB[n_]:=BitLength[n]-1 LSB[n_]:=IntegerExponent[n,2]</lang>
Map[{#,"MSB:",MSB[#],"LSB:",LSB[#]}&,
Join[NestList[(42*#)&,42,5],NestList[(1302*#)&,1302,5]]]//TableForm
42 MSB: 5 LSB: 1
1764 MSB: 10 LSB: 2
74088 MSB: 16 LSB: 3
3111696 MSB: 21 LSB: 4
130691232 MSB: 26 LSB: 5
5489031744 MSB: 32 LSB: 6
1302 MSB: 10 LSB: 1
1695204 MSB: 20 LSB: 2
2207155608 MSB: 31 LSB: 3
2873716601616 MSB: 41 LSB: 4
3741579015304032 MSB: 51 LSB: 5
4871535877925849664 MSB: 62 LSB: 6
PARI/GP
This version uses PARI. These work on arbitrary-length integers; the implementation for wordsize integers would be identical to C's. <lang c>long msb(GEN n) { return expi(n); }
long lsb(GEN n) { return vali(n); }</lang>
This version uses GP. It works on arbitrary-length integers; GP cannot directly work on wordsize integers except in a vecsmall.
<lang parigp>lsb(n)=valuation(n,2);
msb(n)=#binary(n)-1;</lang>
Python
<lang python>def msb(x):
return x.bit_length() - 1
def lsb(x):
return msb(x & -x)
for i in range(6):
x = 42 ** i
print("%10d MSB: %2d LSB: %2d" % (x, msb(x), lsb(x)))
for i in range(6):
x = 1302 ** i
print("%20d MSB: %2d LSB: %2d" % (x, msb(x), lsb(x)))</lang>
- Output:
1 MSB: 0 LSB: 0
42 MSB: 5 LSB: 1
1764 MSB: 10 LSB: 2
74088 MSB: 16 LSB: 3
3111696 MSB: 21 LSB: 4
130691232 MSB: 26 LSB: 5
1 MSB: 0 LSB: 0
1302 MSB: 10 LSB: 1
1695204 MSB: 20 LSB: 2
2207155608 MSB: 31 LSB: 3
2873716601616 MSB: 41 LSB: 4
3741579015304032 MSB: 51 LSB: 5
REXX
<lang rexx>/*REXX program to find the first & last set bit of "int" and "long int".*/
do cycle=1 to 2
base=word(42 1302,cycle); w=digits() /*W is used for nice display*/
call sep '─┬─'
say center(base'**n',w) '│' center('rlwb',4) '│' centre('rupb',4) '│ bits'
call sep '─┼─'
do j=-1
if j<0 then x=0
else x=base**j
if hasE(x) then leave
say right(x,w) '│' right(rlwb(x),4) '│' right(rupb(x),4) '│' bits
end /*j*/
call sep '─┴─'
do 2
say; numeric digits(digits()+digits()); say
end /*2*/
end /*cycle*/
exit /*─────────────────────────────────────subroutines──────────────────────*/ rupb: arg n;call n2b n; if n==0 then return -1; return length(n2b(n))-1 n2b: bits=word(strip(x2b(d2x(arg(1))),'L',0) 0,1); return bits hasE: return pos('E',arg(1))\==0 rlwb: arg n;call n2b n; if n==0 then return 0; return length(n2b(n))-length(strip(bits,'T',0)) sep: say copies('─',w)arg(1)copies('─',4)arg(1)copies('─',4)arg(1) ||,
copies('─',length(n2b(10**digits())))</lang>
output
──────────┬──────┬──────┬───────────────────────────────
42**n │ rlwb │ rupb │ bits
──────────┼──────┼──────┼───────────────────────────────
0 │ 0 │ -1 │ 0
1 │ 0 │ 0 │ 1
42 │ 1 │ 5 │ 101010
1764 │ 2 │ 10 │ 11011100100
74088 │ 3 │ 16 │ 10010000101101000
3111696 │ 4 │ 21 │ 1011110111101100010000
130691232 │ 5 │ 26 │ 111110010100011000010100000
──────────┴──────┴──────┴───────────────────────────────
─────────────────────────────────────┬──────┬──────┬─────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────
1302**n │ rlwb │ rupb │ bits
─────────────────────────────────────┼──────┼──────┼─────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────
0 │ 0 │ -1 │ 0
1 │ 0 │ 0 │ 1
1302 │ 1 │ 10 │ 10100010110
1695204 │ 2 │ 20 │ 110011101110111100100
2207155608 │ 3 │ 31 │ 10000011100011101000010110011000
2873716601616 │ 4 │ 41 │ 101001110100010110110110110111001100010000
3741579015304032 │ 5 │ 51 │ 1101010010101111001001000000000110110011001101100000
4871535877925849664 │ 6 │ 62 │ 100001110011011001011000001001000001010010101110100101001000000
6342739713059456262528 │ 7 │ 72 │ 1010101111101011100110010001000111100000010010111111100111010000110000000
8258247106403412053811456 │ 8 │ 82 │ 11011010100110000000111100100000001110101011000010011010001000101110110000100000000
10752237732537242494062515712 │ 9 │ 93 │ 1000101011111000001010111001110110111101010011111100010111111101101100111001110101011000000000
13999413527763489727269395457024 │ 10 │ 103 │ 10110000101100101000101101110101000100000011010011101110001111100001001111100000100011110110010000000000
18227236413148063624904752885045248 │ 11 │ 113 │ 111000001010101100000100010100010101100000011011010011001110101111101110010001100000011001010001101001100000000000
─────────────────────────────────────┴──────┴──────┴─────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────
Tcl
<lang tcl>proc lwb {x} {
if {$x == 0} {return -1}
set n 0
while {($x&1) == 0} {
set x [expr {$x >> 1}] incr n
} return $n
} proc upb {x} {
if {$x == 0} {return -1}
if {$x < 0} {error "no well-defined max bit for negative numbers"}
set n 0
while {$x != 1} {
set x [expr {$x >> 1}] incr n
} return $n
}</lang> Code to use the above: <lang tcl>package require Tcl 8.6; # For convenient bit string printing
proc powsTo {pow bits} {
set result {}
for {set n 1} {$n < 2**$bits} {set n [expr {$n * $pow}]} {
lappend result $n
} return $result
} proc printPows {pow pows} {
set len [string length [lindex $pows end]]
puts [format "%8s | %*s | LWB | UPB | Bits" "What" $len "Number"]
set n 0
foreach p $pows {
puts [format "%4d**%-2d = %*lld | %3d | %3d | %b" \ $pow $n $len $p [lwb $p] [upb $p] $p] incr n
}
}
puts "Powers of 42 up to machine word size:" printPows 42 [powsTo 42 [expr {$tcl_platform(wordSize) * 8}]] puts "Powers of 1302 up to 128 bits" printPows 1302 [powsTo 1302 128]</lang> Output:
Powers of 42 up to machine word size:
What | Number | LWB | UPB | Bits
42**0 = 1 | 0 | 0 | 1
42**1 = 42 | 1 | 5 | 101010
42**2 = 1764 | 2 | 10 | 11011100100
42**3 = 74088 | 3 | 16 | 10010000101101000
42**4 = 3111696 | 4 | 21 | 1011110111101100010000
42**5 = 130691232 | 5 | 26 | 111110010100011000010100000
Powers of 1302 up to 128 bits
What | Number | LWB | UPB | Bits
1302**0 = 1 | 0 | 0 | 1
1302**1 = 1302 | 1 | 10 | 10100010110
1302**2 = 1695204 | 2 | 20 | 110011101110111100100
1302**3 = 2207155608 | 3 | 31 | 10000011100011101000010110011000
1302**4 = 2873716601616 | 4 | 41 | 101001110100010110110110110111001100010000
1302**5 = 3741579015304032 | 5 | 51 | 1101010010101111001001000000000110110011001101100000
1302**6 = 4871535877925849664 | 6 | 62 | 100001110011011001011000001001000001010010101110100101001000000
1302**7 = 6342739713059456262528 | 7 | 72 | 1010101111101011100110010001000111100000010010111111100111010000110000000
1302**8 = 8258247106403412053811456 | 8 | 82 | 11011010100110000000111100100000001110101011000010011010001000101110110000100000000
1302**9 = 10752237732537242494062515712 | 9 | 93 | 1000101011111000001010111001110110111101010011111100010111111101101100111001110101011000000000
1302**10 = 13999413527763489727269395457024 | 10 | 103 | 10110000101100101000101101110101000100000011010011101110001111100001001111100000100011110110010000000000
1302**11 = 18227236413148063624904752885045248 | 11 | 113 | 111000001010101100000100010100010101100000011011010011001110101111101110010001100000011001010001101001100000000000
1302**12 = 23731861809918778839625988256328912896 | 12 | 124 | 10001110110101001011100011111110101101101100001101011011001001101111110110111011000001001000010001101000010010001000000000000