问题
There seems to be some kind of misconception that this is for a contest. I'm trying to work through an assignment and I've been stuck on this for an hour now.
/*
* isLessOrEqual - if x <= y then return 1, else return 0
* Example: isLessOrEqual(4,5) = 1.
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 24
* Rating: 3
*/
int isLessOrEqual(int x, int y)
{
int greater = (x + (~y + 1))>>31 & 1;
return !(greater)|(!(x^y));
}
I'm only able to use bitwise operators, as instructed in the comments.
I cannot figure out how to solve x <= y;
My thought process is that I can set x as its two's complement (~x +1) and add it with Y. If it is negative, X is greater than Y. Therefore, by negating that I can get the opposite effect.
Similarly, I know that !(x^y) is equivalent to x==y.
However,
doing !(greater)|(!(x^y)) does not return the proper value.
Where am I messing up? I feel like I'm missing a small bit of logic.
回答1:
If x > y, then y - x or (y + (~x + 1)) will be negative, hence the high bit will be 1, otherwise it will be 0. But we want x <= y, which is the negation of this.
/*
* isLessOrEqual - if x <= y then return 1, else return 0
* Example: isLessOrEqual(4,5) = 1.
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 24
* Rating: 3
*/
int isLessOrEqual(int x, int y)
{
return !(((y + (~x + 1)) >> 31) & 1);
}
Even better, drop the shift operator and use a bit mask on the high bit:
int isLessOrEqual(int x, int y)
{
return !((y + (~x + 1)) & 0x80000000);
}
EDIT:
As a commenter pointer out, the above version is susceptible to arithmetic overflow errors. Here is another version that covers the edge cases.
#include <limits>
int isLessOrEqual(int x, int y)
{
static int const vm = std::numeric_limits<int>::max();
static int const sm = ~vm;
return !! ((x & ~y | ~(x ^ y) & ~((y & vm) + ~(x & vm) + 1)) & sm);
}
Explanation: the overall strategy is to treat the sign bit of the inputs as logically distinct from the rest of the bits, the "value bits," and perform the subtraction as in the previous example on just the value bits. In this case, we need only perform the subtraction where the two inputs are either both negative or both non-negative. This avoids the arithmetic overflow condition.
Since the size of int strictly speaking is unknown at run time, we use std::numeric_limits<int>::max() as a convenient mask for the value bits. The mask of the sign bit is simply the bit-wise negation of the value bits.
Turning to the actual expression for <=, we factor out the bit-wise mask sm of the sign bit in each of the sub-expressions and push the operation to the outside of the expression. The first term of the logical expression x & ~y is true when x is negative and y is non-negative. The first factor of the next term ~(x ^ Y) is true when both are negative or both are non-negative. The second factor ~((y & vm) + ~(x & vm) + 1)) is true when y - x is non-negative, in other words x <= y, ignoring the sign bit. The two terms are or'd, so using c++ logical expression syntax we have:
x < 0 && y >= 0 || (x < 0 && y < 0 || x >= 0 && y >= 0) && y - x >= 0
The !! outermost operators convert the raised sign bit to a 1. Finally, here is the Modern C++ templated constexpr version:
template<typename T>
constexpr T isLessOrEqual(T x, T y)
{
using namespace std;
// compile time check that type T makes sense for this function
static_assert(is_integral<T>::value && is_signed<T>::value, "isLessOrEqual requires signed integral params");
T vm = numeric_limits<T>::max();
T sm = ~vm;
return !! ((x & ~y | ~(x ^ y) & ~((y & vm) + ~(x & vm) + 1)) & sm);
}
回答2:
Those functions don't fully work because of the overflow, so that's how I solved the problem. Eh...
int isLessOrEqual(int x, int y) {
int diff_sgn = !(x>>31)^!(y>>31); //is 1 when signs are different
int a = diff_sgn & (x>>31); //diff signs and x is neg, gives 1
int b = !diff_sgn & !((y+(~x+1))>>31); //same signs and difference is pos or = 0, gives 1
int f = a | b;
return f;
}
回答3:
Really enjoyed Yanagar1's answer, which is very easy to understand.
Actually we can remove those shift operators and use De Morgan's laws, which reduce the number of operators from 15 to 11.
long isLessOrEqual(long x, long y) {
long sign = (x ^ y); // highest bit will be 1 if different sign
long diff = sign & x; // highest bit will be 1 if diff sign and neg x
long same = sign | (y + (~x + 1)); // De Morgan's Law with the following ~same
// highest bit will be 0 if same sign and y >= x
long result = !!((diff | ~same) & 0x8000000000000000L); // take highest bit(sign) here
return result;
}
回答4:
Here is my implementation(spend around 3 hours...)
int
isLessOrEqual(int x, int y)
{
int a = y + ~x + 1;
int b = a & 1 << 31 & a; // !b => y >= x, but maybe overflow
int c = !!(x & (1 << 31)) & !(y & (1 << 31)); // y > 0, x < 0
int d = !(x & (1 << 31)) & !!(y & (1 << 31)); // x > 0, y < 0
int mask1 = !c + ~0;
// if y > 0 && x < 0, return 1. else return !b
int ans = ~mask1 & !b | mask1 & 1;
int mask2 = !d + ~0;
// if y < 0 && x > 0, return 0, else return ans
return ~mask2 & ans | mask2 & 0;
}
y - x == y + ~x + 1a & 1 << 31 & ais simplify from!(!(a & (1 << 31)) | !a)
The logic is:
if `y > 0 && x < 0`
return true
if `x > 0 && y < 0`
return false
return y >= x
Why not just y >= x directly? because overflow may happen. So I have to early return to avoid overflow.
回答5:
Inspired by Yanagar1's answer, here's my implementation:
int isLessOrEqual(int x, int y) {
int indicator = !((y + (~x + 1)) >> 31); // negation of the result of y - x, 0 when y < x, -1 when y >= x
int xsign = x >> 31; // -1 when x < 0, 0 when x >= 0
int ysign = y >> 31; // -1 when y < 0, 0 when y >= 0
int xbool = !xsign; // 0 when x < 0, 1 when x >= 0
int ybool = !ysign; // 0 when y < 0, 1 when y >= 0
int result = (!(xbool ^ ybool)) & indicator;
return result | (ybool & !xbool);
}
Explanation: Adding 2's complement negation of x (~x + 1) to y is essentially calculating y - x, then logical negate the sign bit of the result, we can have 0 when y < x, and 1 when y >= x. But there are potential overflow cases (overflow cannot happen when y and -x have opposite signs, that is, when y and x have same signs):
|-----------|------------------------|------------------------|
| | y > 0 | y < 0 |
|-----------|------------------------|------------------------|
| x > 0 | ok | overflow when y = TMin |
|-----------|------------------------|------------------------|
| x < 0 | overflow when x = TMin | ok |
|-----------|------------------------|------------------------|
so we need to be careful when the signs are different.
来源：https://stackoverflow.com/questions/41948852/bitwise-less-than-or-equal-to