Thanks, Tagir!
I was also thinking of how to calculate hashCode quickly but my direction
was wrong. I thought that we can use the formula of the sum of a geometric
progression: Sum(p^k, k = 0..n) = (1-p^n)/(1-p). Unfortunately, this
involves division which doesn't work with the overflow of integers. I
didn't know the trick p^(2n) = (p^n)^2.
Your formulas and implementation look correct.
I would probably rewrite the loop to make it a bit simpler:
for (int mask = n << (Integer.numberOfLeadingZeros(n) + 1); mask != 0; mask
<<= 1) {
sum = sum * (pow + 1);
pow *= pow;
if ((mask & 0x8000_0000) != 0) {
pow *= 31;
sum = sum * 31 + 1;
}
}
But that's just a matter of style.
Great job!
пт, 30 нояб. 2018 г. в 11:02, Tagir Valeev <amae...@gmail.com>:
> Hello!
>
> If you are doing it fast, why not doing it really fast? If you
> deparenthesize and regroup terms, you'll got
> h(e, n) = p ^ n + e * f(n)
> Where h(e, n) is the hashCode of n elements with hashCode of single
> element = e; p = 31 and
> f(n) = Sum(p^k, k = 0..n-1)
>
> Using simple algebraic rules, you'll get:
> p ^ (2n) = (p^n)^2
> f(2n) = f(n) * (p^n + 1)
> p ^ (n+1) = (p^n)*p
> f(n+1) = f(n) * p + 1
>
> Thus the algorithm may look as follows:
>
> public int hashCode() {
> int pow = 1; // -> 31^n
> int sum = 0; // -> Sum(31^k, k = 0..n-1)
> for (int i = Integer.numberOfLeadingZeros(n); i < Integer.SIZE; i++) {
> sum = sum * (pow + 1);
> pow *= pow;
> if (((n << i) & 0x8000_0000) != 0) {
> pow *= 31;
> sum = sum * 31 + 1;
> }
> }
> return pow + sum * (element == null ? 0 : element.hashCode());
> }
>
> It seems reasonable to peel off the n = 0 case, which helps to reduce
> number of multiplications for other cases (also for n = 0 we don't
> need to calculate element hashCode at all):
>
> public int hashCode() {
> if (n == 0) return 1;
> int pow = 31; // -> 31^n
> int sum = 1; // -> Sum(31^k, k = 0..n-1)
> for (int i = Integer.numberOfLeadingZeros(n) + 1; i < Integer.SIZE;
> i++) {
> sum = sum * (pow + 1);
> pow *= pow;
> if (((n << i) & 0x8000_0000) != 0) {
> pow *= 31;
> sum = sum * 31 + 1;
> }
> }
> return pow + sum * (element == null ? 0 : element.hashCode());
> }
>
> Assuming that element hashCode is simple (I used String "foo" as an
> element), I got the following results for different collection sizes:
>
> Benchmark (size) Score Error Units
> hashCodeFast 0 2,299 ± 0,017 ns/op
> hashCodeFast 1 2,731 ± 0,021 ns/op
> hashCodeFast 2 4,073 ± 0,077 ns/op
> hashCodeFast 3 4,315 ± 0,032 ns/op
> hashCodeFast 5 5,470 ± 0,074 ns/op
> hashCodeFast 10 6,904 ± 0,060 ns/op
> hashCodeFast 30 9,102 ± 0,173 ns/op
> hashCodeFast 100 10,093 ± 0,069 ns/op
> hashCodeFast 1000 14,129 ± 0,074 ns/op
> hashCodeFast 10000 17,028 ± 0,249 ns/op
> hashCodeFast 100000 20,795 ± 0,194 ns/op
> hashCodeFast 1000000 23,622 ± 0,264 ns/op
>
> Compared to Zheka's implementation:
>
> Benchmark (size) Score Error Units
> hashCodeZheka 0 2,584 ± 0,024 ns/op
> hashCodeZheka 1 2,868 ± 0,022 ns/op
> hashCodeZheka 2 3,730 ± 0,030 ns/op
> hashCodeZheka 3 4,323 ± 0,027 ns/op
> hashCodeZheka 5 5,285 ± 0,037 ns/op
> hashCodeZheka 10 8,254 ± 0,057 ns/op
> hashCodeZheka 30 24,793 ± 0,218 ns/op
> hashCodeZheka 100 89,017 ± 0,764 ns/op
> hashCodeZheka 1000 923,792 ± 28,194 ns/op
> hashCodeZheka 10000 9157,411 ± 98,902 ns/op
> hashCodeZheka 100000 91705,599 ± 689,299 ns/op
> hashCodeZheka 1000000 919723,545 ± 13092,935 ns/op
>
> So results are quite similar for one-digit counts, but we start
> winning from n = 10 and after that logarithmic algorithm really rocks.
>
> I can file an issue and create a webrev, but I still need a sponsor
> and review for such change. Martin, can you help with this?
>
> With best regards,
> Tagir Valeev.
> On Tue, Nov 27, 2018 at 5:49 PM Martin Buchholz <marti...@google.com>
> wrote:
> >
> > I agree!
> >
> > (but don't have time ...)
> >
> > On Sun, Nov 25, 2018 at 9:01 PM, Zheka Kozlov <orionllm...@gmail.com>
> wrote:
> >
> > > Currently, CopiesList.hashCode() is inherited from AbstractList which:
> > >
> > > - calls hashCode() for each element,
> > > - creates a new Iterator every time.
> > >
> > > However, for Collections.nCopies():
> > >
> > > - All elements are the same. So hashCode() can be called only once.
> > > - An Iterator is unnecessary.
> > >
> > > So, I propose overridding hashCode() implementation for CopiesList:
> > >
> > > @Override
> > > public int hashCode() {
> > > int hashCode = 1;
> > > final int elementHashCode = (element == null) ? 0 :
> element.hashCode();
> > > for (int i = 0; i < n; i++) {
> > > hashCode = 31*hashCode + elementHashCode;
> > > }
> > > return hashCode;
> > > }
> > >
> > > Benchmark:
> > > List<List<String>> list = Collections.nCopies(10_000, new
> > > ArrayList<>(Collections.nCopies(1_000_000, "a")));
> > > long nano = System.nanoTime();
> > > System.out.println(list.hashCode());
> > > System.out.println((System.nanoTime() - nano) / 1_000_000);
> > >
> > > Result:
> > > Old version - ~12 seconds.
> > > New version - ~10 milliseconds.
> > >
>
