Roland Roland - 2 months ago 16
Java Question

How to compute the hash code for a stream in the same way as List.hashCode()

I just realized that implementing the following algorithm to compute the hash code for a stream is not possible using

reduce
. The problem is that the initial seed for the hash code is
1
which is not an identity for the accumulator.

The algorithm for List.hashCode()
:

int hashCode = 1;
for (E e : list)
hashCode = 31*hashCode + (e==null ? 0 : e.hashCode());


You might be tempted to think that the following is correct but it isn't, although it will work if the stream processing is not split up.

List<Object> list = Arrays.asList(1,null, new Object(),4,5,6);
int hashCode = list.stream().map(Objects::hashCode).reduce(1, (a, b) -> 31 * a + b);


It seems that the only sensible way of doing it would be to get the
Iterator
of the
Stream
and do normal sequential processing or collect it to a
List
first.

Answer

While, at the first glance, the hash code algorithm seems to be non-parallelizable due to its non-associativity, it is possible, if we transform the function:

((a * 31 + b) * 31 + c ) * 31 + d

to

a * 31 * 31 * 31 + b * 31 * 31 + c * 31 + d

which basically is

a * 31³ + b * 31² + c * 31¹ + d * 31⁰

or for an arbitrary List of size n:

1 * 31ⁿ + e₀ * 31ⁿ⁻¹ + e₁ * 31ⁿ⁻² + e₂ * 31ⁿ⁻³ +  …  + eₙ₋₃ * 31² + eₙ₋₂ * 31¹ + eₙ₋₁ * 31⁰

with the first 1 being the initial value of the original algorithm and eₓ being the hash code of the list element at index x. While the summands are evaluation order independent now, there’s obviously a dependency to the element’s position, which we can solve by streaming over the indices in the first place, which works for random access lists and arrays, or solve generally, with a collector which tracks the number of encountered objects. The collector can resort to the repeated multiplications for the accumulation and has to resort to the power function only for combining results:

static <T> Collector<T,?,Integer> hashing() {
    return Collector.of(() -> new int[2],
        (a,o)    -> { a[0]=a[0]*31+Objects.hashCode(o); a[1]++; },
        (a1, a2) -> { a1[0]=a1[0]*iPow(31,a2[1])+a2[0]; a1[1]+=a2[1]; return a1; },
        a -> iPow(31,a[1])+a[0]);
}
// derived from http://stackoverflow.com/questions/101439
private static int iPow(int base, int exp) {
    int result = 1;
    for(; exp>0; exp >>= 1, base *= base)
        if((exp & 1)!=0) result *= base;
    return result;
}

 

List<Object> list = Arrays.asList(1,null, new Object(),4,5,6);
int expected = list.hashCode();

int hashCode = list.stream().collect(hashing());
if(hashCode != expected)
    throw new AssertionError();

// works in parallel
hashCode = list.parallelStream().collect(hashing());
if(hashCode != expected)
    throw new AssertionError();

// a method avoiding auto-boxing is more complicated:
int[] result=list.parallelStream().mapToInt(Objects::hashCode)
    .collect(() -> new int[2],
    (a,o)    -> { a[0]=a[0]*31+Objects.hashCode(o); a[1]++; },
    (a1, a2) -> { a1[0]=a1[0]*iPow(31,a2[1])+a2[0]; a1[1]+=a2[1]; });
hashCode = iPow(31,result[1])+result[0];

if(hashCode != expected)
    throw new AssertionError();

// random access lists allow a better solution:
hashCode = IntStream.range(0, list.size()).parallel()
    .map(ix -> Objects.hashCode(list.get(ix))*iPow(31, list.size()-ix-1))
    .sum() + iPow(31, list.size());

if(hashCode != expected)
    throw new AssertionError();
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