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    <h3>Update</h3> <blockquote> <p>Let's say, for example, in the case where we'd have an array that we want to treat as a multiset.</p> </blockquote> <p>So you have to process all the entries as they come, you can't use <code>count</code>, and can't assume the entries come in a known order.</p> <p>The general function I'd consider is</p> <pre><code>int hashCode() { int x = INITIAL_VALUE; for (Object o : this) { x = f(x, o==null ? NULL_HASH : g(o.hashCode())); } return h(x); } </code></pre> <p>Some observations:</p> <ul> <li>As already stated in the other answers, the INITIAL_VALUE doesn't matter much.</li> <li>I wouldn't go for <code>NULL_HASH=0</code> since this would ignore null values.</li> <li>The function <code>g</code> can be used in case you expect the hashes of the members to be in a small range (which can happen in case they are e.g., single characters).</li> <li>The function <code>h</code> can be used to improve the result, which is not very important since this already happens e.g. in <code>HashMap.hash(int)</code>.</li> <li>The function <code>f</code> is the most important one, unfortunately, it's quite limited as it obviously must be both associative and commutative.</li> <li>The function <code>f</code> should be bijective in both arguments, otherwise you'd generate unnecessary collisions.</li> </ul> <p>In no case I'd recommend <code>f(x, y) = x^y</code> since it'd make two occurrences of an element to cancel out. Using addition is better. Something like</p> <pre><code>f(x, y) = x + (2*A*x + 1) * y </code></pre> <p>where <code>A</code> is a constant satisfies all the above conditions. It may be worth it. For <code>A=0</code> it degenerates to addition, using an even <code>A</code> is not good as it shift bits of <code>x*y</code> out. Using <code>A=1</code> is fine, and the expression <code>2*x+1</code> can be computed using a single instruction on the <code>x86</code> architecture. Using a larger odd <code>A</code> might work better in case the hashes of the members are badly distributed.</p> <p>In case you go for a non-trivial <code>hashCode()</code> you should test if it works correctly. You should measure the performance of your program, maybe you'll find simple addition sufficient. Otherwise, I'd for for <code>NULL_HASH=1</code>, <code>g=h=identity</code>, and <code>A=1</code>.</p> <h3>My old answer</h3> <p>It may be for efficiency reasons. Calling <code>count</code> may be expensive for some implementations, but <code>entrySet</code> may be used instead. Still it might be more costly, I can't tell.</p> <p>I did a simple collision benchmark for Guava's hashCode and Rinke's and my own proposals:</p> <pre><code>enum HashCodeMethod { GUAVA { @Override public int hashCode(Multiset&lt;?&gt; multiset) { return multiset.hashCode(); } }, RINKE { @Override public int hashCode(Multiset&lt;?&gt; multiset) { int result = 0; for (final Object o : multiset.elementSet()) { result += (o==null ? 0 : o.hashCode()) * multiset.count(o); } return result; } }, MAAARTIN { @Override public int hashCode(Multiset&lt;?&gt; multiset) { int result = 0; for (final Multiset.Entry&lt;?&gt; e : multiset.entrySet()) { result += (e.getElement()==null ? 0 : e.getElement().hashCode()) * (2*e.getCount()+123); } return result; } } ; public abstract int hashCode(Multiset&lt;?&gt; multiset); } </code></pre> <p>The collision counting code went as follows:</p> <pre><code>private void countCollisions() throws Exception { final String letters1 = "abcdefgh"; final String letters2 = "ABCDEFGH"; final int total = letters1.length() * letters2.length(); for (final HashCodeMethod hcm : HashCodeMethod.values()) { final Multiset&lt;Integer&gt; histogram = HashMultiset.create(); for (final String s1 : Splitter.fixedLength(1).split(letters1)) { for (final String s2 : Splitter.fixedLength(1).split(letters2)) { histogram.add(hcm.hashCode(ImmutableMultiset.of(s1, s2, s2))); } } System.out.println("Collisions " + hcm + ": " + (total-histogram.elementSet().size())); } } </code></pre> <p>and printed</p> <pre><code>Collisions GUAVA: 45 Collisions RINKE: 42 Collisions MAAARTIN: 0 </code></pre> <p>So in this simple example Guava's hashCode performed really bad (45 collisions out of 63 possible). However, I don't claim my example is of much relevance for real life.</p>
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    1. POEfficient hash code for multiset in Java
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    1. COInteresting answer. How did you come up with your hash function? It seems to works great, but what's the math behind it? What would you suggest if getting a count is expensive? (Let's say, for example, in the case where we'd have an array that we want to treat as a multiset.)
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    2. COThe math is just a simple observation: XOR and ADD are quite similar, multiplication makes it better. However, computing `x*y` isn't good for mixing `x` and `y`. Something like `(2*x+1)*(2*xy+1)` works well in general, using bigger constants (in place of the 1) may prevent some collisions nearly for free.
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    3. COIn my above comment it should read `(2*x+1)*(2*y+1)`, but it's not really good, as the result is always odd. However, the multiplication by an odd number is the key here. Using `x+(2*x+1)*y` which is the same as `2*x*y+x+y+1` is fine, multiplying a term by an odd number makes it better in case of small `x` and `y`.
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