Why does the CompletableFuture allOf method do a binary search? - java

I wanted to know if the allOf method of CompletableFuture does polling or goes into a wait state till all the CompletableFutures passed into the method complete their execution.
I looked at the code of the allOf method in IntelliJ and it is doing some sort of binary search.
Please help me to find out what the allOf method of CompletableFuture actually does.
public static CompletableFuture<Void> allOf(CompletableFuture<?>... cfs) {
return andTree(cfs, 0, cfs.length - 1);
}
/** Recursively constructs a tree of completions. */
static CompletableFuture<Void> andTree(CompletableFuture<?>[] cfs, int lo, int hi) {
CompletableFuture<Void> d = new CompletableFuture<Void>();
if (lo > hi) // empty
d.result = NIL;
else {
CompletableFuture<?> a, b;
int mid = (lo + hi) >>> 1;
if ((a = (lo == mid ? cfs[lo] :
andTree(cfs, lo, mid))) == null ||
(b = (lo == hi ? a : (hi == mid+1) ? cfs[hi] :
andTree(cfs, mid+1, hi))) == null)
throw new NullPointerException();
if (!d.biRelay(a, b)) {
BiRelay<?,?> c = new BiRelay<>(d, a, b);
a.bipush(b, c);
c.tryFire(SYNC);
}
}
return d;
}
/** Pushes completion to this and b unless both done. */
final void bipush(CompletableFuture<?> b, BiCompletion<?,?,?> c) {
if (c != null) {
Object r;
while ((r = result) == null && !tryPushStack(c))
lazySetNext(c, null); // clear on failure
if (b != null && b != this && b.result == null) {
Completion q = (r != null) ? c : new CoCompletion(c);
while (b.result == null && !b.tryPushStack(q))
lazySetNext(q, null); // clear on failure
}
}
}
final CompletableFuture<V> tryFire(int mode) {
CompletableFuture<V> d;
CompletableFuture<T> a;
CompletableFuture<U> b;
if ((d = dep) == null ||
!d.orApply(a = src, b = snd, fn, mode > 0 ? null : this))
return null;
dep = null; src = null; snd = null; fn = null;
return d.postFire(a, b, mode);
}

It doesn't do a binary search -- it's building a balanced binary tree with the input futures at the leaves, and inner nodes that each complete when its two children have both completed.
For some reason that is not apparent from the code, the author of the code must have decided it was most efficient to consider allOf(_,_) between exactly two futures to be his primitive operation, and if he's asked for an allOf(...) between more than two futures, he's manufacturing it as a cascade of these binary primitives.
The tree should be balanced for such that no matter what the last future to complete is, there will only be a small number of levels left to collapse before the future at the top can complete. This improves performance in some situations, because it ensures that as much work as possible can be handled before we're completely done, at a point where (if we're lucky) the CPU might just be sitting idle, waiting for something asynchronous to complete.
Balancing the tree is done by having the topmost inner node have about as many leaves under its left child as under its right child -- so both children get about half of the original array, and then the code recursively builds a tree from each half of the array. Splitting in halves can look a bit like the index calculations for a binary search.
The basic structure is obscured slightly by special cases that appear to be designed to
use an optimized code path with fewer allocations when some of the original futures are already completed, and
make sure that the result of allOf(_) with exactly one element will return a fresh CompleteableFuture. For most purposes it would work to just return that single element, but the author must have wanted to ensure that users of the library can rely on the object being fresh, if they are using them as keys in hash maps, or other logic that depends on being able to tell the output from the inputs, and
have only one throw new NullPointerException(); by using ?: and inline assignments instead of honest if statements. This probably produces slightly smaller bytecode at the expense of readability. Cannot be recommended as a style to learn from, unless you personally pay for the storage cost of the resulting bytecode ...

Related

How does my comparator method violate its general contract?

Third-Party Packages
import java.util.Collections;
import java.util.Comparator;
import org.joda.time.DateTime;
My Comparator
public static Comparator<Task> TASK_PRIORITY = new Comparator<Task>() {
public int compare(Task task1, Task task2) {
if (task1 == null && task2 == null) return 0;
if (task1 == null) return +1; //null last
if (task2 == null) return -1; //null last
// Only consider retries after a task is retried 5+ times
if (task1.getRetries() >= 5 || task2.getRetries() >= 5) {
// Primary sort: retry count (ascending)
int retriesCompare = Integer.compare(task1.getRetries(), task2.getRetries());
if (retriesCompare != 0) return retriesCompare;
}
// Secondary sort: creation time (ascending, null first)
int creationCompare = compareTimeNullFirst(task1.getCreationTime(), task2.getCreationTime());
if (creationCompare != 0) return creationCompare;
// Tertiary sort: load time (ascending, null last)
int loadCompare = compareTimeNullLast(task1.getLoadTime(), task2.getLoadTime());
if (loadCompare != 0) return loadCompare;
return 0;
}
};
private static int compareTimeNullLast(DateTime time1, DateTime time2) {
if (time1 == null && time2 == null) return 0;
if (time1 == null) return +1;
if (time2 == null) return -1;
if (time1.isBefore(time2) return -1;
if (time1.isAfter(time2)) return +1;
return 0;
}
private static int compareTimeNullFirst(DateTime time1, DateTime time2) {
if (time1 == null && time2 == null) return 0;
if (time1 == null) return -1;
if (time2 == null) return +1;
if (time1.isBefore(time2) return -1;
if (time1.isAfter(time2)) return +1;
return 0;
}
Using My Comparator
//tasks is a List<Task>
Collections.sort(tasks, TASK_PRIORITY);
My Problem
I sometimes get an IllegalArgumentException for Comparison method violates its general contract!. I can consistently get this Exception thrown with live data running long enough, but I'm not sure how to fix the actual cause of the problem.
My Question
What is wrong with my comparator? (Specifically, which part of the contract am I violating?) How do I fix it without covering up the Exception?
Notes
I'm using Java 7 and cannot upgrade without a major rewrite.
I could possibly cover-up the Exception by setting java.util.Arrays.useLegacyMergeSort to true, but that is not a desirable solution.
I tried to create tests to randomly generate data and verify each of the contract conditions. I wasn't able to get the exception to be thrown.
I tried removing the condition around the retry comparison, but I still eventually got the Exception.
This line throws the exception: Collections.sort(tasks, TASK_PRIORITY);
Let us start at the beginning. My reading of your code is that the logic of your Comparator is sound. (I would have avoided having null Task and DateTime values, but that is not relevant to your problem.)
The other thing that can cause this exception is if the compare method is giving inconsistent results because the Task objects are changing. Indeed, it looks like it is semantically meaningful for (at least) the retry counts to change. If there is another thread that is changing Task the fields that can effect the ordering ... while the current thread is sorting ... that could the IllegalArgumentException.
(Part of the comparison contract is that pairwise ordering does not change while you are sorting the collection.)
You then say this:
I use ImmutableSet.copyOf to copy the list before sorting, and I do that under a read lock in java.util.concurrent.locks.ReadWriteLock.
Copying a collection doesn't make copies of the elements of the collection. It is a shallow copy. So you will end up with two collections that contain the same objects. If the other thread mutates any of the objects (e.g. by increasing retry counts), that could change the ordering of objects.
The locking makes sure that you have consistent copy, but that's not what the problem is.
What is the solution? I can think of a couple:
You could lock something to block all updates to the collections AND the element objects while you copy and sort.
You could deep-copy the collect; i.e. create a new collection containing copies of the elements of the original collection.
You could create light-weight objects that contain a snapshots of the fields of the Task objects that are relevant to sorting; e.g.
public class Key implements Comparable<Key> {
private int retries;
private DateTime creation;
private DateTime load;
private Task task;
public Key(Task task) {
this.task = task;
this.retries = task.getRetryCount();
...
}
public int compareTo(Key other) {
// compare using retries, creation, load
}
}
This has the potential advantages that you are copying less information, and you can go from the sorted collection of Key objects to the original Task objects.
Note all of these alternatives are slower than what you are currently doing. I don't think there is a way to avoid this.

Atomic compareAndSet but with callback?

I know that AtomicReference has compareAndSet, but I feel like what I want to do is this
private final AtomicReference<Boolean> initialized = new AtomicReference<>( false );
...
atomicRef.compareSetAndDo( false, true, () -> {
// stuff that only happens if false
});
this would probably work too, might be better.
atomicRef.compareAndSet( false, () -> {
// stuff that only happens if false
// if I die still false.
return true;
});
I've noticed there's some new functional constructs but I'm not sure if any of them are what I'm looking for.
Can any of the new constructs do this? if so please provide an example.
update
To attempt to simplify my problem, I'm trying to find a less error prone way to guard code in a "do once for object" or (really) lazy initializer fashion, and I know that some developers on my team find compareAndSet confusing.
guard code in a "do once for object"
how exactly to implement that depends on what you want other threads attempting to execute the same thing in the meantime. if you just let them run past the CAS they may observe things in an intermediate state while the one thread that succeeded does its action.
or (really) lazy initializer fashion
that construct is not thread-safe if you're using it for lazy initializers because the "is initialized" boolean may be set to true by one thread and then execute the block while another thread observes the true-state but reads an empty result.
You can use Atomicreference::updateAndGet if multiple concurrent/repeated initialization attempts are acceptable with one object winning in the end and the others being discarded by GC. The update method should be side-effect-free.
Otherwise you should just use the double checked locking pattern with a variable reference field.
Of course you can always package any of these into a higher order function that returns a Runnable or Supplier which you then assign to a final field.
// == FunctionalUtils.java
/** #param mayRunMultipleTimes must be side-effect-free */
public static <T> Supplier<T> instantiateOne(Supplier<T> mayRunMultipleTimes) {
AtomicReference<T> ref = new AtomicReference<>(null);
return () -> {
T val = ref.get(); // fast-path if already initialized
if(val != null)
return val;
return ref.updateAndGet(v -> v == null ? mayRunMultipleTimes.get() : v)
};
}
// == ClassWithLazyField.java
private final Supplier<Foo> lazyInstanceVal = FunctionalUtils.instantiateOne(() -> new Foo());
public Foo getFoo() {
lazyInstanceVal.get();
}
You can easily encapsulate various custom control-flow and locking patterns this way. Here are two of my own..
compareAndSet returns true if the update was done, and false if the actual value was not equal to the expected value.
So just use
if (ref.compareAndSet(expectedValue, newValue)) {
...
}
That said, I don't really understand your examples, since you're passing true and false to a method taking object references as argument. And your second example doesn't do the same thing as the first one. If the second is what you want, I think what you're after is
ref.getAndUpdate(value -> {
if (value.equals(expectedValue)) {
return someNewValue(value);
}
else {
return value;
}
});
You’re over-complicating things. Just because there are now lambda expression, you don’t need to solve everything with lambdas:
private volatile boolean initialized;
…
if(!initialized) synchronized(this) {
if(!initialized) {
// stuff to be done exactly once
initialized=true;
}
}
The double checked locking might not have a good reputation, but for non-static properties, there are little alternatives.
If you consider multiple threads accessing it concurrently in the uninitialized state and want a guaranty that the action runs only once, and that it has completed, before dependent code is executed, an Atomic… object won’t help you.
There’s only one thread that can successfully perform compareAndSet(false,true), but since failure implies that the flag already has the new value, i.e. is initialized, all other threads will proceed as if the “stuff to be done exactly once” has been done while it might still be running. The alternative would be reading the flag first and conditionally perform the stuff and compareAndSet afterwards, but that allows multiple concurrent executions of “stuff”. This is also what happens with updateAndGet or accumulateAndGet and it’s provided function.
To guaranty exactly one execution before proceeding, threads must get blocked, if the “stuff” is currently executed. The code above does this. Note that once the “stuff” has been done, there will be no locking anymore and the performance characteristics of the volatile read are the same as for the Atomic… read.
The only solution which is simpler in programming, is to use a ConcurrentMap:
private final ConcurrentHashMap<String,Boolean> initialized=new ConcurrentHashMap<>();
…
initialized.computeIfAbsent("dummy", ignore -> {
// stuff to do exactly once
return true;
});
It might look a bit oversized, but it provides exactly the required performance characteristics. It will guard the initial computation using synchronized (or well, an implementation dependent exclusion mechanism) but perform a single read with volatile semantics on subsequent queries.
If you want a more lightweight solution, you may stay with the double checked locking shown at the beginning of this answer…
I know this is old, but I've found there is no perfect way to achieve this, more specifically this:
trying to find a less error prone way to guard code in a "do (anything) once..."
I'll add to this "while respecting a happens before behavior." which is required for instantiating singletons in your case.
IMO The best way to achieve this is by means of a synchronized function:
public<T> T transaction(Function<NonSyncObject, T> transaction) {
synchronized (lock) {
return transaction.apply(nonSyncObject);
}
}
This allows to preform atomic "transactions" on the given object.
Other options are double-check spin-locks:
for (;;) {
T t = atomicT.get();
T newT = new T();
if (atomicT.compareAndSet(t, newT)) return;
}
On this one new T(); will get executed repeatedly until the value is set successfully, so it is not really a "do something once".
This would only work on copy on write transactions, and could help on "instantiating objects once" (which in reality is instantiating many but at the end is referencing the same) by tweaking the code.
The final option is a worst performant version of the first one, but this one is a true happens before AND ONCE (as opposed to the double-check spin-lock):
public void doSomething(Runnable r) {
while (!atomicBoolean.compareAndSet(false, true)) {}
// Do some heavy stuff ONCE
r.run();
atomicBoolean.set(false);
}
The reason why the first one is the better option is that it is doing what this one does, but in a more optimized way.
As a side note, in my projects I've actually used the code below (similar to #the8472's answer), that at the time I thought safe, and it may be:
public T get() {
T res = ref.get();
if (res == null) {
res = builder.get();
if (ref.compareAndSet(null, res))
return res;
else
return ref.get();
} else {
return res;
}
}
The thing about this code is that, as the copy on write loop, this one generates multiple instances, one for each contending thread, but only one is cached, the first one, all the other constructions eventually get GC'd.
Looking at the putIfAbsent method I see the benefit is the skipping of 17 lines of code and then a synchronized body:
/** Implementation for put and putIfAbsent */
final V putVal(K key, V value, boolean onlyIfAbsent) {
if (key == null || value == null) throw new NullPointerException();
int hash = spread(key.hashCode());
int binCount = 0;
for (Node<K,V>[] tab = table;;) {
Node<K,V> f; int n, i, fh;
if (tab == null || (n = tab.length) == 0)
tab = initTable();
else if ((f = tabAt(tab, i = (n - 1) & hash)) == null) {
if (casTabAt(tab, i, null,
new Node<K,V>(hash, key, value, null)))
break; // no lock when adding to empty bin
}
else if ((fh = f.hash) == MOVED)
tab = helpTransfer(tab, f);
else {
V oldVal = null;
synchronized (f) {
if (tabAt(tab, i) == f) {
And then the synchronized body itself is another 34 lines:
synchronized (f) {
if (tabAt(tab, i) == f) {
if (fh >= 0) {
binCount = 1;
for (Node<K,V> e = f;; ++binCount) {
K ek;
if (e.hash == hash &&
((ek = e.key) == key ||
(ek != null && key.equals(ek)))) {
oldVal = e.val;
if (!onlyIfAbsent)
e.val = value;
break;
}
Node<K,V> pred = e;
if ((e = e.next) == null) {
pred.next = new Node<K,V>(hash, key,
value, null);
break;
}
}
}
else if (f instanceof TreeBin) {
Node<K,V> p;
binCount = 2;
if ((p = ((TreeBin<K,V>)f).putTreeVal(hash, key,
value)) != null) {
oldVal = p.val;
if (!onlyIfAbsent)
p.val = value;
}
}
}
}
The pro(s) of using a ConcurrentHashMap is that it will undoubtedly work.

How do I lazily concatenate streams?

I'm trying to implement a stream that uses another instance of itself in its implementation. The stream has a few constant elements prepended (with IntStream.concat) to it, so this should work as long as the concatenated stream creates the non-constant part lazily. I think using the StreamSupport.intStream overload taking a Supplier with IntStream.concat (which "creates a lazily concatenated stream") should be lazy enough to only create the second spliterator when elements are demanded from it, but even creating the stream (not evaluating it) overflows the stack. How can I lazily concatenate streams?
I'm attempting to port the streaming prime number sieve from this answer into Java. This sieve uses another instance of itself (ps = postponed_sieve() in the Python code). If I break the initial four constant elements (yield 2; yield 3; yield 5; yield 7;) into their own stream, it's easy to implement the generator as a spliterator:
/**
* based on https://stackoverflow.com/a/10733621/3614835
*/
static class PrimeSpliterator extends Spliterators.AbstractIntSpliterator {
private static final int CHARACTERISTICS = Spliterator.DISTINCT | Spliterator.IMMUTABLE | Spliterator.NONNULL | Spliterator.ORDERED | Spliterator.SORTED;
private final Map<Integer, Supplier<IntStream>> sieve = new HashMap<>();
private final PrimitiveIterator.OfInt postponedSieve = primes().iterator();
private int p, q, c = 9;
private Supplier<IntStream> s;
PrimeSpliterator() {
super(105097564 /* according to Wolfram Alpha */ - 4 /* in prefix */,
CHARACTERISTICS);
//p = next(ps) and next(ps) (that's Pythonic?)
postponedSieve.nextInt();
this.p = postponedSieve.nextInt();
this.q = p*p;
}
#Override
public boolean tryAdvance(IntConsumer action) {
for (; c > 0 /* overflow */; c += 2) {
Supplier<IntStream> maybeS = sieve.remove(c);
if (maybeS != null)
s = maybeS;
else if (c < q) {
action.accept(c);
return true; //continue
} else {
s = () -> IntStream.iterate(q+2*p, x -> x + 2*p);
p = postponedSieve.nextInt();
q = p*p;
}
int m = s.get().filter(x -> !sieve.containsKey(x)).findFirst().getAsInt();
sieve.put(m, s);
}
return false;
}
}
My first attempt at the primes() method returns an IntStream concatenating a constant stream with a new PrimeSpliterator:
public static IntStream primes() {
return IntStream.concat(IntStream.of(2, 3, 5, 7),
StreamSupport.intStream(new PrimeSpliterator()));
}
Calling primes() results in a StackOverflowError because primes() always instantiates a PrimeSpliterator, but PrimeSpliterator's field initializer always calls primes(). However, there's an overload of StreamSupport.intStream that takes a Supplier, which should allow lazily creating the PrimeSpliterator:
public static IntStream primes() {
return IntStream.concat(IntStream.of(2, 3, 5, 7),
StreamSupport.intStream(PrimeSpliterator::new, PrimeSpliterator.CHARACTERISTICS, false));
}
However, I instead get a StackOverflowError with a different backtrace (trimmed, as it repeats). Note that the recursion is entirely in the call to primes() -- the terminal operation iterator() is never invoked on a returned stream.
Exception in thread "main" java.lang.StackOverflowError
at java.util.stream.StreamSpliterators$DelegatingSpliterator$OfInt.<init>(StreamSpliterators.java:582)
at java.util.stream.IntPipeline.lazySpliterator(IntPipeline.java:155)
at java.util.stream.IntPipeline$Head.lazySpliterator(IntPipeline.java:514)
at java.util.stream.AbstractPipeline.spliterator(AbstractPipeline.java:352)
at java.util.stream.IntPipeline.spliterator(IntPipeline.java:181)
at java.util.stream.IntStream.concat(IntStream.java:851)
at com.jeffreybosboom.projecteuler.util.Primes.primes(Primes.java:22)
at com.jeffreybosboom.projecteuler.util.Primes$PrimeSpliterator.<init>(Primes.java:32)
at com.jeffreybosboom.projecteuler.util.Primes$$Lambda$1/834600351.get(Unknown Source)
at java.util.stream.StreamSpliterators$DelegatingSpliterator.get(StreamSpliterators.java:513)
at java.util.stream.StreamSpliterators$DelegatingSpliterator.estimateSize(StreamSpliterators.java:536)
at java.util.stream.Streams$ConcatSpliterator.<init>(Streams.java:713)
at java.util.stream.Streams$ConcatSpliterator$OfPrimitive.<init>(Streams.java:789)
at java.util.stream.Streams$ConcatSpliterator$OfPrimitive.<init>(Streams.java:785)
at java.util.stream.Streams$ConcatSpliterator$OfInt.<init>(Streams.java:819)
at java.util.stream.IntStream.concat(IntStream.java:851)
at com.jeffreybosboom.projecteuler.util.Primes.primes(Primes.java:22)
at com.jeffreybosboom.projecteuler.util.Primes$PrimeSpliterator.<init>(Primes.java:32)
at com.jeffreybosboom.projecteuler.util.Primes$$Lambda$1/834600351.get(Unknown Source)
at java.util.stream.StreamSpliterators$DelegatingSpliterator.get(StreamSpliterators.java:513)
at java.util.stream.StreamSpliterators$DelegatingSpliterator.estimateSize(StreamSpliterators.java:536)
at java.util.stream.Streams$ConcatSpliterator.<init>(Streams.java:713)
at java.util.stream.Streams$ConcatSpliterator$OfPrimitive.<init>(Streams.java:789)
at java.util.stream.Streams$ConcatSpliterator$OfPrimitive.<init>(Streams.java:785)
at java.util.stream.Streams$ConcatSpliterator$OfInt.<init>(Streams.java:819)
at java.util.stream.IntStream.concat(IntStream.java:851)
at com.jeffreybosboom.projecteuler.util.Primes.primes(Primes.java:22)
How can I concatenate streams lazily enough to allow a stream to use another copy of itself in its implementation?
Your apparently assume that the Streams API extends its guarantees of laziness even to the instantiation of spliterators; this is not correct. It expects to be able to instantiate the stream's spliterator at any time before the actual consumption begins, for example just to find out the stream's characteristics and reported size. Consumption only begins by invoking trySplit, tryAdvance, or forEachRemaining.
Having that in mind, you are initializing the postponed sieve earlier than you need it. You don't get to use any of its results until the else if part in tryAdvance. So move the code to the last possible moment which gives correctness:
#Override
public boolean tryAdvance(IntConsumer action) {
for (; c > 0 /* overflow */; c += 2) {
Supplier<IntStream> maybeS = sieve.remove(c);
if (maybeS != null)
s = maybeS;
else {
if (postponedSieve == null) {
postponedSieve = primes().iterator();
postponedSieve.nextInt();
this.p = postponedSieve.nextInt();
this.q = p*p;
}
if (c < q) {
action.accept(c);
return true; //continue
I think that, with this change, even your first attempt at primes() should work.
If you want to stay with your current approach, you could involve the following idiom:
Stream.<Supplier<IntStream>>of(
()->IntStream.of(2, 3, 5, 7),
()->intStream(new PrimeSpliterator()))
.flatMap(Supplier::get);
You may find that this gives you as much laziness as you need.
I like to use Supplier to do that:
return Stream.<Supplier<Stream<WhatEver>>of(
() -> generateStreamOfWhatEverAndChangeSomeState(input, state),
() -> generateStreamOfMoreWhatEversDependendingOnMutatedState(state)
).flatMap(Supplier::get);
Since stream is lazily evaluated the generateStreamOfWhatEverAndChangeSomeState() will finish before generateStreamOfMoreWhatEversDependendingOnMutatedState() start and the state would be updated.
I should note that this is probably not what the designers of Stream had in mind. Idealy a Stream should not change state, only read each item and produce a new item.

first common ancestor of two nodes in a binary tree

I was trying to solve the problem 4.7 from the book cracking the code interview (very cool book!).
Design an algorithm and write code to find the first common ancestor
of two nodes in a binary tree. Avoid storing additional nodes in a
data structure. NOTE: This is not necessarily a binary search tree.
And I came up with this solution which is not even close to the ones provided in the book. I wonder if someone can find any flaws on it?
Solution:
I created a wraper class to hold the first common ancestor (if its found) and 2 booleans to track if a or b was found when recoursively searching the tree. Please read added comments in the code below.
public static void main (String args[]){
NodeTree a, b, head, result; //initialise and fill with data
fillTreeTestData(head);
pickRandomNode(a);
pickRandomNode(b);
result = commonAnsestor(a,b,head);
if(result != null)
System.out.println("First common ansestor "+result);
else
System.out.println("Not found");
}
class TreeNode{
Object value;
TreeNode right, left;
}
class WraperNodeTree{
boolean found_a;
boolean found_b;
NodeTree n;
WraperNodeTree (boolean a, boolean b, NodeTree n){
this.n = n;
this.a = a;
this.b = b;
}
}
static WraperNodeTree commonAnsestor(NodeTree a, NodeTree b, NodeTree current){
// Let's prepare a wraper object
WraperNodeTree wraper = new WraperNodeTree(false, false, null);
// we reached the end
if(current == null) return wraper;
// let's check if current node is either a or b
if(a != null)
wraper.found_a = current.value.equals(a.value);
else if(b != null)
wraper.found_b = current.value.equals(b.value);
else
return wraper; // if both are null we don't need to keep searching recoursively
// if either a or b was found let's stop searching for it for performance
NodeTree to_search_a = wraper.found_a ? null : a;
NodeTree to_search_b = wraper.found_b ? null : b;
// let's search the left
WraperNodeTree wraperLeft = common(to_search_a,to_search_b,current.left);
// if we already have a common ancester just pass it back recoursively
if(wraperLeft.n != null) return wraperLeft;
WraperNodeTree wraperRight = common(to_search_a,to_search_b,current.right);
if(wraperRight.n != null)return wraperRight;
// keep the wraper up to date with what we found so far
wraper.a = wraper.found_a || wraperLeft.found_a || wraperRight.found_a;
wraper.b = wraper.found_b || wraperLeft.found_b || wraperRight.found_b;
// if both a and b were found, let's pass the current node as solution
if(wraper.found_a && wraper.found_b)
wraper.n = current;
return wraper;
}
If it's about finding flaws:
Flaw #1:
I think there are too many typos in your code, which may confuse the interviewer on their first read of the code (and you don't want that!). For example, you write 'NodeTree' and 'TreeNode' interchangeably. Also, you define 'commonAncestor()' and then call 'common()'. Those things make the interviewer confused and make him drift apart from the important thing, which is understanding your way of solving the problem.
Flaw #2: Typos aside, I think another flaw is that this code is difficult to follow. I think one of the reasons is because you have 'return' statements all over the body of your function (at the beginning, at the middle and at the end). This should 'normally' be avoided in favor of readability.
Usually my approach is to organize the code in the following way:
Basic border case checks (which might include return)
Main body of the function (which should NOT return under any conditions)
Last checks and final RETURN
But when you have return statements in the middle, it makes it harder for the reader to imagine the flow.
Flaw #3: I think you're trying to solve two problems with the same function (commonAncestor). You are trying to both search for 'a' and 'b' and also keeping track of the common ancestor. I think if this is an interview question, you could separate those two objectives in favor of simplicity.
For example, consider this code (might not be perfect and need some extra border checks):
/**
* [Assumption]: If we call firstCommonAncestor(a, b, root)
* we TRUST that root contains both a and b.
*
* You can (and should) discuss this
* assumption with your interviewer.
*/
public static Node firstCommonAncestor(Node a, Node b, Node root) {
// If root matches any of the nodes (a or b),
// then root is the first common ancestor
// (because of our assumption)
if(root == a || root == b) return root;
// Search for a and b in both sides
SearchResult leftResult = searchNodes(a, b, root.left);
SearchResult rightResult = searchNodes(a, b, root.right);
// If a and b are on the same side (left or right), then we
// call firstCommonAncestor on that side and that’s it
if(leftResult.aFound && leftResult.bFound)
return firstCommonAncestor(a, b, root.left);
else if(rightResult.aFound && rightResult.bFound)
return firstCommonAncestor(a, b, root.right);
else {
// If a and b are in different sides,
// then we just found the first common ancestor
return root;
}
}
class SearchResult {
boolean aFound, bFound;
}
On the code above, I'm separating the task of actually searching for 'a' and 'b' in a different function called searchNodes, which is fairly easy to implement if your interviewer asks for it. But he might not even do that. And if he does, at that point he already understood your approach, so it's easier now to "make the code a bit more complicated" without confusing the interviewer.
I hope this helps.

comparison of adjacent children in tree does not take place?

What is wrong with this method ? it seems but I am not sure that the comparison of adjacent children in the tree does not take place.
I roughly traced the workings of this algorithm by hand and I think the idea is correct maybe something wrong with the implementation or I have no Idea how recursion works, the second helper (compare) method seems to be the issue
public static int MAX(BST B) {
int m = ((Integer) B.root.data).intValue();
return call(B.root, m);
}
public static int call(node current, int max) {
//first helper method gets the max from two different levels in the tree
if(current == null)
return -1;
if(current.left == null && current.right == null)
return max;
else {
if(((Integer) current.data).intValue()>max)
max = ((Integer) current.data).intValue();
return compare(call(current.left,max),call(current.right,max));
}
}
//second helper method gets the max
static int compare(int m1, int m2) {
if(m1>m2)
return m1;
else
return m2;
}
Since you are searching the entire tree, I'm going to assume that the structure is not properly ordered.
The bug is in your call function with:
if(current.left==null&&current.right==null) return max;
Imagine you have a tree with a root with two leaf nodes (three nodes total). The root has value 3, right has value 2, and left has value 5. The algorithm should return 5, but your code will return 3. This is because you ignore the value of any leaf (a node with no "children") with that line of code. So your code ignores the value 5, in this example, and returns max, which is 3.
You can fix this by returning compare(current.value, max) when left and right are null.
I think (not 100%) that you may have an issue because you only check if BOTH children are null if for example right is null and left is not you will attempt to call the method call on both right and. Perhaps add a case checking if one child is null and if so return call of the non null child.
... I have no Idea how recursion works ...
Recursion means the method you are in gets called from inside itself with some other arguments , and there is some check that exits by returning a value or continues to call itself recursively.
call() is indirectly recursive, as it either exits with a return of -1 or max or it calls itself again with new arguments and continues to do this until it either exits or crashes with an OutOfMemory error as the stack fills up.
This method isn't recursive: It is poorly named though.
static int compare(int m1, int m2) {
if(m1>m2)
return m1;
else
return m2;
}
and could be written ( and renamed ) as
static int min(final int m1, final int m2)
{
return Math.min(m1,m2);
}
or just inlined into
return Math.min(call(current.left,max),call(current.right,max));
either way, you are getting the minimum of the two values, not really comparing them that implies different logic and a different return value.
Either way that method isn't recursive and if the logic of m1 > m2 is appropriate it can't be the problem, more like the input to that function is not what you expect.
Step debugging is a powerful tool and one all experienced developers use every day!

Categories

Resources