Add a flag for bigram presence in the header
This is a cherry-pick of Icb602762 onto jb-dev. Bug: 6355745 Change-Id: Icb602762bb0d81472f024fa491571062ec1fc4e9main
parent
329c8d7bcc
commit
20a6dea1ca
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@ -131,6 +131,7 @@ public class BinaryDictInputOutput {
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// These options need to be the same numeric values as the one in the native reading code.
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private static final int GERMAN_UMLAUT_PROCESSING_FLAG = 0x1;
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private static final int FRENCH_LIGATURE_PROCESSING_FLAG = 0x4;
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private static final int CONTAINS_BIGRAMS_FLAG = 0x8;
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// TODO: Make this value adaptative to content data, store it in the header, and
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// use it in the reading code.
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@ -752,9 +753,12 @@ public class BinaryDictInputOutput {
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/**
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* Makes the 2-byte value for options flags.
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*/
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private static final int makeOptionsValue(final DictionaryOptions options) {
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private static final int makeOptionsValue(final FusionDictionary dictionary) {
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final DictionaryOptions options = dictionary.mOptions;
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final boolean hasBigrams = dictionary.hasBigrams();
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return (options.mFrenchLigatureProcessing ? FRENCH_LIGATURE_PROCESSING_FLAG : 0)
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+ (options.mGermanUmlautProcessing ? GERMAN_UMLAUT_PROCESSING_FLAG : 0);
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+ (options.mGermanUmlautProcessing ? GERMAN_UMLAUT_PROCESSING_FLAG : 0)
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+ (hasBigrams ? CONTAINS_BIGRAMS_FLAG : 0);
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}
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/**
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@ -970,7 +974,7 @@ public class BinaryDictInputOutput {
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headerBuffer.write((byte) (0xFF & version));
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}
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// Options flags
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final int options = makeOptionsValue(dict.mOptions);
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final int options = makeOptionsValue(dict);
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headerBuffer.write((byte) (0xFF & (options >> 8)));
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headerBuffer.write((byte) (0xFF & options));
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if (version >= FIRST_VERSION_WITH_HEADER_SIZE) {
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@ -563,7 +563,7 @@ public class FusionDictionary implements Iterable<Word> {
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* Recursively count the number of nodes in a given branch of the trie.
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*
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* @param node the node to count.
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* @result the number of nodes in this branch.
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* @return the number of nodes in this branch.
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*/
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public static int countNodes(final Node node) {
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int size = 1;
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@ -575,6 +575,32 @@ public class FusionDictionary implements Iterable<Word> {
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return size;
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}
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// Recursively find out whether there are any bigrams.
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// This can be pretty expensive especially if there aren't any (we return as soon
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// as we find one, so it's much cheaper if there are bigrams)
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private static boolean hasBigramsInternal(final Node node) {
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if (null == node) return false;
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for (int i = node.mData.size() - 1; i >= 0; --i) {
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CharGroup group = node.mData.get(i);
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if (null != group.mBigrams) return true;
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if (hasBigramsInternal(group.mChildren)) return true;
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}
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return false;
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}
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/**
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* Finds out whether there are any bigrams in this dictionary.
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*
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* @return true if there is any bigram, false otherwise.
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*/
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// TODO: this is expensive especially for large dictionaries without any bigram.
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// The up side is, this is always accurate and correct and uses no memory. We should
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// find a more efficient way of doing this, without compromising too much on memory
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// and ease of use.
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public boolean hasBigrams() {
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return hasBigramsInternal(mRoot);
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}
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// Historically, the tails of the words were going to be merged to save space.
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// However, that would prevent the code to search for a specific address in log(n)
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// time so this was abandoned.
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