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/**
* Copyright (C) 2013 10gen Inc.
*
* This program is free software: you can redistribute it and/or modify
* it under the terms of the GNU Affero General Public License, version 3,
* as published by the Free Software Foundation.
*
* This program is distributed in the hope that it will be useful,
* but WITHOUT ANY WARRANTY; without even the implied warranty of
* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
* GNU Affero General Public License for more details.
*
* You should have received a copy of the GNU Affero General Public License
* along with this program. If not, see <http://www.gnu.org/licenses/>.
*
* As a special exception, the copyright holders give permission to link the
* code of portions of this program with the OpenSSL library under certain
* conditions as described in each individual source file and distribute
* linked combinations including the program with the OpenSSL library. You
* must comply with the GNU Affero General Public License in all respects
* for all of the code used other than as permitted herein. If you modify
* file(s) with this exception, you may extend this exception to your
* version of the file(s), but you are not obligated to do so. If you do not
* wish to do so, delete this exception statement from your version. If you
* delete this exception statement from all source files in the program,
* then also delete it in the license file.
*/
#define MONGO_LOG_DEFAULT_COMPONENT ::mongo::logger::LogComponent::kQuery
#include "mongo/db/query/plan_enumerator.h"
#include <set>
#include "mongo/db/query/indexability.h"
#include "mongo/db/query/index_tag.h"
#include "mongo/db/query/qlog.h"
#include "mongo/util/log.h"
namespace {
using namespace mongo;
std::string getPathPrefix(std::string path) {
if (mongoutils::str::contains(path, '.')) {
return mongoutils::str::before(path, '.');
}
else {
return path;
}
}
/**
* Returns true if either 'node' or a descendent of 'node'
* is a predicate that is required to use an index.
*/
bool expressionRequiresIndex(const MatchExpression* node) {
return CanonicalQuery::countNodes(node, MatchExpression::GEO_NEAR) > 0
|| CanonicalQuery::countNodes(node, MatchExpression::TEXT) > 0;
}
} // namespace
namespace mongo {
PlanEnumerator::PlanEnumerator(const PlanEnumeratorParams& params)
: _root(params.root),
_indices(params.indices),
_ixisect(params.intersect),
_orLimit(params.maxSolutionsPerOr),
_intersectLimit(params.maxIntersectPerAnd) { }
PlanEnumerator::~PlanEnumerator() {
typedef unordered_map<MemoID, NodeAssignment*> MemoMap;
for (MemoMap::iterator it = _memo.begin(); it != _memo.end(); ++it) {
delete it->second;
}
}
Status PlanEnumerator::init() {
// Fill out our memo structure from the tagged _root.
_done = !prepMemo(_root, PrepMemoContext());
// Dump the tags. We replace them with IndexTag instances.
_root->resetTag();
return Status::OK();
}
std::string PlanEnumerator::dumpMemo() {
mongoutils::str::stream ss;
// Note that this needs to be kept in sync with allocateAssignment which assigns memo IDs.
for (size_t i = 1; i < _memo.size(); ++i) {
ss << "[Node #" << i << "]: " << _memo[i]->toString() << "\n";
}
return ss;
}
string PlanEnumerator::NodeAssignment::toString() const {
if (NULL != pred) {
mongoutils::str::stream ss;
ss << "predicate\n";
ss << "\tfirst indices: [";
for (size_t i = 0; i < pred->first.size(); ++i) {
ss << pred->first[i];
if (i < pred->first.size() - 1)
ss << ", ";
}
ss << "]\n";
ss << "\tpred: " << pred->expr->toString();
ss << "\tindexToAssign: " << pred->indexToAssign;
return ss;
}
else if (NULL != andAssignment) {
mongoutils::str::stream ss;
ss << "AND enumstate counter " << andAssignment->counter;
for (size_t i = 0; i < andAssignment->choices.size(); ++i) {
ss << "\n\tchoice " << i << ":\n";
const AndEnumerableState& state = andAssignment->choices[i];
ss << "\t\tsubnodes: ";
for (size_t j = 0; j < state.subnodesToIndex.size(); ++j) {
ss << state.subnodesToIndex[j] << " ";
}
ss << '\n';
for (size_t j = 0; j < state.assignments.size(); ++j) {
const OneIndexAssignment& oie = state.assignments[j];
ss << "\t\tidx[" << oie.index << "]\n";
for (size_t k = 0; k < oie.preds.size(); ++k) {
ss << "\t\t\tpos " << oie.positions[k]
<< " pred " << oie.preds[k]->toString();
}
}
}
return ss;
}
else if (NULL != arrayAssignment) {
mongoutils::str::stream ss;
ss << "ARRAY SUBNODES enumstate " << arrayAssignment->counter << "/ ONE OF: [ ";
for (size_t i = 0; i < arrayAssignment->subnodes.size(); ++i) {
ss << arrayAssignment->subnodes[i] << " ";
}
ss << "]";
return ss;
}
else {
verify(NULL != orAssignment);
mongoutils::str::stream ss;
ss << "ALL OF: [ ";
for (size_t i = 0; i < orAssignment->subnodes.size(); ++i) {
ss << orAssignment->subnodes[i] << " ";
}
ss << "]";
return ss;
}
}
PlanEnumerator::MemoID PlanEnumerator::memoIDForNode(MatchExpression* node) {
unordered_map<MatchExpression*, MemoID>::iterator it = _nodeToId.find(node);
if (_nodeToId.end() == it) {
error() << "Trying to look up memo entry for node, none found.";
invariant(0);
}
return it->second;
}
bool PlanEnumerator::getNext(MatchExpression** tree) {
if (_done) { return false; }
// Tag with our first solution.
tagMemo(memoIDForNode(_root));
*tree = _root->shallowClone();
tagForSort(*tree);
sortUsingTags(*tree);
_root->resetTag();
QLOG() << "Enumerator: memo just before moving:" << endl << dumpMemo();
_done = nextMemo(memoIDForNode(_root));
return true;
}
//
// Structure creation
//
void PlanEnumerator::allocateAssignment(MatchExpression* expr,
NodeAssignment** assign,
MemoID* id) {
// We start at 1 so that the lookup of any entries not explicitly allocated
// will refer to an invalid memo slot.
size_t newID = _memo.size() + 1;
// Shouldn't be anything there already.
verify(_nodeToId.end() == _nodeToId.find(expr));
_nodeToId[expr] = newID;
verify(_memo.end() == _memo.find(newID));
NodeAssignment* newAssignment = new NodeAssignment();
_memo[newID] = newAssignment;
*assign = newAssignment;
*id = newID;
}
bool PlanEnumerator::prepMemo(MatchExpression* node, PrepMemoContext context) {
PrepMemoContext childContext;
childContext.elemMatchExpr = context.elemMatchExpr;
if (Indexability::nodeCanUseIndexOnOwnField(node)) {
// We only get here if our parent is an OR, an array operator, or we're the root.
// If we have no index tag there are no indices we can use.
if (NULL == node->getTag()) { return false; }
RelevantTag* rt = static_cast<RelevantTag*>(node->getTag());
// In order to definitely use an index it must be prefixed with our field.
// We don't consider notFirst indices here because we must be AND-related to a node
// that uses the first spot in that index, and we currently do not know that
// unless we're in an AND node.
if (0 == rt->first.size()) { return false; }
// We know we can use an index, so grab a memo spot.
size_t myMemoID;
NodeAssignment* assign;
allocateAssignment(node, &assign, &myMemoID);
assign->pred.reset(new PredicateAssignment());
assign->pred->expr = node;
assign->pred->first.swap(rt->first);
return true;
}
else if (Indexability::isBoundsGeneratingNot(node)) {
bool childIndexable = prepMemo(node->getChild(0), childContext);
// If the child isn't indexable then bail out now.
if (!childIndexable) {
return false;
}
// Our parent node, if any exists, will expect a memo entry keyed on 'node'. As such we
// have the node ID for 'node' just point to the memo created for the child that
// actually generates the bounds.
size_t myMemoID;
NodeAssignment* assign;
allocateAssignment(node, &assign, &myMemoID);
OrAssignment* orAssignment = new OrAssignment();
orAssignment->subnodes.push_back(memoIDForNode(node->getChild(0)));
assign->orAssignment.reset(orAssignment);
return true;
}
else if (MatchExpression::OR == node->matchType()) {
// For an OR to be indexed, all its children must be indexed.
for (size_t i = 0; i < node->numChildren(); ++i) {
if (!prepMemo(node->getChild(i), childContext)) {
return false;
}
}
// If we're here we're fully indexed and can be in the memo.
size_t myMemoID;
NodeAssignment* assign;
allocateAssignment(node, &assign, &myMemoID);
OrAssignment* orAssignment = new OrAssignment();
for (size_t i = 0; i < node->numChildren(); ++i) {
orAssignment->subnodes.push_back(memoIDForNode(node->getChild(i)));
}
assign->orAssignment.reset(orAssignment);
return true;
}
else if (Indexability::arrayUsesIndexOnChildren(node)) {
// Add each of our children as a subnode. We enumerate through each subnode one at a
// time until it's exhausted then we move on.
auto_ptr<ArrayAssignment> aa(new ArrayAssignment());
if (MatchExpression::ELEM_MATCH_OBJECT == node->matchType()) {
childContext.elemMatchExpr = node;
}
// For an OR to be indexed, all its children must be indexed.
for (size_t i = 0; i < node->numChildren(); ++i) {
if (prepMemo(node->getChild(i), childContext)) {
aa->subnodes.push_back(memoIDForNode(node->getChild(i)));
}
}
if (0 == aa->subnodes.size()) { return false; }
size_t myMemoID;
NodeAssignment* assign;
allocateAssignment(node, &assign, &myMemoID);
assign->arrayAssignment.reset(aa.release());
return true;
}
else if (MatchExpression::AND == node->matchType()) {
// Map from idx id to children that have a pred over it.
// TODO: The index intersection logic could be simplified if we could iterate over these
// maps in a known order. Currently when iterating over these maps we have to impose an
// ordering on each individual pair of indices in order to make sure that the
// enumeration results are order-independent. See SERVER-12196.
IndexToPredMap idxToFirst;
IndexToPredMap idxToNotFirst;
// Children that aren't predicates, and which do not necessarily need
// to use an index.
vector<MemoID> subnodes;
// Children that aren't predicates, but which *must* use an index.
// (e.g. an OR which contains a TEXT child).
vector<MemoID> mandatorySubnodes;
// A list of predicates contained in the subtree rooted at 'node'
// obtained by traversing deeply through $and and $elemMatch children.
vector<MatchExpression*> indexedPreds;
// Partition the childen into the children that aren't predicates which may or may
// not be indexed ('subnodes'), children that aren't predicates which must use the
// index ('mandatorySubnodes'). and children that are predicates ('indexedPreds').
//
// We have to get the subnodes with mandatory assignments rather than adding the
// mandatory preds to 'indexedPreds'. Adding the mandatory preds directly to
// 'indexedPreds' would lead to problems such as pulling a predicate beneath an OR
// into a set joined by an AND.
if (!partitionPreds(node, childContext, &indexedPreds,
&subnodes, &mandatorySubnodes)) {
return false;
}
if (mandatorySubnodes.size() > 1) {
return false;
}
// There can only be one mandatory predicate (at most one $text, at most one
// $geoNear, can't combine $text/$geoNear).
MatchExpression* mandatoryPred = NULL;
// There could be multiple indices which we could use to satisfy the mandatory
// predicate. Keep the set of such indices. Currently only one text index is
// allowed per collection, but there could be multiple 2d or 2dsphere indices
// available to answer a $geoNear predicate.
set<IndexID> mandatoryIndices;
// Go through 'indexedPreds' and add the predicates to the
// 'idxToFirst' and 'idxToNotFirst' maps.
for (size_t i = 0; i < indexedPreds.size(); ++i) {
MatchExpression* child = indexedPreds[i];
invariant(Indexability::nodeCanUseIndexOnOwnField(child));
RelevantTag* rt = static_cast<RelevantTag*>(child->getTag());
if (expressionRequiresIndex(child)) {
// 'child' is a predicate which *must* be tagged with an index.
// This should include only TEXT and GEO_NEAR preds.
// We expect either 0 or 1 mandatory predicates.
invariant(NULL == mandatoryPred);
// Mandatory predicates are TEXT or GEO_NEAR.
invariant(MatchExpression::TEXT == child->matchType() ||
MatchExpression::GEO_NEAR == child->matchType());
// The mandatory predicate must have a corresponding "mandatory index".
invariant(rt->first.size() != 0 || rt->notFirst.size() != 0);
mandatoryPred = child;
// Find all of the indices that could be used to satisfy the pred,
// and add them to the 'mandatoryIndices' set.
mandatoryIndices.insert(rt->first.begin(), rt->first.end());
mandatoryIndices.insert(rt->notFirst.begin(), rt->notFirst.end());
}
for (size_t j = 0; j < rt->first.size(); ++j) {
idxToFirst[rt->first[j]].push_back(child);
}
for (size_t j = 0 ; j< rt->notFirst.size(); ++j) {
idxToNotFirst[rt->notFirst[j]].push_back(child);
}
}
// If none of our children can use indices, bail out.
if (idxToFirst.empty()
&& (subnodes.size() == 0)
&& (mandatorySubnodes.size() == 0)) {
return false;
}
// At least one child can use an index, so we can create a memo entry.
AndAssignment* andAssignment = new AndAssignment();
size_t myMemoID;
NodeAssignment* nodeAssignment;
allocateAssignment(node, &nodeAssignment, &myMemoID);
// Takes ownership.
nodeAssignment->andAssignment.reset(andAssignment);
// Predicates which must use an index might be buried inside
// a subnode. Handle that case here.
if (1 == mandatorySubnodes.size()) {
AndEnumerableState aes;
aes.subnodesToIndex.push_back(mandatorySubnodes[0]);
andAssignment->choices.push_back(aes);
return true;
}
if (NULL != mandatoryPred) {
// We must have at least one index which can be used to answer 'mandatoryPred'.
invariant(!mandatoryIndices.empty());
enumerateMandatoryIndex(idxToFirst, idxToNotFirst, mandatoryPred,
mandatoryIndices, andAssignment);
return true;
}
enumerateOneIndex(idxToFirst, idxToNotFirst, subnodes, andAssignment);
if (_ixisect) {
enumerateAndIntersect(idxToFirst, idxToNotFirst, subnodes, andAssignment);
}
return true;
}
// Don't know what the node is at this point.
return false;
}
void PlanEnumerator::enumerateMandatoryIndex(const IndexToPredMap& idxToFirst,
const IndexToPredMap& idxToNotFirst,
MatchExpression* mandatoryPred,
const set<IndexID>& mandatoryIndices,
AndAssignment* andAssignment) {
// Generate index assignments for each index in 'mandatoryIndices'. We
// must assign 'mandatoryPred' to one of these indices, but we try all
// possibilities in 'mandatoryIndices' because some might be better than
// others for this query.
for (set<IndexID>::const_iterator indexIt = mandatoryIndices.begin();
indexIt != mandatoryIndices.end();
++indexIt) {
// We have a predicate which *must* be tagged to use an index.
// Get the index entry for the index it should use.
const IndexEntry& thisIndex = (*_indices)[*indexIt];
// Only text, 2d, and 2dsphere index types should be able to satisfy
// mandatory predicates.
invariant(INDEX_TEXT == thisIndex.type ||
INDEX_2D == thisIndex.type ||
INDEX_2DSPHERE == thisIndex.type);
OneIndexAssignment indexAssign;
indexAssign.index = *indexIt;
IndexToPredMap::const_iterator it = idxToFirst.find(*indexIt);
const vector<MatchExpression*>& predsOverLeadingField = it->second;
if (thisIndex.multikey) {
// Special handling for multikey mandatory indices.
if (predsOverLeadingField.end() != std::find(predsOverLeadingField.begin(),
predsOverLeadingField.end(),
mandatoryPred)) {
// The mandatory predicate is over the first field of the index. Assign
// it now.
indexAssign.preds.push_back(mandatoryPred);
indexAssign.positions.push_back(0);
}
else {
// The mandatory pred is notFirst. Assign an arbitrary predicate
// over the first position.
invariant(!predsOverLeadingField.empty());
indexAssign.preds.push_back(predsOverLeadingField[0]);
indexAssign.positions.push_back(0);
// Assign the mandatory predicate at the matching position in the compound
// index. We do this in order to ensure that the mandatory predicate (and not
// some other predicate over the same position in the compound index) gets
// assigned.
//
// The bad thing that could happen otherwise: A non-mandatory predicate gets
// chosen by getMultikeyCompoundablePreds(...) instead of 'mandatoryPred'.
// We would then fail to assign the mandatory predicate, and hence generate
// a bad data access plan.
//
// The mandatory predicate is assigned by calling compound(...) because
// compound(...) has logic for matching up a predicate with the proper
// position in the compound index.
vector<MatchExpression*> mandatoryToCompound;
mandatoryToCompound.push_back(mandatoryPred);
compound(mandatoryToCompound, thisIndex, &indexAssign);
// At this point we have assigned a predicate over the leading field and
// we have assigned the mandatory predicate to a trailing field.
//
// Ex:
// Say we have index {a: 1, b: 1, c: "2dsphere", d: 1}. Also suppose that
// there is a $near predicate over "c", with additional predicates over
// "a", "b", "c", and "d". We will have assigned the $near predicate at
// position 2 and a predicate with path "a" at position 0.
}
// Compound remaining predicates in a multikey-safe way.
IndexToPredMap::const_iterator compIt = idxToNotFirst.find(indexAssign.index);
if (compIt != idxToNotFirst.end()) {
const vector<MatchExpression*>& couldCompound = compIt->second;
vector<MatchExpression*> tryCompound;
getMultikeyCompoundablePreds(indexAssign.preds, couldCompound, &tryCompound);
if (tryCompound.size()) {
compound(tryCompound, thisIndex, &indexAssign);
}
}
}
else {
// For non-multikey, we don't have to do anything too special.
// Just assign all "first" predicates and try to compound like usual.
indexAssign.preds = it->second;
// Since everything in assign.preds prefixes the index, they all go
// at position '0' in the index, the first position.
indexAssign.positions.resize(indexAssign.preds.size(), 0);
// And now we begin compound analysis.
// Find everything that could use assign.index but isn't a pred over
// the first field of that index.
IndexToPredMap::const_iterator compIt = idxToNotFirst.find(indexAssign.index);
if (compIt != idxToNotFirst.end()) {
compound(compIt->second, thisIndex, &indexAssign);
}
}
// The mandatory predicate must be assigned.
invariant(indexAssign.preds.end() != std::find(indexAssign.preds.begin(),
indexAssign.preds.end(),
mandatoryPred));
// Output the assignments for this index.
AndEnumerableState state;
state.assignments.push_back(indexAssign);
andAssignment->choices.push_back(state);
}
}
void PlanEnumerator::enumerateOneIndex(const IndexToPredMap& idxToFirst,
const IndexToPredMap& idxToNotFirst,
const vector<MemoID>& subnodes,
AndAssignment* andAssignment) {
// In the simplest case, an AndAssignment picks indices like a PredicateAssignment. To
// be indexed we must only pick one index
//
// Complications:
//
// Some of our child predicates cannot be answered without an index. As such, the
// indices that those predicates require must always be outputted. We store these
// mandatory index assignments in 'mandatoryIndices'.
//
// Some of our children may not be predicates. We may have ORs (or array operators) as
// children. If one of these subtrees provides an index, the AND is indexed. We store
// these subtree choices in 'subnodes'.
//
// With the above two cases out of the way, we can focus on the remaining case: what to
// do with our children that are leaf predicates.
//
// Guiding principles for index assignment to leaf predicates:
//
// 1. If we assign an index to {x:{$gt: 5}} we should assign the same index to
// {x:{$lt: 50}}. That is, an index assignment should include all predicates
// over its leading field.
//
// 2. If we have the index {a:1, b:1} and we assign it to {a: 5} we should assign it
// to {b:7}, since with a predicate over the first field of the compound index,
// the second field can be bounded as well. We may only assign indices to predicates
// if all fields to the left of the index field are constrained.
// First, add the state of using each subnode.
for (size_t i = 0; i < subnodes.size(); ++i) {
AndEnumerableState aes;
aes.subnodesToIndex.push_back(subnodes[i]);
andAssignment->choices.push_back(aes);
}
// For each FIRST, we assign nodes to it.
for (IndexToPredMap::const_iterator it = idxToFirst.begin(); it != idxToFirst.end(); ++it) {
// The assignment we're filling out.
OneIndexAssignment indexAssign;
// This is the index we assign to.
indexAssign.index = it->first;
const IndexEntry& thisIndex = (*_indices)[it->first];
// If the index is multikey, we only assign one pred to it. We also skip
// compounding. TODO: is this also true for 2d and 2dsphere indices? can they be
// multikey but still compoundable?
if (thisIndex.multikey) {
// TODO: could pick better pred than first but not too worried since we should
// really be isecting indices here. Just take the first pred. We don't assign
// any other preds to this index. The planner will intersect the preds and this
// enumeration strategy is just one index at a time.
indexAssign.preds.push_back(it->second[0]);
indexAssign.positions.push_back(0);
// If there are any preds that could possibly be compounded with this
// index...
IndexToPredMap::const_iterator compIt = idxToNotFirst.find(indexAssign.index);
if (compIt != idxToNotFirst.end()) {
const vector<MatchExpression*>& couldCompound = compIt->second;
vector<MatchExpression*> tryCompound;
// ...select the predicates that are safe to compound and try to
// compound them.
getMultikeyCompoundablePreds(indexAssign.preds, couldCompound, &tryCompound);
if (tryCompound.size()) {
compound(tryCompound, thisIndex, &indexAssign);
}
}
}
else {
// The index isn't multikey. Assign all preds to it. The planner will
// intersect the bounds.
indexAssign.preds = it->second;
// Since everything in assign.preds prefixes the index, they all go
// at position '0' in the index, the first position.
indexAssign.positions.resize(indexAssign.preds.size(), 0);
// Find everything that could use assign.index but isn't a pred over
// the first field of that index.
IndexToPredMap::const_iterator compIt = idxToNotFirst.find(indexAssign.index);
if (compIt != idxToNotFirst.end()) {
compound(compIt->second, thisIndex, &indexAssign);
}
}
AndEnumerableState state;
state.assignments.push_back(indexAssign);
andAssignment->choices.push_back(state);
}
}
void PlanEnumerator::enumerateAndIntersect(const IndexToPredMap& idxToFirst,
const IndexToPredMap& idxToNotFirst,
const vector<MemoID>& subnodes,
AndAssignment* andAssignment) {
// Hardcoded "look at all members of the power set of size 2" search,
// a.k.a. "consider all pairs of indices".
//
// For each unordered pair of indices do the following:
// 0. Impose an ordering (idx1, idx2) using the key patterns.
// (*See note below.)
// 1. Assign predicates which prefix idx1 to idx1.
// 2. Add assigned predicates to a set of predicates---the "already
// assigned set".
// 3. Assign predicates which prefix idx2 to idx2, as long as they
// been assigned to idx1 already. Add newly assigned predicates to
// the "already assigned set".
// 4. Try to assign predicates to idx1 by compounding.
// 5. Add any predicates assigned to idx1 by compounding to the
// "already assigned set",
// 6. Try to assign predicates to idx2 by compounding.
// 7. Determine if we have already assigned all predicates in
// the "already assigned set" to a single index. If so, then
// don't generate an ixisect solution, as compounding will
// be better. Otherwise, output the ixisect assignments.
//
// *NOTE on ordering. Suppose we have two indices A and B, and a
// predicate P1 which is over the prefix of both indices A and B.
// If we order the indices (A, B) then P1 will get assigned to A,
// but if we order the indices (B, A) then P1 will get assigned to
// B. In order to make sure that we get the same result for the unordered
// pair {A, B} we have to begin by imposing an ordering. As a more concrete
// example, if we have indices {x: 1, y: 1} and {x: 1, z: 1} with predicate
// {x: 3}, we want to make sure that {x: 3} gets assigned to the same index
// irrespective of ordering.
size_t sizeBefore = andAssignment->choices.size();
for (IndexToPredMap::const_iterator firstIt = idxToFirst.begin();
firstIt != idxToFirst.end(); ++firstIt) {
const IndexEntry& oneIndex = (*_indices)[firstIt->first];
// 'oneAssign' is used to assign indices and subnodes or to
// make assignments for the first index when it's multikey.
// It is NOT used in the inner loop that considers pairs of
// indices.
OneIndexAssignment oneAssign;
oneAssign.index = firstIt->first;
oneAssign.preds = firstIt->second;
// Since everything in assign.preds prefixes the index, they all go
// at position '0' in the index, the first position.
oneAssign.positions.resize(oneAssign.preds.size(), 0);
// We create a scan per predicate so if we have >1 predicate we'll already
// have at least 2 scans (one predicate per scan as the planner can't
// intersect bounds when the index is multikey), so we stop here.
if (oneIndex.multikey && oneAssign.preds.size() > 1) {
// One could imagine an enormous auto-generated $all query with too many clauses to
// have an ixscan per clause.
static const size_t kMaxSelfIntersections = 10;
if (oneAssign.preds.size() > kMaxSelfIntersections) {
// Only take the first kMaxSelfIntersections preds.
oneAssign.preds.resize(kMaxSelfIntersections);
oneAssign.positions.resize(kMaxSelfIntersections);
}
AndEnumerableState state;
state.assignments.push_back(oneAssign);
andAssignment->choices.push_back(state);
continue;
}
// Output (subnode, firstAssign) pairs.
for (size_t i = 0; i < subnodes.size(); ++i) {
AndEnumerableState indexAndSubnode;
indexAndSubnode.assignments.push_back(oneAssign);
indexAndSubnode.subnodesToIndex.push_back(subnodes[i]);
andAssignment->choices.push_back(indexAndSubnode);
// Limit n^2.
if (andAssignment->choices.size() - sizeBefore > _intersectLimit) {
return;
}
}
// Start looking at all other indices to find one that we want to bundle
// with firstAssign.
IndexToPredMap::const_iterator secondIt = firstIt;
secondIt++;
for (; secondIt != idxToFirst.end(); secondIt++) {
const IndexEntry& firstIndex = (*_indices)[secondIt->first];
const IndexEntry& secondIndex = (*_indices)[secondIt->first];
// Limit n^2.
if (andAssignment->choices.size() - sizeBefore > _intersectLimit) {
return;
}
// If the other index we're considering is multikey with >1 pred, we don't
// want to have it as an additional assignment. Eventually, it1 will be
// equal to the current value of secondIt and we'll assign every pred for
// this mapping to the index.
if (secondIndex.multikey && secondIt->second.size() > 1) {
continue;
}
//
// Step #0:
// Impose an ordering (idx1, idx2) using the key patterns.
//
IndexToPredMap::const_iterator it1, it2;
int ordering = firstIndex.keyPattern.woCompare(secondIndex.keyPattern);
it1 = (ordering > 0) ? firstIt : secondIt;
it2 = (ordering > 0) ? secondIt : firstIt;
const IndexEntry& ie1 = (*_indices)[it1->first];
const IndexEntry& ie2 = (*_indices)[it2->first];
//
// Step #1:
// Assign predicates which prefix firstIndex to firstAssign.
//
OneIndexAssignment firstAssign;
firstAssign.index = it1->first;
firstAssign.preds = it1->second;
// Since everything in assign.preds prefixes the index, they all go
// at position '0' in the index, the first position.
firstAssign.positions.resize(firstAssign.preds.size(), 0);
// We keep track of what preds are assigned to indices either because they
// prefix the index or have been assigned through compounding. We make sure
// that these predicates DO NOT become additional index assignments.
// Example: what if firstAssign is the index (x, y) and we're trying to
// compound? We want to make sure not to compound if the predicate is
// already assigned to index y.
set<MatchExpression*> predsAssigned;
//
// Step #2:
// Add indices assigned in 'firstAssign' to 'predsAssigned'.
//
for (size_t i = 0; i < firstAssign.preds.size(); ++i) {
predsAssigned.insert(firstAssign.preds[i]);
}
//
// Step #3:
// Assign predicates which prefix secondIndex to secondAssign and
// have not already been assigned to firstAssign. Any newly
// assigned predicates are added to 'predsAssigned'.
//
OneIndexAssignment secondAssign;
secondAssign.index = it2->first;
const vector<MatchExpression*>& preds = it2->second;
for (size_t i = 0; i < preds.size(); ++i) {
if (predsAssigned.end() == predsAssigned.find(preds[i])) {
secondAssign.preds.push_back(preds[i]);
secondAssign.positions.push_back(0);
predsAssigned.insert(preds[i]);
}
}
// Every predicate that would use this index is already assigned in
// firstAssign.
if (0 == secondAssign.preds.size()) { continue; }
//
// Step #4:
// Compound on firstAssign, if applicable.
//
IndexToPredMap::const_iterator firstIndexCompound =
idxToNotFirst.find(firstAssign.index);
// Can't compound with multikey indices.
if (!ie1.multikey && firstIndexCompound != idxToNotFirst.end()) {
// We must remove any elements of 'predsAssigned' from consideration.
vector<MatchExpression*> tryCompound;
const vector<MatchExpression*>& couldCompound
= firstIndexCompound->second;
for (size_t i = 0; i < couldCompound.size(); ++i) {
if (predsAssigned.end() == predsAssigned.find(couldCompound[i])) {
tryCompound.push_back(couldCompound[i]);
}
}
if (tryCompound.size()) {
compound(tryCompound, ie1, &firstAssign);
}
}
//
// Step #5:
// Make sure predicates assigned by compounding in step #4 do not get
// assigned again.
//
for (size_t i = 0; i < firstAssign.preds.size(); ++i) {
if (predsAssigned.end() == predsAssigned.find(firstAssign.preds[i])) {
predsAssigned.insert(firstAssign.preds[i]);
}
}
//
// Step #6:
// Compound on firstAssign, if applicable.
//
IndexToPredMap::const_iterator secondIndexCompound =
idxToNotFirst.find(secondAssign.index);
if (!ie2.multikey && secondIndexCompound != idxToNotFirst.end()) {
// We must remove any elements of 'predsAssigned' from consideration.
vector<MatchExpression*> tryCompound;
const vector<MatchExpression*>& couldCompound
= secondIndexCompound->second;
for (size_t i = 0; i < couldCompound.size(); ++i) {
if (predsAssigned.end() == predsAssigned.find(couldCompound[i])) {
tryCompound.push_back(couldCompound[i]);
}
}
if (tryCompound.size()) {
compound(tryCompound, ie2, &secondAssign);
}
}
// Add predicates in 'secondAssign' to the set of all assigned predicates.
for (size_t i = 0; i < secondAssign.preds.size(); ++i) {
if (predsAssigned.end() == predsAssigned.find(secondAssign.preds[i])) {
predsAssigned.insert(secondAssign.preds[i]);
}
}
//
// Step #7:
// Make sure we haven't already assigned this set of predicates by compounding.
// If we have, then bail out for this pair of indices.
//
if (alreadyCompounded(predsAssigned, andAssignment)) {
// There is no need to add either 'firstAssign' or 'secondAssign'
// to 'andAssignment' in this case because we have already performed
// assignments to single indices in enumerateOneIndex(...).
continue;
}
// We're done with this particular pair of indices; output
// the resulting assignments.
AndEnumerableState state;
state.assignments.push_back(firstAssign);
state.assignments.push_back(secondAssign);
andAssignment->choices.push_back(state);
}
}
// TODO: Do we just want one subnode at a time? We can use far more than 2 indices at once
// doing this very easily. If we want to restrict the # of indices the children use, when
// we memoize the subtree above we can restrict it to 1 index at a time. This can get
// tricky if we want both an intersection and a 1-index memo entry, since our state change
// is simple and we don't traverse the memo in any targeted way. Should also verify that
// having a one-to-many mapping of MatchExpression to MemoID doesn't break anything. This
// approach errors on the side of "too much indexing."
for (size_t i = 0; i < subnodes.size(); ++i) {
for (size_t j = i + 1; j < subnodes.size(); ++j) {
AndEnumerableState state;
state.subnodesToIndex.push_back(subnodes[i]);
state.subnodesToIndex.push_back(subnodes[j]);
andAssignment->choices.push_back(state);
}
}
}
bool PlanEnumerator::partitionPreds(MatchExpression* node,
PrepMemoContext context,
vector<MatchExpression*>* indexOut,
vector<MemoID>* subnodesOut,
vector<MemoID>* mandatorySubnodes) {
for (size_t i = 0; i < node->numChildren(); ++i) {
MatchExpression* child = node->getChild(i);
if (Indexability::nodeCanUseIndexOnOwnField(child)) {
RelevantTag* rt = static_cast<RelevantTag*>(child->getTag());
if (NULL != context.elemMatchExpr) {
// If we're in an $elemMatch context, store the
// innermost parent $elemMatch, as well as the
// inner path prefix.
rt->elemMatchExpr = context.elemMatchExpr;
rt->pathPrefix = getPathPrefix(child->path().toString());
}
else {
// We're not an $elemMatch context, so we should store
// the prefix of the full path.
rt->pathPrefix = getPathPrefix(rt->path);
}
// Output this as a pred that can use the index.
indexOut->push_back(child);
}
else if (Indexability::isBoundsGeneratingNot(child)) {
partitionPreds(child, context, indexOut, subnodesOut, mandatorySubnodes);
}
else if (MatchExpression::ELEM_MATCH_OBJECT == child->matchType()) {
PrepMemoContext childContext;
childContext.elemMatchExpr = child;
partitionPreds(child, childContext, indexOut, subnodesOut, mandatorySubnodes);
}
else if (MatchExpression::AND == child->matchType()) {
partitionPreds(child, context, indexOut, subnodesOut, mandatorySubnodes);
}
else {
bool mandatory = expressionRequiresIndex(child);
// Recursively prepMemo for the subnode. We fall through
// to this case for logical nodes other than AND (e.g. OR)
// and for array nodes other than ELEM_MATCH_OBJECT or
// ELEM_MATCH_VALUE (e.g. ALL).
if (prepMemo(child, context)) {
size_t childID = memoIDForNode(child);
// Output the subnode.
if (mandatory) {
mandatorySubnodes->push_back(childID);
}
else {
subnodesOut->push_back(childID);
}
}
else if (mandatory) {
// The subnode is mandatory but cannot be indexed. This means
// that the entire AND cannot be indexed either.
return false;
}
}
}
return true;
}
void PlanEnumerator::getMultikeyCompoundablePreds(const vector<MatchExpression*>& assigned,
const vector<MatchExpression*>& couldCompound,
vector<MatchExpression*>* out) {
// Map from a particular $elemMatch expression to the set of prefixes
// used so far by the predicates inside the $elemMatch. For example,
// {a: {$elemMatch: {b: 1, c: 2}}} would map to the set {'b', 'c'} at
// the end of this function's execution.
//
// NULL maps to the set of prefixes used so far outside of an $elemMatch
// context.
//
// As we iterate over the available indexed predicates, we keep track
// of the used prefixes both inside and outside of an $elemMatch context.
unordered_map<MatchExpression*, set<string> > used;
// Initialize 'used' with the starting predicates in 'assigned'. Begin by
// initializing the top-level scope with the prefix of the full path.
for (size_t i = 0; i < assigned.size(); i++) {
const MatchExpression* assignedPred = assigned[i];
invariant(NULL != assignedPred->getTag());
RelevantTag* usedRt = static_cast<RelevantTag*>(assignedPred->getTag());
set<string> usedPrefixes;
usedPrefixes.insert(getPathPrefix(usedRt->path));
used[NULL] = usedPrefixes;
// If 'assigned' is a predicate inside an $elemMatch, we have to
// add the prefix not only to the top-level context, but also to the
// the $elemMatch context. For example, if 'assigned' is {a: {$elemMatch: {b: 1}}},
// then we will have already added "a" to the set for NULL. We now
// also need to add "b" to the set for the $elemMatch.
if (NULL != usedRt->elemMatchExpr) {
set<string> elemMatchUsed;
// Whereas getPathPrefix(usedRt->path) is the prefix of the full path,
// usedRt->pathPrefix contains the prefix of the portion of the
// path that is inside the $elemMatch. These two prefixes are the same
// in the top-level context, but here must be different because 'usedRt'
// is in an $elemMatch context.
elemMatchUsed.insert(usedRt->pathPrefix);
used[usedRt->elemMatchExpr] = elemMatchUsed;
}
}
for (size_t i = 0; i < couldCompound.size(); ++i) {
invariant(Indexability::nodeCanUseIndexOnOwnField(couldCompound[i]));
RelevantTag* rt = static_cast<RelevantTag*>(couldCompound[i]->getTag());
if (used.end() == used.find(rt->elemMatchExpr)) {
// This is a new $elemMatch that we haven't seen before.
invariant(used.end() != used.find(NULL));
set<string>& topLevelUsed = used.find(NULL)->second;
// If the top-level path prefix of the $elemMatch hasn't been
// used yet, couldCompound[i] is safe to compound.
if (topLevelUsed.end() == topLevelUsed.find(getPathPrefix(rt->path))) {
topLevelUsed.insert(getPathPrefix(rt->path));
set<string> usedPrefixes;
usedPrefixes.insert(rt->pathPrefix);
used[rt->elemMatchExpr] = usedPrefixes;
// Output the predicate.
out->push_back(couldCompound[i]);
}
}
else {
// We've seen this $elemMatch before, or the predicate is
// top-level (not in an $elemMatch context). If the prefix stored
// in the tag has not been used yet, then couldCompound[i] is
// safe to compound.
set<string>& usedPrefixes = used.find(rt->elemMatchExpr)->second;
if (usedPrefixes.end() == usedPrefixes.find(rt->pathPrefix)) {
usedPrefixes.insert(rt->pathPrefix);
// Output the predicate.
out->push_back(couldCompound[i]);
}
}
}
}
bool PlanEnumerator::alreadyCompounded(const set<MatchExpression*>& ixisectAssigned,
const AndAssignment* andAssignment) {
for (size_t i = 0; i < andAssignment->choices.size(); ++i) {
const AndEnumerableState& state = andAssignment->choices[i];
// We cannot have assigned this set of predicates already by
// compounding unless this is an assignment to a single index.
if (state.assignments.size() != 1) {
continue;
}
// If the set of preds in 'ixisectAssigned' is a subset of 'oneAssign.preds',
// then all the preds can be used by compounding on a single index.
const OneIndexAssignment& oneAssign = state.assignments[0];
// If 'ixisectAssigned' is larger than 'oneAssign.preds', then
// it can't be a subset.
if (ixisectAssigned.size() > oneAssign.preds.size()) {
continue;
}
// Check for subset by counting the number of elements in 'oneAssign.preds'
// that are contained in 'ixisectAssigned'. The elements of both 'oneAssign.preds'
// and 'ixisectAssigned' are unique (no repeated elements).
size_t count = 0;
for (size_t j = 0; j < oneAssign.preds.size(); ++j) {
if (ixisectAssigned.end() != ixisectAssigned.find(oneAssign.preds[j])) {
++count;
}
}
if (ixisectAssigned.size() == count) {
return true;
}
// We cannot assign the preds by compounding on 'oneAssign'.
// Move on to the next index.
}
return false;
}
void PlanEnumerator::compound(const vector<MatchExpression*>& tryCompound,
const IndexEntry& thisIndex,
OneIndexAssignment* assign) {
// Let's try to match up the expressions in 'compExprs' with the
// fields in the index key pattern.
BSONObjIterator kpIt(thisIndex.keyPattern);
// Skip the first elt as it's already assigned.
kpIt.next();
// When we compound we store the field number that the predicate
// goes over in order to avoid having to iterate again and compare
// field names.
size_t posInIdx = 0;
while (kpIt.more()) {
BSONElement keyElt = kpIt.next();
++posInIdx;
// Go through 'tryCompound' to see if there is a compoundable
// predicate for 'keyElt'. If there is nothing to compound, then
// simply move on to the next field in the compound index. We
// do not enforce that fields are assigned contiguously from
// right to left, i.e. for compound index {a: 1, b: 1, c: 1}
// it is okay to compound predicates over "a" and "c", skipping "b".
for (size_t j = 0; j < tryCompound.size(); ++j) {
MatchExpression* maybe = tryCompound[j];
// Sigh we grab the full path from the relevant tag.
RelevantTag* rt = static_cast<RelevantTag*>(maybe->getTag());
if (keyElt.fieldName() == rt->path) {
// preds and positions are parallel arrays.
assign->preds.push_back(maybe);
assign->positions.push_back(posInIdx);
}
}
}
}
//
// Structure navigation
//
void PlanEnumerator::tagMemo(size_t id) {
QLOG() << "Tagging memoID " << id << endl;
NodeAssignment* assign = _memo[id];
verify(NULL != assign);
if (NULL != assign->pred) {
PredicateAssignment* pa = assign->pred.get();
verify(NULL == pa->expr->getTag());
verify(pa->indexToAssign < pa->first.size());
pa->expr->setTag(new IndexTag(pa->first[pa->indexToAssign]));
}
else if (NULL != assign->orAssignment) {
OrAssignment* oa = assign->orAssignment.get();
for (size_t i = 0; i < oa->subnodes.size(); ++i) {
tagMemo(oa->subnodes[i]);
}
}
else if (NULL != assign->arrayAssignment) {
ArrayAssignment* aa = assign->arrayAssignment.get();
tagMemo(aa->subnodes[aa->counter]);
}
else if (NULL != assign->andAssignment) {
AndAssignment* aa = assign->andAssignment.get();
verify(aa->counter < aa->choices.size());
const AndEnumerableState& aes = aa->choices[aa->counter];
for (size_t j = 0; j < aes.subnodesToIndex.size(); ++j) {
tagMemo(aes.subnodesToIndex[j]);
}
for (size_t i = 0; i < aes.assignments.size(); ++i) {
const OneIndexAssignment& assign = aes.assignments[i];
for (size_t j = 0; j < assign.preds.size(); ++j) {
MatchExpression* pred = assign.preds[j];
verify(NULL == pred->getTag());
pred->setTag(new IndexTag(assign.index, assign.positions[j]));
}
}
}
else {
verify(0);
}
}
bool PlanEnumerator::nextMemo(size_t id) {
NodeAssignment* assign = _memo[id];
verify(NULL != assign);
if (NULL != assign->pred) {
PredicateAssignment* pa = assign->pred.get();
pa->indexToAssign++;
if (pa->indexToAssign >= pa->first.size()) {
pa->indexToAssign = 0;
return true;
}
return false;
}
else if (NULL != assign->orAssignment) {
OrAssignment* oa = assign->orAssignment.get();
// Limit the number of OR enumerations
oa->counter++;
if (oa->counter >= _orLimit) {
return true;
}
// OR just walks through telling its children to
// move forward.
for (size_t i = 0; i < oa->subnodes.size(); ++i) {
// If there's no carry, we just stop. If there's a carry, we move the next child
// forward.
if (!nextMemo(oa->subnodes[i])) {
return false;
}
}
// If we're here, the last subnode had a carry, therefore the OR has a carry.
return true;
}
else if (NULL != assign->arrayAssignment) {
ArrayAssignment* aa = assign->arrayAssignment.get();
// moving to next on current subnode is OK
if (!nextMemo(aa->subnodes[aa->counter])) { return false; }
// Move to next subnode.
++aa->counter;
if (aa->counter < aa->subnodes.size()) {
return false;
}
aa->counter = 0;
return true;
}
else if (NULL != assign->andAssignment) {
AndAssignment* aa = assign->andAssignment.get();
// One of our subnodes might have to move on to its next enumeration state.
const AndEnumerableState& aes = aa->choices[aa->counter];
for (size_t i = 0; i < aes.subnodesToIndex.size(); ++i) {
if (!nextMemo(aes.subnodesToIndex[i])) {
return false;
}
}
// None of the subnodes had another enumeration state, so we move on to the
// next top-level choice.
++aa->counter;
if (aa->counter < aa->choices.size()) {
return false;
}
aa->counter = 0;
return true;
}
// This shouldn't happen.
verify(0);
return false;
}
} // namespace mongo
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