/* Copyright (C) 2003 MySQL AB This program is free software; you can redistribute it and/or modify it under the terms of the GNU General Public License as published by the Free Software Foundation; either version 2 of the License, or (at your option) any later version. 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 General Public License for more details. You should have received a copy of the GNU General Public License along with this program; if not, write to the Free Software Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA */ #define DBTUX_TREE_CPP #include "Dbtux.hpp" /* * Add entry. Handle the case when there is room for one more. This * is the common case given slack in nodes. */ void Dbtux::treeAdd(Frag& frag, TreePos treePos, TreeEnt ent) { TreeHead& tree = frag.m_tree; NodeHandle node(frag); if (treePos.m_loc != NullTupLoc) { // non-empty tree jam(); selectNode(node, treePos.m_loc); unsigned pos = treePos.m_pos; if (node.getOccup() < tree.m_maxOccup) { // node has room jam(); nodePushUp(node, pos, ent, RNIL); return; } treeAddFull(frag, node, pos, ent); return; } jam(); insertNode(node); nodePushUp(node, 0, ent, RNIL); node.setSide(2); tree.m_root = node.m_loc; } /* * Add entry when node is full. Handle the case when there is g.l.b * node in left subtree with room for one more. It will receive the min * entry of this node. The min entry could be the entry to add. */ void Dbtux::treeAddFull(Frag& frag, NodeHandle lubNode, unsigned pos, TreeEnt ent) { TreeHead& tree = frag.m_tree; TupLoc loc = lubNode.getLink(0); if (loc != NullTupLoc) { // find g.l.b node NodeHandle glbNode(frag); do { jam(); selectNode(glbNode, loc); loc = glbNode.getLink(1); } while (loc != NullTupLoc); if (glbNode.getOccup() < tree.m_maxOccup) { // g.l.b node has room jam(); Uint32 scanList = RNIL; if (pos != 0) { jam(); // add the new entry and return min entry nodePushDown(lubNode, pos - 1, ent, scanList); } // g.l.b node receives min entry from l.u.b node nodePushUp(glbNode, glbNode.getOccup(), ent, scanList); return; } treeAddNode(frag, lubNode, pos, ent, glbNode, 1); return; } treeAddNode(frag, lubNode, pos, ent, lubNode, 0); } /* * Add entry when there is no g.l.b node in left subtree or the g.l.b * node is full. We must add a new left or right child node which * becomes the new g.l.b node. */ void Dbtux::treeAddNode(Frag& frag, NodeHandle lubNode, unsigned pos, TreeEnt ent, NodeHandle parentNode, unsigned i) { NodeHandle glbNode(frag); insertNode(glbNode); // connect parent and child parentNode.setLink(i, glbNode.m_loc); glbNode.setLink(2, parentNode.m_loc); glbNode.setSide(i); Uint32 scanList = RNIL; if (pos != 0) { jam(); // add the new entry and return min entry nodePushDown(lubNode, pos - 1, ent, scanList); } // g.l.b node receives min entry from l.u.b node nodePushUp(glbNode, 0, ent, scanList); // re-balance the tree treeAddRebalance(frag, parentNode, i); } /* * Re-balance tree after adding a node. The process starts with the * parent of the added node. */ void Dbtux::treeAddRebalance(Frag& frag, NodeHandle node, unsigned i) { while (true) { // height of subtree i has increased by 1 int j = (i == 0 ? -1 : +1); int b = node.getBalance(); if (b == 0) { // perfectly balanced jam(); node.setBalance(j); // height change propagates up } else if (b == -j) { // height of shorter subtree increased jam(); node.setBalance(0); // height of tree did not change - done break; } else if (b == j) { // height of longer subtree increased jam(); NodeHandle childNode(frag); selectNode(childNode, node.getLink(i)); int b2 = childNode.getBalance(); if (b2 == b) { jam(); treeRotateSingle(frag, node, i); } else if (b2 == -b) { jam(); treeRotateDouble(frag, node, i); } else { // height of subtree increased so it cannot be perfectly balanced ndbrequire(false); } // height of tree did not increase - done break; } else { ndbrequire(false); } TupLoc parentLoc = node.getLink(2); if (parentLoc == NullTupLoc) { jam(); // root node - done break; } i = node.getSide(); selectNode(node, parentLoc); } } /* * Remove entry. Optimize for nodes with slack. Handle the case when * there is no underflow i.e. occupancy remains at least minOccup. For * interior nodes this is a requirement. For others it means that we do * not need to consider merge of semi-leaf and leaf. */ void Dbtux::treeRemove(Frag& frag, TreePos treePos) { TreeHead& tree = frag.m_tree; unsigned pos = treePos.m_pos; NodeHandle node(frag); selectNode(node, treePos.m_loc); TreeEnt ent; if (node.getOccup() > tree.m_minOccup) { // no underflow in any node type jam(); nodePopDown(node, pos, ent, 0); return; } if (node.getChilds() == 2) { // underflow in interior node jam(); treeRemoveInner(frag, node, pos); return; } // remove entry in semi/leaf nodePopDown(node, pos, ent, 0); if (node.getLink(0) != NullTupLoc) { jam(); treeRemoveSemi(frag, node, 0); return; } if (node.getLink(1) != NullTupLoc) { jam(); treeRemoveSemi(frag, node, 1); return; } treeRemoveLeaf(frag, node); } /* * Remove entry when interior node underflows. There is g.l.b node in * left subtree to borrow an entry from. The max entry of the g.l.b * node becomes the min entry of this node. */ void Dbtux::treeRemoveInner(Frag& frag, NodeHandle lubNode, unsigned pos) { TreeHead& tree = frag.m_tree; TreeEnt ent; // find g.l.b node NodeHandle glbNode(frag); TupLoc loc = lubNode.getLink(0); do { jam(); selectNode(glbNode, loc); loc = glbNode.getLink(1); } while (loc != NullTupLoc); // borrow max entry from semi/leaf Uint32 scanList = RNIL; nodePopDown(glbNode, glbNode.getOccup() - 1, ent, &scanList); nodePopUp(lubNode, pos, ent, scanList); if (glbNode.getLink(0) != NullTupLoc) { jam(); treeRemoveSemi(frag, glbNode, 0); return; } treeRemoveLeaf(frag, glbNode); } /* * Handle semi-leaf after removing an entry. Move entries from leaf to * semi-leaf to bring semi-leaf occupancy above minOccup, if possible. * The leaf may become empty. */ void Dbtux::treeRemoveSemi(Frag& frag, NodeHandle semiNode, unsigned i) { TreeHead& tree = frag.m_tree; ndbrequire(semiNode.getChilds() < 2); TupLoc leafLoc = semiNode.getLink(i); NodeHandle leafNode(frag); selectNode(leafNode, leafLoc); if (semiNode.getOccup() < tree.m_minOccup) { jam(); unsigned cnt = min(leafNode.getOccup(), tree.m_minOccup - semiNode.getOccup()); nodeSlide(semiNode, leafNode, cnt, i); if (leafNode.getOccup() == 0) { // remove empty leaf jam(); treeRemoveNode(frag, leafNode); } } } /* * Handle leaf after removing an entry. If parent is semi-leaf, move * entries to it as in the semi-leaf case. If parent is interior node, * do nothing. */ void Dbtux::treeRemoveLeaf(Frag& frag, NodeHandle leafNode) { TreeHead& tree = frag.m_tree; TupLoc parentLoc = leafNode.getLink(2); if (parentLoc != NullTupLoc) { jam(); NodeHandle parentNode(frag); selectNode(parentNode, parentLoc); unsigned i = leafNode.getSide(); if (parentNode.getLink(1 - i) == NullTupLoc) { // parent is semi-leaf jam(); if (parentNode.getOccup() < tree.m_minOccup) { jam(); unsigned cnt = min(leafNode.getOccup(), tree.m_minOccup - parentNode.getOccup()); nodeSlide(parentNode, leafNode, cnt, i); } } } if (leafNode.getOccup() == 0) { jam(); // remove empty leaf treeRemoveNode(frag, leafNode); } } /* * Remove empty leaf. */ void Dbtux::treeRemoveNode(Frag& frag, NodeHandle leafNode) { TreeHead& tree = frag.m_tree; ndbrequire(leafNode.getChilds() == 0); TupLoc parentLoc = leafNode.getLink(2); unsigned i = leafNode.getSide(); deleteNode(leafNode); if (parentLoc != NullTupLoc) { jam(); NodeHandle parentNode(frag); selectNode(parentNode, parentLoc); parentNode.setLink(i, NullTupLoc); // re-balance the tree treeRemoveRebalance(frag, parentNode, i); return; } // tree is now empty tree.m_root = NullTupLoc; } /* * Re-balance tree after removing a node. The process starts with the * parent of the removed node. */ void Dbtux::treeRemoveRebalance(Frag& frag, NodeHandle node, unsigned i) { while (true) { // height of subtree i has decreased by 1 int j = (i == 0 ? -1 : +1); int b = node.getBalance(); if (b == 0) { // perfectly balanced jam(); node.setBalance(-j); // height of tree did not change - done return; } else if (b == j) { // height of longer subtree has decreased jam(); node.setBalance(0); // height change propagates up } else if (b == -j) { // height of shorter subtree has decreased jam(); // child on the other side NodeHandle childNode(frag); selectNode(childNode, node.getLink(1 - i)); int b2 = childNode.getBalance(); if (b2 == b) { jam(); treeRotateSingle(frag, node, 1 - i); // height of tree decreased and propagates up } else if (b2 == -b) { jam(); treeRotateDouble(frag, node, 1 - i); // height of tree decreased and propagates up } else { jam(); treeRotateSingle(frag, node, 1 - i); // height of tree did not change - done return; } } else { ndbrequire(false); } TupLoc parentLoc = node.getLink(2); if (parentLoc == NullTupLoc) { jam(); // root node - done return; } i = node.getSide(); selectNode(node, parentLoc); } } /* * Single rotation about node 5. One of LL (i=0) or RR (i=1). * * 0 0 * | | * 5 ==> 3 * / \ / \ * 3 6 2 5 * / \ / / \ * 2 4 1 4 6 * / * 1 * * In this change 5,3 and 2 must always be there. 0, 1, 2, 4 and 6 are * all optional. If 4 are there it changes side. */ void Dbtux::treeRotateSingle(Frag& frag, NodeHandle& node, unsigned i) { ndbrequire(i <= 1); /* 5 is the old top node that have been unbalanced due to an insert or delete. The balance is still the old balance before the update. Verify that bal5 is 1 if RR rotate and -1 if LL rotate. */ NodeHandle node5 = node; const TupLoc loc5 = node5.m_loc; const int bal5 = node5.getBalance(); const int side5 = node5.getSide(); ndbrequire(bal5 + (1 - i) == i); /* 3 is the new root of this part of the tree which is to swap place with node 5. For an insert to cause this it must have the same balance as 5. For deletes it can have the balance 0. */ TupLoc loc3 = node5.getLink(i); NodeHandle node3(frag); selectNode(node3, loc3); const int bal3 = node3.getBalance(); /* 2 must always be there but is not changed. Thus we mereley check that it exists. */ ndbrequire(node3.getLink(i) != NullTupLoc); /* 4 is not necessarily there but if it is there it will move from one side of 3 to the other side of 5. For LL it moves from the right side to the left side and for RR it moves from the left side to the right side. This means that it also changes parent from 3 to 5. */ TupLoc loc4 = node3.getLink(1 - i); NodeHandle node4(frag); if (loc4 != NullTupLoc) { jam(); selectNode(node4, loc4); ndbrequire(node4.getSide() == (1 - i) && node4.getLink(2) == loc3); node4.setSide(i); node4.setLink(2, loc5); }//if /* Retrieve the address of 5's parent before it is destroyed */ TupLoc loc0 = node5.getLink(2); /* The next step is to perform the rotation. 3 will inherit 5's parent and side. 5 will become a child of 3 on the right side for LL and on the left side for RR. 5 will get 3 as the parent. It will get 4 as a child and it will be on the right side of 3 for LL and left side of 3 for RR. The final step of the rotate is to check whether 5 originally had any parent. If it had not then 3 is the new root node. We will also verify some preconditions for the change to occur. 1. 3 must have had 5 as parent before the change. 2. 3's side is left for LL and right for RR before change. */ ndbrequire(node3.getLink(2) == loc5); ndbrequire(node3.getSide() == i); node3.setLink(1 - i, loc5); node3.setLink(2, loc0); node3.setSide(side5); node5.setLink(i, loc4); node5.setLink(2, loc3); node5.setSide(1 - i); if (loc0 != NullTupLoc) { jam(); NodeHandle node0(frag); selectNode(node0, loc0); node0.setLink(side5, loc3); } else { jam(); frag.m_tree.m_root = loc3; }//if /* The final step of the change is to update the balance of 3 and 5 that changed places. There are two cases here. The first case is when 3 unbalanced in the same direction by an insert or a delete. In this case the changes will make the tree balanced again for both 3 and 5. The second case only occurs at deletes. In this case 3 starts out balanced. In the figure above this could occur if 4 starts out with a right node and the rotate is triggered by a delete of 6's only child. In this case 5 will change balance but still be unbalanced and 3 will be unbalanced in the opposite direction of 5. */ if (bal3 == bal5) { jam(); node3.setBalance(0); node5.setBalance(0); } else if (bal3 == 0) { jam(); node3.setBalance(-bal5); node5.setBalance(bal5); } else { ndbrequire(false); }//if /* Set node to 3 as return parameter for enabling caller to continue traversing the tree. */ node = node3; } /* * Double rotation about node 6. One of LR (i=0) or RL (i=1). * * 0 0 * | | * 6 ==> 4 * / \ / \ * 2 7 2 6 * / \ / \ / \ * 1 4 1 3 5 7 * / \ * 3 5 * * In this change 6, 2 and 4 must be there, all others are optional. * We will start by proving a Lemma. * Lemma: * The height of the sub-trees 1 and 7 and the maximum height of the * threes from 3 and 5 are all the same. * Proof: * maxheight(3,5) is defined as the maximum height of 3 and 5. * If height(7) > maxheight(3,5) then the AVL condition is ok and we * don't need to perform a rotation. * If height(7) < maxheight(3,5) then the balance of 6 would be at least * -3 which cannot happen in an AVL tree even before a rotation. * Thus we conclude that height(7) == maxheight(3,5) * * The next step is to prove that the height of 1 is equal to maxheight(3,5). * If height(1) - 1 > maxheight(3,5) then we would have * balance in 6 equal to -3 at least which cannot happen in an AVL-tree. * If height(1) - 1 = maxheight(3,5) then we should have solved the * unbalance with a single rotate and not with a double rotate. * If height(1) + 1 = maxheight(3,5) then we would be doing a rotate * with node 2 as the root of the rotation. * If height(1) + k = maxheight(3,5) where k >= 2 then the tree could not have * been an AVL-tree before the insert or delete. * Thus we conclude that height(1) = maxheight(3,5) * * Thus we conclude that height(1) = maxheight(3,5) = height(7). * * Observation: * The balance of node 4 before the rotation can be any (-1, 0, +1). * * The following changes are needed: * Node 6: * 1) Changes parent from 0 -> 4 * 2) 1 - i link stays the same * 3) i side link is derived from 1 - i side link from 4 * 4) Side is set to 1 - i * 5) Balance change: * If balance(4) == 0 then balance(6) = 0 * since height(3) = height(5) = maxheight(3,5) = height(7) * If balance(4) == +1 then balance(6) = 0 * since height(5) = maxheight(3,5) = height(7) * If balance(4) == -1 then balance(6) = 1 * since height(5) + 1 = maxheight(3,5) = height(7) * * Node 2: * 1) Changes parent from 6 -> 4 * 2) i side link stays the same * 3) 1 - i side link is derived from i side link of 4 * 4) Side is set to i (thus not changed) * 5) Balance change: * If balance(4) == 0 then balance(2) = 0 * since height(3) = height(5) = maxheight(3,5) = height(1) * If balance(4) == -1 then balance(2) = 0 * since height(3) = maxheight(3,5) = height(1) * If balance(4) == +1 then balance(6) = 1 * since height(3) + 1 = maxheight(3,5) = height(1) * * Node 4: * 1) Inherits parent from 6 * 2) i side link is 2 * 3) 1 - i side link is 6 * 4) Side is inherited from 6 * 5) Balance(4) = 0 independent of previous balance * Proof: * If height(1) = 0 then only 2, 4 and 6 are involved and then it is * trivially true. * If height(1) >= 1 then we are sure that 1 and 7 exist with the same * height and that if 3 and 5 exist they are of the same height as 1 and * 7 and thus we know that 4 is balanced since newheight(2) = newheight(6). * * If Node 3 exists: * 1) Change parent from 4 to 2 * 2) Change side from i to 1 - i * * If Node 5 exists: * 1) Change parent from 4 to 6 * 2) Change side from 1 - i to i * * If Node 0 exists: * 1) previous link to 6 is replaced by link to 4 on proper side * * Node 1 and 7 needs no changes at all. * * Some additional requires are that balance(2) = - balance(6) = -1/+1 since * otherwise we would do a single rotate. * * The balance(6) is -1 if i == 0 and 1 if i == 1 * */ void Dbtux::treeRotateDouble(Frag& frag, NodeHandle& node, unsigned i) { TreeHead& tree = frag.m_tree; // old top node NodeHandle node6 = node; const TupLoc loc6 = node6.m_loc; // the un-updated balance const int bal6 = node6.getBalance(); const unsigned side6 = node6.getSide(); // level 1 TupLoc loc2 = node6.getLink(i); NodeHandle node2(frag); selectNode(node2, loc2); const int bal2 = node2.getBalance(); // level 2 TupLoc loc4 = node2.getLink(1 - i); NodeHandle node4(frag); selectNode(node4, loc4); const int bal4 = node4.getBalance(); ndbrequire(i <= 1); ndbrequire(bal6 + (1 - i) == i); ndbrequire(bal2 == -bal6); ndbrequire(node2.getLink(2) == loc6); ndbrequire(node2.getSide() == i); ndbrequire(node4.getLink(2) == loc2); // level 3 TupLoc loc3 = node4.getLink(i); TupLoc loc5 = node4.getLink(1 - i); // fill up leaf before it becomes internal if (loc3 == NullTupLoc && loc5 == NullTupLoc) { jam(); if (node4.getOccup() < tree.m_minOccup) { jam(); unsigned cnt = tree.m_minOccup - node4.getOccup(); ndbrequire(cnt < node2.getOccup()); nodeSlide(node4, node2, cnt, i); ndbrequire(node4.getOccup() >= tree.m_minOccup); ndbrequire(node2.getOccup() != 0); } } else { if (loc3 != NullTupLoc) { jam(); NodeHandle node3(frag); selectNode(node3, loc3); node3.setLink(2, loc2); node3.setSide(1 - i); } if (loc5 != NullTupLoc) { jam(); NodeHandle node5(frag); selectNode(node5, loc5); node5.setLink(2, node6.m_loc); node5.setSide(i); } } // parent TupLoc loc0 = node6.getLink(2); NodeHandle node0(frag); // perform the rotation node6.setLink(i, loc5); node6.setLink(2, loc4); node6.setSide(1 - i); node2.setLink(1 - i, loc3); node2.setLink(2, loc4); node4.setLink(i, loc2); node4.setLink(1 - i, loc6); node4.setLink(2, loc0); node4.setSide(side6); if (loc0 != NullTupLoc) { jam(); selectNode(node0, loc0); node0.setLink(side6, loc4); } else { jam(); frag.m_tree.m_root = loc4; } // set balance of changed nodes node4.setBalance(0); if (bal4 == 0) { jam(); node2.setBalance(0); node6.setBalance(0); } else if (bal4 == -bal2) { jam(); node2.setBalance(0); node6.setBalance(bal2); } else if (bal4 == bal2) { jam(); node2.setBalance(-bal2); node6.setBalance(0); } else { ndbrequire(false); } // new top node node = node4; }