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.*/voidDbtux::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 * */voidDbtux::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;}