Editing Persistent data structure (section)

===Efficiency of the generalized persistent data structure===
In order to find the efficiency of the scheme proposed above, we use an argument defined as a credit scheme. The credit represents a currency. For example, the credit can be used to pay for a table. The argument states the following:
*  The creation of one table requires one credit
*  Each call to CREATE-NODE comes with two credits
*  Each call to CHANGE-EDGE comes with one credit
The credit scheme should always satisfy the following invariant: Each row of each active table stores one credit and the table has the same number of credits as the number of rows. Let us confirm that the invariant applies to all the three operations CREATE-NODE, CHANGE-EDGE and CHANGE-LABEL.
*CREATE-NODE: It acquires two credits, one is used to create the table and the other is given to the one row that is added to the table. Thus the invariant is maintained.
*CHANGE-EDGE: There are two cases to consider. The first case occurs when there is still at least one empty row in the table. In this case one credit is used to the newly inserted row. The second case occurs when the table is full. In this case the old table becomes inactive and the <math>d+1</math> credits are transformed to the new table in addition to the one credit acquired from calling the CHANGE-EDGE. So in total we have <math>d+2</math> credits. One credit will be used for the creation of the new table. Another credit will be used for the new row added to the table and the {{mvar|d}} credits left are used for updating the tables of the other vertices that need to point to the new table. We conclude that the invariant is maintained.
*CHANGE-LABEL: It works exactly the same as CHANGE-EDGE.
As a summary, we conclude that having <math>n_{1}</math> calls to CREATE_NODE and <math>n_{2}</math> calls to CHANGE_EDGE will result in the creation of <math>2\cdot n_{1}+n_{2}</math> tables. Since each table has size <math>O(d)</math> without taking into account the recursive calls, then filling in a table requires <math>O(d^{2})</math> where the additional d factor comes from updating the inedges at other nodes. Therefore, the amount of work required to complete a sequence of operations is bounded by the number of tables created multiplied by <math>O(d^{2})</math>. Each access operation can be done in <math>O(\log(d))</math> and there are {{mvar|m}} edge and label operations, thus it requires <math>m\cdot O(\log(d))</math>. We conclude that There exists a data structure that can complete any {{mvar|n}} sequence of CREATE-NODE, CHANGE-EDGE and CHANGE-LABEL in <math>O(n\cdot d^{2})+m\cdot O(\log(d))</math>.