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public HashEntry findEntry(String k) { HashEntry p; /* ----------------- Start at "head" ----------------- */ p = head; /* ========================================= Main search loop ========================================= */ while ( true ) { /* -------------------------------------------- Search RIGHT until you find a LARGER entry -------------------------------------------- */ while ( (p.right.key) != HashEntry.posInf && (p.right.key).compareTo(k) <= 0 ) { p = p.right; } /* --------------------------------- Go down one level if you can... --------------------------------- */ if ( p.down != null ) { p = p.down; } else break; // We reached the LOWEST level... Exit... } return(p); // Nearest p (<= k) } |
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Prelimary result of the run time of the get(k) operation:
# entries visited = # entries visited in level h-1 + # entries visited in level h-2 + .... + # entries visited in level 0 |
So the average running time of the get(k) operation is apprimately equal to:
Avg run time get(k) ~= Avg(h) + Avg(h) × ( avg # right traversals on one level ) ....... (1) |
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Case Frequency Cost -------+-------------+-------- 1 f1 C1 2 f2 C2 .... n fn Cn |
The average cost is equal to:
Avg Cost = f1×C1 + f2×C2 + .... + fn×Cn |
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Goodrich resorts to an approximation....
by looking for:
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n Ҏ[ Skip list has height 3log(n) ] = --- 23log(n) n = --- 2log(n3) n = --- n3 1 = --- n2 |
n Ҏ[ Skip list has height 2log(n) ] = --- 22log(n) n = --- 2log(n2) n = --- n2 1 = --- n |
n Ҏ[ Skip list has height log(n) ] = --- 2log(n) n = --- n = 1 |
That's huge for a probability (it's approximately "certain") !!!
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One down, one more to go !!!.
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Example:
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Note:
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Conclussion:
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How likely is the next tower that you visit has height = i
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So only towers of height ≥ i are in layer i.
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Therefore, the average number consecutive towers of height i in layer i can be computed as follows:
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Avg # right moves = 0 * (0.5) + 1*(0.5)2 + 2*(0.5)3 ... = 1 ................... (2) |
We can use Mathematica to compute this sum:
>> math In[0]:= Sum[ (k-1)*(1/2)^k, {k,1,Infinity}] Out[0]= 1 |
run time get(k) ~= h + h×( # right traversals in 1 level ) ....... (1) (h ~= log(n)) ~= log(n) + log(n)×( # right traversals in 1 level ) ....... (2) (# right traversal in 1 level = # compare op in 1 level) ~= log(n) + log(n)×( 1 ) ~= 2×log(n) ........ (3) |
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