RED-BLACK TREE SEARCH

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RED-BLACK TREE SEARCH THE FOLLOWING METHOD IS IN
tree.h OF THE HEWLETT-PACKARD IMPLEMENTATION
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IMPLEMENTATION OF THE hash_map CLASS
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TIME ESTIMATES LET n count, LET m
length. MAKE THE UNIFORM HASHING ASSUMPTION AND
ASSUME THAT count lt length 0.75.
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THE AVERAGE SIZE OF EACH LIST IS n / m
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FOR THE find METHOD, averageTimeS(n, m) ? n /
2m iterations.
lt 0.75 / 2 SO averageTimeS(n, m) lt A
CONSTANT. averageTimeS(n, m) IS CONSTANT.
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SOLUTION bool marked_for_removal THE
CONSTRUCTOR SETS EACH buckets
marked_for_removal FIELD TO false. insert SETS
marked_for_removal TO false erase SETS
marked_for_removal TO true. SO AFTER THE
INSERTIONS
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THIS SOLUTION LEADS TO ANOTHER PROBLEM SUPPOSE
length 203. insert erase // what was just
inserted (insert AND erase A TOTAL OF
202 TIMES) insert count 1, SO THERE IS NO NEED
TO EXPAND AND REHASH. BUT THERE ARE 202
MARKED-FOR-REMOVALS!
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max 202, min 0, average
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SOLUTION KEEP TRACK OF REMOVALS int count_plus
// count number of removals since last
rehashing void check_for_expansion()
if (count_plus gt int (MAX_RATIO
length)) // Do
not copy the marked_for_removals.
delete temp_buckets
count_plus count // doubling
buckets size // method check_for_expansion
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CLUSTER A SEQUENCE OF NON-EMPTY LOCATIONS
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SOLUTION DOUBLE HASHING, THAT IS, OBTAIN BOTH
INDICES AND OFFSETS BY HASHING   unsigned
long hash_int hash (key) int index hash_int
length, offset hash_int / length NOW THE
OFFSET DEPENDS ON THE KEY, SO DIFFERENT KEYS
WILL USU- ALLY HAVE DIFFERENT OFFSETS, SO NO
MORE PRIMARY CLUSTERING!
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TO GET A NEW INDEX index (index offset)
length
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EXAMPLE length 11 key index
offset 15 4 1 19 8 1 16 5 1 58
3 5 27 5 2 35 2 3 30 8 2 47 3 4 WHERE
WOULD THESE KEYS GO IN buckets?
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index key 0 47 1 2 35 3 58
4 15 5 16 6 7 27 8 19 9 10 30
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PROBLEM WHAT IF OFFSET IS MULTIPLE OF
length? EXAMPLE length 11 key
index offset 15
4 1 19 8 1 16 5 1 58 3 5 27 5 2 35 2 3
47 3 4 246 4 22 // BUT 15 IS AT INDEX
4 FOR KEY 246, NEW INDEX (4 22) 11 4.
OOPS!
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SOLUTION if (offset length 0)
offset 1 ON AVERAGE, offset length
WILL EQUAL 0 ONLY ONCE IN EVERY length TIMES.
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FINAL PROBLEM WHAT IF length HAS SEVERAL
FACTORS? EXAMPLE length 20 key
index offset 20 0
1 25 5 1 30 10 1 35
15 1 110 10 5 // BUT 30 IS AT
INDEX 10 FOR KEY 110, NEW INDEX (10 5) 20
15, WHICH IS OCCUPIED, SO NEW INDEX (15 5)
20, WHICH IS OCCUPIED, SO NEW INDEX ...
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SOLUTION MAKE length A PRIME.
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