This document discusses horizontal fragmentation of database tables. It begins by outlining the basic requirements for horizontal fragmentation, including identifying simple predicates, minterm predicates, minterm selectivity, and access frequencies. It then provides an example PROJECT table and identifies relevant predicates. The document defines minterms and outlines an algorithm for determining minterms for a table. It walks through applying the algorithm to the example PROJECT table to identify the actual horizontal partitions. The summary identifies the key steps and concepts discussed in fragmenting tables horizontally.

1. Unit – 4
Data Distribution Alternatives
Fragmentation

2. Horizontal Fragmentation
• Basic Requirement of Horizontal Fragmentation
1. Find out simple predicate Pr
2. Find out Minterm predicates M
3. Minterm selectivity sel(mi)
4. Access Frequencies acc(qi)
5. Access Frequencies Minterm acc(qi)
1/11/2017 2Prof. Dhaval R. Chandarana

3. Fragmentation Examples
• Example PROJECT table:
• PROJ:
PNO PNAME BUDGET LOC
P1 Instrumentation 150000 Montreal
P2 Database Develop 135000 New York
P3 CAD/CAM 250000 New York
P4 Maintenance 310000 Paris
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4. Predicates
• Predicates
• Appear in the WHERE clause of a query
• Important determiner of fragmentation
• Determine the composition of table fragments
• Let R(A1, A2, …, An) be a relation
• Ai is defined over a domain Di
• We say pi is a simple predicate if it is of the form
• Pi ʘ Value where ʘ ԑ { =, <, >, , , <> }
• Examples:
• PNAME = 'CAD/CAM'
• BUDGET > 200000
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5. Predicates
• Usually, multiple predicates are necessary to describe a selection of
rows in a relation
• Most Boolean combinations can be translated into conjunctive normal
form
• p1 ^ p2 ^ …^ pk
• We attempt to fragment tables according to selection (WHERE clause)
patterns
• A combination of predicates in conjunctive normal form is a Minterm
• Let a set of predicates on a relation be:
Pr = {p1, p2, …, pk }
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6. Minterms
• Let set of minterm predicates be
M = { m1, m2, …, mz }
where M = {mj | mj = ^(pn ԑ Pr) pn}
• Some property equivalences:
• For equality: !(attr = val) = (attr <> val)
• For inequality: !(attr > val) = (attr val)
• It is not necessary to duplicate predicates
• In minterms, one is sufficient
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7. Minterm Examples
• p1: LOC = 'Montreal'
• p2: LOC = 'New York'
• p3: LOC = 'Paris'
• p4: BUDGET > 200000
• p5: BUDGET <= 200000
• m1: LOC = 'New York' ^ BUDGET > 200000
• m2: LOC = 'New York' ^ BUDGET <= 200000
• m3: LOC = 'Paris' ^ BUDGET > 200000
• m4: LOC = 'Paris' ^ BUDGET <= 200000
• m5: LOC = 'Montreal' ^ BUDGET > 200000
• m6: LOC = 'Montreal ' ^ BUDGET <= 200000
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8. Minterm Properties
• Minterm selectivity
• Number of records that satisfy minterm
• sel(m1) = 1; sel(m2) = 1; sel(m4) = 0
• Access frequency by applications and users
• Q = {q1, q2, …, qq} is set of queries
• acc(q1) is frequency of access of query 1
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9. Primary Horizontal Fragmentation
• Using minterms and access frequency, one can generate a horizontal
fragmentation
• Suppose there are w fragments
• Then each relation fragment Ri is given by a formula Fi, where each
formula represents a minterm expression of predicates
Ri = ϭ Fi(R) where 1 <= i <= w
• Examples:
• PROJ1 = BUDGET<=200000 (PROJ)
• PROJ2 = BUDGET>200000 (PROJ)
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10. Algorithm for Determining Minterms
• Rule 1: fragment is partitioned into at least two parts that are
accessed differently by at least one application
• Definitions
• R - relation
• Pr - set of simple predicates
• Pr' - another set of simple predicates
• F - set of minterm fragments
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11. Algorithm for Determining Minterms
• Define set of inferences from the predicates
• Assume val1 and val2 are complimentary and complete the set of values:
• p1: att = val1
• p2: att = val2
• i1: (att = val1) => !(att = val2)
• i2: (att = val2) => !(att = val1)
• set of possible minterms
• m1: (att = val1) ^ (att = val2)
• m2: (att = val1) ^ !(att = val2)
• m3: !(att = val1) ^ (att = val2)
• m4: !(att = val1) ^ !(att = val2)
• m1 and m4 cannot be minterms because they contradict inferences
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12. Calculate Minterms for Table
PHORIZONTAL {
Pr' = COM_MIN(R, Pr)
determine set of minterms M
determine inference set I among Pr'
eliminate contradictory mi's according to I from M
eliminate subsumed minterms
what is left in M is horizontal fragmentation
}
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13. Example
Step 1: Identify relevant predicates
• p1: LOC = 'Montreal'
• p2: LOC = 'New York'
• p3: LOC = 'Paris'
• p4: BUDGET > 200000
• p5: BUDGET <= 200000
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14. Define Full Minterm Set
• m1: LOC = ‘Montreal’
• m2: LOC = ‘New York’
• m3: LOC = ‘Paris’
• m4: BUDGET > 200000
• m5: BUDGET <= 200000
• m6: LOC = ‘Montreal’ ^ LOC = ‘New York’
• m7: LOC = ‘Montreal’ ^ LOC = ‘Paris’
• m8: LOC = ‘Montreal’ ^ BUDGET > 200000
• m9: LOC = ‘Montreal’ ^ BUDGET <= 200000
• m10: LOC = ‘New York’ ^ LOC = ‘Paris’
• m11: LOC = ‘New York’ ^ BUDGET > 200000
• m12: LOC = ‘New York’ ^ BUDGET <= 200000
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15. Define Full Minterm Set
• m13: LOC = ‘Paris’ ^ BUDGET > 200000
• m14: LOC = ‘Paris’ ^ BUDGET <= 200000
• m15: BUDGET > 200000 ^ BUDGET <= 200000
• m16: LOC = ‘Montreal’ ^ LOC = ‘New York’ ^ LOC = ‘Paris’
• m17: LOC = ‘Montreal’ ^ LOC = ‘New York’ ^ BUDGET > 200000
• m18: LOC = ‘Montreal’ ^ LOC = ‘New York’ ^ BUDGET <= 200000
• m19: LOC = ‘Montreal’ ^ LOC = ‘Paris’ ^ BUDGET > 200000
• m20: LOC = ‘Montreal’ ^ LOC = ‘Paris’ ^ BUDGET <= 200000
• m21: LOC = ‘Montreal’ ^ BUDGET > 200000 ^ BUDGET <= 200000
• m22: LOC = ‘New York’ ^ LOC = ‘Paris’ ^ BUDGET > 200000
• m23: LOC = ‘New York’ ^ LOC = ‘Paris’ ^ BUDGET <= 200000
• m24: LOC = ‘New York’ ^ BUDGET > 200000 ^ BUDGET <= 200000
• m25: LOC = ‘Paris’ ^ BUDGET > 200000 ^ BUDGET <= 200000
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16. Define Full Minterm Set
• m26: LOC = ‘Montreal’ ^ LOC = ‘New York’ ^ LOC = ‘Paris’ ^ BUDGET >
200000
• m27: LOC = ‘Montreal’ ^ LOC = ‘New York’ ^ LOC = ‘Paris’ ^ BUDGET <=
200000
• m28: LOC = ‘Montreal’ ^ LOC = ‘New York’ ^ BUDGET > 200000 ^ BUDGET
<= 200000
• m29: LOC = ‘Montreal’ ^ LOC = ‘Paris’ ^ BUDGET > 200000 ^ BUDGET <=
200000
• m30: LOC = ‘New York’ ^ LOC = ‘Paris’ ^ BUDGET > 200000 ^ BUDGET <=
200000
• m31: LOC = ‘Montreal’ ^ LOC = ‘New York’ ^ LOC = ‘Paris’ ^ BUDGET >
200000 ^ BUDGET <= 200000
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17. Define Inferences
• Inferences:
• p1 => ~p2 p3 => ~p1
• p1 => ~p3 p3 => ~p2
• p2 => ~p1 p4 => ~p5
• p2 => ~p3 p5 => ~p4
• Left with only:
• m1: LOC = ‘Montreal’ m8: LOC = ‘Montreal’ ^ BUDGET > 200000
• m2: LOC = ‘New York’ m9: LOC = ‘Montreal’ ^ BUDGET <= 200000
• m3: LOC = ‘Paris’ m12: LOC = ‘New York’ ^ BUDGET <= 200000
• m4: BUDGET > 200000 m13: LOC = ‘Paris’ ^ BUDGET > 200000
• m5: BUDGET <= 200000 m14: LOC = ‘Paris’ ^ BUDGET <= 200000
• After subsumption, only m8, m9, m11, m12, m13, m14 remain
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18. Actual Partitions
• The four actual partitions are: m9, m11, m12, m13
• The two partitions m8 and m14 have no data
PNO PNAME BUDGET LOC
P1 Instrumentation 150000 Montreal
P2 Database Develop 135000 New York
P3 CAD/CAM 250000 New York
P4 Maintenance 310000 Paris
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