Source: 3-PARTITION (ThreePartition)
Target: DYNAMIC STORAGE ALLOCATION (DynamicStorageAllocation)
Motivation: Establishes NP-completeness of DYNAMIC STORAGE ALLOCATION by reduction from 3-PARTITION. The key insight is that partitioning 3m integers into m triples that each sum to B can be encoded as a storage allocation problem: each integer becomes a memory block, and each time slot forces exactly three blocks to fit into memory of size B simultaneously.
Reference: Garey & Johnson, Computers and Intractability, SR2, p.226.
GJ Source Entry
[SR2] DYNAMIC STORAGE ALLOCATION
INSTANCE: Set of items I = {i_1, ..., i_n}, for each i_j a size s(i_j) ∈ Z^+, an arrival time a(i_j) ∈ Z_0^+ and a departure time d(i_j) ∈ Z^+ with a(i_j) < d(i_j), and a storage capacity D ∈ Z^+.
QUESTION: Can items be assigned starting addresses so that no two items with overlapping time intervals share memory locations and all items fit within capacity D?
Reference: Transformation from 3-PARTITION.
Reduction Algorithm
Given a ThreePartition instance with 3m elements A = {a_1, ..., a_{3m}} where each a_i has size s(a_i) with B/4 < s(a_i) < B/2 and sum = m*B, construct a DynamicStorageAllocation instance as follows:
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Items: Create 3m items, one per element. Item i_j has size s(i_j) = s(a_j).
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Time windows: All 3m items share the same time interval: arrival time a(i_j) = 0, departure time d(i_j) = 1.
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Memory capacity: D = B.
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Wait — the above puts all items in one time slot, which cannot work since sum = m*B > B for m > 1. The correct G&J construction uses m time slots:
Corrected construction: Create 3m items, one per element. All items are alive during all m time slots:
- Actually, the standard construction assigns items to staggered time windows to force a grouping. The exact G&J SR2 approach:
- Create 3m items with sizes s(a_j)
- Create m time slots [0,1), [1,2), ..., [m-1,m)
- All items are alive for the entire duration [0, m), so they must all fit simultaneously
This forces items to be packed into D = m*B memory with non-overlapping address ranges. But the 3-partition constraint comes from a more structured assignment.
Standard reduction (Garey & Johnson): The correct reduction uses "filler items" to create m time slots:
- For each time slot t (t = 0, ..., m-1), create 3 "element items" that are alive only during [t, t+1)
- Each element must be assigned to exactly one time slot (this is the partition)
- Memory size D = B forces each time slot to hold items summing to at most B
- Since the total sum is m*B and there are m slots, each slot must sum to exactly B
Simplified direct encoding:
- Items: 3m items, item j has size s(a_j), arrival = 0, departure = m
- Memory size D = m * B
- This is a valid storage allocation instance, but it does not capture the 3-partition constraint directly.
Note: The precise G&J construction for SR2 involves a more subtle gadgetry. The following is the most commonly cited formulation:
Construction (standard)
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Element items: For j = 1, ..., 3m, create item i_j with size s(a_j), arrival time a(i_j) = 0, departure time d(i_j) = m.
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Separator items: For t = 1, ..., m-1, create a separator item sep_t with size D, arrival time a(sep_t) = t - epsilon, departure time d(sep_t) = t + epsilon. (In the integer formulation: arrival = t, departure = t+1, with appropriate discretization.)
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Memory capacity: D = B.
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Correctness: The separator items at each boundary force element items to be "grouped" into time-slot regions. Since each separator occupies the full width D = B, element items cannot straddle a separator boundary. This partitions the 3m items into m groups, each fitting within height B. The 3-partition size bounds (B/4 < s < B/2) ensure exactly 3 items per group.
Solution extraction
From a valid storage allocation, identify which group of 3 elements share each time-slot region. This gives the 3-partition.
Implementation note on construction
Codex verification needed: The derivation's construction requires careful handling of how separators force grouping. The integer time discretization may need adjustment. The implementation should verify the reduction with small instances before trusting the overhead formulas.
Size Overhead
| Target metric (code name) |
Expression |
num_items |
num_elements |
memory_size |
bound |
Note: If separator items are needed, num_items = num_elements + num_groups - 1. The exact overhead depends on the final construction choice.
Implementation Note
DynamicStorageAllocation currently has 0 outgoing reductions (no ILP solver). A DynamicStorageAllocation → ILP rule should be implemented alongside this rule:
- Integer variable addr_j ∈ {0, ..., D - s(i_j)} per item (starting address)
- For each pair (j, k) with overlapping time intervals: either addr_j + s(i_j) ≤ addr_k or addr_k + s(i_k) ≤ addr_j (non-overlap), encoded with big-M binary indicators
- Feasibility objective (Value = Or)
- Overhead:
num_vars = "num_items", constraints = O(num_items^2)
Codex verification needed: The derivation warns that the construction may require an oracle grouping map. The correct G&J construction may differ in details from the sketch above. Verify with concrete instances.
Example
Source (ThreePartition):
A = {4, 5, 6, 4, 6, 5, 3, 5, 7}, m = 3, B = 15
- 9 elements, 3 groups needed
- Each element satisfies B/4 = 3.75 < s(a_i) < B/2 = 7.5 (all elements in range [3, 7] ✓)
- Total sum = 4+5+6+4+6+5+3+5+7 = 45 = 3·15 ✓
Valid 3-partition:
- Group 1: {4, 4, 7} → sum = 15 ✓
- Group 2: {5, 5, 5} → sum = 15 ✓
- Group 3: {6, 6, 3} → sum = 15 ✓
Target (DynamicStorageAllocation):
- 9 items with sizes [4, 5, 6, 4, 6, 5, 3, 5, 7]
- Memory size D = 15
- All items alive during [0, 3) (arrival=0, departure=3)
Solution mapping: Each group's items are stacked vertically in one time-slot region:
- Slot 0: items with sizes 4, 4, 7 → addresses 0, 4, 8 (total height 15)
- Slot 1: items with sizes 5, 5, 5 → addresses 0, 5, 10 (total height 15)
- Slot 2: items with sizes 6, 6, 3 → addresses 0, 6, 12 (total height 15)
No-instance check: If we tried A = {4, 5, 6, 4, 6, 5, 3, 5, 8} with B = 15, sum = 46 ≠ 45, so this is not a valid 3-partition instance. For a valid no-instance: A = {1, 1, 1, 5, 5, 5, 5, 5, 5}, B = 9 — here each triple must sum to 9, but the three 1s cannot pair with the six 5s to make triples of sum 9 (1+5+5=11 > 9). The storage allocation would also fail since no valid packing exists.
Validation Method
- Closed-loop test: reduce ThreePartition to DynamicStorageAllocation, solve target via ILP (DynamicStorageAllocation → ILP), extract grouping from address assignments, verify each group sums to B
- Test with both satisfiable and unsatisfiable instances
- Verify item counts and memory size match overhead formulas
References
- Garey, M. R., & Johnson, D. S. (1979). Computers and Intractability, SR2, p.226.
Source: 3-PARTITION (ThreePartition)
Target: DYNAMIC STORAGE ALLOCATION (DynamicStorageAllocation)
Motivation: Establishes NP-completeness of DYNAMIC STORAGE ALLOCATION by reduction from 3-PARTITION. The key insight is that partitioning 3m integers into m triples that each sum to B can be encoded as a storage allocation problem: each integer becomes a memory block, and each time slot forces exactly three blocks to fit into memory of size B simultaneously.
Reference: Garey & Johnson, Computers and Intractability, SR2, p.226.
GJ Source Entry
Reduction Algorithm
Given a
ThreePartitioninstance with 3m elements A = {a_1, ..., a_{3m}} where each a_i has size s(a_i) with B/4 < s(a_i) < B/2 and sum = m*B, construct aDynamicStorageAllocationinstance as follows:Items: Create 3m items, one per element. Item i_j has size s(i_j) = s(a_j).
Time windows: All 3m items share the same time interval: arrival time a(i_j) = 0, departure time d(i_j) = 1.
Memory capacity: D = B.
Wait — the above puts all items in one time slot, which cannot work since sum = m*B > B for m > 1. The correct G&J construction uses m time slots:
Corrected construction: Create 3m items, one per element. All items are alive during all m time slots:
This forces items to be packed into D = m*B memory with non-overlapping address ranges. But the 3-partition constraint comes from a more structured assignment.
Standard reduction (Garey & Johnson): The correct reduction uses "filler items" to create m time slots:
Simplified direct encoding:
Note: The precise G&J construction for SR2 involves a more subtle gadgetry. The following is the most commonly cited formulation:
Construction (standard)
Element items: For j = 1, ..., 3m, create item i_j with size s(a_j), arrival time a(i_j) = 0, departure time d(i_j) = m.
Separator items: For t = 1, ..., m-1, create a separator item sep_t with size D, arrival time a(sep_t) = t - epsilon, departure time d(sep_t) = t + epsilon. (In the integer formulation: arrival = t, departure = t+1, with appropriate discretization.)
Memory capacity: D = B.
Correctness: The separator items at each boundary force element items to be "grouped" into time-slot regions. Since each separator occupies the full width D = B, element items cannot straddle a separator boundary. This partitions the 3m items into m groups, each fitting within height B. The 3-partition size bounds (B/4 < s < B/2) ensure exactly 3 items per group.
Solution extraction
From a valid storage allocation, identify which group of 3 elements share each time-slot region. This gives the 3-partition.
Implementation note on construction
Codex verification needed: The derivation's construction requires careful handling of how separators force grouping. The integer time discretization may need adjustment. The implementation should verify the reduction with small instances before trusting the overhead formulas.
Size Overhead
num_itemsnum_elementsmemory_sizeboundNote: If separator items are needed,
num_items=num_elements + num_groups - 1. The exact overhead depends on the final construction choice.Implementation Note
DynamicStorageAllocation currently has 0 outgoing reductions (no ILP solver). A DynamicStorageAllocation → ILP rule should be implemented alongside this rule:
num_vars = "num_items", constraints = O(num_items^2)Codex verification needed: The derivation warns that the construction may require an oracle grouping map. The correct G&J construction may differ in details from the sketch above. Verify with concrete instances.
Example
Source (ThreePartition):
A = {4, 5, 6, 4, 6, 5, 3, 5, 7}, m = 3, B = 15
Valid 3-partition:
Target (DynamicStorageAllocation):
Solution mapping: Each group's items are stacked vertically in one time-slot region:
No-instance check: If we tried A = {4, 5, 6, 4, 6, 5, 3, 5, 8} with B = 15, sum = 46 ≠ 45, so this is not a valid 3-partition instance. For a valid no-instance: A = {1, 1, 1, 5, 5, 5, 5, 5, 5}, B = 9 — here each triple must sum to 9, but the three 1s cannot pair with the six 5s to make triples of sum 9 (1+5+5=11 > 9). The storage allocation would also fail since no valid packing exists.
Validation Method
References