COMP 7720 - Online Algorithms
Online Bin Packing
Shahin Kamali
Lecture 22 - Nov. 23rd, 2017
University of Manitoba
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Review & Plan
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Today’s plan
Logistics
Online bin packing with advice
Online Edge-coloring
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Logistics
Assignment 3 is posted
It is due on Saturday November 25th (8 pm).
Project presentation day is on Thursday 30th (next Thursday)
Present the problem and partial results/insights that you have5-7 minutes for individual projects and 7-9 minutes for projects of 2students
Final exam is on Thursday, December 7th (last day of class).
Project final report is due December21st.
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ReserveCritical Algorithm
At the beginning, reserve a space of size 2/3 for critical items
huge items: open a new bin (no other item goes there)critical items: place in a reserve spacemini item: place two of them in the same bin (no other item goesthere)tiny items: apply FF to place in bins with critical or other tiny items
σ = 〈 0.3 0.9 0.6 0.5 0.1 0.1 0.56 0.4 0.3 0.45 0.8 0.51 0.41 0.2 0.1 0.37 0.3 〉
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ReserveCritical algorithm
Theorem
Competitive ratio of ReserveCritical is at most 1.5.
With O(log n) bits of advice, one can achieve a competitive ratio of1.5
Can we improve this?
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RedBlue Algorithm (sketch)
Instead of receiving the number of critical items in O(log n) bits,receive the ratio between critical and tiny bins in the final packingof ReserveCritical
Treat Huge and mini items as before
Place a critical item in the reserved space of a critical bin; if noreserve space exists, open a new bin and declare it as critical
Place a tiny item in non-reserved space of critical bins (using FF)
If no such bin exists, open a new binDeclare the new bin to be a critical or a tiny bin so that the ratiobetween the number of these bins becomes closer to the ratioreceived in advice
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RedBlue Algorithm (sketch)
If the ratio between critical and tiny bins is encoded using k bits ofadvice, RedBlue algorithm has a competitive ratio of at most1.5 + 15
2k/2+1
Theorem
With constant number of bits of advice, one can achieve a com-petitive ratio of (almost) 1.5.
Can we do better?
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Power of Advice of Constant Size
In fact, with a more complicated argument, we can show that withadvice of constant size, one can achieve a competitive ratio of 1.47
Idea: pack items of size larger than 1/3 separately from the rest.
How advice can help in packing items of size larger than 1/3?
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Power of Advice of Constant Size
It is often useful to think of algorithms that ‘complement’ each other
Assume all items are larger than 1/3:
Sbf: All small items (< 1/2) are packed according to BestFit, andeach large item (≥ 1/2) is placed in a new bin.Lbf: All large items are packed according to BestFit, and each smallitem is placed in a new bin.
σ = 〈0.45 0.6 0.75 0.34 0.40 0.56 0.35 0.55 0.50〉
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Power of Advice of Constant Size
Theorem
When all items are larger than 1/3, the better algorithm amongSbf and Lbf has a competitive ratio of 1.39.
With only one bit of advice, one can achieve a competitive ratio of1.39 (when all items are larger than 1/3).
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Lower bound
Advice of size Ω(n) is required to achieve an algorithm with c.r.≤ 9/8
A reduction from binary guessing problem
Consider
σ = 〈0.5 + ε, . . . 0.5 + ε,︸ ︷︷ ︸m green items
a1, a2, a3 . . . , a2m,︸ ︷︷ ︸white items in range(1/3,1/2]
b1, . . . bm︸ ︷︷ ︸complements of smaller white items
〉
The algorithm should ‘guess’ whether each white item is among thelarger half of smaller half of white items!
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Lower bound〈0.51, . . . , 0.51,︸ ︷︷ ︸
m green items
0.42, 0.37, 0.4, 0.39, 0.38, 0.385, 0.388, 0.386︸ ︷︷ ︸2m white items
, 0.63, 0.62, 0.615, 0.614︸ ︷︷ ︸red complements
〉
Guess if an item is among smaller or larger half of white items
open a new bin for smaller half of white items (in anticipation oftheir complements coming in the future)for the larger half of white items, put them with green items
The ‘type’ (being in smaller or larger half) of the white item cannotbe revealed from knowing types of previous white items
For any four mistakes in guessing, at least 1 extra bin is opened
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Lower bound
Theorem
In order to achieve a competitive ratio better than 9/8, adviceof linear size is required
This result can be improved to show that for a competitive ratiobetter than 4− 2
√2 ≈ 1.172, a linear number of bits are required.
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General Picture for Advice Complexity of BinPacking
With Θ(n logN) bits, one can achieve an optimal solutionBoyKamSTACS.
With Θ(log n) bits, one can achieve a competitive ratio of 1.5(better than all online algorithms)
With linear number bits, one can achieve a competitive ratio of 4/3
For a competitive ratio better than 9/8, a linear number of bits arerequired BoyKamSTACS.
With linear number bits, one can achieve a competitive ratio of 1.0
With k ≥ 4 bits, one can get a competitive ratio of 1.5 + 152k/2+1
With Θ(1) bits, one can get a competitive ratio of 1.4702
For a competitive ratio better than 7/6, a linear number of bits arerequired
For a competitive ratio better than 4− 2√
2 ≈ 1.172, a linearnumber of bits are required
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General Picture for Advice Complexity of BinPacking
12
adv
ice
siz
e
56
78
10
4
9
advice size
competitive ratio
1.7
1.58171.5403
1.172
7/6 1.666
9/8 = 1.125
1.5
1.47012
Θ (log n) Θ (n)
4/3 1.333
Ω (n) Θ (n log N)16big
constant
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Online Edge Coloring in Graphs
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Problem Definition
In edge coloring, the goal is to color edges of a graph with minimumnumber of colors
No two adjacent edges (edges sharing an endpoint) should have thesame color
In the offline setting, the problem is NP-hard!
For a graph of degree ∆, at least ∆ and at most ∆ + 1 colors arerequired (Vizing theorem)
this implies that cost(Opt) ≈ ∆
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Problem Definition
In the online setting, edges arrive one by one, and an algorithmshould take an irrevocable decision on coloring the edges withoutany knowledge about future edges (or how graph looks).
For example, Greedy family of algorithms maintain a set of colorsand use them, if possible, before requesting a new coloring
cost of Opt is 3
Cost of Greedy is 4, which is not optimal
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Greedy Algorithm
For a graph of degree ∆, a Greedy algorithm uses at most 2∆− 1colors.
Assume otherwise, so, there is an edge e with color 2∆.Assume the two endpoints of e have degrees a and b before addinge (we have a, b < ∆).Greedy uses at most a and b distinct colors for edges adjacent to e.So, e can be colored using color a + b + 1 < 2∆ → contradiction.
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Greedy algorithm
Theorem
Greedy has a competitive ratio of at most 2.
For any graph of degree ∆, cost of Opt is at least ∆.
Cost of greedy is at most 2∆− 1.
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Lower Bound
Theorem
No deterministic online algorithm can have a competitive ratiobetter than 2.
This implies that greedy algorithms are the best deterministicalgorithm
We see the proof in the next class
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