An Improved FPTAS for 0-1 Knapsack Ce Jin Tsinghua University 0-1 - - PowerPoint PPT Presentation

An Improved FPTAS for 0-1 Knapsack

Ce Jin Tsinghua University

𝑜 items Each item 𝑗 has weight 0 < 𝑥𝑗 ≤ 𝑋 and profit 𝑞𝑗 > 0

Given:

• 𝑥 𝐽 ≔ σ𝑗∈𝐽 𝑥𝑗 ≤ 𝑋
• 𝑞 𝐽 ≔ σ𝑗∈𝐽 𝑞𝑗 is maximized

Find a subset of items 𝐽 ⊆ 𝑜 such that:

0-1 Knapsack Problem

Knapsack capacity 𝑋 > 0

𝑜 items Each item 𝑗 has weight 0 < 𝑥𝑗 ≤ 𝑋 and profit 𝑞𝑗 > 0

Given:

• 𝑥 𝐽 ≔ σ𝑗∈𝐽 𝑥𝑗 ≤ 𝑋
• 𝑞 𝐽 ≔ σ𝑗∈𝐽 𝑞𝑗 is maximized

Find a subset of items 𝐽 ⊆ 𝑜 such that: A well-known NP-hard problem

0-1 Knapsack Problem

Knapsack capacity 𝑋 > 0

FPTAS for 0-1 Knapsack

OPT = optimal total profit 𝑞(𝐽) For 𝜁 > 0, find a subset 𝐽 ⊆ 𝑜 such that:

• 𝑥 𝐽 ≤ 𝑋
• 𝑞 𝐽 ≥

OPT 1+𝜁

Solvable in poly 𝑜,

1 𝜁 time

Prior Work

• ෨

𝑃(𝑜3/𝜁) (textbook algorithm)

• ෨

𝑃(𝑜 + 𝜁−4) [Ibarra and Kim, 1975]

• ෨

𝑃(𝑜 + 𝜁−3) [Kellerer and Pferschy, 2004]

• ෨

𝑃(𝑜 + 𝜁−2.5) [Rhee, 2015]

• ෩

𝑷(𝒐 + 𝜻−𝟑.𝟓) [Chan, 2018]

FPTAS for 0-1 Knapsack:

Prior Work

• ෨

𝑃(𝑜3/𝜁) (textbook algorithm)

• ෨

𝑃(𝑜 + 𝜁−4) [Ibarra and Kim, 1975]

• ෨

𝑃(𝑜 + 𝜁−3) [Kellerer and Pferschy, 2004]

• ෨

𝑃(𝑜 + 𝜁−2.5) [Rhee, 2015]

• ෩

𝑷(𝒐 + 𝜻−𝟑.𝟓) [Chan, 2018]

• ෩

𝑷(𝒐 + 𝜻−𝟑.𝟑𝟔) (this work) FPTAS for 0-1 Knapsack:

No 𝑃 𝑜 + 𝜁−1 2−𝜀 time algorithm,

unless 𝐧𝐣𝐨, + convolution has truly subquadratic algo [Cygan, Mucha, Węgrzycki, and Włodarczyk, 2017]

Conditional Lower Bound:

A Special Case

FPTAS for Subset Sum (𝑞𝑗 = 𝑥𝑗):

• ෨

𝑃 min 𝒐 + 𝜻−𝟑, 𝑜𝜁−1 [Kellerer, Mansini, Pferschy, and Speranza, 2003]

A Special Case

FPTAS for Subset Sum (𝑞𝑗 = 𝑥𝑗):

• ෨

𝑃 min 𝒐 + 𝜻−𝟑, 𝑜𝜁−1 [Kellerer, Mansini, Pferschy, and Speranza, 2003]

Our 0-1 Knapsack algorithm utilizes this result

• Our ෩

𝑷(𝒐 + 𝜻−𝟑.𝟑𝟔) algorithm builds on Chan’s ෩ 𝑷(𝒐 + 𝜻−𝟑.𝟓) algorithm

• Two new ideas
• 1. Extending Chan’s number-theoretic

technique from two levels to multiple levels.

• Our ෩

𝑷(𝒐 + 𝜻−𝟑.𝟑𝟔) algorithm builds on Chan’s ෩ 𝑷(𝒐 + 𝜻−𝟑.𝟓) algorithm

• Two new ideas
• 1. Extending Chan’s number-theoretic

technique from two levels to multiple levels.

• 2. A greedy argument ⇒ less computation

spent on cheap items (small unit profit 𝑞𝑗/𝑥𝑗)

• Our ෩

𝑷(𝒐 + 𝜻−𝟑.𝟑𝟔) algorithm builds on Chan’s ෩ 𝑷(𝒐 + 𝜻−𝟑.𝟓) algorithm

• Two new ideas
• 1. Extending Chan’s number-theoretic

technique from two levels to multiple levels.

• 2. A greedy argument ⇒ less computation

spent on cheap items (small unit profit 𝑞𝑗/𝑥𝑗)

This talk

• Our ෩

𝑷(𝒐 + 𝜻−𝟑.𝟑𝟔) algorithm builds on Chan’s ෩ 𝑷(𝒐 + 𝜻−𝟑.𝟓) algorithm

• Two new ideas
• 1. Extending Chan’s number-theoretic

technique from two levels to multiple levels.

• 2. A greedy argument ⇒ less computation

spent on cheap items (small unit profit 𝑞𝑗/𝑥𝑗)

This talk

Using Chan’s Lemmas as blackboxes

෩ 𝑷(𝒐 + 𝜻−𝟑.𝟒𝟒𝟒) algo

Preliminaries (based

• n [Chan, 2018])

Preliminaries

• Assume 𝑜 ≤ poly 𝜁−1 .
• Too cheap items (𝑞𝑗 <

𝜁 𝑜 max 𝑘

𝑞𝑘) are discarded at the beginning (loss ≤ 𝜁 ⋅ OPT) So

max 𝑞𝑘 min 𝑞𝑘 ≤ poly 𝜁−1

𝑜 items, capacity = 𝑋 weight 0 < 𝑥𝑗 ≤ 𝑋 and profit 𝑞𝑗 > 0

Preliminaries

𝑜 ≤ poly 𝜁−1 items profit max 𝑞𝑘/min 𝑞𝑘 ≤ poly 𝜁−1 weight 0 < 𝑥𝑗 ≤ 𝑋

• “Profit function” (defined over real 𝑦 ≥ 0)

𝑔

𝐽 𝑦 = max 𝑞 𝐾 : 𝐾 ⊆ 𝐽, 𝑥 𝐾 ≤ 𝑦

Preliminaries

𝑜 ≤ poly 𝜁−1 items profit max 𝑞𝑘/min 𝑞𝑘 ≤ poly 𝜁−1 weight 0 < 𝑥𝑗 ≤ 𝑋

• “Profit function” (defined over real 𝑦 ≥ 0)

𝑔

𝐽 𝑦 = max 𝑞 𝐾 : 𝐾 ⊆ 𝐽, 𝑥 𝐾 ≤ 𝑦

• Task: compute a 1 + 𝜁 -approximation of profit

function 𝑔

𝐽,

෩ 𝑔

𝐽 𝑦 ≤ 𝑔 𝐽 𝑦 ≤ 1 + 𝜁 ෩

𝑔

𝐽 𝑦 , ∀𝑦 ≥ 0

Preliminaries

𝑜 ≤ poly 𝜁−1 items profit max 𝑞𝑘/min 𝑞𝑘 ≤ poly 𝜁−1 weight 0 < 𝑥𝑗 ≤ 𝑋

• “Profit function” (defined over real 𝑦 ≥ 0)

𝑔

𝐽 𝑦 = max 𝑞 𝐾 : 𝐾 ⊆ 𝐽, 𝑥 𝐾 ≤ 𝑦

• Task: compute a 1 + 𝜁 -approximation of profit

function 𝑔

𝐽,

෩ 𝑔

𝐽 𝑦 ≤ 𝑔 𝐽 𝑦 ≤ 1 + 𝜁 ෩

𝑔

𝐽 𝑦 , ∀𝑦 ≥ 0

• For disjoint sets 𝐽, 𝐾 of items,

𝑔

𝐽∪𝐾 𝑦 = 𝑔 𝐽 ⊕ 𝑔 𝐾

𝑦 ≔ max

0≤𝑧≤𝑦 𝑔 𝐽 𝑧 + 𝑔 𝐾 𝑦 − 𝑧

Preliminaries

𝑜 ≤ poly 𝜁−1 items profit max 𝑞𝑘/min 𝑞𝑘 ≤ poly 𝜁−1 weight 0 < 𝑥𝑗 ≤ 𝑋

• “Profit function” (defined over real 𝑦 ≥ 0)

𝑔

𝐽 𝑦 = max 𝑞 𝐾 : 𝐾 ⊆ 𝐽, 𝑥 𝐾 ≤ 𝑦

• Task: compute a 1 + 𝜁 -approximation of profit

function 𝑔

𝐽,

෩ 𝑔

𝐽 𝑦 ≤ 𝑔 𝐽 𝑦 ≤ 1 + 𝜁 ෩

𝑔

𝐽 𝑦 , ∀𝑦 ≥ 0

• For disjoint sets 𝐽, 𝐾 of items,

𝑔

𝐽∪𝐾 𝑦 = 𝑔 𝐽 ⊕ 𝑔 𝐾

𝑦 ≔ max

0≤𝑧≤𝑦 𝑔 𝐽 𝑧 + 𝑔 𝐾 𝑦 − 𝑧

• ෩

𝑔

𝐽 ⊕ ෩

𝑔

𝐾 is a 1 + 𝜁 -approximation of 𝑔 𝐽∪𝐾

Preliminaries

𝑜 ≤ poly 𝜁−1 items profit max 𝑞𝑘/min 𝑞𝑘 ≤ poly 𝜁−1

• A nondecreasing step function 𝑔 has a

(1 + 𝜁)-approx. with only ෨ 𝑃 1/𝜁 steps (by rounding down to powers of (1 + 𝜁))

Preliminaries

• A nondecreasing step function 𝑔 has a

(1 + 𝜁)-approx. with only ෨ 𝑃 1/𝜁 steps (by rounding down to powers of (1 + 𝜁))

• “Merging Lemma”: Computing (a 1 + 𝜁 -
• approx. of) 𝑔

1 ⊕ ⋯ ⊕ 𝑔 𝑛 takes ෨

𝑃 𝑛/𝜁2 time. (log 𝑛 depth binary tree. 𝜁′ ≔ 𝜁/ log 𝑛)

𝑔 ⊕ 𝑕 ≔ max

0≤𝑧≤𝑦 𝑔 𝑧 + 𝑕 𝑦 − 𝑧

• Divide items into 𝑃 log 1/𝜁 groups

(group 𝑙: 𝑞𝑗 ∈ [2𝑙, 2𝑙+1])

Preliminaries

𝑜 ≤ poly 𝜁−1 items profit max 𝑞𝑘/min 𝑞𝑘 ≤ poly 𝜁−1 weight 0 < 𝑥𝑗 ≤ 𝑋 Merging 𝑔

1 ⊕ ⋯ ⊕ 𝑔 𝑛: ෨

𝑃 𝑛/𝜁2 time.

• Divide items into 𝑃 log 1/𝜁 groups

(group 𝑙: 𝑞𝑗 ∈ [2𝑙, 2𝑙+1])

• Compute all 𝑔

𝑙 and merge them in

෨ 𝑃(1/𝜁2) time

Preliminaries

𝑜 ≤ poly 𝜁−1 items profit max 𝑞𝑘/min 𝑞𝑘 ≤ poly 𝜁−1 weight 0 < 𝑥𝑗 ≤ 𝑋 Merging 𝑔

1 ⊕ ⋯ ⊕ 𝑔 𝑛: ෨

𝑃 𝑛/𝜁2 time.

• Divide items into 𝑃 log 1/𝜁 groups

(group 𝑙: 𝑞𝑗 ∈ [2𝑙, 2𝑙+1])

• Compute all 𝑔

𝑙 and merge them in

෨ 𝑃(1/𝜁2) time

• Now assume 𝑞𝑗 ∈ [1,2]

Preliminaries

𝑜 ≤ poly 𝜁−1 items profit max 𝑞𝑘/min 𝑞𝑘 ≤ poly 𝜁−1 weight 0 < 𝑥𝑗 ≤ 𝑋 Merging 𝑔

1 ⊕ ⋯ ⊕ 𝑔 𝑛: ෨

𝑃 𝑛/𝜁2 time.

• Simple greedy (sort by unit profit

𝑞1 𝑥1 ≥ 𝑞2 𝑥2 ≥ ⋯)

approximates with additive error 𝑒 ≤ max 𝑞𝑗 = 𝑃(1)

profit 𝑞𝑗 ∈ 1,2 weight 0 < 𝑥𝑗 ≤ 𝑋

Preliminaries

• Simple greedy (sort by unit profit

𝑞1 𝑥1 ≥ 𝑞2 𝑥2 ≥ ⋯)

approximates with additive error 𝑒 ≤ max 𝑞𝑗 = 𝑃(1)

• For 𝑔 𝑥 ≥ Ω(𝜁−1), this is 1 + 𝑃(𝜁)

multiplicative approx.

profit 𝑞𝑗 ∈ 1,2 weight 0 < 𝑥𝑗 ≤ 𝑋

Preliminaries

• Simple greedy (sort by unit profit

𝑞1 𝑥1 ≥ 𝑞2 𝑥2 ≥ ⋯)

approximates with additive error 𝑒 ≤ max 𝑞𝑗 = 𝑃(1)

• For 𝑔 𝑥 ≥ Ω(𝜁−1), this is 1 + 𝑃(𝜁)

multiplicative approx.

• Only need to 1 + 𝑃(𝜁) approximate

𝐧𝐣𝐨 𝑪, 𝒈𝑱 for 𝑪 = 𝑷 𝜻−𝟐 !

profit 𝑞𝑗 ∈ 1,2 weight 0 < 𝑥𝑗 ≤ 𝑋

Preliminaries

• Round 𝑞𝑗 down to 1,1 + 𝜁, 1 + 2𝜁, … , 2 − 𝜁

1 + 𝜁 multiplicative error

• Only 𝑛 = 𝑃(1/𝜁) different 𝑞𝑗’s!

Preliminaries

𝑞𝑗 ∈ 1,2

• Round 𝑞𝑗 down to 1,1 + 𝜁, 1 + 2𝜁, … , 2 − 𝜁

1 + 𝜁 multiplicative error

• Only 𝑛 = 𝑃(1/𝜁) different 𝑞𝑗’s!
• Collect all items with the same profit 𝒒.

Then 𝑔

𝑞 can be computed by simple

greedy (sort 𝑥1 ≤ 𝑥2 ≤ ⋯)

Preliminaries

𝑞𝑗 ∈ 1,2

Recap

𝑞𝑗 ∈ 1,2 Profit functions 𝑔

1, … , 𝑔 𝑛 obtained by simple

greedy (one for every 𝑞𝑗) (𝑛 = 𝑃(1/𝜁) ) Task: 1 + 𝑃 𝜁 approximate min 𝐶, 𝑔

1 ⊕ ⋯ ⊕ 𝑔 𝑛

(𝐶 = 𝑃(𝜁−1)) Lemma: Merging 𝑔

1 ⊕ ⋯ ⊕ 𝑔 𝑛 in ෨

𝑃 𝑛/𝜁2 time.

(Immediately gives ෨

𝑃(𝑜 + 𝜁−3) algo)

𝑜 items with 𝑛 = 𝑃(𝜁−1) distinct profit values 𝑞𝑗 ∈ 1,2 Task: 1 + 𝑃 𝜁 approximate min 𝐶, 𝑔

1 ⊕ ⋯ ⊕ 𝑔 𝑛

(𝐶 = 𝑃(𝜁−1))

෨ 𝑃 𝜁−1 𝐶𝑛 algo

(faster when 𝑪 small)

෨ 𝑃(𝜁−4/3𝑜 + 𝜁−2) algo

(faster when 𝒐 small)

Chan’s results

A Greedy Lemma

• the items can be divided into two groups

𝐼, 𝑀 with a large enough gap between their unit profits, max

ℓ∈𝑀 𝑞ℓ/𝑥ℓ ≤ 1 − 𝛽 ⋅ min ℎ∈𝐼 𝑞ℎ/𝑥ℎ

• and group 𝐼 is large enough,

σℎ∈𝐼 𝑥ℎ > 𝑋

𝑞𝑗 ∈ 1,2 capacity = 𝑋

If Then

• In an optimal solution, 𝑀-items contribute

total profit ≤ 2/𝛽

𝑞𝑗 ∈ 1,2 capacity = 𝑋 max

ℓ∈𝑀 𝑞ℓ/𝑥ℓ ≤ 1 − 𝛽 ⋅ min ℎ∈𝐼 𝑞ℎ/𝑥ℎ

σℎ∈𝐼 𝑥ℎ > 𝑋

• Suppose the optimal solution is

𝑔

𝐼∪𝑀 𝑋 = 𝑔 𝐼 𝑋 − 𝑋 𝑀 + 𝑔 𝑀 𝑋 𝑀

𝑋 − 𝑋

𝑀

𝑋

𝑀

𝐼 𝑀

𝑞𝑗 ∈ 1,2 capacity = 𝑋 max

ℓ∈𝑀 𝑞ℓ/𝑥ℓ ≤ 1 − 𝛽 ⋅ min ℎ∈𝐼 𝑞ℎ/𝑥ℎ

σℎ∈𝐼 𝑥ℎ > 𝑋

• Remove all 𝑀-items, and insert more 𝐼-items

(denoted by 𝐿) to fill in the 𝑋

𝑀 space

𝑋 − 𝑋

𝑀

𝑋

𝑀

𝐼 𝑀

𝑋 − 𝑋

𝑀

𝑋

𝑀

𝑋

𝐿

𝑞𝑗 ∈ 1,2 capacity = 𝑋 max

ℓ∈𝑀 𝑞ℓ/𝑥ℓ ≤ 1 − 𝛽 ⋅ min ℎ∈𝐼 𝑞ℎ/𝑥ℎ

σ𝒊∈𝑰 𝒙𝒊 > 𝑿

• Remove all 𝑀-items, and insert more 𝐼-items

(denoted by 𝐿) to fill in the 𝑋

𝑀 space

• Until 𝑋

𝑀 − 𝑋 𝐿 < maxℎ∈𝐼 𝑥ℎ

𝐼 𝑀

𝑋 − 𝑋

𝑀

𝑋

𝑀

𝑋

𝐿

𝒒𝒋 ∈ 𝟐, 𝟑 capacity = 𝑋 max

ℓ∈𝑀 𝑞ℓ/𝑥ℓ ≤ 1 − 𝛽 ⋅ 𝐧𝐣𝐨 𝒊∈𝑰 𝒒𝒊/𝒙𝒊 = 1 − 𝛽 ⋅ 𝒓

σℎ∈𝐼 𝑥ℎ > 𝑋

• Remove all 𝑀-items, and insert more 𝐼-items

(denoted by 𝐿) to fill in the 𝑋

𝑀 space

• Until 𝑋

𝑀 − 𝑋 𝐿 < maxℎ∈𝐼 𝑥ℎ ≤ 𝟑/𝒓

𝐼 𝑀

𝑋 − 𝑋

𝑀

𝑋

𝑀

𝑋

𝐿

𝑞𝑗 ∈ 1,2 capacity = 𝑋 max

ℓ∈𝑀 𝑞ℓ/𝑥ℓ ≤ 1 − 𝛽 ⋅ 𝐧𝐣𝐨 𝒊∈𝑰 𝒒𝒊/𝒙𝒊 = 1 − 𝛽 ⋅ 𝒓

σℎ∈𝐼 𝑥ℎ > 𝑋 𝑋

𝑀 − 𝑋 𝐿 < 2/𝑟

𝑋 − 𝑋

𝑀

𝑋

𝑀

• ptimal sol:

𝑔

𝐼∪𝑀 𝑋 = 𝑔 𝐼 𝑋 − 𝑋 𝑀 + 𝑔 𝑀 𝑋 𝑀

tot profit ≥ 𝑔

𝐼 𝑋 − 𝑋 𝑀 + 𝑟𝑋 𝐿

𝑋 − 𝑋

𝑀

𝑋

𝑀

𝑋

𝐿

𝑞𝑗 ∈ 1,2 capacity = 𝑋 max

ℓ∈𝑀 𝑞ℓ/𝑥ℓ ≤ 1 − 𝛽 ⋅ 𝐧𝐣𝐨 𝒊∈𝑰 𝒒𝒊/𝒙𝒊 = 1 − 𝛽 ⋅ 𝒓

σℎ∈𝐼 𝑥ℎ > 𝑋 𝑋

𝑀 − 𝑋 𝐿 < 2/𝑟

𝑋 − 𝑋

𝑀

𝑋

𝑀

• ptimal sol:

𝑔

𝐼∪𝑀 𝑋 = 𝑔 𝐼 𝑋 − 𝑋 𝑀 + 𝑔 𝑀 𝑋 𝑀

tot profit ≥ 𝑔

𝐼 𝑋 − 𝑋 𝑀 + 𝑟𝑋 𝐿

𝒓𝑿𝑳 ≤ 𝒈𝑴 𝑿𝑴

𝑋 − 𝑋

𝑀

𝑋

𝑀

𝑋

𝐿

𝑞𝑗 ∈ 1,2 capacity = 𝑋 max

ℓ∈𝑀 𝑞ℓ/𝑥ℓ ≤ 1 − 𝛽 ⋅ 𝐧𝐣𝐨 𝒊∈𝑰 𝒒𝒊/𝒙𝒊 = 1 − 𝛽 ⋅ 𝒓

σℎ∈𝐼 𝑥ℎ > 𝑋 𝑿𝑴 − 𝑿𝑳 < 𝟑/𝒓 𝑋 − 𝑋

𝑀

𝑋

𝑀

• ptimal sol:

𝑔

𝐼∪𝑀 𝑋 = 𝑔 𝐼 𝑋 − 𝑋 𝑀 + 𝑔 𝑀 𝑋 𝑀

tot profit ≥ 𝑔

𝐼 𝑋 − 𝑋 𝑀 + 𝑟𝑋 𝐿

𝒓𝑿𝑳 ≤ 𝒈𝑴 𝑿𝑴

1 − 𝛽 ⋅ 𝑟 ⋅ 𝑋

𝑀 ≥ 𝑔 𝑀 𝑋 𝑀

𝑔

𝑀 𝑋 𝑀 ≥ 𝑟𝑋 𝐿

𝑟𝑋

𝐿 > 𝑟𝑋 𝑀 − 2

↓ 1 − 𝛽 ⋅ 𝑟𝑋

𝑀 > 𝑟𝑋 𝑀 − 2

𝛽𝑟𝑋

𝑀 < 2

𝑔

𝑀 𝑋 𝑀 < 2/𝛽

𝑋 − 𝑋

𝑀

𝑋

𝑀

𝑋

𝐿

𝑞𝑗 ∈ 1,2 capacity = 𝑋 max

ℓ∈𝑀 𝑞ℓ/𝑥ℓ ≤ 1 − 𝛽 ⋅ 𝐧𝐣𝐨 𝒊∈𝑰 𝒒𝒊/𝒙𝒊 = 1 − 𝛽 ⋅ 𝒓

σℎ∈𝐼 𝑥ℎ > 𝑋 𝑿𝑴 − 𝑿𝑳 < 𝟑/𝒓 𝑋 − 𝑋

𝑀

𝑋

𝑀

• ptimal sol:

𝑔

𝐼∪𝑀 𝑋 = 𝑔 𝐼 𝑋 − 𝑋 𝑀 + 𝑔 𝑀 𝑋 𝑀

tot profit ≥ 𝑔

𝐼 𝑋 − 𝑋 𝑀 + 𝑟𝑋 𝐿

𝒓𝑿𝑳 ≤ 𝒈𝑴 𝑿𝑴

1 − 𝛽 ⋅ 𝑟 ⋅ 𝑋

𝑀 ≥ 𝑔 𝑀 𝑋 𝑀

𝑔

𝑀 𝑋 𝑀 ≥ 𝑟𝑋 𝐿

𝑟𝑋

𝐿 > 𝑟𝑋 𝑀 − 2

↓ 1 − 𝛽 ⋅ 𝑟𝑋

𝑀 > 𝑟𝑋 𝑀 − 2

↓ 𝛽𝑟𝑋

𝑀 < 2

𝑔

𝑀 𝑋 𝑀 < 2/𝛽

𝑋 − 𝑋

𝑀

𝑋

𝑀

𝑋

𝐿

𝑞𝑗 ∈ 1,2 capacity = 𝑋 max

ℓ∈𝑀 𝑞ℓ/𝑥ℓ ≤ 1 − 𝛽 ⋅ 𝐧𝐣𝐨 𝒊∈𝑰 𝒒𝒊/𝒙𝒊 = 1 − 𝛽 ⋅ 𝒓

σℎ∈𝐼 𝑥ℎ > 𝑋 𝑿𝑴 − 𝑿𝑳 < 𝟑/𝒓 𝑋 − 𝑋

𝑀

𝑋

𝑀

• ptimal sol:

𝑔

𝐼∪𝑀 𝑋 = 𝑔 𝐼 𝑋 − 𝑋 𝑀 + 𝑔 𝑀 𝑋 𝑀

tot profit ≥ 𝑔

𝐼 𝑋 − 𝑋 𝑀 + 𝑟𝑋 𝐿

𝒓𝑿𝑳 ≤ 𝒈𝑴 𝑿𝑴

𝟐 − 𝜷 ⋅ 𝒓 ⋅ 𝑿𝑴 ≥ 𝒈𝑴 𝑿𝑴

𝑔

𝑀 𝑋 𝑀 ≥ 𝑟𝑋 𝐿

𝑟𝑋

𝐿 > 𝑟𝑋 𝑀 − 2

↓ 1 − 𝛽 ⋅ 𝑟𝑋

𝑀 > 𝑟𝑋 𝑀 − 2

↓ 𝜷𝒓𝑿𝑴 < 𝟑 ↓

𝒈𝑴 𝑿𝑴 < 𝟑/𝜷

Recap

If: max

ℓ∈𝑀 𝑞ℓ/𝑥ℓ ≤ 1 − 𝛽 ⋅ min ℎ∈𝐼 𝑞ℎ/𝑥ℎ

σℎ∈𝐼 𝑥ℎ > 𝑋 Then: In optimal solution of 𝑔

𝐼∪𝑀(𝑋), 𝑀-items

contribute total profit ≤ 2/𝛽

Task: 1 + 𝑃 𝜁 approximate min 𝐶, 𝑔

𝐽 , 𝐶 = 𝑃(𝜁−1)

𝑛 = 𝑃(𝜁−1) distinct values 𝑞𝑗 ∈ 1,2

෨ 𝑃 𝜁−1 𝐶𝑛 algo (Chan)

(faster when 𝑪 small)

෨ 𝑃(𝜁−4/3𝑜 + 𝜁−2) algo (Chan)

(faster when 𝒐 small)

Improved Algorithm

• Sort the items by 𝑞𝑗/𝑥𝑗, and divide into

three groups

𝑞𝑗 ∈ 1,2

Improved Algorithm

• Sort the items by 𝑞𝑗/𝑥𝑗, and divide into

three groups 𝐼: top Θ 𝜁−1 items

𝑞𝑗 ∈ 1,2

Improved Algorithm

• Sort the items by 𝑞𝑗/𝑥𝑗, and divide into

three groups 𝐼: top Θ 𝜁−1 items 𝑀: (1 − 𝛽) multiplicative gap

𝒓 (𝟐 − 𝜷)𝒓 𝑞𝑗 ∈ 1,2

Improved Algorithm

• Sort the items by 𝑞𝑗/𝑥𝑗, and divide into

three groups 𝐼: top Θ 𝜁−1 items 𝑀: (1 − 𝛽) multiplicative gap

𝑟 (1 − 𝛽)𝑟

𝑁: items in between

𝑞𝑗 ∈ 1,2

Improved Algorithm

𝑟 (1 − 𝛽)𝑟 𝐼 = Θ(𝜁−1) 𝐼 𝑁 𝑀

If: max

ℓ∈𝑀 𝑞ℓ/𝑥ℓ ≤ 1 − 𝛽 ⋅ min ℎ∈𝐼 𝑞ℎ/𝑥ℎ

σℎ∈𝐼 𝑥ℎ > 𝑋 Then: In optimal solution of 𝑔

𝐼∪𝑀(𝑋), 𝑀-items

contribute total profit ≤ 2/𝛽 Task: 1 + 𝑃 𝜁 approximate min 𝑃(𝜁−1), 𝑔

𝐽

𝑞𝑗 ∈ 1,2

If: max

ℓ∈𝑀 𝑞ℓ/𝑥ℓ ≤ 1 − 𝛽 ⋅ min ℎ∈𝐼 𝑞ℎ/𝑥ℎ

σℎ∈𝐼 𝑥ℎ > 𝑋 Then: In optimal solution of 𝑔

𝐼∪𝑀(𝑋), 𝑀-items

contribute total profit ≤ 2/𝛽

Improved Algorithm

𝑟 (1 − 𝛽)𝑟 𝐼 = Θ(𝜁−1) 𝐼 𝑁 𝑀 𝑞𝑗 ∈ 1,2

Task: 1 + 𝑃 𝜁 approximate min 𝑃(𝜁−1), 𝑔

𝐽

Improved Algorithm

𝑟 (1 − 𝛽)𝑟 𝑰 = 𝚰(𝜻−𝟐) 𝐼 𝑁 𝑀

(If 𝒈𝑱 𝑿 < 𝑷(𝜻−𝟐) then 𝑿 < 𝒙(𝑰)) Task: 1 + 𝑃 𝜁 approximate 𝐧𝐣𝐨 𝑷(𝜻−𝟐), 𝒈𝑱 If: max

ℓ∈𝑀 𝑞ℓ/𝑥ℓ ≤ 1 − 𝛽 ⋅ min ℎ∈𝐼 𝑞ℎ/𝑥ℎ

σℎ∈𝐼 𝑥ℎ > 𝑋 Then: In optimal solution of 𝑔

𝐼∪𝑀(𝑋), 𝑀-items

contribute total profit ≤ 2/𝛽

𝒒𝒋 ∈ 𝟐, 𝟑

Improved Algorithm

𝑟 (1 − 𝛽)𝑟 𝐼 = Θ(𝜁−1) 𝐼 𝑁 𝑀

Task: 1 + 𝑃 𝜁 approximate min 𝑃(𝜁−1), 𝑔

𝐽

If: max

ℓ∈𝑀 𝑞ℓ/𝑥ℓ ≤ 1 − 𝛽 ⋅ min ℎ∈𝐼 𝑞ℎ/𝑥ℎ

σℎ∈𝐼 𝑥ℎ > 𝑋 Then: In optimal solution of 𝑔

𝐼∪𝑀(𝑋), 𝑀-items

contribute total profit ≤ 2/𝛽

𝑞𝑗 ∈ 1,2 𝐧𝐣𝐨{𝒈𝑴 , 𝟑/𝜷}

Improved Algorithm

𝑟 (1 − 𝛽)𝑟 𝐼 = Θ(𝜁−1)

𝐼

𝑁

𝑴

min{𝑔

𝑀 , 2/𝛽}

෨ 𝑃 𝜁−1 𝐶𝑛 algo (Chan)

(faster when 𝑪 small)

෨ 𝑃(𝜁−4/3𝑜 + 𝜁−2) algo (Chan)

(faster when 𝒐 small)

Let 𝜷 = 𝜻𝟑/𝟒 ෩ 𝑷(𝜻−𝟖/𝟒) ෩ 𝑷(𝜻−𝟖/𝟒)

𝒐 ≔ 𝑰 ≈ 𝟐/𝜻 𝑪 ≔ 𝟑/𝜷 𝑛 ≔ 𝑃 𝜁−1

Improved Algorithm

𝑟 (1 − 𝛽)𝑟

𝐼

𝑁

𝑴

Let 𝜷 = 𝜻𝟑/𝟒

෨ 𝑃 𝜁−7/3 -time ෨ 𝑃(𝜁−7/3)-time

Group 𝑁: Round 𝑞𝑗/𝑥𝑗 down to powers of (1 + 𝜁)

Improved Algorithm

𝑟 (1 − 𝛽)𝑟

𝐼

𝑁

𝑴

Let 𝜷 = 𝜻𝟑/𝟒

෨ 𝑃 𝜁−7/3 -time ෨ 𝑃(𝜁−7/3)-time

Group 𝑁: Round 𝑞𝑗/𝑥𝑗 down to powers of (1 + 𝜁) Only log1+𝜁

1 1−𝛽 ≈ 𝛽/𝜁 = 𝜻−𝟐/𝟒

distinct values of 𝑞𝑗/𝑥𝑗

Improved Algorithm

𝑟 (1 − 𝛽)𝑟

𝐼

𝑁

𝑴

Let 𝜷 = 𝜻𝟑/𝟒

෨ 𝑃 𝜁−7/3 -time ෨ 𝑃(𝜁−7/3)-time

Group 𝑁: Round 𝑞𝑗/𝑥𝑗 down to powers of (1 + 𝜁) Only log1+𝜁

1 1−𝛽 ≈ 𝛽/𝜁 = 𝜻−𝟐/𝟒

distinct values of 𝑞𝑗/𝑥𝑗 Items with the same 𝑞𝑗/𝑥𝑗 (profit ∝ weight ): a Subset Sum instance

෨ 𝑃(𝑜 + 𝜁−2) time [KMPS03]

Improved Algorithm

𝑟 (1 − 𝛽)𝑟

𝐼

𝑁

𝑴

Let 𝜷 = 𝜻𝟑/𝟒

෨ 𝑃 𝜁−7/3 -time ෨ 𝑃(𝜁−7/3)-time

Group 𝑁: Round 𝑞𝑗/𝑥𝑗 down to powers of (1 + 𝜁) Only log1+𝜁

1 1−𝛽 ≈ 𝛽/𝜁 = 𝜻−𝟐/𝟒

distinct values of 𝑞𝑗/𝑥𝑗 Items with the same 𝑞𝑗/𝑥𝑗 (profit ∝ weight ): a Subset Sum instance

෨ 𝑃(𝑜 + 𝜁−2) time [KMPS03]

Merge 𝜻−𝟐/𝟒 groups

Improved Algorithm

𝑟 (1 − 𝛽)𝑟

𝐼

𝑁

𝑴

Let 𝜷 = 𝜻𝟑/𝟒

෨ 𝑃 𝜁−7/3 -time ෨ 𝑃(𝜁−7/3)-time

Group 𝑁: Round 𝑞𝑗/𝑥𝑗 down to powers of (1 + 𝜁) Only log1+𝜁

1 1−𝛽 ≈ 𝛽/𝜁 = 𝜻−𝟐/𝟒

distinct values of 𝑞𝑗/𝑥𝑗 Items with the same 𝑞𝑗/𝑥𝑗 (profit ∝ weight ): a Subset Sum instance

෨ 𝑃(𝑜 + 𝜁−2) time [KMPS03] 𝑜 + 𝜻−𝟐/𝟒 ⋅ 𝜁−2

Merge 𝜻−𝟐/𝟒 groups

Improved Algorithm

𝑟 (1 − 𝛽)𝑟

𝐼

𝑁

𝑴

Let 𝜷 = 𝜻𝟑/𝟒

෨ 𝑃 𝜁−7/3 -time ෨ 𝑃(𝜁−7/3)-time

Group 𝑁: Round 𝑞𝑗/𝑥𝑗 down to powers of (1 + 𝜁) Only log1+𝜁

1 1−𝛽 ≈ 𝛽/𝜁 = 𝜻−𝟐/𝟒

distinct values of 𝑞𝑗/𝑥𝑗 Items with the same 𝑞𝑗/𝑥𝑗 (profit ∝ weight ): a Subset Sum instance

෨ 𝑃(𝑜 + 𝜁−2) time [KMPS03] 𝑜 + 𝜻−𝟐/𝟒 ⋅ 𝜁−2

Merge 𝜻−𝟐/𝟒 groups

Total time: ෩ 𝑷 𝒐 + 𝜻−𝟖/𝟒

Further improvement

෨ 𝑃 𝜁−1 𝐶𝑛 (Chan)

(faster when 𝑪 small)

෨ 𝑃(𝜁−4/3𝑜 + 𝜁−2) (Chan)

(faster when 𝒐 small)

෨ 𝑃(𝜁−3/2𝑜3/4 + 𝑜 + 𝜁−2)

(faster when 𝒐 small)

෨ 𝑃 𝜁−4/3𝐶1/3𝑛2/3

(𝑛2 ≫ 𝜁−2/𝐶)

(faster when 𝑪 small)

෨ 𝑃(𝑜 + 𝜁−9/4)

Greedy argument (this talk) extending Chan’s techniques from two levels to multiple levels

Open problem

Unbounded Knapsack (each item has infinitely many copies) which easily reduces to 0-1 Knapsack: ෨ 𝑃(𝑜 + 𝜁−2) [Jansen and Kraft, 2015]

Improve 0-1 Knapsack to ෨ 𝑃(𝑜 + 𝜁−2) time?

෨ 𝑃 min 𝒐 + 𝜻−𝟑, 𝑜𝜁−1 [Kellerer, Mansini, Pferschy, and Speranza, 2003] Subset Sum:

Thank you!