' \ufeffPlease help with the questions '

Question

Manu · Accepted Answer

Step 1: 12. (10 points) In class we saw a poly-time reduction from 3D Matching to Subset Sum. Recall that the 3D Matching problem gives you groups of three (aka “triples”) to choose from, where each triple has one element from each of three different sets, 𝑋 X , 𝑌 Y , and 𝑍 Z , and then asks whether a subset 𝑀 M of the given triples 𝑇 T can be chosen so that each element of 𝑋 ∪ 𝑌 ∪ 𝑍 X ∪ Y ∪ Z is in exactly one triple of 𝑀 M , thus forming a so-called “perfect 3D matching” . Solution :  3D Matching Problem Input : Three sets 𝑋 X , 𝑌 Y , and 𝑍 Z , and a collection of triples 𝑇 ⊆ 𝑋 × 𝑌 × 𝑍 T ⊆ X × Y × Z . Output : A subset 𝑀 ⊆ 𝑇 M ⊆ T such that every element of 𝑋 ∪ 𝑌 ∪ 𝑍 X ∪ Y ∪ Z is included in exactly one triple of 𝑀 M . This is called a "perfect 3D matching." Subset Sum Problem Input : A set of integers 𝑆 S and an integer 𝐾 K . Output : Determine if there exists a subset of 𝑆 S whose sum equals 𝐾 K . Reduction from 3D Matching to Subset Sum Constructing the Set and Target for Subset Sum : Each element in 𝑋 X , 𝑌 Y , and 𝑍 Z is represented by a unique binary vector. Each triple in 𝑇 T is represented by a vector that combines the vectors of its elements. The target sum 𝐾 K is a vector where each bit position has a value equal to 1. Representing Elements and Triples : Assign each element in 𝑋 ∪ 𝑌 ∪ 𝑍 X ∪ Y ∪ Z a unique position in the binary vector. For each triple ( 𝑥 , 𝑦 , 𝑧 ) ( x , y , z ) , construct a binary vector where the positions corresponding to 𝑥 x , 𝑦 y , and 𝑧 z are set to 1. Constructing the Binary Vectors : Suppose 𝑋 = { 𝑥 1 , 𝑥 2 , … , 𝑥 𝑛 } X = { x 1 ​ , x 2 ​ , … , x n ​ } , 𝑌 = { 𝑦 1 , 𝑦 2 , … , 𝑦 𝑛 } Y = { y 1 ​ , y 2 ​ , … , y n ​ } , and 𝑍 = { 𝑧 1 , 𝑧 2 , … , 𝑧 𝑛 } Z = { z 1 ​ , z 2 ​ , … , z n ​ } . Each 𝑥 𝑖 x i ​ , 𝑦 𝑖 y i ​ , and 𝑧 𝑖 z i ​ corresponds to a unique position in an 𝑛 n -dimensional binary vector. For each triple ( 𝑥 𝑖 , 𝑦 𝑗 , 𝑧 𝑘 ) ∈ 𝑇 ( x i ​ , y j ​ , z k ​ ) ∈ T , create a binary vector of length 3 𝑛 3 n with 1s at the positions corresponding to 𝑥 𝑖 x i ​ , 𝑦 𝑗 y j ​ , and 𝑧 𝑘 z k ​ . Example Construction : If 𝑋 = { 𝑥 1 , 𝑥 2 } X = { x 1 ​ , x 2 ​ } , 𝑌 = { 𝑦 1 , 𝑦 2 } Y = { y 1 ​ , y 2 ​ } , 𝑍 = { 𝑧 1 , 𝑧 2 } Z = { z 1 ​ , z 2 ​ } , and 𝑇 = { ( 𝑥 1 , 𝑦 1 , 𝑧 1 ) , ( 𝑥 1 , 𝑦 2 , 𝑧 2 ) , ( 𝑥 2 , 𝑦 1 , 𝑧 2 ) , ( 𝑥 2 , 𝑦 2 , 𝑧 1 ) } T = {( x 1 ​ , y 1 ​ , z 1 ​ ) , ( x 1 ​ , y 2 ​ , z 2 ​ ) , ( x 2 ​ , y 1 ​ , z 2 ​ ) , ( x 2 ​ , y 2 ​ , z 1 ​ )} : Binary vectors: 𝑥 1 x 1 ​ -> 100000, 𝑦 1 y 1 ​ -> 010000, 𝑧 1 z 1 ​ -> 001000 𝑥 2 x 2 ​ -> 000100, 𝑦 2 y 2 ​ -> 000010, 𝑧 2 z 2 ​ -> 000001 Triples represented as binary vectors: ( 𝑥 1 , 𝑦 1 , 𝑧 1 ) ( x 1 ​ , y 1 ​ , z 1 ​ ) -> 111000 ( 𝑥 1 , 𝑦 2 , 𝑧 2 ) ( x 1 ​ , y 2 ​ , z 2 ​ ) -> 100011 ( 𝑥 2 , 𝑦 1 , 𝑧 2 ) ( x 2 ​ , y 1 ​ , z 2 ​ ) -> 010101 ( 𝑥 2 , 𝑦 2 , 𝑧 1 ) ( x 2 ​ , y 2 ​ , z 1 ​ ) -> 001110 Target Vector : The target vector 𝐾 K is 111111, indicating that each position must be covered exactly once by the sum of the chosen vectors. Subset Sum Solution : Find a subset of the vectors representing the triples such that their sum equals the target vector 𝐾 K . Explanation: The reduction from 3D Matching to Subset Sum involves representing each element and triple as binary vectors and finding a subset of these vectors whose sum matches the target vector. This reduction shows that solving the 3D Matching problem can be translated into solving the Subset Sum problem in polynomial time. Step 2: (a) Complete the blanks in the figure/slide and table above to illustrate the reduction of the above 3D matching instance into a corresponding subset sum instance. To complete the blanks in the figure and the table, we need to ensure that each row in the table accurately represents the triple's binary encoding as per the rules provided. Rules Recap For each triple 𝑡 = ( 𝑥 𝑖 , 𝑦 𝑗 , 𝑧 𝑘 ) t = ( x i ​ , y j ​ , z k ​ ) , create an integer 𝑤 𝑡 w t ​ with 3 𝑛 3 n digits. Position 𝑖 i for 𝑥 𝑖 x i ​ Position 𝑛 + 𝑗 n + j for 𝑦 𝑗 y j ​ Position 2 𝑛 + 𝑘 2 n + k for 𝑧 𝑘 z k ​ Where 𝑛 = 4 n = 4 (since ∣ 𝑋 ∣ = ∣ 𝑌 ∣ = ∣ 𝑍 ∣ = 4 ∣ X ∣ = ∣ Y ∣ = ∣ Z ∣ = 4 ). Target Vector 𝑊 W The target vector 𝑊 W is 111111111111 111111111111 (binary). Completing the Table Let’s start with the incomplete entries and verify the existing ones: Triple ( 𝑥 1 , 𝑦 2 , 𝑧 3 ) ( x 1 ​ , y 2 ​ , z 3 ​ ) Position of 𝑥 1 x 1 ​ : 1 Position of 𝑦 2 y 2 ​ : 4 + 2 = 6 Position of 𝑧 3 z 3 ​ : 8 + 3 = 11 Binary Vector: 100000100010 Verify: 100001000010 Triple ( 𝑥 4 , 𝑦 4 , 𝑧 1 ) ( x 4 ​ , y 4 ​ , z 1 ​ ) Position of 𝑥 4 x 4 ​ : 4 Position of 𝑦 4 y 4 ​ : 4 + 4 = 8 Position of 𝑧 1 z 1 ​ : 8 + 1 = 9 Binary Vector: 000100010001 Verify: 000110010001 Triple ( 𝑥 1 , 𝑦 1 , 𝑧 2 ) ( x 1 ​ , y 1 ​ , z 2 ​ ) Position of 𝑥 1 x 1 ​ : 1 Position of 𝑦 1 y 1 ​ : 4 + 1 = 5 Position of 𝑧 2 z 2 ​ : 8 + 2 = 10 Binary Vector: 100010001000 Verify: 100000100100 Triple ( 𝑥 2 , 𝑦 2 , 𝑧 1 ) ( x 2 ​ , y 2 ​ , z 1 ​ ) Position of 𝑥 2 x 2 ​ : 2 Position of 𝑦 2 y 2 ​ : 4 + 2 = 6 Po

AI tools

Study help

Free tools

Earn & more

Computer science

Computer science

Manu

Answer

Extracted Transcript from Images

AI

Study help

Earn & more

Company