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Problem 1 : Longest Stable Subsequence Consider a list of numbers [a?, a?, ···, a???]. Our goal is to find the the longest stable subsequence: [a??, a??, ···, a??] which is a sub-list of the original list that selects elements at indices i?, i?, ..., i? from the original list such that 1. i? < i? < ··· < i?; 2. a???? ? 1 ? a?? ? a???? + 1. We can also write this as |a???? ? a??| ? 1. I.e, each element of the subsequence must be within ±1 or equal to the previous element. 3. The length of the subsequence k is maximized. Example Consider the list [1, 4, 2, ?2, 0, ?1, 2, 3]. There are many "stable subsequences": • [1, 0, ?1] • [1, 2, 2, 3] • [4, 3] The longest stable subsequence is [1, 2, 2, 3] of length 4. Note that each element of the subsequence is at most 1 away from the previous element. The goal of this problem is to formulate a dynamic programming solution to find the length of the longest stable subsequence and the subsequence itself. A: Write a Recurrence With Base Case Let n be the length of the original array [a?, ..., a???]. Define LSSLength(i, a?) to be the length of the longest stable subsequence for the subarray from [a?, ..., a???] (note that a? is included) with the additional constraint that the first element in the subsequence chosen (let us call it a??) must satisfy |a?? ? a?| ? 1 . Notes 0. 0 ? i ? n. i = n denotes the empty subarray. 1. a? represents a previous choice we have made before encountering the current subproblem. It is made an argument of the recurrence to ensure that the subsequent choice made from [a?, ..., a???] satisfies |a ? a?| ? 1. 2. We will use the special value a? = None to denote that no such element a? has been chosen. Fill out the missing portion of the recurrence and base cases. We will not grade your answer below. Instead please use it as a guide to complete the code for the recurrence and pass the test cases provided. LSSLength(i, a?) = { ?? i = n # Base Case when subarray is empty { LSSLength(i + 1, a?) i < n and a? ? None and |a? ? a?| > 1 # We cannot choose a[i], skip it and move right along { max(??? + 1, ???) i < n and (a? = None or |a? ? a?| ? 1) # Choose maximum of two options: take a[i] or skip a[i] YOUR ANSWER HERE

          Problem 1 : Longest Stable Subsequence
Consider a list of numbers [a?, a?, ···, a???]. Our goal is to find the the longest stable subsequence: [a??, a??, ···, a??] which is a sub-list of the original list that selects elements at indices i?, i?, ..., i? from the original list such that

1. i? < i? < ··· < i?;
2. a???? ? 1 ? a?? ? a???? + 1. We can also write this as |a???? ? a??| ? 1. I.e, each element of the subsequence must be within ±1 or equal to the previous element.
3. The length of the subsequence k is maximized.

Example
Consider the list [1, 4, 2, ?2, 0, ?1, 2, 3]. There are many "stable subsequences":
• [1, 0, ?1]
• [1, 2, 2, 3]
• [4, 3]

The longest stable subsequence is [1, 2, 2, 3] of length 4. Note that each element of the subsequence is at most 1 away from the previous element.

The goal of this problem is to formulate a dynamic programming solution to find the length of the longest stable subsequence and the subsequence itself.

A: Write a Recurrence With Base Case

Let n be the length of the original array [a?, ..., a???]. Define
LSSLength(i, a?)
to be the length of the longest stable subsequence for the subarray from [a?, ..., a???] (note that a? is included) with the additional constraint that the first element in the subsequence chosen (let us call it a??) must satisfy
|a?? ? a?| ? 1
.
Notes
0. 0 ? i ? n. i = n denotes the empty subarray.
1. a? represents a previous choice we have made before encountering the current subproblem. It is made an argument of the recurrence to ensure that the subsequent choice made from [a?, ..., a???] satisfies |a ? a?| ? 1.
2. We will use the special value a? = None to denote that no such element a? has been chosen.

Fill out the missing portion of the recurrence and base cases. We will not grade your answer below. Instead please use it as a guide to complete the code for the recurrence and pass the test cases provided.

LSSLength(i, a?) = 
{ ?? i = n # Base Case when subarray is empty
{ LSSLength(i + 1, a?) i < n and a? ? None and |a? ? a?| > 1 # We cannot choose a[i], skip it and move right along
{ max(??? + 1, ???) i < n and (a? = None or |a? ? a?| ? 1) # Choose maximum of two options: take a[i] or skip a[i]

YOUR ANSWER HERE

Problem 1 : Longest Stable Subsequence
Consider a list of numbers [a?, a?, ···, a???]. Our goal is to find the the longest stable subsequence: [a??, a??, ···, a??] which is a sub-list of the original list that selects elements at indices i?, i?, ..., i? from the original list such that

1. i? < i? < ··· < i?;
2. a???? ? 1 ? a?? ? a???? + 1. We can also write this as |a???? ? a??| ? 1. I.e, each element of the subsequence must be within ±1 or equal to the previous element.
3. The length of the subsequence k is maximized.

Example
Consider the list [1, 4, 2, ?2, 0, ?1, 2, 3]. There are many "stable subsequences":
• [1, 0, ?1]
• [1, 2, 2, 3]
• [4, 3]

The longest stable subsequence is [1, 2, 2, 3] of length 4. Note that each element of the subsequence is at most 1 away from the previous element.

The goal of this problem is to formulate a dynamic programming solution to find the length of the longest stable subsequence and the subsequence itself.

A: Write a Recurrence With Base Case

Let n be the length of the original array [a?, ..., a???]. Define
LSSLength(i, a?)
to be the length of the longest stable subsequence for the subarray from [a?, ..., a???] (note that a? is included) with the additional constraint that the first element in the subsequence chosen (let us call it a??) must satisfy
|a?? ? a?| ? 1
.
Notes
0. 0 ? i ? n. i = n denotes the empty subarray.
1. a? represents a previous choice we have made before encountering the current subproblem. It is made an argument of the recurrence to ensure that the subsequent choice made from [a?, ..., a???] satisfies |a ? a?| ? 1.
2. We will use the special value a? = None to denote that no such element a? has been chosen.

Fill out the missing portion of the recurrence and base cases. We will not grade your answer below. Instead please use it as a guide to complete the code for the recurrence and pass the test cases provided.

LSSLength(i, a?) =
?? i = n # Base Case when subarray is empty
LSSLength(i + 1, a?) i < n and a? ? None and |a? ? a?| > 1 # We cannot choose a[i], skip it and move right along
max(??? + 1, ???) i < n and (a? = None or |a? ? a?| ? 1) # Choose maximum of two options: take a[i] or skip a[i]

YOUR ANSWER HERE

Added by Patricia B.

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