Exercise 1: Write a function (permutations lst)
taking a list lst
and returning all its permutations. E.g., (permutations '(1 2 3)) => ((1 2 3) (2 1 3) (2 3 1) (1 3 2) (3 1 2) (3 2 1))
.
Hint: Suppose that we have all permutations of a list of length $n$, and we want to build all permutations of its extension by an element. To do that, it suffices to take the element and interleave it in all possible ways into all the permutations of length $n$. For instance, ((2 3) (3 2))
are all permutations of the list (2 3)
. If we want to compute all permutations of (1 2 3)
, we take each permutation of length 2 and interleave the element 1
into it as follows:
(2 3) => ((1 2 3) (2 1 3) (2 3 1)) (3 2) => ((1 3 2) (3 1 2) (3 2 1))Appending all these lists gives us the desired permutations of
(1 2 3)
.
Write first a function interleave
taking an element, a list, and returning all possible ways of inserting the element into the list.
Using this function, devise the function permutations
using the recursion on the length of lst
.
Exercise 2: Binary decision trees represent Boolean functions, i.e., functions from $\{0,1\}^n$ to $\{0,1\}$. Let $f(x_1,\ldots,x_n)$ be a Boolean function. The corresponding binary decision tree is created as follows:
Each path from the root to a leaf encodes an evaluation of input variables. If the path in an internal node $x_i$ goes to the left, the variable $x_i$ is evaluated by $0$. If to the right, it is evaluated by $1$. The leaf in the path represents the value $f(x_1,\ldots,x_n)$ for the evaluation defined by the path. Example of a Boolean function and its binary decision tree:
We will represent the inner variable nodes as Racket structures:
(struct node (var left right) #:transparent)For instance, the above tree is represented as follows:
(define bool-tree (node 'x1 (node 'x2 (node 'x3 1 0) (node 'x3 0 1)) (node 'x2 (node 'x3 0 0) (node 'x3 1 1))))
Your task is to implement two functions. The first one (evaluate tree vals)
takes a binary decision tree tree
representing a Boolean function $f(x_1,\ldots,x_n)$, a list vals
of values of variables $x_1,\ldots,x_n$ and returns $f(x_1,\ldots,x_n)$. E.g.
(evaluate bool-tree '(1 0 1)) => 0 (evaluate bool-tree '(0 1 1)) => 1The second function
(satisficing-evaluations tree)
takes a binary decision tree tree
representing a Boolean function $f(x_1,\ldots,x_n)$ and returns all its satisficing evaluations, i.e., those for which $f(x_1,\ldots,x_n)=1$. To represent a variable assignment, we introduce the following structure:
(struct assignment (var val) #:transparent)An evaluation is a list of assignments for all variables occurring in the tree. Thus the output of
satisficing-evaluations
might look as follows:
(satisficing-evaluations bool-tree) => (list (list (assignment 'x1 0) (assignment 'x2 0) (assignment 'x3 0)) (list (assignment 'x1 0) (assignment 'x2 1) (assignment 'x3 1)) (list (assignment 'x1 1) (assignment 'x2 1) (assignment 'x3 0)) (list (assignment 'x1 1) (assignment 'x2 1) (assignment 'x3 1)))
We devise two versions of evaluate
. The first is the recursive function consuming consecutively values of $x_1,\ldots,x_n$ and, based on its value, recursively evaluates either the left or right subtree. Once all the values are consumed, we should be in a leaf specifying the value of $f(x_1,\ldots,x_n)$.
The second version uses higher-order functions. It converts the list of values of $x_1,\ldots,x_n$ into the list of functions node-left
, node-right
corresponding to the path defined by vals
. Finally, it applies their composition to tree
.
The function satisficing-evaluations
is a recursive function using an accumulator ev
, keeping partial evaluation as we traverse the tree.
It recursively finds all satisficing evaluations of the left and right subtree, extends them by $0$ (resp. $1$) if they come from left (resp. right), and append them together.
Task 1: Write a function (sub-seq lst)
taking a list lst
and returning a list of all its sublists/subsequences. E.g.
(sub-seq '(1 2 3)) => (() (3) (2) (2 3) (1) (1 3) (1 2) (1 2 3))
Hint: Code it as a recursive function using the following facts. 1) There is only a single subsequence of the empty list, namely the empty list. 2) Subsequences of $(x_1,x_2,\ldots,x_n)$ are just subsequences of $(x_2,\ldots,x_n)$ together with subsequences starting with $x_1$ and following by a subsequence of $(x_2,\ldots,x_n)$.
Task 2: Consider a binary tree representing a tournament. Each internal node corresponds to a match. We represent it as the following structure:
(struct mtch (winner left right) #:transparent)Leaves are of the form
'<team>
. E.g.
(define tour (mtch 'F (mtch 'D (mtch 'A 'A 'B) (mtch 'D 'C 'D)) (mtch 'F (mtch 'F 'E 'F) (mtch 'G 'G 'H))))represents the following tree:
F / \ / \ / \ / \ / \ D F / \ / \ / \ / \ A D F G / \ / \ / \ / \ A B C D E F G H
Write a function (beaten-teams tree)
taking a binary tournament tree and outputting the list of beaten teams by the winner. E.g., (beaten-teams tour) ⇒ (E G D)
.
Hint: Code it as a recursive function starting in the root defining the tournament winner. Then follow the path labelled by the winner and collects the beaten teams along the path to an accumulator. You can use nested patterns in pattern matching to find out the losers.