How to resolve the algorithm Display a linear combination step by step in the Racket programming language

Published on 12 May 2024 09:40 PM
#Racket

How to resolve the algorithm Display a linear combination step by step in the Racket programming language

Table of Contents

Problem Statement

Display a finite linear combination in an infinite vector basis

(

e

1

,

e

2

, … )

{\displaystyle (e_{1},e_{2},\ldots )}

. Write a function that, when given a finite list of scalars

(

α

1

,

α

2

, … )

{\displaystyle (\alpha ^{1},\alpha ^{2},\ldots )}

, creates a string representing the linear combination

∑

i

α

i

e

i

{\displaystyle \sum {i}\alpha ^{i}e{i}}

in an explicit format often used in mathematics, that is: where

α

i

k

≠ 0

{\displaystyle \alpha ^{i_{k}}\neq 0}

The output must comply to the following rules:

Show here output for the following lists of scalars:

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Display a linear combination step by step in the Racket programming language

Source code in the racket programming language

#lang racket/base
(require racket/match racket/string)

(define (linear-combination->string es)
  (let inr ((es es) (i 1) (rv ""))
    (match* (es rv)
      [((list) "") "0"]
      [((list) rv) rv]
      [((list (? zero?) t ...) rv)
       (inr t (add1 i) rv)]
      [((list n t ...) rv)
       (define ±n
         (match* (n rv)
           ;; zero is handled above
           [(1 "") ""]
           [(1 _) "+"]
           [(-1 _) "-"]
           [((? positive? n) (not "")) (format "+~a*" n)]
           [(n _) (format "~a*" n)]))
       (inr t (add1 i) (string-append rv ±n "e("(number->string i)")"))])))

(for-each
 (compose displayln linear-combination->string)
 '((1 2 3)
   (0 1 2 3)
   (1 0 3 4)
   (1 2 0)
   (0 0 0)
   (0)
   (1 1 1)
   (-1 -1 -1)
   (-1 -2 0 -3)
   (-1)))

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