Matrix algebra operations using recursion
This post is inspired by a couple of exercises from the classical book, SICP. I found them pretty interesting as they were just using recursion & some common list operations to multiply matrices !! I also wanted to try out the Jupyter notebook kernel for MIT Scheme.
Representation & Problem Statement
Suppose we represent vectors v = ( vi ) as sequences of numbers, and matrices m = ( mij ) as sequences of vectors (the rows of the matrix). For example, the matrix
1 | 2 | 3 | 4 |
5 | 6 | 7 | 8 |
9 | 10 | 11 | 12 |
is represented as the sequence ((1 2 3 4) (5 6 7 8) (9 10 11 12))
. With this representation, we can use sequence operations to concisely express the basic matrix and vector operations.
We will look at the following 4 basic operations on matrices:
(dot-product v w)
returns the sum ∑i = vi wi(matrix-*-vector m v)
returns the vectort
, where ti = ∑j mij vj(transpose m)
returns the matrixn
, where nij = mji(matrix-*-matrix m n)
returns the matrixp
, where pij = ∑k mik nkj
Dot product of 2 vectors in this notation can be done by using 2 higher order functions, map
and fold
, both of which are implemented using recursion.
Implementation of map
:
(define (map proc items)
(if (null? items)
nil
(cons (proc (car items))
(map proc (cdr items)))))
Implementation of fold-right
:
(define (fold-right op initial sequence)
(if (null? sequence)
initial
(op (car sequence)
(fold-right op initial (cdr sequence)))))
We can use either fold-left
or fold-right
for dot-product
.
;; Define dot product of 2 vectors of equal length
(define (dot-product v w)
(fold-right + 0 (map * v w))
)
;; testing our function
(define vec1 (list 1 2 3) )
(define vec2 (list 1 1 1) )
(dot-product vec1 vec2)
6
Calculating a dot product was really easy with a couple of higher order functions! Let’s work with matrices now. We will now right a function to multiply a matrix and a vector:
(define (matrix-*-vector m v)
(map (lambda (m-row)(dot-product m-row v) )
m)
)
;; testing the function
(define mat1 (list (list 1 0 0) (list 0 1 0) (list 0 0 1)))
(matrix-*-vector mat1 vec1)
(1 2 3)
Let’s look at transpose now! For this, we will need to implement a helper function, accumulate-n
, which is similar to fold
except that it takes as its third argument a sequence of sequences, which are all assumed to have the same number of elements.
;; Defining helper functions for transpose to
;; apply the operation op to combine all the first elements of the sequences,
;; all the second elements of the sequences, and so on,
;; and returns a sequence of the results.
(define (accumulate-n op init seqs)
(if (null? (car seqs))
'()
(cons (fold-right op init (map car seqs))
(accumulate-n op init (map cdr seqs))
)
)
)
(define (transpose mat)
(accumulate-n cons '() mat)
)
;; testing transpose
(define mat2 (list (list 1 2 3) (list 4 5 6) (list 7 8 9)))
(transpose mat2)
((1 4 7) (2 5 8) (3 6 9))
Now, let’s use this transpose function to do matrix multiplication:
; Matrix multiplication
(define (matrix-*-matrix m n)
(let ((n-cols (transpose n)))
(map (lambda (m-row)(matrix-*-vector n-cols m-row))
m)
)
)
;; For testing
(matrix-*-matrix mat2 mat1)
((1 2 3) (4 5 6) (7 8 9))
We can even write recursive procedures to create some special kinds of matrices, for eg, an identity matrix. Here is a recursive procudure to create an identity matrix of length n
:
;; Create an identity matrix of length n
; N -> [List-of [List-of N]]
(define (identityM n)
(letrec ;; Documentation for letrec : https://groups.csail.mit.edu/mac/ftpdir/scheme-7.4/doc-html/scheme_3.html
(
;; N -> [List-of [List-of N]]
(generate-matrix (lambda (row)
(cond
((= row 0) '())
(else (cons (generate-row row n)
(generate-matrix (- row 1)))))))
;; N N -> [List-of N]
(generate-row (lambda (row col)
;; col goes from column n to 0
(cond
((= col 0) '())
(else (cons (if (= row col) 1 0)
(generate-row row (- col 1)))))))
)
(generate-matrix n)
)
)
(identityM 3)
((1 0 0) (0 1 0) (0 0 1))
If you found these functions interesting, I’de definitely encourage to go read SICP. I wrote about why I’m reading SICP here.