Determinant and permanent: Difference between revisions
Added Common Lisp
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return 0;
}</lang>
=={{header|CommonLisp}}==
A recursive version, no libraries required, it doesn't use much consing, only for the list of columns to skip
<lang lisp>
(defun determinant (rows &optional (skip-cols nil))
(let* ((result 0) (sgn -1))
(dotimes (col (length (car rows)) result)
(unless (member col skip-cols)
(if (null (cdr rows))
(return-from determinant (elt (car rows) col))
(incf result (* (setq sgn (- sgn)) (elt (car rows) col) (determinant (cdr rows) (cons col skip-cols)))) )))))
(defun permanent (rows &optional (skip-cols nil))
(let* ((result 0))
(dotimes (col (length (car rows)) result)
(unless (member col skip-cols)
(if (null (cdr rows))
(return-from permanent (elt (car rows) col))
(incf result (* (elt (car rows) col) (permanent (cdr rows) (cons col skip-cols)))) )))))
Test using the first set of definitions (from task description):
(setq m2
'((1 2)
(3 4)))
(setq m3
'((-2 2 -3)
(-1 1 3)
( 2 0 -1)))
(setq m4
'(( 1 2 3 4)
( 4 5 6 7)
( 7 8 9 10)
(10 11 12 13)))
(setq m5
'(( 0 1 2 3 4)
( 5 6 7 8 9)
(10 11 12 13 14)
(15 16 17 18 19)
(20 21 22 23 24)))
(dolist (m (list m2 m3 m4 m5))
(format t "~a determinant: ~a, permanent: ~a~%" m (determinant m) (permanent m)) )
</lang>
{{out}}
<pre>((1 2) (3 4)) determinant: -2, permanent: 10
((-2 2 -3) (-1 1 3) (2 0 -1)) determinant: 18, permanent: 10
((1 2 3 4) (4 5 6 7) (7 8 9 10) (10 11 12 13)) determinant: 0, permanent: 29556
((0 1 2 3 4) (5 6 7 8 9) (10 11 12 13 14) (15 16 17 18 19) (20 21 22 23 24)) determinant: 0, permanent: 6778800
</pre>
=={{header|D}}==
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