LdBeth

It seems the `fermata` attribute would cause tempo change in MIDI output, however when combined with changing tempo in `def-score`, the restored level is the last value in the `:tempo` list, which I guess is a bug. the problem can be demonstrated by the following code snippet: (setf ns '((q a4b4d5fs5cs6 mp arp e b5 e5 q fs4g4b4d5a5 arp d5) (q c3g3d4e4b4 arp h. a5b5d6 #|fermata|#) ;; uncomment this ((leg e b2 fs3 s a3 e d4 s_h e4)) ;; the tempo starts here will be whichever is the last ((leg e b2 fs3 s a3 e d4 s_h e4)) ;; in the `:tempo' list, which is 164 ((leg e d4fs4a4 mf marc d4 e4 cs4e4gs4 marc cs4 e4 d4fs4cs5 marc d4 leg)) ((leg e e4 d4fs4a4 marc d4 e4 cs4e4gs4 marc cs4 d4fs4cs5 marc d4)))) (def-score score (:key-signature '(d major) :time-signature '(4 4) :tempo '((q :rit 74 70 1/32 :bars 2) (70 2) (164 3) ) :layout (list '(:treble notes)) ) (notes :omn ns)) My current version is 2.2.26652.

I guess it is because combinations like (a3 b3 c3 b3) are acceptable but won't be available with combination

There's a count-repeat function. ? (count-repeat '(a3 b3 b3 b3)) count-repeat (1 3) Thus (let ((*do-verbose* nil)) (remove-if (lambda (x) (> (apply #'max (count-repeat x)) 1)) ;; or replace apply with (reduce #'max (count-repeat x)), essentially does the same (combination2 4 '(c3 d3 e3 f3 g3 a3 b3)))) ((a3 b3 a3 b3) (g3 b3 a3 b3) (f3 b3 a3 b3) (e3 b3 a3 b3) (d3 b3 a3 b3) (c3 b3 a3 b3) (b3 g3 a3 b3) (a3 g3 a3 b3) (f3 g3 a3 b3) (e3 g3 a3 b3) (d3 g3 a3 b3) (c3 g3 a3 b3) (b3 f3 a3 b3) (a3 f3 a3 b3) (g3 f3 a3 b3) (e3 f3 a3 b3) (d3 f3 a3 b3) (c3 f3 a3 b3) (b3 e3 a3 b3) (a3 e3 a3 b3) (g3 e3 a3 b3) (f3 e3 a3 b3) (d3 e3 a3 b3) (c3 e3 a3 b3) (b3 d3 a3 b3) (a3 d3 a3 b3) (g3 d3 a3 b3) (f3 d3 a3 b3) (e3 d3 a3 b3) (c3 d3 a3 b3) (b3 c3 a3 b3) ...

you can work with Common Lisp arrays to reduce memory usage. Here's a very direct translation of Algorithm T from http://www.kcats.org/csci/464/doc/knuth/fascicles/fasc2b.pdf which precomputes the permutation table of n distinct elements. ;; We needs to use the factorial function (defun factorial (x) ...) (defun algorithm-t (n &aux (num (factorial n))) ;; The `(integer 1 ,n) declaration will make sure CCL allocates far less memory ;; then using 'integer or 'fixnum. (let ((table (make-array num :element-type `(integer 1 ,n) :initial-element 0)) (m 2) j k d) (setf d (floor num 2) (aref table d) 1) (loop until (= m n) do (progn (incf m) (setf d (floor d m)) (setf k 0) (prog () t3 (setf k (+ k d) j (- m 1)) (loop while (> j 0) do (setf (aref table k) j j (1- j) k (+ k d))) (incf (aref table k)) (setf k (+ k d)) (loop while (< j (1- m)) do (progn (incf j) (setf (aref table k) j k (+ k d)))) (if (< k num) (go t3))) ) finally (return table)))) ;; CL-USER> (algorithm-t 4) ;; #(0 3 2 1 3 1 2 3 1 3 2 1 3 1 2 3 1 3 2 1 3 1 2 3) ;; Since Knuth uses 1 as index origin but CL uses 0, there's one extra element in the computated array. ;; here's a warper (defun n-th-permutation (n array table &aux tmp (array (copy-seq array))) (loop for k from 1 to n do (setf tmp (aref array (1- (aref table k))) (aref array (1- (aref table k))) (aref array (aref table k)) (aref array (aref table k)) tmp) finally (return array))) ;; usage: for a list of 5 pitchs '(c4 e4 f4 d4 g4) ;; first converts to a array > (setf tmp (coerce '(c4 e4 f4 d4 g4) 'array)) #(C4 E4 F4 D4 G4) ;; compute the permutation table, it could take a while for a "big" value such as 12 > (setf n-table (algorithm-t 5)) #(0 4 3 2 1 4 1 2 3 4 ... ;; then apply the n-th-permutation > (loop for i from 1 to 15 by 2 collect (n-th-permutation i tmp n-table)) ;Compiler warnings : ; In an anonymous lambda form: Undeclared free variable TMP ; In an anonymous lambda form: Undeclared free variable N-TABLE (#(C4 E4 F4 G4 D4) #(C4 G4 E4 F4 D4) #(G4 C4 E4 D4 F4) #(C4 E4 G4 D4 F4) #(C4 E4 D4 F4 G4) #(C4 D4 E4 G4 F4) #(C4 G4 D4 E4 F4) #(G4 D4 C4 E4 F4))