Left factorials: Difference between revisions

Content added Content deleted

Inline

@@ Line 3,208: / Line 3,208: @@
 =={{header|Sidef}}==
+Built-in:
+<lang ruby>say 20.of { .left_factorial }</lang>
 Straightforward:
-<lang ruby>func left_fact(k) {
+<lang ruby>func left_factorial(n) {
-    ^k -> map {|n| n! } -> sum
+    ^n -> sum { _! }
 }</lang>
-Memory efficient with ''Range.reduce()'':
+Alternatively, using ''Range.reduce()'':
-<lang ruby>func left_fact(k) {
+<lang ruby>func left_factorial(n) {
-    ^k -> reduce({ |a,b| a + b! }, 0)
+    ^n -> reduce({ |a,b| a + b! }, 0)
 }</lang>
-A much faster approach:
+A faster approach:
-<lang ruby>func left_fact(n) {
+<lang ruby>func left_factorial(n) {
     static cached    = 0
     static factorial = 1
     static leftfact  = 0
     if (n < cached) {
         cached    = 0
@@ Line 3,229: / Line 3,232: @@
         leftfact  = 0
     }
     while (n > cached) {
         leftfact  += factorial
         factorial *= ++cached
     }
     leftfact
 }</lang>
 Completing the task:
-<lang ruby>for i (0..10, 20..110 `by` 10) {
+<lang ruby>for n in (0..10, 20..110 `by` 10) {
-    printf("!%d  = %s\n", i, left_fact(i))
+    printf("!%d  = %s\n", n, left_factorial(n))
 }
-for i (1000..10000 `by` 1000) {
+for n in (1000..10000 `by` 1000) {
-    printf("!%d has %d digits.\n", i, left_fact(i).len)
+    printf("!%d has %d digits.\n", n, left_factorial(n).len)
 }</lang>