Executable library
The general idea behind an executable library is to create a library that when used as a library does one thing; but has the ability to be run directly via command line. Thus the API comes with a CLI in the very same source code file.
You are encouraged to solve this task according to the task description, using any language you may know.
Task detail
- Create a library/module/dll/shared object/... for a programming language that contains a function/method called hailstone that is a function taking a positive integer and returns the Hailstone sequence for that number.
- The library, when executed directly should satisfy the remaining requirements of the Hailstone sequence task:
- 2. Use the routine to show that the hailstone sequence for the number 27 has 112 elements starting with 27, 82, 41, 124 and ending with 8, 4, 2, 1
- 3. Show the number less than 100,000 which has the longest hailstone sequence together with that sequence's length.
- Create a second executable to calculate the following:
- Use the library's hailstone function, in the standard manner, (or document how this use deviates from standard use of a library), together with extra code in this executable, to find the hailstone length returned most often for 1 ≤ n < 100,000.
- Explain any extra setup/run steps needed to complete the task.
Notes:
- It is assumed that for a language that overwhelmingly ships in a compiled form, such as C, the library must also be an executable and the compiled user of that library is to do so without changing the compiled library. I.e. the compile tool-chain is assumed not to be present in the runtime environment.
- Interpreters are present in the runtime environment.
Ada
In Ada, any parameterless procedure can either run as a (stand-alone) main program, or can be called from another program like a library function. For the task at hand, this appears useful -- except for the following two obstacles:
1. There are neither ingoing parameters into a parameterless procedure, nor is there a return value.
2. The procedure does not know how it has been called: is it running as a main program, or has it been called from another program?
To overcome the first obstacle, we implement a very simplistic parameter passing mechanism in a package Parameter (parameter.ads): The global variable Parameter.X will hold the ingoing parameter, the other global variable Parameter.Y will take the return value. To overcome the second obstacle, we ensure that Parameter.X is 0 by default.
<lang Ada>package Parameter is
X: Natural := 0; Y: Natural;
end Parameter;</lang>
Now comes our parameterless procedure Hailstone (hailstone.adb). Note that we are using the the package Hailstones (hailstones.adb/hailstones.ads) from Hailstone sequence#Alternative method to perform the real computation.
<lang Ada>with Ada.Text_IO, Parameter, Hailstones;
procedure Hailstone is
-- if Parameter.X > 0, the length of Hailstone(Parameter.X) -- is computed and written into Parameter.Y
-- if Parameter.X = 0, Hailstone(27) and N <= 100_000 with maximal -- Hailstone(N) are computed and printed.
procedure Show_Sequence(N: Natural) is
Seq: Hailstones.Integer_Sequence := Hailstones.Create_Sequence(N);
begin
Ada.Text_IO.Put("Hailstone(" & Integer'Image(N) & " ) = (");
if Seq'Length < 8 then
for I in Seq'First .. Seq'Last-1 loop
Ada.Text_IO.Put(Integer'Image(Seq(I)) & ",");
end loop;
else
for I in Seq'First .. Seq'First+3 loop
Ada.Text_IO.Put(Integer'Image(Seq(I)) & ",");
end loop;
Ada.Text_IO.Put(" ...,");
for I in Seq'Last-3 .. Seq'Last-1 loop
Ada.Text_IO.Put(Integer'Image(Seq(I)) &",");
end loop;
end if;
Ada.Text_IO.Put_Line(Integer'Image(Seq(Seq'Last)) & " ); Length: " &
Integer'Image(seq'Length));
end Show_Sequence;
begin
if Parameter.X>0 then
Parameter.Y := Hailstones.Create_Sequence(Parameter.X)'Length;
else
Show_Sequence(27);
declare
Longest: Natural := 0;
Longest_Length: Natural := 0;
begin
for I in 2 .. 100_000 loop
if Hailstones.Create_Sequence(I)'Length > Longest_Length then
Longest := I;
Longest_Length := Hailstones.Create_Sequence(I)'Length;
end if;
end loop;
Ada.Text_IO.Put("Longest<=100_000: ");
Show_Sequence(Longest);
end;
end if;
end Hailstone;</lang>
If we compile this and run it, we get the following output:
> ./hailstone Hailstone( 27 ) = ( 27, 82, 41, 124, ..., 8, 4, 2, 1 ); Length: 112 Longest<=100_000: Hailstone( 77031 ) = ( 77031, 231094, 115547, 346642, ..., 8, 4, 2, 1 ); Length: 351
To use the same procedure like a library function, we need a specification (file hailstone.ads), that essentially repeats the parameter profile. As our procedure is actually parameterless, this specification is more than trivial.
<lang Ada>procedure Hailstone;</lang>
Finally, we write another parameterless procedure (hailstone_test.adb), that will call the procedure Hailstone. Note that we must change the Parameter.X to a value > 0 before calling Hailstone, otherwise, Hailstone would act as if it where the main program.
<lang Ada>with Hailstone, Parameter, Ada.Text_IO;
procedure Hailstone_Test is
Counts: array (1 .. 100_000) of Natural := (others => 0); Max_Count: Natural := 0; Most_Common: Positive := Counts'First; Length: Natural renames Parameter.Y; Sample: Natural := 0;
begin
for I in Counts'Range loop
Parameter.X := I;
Hailstone; -- compute the length of Hailstone(I)
Counts(Length) := Counts(Length)+1;
end loop;
for I in Counts'Range loop
if Counts(I) > Max_Count then
Max_Count := Counts(I);
Most_Common := I;
end if;
end loop;
Ada.Text_IO.Put_Line("Most frequent length:"
& Integer'Image(Most_Common)
& ";" & Integer'Image(Max_Count)
& " sequences of that length.");
for I in Counts'Range loop
Parameter.X := I;
Hailstone; -- compute the length of Hailstone(I)
if Length = Most_Common then
Sample := I;
exit;
end if;
end loop;
Ada.Text_IO.Put_Line("The first such sequence: Hailstone("
& Integer'Image(Sample) & " ).");
end Hailstone_Test;</lang>.
Compiling and running this gives the following output:
> ./hailstone_test Most frequent length: 72; 1467 sequences of that length. The first such sequence: Hailstone( 444 ).
Note that using global variables for parameter and return value passing works here, but is bad programming practice. Ada is a compiled language, and it is not clear how useful an executable library written in a compiled language is, anyway.
In fact, except for the constraints imposed by this task, there is no reason to ask the procedure Hailstone for the length of a Hailstone sequence -- solid software engineering practice would require to directly call the parameterized function Hailstones.Create_Sequence.
AutoHotkey
First we create the library, hailstone.ahk: <lang AHK>#NoEnv SetBatchLines, -1
- Check if we're executed directly
If (A_LineFile = A_ScriptFullPath){ h27 := hailstone(27) MsgBox % "Length of hailstone 27: " (m := h27.MaxIndex()) "`nStarts with " . h27[1] ", " h27[2] ", " h27[3] ", " h27[4] . "`nEnds with " . h27[m-3] ", " h27[m-2] ", " h27[m-1] ", " h27[m]
Loop 100000 { h := hailstone(A_Index) If (h.MaxIndex() > m) m := h.MaxIndex(), longest := A_Index } MsgBox % "Longest hailstone is that of " longest " with a length of " m "!" }
hailstone(n){
out := [n]
Loop
n := n & 1 ? n*3+1 : n//2, out.insert(n)
until n=1
return out
}</lang>
Running this directly gives the output:
Length of hailstone 27: 112 Starts with 27, 82, 41, 124 Ends with 8, 4, 2, 1 Longest hailstone is that of 77031 with a length of 351!
Then we can create a file (test.ahk) that uses the library (note the #Include line): <lang AHK>#NoEnv
- Include %A_ScriptDir%\hailstone.ahk
SetBatchLines -1
col := Object(), highestCount := 0
Loop 100000 { length := hailstone(A_Index).MaxIndex() if not col[length] col[length] := 0 col[length]++ } for length, count in col if (count > highestCount) highestCount := count, highestN := length MsgBox % "the most common length was " highestN "; it occurred " highestCount " times."</lang> Running this does not trigger the output of the hailstone.ahk, instead it outputs this:
the most common length was 72; it occurred 1467 times.
BBC BASIC
To meet the terms of this task the BBC BASIC run-time engine bbcwrun.exe must be installed on the target PC and the file extension .bbc must be associated with this executable. This is normally the case when BBC BASIC for Windows has been installed.
Library
This must be saved as the file HAILSTONE.BBC. It may be used as a library (see below) or executed directly. <lang bbcbasic> seqlen% = FNhailstone(27)
PRINT "Sequence length for 27 is "; seqlen%
maxlen% = 0
FOR number% = 2 TO 100000
seqlen% = FNhailstone(number%)
IF seqlen% > maxlen% THEN
maxlen% = seqlen%
maxnum% = number%
ENDIF
NEXT
PRINT "The number with the longest hailstone sequence is " ; maxnum%
PRINT "Its sequence length is " ; maxlen%
END
DEF FNhailstone(N%)
LOCAL L%
WHILE N% <> 1
IF N% AND 1 THEN N% = 3 * N% + 1 ELSE N% DIV= 2
L% += 1
ENDWHILE
= L% + 1</lang>
- Output:
Sequence length for 27 is 112 The number with the longest hailstone sequence is 77031 Its sequence length is 351
Client
This uses the above program as a library: <lang bbcbasic> INSTALL "HAILSTONE"
DIM freq%(351)
FOR number% = 2 TO 100000
seqlen% = FNhailstone(number%)
freq%(seqlen%) += 1
NEXT
max% = 0
FOR i% = 0 TO 351
IF freq%(i%) > max% THEN
max% = freq%(i%)
mostcommon% = i%
ENDIF
NEXT
PRINT "The most common sequence length is " ; mostcommon%
PRINT "It occurs " ; max% " times"
END</lang>
- Output:
The most common sequence length is 72 It occurs 1467 times
C
Solution for Linux/GCC. First, header file hailstone.h: <lang C>#ifndef HAILSTONE
- define HAILSTONE
long hailstone(long, long**); void free_sequence(long *);
- endif/*HAILSTONE*/</lang>
Then the lib source code hailstone.c (actual name doesn't matter): <lang C>#include <stdio.h>
- include <stdlib.h>
long hailstone(long n, long **seq) {
long len = 0, buf_len = 4;
if (seq)
*seq = malloc(sizeof(long) * buf_len);
while (1) {
if (seq) {
if (len >= buf_len) {
buf_len *= 2;
*seq = realloc(*seq, sizeof(long) * buf_len);
}
(*seq)[len] = n;
}
len ++;
if (n == 1) break;
if (n & 1) n = 3 * n + 1;
else n >>= 1;
}
return len;
}
void free_sequence(long * s) { free(s); }
const char my_interp[] __attribute__((section(".interp"))) = "/lib/ld-linux.so.2"; /* "ld-linux.so.2" should be whatever you use on your platform */
int hail_main() /* entry point when running along, see compiler command line */ {
long i, *seq;
long len = hailstone(27, &seq);
printf("27 has %ld numbers in sequence:\n", len);
for (i = 0; i < len; i++) {
printf("%ld ", seq[i]);
}
printf("\n");
free_sequence(seq);
exit(0);
}</lang> A program to use the lib (I call it test.c): <lang C>#include <stdio.h>
- include "hailstone.h"
int main() {
long i, longest, longest_i, len;
longest = 0;
for (i = 1; i < 100000; i++) {
len = hailstone(i, 0);
if (len > longest) {
longest_i = i;
longest = len;
}
}
printf("Longest sequence at %ld, length %ld\n", longest_i, longest);
return 0;
}</lang>
Building the lib: gcc -Wall -W -fPIC -shared -o libhail.so hailstone.c -lc -Wl,-e,hail_main
Building the test.c code: gcc -Wall -L. -lhail test.c -o hailtest
Running the lib:
% ./libhail.so 27 has 112 numbers in sequence: 27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274....
Running the program:
% LD_LIBRARY_PATH=. ./hailtest Longest sequence at 77031, length 351
For a serious library the libhail.so would have been put into a system lib dir, but for now we'll just leave it in the same directory, so to run the program, we need to give additional hints to tell it where to find the lib: LD_LIBRARY_PATH=. ./hailtest
Clojure
In a project's source directory, you might have a subfolder named rosetta_code with two Clojure source files:
.
├── clojure.jar
└── rosetta_code
├── frequent_hailstone_lengths.clj
└── hailstone_sequence.clj
The file hailstone_sequence.clj is an executable library:
<lang clojure>(ns rosetta-code.hailstone-sequence)
(defn next-in-hailstone
"Returns the next number in the Hailstone sequence that starts with x.
If x is less than 2, returns nil."
[x]
(when (> x 1)
(if (even? x)
(/ x 2)
(inc (* 3 x)))))
(defn hailstone-seq
"Returns a lazy Hailstone sequence starting with the number x."
[x]
(take-while some?
(iterate next-in-hailstone x)))
(defn -main [& args]
(let [h27 (hailstone-seq 27)]
(printf "The Hailstone sequence starting at 27 contains %s elements:\n%s ... %s.\n"
(count h27)
(vec (take 4 h27))
(vec (take-last 4 h27)))
(let [[number length] (apply max-key second
(map (fn [x] [x (count (hailstone-seq x))])
(range 100000)))]
(printf "The number %s has the longest Hailstone sequence under 100000, of length %s.\n"
number length))))</lang>
You can run it from the command line (clojure.jar is the jar that contains the Clojure language):
$ java -cp clojure.jar clojure.main -m rosetta-code.hailstone-sequence The Hailstone sequence starting at 27 contains 112 elements: [27 82 41 124] ... [8 4 2 1]. The number 77031 has the longest Hailstone sequence under 100000, of length 351.
You can also use its functions from other programs. The file frequent_hailstone_lengths.clj:
<lang clojure>(ns rosetta-code.frequent-hailstone-lengths
(:require [rosetta-code.hailstone-sequence
:refer [hailstone-seq]]))
(defn -main [& args]
(let [frequencies (apply merge-with +
(for [x (range 1 100000)]
{(count (hailstone-seq x)) 1}))
[most-frequent-length frequency]
(apply max-key val (seq frequencies))]
(printf (str "The most frequent Hailstone sequence length for numbers under 100000 is %s,"
" with a frequency of %s.\n")
most-frequent-length frequency)))</lang>
You can run it from the command line:
$ java -cp clojure.jar clojure.main -m rosetta-code.frequent-hailstone-lengths The most frequent Hailstone sequence length for numbers under 100000 is 72, with a frequency of 1467.
Clojure also supports ahead-of-time compilation to class files, and the standard packaging tools Leiningen and Boot provide meta-commands to compile and package Clojure programs into stand-alone runnable jars. At this point, Clojure has more or less the same executable library story as Java (class files can be named as the main entry point, and they can be included by other classes that import them).
Déjà Vu
The library, named hailstone.deja:
<lang dejavu>local hailstone:
swap [ over ]
while < 1 dup:
if % over 2:
#odd
++ * 3
else:
#even
/ swap 2
swap push-through rot dup
drop
if = (name) :(main): local :h27 hailstone 27 !. = 112 len h27 !. = 27 h27! 0 !. = 82 h27! 1 !. = 41 h27! 2 !. = 124 h27! 3 !. = 8 h27! 108 !. = 4 h27! 109 !. = 2 h27! 110 !. = 1 h27! 111
local :max 0 local :maxlen 0 for i range 1 99999: dup len hailstone i if < maxlen: set :maxlen set :max i else: drop !print( "number: " to-str max ", length: " to-str maxlen ) else: @hailstone</lang>
The client: <lang dejavu>!import!hailstone
local :counts {} set-default counts 0 for i range 1 99999: set-to counts swap ++ counts! dup len hailstone i
local :maxlen 0 for k in keys counts: if < maxlen counts! k: set :maxlen counts! k !print( "Maximum length: " to-str maxlen ) </lang>
Factor
An executable library is a vocabulary with a main entry point.
This vocabulary, rosetta.hailstone, exports the word hailstone, but also uses MAIN: to declare a main entry point.
<lang factor>! rosetta/hailstone/hailstone.factor USING: arrays io kernel math math.ranges prettyprint sequences vectors ; IN: rosetta.hailstone
- hailstone ( n -- seq )
[ 1vector ] keep
[ dup 1 number= ]
[
dup even? [ 2 / ] [ 3 * 1 + ] if
2dup swap push
] until
drop ;
<PRIVATE
- main ( -- )
27 hailstone dup dup "The hailstone sequence from 27:" print " has length " write length . " starts with " write 4 head [ unparse ] map ", " join print " ends with " write 4 tail* [ unparse ] map ", " join print
! Maps n => { length n }, and reduces to longest Hailstone sequence.
1 100000 [a,b)
[ [ hailstone length ] keep 2array ]
[ [ [ first ] bi@ > ] most ] map-reduce
first2
"The hailstone sequence from " write pprint
" has length " write pprint "." print ;
PRIVATE>
MAIN: main</lang>
There are two ways to run this program:
- Inside Factor, from its listener:
"rosetta.hailstone" run - Outside Factor, from some shell:
./factor -run=rosetta.hailstone
$ ./factor -run=rosetta.hailstone Loading resource:work/rosetta/hailstone/hailstone.factor The hailstone sequence from 27: has length 112 starts with 27, 82, 41, 124 ends with 8, 4, 2, 1 The hailstone sequence from 77031 has length 351.
Any other Factor program can also use rosetta.hailstone as a regular vocabulary. This program only uses the word hailstone from that vocabulary, and never calls the main entry point of rosetta.hailstone.
<lang factor>! rosetta/hailstone/length/length.factor USING: assocs kernel io math math.ranges prettyprint
rosetta.hailstone sequences ;
IN: rosetta.hailstone.length
<PRIVATE
- f>0 ( object/f -- object/0 )
dup [ drop 0 ] unless ;
- max-value ( pair1 pair2 -- pair )
[ [ second ] bi@ > ] most ;
- main ( -- )
H{ } clone ! Maps sequence length => count.
1 100000 [a,b) [
hailstone length ! Find sequence length.
over [ f>0 1 + ] change-at ! Add 1 to count.
] each
! Find the length-count pair with the highest count.
>alist unclip-slice [ max-value ] reduce
first2 swap
"Among Hailstone sequences from 1 <= n < 100000," print
"there are " write pprint
" sequences of length " write pprint "." print ;
PRIVATE>
MAIN: main</lang>
$ ./factor -run=rosetta.hailstone.length Loading resource:work/rosetta/hailstone/length/length.factor Loading resource:work/rosetta/hailstone/hailstone.factor Among Hailstone sequences from 1 <= n < 100000, there are 72 sequences of length 1467.
Go
To be directly executable, a Go application must contain a package called 'main' which includes a main() function from which execution begins automatically. Any other application is regarded as a library.
However, a package can consist of not just one but several source code files contained in the same directory.
Consequently, Go can emulate an executable library by splitting package 'main' into two files:
1. A file including the library code which includes (say) a libMain() function but no main() function as such.
2. A file including a main() function which calls libMain().
If these two files are 'run' as a package, then they can mimic executing the library code directly.
To use the library from another executable application, all that is needed is to copy the first file to the application's directory and then 'run' the files as a package. Care would be needed to ensure that there are no name clashes in the 'combined' package.
Note that, although the 'go run' command may look like it's launching an interpreter, it does in fact build an application from the package files in a temporary directory and then executes it in a single step.
To complete the task, first create these two files in the 'modulino' directory:
<lang go>// modulino.go package main
import "fmt"
// Function borrowed from Hailstone sequence task. // 1st arg is the number to generate the sequence for. // 2nd arg is a slice to recycle, to reduce garbage. func hailstone(n int, recycle []int) []int {
s := append(recycle[:0], n)
for n > 1 {
if n&1 == 0 {
n = n / 2
} else {
n = 3*n + 1
}
s = append(s, n)
}
return s
}
func libMain() {
seq := hailstone(27, nil)
fmt.Println("\nHailstone sequence for the number 27:")
fmt.Println(" has", len(seq), "elements")
fmt.Println(" starts with", seq[0:4])
fmt.Println(" ends with", seq[len(seq)-4:])
var longest, length int
for i := 1; i < 100000; i++ {
if le := len(hailstone(i, nil)); le > length {
longest = i
length = le
}
}
fmt.Printf("\n%d has the longest Hailstone sequence, its length being %d.\n", longest, length)
}</lang>
<lang go>// modulino_main.go package main
func main() {
libMain()
}</lang>
To emulate an executable library:
- Output:
$ go run modulino Hailstone sequence for the number 27: has 112 elements starts with [27 82 41 124] ends with [8 4 2 1] 77031 has the longest Hailstone sequence, its length being 351.
Now create this file in the 'hailstone' directory:
<lang go>// hailstone.go package main
import "fmt"
func main() {
freq := make(map[int]int)
for i := 1; i < 100000; i++ {
freq[len(hailstone(i, nil))]++
}
var mk, mv int
for k, v := range freq {
if v > mv {
mk = k
mv = v
}
}
fmt.Printf("\nThe Hailstone length returned most is %d, which occurs %d times.\n", mk, mv)
}</lang>
and copy modulino.go to the 'hailstone' directory. The library can then be used in the 'normal' way:
- Output:
$ go run hailstone The Hailstone length returned most is 72, which occurs 1467 times.
Io
HailStone.io <lang io>HailStone := Object clone HailStone sequence := method(n,
if(n < 1, Exception raise("hailstone: expect n >= 1 not #{n}" interpolate))
n = n floor // make sure integer value
stones := list(n)
while (n != 1,
n = if(n isEven, n/2, 3*n + 1)
stones append(n)
)
stones
)
if( isLaunchScript,
out := HailStone sequence(27)
writeln("hailstone(27) has length ",out size,": ",
out slice(0,4) join(" ")," ... ",out slice(-4) join(" "))
maxSize := 0
maxN := 0
for(n, 1, 100000-1,
out = HailStone sequence(n)
if(out size > maxSize,
maxSize = out size
maxN = n
)
)
writeln("For numbers < 100,000, ", maxN,
" has the longest sequence of ", maxSize, " elements.")
)</lang> client.io <lang io>counts := Map clone for(n, 1, 100000-1,
out := HailStone sequence(n) key := out size asCharacter counts atPut(key, counts atIfAbsentPut(key, 0) + 1)
)
maxCount := counts values max lengths := list() counts foreach(k,v,
if(v == maxCount, lengths append(k at(0)))
)
if(lengths size == 1,
writeln("The most frequent sequence length for n < 100,000 is ",lengths at(0),
" occurring ",maxCount," times.")
,
writeln("The most frequent sequence lengths for n < 100,000 are:\n",
lengths join(",")," occurring ",maxCount," times each.")
)</lang> Terminal Session
$ io HailStone.io hailstone(27) has length 112: 27 82 41 124 ... 8 4 2 1 For numbers < 100,000, 77031 has the longest sequence of 351 elements. $ io client.io The most frequent sequence length for n < 100,000 is 72 occurring 1467 times. $
Io has a built-in auto importer that will attempt to automatically
import a file with the same name (plus "io" extension) as an undefined
prototype object when it is first referenced. To be imported the first letter of the
prototype name must also be upper case which by convention indicates a
prototype that has the same role as a class in class based object-oriented
languages. The Importer's default search path is the current working
directory, but you can add search paths using Importer addSearchPath().
J
This is the executable library:
<lang j>hailseq=: -:`(1 3&p.)@.(2&|) ^:(1 ~: ]) ^:a:"0 9!:29]1 9!:27'main 0' main=:3 :0
smoutput 'Hailstone sequence for the number 27' smoutput hailseq 27 smoutput smoutput 'Finding number with longest hailstone sequence which is' smoutput 'less than 100000 (and finding that sequence length):' smoutput (I.@(= >./),>./) #@hailseq i.1e5
)</lang>
Running it might look like this:
<lang j> load jpath '~temp/hailseq.ijs' Hailstone sequence for the number 27 27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 ... Finding number with longest hailstone sequence which is less than 100000 (and finding that sequence length): 77031 351</lang>
This is the program which uses the library part of that executable library:
<lang j>require '~temp/hailseq.ijs' 9!:29]1 9!:27'main 0' main=:3 :0
smoutput 'Finding most frequent hailstone sequence length for'
smoutput 'Hailstone sequences for whole numbers less than 100000:'
smoutput {:{.\:~ (#/.~,.~.) #@hailseq }.i.1e5
) </lang>
Running it might look like this: <lang j> load jpath '~temp/66.ijs' Finding most frequent hailstone sequence length for Hailstone sequences for whole numbers less than 100000 72</lang>
Notes: 9!:29]1 tells the interpeter to run a phrase. 9!:27'phrase' tells the interpeter the phrase to execute. (9!: means, in essence: standard library number 9, and 9!:29 identifies a specific entry point in that library.)
In 66.ijs we can not use the presence of 9!:29]1 from hailseq.ijs because hailseq.ijs was loaded with require which means that if it had already been loaded it will not be loaded again.
(And, 66 here is just an arbitrary temporary file name.)
Java
Java files are compiled into bytecode class files. The CLASSPATH will specify the location of class files.
Library
<lang java> import java.util.ArrayList; import java.util.List;
// task 1 public class HailstoneSequence {
public static void main(String[] args) {
// task 2
int n = 27;
List<Long> sequence27 = hailstoneSequence(n);
System.out.printf("Hailstone sequence for %d has a length of %d:%nhailstone(%d) = %s%n", n, sequence27.size(), n, sequence27);
// task 3
int maxN = 0;
int maxLength = 0;
for ( int i = 1 ; i < 100_000 ; i++ ) {
int seqLength = hailstoneSequence(i).size();
if ( seqLength > maxLength ) {
maxLength = seqLength;
maxN = i;
}
}
System.out.printf("Longest hailstone sequence less than 100,000: hailstone(%d).length() = %d", maxN, maxLength);
}
public static List<Long> hailstoneSequence(long n) {
if ( n <= 0 ) {
throw new IllegalArgumentException("Must be grater than or equal to zero.");
}
List<Long> sequence = new ArrayList<>();
sequence.add(n);
while ( n > 1 ) {
if ( (n & 1) == 0 ) {
n /= 2;
}
else {
n = 3 * n + 1;
}
sequence.add(n);
}
return sequence;
}
} </lang>
- Output:
Hailstone sequence for 27 has a length of 112: hailstone(27) = [27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1] Longest hailstone sequence less than 100,000: hailstone(77031).length() = 351
Client
<lang java> import java.util.HashMap; import java.util.Map; import java.util.stream.IntStream;
public class ExecutableLibrary {
public static void main(String[] args) {
Map<Integer,Integer> lengthMap = new HashMap<>();
IntStream.range(1, 100_000)
.map(n -> HailstoneSequence.hailstoneSequence(n).size())
.forEach(len -> lengthMap.merge(len, 1, (v1, v2) -> v1 + v2));
int mostOften = lengthMap.values()
.stream()
.mapToInt(x -> x)
.max().orElse(-1);
System.out.printf("The most frequent hailstone length for n < 100,000 is %d.%n", mostOften);
}
} </lang>
- Output:
The most frequent hailstone length for n < 100,000 is 1467.
Julia
A Julia module can check to see if it is also the file run from the command lime by checking the PROGRAM_NAME variable. <lang julia>
- in file hailstone.jl ###############
module Hailstone
function hailstone(n)
ret = [n]
while n > 1
if n & 1 > 0
n = 3n + 1
else
n = Int(n//2)
end
append!(ret, n)
end
return ret
end
export hailstone
end
if PROGRAM_FILE == "hailstone.jl"
using Hailstone
h = hailstone(27)
n = length(h)
println("The sequence of hailstone(27) is:\n $h.\nThis sequence is of length $n. It starts with $(h[1:4]) and ends with $(h[n-3:end]).")
end
- in file moduletest.jl ####################
include("hailstone.jl") using Hailstone function countstones(mi, mx)
lengths2occurences = Dict()
mostfreq = mi
maxcount = 1
for i in mi:mx
h = hailstone(i)
n = length(h)
if haskey(lengths2occurences, n)
newoccurences = lengths2occurences[n] + 1
if newoccurences > maxcount
maxcount = newoccurences
mostfreq = n
end
lengths2occurences[n] = newoccurences
else
lengths2occurences[n] = 1
end
end
mostfreq, maxcount
end
nlen, cnt = countstones(1,99999)
print("The most common hailstone sequence length for hailstone(n) for 1 <= n < 100000 is $nlen, which occurs $cnt times.") </lang>
- Output:
Running hailstone.jl standalone: The sequence of hailstone(27) is:
[27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1].This sequence is of length 112. It starts with [27, 82, 41, 124] and ends with [8, 4, 2, 1]. Running the program that uses the Hailstone module: The most common hailstone sequence length for hailstone(n) for 1 <= n < 100000 is 72, which occurs 1467 times.
Limbo
There's no real difference in compilation or output
for libraries versus commands in Inferno;
commands (by convention) are expected to define an init() function that accepts a reference to a graphical context and a list of strings (i.e., the argument list) in order to satisy the type-checker.
So this task is fairly simple.
First, execlib.b looks like this:
<lang Limbo>implement Execlib;
include "sys.m"; sys: Sys; include "draw.m";
Execlib: module { init: fn(ctxt: ref Draw->Context, args: list of string); hailstone: fn(i: big): list of big; };
init(nil: ref Draw->Context, nil: list of string) { sys = load Sys Sys->PATH;
seq := hailstone(big 27); l := len seq;
sys->print("hailstone(27): "); for(i := 0; i < 4; i++) { sys->print("%bd, ", hd seq); seq = tl seq; } sys->print("⋯");
while(len seq > 4) seq = tl seq;
while(seq != nil) { sys->print(", %bd", hd seq); seq = tl seq; } sys->print(" (length %d)\n", l);
max := 1; maxn := big 1; for(n := big 2; n < big 100000; n++) { cur := len hailstone(n); if(cur > max) { max = cur; maxn = n; } } sys->print("hailstone(%bd) has length %d\n", maxn, max); }
hailstone(i: big): list of big { if(i == big 1) return big 1 :: nil; if(i % big 2 == big 0) return i :: hailstone(i / big 2); return i :: hailstone(big 3 * i + big 1); } </lang>
And execsexeclib.b (which executes execlib) looks like this:
<lang Limbo>implement ExecsExeclib;
include "sys.m"; sys: Sys; include "draw.m";
ExecsExeclib: module { init: fn(ctxt: ref Draw->Context, args: list of string); };
- Usually, this would be placed into something like "execlib.m",
- but it's fine here.
Execlib: module { hailstone: fn(i: big): list of big; };
init(nil: ref Draw->Context, nil: list of string) { sys = load Sys Sys->PATH; # This program expects that the result of compiling Execlib is execlib.dis, # so you'll need to adjust this line if you used a different filename. lib := load Execlib "execlib.dis"; if(lib == nil) die("Couldn't load execlib.dis");
counts := array[352] of { * => 0 }; for(i := 1; i < 100000; i++) { counts[len lib->hailstone(big i)]++; }
max := 0; maxi := 0; for(i = 1; i < len counts; i++) { if(counts[i] > max) { max = counts[i]; maxi = i; } } sys->print("The most common sequence length is %d (encountered %d times)\n", maxi, max); }
die(s: string) { sys->fprint(sys->fildes(2), "runls: %s: %r", s); raise "fail:errors"; } </lang>
- Output:
% apply {limbo $1} *execlib.b
% ./execlib
hailstone(27): 27, 82, 41, 124, ⋯, 8, 4, 2, 1 (length 112)
hailstone(77031) has length 351
% ./execsexeclib
The most common sequence length is 72 (encountered 1467 times)
Lua
The library is called hailstone.lua: <lang Lua>#!/usr/bin/env luajit bit32=bit32 or bit local lib={ hailstone=function(n) local seq={n} while n>1 do n=bit32.band(n,1)==1 and 3*n+1 or n/2 seq[#seq+1]=n end return seq end }
if arg[0] and arg[0]:match("hailstone.lua") then local function printf(fmt, ...) io.write(string.format(fmt, ...)) end local seq=lib.hailstone(27) printf("27 has %d numbers in sequence:\n",#seq) for _,i in ipairs(seq) do printf("%d ", i) end printf("\n") else return lib end </lang>
The executable is called hailtest.lua:
<lang Lua>#!/usr/bin/env luajit local lib=require"hailstone" local longest,longest_i=0,0 for i=1,1e5 do local len=#lib.hailstone(i) if len>longest then longest_i=i longest=len end end print("Longest sequence at "..longest_i..", length "..longest) </lang>
- Output:
> ./hailstone.lua 27 has 112 numbers in sequence: 27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 … 8 4 2 1 > ./hailtest.lua Longest sequence at 77031, length 351
Mathematica
Library
<lang Mathematica>hailstone[1] = {1}; hailstone[n_] :=
hailstone[n] = Prepend[hailstone[If[EvenQ[n], n/2, 3 n + 1]], n];
If[$ScriptCommandLine1 == $Input,
val = hailstone[27]; Print["hailstone(27) starts with ", val;; 4, ", ends with ", val-4 ;;, ", and has length ", Length[val], "."]; val = MaximalBy[Range[99999], Length@*hailstone]1; Print[val, " has the longest hailstone sequence with length ", Length[hailstone[val]], "."]];</lang>
- Output:
hailstone(27) starts with {27, 82, 41, 124}, ends with {8, 4, 2, 1}, and has length 112.
77031 has the longest hailstone sequence with length 351.
Client
This assumes that the library is named hailstone.m. <lang Mathematica><< hailstone.m; Print["The most common hailstone length is ",
Commonest[Length@*hailstone /@ Range[99999]]1, "."];</lang>
- Output:
The most common hailstone length is 72.
Nanoquery
The global boolean value 'main' is true is the program is being executed directly and false if it is being imported as a library. <lang Nanoquery>def hailstone(n) seq = list()
while (n > 1) append seq n if (n % 2)=0 n = int(n / 2) else n = int((3 * n) + 1) end end append seq n return seq end
if main h = hailstone(27) println "hailstone(27)" println "total elements: " + len(hailstone(27)) print h[0] + ", " + h[1] + ", " + h[2] + ", " + h[3] + ", ..., " println h[-4] + ", " + h[-3] + ", " + h[-2] + ", " + h[-1]
max = 0 maxLoc = 0 for i in range(1,99999) result = len(hailstone(i)) if (result > max) max = result maxLoc = i end end print "\nThe number less than 100,000 with the longest sequence is " println maxLoc + " with a length of " + max end</lang>
NetRexx
The NetRexx compiler can generate Java classes and in common with all Java classes, public methods within each class are available for use by other programs. Packaging a class in a JAR file effectively crates a library that can be used by any other Java program. If this file is constructed correctly it can also by delivered as an "executable JAR file" which can be launched via the -jar switch of the java command. The following command can be used to package the NetRexx Hailstone Sequence sample as an executable JAR file:
$ jar cvfe RHailstoneSequence.jar RHailstoneSequence RHailstoneSequence.class added manifest adding: RHailstoneSequence.class(in = 2921) (out= 1567)(deflated 46%)
Here, the e switch causes the jar program to add a Main-Class property to the generated jar manifest which now contains the following:
Manifest-Version: 1.0 Created-By: 1.7.0_15 (Oracle Corporation) Main-Class: RHailstoneSequence
With this Main-Class property present, launching the program via java -jar will cause Java to attempt to execute the main() method of the program specified above (RHailstoneSequence):
$ java -jar RHailstoneSequence.jar The number 27 has a hailstone sequence comprising 112 elements its first four elements are: 27 82 41 124 and last four elements are: 8 4 2 1 The number 77031 has the longest hailstone sequence in the range 1 to 99999 with a sequence length of 351
Using this JAR file as a library, the following program can use the hailstone(N) method to complete the task:
<lang NetRexx>/* NetRexx */ options replace format comments java crossref symbols nobinary
import RHailstoneSequence
runSample(arg) return
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ method runSample(arg) private static
parse arg beginNum endNum .
if beginNum = | beginNum = '.' then beginNum = 1
if endNum = | endNum = '.' then endNum = 100000
if beginNum > endNum then signal IllegalArgumentException('Gronk!')
-- collect sequences hailstones = 0 loop hn = beginNum while hn < endNum hslist = RHailstoneSequence.hailstone(hn) hscount = hslist.words() hailstones[hscount] = hailstones[hscount] + 1 end hn
-- locate most common
mostOftenNum = 0
mostOftenCount = 0
loop hn = beginNum while hn < endNum
if hailstones[hn] > mostOftenCount then do
mostOftenCount = hailstones[hn]
mostOftenNum = hn
end
end hn
say 'The length of hailstone sequence that is most common in the range' beginNum '<= N <' endNum 'is' mostOftenNum'. It occurs' mostOftenCount 'times.' return
</lang>
The program can then be launched with the java command. In this sample the JAR file is included via the -cp switch:
- Output:
$ java -cp .:RHailstoneSequence.jar RHailstoneSequenceUser The length of hailstone sequence that is most common in the range 1 <= N < 100000 is 72. It occurs 1467 times.
Nim
<lang nim>proc hailstone*(n): auto =
result = @[n]
var n = n
while n > 1:
if (n and 1) == 1:
n = 3 * n + 1
else:
n = n div 2
result.add n
when isMainModule:
let h = hailstone 27
assert h.len == 112 and h[0..3] == @[27,82,41,124] and h[h.high-3..h.high] == @[8,4,2,1]
var m, mi = 0
for i in 1 .. <100_000:
let n = hailstone(i).len
if n > m:
m = n
mi = i
echo "Maximum length ", m, " was found for hailstone(", mi, ") for numbers <100,000"</lang>
In Nim the value isMainModule is set at compiletime, and we can use it with when (a compiletime if).
Library Importing Example <lang nim>import hailstone, tables
var t = initCountTable[int]()
for i in 1 .. <100_000:
t.inc(hailstone(i).len)
let (val, cnt) = t.largest() echo "The length of hailstone sequence that is most common for" echo "hailstone(n) where 1<=n<100000, is ", val, ". It occurs ", cnt, " times."</lang>
- Output:
The length of hailstone sequence that is most common for hailstone(n) where 1<=n<100000, is 72. It occurs 1467 times.
PARI/GP
Linux solution (i386 and x86_64).
Library file hailstone.c: <lang C>#include <pari/pari.h>
- define HAILSTONE1 "n=1;print1(%d,\": \");apply(x->while(x!=1,if(x/2==x\\2,x/=2,x=x*3+1);n++;print1(x,\", \")),%d);print(\"(\",n,\")\n\")"
- define HAILSTONE2 "m=n=0;for(i=2,%d,h=1;apply(x->while(x!=1,if(x/2==x\\2,x/=2,x=x*3+1);h++),i);if(m<h,m=h;n=i));print(n,\": \",m)"
void hailstone1(int x) {
char buf[1024];
snprintf(buf, sizeof(buf), HAILSTONE1, x, x);
pari_init(1000000, 2);
geval(strtoGENstr(buf));
pari_close();
}
void hailstone2(int range) {
char buf[1024];
snprintf(buf, sizeof(buf), HAILSTONE2, range);
pari_init(1000000, 2);
geval(strtoGENstr(buf));
pari_close();
}
- if __i386__
const char _hail[] __attribute__((section(".interp"))) = "/lib/ld-linux.so.2";
- else // __x86_64__
const char _hail[] __attribute__((section(".interp"))) = "/lib64/ld-linux-x86-64.so.2";
- endif
int main(void) {
hailstone1(27); hailstone2(100000);
exit(0);
}</lang> Compile hailstone.c: gcc -O2 -Wall hailstone.c -fPIC -shared -o libhailstone.so -lpari -Wl,-e,main
Execute library: ./libhailstone.so:
27: 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1, (112) 77031: 351
Main program linking libhailstone.so... main.c: <lang C>void hailstone1(int);
int main(void) {
hailstone1(27);
return 0;
}</lang> Compile main.c: gcc -O2 -Wall main.c -o main -L. libhailstone.so
Execute main program: LD_LIBRARY_PATH=. ./main :
27: 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1, (112)
Perl
Lib package in file Hailstone.pm:
<lang perl>package Hailstone;
sub seq { my $x = shift; $x == 1 ? (1) : ($x & 1)? ($x, seq($x * 3 + 1)) : ($x, seq($x / 2)) }
my %cache = (1 => 1); sub len { my $x = shift; $cache{$x} //= 1 + ( $x & 1 ? len($x * 3 + 1) : len($x / 2)) }
unless (caller) {
for (1 .. 100_000) {
my $l = len($_);
($m, $len) = ($_, $l) if $l > $len;
}
print "seq of 27 - $cache{27} elements: @{[seq(27)]}\n";
print "Longest sequence is for $m: $len\n";
}
1;</lang>
Main program in file test.pl:<lang perl>use Hailstone;
use strict;
use warnings;
my %seqs; for (1 .. 100_000) {
$seqs{Hailstone::len($_)}++;
}
my ($most_frequent) = sort {$seqs{$b} <=> $seqs{$a}} keys %seqs; print "Most frequent length: $most_frequent ($seqs{$most_frequent} occurrences)\n";</lang>
Running the lib:
% perl Hailstone.pm seq of 27 - 112 elements: 27 82 41 124 62 31 94 47 142 ... 10 5 16 8 4 2 1 Longest sequence is for 77031: 351
Running the main program:
% perl test.pl Most frequent length: 72 (1467 occurrences)
Phix
Simply test whether include_file() returns 1. hail.exw: <lang Phix>--global (if you want to be able to call this from test.exw) function hailstone(atom n) sequence s = {n}
while n!=1 do
if remainder(n,2)=0 then
n /= 2
else
n = 3*n+1
end if
s &= n
end while
return s
end function
global function hailstone_count(atom n) integer count = 1
while n!=1 do
if remainder(n,2)=0 then
n /= 2
else
n = 3*n+1
end if
count += 1
end while
return count
end function
if include_file()==1 then
sequence s = hailstone(27)
integer ls = length(s)
s[5..-5] = {".."}
puts(1,"hailstone(27) = ")
? s
printf(1,"length = %d\n\n",ls)
integer hmax = 1, imax = 1,count
for i=2 to 1e5-1 do
count = hailstone_count(i)
if count>hmax then
hmax = count
imax = i
end if
end for
printf(1,"The longest hailstone sequence under 100,000 is %d with %d elements.\n",{imax,hmax})
end if</lang>
- Output:
hailstone(27) = {27,82,41,124,"..",8,4,2,1}
length = 112
The longest hailstone sequence under 100,000 is 77031 with 351 elements.
test.exw: <lang Phix>include hail.exw
sequence counts = {} for i=1 to 100000 do
integer l = hailstone_count(i)
if length(counts)<l then
counts &= repeat(0,l-length(counts))
end if
counts[l] += 1
end for printf(1,"The hailstone length returned most often between 1 and 100,000 is %d.\n",{largest(counts,1)})</lang>
- Output:
The hailstone length returned most often between 1 and 100,000 is 72.
PicoLisp
There is no formal difference between libraries and other executable files in PicoLisp. Any function in a library can be called from the command line by prefixing it with '-'. Create an executable file (chmod +x) "hailstone.l": <lang PicoLisp>#!/usr/bin/picolisp /usr/lib/picolisp/lib.l
(de hailstone (N)
(make
(until (= 1 (link N))
(setq N
(if (bit? 1 N)
(inc (* N 3))
(/ N 2) ) ) ) ) )
(de hailtest ()
(let L (hailstone 27)
(test 112 (length L))
(test (27 82 41 124) (head 4 L))
(test (8 4 2 1) (tail 4 L)) )
(let N (maxi '((N) (length (hailstone N))) (range 1 100000))
(test 77031 N)
(test 351 (length (hailstone N))) )
(println 'OK)
(bye) )</lang>
and an executable file (chmod +x) "test.l": <lang PicoLisp>#!/usr/bin/picolisp /usr/lib/picolisp/lib.l
(load "hailstone.l")
(let Len NIL
(for N 100000
(accu 'Len (length (hailstone N)) 1) )
(let M (maxi cdr Len)
(prinl "The hailstone length returned most often is " (car M))
(prinl "It is returned " (cdr M) " times") ) )
(bye)</lang> Test:
$ ./hailstone.l -hailtest OK $ ./test.l The hailstone length returned most often is 72 It is returned 1467 times
Pike
any Pike source file is a class and can be instantiated as an object.
to executable a Pike file it needs a main() function.
Pike modules are classes instantiated at compile time. below we demonstrate both forms:
to use the library as a class, save it as HailStone.pike to use it as a module, save it as Hailstone.pmod
both can be used as an executable. <lang Pike>#!/usr/bin/env pike
int next(int n) {
if (n==1)
return 0;
if (n%2)
return 3*n+1;
else
return n/2;
}
array(int) hailstone(int n) {
array seq = ({ n });
while (n=next(n))
seq += ({ n });
return seq;
}
void main() {
array(int) two = hailstone(27);
if (equal(two[0..3], ({ 27, 82, 41, 124 })) && equal(two[<3..], ({ 8,4,2,1 })))
write("sizeof(({ %{%d, %}, ... %{%d, %} }) == %d\n", two[0..3], two[<3..], sizeof(two));
mapping longest = ([ "length":0, "start":0 ]);
foreach(allocate(100000); int start; )
{
int length = sizeof(hailstone(start));
if (length > longest->length)
{
longest->length = length;
longest->start = start;
}
}
write("longest sequence starting at %d has %d elements\n", longest->start, longest->length);
}</lang>
if run directly we get:
$ pike hailstone.pike
sizeof(({ 27, 82, 41, 124, , ... 8, 4, 2, 1, }) == 112
longest sequence starting at 77031 has 351 elements
to use it as a class we need to instantiate an object. note that the . in .HailStone only signifies calling a class or module from the current directory. the analyze function is identical in both examples: <lang Pike>void main() {
.HailStone HailStone = .HailStone();
mapping long = ([]);
foreach (allocate(100000); int start; )
long[sizeof(HailStone->hailstone(start))]++;
analyze(long);
}
void analyze(mapping long) {
mapping max = ([ "count":0, "length":0 ]);
foreach (long; int length; int count)
{
if (count > max->count)
{
max->length = length;
max->count = count;
}
}
write("most common length %d appears %d times\n", max->length, max->count);
}</lang>
a module is already instantiated so we can use it directly. like above the initial . in .Hailstone.hailstone only signifies the current directory, the second . is a member reference resolved at compile time. <lang Pike>void main() {
mapping long = ([]);
foreach (allocate(100000); int start; )
long[sizeof(.Hailstone.hailstone(start))]++;
analyze(long);
}
void analyze(mapping long) {
mapping max = ([ "count":0, "length":0 ]);
foreach (long; int length; int count)
{
if (count > max->count)
{
max->length = length;
max->count = count;
}
}
write("most common length %d appears %d times\n", max->length, max->count);
}</lang>
- Output:
for both examples
most common length 72 appears 1467 times
Python
Executable libraries are common in Python. The Python entry for Hailstone sequence is already written in the correct manner.
The entry is copied below and, for this task needs to be in a file called hailstone.py:
<lang python>def hailstone(n):
seq = [n]
while n>1:
n = 3*n + 1 if n & 1 else n//2
seq.append(n)
return seq
if __name__ == '__main__':
h = hailstone(27)
assert len(h)==112 and h[:4]==[27, 82, 41, 124] and h[-4:]==[8, 4, 2, 1]
print("Maximum length %i was found for hailstone(%i) for numbers <100,000" %
max((len(hailstone(i)), i) for i in range(1,100000)))</lang>
In the case of the Python language the interpreter maintains a module level variable called __name__. If the file hailstone.py is imported (as import hailstone), then the __name__ variable is set to the import name of 'hailstone' and the if __name__ == '__main__' expression would then be false, and only the hailstone function is available to the importer.
If the same file hailstone.py is run, (as maybe python hailstone.py; or maybe double-clicking the hailstone.py file), then the __name__ variable is set to the special name of '__main__' and the if __name__ == '__main__' expression would then be true causing its block of code to be executed.
Library importing executable
The second executable is the file common_hailstone_length.py with this content:
<lang python>from collections import Counter
def function_length_frequency(func, hrange):
return Counter(len(func(n)) for n in hrange).most_common()
if __name__ == '__main__':
from executable_hailstone_library import hailstone
upto = 100000
hlen, freq = function_length_frequency(hailstone, range(1, upto))[0]
print("The length of hailstone sequence that is most common for\n"
"hailstone(n) where 1<=n<%i, is %i. It occurs %i times."
% (upto, hlen, freq))</lang>
Both files could be in the same directory. (That is the easiest way to make the library known to its importer for this example)
- Output:
On executing the file common_hailstone_length.py it loads the library and produces the following result:
The length of hailstone sequence that is most common for hailstone(n) where 1<=n<100000, is 72. It occurs 1467 times
Note that the file common_hailstone_length.py is itself written as an executable library. When imported it makes function_length_frequency available to the importer.
Other examples
- The Python Prime decomposition entry of Least common multiple employs Prime decomposition#Python as an executable library.
- Names_to_numbers#Python uses Number_names#Python as an executable library.
Racket
When Racket runs a file (with racket some-file) it executes its toplevel expressions, and then it runs a submodule named main if there is one. When a file is used as a library (with require), the toplevel expressions are executed as well, but the main is not executed. The idea is that toplevel expressions might be used to initialize state that the library needs -- a good example here is the initialization of the memoization hash table. (Note that this is better than the common hacks of check-the-loaded-script-name, since it is robust against failures due to symlinks, case normalization, etc etc.)
We start with a "hs.rkt" file that has the exact code from the Hailstone sequence#Racket solution, except that the hailstone function is now provided, and the demonstration printout is pushed into a main submodule: <lang Racket>
- lang racket
(provide hailstone) (define hailstone
(let ([t (make-hasheq)])
(hash-set! t 1 '(1))
(λ(n) (hash-ref! t n
(λ() (cons n (hailstone (if (even? n) (/ n 2) (+ (* 3 n) 1)))))))))
(module+ main
(define h27 (hailstone 27))
(printf "h(27) = ~s, ~s items\n"
`(,@(take h27 4) ... ,@(take-right h27 4))
(length h27))
(define N 100000)
(define longest
(for/fold ([m #f]) ([i (in-range 1 (add1 N))])
(define h (hailstone i))
(if (and m (> (cdr m) (length h))) m (cons i (length h)))))
(printf "for x<=~s, ~s has the longest sequence with ~s items\n"
N (car longest) (cdr longest)))
</lang>
Running it directly produces the same output as Hailstone sequence#Racket:
$ racket hs.rkt first 4 elements of h(27): '(27 82 41 124) last 4 elements of h(27): '(8 4 2 1) x < 10000 such that h(x) gives the longest sequence: 351
And now this can be used from a second source file, "hsfreq.rkt" as a library: <lang Racket>
- lang racket
(require "hs.rkt") (define N 100000) (define t (make-hasheq)) (define best
(for/fold ([best #f]) ([i (in-range 1 (add1 N))]) (define len (length (hailstone i))) (define freq (add1 (hash-ref t len 0))) (hash-set! t len freq) (if (and best (> (car best) freq)) best (cons freq len))))
(printf "Most frequent sequence length for x<=~s: ~s, appearing ~s times\n" N
(cdr best) (car best))
</lang>
$ racket hsfreq.rkt Most frequent sequence length for x<=100000: 72, appearing 1467 times
Raku
(formerly Perl 6) The library can be written as a module: <lang perl6>module Hailstone {
our sub hailstone($n) is export {
$n, { $_ %% 2 ?? $_ div 2 !! $_ * 3 + 1 } ... 1
}
}
sub MAIN {
say "hailstone(27) = {.[^4]} [...] {.[*-4 .. *-1]}" given Hailstone::hailstone 27;
}</lang>
It can be run with: <lang shell>$ raku Hailstone.rakumod</lang>
- Output:
hailstone(27) = 27 82 41 124 [...] 8 4 2 1
It can then be used with a program such as: <lang perl6>use Hailstone; my %score; (1 .. 100_000).race.map: { %score{hailstone($_).elems}++ }; say "Most common length is {.key}, occurring {.value} times." given max :by(*.value), %score;</lang>
Called with a command line as:
$ RAKULIB=. raku test-hailstone.raku
The environment variable RAKULIB might be necessary if the file Hailstone.rakumod is not in the standard library path for Raku.
REXX
task 1
The following REXX subroutine (or function, as it returns a value)
is normally stored in a folder that the REXX interpreter
searches first for subroutine/function call/invokes.
If not there, the REXX interpreter normally checks the
current drive (or default disk), and then through some sort of heirarchy --- depending upon the particular REXX interpreter and operating system.
On Microsoft Windows systems using Regina, PC/REXX, Personal REXX, R4, or ROO, the program name is normally the function name with a file extension of REX (but that isn't a strict requirement or rule, each REXX interpreter has multiple file extensions that are supported).
On VM/CMS systems, the filetype (the file extension) is normally EXEC. If however, the REXX program was previously EXECLOADed, it may have a different name (identity) assigned to it.
The following program (function) is named: HAILSTONE.REX (the case doesn't matter for Microsoft Windows systems).
All REXX interpreters support subroutines/functions being on the current drive (CD), default disk (or MDISK in the case of CMS), or the equivalent.
<lang rexx>/*REXX program returns the hailstone (Collatz) sequence for any integer.*/
numeric digits 20 /*ensure enough digits for mult. */
parse arg n 1 s /*N & S assigned to the first arg*/
do while n\==1 /*loop while N isn't unity. */
if n//2 then n=n*3+1 /*if N is odd, calc: 3*n +1 */
else n=n%2 /* " " " even, perform fast ÷ */
s=s n /*build a sequence list (append).*/
end /*while*/
return s</lang>
task 2, 3
The following program is named: : HAIL_PGM.REX and is stored in the current directory. <lang rexx>/*REXX pgm tests a number and a range for hailstone (Collatz) sequences.*/ parse arg x .; if x== then x=27 /*get the optional first argument*/
$=hailstone(x) /*═════════════task 2════════════*/
- =words($) /*number of numbers in sequence. */
say x 'has a hailstone sequence of' # 'and starts with: ' subword($,1,4),
' and ends with:' subword($,#-3)
say w=0; do j=1 for 99999 /*═════════════task 3════════════*/
$=hailstone(j); #=words($) /*obtain the hailstone sequence. */
if #<=w then iterate /*Not big 'nuff? Then keep going.*/
bigJ=j; w=# /*remember what # has biggest HS.*/
end /*j*/
say '(between 1──►99,999) ' bigJ 'has the longest hailstone sequence:' w
/*stick a fork in it, we're done.*/</lang>
- Output:
27 has a hailstone sequence of 112 and starts with: 27 82 41 124 and ends with: 8 4 2 1 (between 1──►99,999) 77031 has the longest hailstone sequence: 351
task 4
The following program is named: HAIL_POP.REX and is stored in the current directory. <lang rexx>/*REXX pgm finds the most common (popular) hailstone sequence length. */ parse arg z .; if z== then z=99999 /*get the optional first argument*/ !.=0 w=0; do j=1 for z /*═════════════task 4════════════*/
#=words(hailstone(j)) /*obtain hailstone sequence count*/
!.# = !.# + 1 /*add unity to popularity count. */
end /*j*/
occ=0; p=0
do k=1 for z
if !.k>occ then do; occ=!.k; p=k; end
end /*p*/
say '(between 1──►'z") " p, ' is the most common hailstone sequence length (with' occ "occurrences)."
/*stick a fork in it, we're done.*/</lang>
- Output:
(between 1──►99999) 72 is the most common hailstone sequence length (with 1467 occurrences).
task 5
To run a REXX program, it depends on the REXX interpretor and which operating system is being used (and what options where used when the REXX interpreter was installed/set up).
On a VM/CMS system, you could enter either of:
- HAILSTONE
- EXEC HAILSTONE
to execute the HAILSTONE EXEC A program (there are also other ways to invoke it).
On a Microsoft Windows system, you could enter any of:
- HAILSTONE.REX
- HAILSTONE
- xxx HAILSTONE.REX
- xxx HAILSTONE
where xxx is the name of the REXX interpreter, and if installed under a Microsoft Windows (Next family), the file extension and/or the REXX interpreter can be omitted.
Ruby
An executable library checks __FILE__ == $0. Here, __FILE__ is the path of the current source file, and $0 is the path of the current executable. If __FILE__ == $0, then the current source file is the executable, else the current source file is a library for some other executable.
- __FILE__ == $0 also works with older versions of Ruby, but this Hailstone example calls new methods in Ruby 1.8.7.
This is hailstone.rb, a modification of Hailstone sequence#Ruby as an executable library.
<lang ruby># hailstone.rb module Hailstone
module_function
def hailstone n
seq = [n]
until n == 1
n = (n.even?) ? (n / 2) : (3 * n + 1)
seq << n
end
seq
end
end
if __FILE__ == $0
include Hailstone
# for n = 27, show sequence length and first and last 4 elements hs27 = hailstone 27 p [hs27.length, hs27[0..3], hs27[-4..-1]]
# find the longest sequence among n less than 100,000
n, len = (1 ... 100_000) .collect {|n|
[n, hailstone(n).length]} .max_by {|n, len| len}
puts "#{n} has a hailstone sequence length of #{len}"
puts "the largest number in that sequence is #{hailstone(n).max}"
end</lang>
- It runs like any Ruby program:
$ ruby scratch.rb [112, [27, 82, 41, 124], [8, 4, 2, 1]] 77031 has a hailstone sequence length of 351 the largest number in that sequence is 21933016
This is hsfreq.rb, which requires hailstone.rb as a library.
<lang ruby># hsfreq.rb require 'hailstone'
h = Hash.new(0) last = 99_999 (1..last).each {|n| h[Hailstone.hailstone(n).length] += 1} length, count = h.max_by {|length, count| count}
puts "Given the hailstone sequences from 1 to #{last}," puts "the most common sequence length is #{length}," puts "with #{count} such sequences."</lang>
As with any library, hailstone.rb must be in $:, the search path for libraries.
One way is to leave hailstone.rb in the current directory and run ruby -I. hsfreq.rb.
(Ruby older than 1.9.2 also searches the current directory by default.)
$ ruby -I. hsfreq.rb Given the hailstone sequences from 1 to 99999, the most common sequence length is 72, with 1467 such sequences.
Scala
In Scala it is possible to combine several "main"s (mixed-in by the App trait) in one file (e.g. HailstoneSequence.scala): <lang Scala> object HailstoneSequence extends App { // Show it all, default number is 27.
def hailstone(n: Int): LazyList[Int] = n #:: (if (n == 1) LazyList.empty else hailstone(if (n % 2 == 0) n / 2 else n * 3 + 1))
Hailstone.details(args.headOption.map(_.toInt).getOrElse(27)) HailTest.main(Array())
}
object Hailstone extends App { // Compute a given or default number to Hailstone sequence
def details(nr: Int): Unit = {
val collatz = HailstoneSequence.hailstone(nr)
println(s"Use the routine to show that the hailstone sequence for the number: $nr.")
println(collatz.toList)
println(s"It has ${collatz.length} elements.")
}
details(args.headOption.map(_.toInt).getOrElse(27))
}
object HailTest extends App { // Compute only the < 100000 test
println( "Compute the number < 100,000, which has the longest hailstone sequence with that sequence's length.") val (n, len) = (1 until 100000).map(n => (n, HailstoneSequence.hailstone(n).length)).maxBy(_._2) println(s"Longest hailstone sequence length= $len occurring with number $n.")
} </lang>
Steps:
1. First let the compiler process the source file:
C:\Users\FransAdm\Documents>scalac HailstoneSequence.scala
2. Run the Hailstone function with a parameter:
C:\Users\FransAdm\Documents>scala Hailstone 42 Use the routine to show that the hailstone sequence for the number: 42. List(42, 21, 64, 32, 16, 8, 4, 2, 1) It has 9 elements.
3. Run the combined function and < 100000 test:
C:\Users\FransAdm\Documents>scala HailstoneSequence 27 Use the routine to show that the hailstone sequence for the number: 27. List(27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80 , 40, 20, 10, 5, 16, 8, 4, 2, 1) It has 112 elements. Compute the number < 100,000, which has the longest hailstone sequence with that sequence's length. Longest hailstone sequence length= 351 occurring with number 77031.
4. Finally do only the callable < 100000 test
C:\Users\FransAdm\Documents>scala HailTest Compute the number < 100,000, which has the longest hailstone sequence with that sequence's length. Longest hailstone sequence length= 351 occurring with number 77031. C:\Users\FransAdm\Documents>
Sidef
Library saved as Hailstone.sm <lang ruby>func hailstone(n) {
gather {
while (n > 1) {
take(n)
n = (n.is_even ? n/2 : (3*n + 1))
}
take(1)
}
} if (__FILE__ == __MAIN__) { # true when not imported
var seq = hailstone(27)
say "hailstone(27) - #{seq.len} elements: #{seq.ft(0, 3)} [...] #{seq.ft(-4)}"
var n = 0
var max = 0
100_000.times { |i|
var seq = hailstone(i)
if (seq.len > max) {
max = seq.len
n = i
}
}
say "Longest sequence is for #{n}: #{max}"
}</lang>
It can be run with: <lang shell>$ sidef Hailstone.sm</lang>
It can then be used with a program such as: <lang ruby>include Hailstone var score = Hash() 100_000.times { |i| score{ Hailstone::hailstone(i).len } := 0 ++ } var k = score.keys.max_by {|k| score{k} } say "Most common length is #{k}, occurring #{score{k}} times"</lang>
Called with a command line as: <lang shell>$ sidef test_hailstone.sf</lang>
The library is searched in the directories specified in the SIDEF_INC environment variable, which defaults to the current directory.
Tcl
The standard idiom for detecting whether a script is being loaded as a library or run directly is to compare the result of info script (which describes the name of the currently sourced script file) and the global argv0 variable (which holds the name of the main script).
<lang tcl>### In the file hailstone.tcl ###
package provide hailstone 1.0
proc hailstone n {
while 1 {
lappend seq $n if {$n == 1} {return $seq} set n [expr {$n & 1 ? $n*3+1 : $n/2}]
}
}
- If directly executed, run demo code
if {[info script] eq $::argv0} {
set h27 [hailstone 27] puts "h27 len=[llength $h27]" puts "head4 = [lrange $h27 0 3]" puts "tail4 = [lrange $h27 end-3 end]"
set maxlen [set max 0]
for {set i 1} {$i<100000} {incr i} {
set l [llength [hailstone $i]] if {$l>$maxlen} {set maxlen $l;set max $i}
} puts "max is $max, with length $maxlen"
}</lang>
To make the package locatable, run this Tcl script in the same directory which builds the index file: <lang tcl>pkg_mkIndex .</lang>
Using the above code as a library then just requires that we tell the script the location of the additional library directory by adding it to the global auto_path variable; it is unnecessary if the script is installed in one of the standard locations (a fairly long list that depends on the installation):
<lang tcl>#!/usr/bin/tclsh8.6
package require Tcl 8.6 ;# For [lsort -stride] option
lappend auto_path . ;# Or wherever it is located
package require hailstone 1.0
- Construct a histogram of length frequencies
set histogram {} for {set n 1} {$n < 100000} {incr n} {
dict incr histogram [llength [hailstone $n]]
}
- Identify the most common length by sorting...
set sortedHist [lsort -decreasing -stride 2 -index 1 $histogram] lassign $sortedHist mostCommonLength freq
puts "most common length is $mostCommonLength, with frequency $freq"</lang>
zkl
zkl is a system of VM, compiler, etc (ala Python) so no need for compiling/linking etc.
File hailstone.zkl: <lang zkl>fcn collatz(n,z=L()){ z.append(n); if(n==1) return(z);
if(n.isEven) return(self.fcn(n/2,z)); return(self.fcn(n*3+1,z)) }
h27:=collatz(27); println("Hailstone(27)-->",h27[0,4].concat(","),"...", h27[-4,*].concat(",")," length ",h27.len());
[2..0d100_000].pump(Void, // loop n from 2 to 100,000
collatz, // generate Collatz sequence(n)
fcn(c,n){ // if new longest sequence, save length/C, return longest
if(c.len()>n[0]) n.clear(c.len(),c[0]); n}.fp1(L(0,0)))
.println();</lang> File hailstone2.zkl: <lang zkl>#!/home/craigd/Bin/zkl collatz:=Import("hailstone",False,False,False).collatz; // don't run constructor d:=Dictionary(); [2..0d100_000].pump(Void, // loop n from 2 to 100,000
collatz, // generate Collatz sequence(n)
'wrap(c){ d.incV(c.len()) } // save length
); println("Number of different lengths: ",d.len()); longest:=(0).max(d.values); mostFreqLen:=d.filter1('wrap([(k,v)]){ v==longest })[0]; println("Most frequent length: %d; %d sequences of that length."
.fmt(mostFreqLen,longest));</lang>
- Output:
$ zkl hailstone Hailstone(27)-->27,82,41,124...8,4,2,1 length 112 L(351,77031) $ ./hailstone2.zkl Number of different lengths: 314 Most frequent length: 72; 1467 sequences of that length. $