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http://rosettacode.org/wiki/Babbage_problem | Babbage problem |
Charles Babbage, looking ahead to the sorts of problems his Analytical Engine would be able to solve, gave this example:
What is the smallest positive integer whose square ends in the digits 269,696?
— Babbage, letter to Lord Bowden, 1837; see Hollingdale and Tootill, Electronic Computers, second edition, 1970, p. 125.
He thought the answer might be 99,736, whose square is 9,947,269,696; but he couldn't be certain.
Task[edit]
The task is to find out if Babbage had the right answer — and to do so, as far as your language allows it, in code that Babbage himself would have been able to read and understand.
As Babbage evidently solved the task with pencil and paper, a similar efficient solution is preferred.
For these purposes, Charles Babbage may be taken to be an intelligent person, familiar with mathematics and with the idea of a computer; he has written the first drafts of simple computer programmes in tabular form. [Babbage Archive Series L].
Motivation
The aim of the task is to write a program that is sufficiently clear and well-documented for such a person to be able to read it and be confident that it does indeed solve the specified problem.
| #Processing | Processing | // Lines that begin with two slashes, thus, are comments: they
// will be ignored by the machine.
// First we must declare a variable, n, suitable to store an integer:
int n;
// Each statement we address to the machine must end with a semicolon.
// To begin with, the value of n will be zero:
n = 0;
// Now we must repeatedly increase it by one, checking each time to see
// whether its square ends in 269,696.
// We shall do this by seeing whether the remainder, when n squared
// is divided by one million, is equal to 269,696.
do {
n = n + 1;
} while (n * n % 1000000 != 269696);
// To read this formula, it is necessary to know the following
// elements of the notation:
// * means 'multiplied by'
// % means 'modulo', or remainder after division
// != means 'is not equal to'
// Now that we have our result, we need to display it.
// println is short for 'print line'
println(n); |
http://rosettacode.org/wiki/Approximate_equality | Approximate equality | Sometimes, when testing whether the solution to a task (for example, here on Rosetta Code) is correct, the
difference in floating point calculations between different language implementations becomes significant.
For example, a difference between 32 bit and 64 bit floating point calculations may appear by
about the 8th significant digit in base 10 arithmetic.
Task
Create a function which returns true if two floating point numbers are approximately equal.
The function should allow for differences in the magnitude of numbers, so that, for example,
100000000000000.01 may be approximately equal to 100000000000000.011,
even though 100.01 is not approximately equal to 100.011.
If the language has such a feature in its standard library, this may be used instead of a custom function.
Show the function results with comparisons on the following pairs of values:
100000000000000.01, 100000000000000.011 (note: should return true)
100.01, 100.011 (note: should return false)
10000000000000.001 / 10000.0, 1000000000.0000001000
0.001, 0.0010000001
0.000000000000000000000101, 0.0
sqrt(2) * sqrt(2), 2.0
-sqrt(2) * sqrt(2), -2.0
3.14159265358979323846, 3.14159265358979324
Answers should be true for the first example and false in the second, so that just rounding the numbers to a fixed number of decimals should not be enough. Otherwise answers may vary and still be correct. See the Python code for one type of solution.
| #Groovy | Groovy | class Approximate {
private static boolean approxEquals(double value, double other, double epsilon) {
return Math.abs(value - other) < epsilon
}
private static void test(double a, double b) {
double epsilon = 1e-18
System.out.printf("%f, %f => %s\n", a, b, approxEquals(a, b, epsilon))
}
static void main(String[] args) {
test(100000000000000.01, 100000000000000.011)
test(100.01, 100.011)
test(10000000000000.001 / 10000.0, 1000000000.0000001000)
test(0.001, 0.0010000001)
test(0.000000000000000000000101, 0.0)
test(Math.sqrt(2.0) * Math.sqrt(2.0), 2.0)
test(-Math.sqrt(2.0) * Math.sqrt(2.0), -2.0)
test(3.14159265358979323846, 3.14159265358979324)
}
} |
http://rosettacode.org/wiki/Approximate_equality | Approximate equality | Sometimes, when testing whether the solution to a task (for example, here on Rosetta Code) is correct, the
difference in floating point calculations between different language implementations becomes significant.
For example, a difference between 32 bit and 64 bit floating point calculations may appear by
about the 8th significant digit in base 10 arithmetic.
Task
Create a function which returns true if two floating point numbers are approximately equal.
The function should allow for differences in the magnitude of numbers, so that, for example,
100000000000000.01 may be approximately equal to 100000000000000.011,
even though 100.01 is not approximately equal to 100.011.
If the language has such a feature in its standard library, this may be used instead of a custom function.
Show the function results with comparisons on the following pairs of values:
100000000000000.01, 100000000000000.011 (note: should return true)
100.01, 100.011 (note: should return false)
10000000000000.001 / 10000.0, 1000000000.0000001000
0.001, 0.0010000001
0.000000000000000000000101, 0.0
sqrt(2) * sqrt(2), 2.0
-sqrt(2) * sqrt(2), -2.0
3.14159265358979323846, 3.14159265358979324
Answers should be true for the first example and false in the second, so that just rounding the numbers to a fixed number of decimals should not be enough. Otherwise answers may vary and still be correct. See the Python code for one type of solution.
| #Haskell | Haskell | class (Num a, Ord a, Eq a) => AlmostEq a where
eps :: a
infix 4 ~=
(~=) :: AlmostEq a => a -> a -> Bool
a ~= b = or [ a == b
, abs (a - b) < eps * abs(a + b)
, abs (a - b) < eps ]
instance AlmostEq Int where eps = 0
instance AlmostEq Integer where eps = 0
instance AlmostEq Double where eps = 1e-14
instance AlmostEq Float where eps = 1e-5 |
http://rosettacode.org/wiki/Balanced_brackets | Balanced brackets | Task:
Generate a string with N opening brackets [ and with N closing brackets ], in some arbitrary order.
Determine whether the generated string is balanced; that is, whether it consists entirely of pairs of opening/closing brackets (in that order), none of which mis-nest.
Examples
(empty) OK
[] OK
[][] OK
[[][]] OK
][ NOT OK
][][ NOT OK
[]][[] NOT OK
| #CLU | CLU | % This program needs the random number generator from
% "misc.lib" that comes with PCLU.
shuffle = proc [T: type] (a: array[t])
aT = array[t]
for i: int in int$from_to_by(aT$size(a)-1,0,-1) do
x: int := aT$low(a) + i
y: int := aT$low(a) + random$next(i+1)
temp: T := a[x]
a[x] := a[y]
a[y] := temp
end
end shuffle
brackets = proc (n: int) returns (string)
br: array[char] := array[char]$[]
for i: int in int$from_to(1,2*n) do
if i<=n then array[char]$addh(br, '[')
else array[char]$addh(br, ']')
end
end
shuffle[char](br)
return(string$ac2s(br))
end brackets
balanced = proc (br: string) returns (bool)
depth: int := 0
for c: char in string$chars(br) do
if c='[' then depth := depth + 1
elseif c=']' then depth := depth - 1
end
if depth<0 then return(false) end
end
return(depth = 0)
end balanced
start_up = proc ()
po: stream := stream$primary_output()
d: date := now()
random$seed(d.second + 60*(d.minute + 60*d.hour))
for size: int in int$from_to(0, 10) do
b: string := brackets(size)
stream$puts(po, "\"" || b || "\": ")
if balanced(b) then
stream$putl(po, "balanced")
else
stream$putl(po, "not balanced")
end
end
end start_up |
http://rosettacode.org/wiki/Append_a_record_to_the_end_of_a_text_file | Append a record to the end of a text file | Many systems offer the ability to open a file for writing, such that any data written will be appended to the end of the file. Further, the file operations will always adjust the position pointer to guarantee the end of the file, even in a multitasking environment.
This feature is most useful in the case of log files, where many jobs may be appending to the log file at the same time, or where care must be taken to avoid concurrently overwriting the same record from another job.
Task
Given a two record sample for a mythical "passwd" file:
Write these records out in the typical system format.
Ideally these records will have named fields of various types.
Close the file, then reopen the file for append.
Append a new record to the file and close the file again.
Take appropriate care to avoid concurrently overwrites from another job.
Open the file and demonstrate the new record has indeed written to the end.
Source record field types and contents.
account
password
UID
GID
fullname,office,extension,homephone,email
directory
shell
string
string
int
int
struct(string,string,string,string,string)
string
string
jsmith
x
1001
1000
Joe Smith,Room 1007,(234)555-8917,(234)555-0077,[email protected]
/home/jsmith
/bin/bash
jdoe
x
1002
1000
Jane Doe,Room 1004,(234)555-8914,(234)555-0044,[email protected]
/home/jdoe
/bin/bash
Record to be appended.
account
password
UID
GID
fullname,office,extension,homephone,email
directory
shell
string
string
int
int
struct(string,string,string,string,string)
string
string
xyz
x
1003
1000
X Yz,Room 1003,(234)555-8913,(234)555-0033,[email protected]
/home/xyz
/bin/bash
Resulting file format: should mimic Linux's /etc/passwd file format with particular attention to the "," separator used in the GECOS field. But if the specific language has a particular or unique format of storing records in text file, then this format should be named and demonstrated with an additional example.
Expected output:
Appended record: xyz:x:1003:1000:X Yz,Room 1003,(234)555-8913,(234)555-0033,[email protected]:/home/xyz:/bin/bash
Finally: Provide a summary of the language's "append record" capabilities in a table. eg.
Append Capabilities.
Data Representation
IO
Library
Append
Possible
Automatic
Append
Multi-tasking
Safe
In core
On disk
C struct
CSV text file
glibc/stdio
☑
☑
☑ (Not all, eg NFS)
Alternatively: If the language's appends can not guarantee its writes will always append, then note this restriction in the table. If possible, provide an actual code example (possibly using file/record locking) to guarantee correct concurrent appends.
| #Batch_File | Batch File |
@echo off
(
echo jsmith:x:1001:1000:Joe Smith,Room 1007,^(234^)555-8917,^(234^)555-0077,[email protected]:/home/jsmith:/bin/bash
echo jdoe:x:1002:1000:Jane Doe,Room 1004,^(234^)555-8914,^(234^)555-0044,[email protected]:/home/jdoe:/bin/bash
) > append.txt
echo Current contents of append.txt:
type append.txt
echo.
echo xyz:x:1003:1000:X Yz,Room 1003,^(234^)555-8913,^(234^)555-0033,[email protected]:/home/xyz:/bin/bash >> append.txt
echo New contents of append.txt:
type append.txt
pause>nul
|
http://rosettacode.org/wiki/Append_a_record_to_the_end_of_a_text_file | Append a record to the end of a text file | Many systems offer the ability to open a file for writing, such that any data written will be appended to the end of the file. Further, the file operations will always adjust the position pointer to guarantee the end of the file, even in a multitasking environment.
This feature is most useful in the case of log files, where many jobs may be appending to the log file at the same time, or where care must be taken to avoid concurrently overwriting the same record from another job.
Task
Given a two record sample for a mythical "passwd" file:
Write these records out in the typical system format.
Ideally these records will have named fields of various types.
Close the file, then reopen the file for append.
Append a new record to the file and close the file again.
Take appropriate care to avoid concurrently overwrites from another job.
Open the file and demonstrate the new record has indeed written to the end.
Source record field types and contents.
account
password
UID
GID
fullname,office,extension,homephone,email
directory
shell
string
string
int
int
struct(string,string,string,string,string)
string
string
jsmith
x
1001
1000
Joe Smith,Room 1007,(234)555-8917,(234)555-0077,[email protected]
/home/jsmith
/bin/bash
jdoe
x
1002
1000
Jane Doe,Room 1004,(234)555-8914,(234)555-0044,[email protected]
/home/jdoe
/bin/bash
Record to be appended.
account
password
UID
GID
fullname,office,extension,homephone,email
directory
shell
string
string
int
int
struct(string,string,string,string,string)
string
string
xyz
x
1003
1000
X Yz,Room 1003,(234)555-8913,(234)555-0033,[email protected]
/home/xyz
/bin/bash
Resulting file format: should mimic Linux's /etc/passwd file format with particular attention to the "," separator used in the GECOS field. But if the specific language has a particular or unique format of storing records in text file, then this format should be named and demonstrated with an additional example.
Expected output:
Appended record: xyz:x:1003:1000:X Yz,Room 1003,(234)555-8913,(234)555-0033,[email protected]:/home/xyz:/bin/bash
Finally: Provide a summary of the language's "append record" capabilities in a table. eg.
Append Capabilities.
Data Representation
IO
Library
Append
Possible
Automatic
Append
Multi-tasking
Safe
In core
On disk
C struct
CSV text file
glibc/stdio
☑
☑
☑ (Not all, eg NFS)
Alternatively: If the language's appends can not guarantee its writes will always append, then note this restriction in the table. If possible, provide an actual code example (possibly using file/record locking) to guarantee correct concurrent appends.
| #C | C | #include <stdio.h>
#include <string.h>
/* note that UID & GID are of type "int" */
typedef const char *STRING;
typedef struct{STRING fullname, office, extension, homephone, email; } gecos_t;
typedef struct{STRING account, password; int uid, gid; gecos_t gecos; STRING directory, shell; } passwd_t;
#define GECOS_FMT "%s,%s,%s,%s,%s"
#define PASSWD_FMT "%s:%s:%d:%d:"GECOS_FMT":%s:%s"
passwd_t passwd_list[]={
{"jsmith", "x", 1001, 1000, /* UID and GID are type int */
{"Joe Smith", "Room 1007", "(234)555-8917", "(234)555-0077", "[email protected]"},
"/home/jsmith", "/bin/bash"},
{"jdoe", "x", 1002, 1000,
{"Jane Doe", "Room 1004", "(234)555-8914", "(234)555-0044", "[email protected]"},
"/home/jdoe", "/bin/bash"}
};
main(){
/****************************
* Create a passwd text file *
****************************/
FILE *passwd_text=fopen("passwd.txt", "w");
int rec_num;
for(rec_num=0; rec_num < sizeof passwd_list/sizeof(passwd_t); rec_num++)
fprintf(passwd_text, PASSWD_FMT"\n", passwd_list[rec_num]);
fclose(passwd_text);
/********************************
* Load text ready for appending *
********************************/
passwd_text=fopen("passwd.txt", "a+");
char passwd_buf[BUFSIZ]; /* warning: fixed length */
passwd_t new_rec =
{"xyz", "x", 1003, 1000, /* UID and GID are type int */
{"X Yz", "Room 1003", "(234)555-8913", "(234)555-0033", "[email protected]"},
"/home/xyz", "/bin/bash"};
sprintf(passwd_buf, PASSWD_FMT"\n", new_rec);
/* An atomic append without a file lock,
Note: wont work on some file systems, eg NFS */
write(fileno(passwd_text), passwd_buf, strlen(passwd_buf));
close(passwd_text);
/***********************************************
* Finally reopen and check record was appended *
***********************************************/
passwd_text=fopen("passwd.txt", "r");
while(!feof(passwd_text))
fscanf(passwd_text, "%[^\n]\n", passwd_buf, "\n");
if(strstr(passwd_buf, "xyz"))
printf("Appended record: %s\n", passwd_buf);
} |
http://rosettacode.org/wiki/Associative_array/Creation | Associative array/Creation | Task
The goal is to create an associative array (also known as a dictionary, map, or hash).
Related tasks:
Associative arrays/Iteration
Hash from two arrays
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #Aikido | Aikido |
var names = {} // empty map
names["foo"] = "bar"
names[3] = 4
// initialized map
var names2 = {"foo": bar, 3:4}
// lookup map
var name = names["foo"]
if (typeof(name) == "none") {
println ("not found")
} else {
println (name)
}
// remove from map
delete names["foo"]
|
http://rosettacode.org/wiki/Anti-primes | Anti-primes | The anti-primes
(or highly composite numbers, sequence A002182 in the OEIS)
are the natural numbers with more factors than any smaller than itself.
Task
Generate and show here, the first twenty anti-primes.
Related tasks
Factors of an integer
Sieve of Eratosthenes
| #AArch64_Assembly | AArch64 Assembly |
/* ARM assembly AARCH64 Raspberry PI 3B */
/* program antiprime64.s */
/************************************/
/* Constantes */
/************************************/
.include "../includeConstantesARM64.inc"
.equ NMAXI, 20
.equ MAXLINE, 5
/*********************************/
/* Initialized data */
/*********************************/
.data
sMessResult: .asciz " @ "
szCarriageReturn: .asciz "\n"
/*********************************/
/* UnInitialized data */
/*********************************/
.bss
sZoneConv: .skip 24
/*********************************/
/* code section */
/*********************************/
.text
.global main
main: // entry of program
ldr x3,qNMaxi // load limit
mov x5,#0 // maxi
mov x6,#0 // result counter
mov x7,#0 // display counter
mov x4,#1 // number begin
1:
mov x0,x4 // number
bl decFactor // compute number factors
cmp x0,x5 // maxi ?
cinc x4,x4,le // no -> increment indice
//addle x4,x4,#1 // no -> increment indice
ble 1b // and loop
mov x5,x0
mov x0,x4
bl displayResult
add x7,x7,#1 // increment display counter
cmp x7,#MAXLINE // line maxi ?
blt 2f
mov x7,#0
ldr x0,qAdrszCarriageReturn
bl affichageMess // display message
2:
add x6,x6,#1 // increment result counter
add x4,x4,#1 // increment number
cmp x6,x3 // end ?
blt 1b
100: // standard end of the program
mov x0, #0 // return code
mov x8,EXIT
svc #0 // perform the system call
qAdrszCarriageReturn: .quad szCarriageReturn
qNMaxi: .quad NMAXI
/***************************************************/
/* display message number */
/***************************************************/
/* x0 contains number 1 */
/* x1 contains number 2 */
displayResult:
stp x1,lr,[sp,-16]! // save registers
ldr x1,qAdrsZoneConv
bl conversion10 // call décimal conversion
ldr x0,qAdrsMessResult
ldr x1,qAdrsZoneConv // insert conversion in message
bl strInsertAtCharInc
bl affichageMess // display message
ldp x1,lr,[sp],16 // restaur registers
ret
qAdrsMessResult: .quad sMessResult
qAdrsZoneConv: .quad sZoneConv
/***************************************************/
/* compute factors sum */
/***************************************************/
/* x0 contains the number */
decFactor:
stp x1,lr,[sp,-16]! // save registers
stp x2,x3,[sp,-16]! // save registers
stp x4,x5,[sp,-16]! // save registers
mov x5,#0 // init number factors
mov x4,x0 // save number
mov x1,#1 // start factor -> divisor
1:
mov x0,x4 // dividende
udiv x2,x0,x1
msub x3,x2,x1,x0
cmp x1,x2 // divisor > quotient ?
bgt 3f
cmp x3,#0 // remainder = 0 ?
bne 2f
add x5,x5,#1 // increment counter factors
cmp x1,x2 // divisor = quotient ?
beq 3f // yes -> end
add x5,x5,#1 // no -> increment counter factors
2:
add x1,x1,#1 // increment factor
b 1b // and loop
3:
mov x0,x5 // return counter
ldp x4,x5,[sp],16 // restaur registers
ldp x2,x3,[sp],16 // restaur registers
ldp x1,lr,[sp],16 // restaur registers
ret
/***************************************************/
/* ROUTINES INCLUDE */
/***************************************************/
.include "../includeARM64.inc"
|
http://rosettacode.org/wiki/Anti-primes | Anti-primes | The anti-primes
(or highly composite numbers, sequence A002182 in the OEIS)
are the natural numbers with more factors than any smaller than itself.
Task
Generate and show here, the first twenty anti-primes.
Related tasks
Factors of an integer
Sieve of Eratosthenes
| #Action.21 | Action! | BYTE FUNC CountDivisors(INT a)
INT i
BYTE prod,count
prod=1 count=0
WHILE a MOD 2=0
DO
count==+1
a==/2
OD
prod==*(1+count)
i=3
WHILE i*i<=a
DO
count=0
WHILE a MOD i=0
DO
count==+1
a==/i
OD
prod==*(1+count)
i==+2
OD
IF a>2 THEN
prod==*2
FI
RETURN (prod)
PROC Main()
BYTE toFind=[20],found=[0],count,max=[0]
INT i=[1]
PrintF("The first %B Anti-primes are:%E",toFind)
WHILE found<toFind
DO
count=CountDivisors(i)
IF count>max THEN
max=count
found==+1
PrintI(i)
IF found<toFind THEN
Print(", ")
FI
FI
i==+1
OD
RETURN |
http://rosettacode.org/wiki/Atomic_updates | Atomic updates |
Task
Define a data type consisting of a fixed number of 'buckets', each containing a nonnegative integer value, which supports operations to:
get the current value of any bucket
remove a specified amount from one specified bucket and add it to another, preserving the total of all bucket values, and clamping the transferred amount to ensure the values remain non-negative
In order to exercise this data type, create one set of buckets, and start three concurrent tasks:
As often as possible, pick two buckets and make their values closer to equal.
As often as possible, pick two buckets and arbitrarily redistribute their values.
At whatever rate is convenient, display (by any means) the total value and, optionally, the individual values of each bucket.
The display task need not be explicit; use of e.g. a debugger or trace tool is acceptable provided it is simple to set up to provide the display.
This task is intended as an exercise in atomic operations. The sum of the bucket values must be preserved even if the two tasks attempt to perform transfers simultaneously, and a straightforward solution is to ensure that at any time, only one transfer is actually occurring — that the transfer operation is atomic.
| #FreeBASIC | FreeBASIC | Randomize Timer
Dim Shared As Uinteger cubo(1 To 10), a, i
For i As Uinteger = 1 To 10
cubo(i) = Int(Rnd * 90)
Next i
Function Display(cadena As String) As Uinteger
Dim As Uinteger valor
Print cadena; Spc(2);
For i As Uinteger = 1 To 10
valor += cubo(i)
Print Using "###"; cubo(i);
Next i
Print " Total:"; valor
Return valor
End Function
Sub Flatten(f As Uinteger)
Dim As Uinteger f1 = Int((f / 10) + .5), f2
For i As Uinteger = 1 To 10
cubo(i) = f1
f2 += f1
Next i
cubo(10) += f - f2
End Sub
Sub Transfer(a1 As Uinteger, a2 As Uinteger)
Dim As Uinteger temp = Int(Rnd * cubo(a1))
cubo(a1) -= temp
cubo(a2) += temp
End Sub
a = Display(" Display:") ' show original array
Flatten(a) ' flatten the array
a = Display(" Flatten:") ' show flattened array
Transfer(3, 5) ' transfer some amount from 3 to 5
Display(" 19 from 3 to 5:") ' show transfer array
Sleep |
http://rosettacode.org/wiki/Assertions | Assertions | Assertions are a way of breaking out of code when there is an error or an unexpected input.
Some languages throw exceptions and some treat it as a break point.
Task
Show an assertion in your language by asserting that an integer variable is equal to 42.
| #Dyalect | Dyalect | var x = 42
assert(42, x) |
http://rosettacode.org/wiki/Assertions | Assertions | Assertions are a way of breaking out of code when there is an error or an unexpected input.
Some languages throw exceptions and some treat it as a break point.
Task
Show an assertion in your language by asserting that an integer variable is equal to 42.
| #E | E | require(a == 42) # default message, "Required condition failed"
require(a == 42, "The Answer is Wrong.") # supplied message
require(a == 42, fn { `Off by ${a - 42}.` }) # computed only on failure |
http://rosettacode.org/wiki/Assertions | Assertions | Assertions are a way of breaking out of code when there is an error or an unexpected input.
Some languages throw exceptions and some treat it as a break point.
Task
Show an assertion in your language by asserting that an integer variable is equal to 42.
| #EchoLisp | EchoLisp |
(assert (integer? 42)) → #t ;; success returns true
;; error and return to top level if not true;
(assert (integer? 'quarante-deux))
⛔ error: assert : assertion failed : (#integer? 'quarante-deux)
;; assertion with message (optional)
(assert (integer? 'quarante-deux) "☝️ expression must evaluate to the integer 42")
💥 error: ☝️ expression must evaluate to the integer 42 : assertion failed : (#integer? 'quarante-deux)
|
http://rosettacode.org/wiki/Apply_a_callback_to_an_array | Apply a callback to an array | Task
Take a combined set of elements and apply a function to each element.
| #Ada | Ada | with Ada.Text_Io;
with Ada.Integer_text_IO;
procedure Call_Back_Example is
-- Purpose: Apply a callback to an array
-- Output: Prints the squares of an integer array to the console
-- Define the callback procedure
procedure Display(Location : Positive; Value : Integer) is
begin
Ada.Text_Io.Put("array(");
Ada.Integer_Text_Io.Put(Item => Location, Width => 1);
Ada.Text_Io.Put(") = ");
Ada.Integer_Text_Io.Put(Item => Value * Value, Width => 1);
Ada.Text_Io.New_Line;
end Display;
-- Define an access type matching the signature of the callback procedure
type Call_Back_Access is access procedure(L : Positive; V : Integer);
-- Define an unconstrained array type
type Value_Array is array(Positive range <>) of Integer;
-- Define the procedure performing the callback
procedure Map(Values : Value_Array; Worker : Call_Back_Access) is
begin
for I in Values'range loop
Worker(I, Values(I));
end loop;
end Map;
-- Define and initialize the actual array
Sample : Value_Array := (5,4,3,2,1);
begin
Map(Sample, Display'access);
end Call_Back_Example; |
http://rosettacode.org/wiki/Averages/Mode | Averages/Mode | Task[edit]
Write a program to find the mode value of a collection.
The case where the collection is empty may be ignored. Care must be taken to handle the case where the mode is non-unique.
If it is not appropriate or possible to support a general collection, use a vector (array), if possible. If it is not appropriate or possible to support an unspecified value type, use integers.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #M2000_Interpreter | M2000 Interpreter |
Module Checkit {
\\ find mode
Function GetMode {
Inventory N
Inventory ALLMODES
m=1
While not empty {
if islet then {
Read A$
if Exist(N, A$) then {
k=Eval(N)
k++
if m=k then {
Append ALLMODES, A$
}
if m<k then m=k : Clear ALLMODES : Append ALLMODES, A$
return N, A$:=k
} Else Append N, A$:=1 : if m=1 then Append ALLMODES, A$
} else {
Read A
if Exist(N, A) then {
k=Eval(N)
k++
if m=k then {
Append ALLMODES, A
}
if m<k then m=k : Clear ALLMODES : Append ALLMODES, A
return N, A:=k
} Else Append N, A:=1 : if m=1 then Append ALLMODES, A
}
}
=ALLMODES
}
Print GetMode(1, 2, 3, 1, 2, 4, 2, 5, 2, 3, 3, 1, 3, 6) ' print 2 3
Dim A()
A()=(1, 2, 3, 1, 2, 4, 2, 5, 2, 3, 3, 1, 3, 6)
\\ get a pointer from A
m=A()
Print GetMode(!m) ' print 2 3
z=stack:=1, 2,"B", 3, 1, 2, "B", 4, 2, 5,"B", 2, 3, 3, 1, 3, 6, "B"
Print GetMode(!z) ' print 2 3 B
}
Checkit
|
http://rosettacode.org/wiki/Averages/Mode | Averages/Mode | Task[edit]
Write a program to find the mode value of a collection.
The case where the collection is empty may be ignored. Care must be taken to handle the case where the mode is non-unique.
If it is not appropriate or possible to support a general collection, use a vector (array), if possible. If it is not appropriate or possible to support an unspecified value type, use integers.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Maple | Maple | Statistics:-Mode([1, 2.1, 2.1, 3]);
Statistics:-Mode([1, 2.1, 2.1, 3.2, 3.2, 5]); |
http://rosettacode.org/wiki/Associative_array/Iteration | Associative array/Iteration | Show how to iterate over the key-value pairs of an associative array, and print each pair out.
Also show how to iterate just over the keys, or the values, if there is a separate way to do that in your language.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #BASIC256 | BASIC256 | REM Store some values with their keys:
PROCputdict(mydict$, "FF0000", "red")
PROCputdict(mydict$, "00FF00", "green")
PROCputdict(mydict$, "0000FF", "blue")
REM Iterate through the dictionary:
i% = 1
REPEAT
i% = FNdict(mydict$, i%, v$, k$)
PRINT v$, k$
UNTIL i% = 0
END
DEF PROCputdict(RETURN dict$, value$, key$)
IF dict$ = "" dict$ = CHR$(0)
dict$ += key$ + CHR$(1) + value$ + CHR$(0)
ENDPROC
DEF FNdict(dict$, I%, RETURN value$, RETURN key$)
LOCAL J%, K%
J% = INSTR(dict$, CHR$(1), I%)
K% = INSTR(dict$, CHR$(0), J%)
value$ = MID$(dict$, I%+1, J%-I%-1)
key$ = MID$(dict$, J%+1, K%-J%-1)
IF K% >= LEN(dict$) THEN K% = 0
= K% |
http://rosettacode.org/wiki/Associative_array/Iteration | Associative array/Iteration | Show how to iterate over the key-value pairs of an associative array, and print each pair out.
Also show how to iterate just over the keys, or the values, if there is a separate way to do that in your language.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #BBC_BASIC | BBC BASIC | REM Store some values with their keys:
PROCputdict(mydict$, "FF0000", "red")
PROCputdict(mydict$, "00FF00", "green")
PROCputdict(mydict$, "0000FF", "blue")
REM Iterate through the dictionary:
i% = 1
REPEAT
i% = FNdict(mydict$, i%, v$, k$)
PRINT v$, k$
UNTIL i% = 0
END
DEF PROCputdict(RETURN dict$, value$, key$)
IF dict$ = "" dict$ = CHR$(0)
dict$ += key$ + CHR$(1) + value$ + CHR$(0)
ENDPROC
DEF FNdict(dict$, I%, RETURN value$, RETURN key$)
LOCAL J%, K%
J% = INSTR(dict$, CHR$(1), I%)
K% = INSTR(dict$, CHR$(0), J%)
value$ = MID$(dict$, I%+1, J%-I%-1)
key$ = MID$(dict$, J%+1, K%-J%-1)
IF K% >= LEN(dict$) THEN K% = 0
= K% |
http://rosettacode.org/wiki/Apply_a_digital_filter_(direct_form_II_transposed) | Apply a digital filter (direct form II transposed) | Digital filters are used to apply a mathematical operation to a sampled signal. One of the common formulations is the "direct form II transposed" which can represent both infinite impulse response (IIR) and finite impulse response (FIR) filters, as well as being more numerically stable than other forms. [1]
Task
Filter a signal using an order 3 low-pass Butterworth filter. The coefficients for the filter are a=[1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17] and b = [0.16666667, 0.5, 0.5, 0.16666667]
The signal that needs filtering is the following vector: [-0.917843918645, 0.141984778794, 1.20536903482, 0.190286794412, -0.662370894973, -1.00700480494, -0.404707073677 ,0.800482325044, 0.743500089861, 1.01090520172, 0.741527555207, 0.277841675195, 0.400833448236, -0.2085993586, -0.172842103641, -0.134316096293, 0.0259303398477, 0.490105989562, 0.549391221511, 0.9047198589]
| #Common_Lisp | Common Lisp | (defparameter a #(1.00000000L0 -2.77555756L-16 3.33333333L-01 -1.85037171L-17))
(defparameter b #(0.16666667L0 0.50000000L0 0.50000000L0 0.16666667L0))
(defparameter s #(-0.917843918645 0.141984778794 1.20536903482 0.190286794412 -0.662370894973
-1.00700480494 -0.404707073677 0.800482325044 0.743500089861 1.01090520172
0.741527555207 0.277841675195 0.400833448236 -0.2085993586 -0.172842103641
-0.134316096293 0.0259303398477 0.490105989562 0.549391221511 0.9047198589))
(loop with out = (make-array (length s) :initial-element 0.0D0)
for i below (length s)
do (setf (svref out i)
(/ (- (loop for j below (length b)
when (>= i j) sum (* (svref b j) (svref s (- i j))))
(loop for j below (length a)
when (>= i j) sum (* (svref a j) (svref out (- i j)))))
(svref a 0)))
(format t "~%~16,8F" (svref out i))) |
http://rosettacode.org/wiki/Apply_a_digital_filter_(direct_form_II_transposed) | Apply a digital filter (direct form II transposed) | Digital filters are used to apply a mathematical operation to a sampled signal. One of the common formulations is the "direct form II transposed" which can represent both infinite impulse response (IIR) and finite impulse response (FIR) filters, as well as being more numerically stable than other forms. [1]
Task
Filter a signal using an order 3 low-pass Butterworth filter. The coefficients for the filter are a=[1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17] and b = [0.16666667, 0.5, 0.5, 0.16666667]
The signal that needs filtering is the following vector: [-0.917843918645, 0.141984778794, 1.20536903482, 0.190286794412, -0.662370894973, -1.00700480494, -0.404707073677 ,0.800482325044, 0.743500089861, 1.01090520172, 0.741527555207, 0.277841675195, 0.400833448236, -0.2085993586, -0.172842103641, -0.134316096293, 0.0259303398477, 0.490105989562, 0.549391221511, 0.9047198589]
| #D | D | import std.stdio;
alias T = real;
alias AT = T[];
AT filter(const AT a, const AT b, const AT signal) {
AT result = new T[signal.length];
foreach (int i; 0..signal.length) {
T tmp = 0.0;
foreach (int j; 0..b.length) {
if (i-j<0) continue;
tmp += b[j] * signal[i-j];
}
foreach (int j; 1..a.length) {
if (i-j<0) continue;
tmp -= a[j] * result[i-j];
}
tmp /= a[0];
result[i] = tmp;
}
return result;
}
void main() {
AT a = [1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17];
AT b = [0.16666667, 0.5, 0.5, 0.16666667];
AT signal = [
-0.917843918645, 0.141984778794, 1.20536903482, 0.190286794412,
-0.662370894973, -1.00700480494, -0.404707073677, 0.800482325044,
0.743500089861, 1.01090520172, 0.741527555207, 0.277841675195,
0.400833448236, -0.2085993586, -0.172842103641, -0.134316096293,
0.0259303398477, 0.490105989562, 0.549391221511, 0.9047198589
];
AT result = filter(a,b,signal);
foreach (i; 0..result.length) {
writef("% .8f", result[i]);
if ((i+1)%5 != 0) {
write(", ");
} else {
writeln;
}
}
} |
http://rosettacode.org/wiki/Averages/Arithmetic_mean | Averages/Arithmetic mean | Task[edit]
Write a program to find the mean (arithmetic average) of a numeric vector.
In case of a zero-length input, since the mean of an empty set of numbers is ill-defined, the program may choose to behave in any way it deems appropriate, though if the programming language has an established convention for conveying math errors or undefined values, it's preferable to follow it.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #BQN | BQN | Avg ← +´÷≠
Avg 1‿2‿3‿4 |
http://rosettacode.org/wiki/Averages/Arithmetic_mean | Averages/Arithmetic mean | Task[edit]
Write a program to find the mean (arithmetic average) of a numeric vector.
In case of a zero-length input, since the mean of an empty set of numbers is ill-defined, the program may choose to behave in any way it deems appropriate, though if the programming language has an established convention for conveying math errors or undefined values, it's preferable to follow it.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #C | C | #include <stdio.h>
double mean(double *v, int len)
{
double sum = 0;
int i;
for (i = 0; i < len; i++)
sum += v[i];
return sum / len;
}
int main(void)
{
double v[] = {1, 2, 2.718, 3, 3.142};
int i, len;
for (len = 5; len >= 0; len--) {
printf("mean[");
for (i = 0; i < len; i++)
printf(i ? ", %g" : "%g", v[i]);
printf("] = %g\n", mean(v, len));
}
return 0;
} |
http://rosettacode.org/wiki/Associative_array/Merging | Associative array/Merging | Task
Define two associative arrays, where one represents the following "base" data:
Key
Value
"name"
"Rocket Skates"
"price"
12.75
"color"
"yellow"
And the other represents "update" data:
Key
Value
"price"
15.25
"color"
"red"
"year"
1974
Merge these into a new associative array that contains every key found in either of the source ones. Each key should map to the value in the second (update) table if that exists, or else to the value in the first (base) table. If possible, do this in a way that does not mutate the original two associative arrays. Obviously this should be done in a way that would work for any data, not just the specific data given here, but in this example the result should be:
Key
Value
"name"
"Rocket Skates"
"price"
15.25
"color"
"red"
"year"
1974
| #Lua | Lua | base = {name="Rocket Skates", price=12.75, color="yellow"}
update = {price=15.25, color="red", year=1974}
--[[ clone the base data ]]--
result = {}
for key,val in pairs(base) do
result[key] = val
end
--[[ copy in the update data ]]--
for key,val in pairs(update) do
result[key] = val
end
--[[ print the result ]]--
for key,val in pairs(result) do
print(string.format("%s: %s", key, val))
end |
http://rosettacode.org/wiki/Associative_array/Merging | Associative array/Merging | Task
Define two associative arrays, where one represents the following "base" data:
Key
Value
"name"
"Rocket Skates"
"price"
12.75
"color"
"yellow"
And the other represents "update" data:
Key
Value
"price"
15.25
"color"
"red"
"year"
1974
Merge these into a new associative array that contains every key found in either of the source ones. Each key should map to the value in the second (update) table if that exists, or else to the value in the first (base) table. If possible, do this in a way that does not mutate the original two associative arrays. Obviously this should be done in a way that would work for any data, not just the specific data given here, but in this example the result should be:
Key
Value
"name"
"Rocket Skates"
"price"
15.25
"color"
"red"
"year"
1974
| #Mathematica_.2F_Wolfram_Language | Mathematica / Wolfram Language | a1 = <|"name" -> "Rocket Skates", "price" -> 12.75, "color" -> "yellow"|>;
a2 = <|"price" -> 15.25, "color" -> "red", "year" -> 1974|>;
Merge[{a1, a2}, Last] |
http://rosettacode.org/wiki/Associative_array/Merging | Associative array/Merging | Task
Define two associative arrays, where one represents the following "base" data:
Key
Value
"name"
"Rocket Skates"
"price"
12.75
"color"
"yellow"
And the other represents "update" data:
Key
Value
"price"
15.25
"color"
"red"
"year"
1974
Merge these into a new associative array that contains every key found in either of the source ones. Each key should map to the value in the second (update) table if that exists, or else to the value in the first (base) table. If possible, do this in a way that does not mutate the original two associative arrays. Obviously this should be done in a way that would work for any data, not just the specific data given here, but in this example the result should be:
Key
Value
"name"
"Rocket Skates"
"price"
15.25
"color"
"red"
"year"
1974
| #MiniScript | MiniScript | base = {"name":"Rocket Skates", "price":12.75, "color":"yellow"}
update = {"price":15.25, "color":"red", "year":1974}
result = base + update
print result |
http://rosettacode.org/wiki/Average_loop_length | Average loop length | Let f be a uniformly-randomly chosen mapping from the numbers 1..N to the numbers 1..N (note: not necessarily a permutation of 1..N; the mapping could produce a number in more than one way or not at all). At some point, the sequence 1, f(1), f(f(1))... will contain a repetition, a number that occurring for the second time in the sequence.
Task
Write a program or a script that estimates, for each N, the average length until the first such repetition.
Also calculate this expected length using an analytical formula, and optionally compare the simulated result with the theoretical one.
This problem comes from the end of Donald Knuth's Christmas tree lecture 2011.
Example of expected output:
N average analytical (error)
=== ========= ============ =========
1 1.0000 1.0000 ( 0.00%)
2 1.4992 1.5000 ( 0.05%)
3 1.8784 1.8889 ( 0.56%)
4 2.2316 2.2188 ( 0.58%)
5 2.4982 2.5104 ( 0.49%)
6 2.7897 2.7747 ( 0.54%)
7 3.0153 3.0181 ( 0.09%)
8 3.2429 3.2450 ( 0.07%)
9 3.4536 3.4583 ( 0.14%)
10 3.6649 3.6602 ( 0.13%)
11 3.8091 3.8524 ( 1.12%)
12 3.9986 4.0361 ( 0.93%)
13 4.2074 4.2123 ( 0.12%)
14 4.3711 4.3820 ( 0.25%)
15 4.5275 4.5458 ( 0.40%)
16 4.6755 4.7043 ( 0.61%)
17 4.8877 4.8579 ( 0.61%)
18 4.9951 5.0071 ( 0.24%)
19 5.1312 5.1522 ( 0.41%)
20 5.2699 5.2936 ( 0.45%)
| #Mathematica_.2F_Wolfram_Language | Mathematica / Wolfram Language | Grid@Prepend[
Table[{n, #[[1]], #[[2]],
Row[{Round[10000 Abs[#[[1]] - #[[2]]]/#[[2]]]/100., "%"}]} &@
N[{Mean[Array[
Length@NestWhileList[#, 1, UnsameQ[##] &, All] - 1 &[# /.
MapIndexed[#2[[1]] -> #1 &,
RandomInteger[{1, n}, n]] &] &, 10000]],
Sum[n! n^(n - k - 1)/(n - k)!, {k, n}]/n^(n - 1)}, 5], {n, 1,
20}], {"N", "average", "analytical", "error"}] |
http://rosettacode.org/wiki/Average_loop_length | Average loop length | Let f be a uniformly-randomly chosen mapping from the numbers 1..N to the numbers 1..N (note: not necessarily a permutation of 1..N; the mapping could produce a number in more than one way or not at all). At some point, the sequence 1, f(1), f(f(1))... will contain a repetition, a number that occurring for the second time in the sequence.
Task
Write a program or a script that estimates, for each N, the average length until the first such repetition.
Also calculate this expected length using an analytical formula, and optionally compare the simulated result with the theoretical one.
This problem comes from the end of Donald Knuth's Christmas tree lecture 2011.
Example of expected output:
N average analytical (error)
=== ========= ============ =========
1 1.0000 1.0000 ( 0.00%)
2 1.4992 1.5000 ( 0.05%)
3 1.8784 1.8889 ( 0.56%)
4 2.2316 2.2188 ( 0.58%)
5 2.4982 2.5104 ( 0.49%)
6 2.7897 2.7747 ( 0.54%)
7 3.0153 3.0181 ( 0.09%)
8 3.2429 3.2450 ( 0.07%)
9 3.4536 3.4583 ( 0.14%)
10 3.6649 3.6602 ( 0.13%)
11 3.8091 3.8524 ( 1.12%)
12 3.9986 4.0361 ( 0.93%)
13 4.2074 4.2123 ( 0.12%)
14 4.3711 4.3820 ( 0.25%)
15 4.5275 4.5458 ( 0.40%)
16 4.6755 4.7043 ( 0.61%)
17 4.8877 4.8579 ( 0.61%)
18 4.9951 5.0071 ( 0.24%)
19 5.1312 5.1522 ( 0.41%)
20 5.2699 5.2936 ( 0.45%)
| #Nim | Nim | import random, math, strformat
randomize()
const
maxN = 20
times = 1_000_000
proc factorial(n: int): float =
result = 1
for i in 1 .. n:
result *= i.float
proc expected(n: int): float =
for i in 1 .. n:
result += factorial(n) / pow(n.float, i.float) / factorial(n - i)
proc test(n, times: int): int =
for i in 1 .. times:
var
x = 1
bits = 0
while (bits and x) == 0:
inc result
bits = bits or x
x = 1 shl rand(n - 1)
echo " n\tavg\texp.\tdiff"
echo "-------------------------------"
for n in 1 .. maxN:
let cnt = test(n, times)
let avg = cnt.float / times
let theory = expected(n)
let diff = (avg / theory - 1) * 100
echo fmt"{n:2} {avg:8.4f} {theory:8.4f} {diff:6.3f}%" |
http://rosettacode.org/wiki/Average_loop_length | Average loop length | Let f be a uniformly-randomly chosen mapping from the numbers 1..N to the numbers 1..N (note: not necessarily a permutation of 1..N; the mapping could produce a number in more than one way or not at all). At some point, the sequence 1, f(1), f(f(1))... will contain a repetition, a number that occurring for the second time in the sequence.
Task
Write a program or a script that estimates, for each N, the average length until the first such repetition.
Also calculate this expected length using an analytical formula, and optionally compare the simulated result with the theoretical one.
This problem comes from the end of Donald Knuth's Christmas tree lecture 2011.
Example of expected output:
N average analytical (error)
=== ========= ============ =========
1 1.0000 1.0000 ( 0.00%)
2 1.4992 1.5000 ( 0.05%)
3 1.8784 1.8889 ( 0.56%)
4 2.2316 2.2188 ( 0.58%)
5 2.4982 2.5104 ( 0.49%)
6 2.7897 2.7747 ( 0.54%)
7 3.0153 3.0181 ( 0.09%)
8 3.2429 3.2450 ( 0.07%)
9 3.4536 3.4583 ( 0.14%)
10 3.6649 3.6602 ( 0.13%)
11 3.8091 3.8524 ( 1.12%)
12 3.9986 4.0361 ( 0.93%)
13 4.2074 4.2123 ( 0.12%)
14 4.3711 4.3820 ( 0.25%)
15 4.5275 4.5458 ( 0.40%)
16 4.6755 4.7043 ( 0.61%)
17 4.8877 4.8579 ( 0.61%)
18 4.9951 5.0071 ( 0.24%)
19 5.1312 5.1522 ( 0.41%)
20 5.2699 5.2936 ( 0.45%)
| #Oberon-2 | Oberon-2 |
MODULE AvgLoopLen;
(* Oxford Oberon-2 *)
IMPORT Random, Out;
PROCEDURE Fac(n: INTEGER; f: REAL): REAL;
BEGIN
IF n = 0 THEN
RETURN f
ELSE
RETURN Fac(n - 1,n*f)
END
END Fac;
PROCEDURE Power(n,i: INTEGER): REAL;
VAR
p: REAL;
BEGIN
p := 1.0;
WHILE i > 0 DO p := p * n; DEC(i) END;
RETURN p
END Power;
PROCEDURE Abs(x: REAL): REAL;
BEGIN
IF x < 0 THEN RETURN -x ELSE RETURN x END
END Abs;
PROCEDURE Analytical(n: INTEGER): REAL;
VAR
i: INTEGER;
res: REAL;
BEGIN
res := 0.0;
FOR i := 1 TO n DO
res := res + (Fac(n,1.0) / Power(n,i) / Fac(n - i,1.0));
END;
RETURN res
END Analytical;
PROCEDURE Averages(n: INTEGER): REAL;
CONST
times = 100000;
VAR
rnds: SET;
r,count,i: INTEGER;
BEGIN
count := 0; i := 0;
WHILE i < times DO
rnds := {};
LOOP
r := Random.Roll(n);
IF r IN rnds THEN EXIT ELSE INCL(rnds,r); INC(count) END
END;
INC(i)
END;
RETURN count / times
END Averages;
VAR
i: INTEGER;
av,an,df: REAL;
BEGIN
Random.Randomize;
Out.String(" Averages Analytical Diff% ");Out.Ln;
FOR i := 1 TO 20 DO
Out.Int(i,3); Out.String(": ");
av := Averages(i);an := Analytical(i);df := Abs(av - an) / an * 100.0;
Out.Fixed(av,10,4);Out.Fixed(an,11,4);Out.Fixed(df,10,4);Out.Ln
END
END AvgLoopLen.
|
http://rosettacode.org/wiki/Averages/Simple_moving_average | Averages/Simple moving average | Computing the simple moving average of a series of numbers.
Task[edit]
Create a stateful function/class/instance that takes a period and returns a routine that takes a number as argument and returns a simple moving average of its arguments so far.
Description
A simple moving average is a method for computing an average of a stream of numbers by only averaging the last P numbers from the stream, where P is known as the period.
It can be implemented by calling an initialing routine with P as its argument, I(P), which should then return a routine that when called with individual, successive members of a stream of numbers, computes the mean of (up to), the last P of them, lets call this SMA().
The word stateful in the task description refers to the need for SMA() to remember certain information between calls to it:
The period, P
An ordered container of at least the last P numbers from each of its individual calls.
Stateful also means that successive calls to I(), the initializer, should return separate routines that do not share saved state so they could be used on two independent streams of data.
Pseudo-code for an implementation of SMA is:
function SMA(number: N):
stateful integer: P
stateful list: stream
number: average
stream.append_last(N)
if stream.length() > P:
# Only average the last P elements of the stream
stream.delete_first()
if stream.length() == 0:
average = 0
else:
average = sum( stream.values() ) / stream.length()
return average
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #REXX | REXX | /*REXX program illustrates and displays a simple moving average using a constructed list*/
parse arg p q n . /*obtain optional arguments from the CL*/
if p=='' | p=="," then p= 3 /*Not specified? Then use the default.*/
if q=='' | q=="," then q= 5 /* " " " " " " */
if n=='' | n=="," then n= 10 /* " " " " " " */
@.= 0 /*default value, only needed for odd N.*/
do j=1 for n%2; @.j= j /*build 1st half of list, increasing #s*/
end /*j*/
do k=n%2 by -1 to 1; @.j= k; j= j+1 /* " 2nd " " " decreasing " */
end /*k*/
say ' number ' " SMA with period" p' ' " SMA with period" q
say ' ──────── ' "───────────────────" '───────────────────'
pad=' '
do m=1 for n; say center(@.m, 10) pad left(SMA(p, m), 19) left(SMA(q, m), 19)
end /*m*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
SMA: procedure expose @.; parse arg p,j; i= 0 ; $= 0
do k=max(1, j-p+1) to j+p for p while k<=j; i= i + 1; $= $ + @.k
end /*k*/
return $/i /*SMA ≡ simple moving average. */ |
http://rosettacode.org/wiki/Attractive_numbers | Attractive numbers | A number is an attractive number if the number of its prime factors (whether distinct or not) is also prime.
Example
The number 20, whose prime decomposition is 2 × 2 × 5, is an attractive number because the number of its prime factors (3) is also prime.
Task
Show sequence items up to 120.
Reference
The OEIS entry: A063989: Numbers with a prime number of prime divisors.
| #Go | Go | package main
import "fmt"
func isPrime(n int) bool {
switch {
case n < 2:
return false
case n%2 == 0:
return n == 2
case n%3 == 0:
return n == 3
default:
d := 5
for d*d <= n {
if n%d == 0 {
return false
}
d += 2
if n%d == 0 {
return false
}
d += 4
}
return true
}
}
func countPrimeFactors(n int) int {
switch {
case n == 1:
return 0
case isPrime(n):
return 1
default:
count, f := 0, 2
for {
if n%f == 0 {
count++
n /= f
if n == 1 {
return count
}
if isPrime(n) {
f = n
}
} else if f >= 3 {
f += 2
} else {
f = 3
}
}
return count
}
}
func main() {
const max = 120
fmt.Println("The attractive numbers up to and including", max, "are:")
count := 0
for i := 1; i <= max; i++ {
n := countPrimeFactors(i)
if isPrime(n) {
fmt.Printf("%4d", i)
count++
if count % 20 == 0 {
fmt.Println()
}
}
}
fmt.Println()
} |
http://rosettacode.org/wiki/Averages/Mean_time_of_day | Averages/Mean time of day | Task[edit]
A particular activity of bats occurs at these times of the day:
23:00:17, 23:40:20, 00:12:45, 00:17:19
Using the idea that there are twenty-four hours in a day,
which is analogous to there being 360 degrees in a circle,
map times of day to and from angles;
and using the ideas of Averages/Mean angle
compute and show the average time of the nocturnal activity
to an accuracy of one second of time.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Lua | Lua |
local times = {"23:00:17","23:40:20","00:12:45","00:17:19"}
-- returns time converted to a radian format
local function timeToAngle(str)
local h,m,s = str:match("(..):(..):(..)")
return (h + m / 60 + s / 3600)/12 * math.pi
end
-- computes the mean of the angles inside a list
local function meanAngle(angles)
local sumSin,sumCos = 0,0
for k,v in pairs(angles) do
sumSin = sumSin + math.sin(v)
sumCos = sumCos + math.cos(v)
end
return math.atan2(sumSin,sumCos)
end
-- converts and angle back to a time string
local function angleToTime(angle)
local abs = angle % (math.pi * 2)
local time = abs / math.pi * 12
local h = math.floor(time)
local m = math.floor(time * 60) % 60
local s = math.floor(time * 3600) % 60
return string.format("%02d:%02d:%02d", h, m, s)
end
-- convert times to angles
for k,v in pairs(times) do
times[k] = timeToAngle(v)
end
print(angleToTime(meanAngle(times)))
|
http://rosettacode.org/wiki/AVL_tree | AVL tree |
This page uses content from Wikipedia. The original article was at AVL tree. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)
In computer science, an AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; at no time do they differ by more than one because rebalancing is done ensure this is the case. Lookup, insertion, and deletion all take O(log n) time in both the average and worst cases, where n is the number of nodes in the tree prior to the operation. Insertions and deletions may require the tree to be rebalanced by one or more tree rotations. Note the tree of nodes comprise a set, so duplicate node keys are not allowed.
AVL trees are often compared with red-black trees because they support the same set of operations and because red-black trees also take O(log n) time for the basic operations. Because AVL trees are more rigidly balanced, they are faster than red-black trees for lookup-intensive applications. Similar to red-black trees, AVL trees are height-balanced, but in general not weight-balanced nor μ-balanced; that is, sibling nodes can have hugely differing numbers of descendants.
Task
Implement an AVL tree in the language of choice, and provide at least basic operations.
| #Objeck | Objeck | class AVLNode {
@key : Int;
@balance : Int;
@height : Int;
@left : AVLNode;
@right : AVLNode;
@above : AVLNode;
New(key : Int, above : AVLNode) {
@key := key;
@above := above;
}
method : public : GetKey() ~ Int {
return @key;
}
method : public : GetLeft() ~ AVLNode {
return @left;
}
method : public : GetRight() ~ AVLNode {
return @right;
}
method : public : GetAbove() ~ AVLNode {
return @above;
}
method : public : GetBalance() ~ Int {
return @balance;
}
method : public : GetHeight() ~ Int {
return @height;
}
method : public : SetBalance(balance : Int) ~ Nil {
@balance := balance;
}
method : public : SetHeight(height : Int) ~ Nil {
@height := height;
}
method : public : SetAbove(above : AVLNode) ~ Nil {
@above := above;
}
method : public : SetLeft(left : AVLNode) ~ Nil {
@left := left;
}
method : public : SetRight(right : AVLNode) ~ Nil {
@right := right;
}
method : public : SetKey(key : Int) ~ Nil {
@key := key;
}
}
class AVLTree {
@root : AVLNode;
New() {}
method : public : Insert(key : Int) ~ Bool {
if(@root = Nil) {
@root := AVLNode->New( key, Nil);
return true;
};
n := @root;
while(true) {
if(n->GetKey() = key) {
return false;
};
parent := n;
goLeft := n->GetKey() > key;
n := goLeft ? n->GetLeft() : n->GetRight();
if(n = Nil) {
if(goLeft) {
parent->SetLeft(AVLNode->New( key, parent));
} else {
parent->SetRight(AVLNode->New( key, parent));
};
Rebalance(parent);
break;
};
};
return true;
}
method : Delete(node : AVLNode) ~ Nil {
if (node->GetLeft() = Nil & node->GetRight() = Nil) {
if (node ->GetAbove() = Nil) {
@root := Nil;
} else {
parent := node ->GetAbove();
if (parent->GetLeft() = node) {
parent->SetLeft(Nil);
} else {
parent->SetRight(Nil);
};
Rebalance(parent);
};
return;
};
if (node->GetLeft() <> Nil) {
child := node->GetLeft();
while (child->GetRight() <> Nil) {
child := child->GetRight();
};
node->SetKey(child->GetKey());
Delete(child);
} else {
child := node->GetRight();
while (child->GetLeft() <> Nil) {
child := child->GetLeft();
};
node->SetKey(child->GetKey());
Delete(child);
};
}
method : public : Delete(delKey : Int) ~ Nil {
if (@root = Nil) {
return;
};
child := @root;
while (child <> Nil) {
node := child;
child := delKey >= node->GetKey() ? node->GetRight() : node->GetLeft();
if (delKey = node->GetKey()) {
Delete(node);
return;
};
};
}
method : Rebalance(n : AVLNode) ~ Nil {
SetBalance(n);
if (n->GetBalance() = -2) {
if (Height(n->GetLeft()->GetLeft()) >= Height(n->GetLeft()->GetRight())) {
n := RotateRight(n);
}
else {
n := RotateLeftThenRight(n);
};
} else if (n->GetBalance() = 2) {
if(Height(n->GetRight()->GetRight()) >= Height(n->GetRight()->GetLeft())) {
n := RotateLeft(n);
}
else {
n := RotateRightThenLeft(n);
};
};
if(n->GetAbove() <> Nil) {
Rebalance(n->GetAbove());
} else {
@root := n;
};
}
method : RotateLeft(a : AVLNode) ~ AVLNode {
b := a->GetRight();
b->SetAbove(a->GetAbove());
a->SetRight(b->GetLeft());
if(a->GetRight() <> Nil) {
a->GetRight()->SetAbove(a);
};
b->SetLeft(a);
a->SetAbove(b);
if (b->GetAbove() <> Nil) {
if (b->GetAbove()->GetRight() = a) {
b->GetAbove()->SetRight(b);
} else {
b->GetAbove()->SetLeft(b);
};
};
SetBalance(a);
SetBalance(b);
return b;
}
method : RotateRight(a : AVLNode) ~ AVLNode {
b := a->GetLeft();
b->SetAbove(a->GetAbove());
a->SetLeft(b->GetRight());
if (a->GetLeft() <> Nil) {
a->GetLeft()->SetAbove(a);
};
b->SetRight(a);
a->SetAbove(b);
if (b->GetAbove() <> Nil) {
if (b->GetAbove()->GetRight() = a) {
b->GetAbove()->SetRight(b);
} else {
b->GetAbove()->SetLeft(b);
};
};
SetBalance(a);
SetBalance(b);
return b;
}
method : RotateLeftThenRight(n : AVLNode) ~ AVLNode {
n->SetLeft(RotateLeft(n->GetLeft()));
return RotateRight(n);
}
method : RotateRightThenLeft(n : AVLNode) ~ AVLNode {
n->SetRight(RotateRight(n->GetRight()));
return RotateLeft(n);
}
method : SetBalance(n : AVLNode) ~ Nil {
Reheight(n);
n->SetBalance(Height(n->GetRight()) - Height(n->GetLeft()));
}
method : Reheight(node : AVLNode) ~ Nil {
if(node <> Nil) {
node->SetHeight(1 + Int->Max(Height(node->GetLeft()), Height(node->GetRight())));
};
}
method : Height(n : AVLNode) ~ Int {
if(n = Nil) {
return -1;
};
return n->GetHeight();
}
method : public : PrintBalance() ~ Nil {
PrintBalance(@root);
}
method : PrintBalance(n : AVLNode) ~ Nil {
if (n <> Nil) {
PrintBalance(n->GetLeft());
balance := n->GetBalance();
"{$balance} "->Print();
PrintBalance(n->GetRight());
};
}
}
class Test {
function : Main(args : String[]) ~ Nil {
tree := AVLTree->New();
"Inserting values 1 to 10"->PrintLine();
for(i := 1; i < 10; i+=1;) {
tree->Insert(i);
};
"Printing balance: "->Print();
tree->PrintBalance();
}
}
|
http://rosettacode.org/wiki/Averages/Mean_angle | Averages/Mean angle | When calculating the average or mean of an angle one has to take into account how angles wrap around so that any angle in degrees plus any integer multiple of 360 degrees is a measure of the same angle.
If one wanted an average direction of the wind over two readings where the first reading was of 350 degrees and the second was of 10 degrees then the average of the numbers is 180 degrees, whereas if you can note that 350 degrees is equivalent to -10 degrees and so you have two readings at 10 degrees either side of zero degrees leading to a more fitting mean angle of zero degrees.
To calculate the mean angle of several angles:
Assume all angles are on the unit circle and convert them to complex numbers expressed in real and imaginary form.
Compute the mean of the complex numbers.
Convert the complex mean to polar coordinates whereupon the phase of the complex mean is the required angular mean.
(Note that, since the mean is the sum divided by the number of numbers, and division by a positive real number does not affect the angle, you can also simply compute the sum for step 2.)
You can alternatively use this formula:
Given the angles
α
1
,
…
,
α
n
{\displaystyle \alpha _{1},\dots ,\alpha _{n}}
the mean is computed by
α
¯
=
atan2
(
1
n
⋅
∑
j
=
1
n
sin
α
j
,
1
n
⋅
∑
j
=
1
n
cos
α
j
)
{\displaystyle {\bar {\alpha }}=\operatorname {atan2} \left({\frac {1}{n}}\cdot \sum _{j=1}^{n}\sin \alpha _{j},{\frac {1}{n}}\cdot \sum _{j=1}^{n}\cos \alpha _{j}\right)}
Task[edit]
write a function/method/subroutine/... that given a list of angles in degrees returns their mean angle.
(You should use a built-in function if you have one that does this for degrees or radians).
Use the function to compute the means of these lists of angles (in degrees):
[350, 10]
[90, 180, 270, 360]
[10, 20, 30]
Show your output here.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Kotlin | Kotlin | // version 1.0.5-2
fun meanAngle(angles: DoubleArray): Double {
val sinSum = angles.sumByDouble { Math.sin(it * Math.PI / 180.0) }
val cosSum = angles.sumByDouble { Math.cos(it * Math.PI / 180.0) }
return Math.atan2(sinSum / angles.size, cosSum / angles.size) * 180.0 / Math.PI
}
fun main(args: Array<String>) {
val angles1 = doubleArrayOf(350.0, 10.0)
val angles2 = doubleArrayOf(90.0, 180.0, 270.0, 360.0)
val angles3 = doubleArrayOf(10.0, 20.0, 30.0)
val fmt = "%.2f degrees" // format results to 2 decimal places
println("Mean for angles 1 is ${fmt.format(meanAngle(angles1))}")
println("Mean for angles 2 is ${fmt.format(meanAngle(angles2))}")
println("Mean for angles 3 is ${fmt.format(meanAngle(angles3))}")
} |
http://rosettacode.org/wiki/Averages/Mean_angle | Averages/Mean angle | When calculating the average or mean of an angle one has to take into account how angles wrap around so that any angle in degrees plus any integer multiple of 360 degrees is a measure of the same angle.
If one wanted an average direction of the wind over two readings where the first reading was of 350 degrees and the second was of 10 degrees then the average of the numbers is 180 degrees, whereas if you can note that 350 degrees is equivalent to -10 degrees and so you have two readings at 10 degrees either side of zero degrees leading to a more fitting mean angle of zero degrees.
To calculate the mean angle of several angles:
Assume all angles are on the unit circle and convert them to complex numbers expressed in real and imaginary form.
Compute the mean of the complex numbers.
Convert the complex mean to polar coordinates whereupon the phase of the complex mean is the required angular mean.
(Note that, since the mean is the sum divided by the number of numbers, and division by a positive real number does not affect the angle, you can also simply compute the sum for step 2.)
You can alternatively use this formula:
Given the angles
α
1
,
…
,
α
n
{\displaystyle \alpha _{1},\dots ,\alpha _{n}}
the mean is computed by
α
¯
=
atan2
(
1
n
⋅
∑
j
=
1
n
sin
α
j
,
1
n
⋅
∑
j
=
1
n
cos
α
j
)
{\displaystyle {\bar {\alpha }}=\operatorname {atan2} \left({\frac {1}{n}}\cdot \sum _{j=1}^{n}\sin \alpha _{j},{\frac {1}{n}}\cdot \sum _{j=1}^{n}\cos \alpha _{j}\right)}
Task[edit]
write a function/method/subroutine/... that given a list of angles in degrees returns their mean angle.
(You should use a built-in function if you have one that does this for degrees or radians).
Use the function to compute the means of these lists of angles (in degrees):
[350, 10]
[90, 180, 270, 360]
[10, 20, 30]
Show your output here.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Liberty_BASIC | Liberty BASIC | global Pi
Pi =3.1415926535
print "Mean Angle( "; chr$( 34); "350,10"; chr$( 34); ") = "; using( "###.#", meanAngle( "350,10")); " degrees." ' 0
print "Mean Angle( "; chr$( 34); "90,180,270,360"; chr$( 34); ") = "; using( "###.#", meanAngle( "90,180,270,360")); " degrees." ' -90
print "Mean Angle( "; chr$( 34); "10,20,30"; chr$( 34); ") = "; using( "###.#", meanAngle( "10,20,30")); " degrees." ' 20
end
function meanAngle( angles$)
term =1
while word$( angles$, term, ",") <>""
angle =val( word$( angles$, term, ","))
sumSin = sumSin +sin( degRad( angle))
sumCos = sumCos +cos( degRad( angle))
term =term +1
wend
meanAngle= radDeg( atan2( sumSin, sumCos))
if abs( sumSin) +abs( sumCos) <0.001 then notice "Not Available." +_
chr$( 13) +"(" +angles$ +")" +_
chr$( 13) +"Result nearly equals zero vector." +_
chr$( 13) +"Displaying '666'.": meanAngle =666
end function
function degRad( theta)
degRad =theta *Pi /180
end function
function radDeg( theta)
radDeg =theta *180 /Pi
end function
function atan2( y, x)
if x >0 then at =atn( y /x)
if y >=0 and x <0 then at =atn( y /x) +pi
if y <0 and x <0 then at =atn( y /x) -pi
if y >0 and x =0 then at = pi /2
if y <0 and x =0 then at = 0 -pi /2
if y =0 and x =0 then notice "undefined": end
atan2 =at
end function |
http://rosettacode.org/wiki/Averages/Median | Averages/Median | Task[edit]
Write a program to find the median value of a vector of floating-point numbers.
The program need not handle the case where the vector is empty, but must handle the case where there are an even number of elements. In that case, return the average of the two middle values.
There are several approaches to this. One is to sort the elements, and then pick the element(s) in the middle.
Sorting would take at least O(n logn). Another approach would be to build a priority queue from the elements, and then extract half of the elements to get to the middle element(s). This would also take O(n logn). The best solution is to use the selection algorithm to find the median in O(n) time.
See also
Quickselect_algorithm
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #EasyLang | EasyLang | func quickselect k . list[] res .
#
subr partition
swap list[(left + right) / 2] list[left]
mid = left
for i = left + 1 to right
if list[i] < list[left]
mid += 1
swap list[i] list[mid]
.
.
swap list[left] list[mid]
.
left = 0
right = len list[] - 1
while left < right
call partition
if mid < k
left = mid + 1
elif mid > k
right = mid - 1
else
left = right
.
.
res = list[k]
.
func median . list[] res .
h = len list[] / 2
call quickselect h list[] res
if len list[] mod 2 = 0
call quickselect h - 1 list[] h
res = (res + h) / 2
.
.
test[] = [ 4.1 5.6 7.2 1.7 9.3 4.4 3.2 ]
call median test[] med
print med
test[] = [ 4.1 7.2 1.7 9.3 4.4 3.2 ]
call median test[] med
print med |
http://rosettacode.org/wiki/Averages/Median | Averages/Median | Task[edit]
Write a program to find the median value of a vector of floating-point numbers.
The program need not handle the case where the vector is empty, but must handle the case where there are an even number of elements. In that case, return the average of the two middle values.
There are several approaches to this. One is to sort the elements, and then pick the element(s) in the middle.
Sorting would take at least O(n logn). Another approach would be to build a priority queue from the elements, and then extract half of the elements to get to the middle element(s). This would also take O(n logn). The best solution is to use the selection algorithm to find the median in O(n) time.
See also
Quickselect_algorithm
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #EchoLisp | EchoLisp |
(define (median L) ;; O(n log(n))
(set! L (vector-sort! < (list->vector L)))
(define dim (// (vector-length L) 2))
(if (integer? dim)
(// (+ [L dim] [L (1- dim)]) 2)
[L (floor dim)]))
(median '( 3 4 5))
→ 4
(median '(6 5 4 3))
→ 4.5
(median (iota 10000))
→ 4999.5
(median (iota 10001))
→ 5000
|
http://rosettacode.org/wiki/Averages/Pythagorean_means | Averages/Pythagorean means | Task[edit]
Compute all three of the Pythagorean means of the set of integers 1 through 10 (inclusive).
Show that
A
(
x
1
,
…
,
x
n
)
≥
G
(
x
1
,
…
,
x
n
)
≥
H
(
x
1
,
…
,
x
n
)
{\displaystyle A(x_{1},\ldots ,x_{n})\geq G(x_{1},\ldots ,x_{n})\geq H(x_{1},\ldots ,x_{n})}
for this set of positive integers.
The most common of the three means, the arithmetic mean, is the sum of the list divided by its length:
A
(
x
1
,
…
,
x
n
)
=
x
1
+
⋯
+
x
n
n
{\displaystyle A(x_{1},\ldots ,x_{n})={\frac {x_{1}+\cdots +x_{n}}{n}}}
The geometric mean is the
n
{\displaystyle n}
th root of the product of the list:
G
(
x
1
,
…
,
x
n
)
=
x
1
⋯
x
n
n
{\displaystyle G(x_{1},\ldots ,x_{n})={\sqrt[{n}]{x_{1}\cdots x_{n}}}}
The harmonic mean is
n
{\displaystyle n}
divided by the sum of the reciprocal of each item in the list:
H
(
x
1
,
…
,
x
n
)
=
n
1
x
1
+
⋯
+
1
x
n
{\displaystyle H(x_{1},\ldots ,x_{n})={\frac {n}{{\frac {1}{x_{1}}}+\cdots +{\frac {1}{x_{n}}}}}}
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Icon_and_Unicon | Icon and Unicon | link numbers # for a/g/h means
procedure main()
every put(x := [], 1 to 10)
writes("x := [ "); every writes(!x," "); write("]")
write("Arithmetic mean:", a := amean!x)
write("Geometric mean:",g := gmean!x)
write("Harmonic mean:", h := hmean!x)
write(" a >= g >= h is ", if a >= g >= h then "true" else "false")
end
|
http://rosettacode.org/wiki/Balanced_ternary | Balanced ternary | Balanced ternary is a way of representing numbers. Unlike the prevailing binary representation, a balanced ternary integer is in base 3, and each digit can have the values 1, 0, or −1.
Examples
Decimal 11 = 32 + 31 − 30, thus it can be written as "++−"
Decimal 6 = 32 − 31 + 0 × 30, thus it can be written as "+−0"
Task
Implement balanced ternary representation of integers with the following:
Support arbitrarily large integers, both positive and negative;
Provide ways to convert to and from text strings, using digits '+', '-' and '0' (unless you are already using strings to represent balanced ternary; but see requirement 5).
Provide ways to convert to and from native integer type (unless, improbably, your platform's native integer type is balanced ternary). If your native integers can't support arbitrary length, overflows during conversion must be indicated.
Provide ways to perform addition, negation and multiplication directly on balanced ternary integers; do not convert to native integers first.
Make your implementation efficient, with a reasonable definition of "efficient" (and with a reasonable definition of "reasonable").
Test case With balanced ternaries a from string "+-0++0+", b from native integer -436, c "+-++-":
write out a, b and c in decimal notation;
calculate a × (b − c), write out the result in both ternary and decimal notations.
Note: The pages generalised floating point addition and generalised floating point multiplication have code implementing arbitrary precision floating point balanced ternary.
| #OCaml | OCaml | type btdigit = Pos | Zero | Neg
type btern = btdigit list
let to_string n =
String.concat ""
(List.rev_map (function Pos -> "+" | Zero -> "0" | Neg -> "-") n)
let from_string s =
let sl = ref [] in
let digit = function '+' -> Pos | '-' -> Neg | '0' -> Zero
| _ -> failwith "invalid digit" in
String.iter (fun c -> sl := (digit c) :: !sl) s; !sl
let rec to_int = function
| [Zero] | [] -> 0
| Pos :: t -> 1 + 3 * to_int t
| Neg :: t -> -1 + 3 * to_int t
| Zero :: t -> 3 * to_int t
let rec from_int n =
if n = 0 then [] else
match n mod 3 with
| 0 -> Zero :: from_int (n/3)
| 1 | -2 -> Pos :: from_int ((n-1)/3)
| 2 | -1 -> Neg :: from_int ((n+1)/3)
let rec (+~) n1 n2 = match (n1,n2) with
| ([], a) | (a,[]) -> a
| (Pos::t1, Neg::t2) | (Neg::t1, Pos::t2) | (Zero::t1, Zero::t2) ->
let sum = t1 +~ t2 in if sum = [] then [] else Zero :: sum
| (Pos::t1, Pos::t2) -> Neg :: t1 +~ t2 +~ [Pos]
| (Neg::t1, Neg::t2) -> Pos :: t1 +~ t2 +~ [Neg]
| (Zero::t1, h::t2) | (h::t1, Zero::t2) -> h :: t1 +~ t2
let neg = List.map (function Pos -> Neg | Neg -> Pos | Zero -> Zero)
let (-~) a b = a +~ (neg b)
let rec ( *~) n1 = function
| [] -> []
| [Pos] -> n1
| [Neg] -> neg n1
| Pos::t -> (Zero :: t *~ n1) +~ n1
| Neg::t -> (Zero :: t *~ n1) -~ n1
| Zero::t -> Zero :: t *~ n1
let a = from_string "+-0++0+"
let b = from_int (-436)
let c = from_string "+-++-"
let d = a *~ (b -~ c)
let _ =
Printf.printf "a = %d\nb = %d\nc = %d\na * (b - c) = %s = %d\n"
(to_int a) (to_int b) (to_int c) (to_string d) (to_int d); |
http://rosettacode.org/wiki/Babbage_problem | Babbage problem |
Charles Babbage, looking ahead to the sorts of problems his Analytical Engine would be able to solve, gave this example:
What is the smallest positive integer whose square ends in the digits 269,696?
— Babbage, letter to Lord Bowden, 1837; see Hollingdale and Tootill, Electronic Computers, second edition, 1970, p. 125.
He thought the answer might be 99,736, whose square is 9,947,269,696; but he couldn't be certain.
Task[edit]
The task is to find out if Babbage had the right answer — and to do so, as far as your language allows it, in code that Babbage himself would have been able to read and understand.
As Babbage evidently solved the task with pencil and paper, a similar efficient solution is preferred.
For these purposes, Charles Babbage may be taken to be an intelligent person, familiar with mathematics and with the idea of a computer; he has written the first drafts of simple computer programmes in tabular form. [Babbage Archive Series L].
Motivation
The aim of the task is to write a program that is sufficiently clear and well-documented for such a person to be able to read it and be confident that it does indeed solve the specified problem.
| #Prolog | Prolog | :- use_module(library(clpfd)).
babbage_(B, B, Sq) :-
B * B #= Sq,
number_chars(Sq, R),
append(_, ['2','6','9','6','9','6'], R).
babbage_(B, R, Sq) :-
N #= B + 1,
babbage_(N, R, Sq).
babbage :-
once(babbage_(1, Num, Square)),
format('lowest number is ~p which squared becomes ~p~n', [Num, Square]). |
http://rosettacode.org/wiki/Approximate_equality | Approximate equality | Sometimes, when testing whether the solution to a task (for example, here on Rosetta Code) is correct, the
difference in floating point calculations between different language implementations becomes significant.
For example, a difference between 32 bit and 64 bit floating point calculations may appear by
about the 8th significant digit in base 10 arithmetic.
Task
Create a function which returns true if two floating point numbers are approximately equal.
The function should allow for differences in the magnitude of numbers, so that, for example,
100000000000000.01 may be approximately equal to 100000000000000.011,
even though 100.01 is not approximately equal to 100.011.
If the language has such a feature in its standard library, this may be used instead of a custom function.
Show the function results with comparisons on the following pairs of values:
100000000000000.01, 100000000000000.011 (note: should return true)
100.01, 100.011 (note: should return false)
10000000000000.001 / 10000.0, 1000000000.0000001000
0.001, 0.0010000001
0.000000000000000000000101, 0.0
sqrt(2) * sqrt(2), 2.0
-sqrt(2) * sqrt(2), -2.0
3.14159265358979323846, 3.14159265358979324
Answers should be true for the first example and false in the second, so that just rounding the numbers to a fixed number of decimals should not be enough. Otherwise answers may vary and still be correct. See the Python code for one type of solution.
| #J | J |
NB. default comparison tolerance matches the python result
".;._2]0 :0
100000000000000.01 = 100000000000000.011
100.01 = 100.011
(10000000000000.001 % 10000.0) = 1000000000.0000001000
0.001 = 0.0010000001
0.000000000000000000000101 = 0.0
(= ([: *~ %:)) 2 NB. sqrt(2)*sqrt(2)
((= -)~ ([: (* -) %:)) 2 NB. -sqrt(2) * sqrt(2), -2.0
3.14159265358979323846 = 3.14159265358979324
)
1 0 1 0 0 1 1 1
NB. tolerance of 1e_12 matches the python result
".;._2]0 :0[CT=:1e_12
100000000000000.01 =!.CT 100000000000000.011
100.01 =!.CT 100.011
(10000000000000.001 % 10000.0) =!.CT 1000000000.0000001000
0.001 =!.CT 0.0010000001
0.000000000000000000000101 =!.CT 0.0
(=!.CT ([: *~ %:)) 2 NB. sqrt(2)*sqrt(2)
((=!.CT -)~ ([: (* -) %:)) 2 NB. -sqrt(2) * sqrt(2), -2.0
3.14159265358979323846 =!.CT 3.14159265358979324
)
1 0 1 0 0 1 1 1
NB. tight tolerance
".;._2]0 :0[CT=:1e_18
100000000000000.01 =!.CT 100000000000000.011
100.01 =!.CT 100.011
(10000000000000.001 % 10000.0) =!.CT 1000000000.0000001000
0.001 =!.CT 0.0010000001
0.000000000000000000000101 =!.CT 0.0
(=!.CT ([: *~ %:)) 2 NB. sqrt(2)*sqrt(2)
((=!.CT -)~ ([: (* -) %:)) 2 NB. -sqrt(2) * sqrt(2), -2.0
3.14159265358979323846 =!.CT 3.14159265358979324
)
1 0 0 0 0 0 0 1
2 (=!.1e_8) 9
|domain error
| 2(= !.1e_8)9
|
http://rosettacode.org/wiki/Approximate_equality | Approximate equality | Sometimes, when testing whether the solution to a task (for example, here on Rosetta Code) is correct, the
difference in floating point calculations between different language implementations becomes significant.
For example, a difference between 32 bit and 64 bit floating point calculations may appear by
about the 8th significant digit in base 10 arithmetic.
Task
Create a function which returns true if two floating point numbers are approximately equal.
The function should allow for differences in the magnitude of numbers, so that, for example,
100000000000000.01 may be approximately equal to 100000000000000.011,
even though 100.01 is not approximately equal to 100.011.
If the language has such a feature in its standard library, this may be used instead of a custom function.
Show the function results with comparisons on the following pairs of values:
100000000000000.01, 100000000000000.011 (note: should return true)
100.01, 100.011 (note: should return false)
10000000000000.001 / 10000.0, 1000000000.0000001000
0.001, 0.0010000001
0.000000000000000000000101, 0.0
sqrt(2) * sqrt(2), 2.0
-sqrt(2) * sqrt(2), -2.0
3.14159265358979323846, 3.14159265358979324
Answers should be true for the first example and false in the second, so that just rounding the numbers to a fixed number of decimals should not be enough. Otherwise answers may vary and still be correct. See the Python code for one type of solution.
| #Java | Java | public class Approximate {
private static boolean approxEquals(double value, double other, double epsilon) {
return Math.abs(value - other) < epsilon;
}
private static void test(double a, double b) {
double epsilon = 1e-18;
System.out.printf("%f, %f => %s\n", a, b, approxEquals(a, b, epsilon));
}
public static void main(String[] args) {
test(100000000000000.01, 100000000000000.011);
test(100.01, 100.011);
test(10000000000000.001 / 10000.0, 1000000000.0000001000);
test(0.001, 0.0010000001);
test(0.000000000000000000000101, 0.0);
test(Math.sqrt(2.0) * Math.sqrt(2.0), 2.0);
test(-Math.sqrt(2.0) * Math.sqrt(2.0), -2.0);
test(3.14159265358979323846, 3.14159265358979324);
}
} |
http://rosettacode.org/wiki/Balanced_brackets | Balanced brackets | Task:
Generate a string with N opening brackets [ and with N closing brackets ], in some arbitrary order.
Determine whether the generated string is balanced; that is, whether it consists entirely of pairs of opening/closing brackets (in that order), none of which mis-nest.
Examples
(empty) OK
[] OK
[][] OK
[[][]] OK
][ NOT OK
][][ NOT OK
[]][[] NOT OK
| #COBOL | COBOL | IDENTIFICATION DIVISION.
PROGRAM-ID. test-balanced-brackets.
DATA DIVISION.
WORKING-STORAGE SECTION.
01 True-Val CONSTANT 0.
01 False-Val CONSTANT 1.
LOCAL-STORAGE SECTION.
01 current-time PIC 9(10).
01 bracket-type PIC 9.
88 add-open-bracket VALUE 1.
01 bracket-string-area.
03 bracket-string PIC X(10) OCCURS 10 TIMES.
01 i PIC 999.
01 j PIC 999.
PROCEDURE DIVISION.
*> Seed RANDOM().
MOVE FUNCTION CURRENT-DATE (7:10) TO current-time
MOVE FUNCTION RANDOM(current-time) TO current-time
*> Generate random strings of brackets.
PERFORM VARYING i FROM 1 BY 1 UNTIL 10 < i
PERFORM VARYING j FROM 1 BY 1 UNTIL i < j
COMPUTE bracket-type =
FUNCTION REM(FUNCTION RANDOM * 1000, 2)
IF add-open-bracket
MOVE "[" TO bracket-string (i) (j:1)
ELSE
MOVE "]" TO bracket-string (i) (j:1)
END-IF
END-PERFORM
END-PERFORM
*> Display if the strings are balanced or not.
PERFORM VARYING i FROM 1 BY 1 UNTIL 10 < i
CALL "check-if-balanced" USING bracket-string (i)
IF RETURN-CODE = True-Val
DISPLAY FUNCTION TRIM(bracket-string (i))
" is balanced."
ELSE
DISPLAY FUNCTION TRIM(bracket-string (i))
" is not balanced."
END-IF
END-PERFORM
GOBACK
.
END PROGRAM test-balanced-brackets.
IDENTIFICATION DIVISION.
PROGRAM-ID. check-if-balanced.
DATA DIVISION.
WORKING-STORAGE SECTION.
01 True-Val CONSTANT 0.
01 False-Val CONSTANT 1.
LOCAL-STORAGE SECTION.
01 nesting-level PIC S999.
01 i PIC 999.
LINKAGE SECTION.
01 bracket-string PIC X(100).
PROCEDURE DIVISION USING bracket-string.
PERFORM VARYING i FROM 1 BY 1
UNTIL (100 < i)
OR (bracket-string (i:1) = SPACE)
OR (nesting-level < 0)
IF bracket-string (i:1) = "["
ADD 1 TO nesting-level
ELSE
SUBTRACT 1 FROM nesting-level
IF nesting-level < 0
MOVE False-Val TO RETURN-CODE
GOBACK
END-IF
END-IF
END-PERFORM
IF nesting-level = 0
MOVE True-Val TO RETURN-CODE
ELSE
MOVE False-Val TO RETURN-CODE
END-IF
GOBACK
.
END PROGRAM check-if-balanced. |
http://rosettacode.org/wiki/Append_a_record_to_the_end_of_a_text_file | Append a record to the end of a text file | Many systems offer the ability to open a file for writing, such that any data written will be appended to the end of the file. Further, the file operations will always adjust the position pointer to guarantee the end of the file, even in a multitasking environment.
This feature is most useful in the case of log files, where many jobs may be appending to the log file at the same time, or where care must be taken to avoid concurrently overwriting the same record from another job.
Task
Given a two record sample for a mythical "passwd" file:
Write these records out in the typical system format.
Ideally these records will have named fields of various types.
Close the file, then reopen the file for append.
Append a new record to the file and close the file again.
Take appropriate care to avoid concurrently overwrites from another job.
Open the file and demonstrate the new record has indeed written to the end.
Source record field types and contents.
account
password
UID
GID
fullname,office,extension,homephone,email
directory
shell
string
string
int
int
struct(string,string,string,string,string)
string
string
jsmith
x
1001
1000
Joe Smith,Room 1007,(234)555-8917,(234)555-0077,[email protected]
/home/jsmith
/bin/bash
jdoe
x
1002
1000
Jane Doe,Room 1004,(234)555-8914,(234)555-0044,[email protected]
/home/jdoe
/bin/bash
Record to be appended.
account
password
UID
GID
fullname,office,extension,homephone,email
directory
shell
string
string
int
int
struct(string,string,string,string,string)
string
string
xyz
x
1003
1000
X Yz,Room 1003,(234)555-8913,(234)555-0033,[email protected]
/home/xyz
/bin/bash
Resulting file format: should mimic Linux's /etc/passwd file format with particular attention to the "," separator used in the GECOS field. But if the specific language has a particular or unique format of storing records in text file, then this format should be named and demonstrated with an additional example.
Expected output:
Appended record: xyz:x:1003:1000:X Yz,Room 1003,(234)555-8913,(234)555-0033,[email protected]:/home/xyz:/bin/bash
Finally: Provide a summary of the language's "append record" capabilities in a table. eg.
Append Capabilities.
Data Representation
IO
Library
Append
Possible
Automatic
Append
Multi-tasking
Safe
In core
On disk
C struct
CSV text file
glibc/stdio
☑
☑
☑ (Not all, eg NFS)
Alternatively: If the language's appends can not guarantee its writes will always append, then note this restriction in the table. If possible, provide an actual code example (possibly using file/record locking) to guarantee correct concurrent appends.
| #C.23 | C# | using System;
using System.IO;
namespace AppendPwdRosetta
{
class PasswordRecord
{
public string account, password, fullname, office, extension, homephone, email, directory, shell;
public int UID, GID;
public PasswordRecord(string account, string password, int UID, int GID, string fullname, string office, string extension, string homephone,
string email, string directory, string shell)
{
this.account = account; this.password = password; this.UID = UID; this.GID = GID; this.fullname = fullname; this.office = office;
this.extension = extension; this.homephone = homephone; this.email = email; this.directory = directory; this.shell = shell;
}
public override string ToString()
{
var gecos = string.Join(",", new string[] { fullname, office, extension, homephone, email });
return string.Join(":", new string[] { account, password, UID.ToString(), GID.ToString(), gecos, directory, shell });
}
}
class Program
{
static void Main(string[] args)
{
var jsmith = new PasswordRecord("jsmith", "x", 1001, 1000, "Joe Smith", "Room 1007", "(234)555-8917", "(234)555-0077", "[email protected]",
"/home/jsmith", "/bin/bash");
var jdoe = new PasswordRecord("jdoe", "x", 1002, 1000, "Jane Doe", "Room 1004", "(234)555-8914", "(234)555-0044", "[email protected]", "/home/jdoe",
"/bin/bash");
var xyz = new PasswordRecord("xyz", "x", 1003, 1000, "X Yz", "Room 1003", "(234)555-8913", "(234)555-0033", "[email protected]", "/home/xyz", "/bin/bash");
// Write these records out in the typical system format.
File.WriteAllLines("passwd.txt", new string[] { jsmith.ToString(), jdoe.ToString() });
// Append a new record to the file and close the file again.
File.AppendAllText("passwd.txt", xyz.ToString());
// Open the file and demonstrate the new record has indeed written to the end.
string[] lines = File.ReadAllLines("passwd.txt");
Console.WriteLine("Appended record: " + lines[2]);
}
}
}
|
http://rosettacode.org/wiki/Associative_array/Creation | Associative array/Creation | Task
The goal is to create an associative array (also known as a dictionary, map, or hash).
Related tasks:
Associative arrays/Iteration
Hash from two arrays
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #Aime | Aime | record r; |
http://rosettacode.org/wiki/Anti-primes | Anti-primes | The anti-primes
(or highly composite numbers, sequence A002182 in the OEIS)
are the natural numbers with more factors than any smaller than itself.
Task
Generate and show here, the first twenty anti-primes.
Related tasks
Factors of an integer
Sieve of Eratosthenes
| #Ada | Ada | with Ada.Text_IO; use Ada.Text_IO;
procedure Antiprimes is
function Count_Divisors (N : Integer) return Integer is
Count : Integer := 1;
begin
for i in 1 .. N / 2 loop
if N mod i = 0 then
Count := Count + 1;
end if;
end loop;
return Count;
end Count_Divisors;
Results : array (1 .. 20) of Integer;
Candidate : Integer := 1;
Divisors : Integer;
Max_Divisors : Integer := 0;
begin
for i in Results'Range loop
loop
Divisors := Count_Divisors (Candidate);
if Max_Divisors < Divisors then
Results (i) := Candidate;
Max_Divisors := Divisors;
exit;
end if;
Candidate := Candidate + 1;
end loop;
end loop;
Put_Line ("The first 20 anti-primes are:");
for I in Results'Range loop
Put (Integer'Image (Results (I)));
end loop;
New_Line;
end Antiprimes; |
http://rosettacode.org/wiki/Anti-primes | Anti-primes | The anti-primes
(or highly composite numbers, sequence A002182 in the OEIS)
are the natural numbers with more factors than any smaller than itself.
Task
Generate and show here, the first twenty anti-primes.
Related tasks
Factors of an integer
Sieve of Eratosthenes
| #ALGOL_68 | ALGOL 68 | BEGIN # find some anti-primes: numbers with more divisors than the #
# previous numbers #
INT max number := 10 000;
INT max divisors := 0;
# construct a table of the divisor counts #
[ 1 : max number ]INT ndc; FOR i FROM 1 TO UPB ndc DO ndc[ i ] := 1 OD;
FOR i FROM 2 TO UPB ndc DO
FOR j FROM i BY i TO UPB ndc DO ndc[ j ] +:= 1 OD
OD;
# show the numbers with more divisors than their predecessors #
INT a count := 0;
FOR i TO UPB ndc WHILE a count < 20 DO
INT divisor count = ndc[ i ];
IF divisor count > max divisors THEN
print( ( " ", whole( i, 0 ) ) );
max divisors := divisor count;
a count +:= 1
FI
OD;
print( ( newline ) )
END |
http://rosettacode.org/wiki/Atomic_updates | Atomic updates |
Task
Define a data type consisting of a fixed number of 'buckets', each containing a nonnegative integer value, which supports operations to:
get the current value of any bucket
remove a specified amount from one specified bucket and add it to another, preserving the total of all bucket values, and clamping the transferred amount to ensure the values remain non-negative
In order to exercise this data type, create one set of buckets, and start three concurrent tasks:
As often as possible, pick two buckets and make their values closer to equal.
As often as possible, pick two buckets and arbitrarily redistribute their values.
At whatever rate is convenient, display (by any means) the total value and, optionally, the individual values of each bucket.
The display task need not be explicit; use of e.g. a debugger or trace tool is acceptable provided it is simple to set up to provide the display.
This task is intended as an exercise in atomic operations. The sum of the bucket values must be preserved even if the two tasks attempt to perform transfers simultaneously, and a straightforward solution is to ensure that at any time, only one transfer is actually occurring — that the transfer operation is atomic.
| #Go | Go | package main
import (
"fmt"
"math/rand"
"sync"
"time"
)
const nBuckets = 10
type bucketList struct {
b [nBuckets]int // bucket data specified by task
// transfer counts for each updater, not strictly required by task but
// useful to show that the two updaters get fair chances to run.
tc [2]int
sync.Mutex // synchronization
}
// Updater ids, to track number of transfers by updater.
// these can index bucketlist.tc for example.
const (
idOrder = iota
idChaos
)
const initialSum = 1000 // sum of all bucket values
// Constructor.
func newBucketList() *bucketList {
var bl bucketList
// Distribute initialSum across buckets.
for i, dist := nBuckets, initialSum; i > 0; {
v := dist / i
i--
bl.b[i] = v
dist -= v
}
return &bl
}
// method 1 required by task, get current value of a bucket
func (bl *bucketList) bucketValue(b int) int {
bl.Lock() // lock before accessing data
r := bl.b[b]
bl.Unlock()
return r
}
// method 2 required by task
func (bl *bucketList) transfer(b1, b2, a int, ux int) {
// Get access.
bl.Lock()
// Clamping maintains invariant that bucket values remain nonnegative.
if a > bl.b[b1] {
a = bl.b[b1]
}
// Transfer.
bl.b[b1] -= a
bl.b[b2] += a
bl.tc[ux]++ // increment transfer count
bl.Unlock()
}
// additional useful method
func (bl *bucketList) snapshot(s *[nBuckets]int, tc *[2]int) {
bl.Lock()
*s = bl.b
*tc = bl.tc
bl.tc = [2]int{} // clear transfer counts
bl.Unlock()
}
var bl = newBucketList()
func main() {
// Three concurrent tasks.
go order() // make values closer to equal
go chaos() // arbitrarily redistribute values
buddha() // display total value and individual values of each bucket
}
// The concurrent tasks exercise the data operations by calling bucketList
// methods. The bucketList methods are "threadsafe", by which we really mean
// goroutine-safe. The conconcurrent tasks then do no explicit synchronization
// and are not responsible for maintaining invariants.
// Exercise 1 required by task: make values more equal.
func order() {
r := rand.New(rand.NewSource(time.Now().UnixNano()))
for {
b1 := r.Intn(nBuckets)
b2 := r.Intn(nBuckets - 1)
if b2 >= b1 {
b2++
}
v1 := bl.bucketValue(b1)
v2 := bl.bucketValue(b2)
if v1 > v2 {
bl.transfer(b1, b2, (v1-v2)/2, idOrder)
} else {
bl.transfer(b2, b1, (v2-v1)/2, idOrder)
}
}
}
// Exercise 2 required by task: redistribute values.
func chaos() {
r := rand.New(rand.NewSource(time.Now().Unix()))
for {
b1 := r.Intn(nBuckets)
b2 := r.Intn(nBuckets - 1)
if b2 >= b1 {
b2++
}
bl.transfer(b1, b2, r.Intn(bl.bucketValue(b1)+1), idChaos)
}
}
// Exercise 3 requred by task: display total.
func buddha() {
var s [nBuckets]int
var tc [2]int
var total, nTicks int
fmt.Println("sum ---updates--- mean buckets")
tr := time.Tick(time.Second / 10)
for {
<-tr
bl.snapshot(&s, &tc)
var sum int
for _, l := range s {
if l < 0 {
panic("sob") // invariant not preserved
}
sum += l
}
// Output number of updates per tick and cummulative mean
// updates per tick to demonstrate "as often as possible"
// of task exercises 1 and 2.
total += tc[0] + tc[1]
nTicks++
fmt.Printf("%d %6d %6d %7d %3d\n", sum, tc[0], tc[1], total/nTicks, s)
if sum != initialSum {
panic("weep") // invariant not preserved
}
}
} |
http://rosettacode.org/wiki/Assertions | Assertions | Assertions are a way of breaking out of code when there is an error or an unexpected input.
Some languages throw exceptions and some treat it as a break point.
Task
Show an assertion in your language by asserting that an integer variable is equal to 42.
| #ECL | ECL |
ASSERT(a = 42,'A is not 42!',FAIL); |
http://rosettacode.org/wiki/Assertions | Assertions | Assertions are a way of breaking out of code when there is an error or an unexpected input.
Some languages throw exceptions and some treat it as a break point.
Task
Show an assertion in your language by asserting that an integer variable is equal to 42.
| #Eiffel | Eiffel | class MAIN
creation main
feature main is
local
test: TEST;
do
create test;
io.read_integer;
test.assert(io.last_integer);
end
end |
http://rosettacode.org/wiki/Assertions | Assertions | Assertions are a way of breaking out of code when there is an error or an unexpected input.
Some languages throw exceptions and some treat it as a break point.
Task
Show an assertion in your language by asserting that an integer variable is equal to 42.
| #Elixir | Elixir | ExUnit.start
defmodule AssertionTest do
use ExUnit.Case
def return_5, do: 5
test "not equal" do
assert 42 == return_5
end
end |
http://rosettacode.org/wiki/Apply_a_callback_to_an_array | Apply a callback to an array | Task
Take a combined set of elements and apply a function to each element.
| #Aime | Aime | void
map(list l, void (*fp)(object))
{
l.ucall(fp, 0);
}
void
out(object o)
{
o_(o, "\n");
}
integer
main(void)
{
list(0, 1, 2, 3).map(out);
return 0;
} |
http://rosettacode.org/wiki/Apply_a_callback_to_an_array | Apply a callback to an array | Task
Take a combined set of elements and apply a function to each element.
| #ALGOL_68 | ALGOL 68 | PROC call back proc = (INT location, INT value)VOID:
(
printf(($"array["g"] = "gl$, location, value))
);
PROC map = (REF[]INT array, PROC (INT,INT)VOID call back)VOID:
(
FOR i FROM LWB array TO UPB array DO
call back(i, array[i])
OD
);
main:
(
[4]INT array := ( 1, 4, 9, 16 );
map(array, call back proc)
) |
http://rosettacode.org/wiki/Averages/Mode | Averages/Mode | Task[edit]
Write a program to find the mode value of a collection.
The case where the collection is empty may be ignored. Care must be taken to handle the case where the mode is non-unique.
If it is not appropriate or possible to support a general collection, use a vector (array), if possible. If it is not appropriate or possible to support an unspecified value type, use integers.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Mathematica_.2F_Wolfram_Language | Mathematica / Wolfram Language | Commonest[{b, a, c, 2, a, b, 1, 2, 3}]
Commonest[{1, 3, 2, 3}] |
http://rosettacode.org/wiki/Averages/Mode | Averages/Mode | Task[edit]
Write a program to find the mode value of a collection.
The case where the collection is empty may be ignored. Care must be taken to handle the case where the mode is non-unique.
If it is not appropriate or possible to support a general collection, use a vector (array), if possible. If it is not appropriate or possible to support an unspecified value type, use integers.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #MATLAB | MATLAB | function modeValue = findmode(setOfValues)
modeValue = mode(setOfValues);
end |
http://rosettacode.org/wiki/Associative_array/Iteration | Associative array/Iteration | Show how to iterate over the key-value pairs of an associative array, and print each pair out.
Also show how to iterate just over the keys, or the values, if there is a separate way to do that in your language.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #Bracmat | Bracmat | ( new$hash:?myhash
& (myhash..insert)$(title."Some title")
& (myhash..insert)$(formula.a+b+x^7)
& (myhash..insert)$(fruit.apples oranges kiwis)
& (myhash..insert)$(meat.)
& (myhash..insert)$(fruit.melons bananas)
& (myhash..remove)$formula
& (myhash..insert)$(formula.x^2+y^2)
& (myhash..forall)
$ (
= key value
. whl
' ( !arg:(?key.?value) ?arg
& put$("key:" !key "\nvalue:" !value \n)
)
& put$\n
)
); |
http://rosettacode.org/wiki/Associative_array/Iteration | Associative array/Iteration | Show how to iterate over the key-value pairs of an associative array, and print each pair out.
Also show how to iterate just over the keys, or the values, if there is a separate way to do that in your language.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #Brat | Brat | h = [ hello: 1 world: 2 :! : 3]
#Iterate over key, value pairs
h.each { k, v |
p "Key: #{k} Value: #{v}"
}
#Iterate over keys
h.each_key { k |
p "Key: #{k}"
}
#Iterate over values
h.each_value { v |
p "Value: #{v}"
} |
http://rosettacode.org/wiki/Apply_a_digital_filter_(direct_form_II_transposed) | Apply a digital filter (direct form II transposed) | Digital filters are used to apply a mathematical operation to a sampled signal. One of the common formulations is the "direct form II transposed" which can represent both infinite impulse response (IIR) and finite impulse response (FIR) filters, as well as being more numerically stable than other forms. [1]
Task
Filter a signal using an order 3 low-pass Butterworth filter. The coefficients for the filter are a=[1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17] and b = [0.16666667, 0.5, 0.5, 0.16666667]
The signal that needs filtering is the following vector: [-0.917843918645, 0.141984778794, 1.20536903482, 0.190286794412, -0.662370894973, -1.00700480494, -0.404707073677 ,0.800482325044, 0.743500089861, 1.01090520172, 0.741527555207, 0.277841675195, 0.400833448236, -0.2085993586, -0.172842103641, -0.134316096293, 0.0259303398477, 0.490105989562, 0.549391221511, 0.9047198589]
| #FreeBASIC | FreeBASIC | Sub Filtro(a() As Double, b() As Double, senal() As Double, resultado() As Double)
Dim As Integer j, k
Dim As Double tmp
For j = 0 To Ubound(senal)
tmp = 0
For k = 0 To Ubound(b)
If (j-k < 0) Then Continue For
tmp = tmp + b(k) * senal(j-k)
Next k
For k = 0 To Ubound(a)
If (j-k < 0) Then Continue For
tmp = tmp - a(k) * resultado(j-k)
Next k
tmp /= a(0)
resultado(j) = tmp
Next j
End Sub
Dim Shared As Double a(4), b(4), senal(20), resultado(20)
Dim As Integer i
For i = 0 To 3 : Read a(i) : Next i
For i = 0 To 3 : Read b(i) : Next i
For i = 0 To 19 : Read senal(i) : Next i
Filtro(a(),b(),senal(),resultado())
For i = 0 To 19
Print Using "###.########"; resultado(i);
If (i+1) Mod 5 <> 0 Then
Print ", ";
Else
Print
End If
Next i
'' a()
Data 1, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17
'' b()
Data 0.16666667, 0.5, 0.5, 0.16666667
'' senal()
Data -0.917843918645, 0.141984778794, 1.20536903482, 0.190286794412
Data -0.662370894973, -1.00700480494, -0.404707073677, 0.800482325044
Data 0.743500089861, 1.01090520172, 0.741527555207, 0.277841675195
Data 0.400833448236, -0.2085993586, -0.172842103641, -0.134316096293
Data 0.0259303398477, 0.490105989562, 0.549391221511, 0.9047198589
Sleep |
http://rosettacode.org/wiki/Averages/Arithmetic_mean | Averages/Arithmetic mean | Task[edit]
Write a program to find the mean (arithmetic average) of a numeric vector.
In case of a zero-length input, since the mean of an empty set of numbers is ill-defined, the program may choose to behave in any way it deems appropriate, though if the programming language has an established convention for conveying math errors or undefined values, it's preferable to follow it.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #C.23 | C# | using System;
using System.Linq;
class Program
{
static void Main()
{
Console.WriteLine(new[] { 1, 2, 3 }.Average());
}
} |
http://rosettacode.org/wiki/Averages/Arithmetic_mean | Averages/Arithmetic mean | Task[edit]
Write a program to find the mean (arithmetic average) of a numeric vector.
In case of a zero-length input, since the mean of an empty set of numbers is ill-defined, the program may choose to behave in any way it deems appropriate, though if the programming language has an established convention for conveying math errors or undefined values, it's preferable to follow it.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #C.2B.2B | C++ | #include <vector>
double mean(const std::vector<double>& numbers)
{
if (numbers.size() == 0)
return 0;
double sum = 0;
for (std::vector<double>::iterator i = numbers.begin(); i != numbers.end(); i++)
sum += *i;
return sum / numbers.size();
} |
http://rosettacode.org/wiki/Associative_array/Merging | Associative array/Merging | Task
Define two associative arrays, where one represents the following "base" data:
Key
Value
"name"
"Rocket Skates"
"price"
12.75
"color"
"yellow"
And the other represents "update" data:
Key
Value
"price"
15.25
"color"
"red"
"year"
1974
Merge these into a new associative array that contains every key found in either of the source ones. Each key should map to the value in the second (update) table if that exists, or else to the value in the first (base) table. If possible, do this in a way that does not mutate the original two associative arrays. Obviously this should be done in a way that would work for any data, not just the specific data given here, but in this example the result should be:
Key
Value
"name"
"Rocket Skates"
"price"
15.25
"color"
"red"
"year"
1974
| #Nim | Nim | import tables
let t1 = {"name": "Rocket Skates", "price": "12.75", "color": "yellow"}.toTable
let t2 = {"price": "15.25", "color": "red", "year": "1974"}.toTable
var t3 = t1 # Makes a copy.
for key, value in t2.pairs:
t3[key] = value
echo t3 |
http://rosettacode.org/wiki/Associative_array/Merging | Associative array/Merging | Task
Define two associative arrays, where one represents the following "base" data:
Key
Value
"name"
"Rocket Skates"
"price"
12.75
"color"
"yellow"
And the other represents "update" data:
Key
Value
"price"
15.25
"color"
"red"
"year"
1974
Merge these into a new associative array that contains every key found in either of the source ones. Each key should map to the value in the second (update) table if that exists, or else to the value in the first (base) table. If possible, do this in a way that does not mutate the original two associative arrays. Obviously this should be done in a way that would work for any data, not just the specific data given here, but in this example the result should be:
Key
Value
"name"
"Rocket Skates"
"price"
15.25
"color"
"red"
"year"
1974
| #Objective-C | Objective-C | #import <Foundation/Foundation.h>
int main(void) {
@autoreleasepool {
NSDictionary *base = @{@"name": @"Rocket Skates", @"price": @12.75, @"color": @"yellow"};
NSDictionary *update = @{@"price": @15.25, @"color": @"red", @"year": @1974};
NSMutableDictionary *result = [[NSMutableDictionary alloc] initWithDictionary:base];
[result addEntriesFromDictionary:update];
NSLog(@"%@", result);
}
return 0;
} |
http://rosettacode.org/wiki/Associative_array/Merging | Associative array/Merging | Task
Define two associative arrays, where one represents the following "base" data:
Key
Value
"name"
"Rocket Skates"
"price"
12.75
"color"
"yellow"
And the other represents "update" data:
Key
Value
"price"
15.25
"color"
"red"
"year"
1974
Merge these into a new associative array that contains every key found in either of the source ones. Each key should map to the value in the second (update) table if that exists, or else to the value in the first (base) table. If possible, do this in a way that does not mutate the original two associative arrays. Obviously this should be done in a way that would work for any data, not just the specific data given here, but in this example the result should be:
Key
Value
"name"
"Rocket Skates"
"price"
15.25
"color"
"red"
"year"
1974
| #OCaml | OCaml |
type ty =
| TFloat of float
| TInt of int
| TString of string
type key = string
type assoc = string * ty
let string_of_ty : ty -> string = function
| TFloat x -> string_of_float x
| TInt i -> string_of_int i
| TString s -> s
let print_pair key el =
Printf.printf "%s: %s\n" key (string_of_ty el)
;;
|
http://rosettacode.org/wiki/Average_loop_length | Average loop length | Let f be a uniformly-randomly chosen mapping from the numbers 1..N to the numbers 1..N (note: not necessarily a permutation of 1..N; the mapping could produce a number in more than one way or not at all). At some point, the sequence 1, f(1), f(f(1))... will contain a repetition, a number that occurring for the second time in the sequence.
Task
Write a program or a script that estimates, for each N, the average length until the first such repetition.
Also calculate this expected length using an analytical formula, and optionally compare the simulated result with the theoretical one.
This problem comes from the end of Donald Knuth's Christmas tree lecture 2011.
Example of expected output:
N average analytical (error)
=== ========= ============ =========
1 1.0000 1.0000 ( 0.00%)
2 1.4992 1.5000 ( 0.05%)
3 1.8784 1.8889 ( 0.56%)
4 2.2316 2.2188 ( 0.58%)
5 2.4982 2.5104 ( 0.49%)
6 2.7897 2.7747 ( 0.54%)
7 3.0153 3.0181 ( 0.09%)
8 3.2429 3.2450 ( 0.07%)
9 3.4536 3.4583 ( 0.14%)
10 3.6649 3.6602 ( 0.13%)
11 3.8091 3.8524 ( 1.12%)
12 3.9986 4.0361 ( 0.93%)
13 4.2074 4.2123 ( 0.12%)
14 4.3711 4.3820 ( 0.25%)
15 4.5275 4.5458 ( 0.40%)
16 4.6755 4.7043 ( 0.61%)
17 4.8877 4.8579 ( 0.61%)
18 4.9951 5.0071 ( 0.24%)
19 5.1312 5.1522 ( 0.41%)
20 5.2699 5.2936 ( 0.45%)
| #PARI.2FGP | PARI/GP | expected(n)=sum(i=1,n,n!/(n-i)!/n^i,0.);
test(n, times)={
my(ct);
for(i=1,times,
my(x=1,bits);
while(!bitand(bits,x),ct++; bits=bitor(bits,x); x = 1<<random(n))
);
ct
};
TIMES=1000000;
{for(n=1,20,
my(cnt=test(n, TIMES),avg=cnt/TIMES,ex=expected(n),diff=(avg/ex-1)*100.);
print(n"\t"avg*1."\t"ex*1."\t"diff);
)} |
http://rosettacode.org/wiki/Average_loop_length | Average loop length | Let f be a uniformly-randomly chosen mapping from the numbers 1..N to the numbers 1..N (note: not necessarily a permutation of 1..N; the mapping could produce a number in more than one way or not at all). At some point, the sequence 1, f(1), f(f(1))... will contain a repetition, a number that occurring for the second time in the sequence.
Task
Write a program or a script that estimates, for each N, the average length until the first such repetition.
Also calculate this expected length using an analytical formula, and optionally compare the simulated result with the theoretical one.
This problem comes from the end of Donald Knuth's Christmas tree lecture 2011.
Example of expected output:
N average analytical (error)
=== ========= ============ =========
1 1.0000 1.0000 ( 0.00%)
2 1.4992 1.5000 ( 0.05%)
3 1.8784 1.8889 ( 0.56%)
4 2.2316 2.2188 ( 0.58%)
5 2.4982 2.5104 ( 0.49%)
6 2.7897 2.7747 ( 0.54%)
7 3.0153 3.0181 ( 0.09%)
8 3.2429 3.2450 ( 0.07%)
9 3.4536 3.4583 ( 0.14%)
10 3.6649 3.6602 ( 0.13%)
11 3.8091 3.8524 ( 1.12%)
12 3.9986 4.0361 ( 0.93%)
13 4.2074 4.2123 ( 0.12%)
14 4.3711 4.3820 ( 0.25%)
15 4.5275 4.5458 ( 0.40%)
16 4.6755 4.7043 ( 0.61%)
17 4.8877 4.8579 ( 0.61%)
18 4.9951 5.0071 ( 0.24%)
19 5.1312 5.1522 ( 0.41%)
20 5.2699 5.2936 ( 0.45%)
| #Perl | Perl | use List::Util qw(sum reduce);
sub find_loop {
my($n) = @_;
my($r,@seen);
while () { $seen[$r] = $seen[($r = int(1+rand $n))] ? return sum @seen : 1 }
}
print " N empiric theoric (error)\n";
print "=== ========= ============ =========\n";
my $MAX = 20;
my $TRIALS = 1000;
for my $n (1 .. $MAX) {
my $empiric = ( sum map { find_loop($n) } 1..$TRIALS ) / $TRIALS;
my $theoric = sum map { (reduce { $a*$b } $_**2, ($n-$_+1)..$n ) / $n ** ($_+1) } 1..$n;
printf "%3d %9.4f %12.4f (%5.2f%%)\n",
$n, $empiric, $theoric, 100 * ($empiric - $theoric) / $theoric;
} |
http://rosettacode.org/wiki/Averages/Simple_moving_average | Averages/Simple moving average | Computing the simple moving average of a series of numbers.
Task[edit]
Create a stateful function/class/instance that takes a period and returns a routine that takes a number as argument and returns a simple moving average of its arguments so far.
Description
A simple moving average is a method for computing an average of a stream of numbers by only averaging the last P numbers from the stream, where P is known as the period.
It can be implemented by calling an initialing routine with P as its argument, I(P), which should then return a routine that when called with individual, successive members of a stream of numbers, computes the mean of (up to), the last P of them, lets call this SMA().
The word stateful in the task description refers to the need for SMA() to remember certain information between calls to it:
The period, P
An ordered container of at least the last P numbers from each of its individual calls.
Stateful also means that successive calls to I(), the initializer, should return separate routines that do not share saved state so they could be used on two independent streams of data.
Pseudo-code for an implementation of SMA is:
function SMA(number: N):
stateful integer: P
stateful list: stream
number: average
stream.append_last(N)
if stream.length() > P:
# Only average the last P elements of the stream
stream.delete_first()
if stream.length() == 0:
average = 0
else:
average = sum( stream.values() ) / stream.length()
return average
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Ring | Ring |
load "stdlib.ring"
decimals(8)
maxperiod = 20
nums = newlist(maxperiod,maxperiod)
accum = list(maxperiod)
index = list(maxperiod)
window = list(maxperiod)
for i = 1 to maxperiod
index[i] = 1
accum[i] = 0
window[i] = 0
next
for i = 1 to maxperiod
for j = 1 to maxperiod
nums[i][j] = 0
next
next
for n = 1 to 5
see "number = " + n + " sma3 = " + left((string(sma(n,3)) + " "),9) + " sma5 = " + sma(n,5) + nl
next
for n = 5 to 1 step -1
see "number = " + n + " sma3 = " + left((string(sma(n,3)) + " "),9) + " sma5 = " + sma(n,5) + nl
next
see nl
func sma number, period
accum[period] += number - nums[period][index[period]]
nums[period][index[period]] = number
index[period]= (index[period] + 1) % period + 1
if window[period]<period window[period] += 1 ok
return (accum[period] / window[period])
|
http://rosettacode.org/wiki/Attractive_numbers | Attractive numbers | A number is an attractive number if the number of its prime factors (whether distinct or not) is also prime.
Example
The number 20, whose prime decomposition is 2 × 2 × 5, is an attractive number because the number of its prime factors (3) is also prime.
Task
Show sequence items up to 120.
Reference
The OEIS entry: A063989: Numbers with a prime number of prime divisors.
| #Groovy | Groovy | class AttractiveNumbers {
static boolean isPrime(int n) {
if (n < 2) return false
if (n % 2 == 0) return n == 2
if (n % 3 == 0) return n == 3
int d = 5
while (d * d <= n) {
if (n % d == 0) return false
d += 2
if (n % d == 0) return false
d += 4
}
return true
}
static int countPrimeFactors(int n) {
if (n == 1) return 0
if (isPrime(n)) return 1
int count = 0, f = 2
while (true) {
if (n % f == 0) {
count++
n /= f
if (n == 1) return count
if (isPrime(n)) f = n
} else if (f >= 3) f += 2
else f = 3
}
}
static void main(String[] args) {
final int max = 120
printf("The attractive numbers up to and including %d are:\n", max)
int count = 0
for (int i = 1; i <= max; ++i) {
int n = countPrimeFactors(i)
if (isPrime(n)) {
printf("%4d", i)
if (++count % 20 == 0) println()
}
}
println()
}
} |
http://rosettacode.org/wiki/Averages/Mean_time_of_day | Averages/Mean time of day | Task[edit]
A particular activity of bats occurs at these times of the day:
23:00:17, 23:40:20, 00:12:45, 00:17:19
Using the idea that there are twenty-four hours in a day,
which is analogous to there being 360 degrees in a circle,
map times of day to and from angles;
and using the ideas of Averages/Mean angle
compute and show the average time of the nocturnal activity
to an accuracy of one second of time.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Mathematica_.2F_Wolfram_Language | Mathematica / Wolfram Language | meanTime[list_] :=
StringJoin@
Riffle[ToString /@
Floor@{Mod[24 #, 24], Mod[24*60 #, 60], Mod[24*60*60 #, 60]} &[
Arg[Mean[
Exp[FromDigits[ToExpression@StringSplit[#, ":"], 60] & /@
list/(24*60*60) 2 Pi I]]]/(2 Pi)], ":"];
meanTime[{"23:00:17", "23:40:20", "00:12:45", "00:17:19"}] |
http://rosettacode.org/wiki/Averages/Mean_time_of_day | Averages/Mean time of day | Task[edit]
A particular activity of bats occurs at these times of the day:
23:00:17, 23:40:20, 00:12:45, 00:17:19
Using the idea that there are twenty-four hours in a day,
which is analogous to there being 360 degrees in a circle,
map times of day to and from angles;
and using the ideas of Averages/Mean angle
compute and show the average time of the nocturnal activity
to an accuracy of one second of time.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #MATLAB_.2F_Octave | MATLAB / Octave | function t = mean_time_of_day(t)
c = pi/(12*60*60);
for k=1:length(t)
a = sscanf(t{k},'%d:%d:%d');
phi(k) = (a(1)*3600+a(2)*60+a(3));
end;
d = angle(mean(exp(i*phi*c)))/(2*pi); % days
if (d<0) d += 1;
t = datestr(d,"HH:MM:SS");
end; |
http://rosettacode.org/wiki/AVL_tree | AVL tree |
This page uses content from Wikipedia. The original article was at AVL tree. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)
In computer science, an AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; at no time do they differ by more than one because rebalancing is done ensure this is the case. Lookup, insertion, and deletion all take O(log n) time in both the average and worst cases, where n is the number of nodes in the tree prior to the operation. Insertions and deletions may require the tree to be rebalanced by one or more tree rotations. Note the tree of nodes comprise a set, so duplicate node keys are not allowed.
AVL trees are often compared with red-black trees because they support the same set of operations and because red-black trees also take O(log n) time for the basic operations. Because AVL trees are more rigidly balanced, they are faster than red-black trees for lookup-intensive applications. Similar to red-black trees, AVL trees are height-balanced, but in general not weight-balanced nor μ-balanced; that is, sibling nodes can have hugely differing numbers of descendants.
Task
Implement an AVL tree in the language of choice, and provide at least basic operations.
| #Objective-C | Objective-C |
@implementation AVLTree
-(BOOL)insertWithKey:(NSInteger)key {
if (self.root == nil) {
self.root = [[AVLTreeNode alloc]initWithKey:key andParent:nil];
} else {
AVLTreeNode *n = self.root;
AVLTreeNode *parent;
while (true) {
if (n.key == key) {
return false;
}
parent = n;
BOOL goLeft = n.key > key;
n = goLeft ? n.left : n.right;
if (n == nil) {
if (goLeft) {
parent.left = [[AVLTreeNode alloc]initWithKey:key andParent:parent];
} else {
parent.right = [[AVLTreeNode alloc]initWithKey:key andParent:parent];
}
[self rebalanceStartingAtNode:parent];
break;
}
}
}
return true;
}
-(void)rebalanceStartingAtNode:(AVLTreeNode*)n {
[self setBalance:@[n]];
if (n.balance == -2) {
if ([self height:(n.left.left)] >= [self height:n.left.right]) {
n = [self rotateRight:n];
} else {
n = [self rotateLeftThenRight:n];
}
} else if (n.balance == 2) {
if ([self height:n.right.right] >= [self height:n.right.left]) {
n = [self rotateLeft:n];
} else {
n = [self rotateRightThenLeft:n];
}
}
if (n.parent != nil) {
[self rebalanceStartingAtNode:n.parent];
} else {
self.root = n;
}
}
-(AVLTreeNode*)rotateRight:(AVLTreeNode*)a {
AVLTreeNode *b = a.left;
b.parent = a.parent;
a.left = b.right;
if (a.left != nil) {
a.left.parent = a;
}
b.right = a;
a.parent = b;
if (b.parent != nil) {
if (b.parent.right == a) {
b.parent.right = b;
} else {
b.parent.left = b;
}
}
[self setBalance:@[a,b]];
return b;
}
-(AVLTreeNode*)rotateLeftThenRight:(AVLTreeNode*)n {
n.left = [self rotateLeft:n.left];
return [self rotateRight:n];
}
-(AVLTreeNode*)rotateRightThenLeft:(AVLTreeNode*)n {
n.right = [self rotateRight:n.right];
return [self rotateLeft:n];
}
-(AVLTreeNode*)rotateLeft:(AVLTreeNode*)a {
//set a's right node as b
AVLTreeNode* b = a.right;
//set b's parent as a's parent (which could be nil)
b.parent = a.parent;
//in case b had a left child transfer it to a
a.right = b.left;
// after changing a's reference to the right child, make sure the parent is set too
if (a.right != nil) {
a.right.parent = a;
}
// switch a over to the left to be b's left child
b.left = a;
a.parent = b;
if (b.parent != nil) {
if (b.parent.right == a) {
b.parent.right = b;
} else {
b.parent.right = b;
}
}
[self setBalance:@[a,b]];
return b;
}
-(void) setBalance:(NSArray*)nodesArray {
for (AVLTreeNode* n in nodesArray) {
n.balance = [self height:n.right] - [self height:n.left];
}
}
-(int)height:(AVLTreeNode*)n {
if (n == nil) {
return -1;
}
return 1 + MAX([self height:n.left], [self height:n.right]);
}
-(void)printKey:(AVLTreeNode*)n {
if (n != nil) {
[self printKey:n.left];
NSLog(@"%ld", n.key);
[self printKey:n.right];
}
}
-(void)printBalance:(AVLTreeNode*)n {
if (n != nil) {
[self printBalance:n.left];
NSLog(@"%ld", n.balance);
[self printBalance:n.right];
}
}
@end
-- test
int main(int argc, const char * argv[]) {
@autoreleasepool {
AVLTree *tree = [AVLTree new];
NSLog(@"inserting values 1 to 6");
[tree insertWithKey:1];
[tree insertWithKey:2];
[tree insertWithKey:3];
[tree insertWithKey:4];
[tree insertWithKey:5];
[tree insertWithKey:6];
NSLog(@"printing balance: ");
[tree printBalance:tree.root];
NSLog(@"printing key: ");
[tree printKey:tree.root];
}
return 0;
}
|
http://rosettacode.org/wiki/Averages/Mean_angle | Averages/Mean angle | When calculating the average or mean of an angle one has to take into account how angles wrap around so that any angle in degrees plus any integer multiple of 360 degrees is a measure of the same angle.
If one wanted an average direction of the wind over two readings where the first reading was of 350 degrees and the second was of 10 degrees then the average of the numbers is 180 degrees, whereas if you can note that 350 degrees is equivalent to -10 degrees and so you have two readings at 10 degrees either side of zero degrees leading to a more fitting mean angle of zero degrees.
To calculate the mean angle of several angles:
Assume all angles are on the unit circle and convert them to complex numbers expressed in real and imaginary form.
Compute the mean of the complex numbers.
Convert the complex mean to polar coordinates whereupon the phase of the complex mean is the required angular mean.
(Note that, since the mean is the sum divided by the number of numbers, and division by a positive real number does not affect the angle, you can also simply compute the sum for step 2.)
You can alternatively use this formula:
Given the angles
α
1
,
…
,
α
n
{\displaystyle \alpha _{1},\dots ,\alpha _{n}}
the mean is computed by
α
¯
=
atan2
(
1
n
⋅
∑
j
=
1
n
sin
α
j
,
1
n
⋅
∑
j
=
1
n
cos
α
j
)
{\displaystyle {\bar {\alpha }}=\operatorname {atan2} \left({\frac {1}{n}}\cdot \sum _{j=1}^{n}\sin \alpha _{j},{\frac {1}{n}}\cdot \sum _{j=1}^{n}\cos \alpha _{j}\right)}
Task[edit]
write a function/method/subroutine/... that given a list of angles in degrees returns their mean angle.
(You should use a built-in function if you have one that does this for degrees or radians).
Use the function to compute the means of these lists of angles (in degrees):
[350, 10]
[90, 180, 270, 360]
[10, 20, 30]
Show your output here.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Logo | Logo | to mean_angle :angles
local "avgsin
make "avgsin quotient apply "sum map "sin :angles count :angles
local "avgcos
make "avgcos quotient apply "sum map "cos :angles count :angles
output (arctan :avgcos :avgsin)
end
foreach [[350 10] [90 180 270 360] [10 20 30]] [
print (sentence [The average of \(] ? [\) is] (mean_angle ?))
]
bye
|
http://rosettacode.org/wiki/Averages/Mean_angle | Averages/Mean angle | When calculating the average or mean of an angle one has to take into account how angles wrap around so that any angle in degrees plus any integer multiple of 360 degrees is a measure of the same angle.
If one wanted an average direction of the wind over two readings where the first reading was of 350 degrees and the second was of 10 degrees then the average of the numbers is 180 degrees, whereas if you can note that 350 degrees is equivalent to -10 degrees and so you have two readings at 10 degrees either side of zero degrees leading to a more fitting mean angle of zero degrees.
To calculate the mean angle of several angles:
Assume all angles are on the unit circle and convert them to complex numbers expressed in real and imaginary form.
Compute the mean of the complex numbers.
Convert the complex mean to polar coordinates whereupon the phase of the complex mean is the required angular mean.
(Note that, since the mean is the sum divided by the number of numbers, and division by a positive real number does not affect the angle, you can also simply compute the sum for step 2.)
You can alternatively use this formula:
Given the angles
α
1
,
…
,
α
n
{\displaystyle \alpha _{1},\dots ,\alpha _{n}}
the mean is computed by
α
¯
=
atan2
(
1
n
⋅
∑
j
=
1
n
sin
α
j
,
1
n
⋅
∑
j
=
1
n
cos
α
j
)
{\displaystyle {\bar {\alpha }}=\operatorname {atan2} \left({\frac {1}{n}}\cdot \sum _{j=1}^{n}\sin \alpha _{j},{\frac {1}{n}}\cdot \sum _{j=1}^{n}\cos \alpha _{j}\right)}
Task[edit]
write a function/method/subroutine/... that given a list of angles in degrees returns their mean angle.
(You should use a built-in function if you have one that does this for degrees or radians).
Use the function to compute the means of these lists of angles (in degrees):
[350, 10]
[90, 180, 270, 360]
[10, 20, 30]
Show your output here.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Lua | Lua | function meanAngle (angleList)
local sumSin, sumCos = 0, 0
for i, angle in pairs(angleList) do
sumSin = sumSin + math.sin(math.rad(angle))
sumCos = sumCos + math.cos(math.rad(angle))
end
local result = math.deg(math.atan2(sumSin, sumCos))
return string.format("%.2f", result)
end
print(meanAngle({350, 10}))
print(meanAngle({90, 180, 270, 360}))
print(meanAngle({10, 20, 30})) |
http://rosettacode.org/wiki/Averages/Median | Averages/Median | Task[edit]
Write a program to find the median value of a vector of floating-point numbers.
The program need not handle the case where the vector is empty, but must handle the case where there are an even number of elements. In that case, return the average of the two middle values.
There are several approaches to this. One is to sort the elements, and then pick the element(s) in the middle.
Sorting would take at least O(n logn). Another approach would be to build a priority queue from the elements, and then extract half of the elements to get to the middle element(s). This would also take O(n logn). The best solution is to use the selection algorithm to find the median in O(n) time.
See also
Quickselect_algorithm
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Elena | Elena | import system'routines;
import system'math;
import extensions;
extension op
{
get Median()
{
var sorted := self.ascendant();
var len := sorted.Length;
if (len == 0)
{
^ nil
}
else
{
var middleIndex := len / 2;
if (len.mod:2 == 0)
{
^ (sorted[middleIndex - 1] + sorted[middleIndex]) / 2
}
else
{
^ sorted[middleIndex]
}
}
}
}
public program()
{
var a1 := new real[]{4.1r, 5.6r, 7.2r, 1.7r, 9.3r, 4.4r, 3.2r};
var a2 := new real[]{4.1r, 7.2r, 1.7r, 9.3r, 4.4r, 3.2r};
console.printLine("median of (",a1.asEnumerable(),") is ",a1.Median);
console.printLine("median of (",a2.asEnumerable(),") is ",a2.Median);
console.readChar()
} |
http://rosettacode.org/wiki/Averages/Median | Averages/Median | Task[edit]
Write a program to find the median value of a vector of floating-point numbers.
The program need not handle the case where the vector is empty, but must handle the case where there are an even number of elements. In that case, return the average of the two middle values.
There are several approaches to this. One is to sort the elements, and then pick the element(s) in the middle.
Sorting would take at least O(n logn). Another approach would be to build a priority queue from the elements, and then extract half of the elements to get to the middle element(s). This would also take O(n logn). The best solution is to use the selection algorithm to find the median in O(n) time.
See also
Quickselect_algorithm
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Elixir | Elixir | defmodule Average do
def median([]), do: nil
def median(list) do
len = length(list)
sorted = Enum.sort(list)
mid = div(len, 2)
if rem(len,2) == 0, do: (Enum.at(sorted, mid-1) + Enum.at(sorted, mid)) / 2,
else: Enum.at(sorted, mid)
end
end
median = fn list -> IO.puts "#{inspect list} => #{inspect Average.median(list)}" end
median.([])
Enum.each(1..6, fn i ->
(for _ <- 1..i, do: :rand.uniform(6)) |> median.()
end) |
http://rosettacode.org/wiki/Averages/Pythagorean_means | Averages/Pythagorean means | Task[edit]
Compute all three of the Pythagorean means of the set of integers 1 through 10 (inclusive).
Show that
A
(
x
1
,
…
,
x
n
)
≥
G
(
x
1
,
…
,
x
n
)
≥
H
(
x
1
,
…
,
x
n
)
{\displaystyle A(x_{1},\ldots ,x_{n})\geq G(x_{1},\ldots ,x_{n})\geq H(x_{1},\ldots ,x_{n})}
for this set of positive integers.
The most common of the three means, the arithmetic mean, is the sum of the list divided by its length:
A
(
x
1
,
…
,
x
n
)
=
x
1
+
⋯
+
x
n
n
{\displaystyle A(x_{1},\ldots ,x_{n})={\frac {x_{1}+\cdots +x_{n}}{n}}}
The geometric mean is the
n
{\displaystyle n}
th root of the product of the list:
G
(
x
1
,
…
,
x
n
)
=
x
1
⋯
x
n
n
{\displaystyle G(x_{1},\ldots ,x_{n})={\sqrt[{n}]{x_{1}\cdots x_{n}}}}
The harmonic mean is
n
{\displaystyle n}
divided by the sum of the reciprocal of each item in the list:
H
(
x
1
,
…
,
x
n
)
=
n
1
x
1
+
⋯
+
1
x
n
{\displaystyle H(x_{1},\ldots ,x_{n})={\frac {n}{{\frac {1}{x_{1}}}+\cdots +{\frac {1}{x_{n}}}}}}
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #IS-BASIC | IS-BASIC | 100 PROGRAM "Averages.bas"
110 NUMERIC ARR(1 TO 10)
120 FOR I=LBOUND(ARR) TO UBOUND(ARR)
130 LET ARR(I)=I
140 NEXT
150 PRINT "Arithmetic mean =";ARITHM(ARR)
160 PRINT "Geometric mean =";GEOMETRIC(ARR)
170 PRINT "Harmonic mean =";HARMONIC(ARR)
180 DEF ARITHM(REF A)
190 LET T=0
200 FOR I=LBOUND(A) TO UBOUND(A)
210 LET T=T+A(I)
220 NEXT
230 LET ARITHM=T/SIZE(A)
240 END DEF
250 DEF GEOMETRIC(REF A)
260 LET T=1
270 FOR I=LBOUND(A) TO UBOUND(A)
280 LET T=T*A(I)
290 NEXT
300 LET GEOMETRIC=T^(1/SIZE(A))
310 END DEF
320 DEF HARMONIC(REF A)
330 LET T=0
340 FOR I=LBOUND(A) TO UBOUND(A)
350 LET T=T+(1/A(I))
360 NEXT
370 LET HARMONIC=SIZE(A)/T
380 END DEF |
http://rosettacode.org/wiki/Balanced_ternary | Balanced ternary | Balanced ternary is a way of representing numbers. Unlike the prevailing binary representation, a balanced ternary integer is in base 3, and each digit can have the values 1, 0, or −1.
Examples
Decimal 11 = 32 + 31 − 30, thus it can be written as "++−"
Decimal 6 = 32 − 31 + 0 × 30, thus it can be written as "+−0"
Task
Implement balanced ternary representation of integers with the following:
Support arbitrarily large integers, both positive and negative;
Provide ways to convert to and from text strings, using digits '+', '-' and '0' (unless you are already using strings to represent balanced ternary; but see requirement 5).
Provide ways to convert to and from native integer type (unless, improbably, your platform's native integer type is balanced ternary). If your native integers can't support arbitrary length, overflows during conversion must be indicated.
Provide ways to perform addition, negation and multiplication directly on balanced ternary integers; do not convert to native integers first.
Make your implementation efficient, with a reasonable definition of "efficient" (and with a reasonable definition of "reasonable").
Test case With balanced ternaries a from string "+-0++0+", b from native integer -436, c "+-++-":
write out a, b and c in decimal notation;
calculate a × (b − c), write out the result in both ternary and decimal notations.
Note: The pages generalised floating point addition and generalised floating point multiplication have code implementing arbitrary precision floating point balanced ternary.
| #Perl | Perl | use strict;
use warnings;
my @d = qw( 0 + - );
my @v = qw( 0 1 -1 );
sub to_bt {
my $n = shift;
my $b = '';
while( $n ) {
my $r = $n%3;
$b .= $d[$r];
$n -= $v[$r];
$n /= 3;
}
return scalar reverse $b;
}
sub from_bt {
my $n = 0;
for( split //, shift ) { # Horner
$n *= 3;
$n += "${_}1" if $_;
}
return $n;
}
my %addtable = (
'-0' => [ '-', '' ],
'+0' => [ '+', '' ],
'+-' => [ '0', '' ],
'00' => [ '0', '' ],
'--' => [ '+', '-' ],
'++' => [ '-', '+' ],
);
sub add {
my ($b1, $b2) = @_;
return ($b1 or $b2 ) unless ($b1 and $b2);
my $d = $addtable{ join '', sort substr( $b1, -1, 1, '' ), substr( $b2, -1, 1, '' ) };
return add( add($b1, $d->[1]), $b2 ).$d->[0];
}
sub unary_minus {
my $b = shift;
$b =~ tr/-+/+-/;
return $b;
}
sub subtract {
my ($b1, $b2) = @_;
return add( $b1, unary_minus $b2 );
}
sub mult {
my ($b1, $b2) = @_;
my $r = '0';
for( reverse split //, $b2 ){
$r = add $r, $b1 if $_ eq '+';
$r = subtract $r, $b1 if $_ eq '-';
$b1 .= '0';
}
$r =~ s/^0+//;
return $r;
}
my $a = "+-0++0+";
my $b = to_bt( -436 );
my $c = "+-++-";
my $d = mult( $a, subtract( $b, $c ) );
printf " a: %14s %10d\n", $a, from_bt( $a );
printf " b: %14s %10d\n", $b, from_bt( $b );
printf " c: %14s %10d\n", $c, from_bt( $c );
printf "a*(b-c): %14s %10d\n", $d, from_bt( $d );
|
http://rosettacode.org/wiki/Babbage_problem | Babbage problem |
Charles Babbage, looking ahead to the sorts of problems his Analytical Engine would be able to solve, gave this example:
What is the smallest positive integer whose square ends in the digits 269,696?
— Babbage, letter to Lord Bowden, 1837; see Hollingdale and Tootill, Electronic Computers, second edition, 1970, p. 125.
He thought the answer might be 99,736, whose square is 9,947,269,696; but he couldn't be certain.
Task[edit]
The task is to find out if Babbage had the right answer — and to do so, as far as your language allows it, in code that Babbage himself would have been able to read and understand.
As Babbage evidently solved the task with pencil and paper, a similar efficient solution is preferred.
For these purposes, Charles Babbage may be taken to be an intelligent person, familiar with mathematics and with the idea of a computer; he has written the first drafts of simple computer programmes in tabular form. [Babbage Archive Series L].
Motivation
The aim of the task is to write a program that is sufficiently clear and well-documented for such a person to be able to read it and be confident that it does indeed solve the specified problem.
| #PureBasic | PureBasic | EnableExplicit
Macro putresult(n)
If OpenConsole("Babbage_problem")
PrintN("The smallest number whose square ends in 269696 is " + Str(n))
Input()
EndIf
EndMacro
CompilerIf #PB_Processor_x64
#MAXINT = 1 << 63 - 1
CompilerElseIf #PB_Processor_x86
#MAXINT = 1 << 31 - 1
CompilerEndIf
#GOAL = 269696
#DIV = 1000000
Define n.i, q.i = Int(Sqr(#MAXINT))
For n = 2 To q Step 2
If (n*n) % #DIV = #GOAL : putresult(n) : Break : EndIf
Next |
http://rosettacode.org/wiki/Approximate_equality | Approximate equality | Sometimes, when testing whether the solution to a task (for example, here on Rosetta Code) is correct, the
difference in floating point calculations between different language implementations becomes significant.
For example, a difference between 32 bit and 64 bit floating point calculations may appear by
about the 8th significant digit in base 10 arithmetic.
Task
Create a function which returns true if two floating point numbers are approximately equal.
The function should allow for differences in the magnitude of numbers, so that, for example,
100000000000000.01 may be approximately equal to 100000000000000.011,
even though 100.01 is not approximately equal to 100.011.
If the language has such a feature in its standard library, this may be used instead of a custom function.
Show the function results with comparisons on the following pairs of values:
100000000000000.01, 100000000000000.011 (note: should return true)
100.01, 100.011 (note: should return false)
10000000000000.001 / 10000.0, 1000000000.0000001000
0.001, 0.0010000001
0.000000000000000000000101, 0.0
sqrt(2) * sqrt(2), 2.0
-sqrt(2) * sqrt(2), -2.0
3.14159265358979323846, 3.14159265358979324
Answers should be true for the first example and false in the second, so that just rounding the numbers to a fixed number of decimals should not be enough. Otherwise answers may vary and still be correct. See the Python code for one type of solution.
| #jq | jq | # Return whether the two numbers `a` and `b` are close.
# Closeness is determined by the `epsilon` parameter -
# the numbers are considered close if the difference between them
# is no more than epsilon * max(abs(a), abs(b)).
def isclose(a; b; epsilon):
((a - b) | fabs) <= (([(a|fabs), (b|fabs)] | max) * epsilon);
def lpad($len; $fill): tostring | ($len - length) as $l | ($fill * $l)[:$l] + .;
def lpad: lpad(20; " ");
# test values
def tv: [
{x: 100000000000000.01, y: 100000000000000.011 },
{x: 100.01, y: 100.011 },
{x: (10000000000000.001 / 10000.0), y: 1000000000.0000001000 },
{x: 0.001, y: 0.0010000001 },
{x: 0.000000000000000000000101, y: 0.0 },
{x: ((2|sqrt) * (2|sqrt)), y: 2.0 },
{x: (-(2|sqrt) * (2|sqrt)), y: -2.0 },
{x: 3.14159265358979323846, y: 3.14159265358979324 }
]
;
tv[] | "\(.x|lpad) \(if isclose(.x; .y; 1.0e-9) then " ≈ " else " ≉ " end) \(.y|lpad)"
|
http://rosettacode.org/wiki/Approximate_equality | Approximate equality | Sometimes, when testing whether the solution to a task (for example, here on Rosetta Code) is correct, the
difference in floating point calculations between different language implementations becomes significant.
For example, a difference between 32 bit and 64 bit floating point calculations may appear by
about the 8th significant digit in base 10 arithmetic.
Task
Create a function which returns true if two floating point numbers are approximately equal.
The function should allow for differences in the magnitude of numbers, so that, for example,
100000000000000.01 may be approximately equal to 100000000000000.011,
even though 100.01 is not approximately equal to 100.011.
If the language has such a feature in its standard library, this may be used instead of a custom function.
Show the function results with comparisons on the following pairs of values:
100000000000000.01, 100000000000000.011 (note: should return true)
100.01, 100.011 (note: should return false)
10000000000000.001 / 10000.0, 1000000000.0000001000
0.001, 0.0010000001
0.000000000000000000000101, 0.0
sqrt(2) * sqrt(2), 2.0
-sqrt(2) * sqrt(2), -2.0
3.14159265358979323846, 3.14159265358979324
Answers should be true for the first example and false in the second, so that just rounding the numbers to a fixed number of decimals should not be enough. Otherwise answers may vary and still be correct. See the Python code for one type of solution.
| #Julia | Julia | testvalues = [[100000000000000.01, 100000000000000.011],
[100.01, 100.011],
[10000000000000.001 / 10000.0, 1000000000.0000001000],
[0.001, 0.0010000001],
[0.000000000000000000000101, 0.0],
[sqrt(2) * sqrt(2), 2.0],
[-sqrt(2) * sqrt(2), -2.0],
[3.14159265358979323846, 3.14159265358979324]]
for (x, y) in testvalues
println(rpad(x, 21), " ≈ ", lpad(y, 22), ": ", x ≈ y)
end
|
http://rosettacode.org/wiki/Approximate_equality | Approximate equality | Sometimes, when testing whether the solution to a task (for example, here on Rosetta Code) is correct, the
difference in floating point calculations between different language implementations becomes significant.
For example, a difference between 32 bit and 64 bit floating point calculations may appear by
about the 8th significant digit in base 10 arithmetic.
Task
Create a function which returns true if two floating point numbers are approximately equal.
The function should allow for differences in the magnitude of numbers, so that, for example,
100000000000000.01 may be approximately equal to 100000000000000.011,
even though 100.01 is not approximately equal to 100.011.
If the language has such a feature in its standard library, this may be used instead of a custom function.
Show the function results with comparisons on the following pairs of values:
100000000000000.01, 100000000000000.011 (note: should return true)
100.01, 100.011 (note: should return false)
10000000000000.001 / 10000.0, 1000000000.0000001000
0.001, 0.0010000001
0.000000000000000000000101, 0.0
sqrt(2) * sqrt(2), 2.0
-sqrt(2) * sqrt(2), -2.0
3.14159265358979323846, 3.14159265358979324
Answers should be true for the first example and false in the second, so that just rounding the numbers to a fixed number of decimals should not be enough. Otherwise answers may vary and still be correct. See the Python code for one type of solution.
| #Kotlin | Kotlin | import kotlin.math.abs
import kotlin.math.sqrt
fun approxEquals(value: Double, other: Double, epsilon: Double): Boolean {
return abs(value - other) < epsilon
}
fun test(a: Double, b: Double) {
val epsilon = 1e-18
println("$a, $b => ${approxEquals(a, b, epsilon)}")
}
fun main() {
test(100000000000000.01, 100000000000000.011)
test(100.01, 100.011)
test(10000000000000.001 / 10000.0, 1000000000.0000001000)
test(0.001, 0.0010000001)
test(0.000000000000000000000101, 0.0)
test(sqrt(2.0) * sqrt(2.0), 2.0)
test(-sqrt(2.0) * sqrt(2.0), -2.0)
test(3.14159265358979323846, 3.14159265358979324)
} |
http://rosettacode.org/wiki/Balanced_brackets | Balanced brackets | Task:
Generate a string with N opening brackets [ and with N closing brackets ], in some arbitrary order.
Determine whether the generated string is balanced; that is, whether it consists entirely of pairs of opening/closing brackets (in that order), none of which mis-nest.
Examples
(empty) OK
[] OK
[][] OK
[[][]] OK
][ NOT OK
][][ NOT OK
[]][[] NOT OK
| #CoffeeScript | CoffeeScript |
isBalanced = (brackets) ->
openCount = 0
for bracket in brackets
openCount += if bracket is '[' then 1 else -1
return false if openCount < 0
openCount is 0
bracketsCombinations = (n) ->
for i in [0...Math.pow 2, n]
str = i.toString 2
str = '0' + str while str.length < n
str.replace(/0/g, '[').replace(/1/g, ']')
for brackets in bracketsCombinations 4
console.log brackets, isBalanced brackets
|
http://rosettacode.org/wiki/Append_a_record_to_the_end_of_a_text_file | Append a record to the end of a text file | Many systems offer the ability to open a file for writing, such that any data written will be appended to the end of the file. Further, the file operations will always adjust the position pointer to guarantee the end of the file, even in a multitasking environment.
This feature is most useful in the case of log files, where many jobs may be appending to the log file at the same time, or where care must be taken to avoid concurrently overwriting the same record from another job.
Task
Given a two record sample for a mythical "passwd" file:
Write these records out in the typical system format.
Ideally these records will have named fields of various types.
Close the file, then reopen the file for append.
Append a new record to the file and close the file again.
Take appropriate care to avoid concurrently overwrites from another job.
Open the file and demonstrate the new record has indeed written to the end.
Source record field types and contents.
account
password
UID
GID
fullname,office,extension,homephone,email
directory
shell
string
string
int
int
struct(string,string,string,string,string)
string
string
jsmith
x
1001
1000
Joe Smith,Room 1007,(234)555-8917,(234)555-0077,[email protected]
/home/jsmith
/bin/bash
jdoe
x
1002
1000
Jane Doe,Room 1004,(234)555-8914,(234)555-0044,[email protected]
/home/jdoe
/bin/bash
Record to be appended.
account
password
UID
GID
fullname,office,extension,homephone,email
directory
shell
string
string
int
int
struct(string,string,string,string,string)
string
string
xyz
x
1003
1000
X Yz,Room 1003,(234)555-8913,(234)555-0033,[email protected]
/home/xyz
/bin/bash
Resulting file format: should mimic Linux's /etc/passwd file format with particular attention to the "," separator used in the GECOS field. But if the specific language has a particular or unique format of storing records in text file, then this format should be named and demonstrated with an additional example.
Expected output:
Appended record: xyz:x:1003:1000:X Yz,Room 1003,(234)555-8913,(234)555-0033,[email protected]:/home/xyz:/bin/bash
Finally: Provide a summary of the language's "append record" capabilities in a table. eg.
Append Capabilities.
Data Representation
IO
Library
Append
Possible
Automatic
Append
Multi-tasking
Safe
In core
On disk
C struct
CSV text file
glibc/stdio
☑
☑
☑ (Not all, eg NFS)
Alternatively: If the language's appends can not guarantee its writes will always append, then note this restriction in the table. If possible, provide an actual code example (possibly using file/record locking) to guarantee correct concurrent appends.
| #C.2B.2B | C++ | #include <iostream>
#include <fstream>
#include <string>
#include <vector>
std::ostream& operator<<(std::ostream& out, const std::string s) {
return out << s.c_str();
}
struct gecos_t {
std::string fullname, office, extension, homephone, email;
friend std::ostream& operator<<(std::ostream&, const gecos_t&);
};
std::ostream& operator<<(std::ostream& out, const gecos_t& g) {
return out << g.fullname << ',' << g.office << ',' << g.extension << ',' << g.homephone << ',' << g.email;
}
struct passwd_t {
std::string account, password;
int uid, gid;
gecos_t gecos;
std::string directory, shell;
passwd_t(const std::string& a, const std::string& p, int u, int g, const gecos_t& ge, const std::string& d, const std::string& s)
: account(a), password(p), uid(u), gid(g), gecos(ge), directory(d), shell(s)
{
//empty
}
friend std::ostream& operator<<(std::ostream&, const passwd_t&);
};
std::ostream& operator<<(std::ostream& out, const passwd_t& p) {
return out << p.account << ':' << p.password << ':' << p.uid << ':' << p.gid << ':' << p.gecos << ':' << p.directory << ':' << p.shell;
}
std::vector<passwd_t> passwd_list{
{
"jsmith", "x", 1001, 1000,
{"Joe Smith", "Room 1007", "(234)555-8917", "(234)555-0077", "[email protected]"},
"/home/jsmith", "/bin/bash"
},
{
"jdoe", "x", 1002, 1000,
{"Jane Doe", "Room 1004", "(234)555-8914", "(234)555-0044", "[email protected]"},
"/home/jdoe", "/bin/bash"
}
};
int main() {
// Write the first two records
std::ofstream out_fd("passwd.txt");
for (size_t i = 0; i < passwd_list.size(); ++i) {
out_fd << passwd_list[i] << '\n';
}
out_fd.close();
// Append the third record
out_fd.open("passwd.txt", std::ios::app);
out_fd << passwd_t("xyz", "x", 1003, 1000, { "X Yz", "Room 1003", "(234)555-8913", "(234)555-0033", "[email protected]" }, "/home/xyz", "/bin/bash") << '\n';
out_fd.close();
// Verify the record was appended
std::ifstream in_fd("passwd.txt");
std::string line, temp;
while (std::getline(in_fd, temp)) {
// the last line of the file is empty, make sure line contains the last record
if (!temp.empty()) {
line = temp;
}
}
if (line.substr(0, 4) == "xyz:") {
std::cout << "Appended record: " << line << '\n';
} else {
std::cout << "Failed to find the expected record appended.\n";
}
return 0;
} |
http://rosettacode.org/wiki/Associative_array/Creation | Associative array/Creation | Task
The goal is to create an associative array (also known as a dictionary, map, or hash).
Related tasks:
Associative arrays/Iteration
Hash from two arrays
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #ALGOL_68 | ALGOL 68 | main:(
MODE COLOR = BITS;
FORMAT color repr = $"16r"16r6d$;
# This is an associative array which maps strings to ints #
MODE ITEM = STRUCT(STRING key, COLOR value);
REF[]ITEM color map items := LOC[0]ITEM;
PROC color map find = (STRING color)REF COLOR:(
REF COLOR out;
# linear search! #
FOR index FROM LWB key OF color map items TO UPB key OF color map items DO
IF color = key OF color map items[index] THEN
out := value OF color map items[index]; GO TO found
FI
OD;
NIL EXIT
found:
out
);
PROC color map = (STRING color)REF COLOR:(
REF COLOR out = color map find(color);
IF out :=: REF COLOR(NIL) THEN # extend color map array #
HEAP[UPB key OF color map items + 1]ITEM color map append;
color map append[:UPB key OF color map items] := color map items;
color map items := color map append;
value OF (color map items[UPB value OF color map items] := (color, 16r000000)) # black #
ELSE
out
FI
);
# First, populate it with some values #
color map("red") := 16rff0000;
color map("green") := 16r00ff00;
color map("blue") := 16r0000ff;
color map("my favourite color") := 16r00ffff;
# then, get some values out #
COLOR color := color map("green"); # color gets 16r00ff00 #
color := color map("black"); # accessing unassigned values assigns them to 16r0 #
# get some value out without accidentally inserting new ones #
REF COLOR value = color map find("green");
IF value :=: REF COLOR(NIL) THEN
put(stand error, ("color not found!", new line))
ELSE
printf(($"green: "f(color repr)l$, value))
FI;
# Now I changed my mind about my favourite color, so change it #
color map("my favourite color") := 16r337733;
# print out all defined colors #
FOR index FROM LWB color map items TO UPB color map items DO
ITEM item = color map items[index];
putf(stand error, ($"color map("""g""") = "f(color repr)l$, item))
OD;
FORMAT fmt;
FORMAT comma sep = $"("n(UPB color map items-1)(f(fmt)", ")f(fmt)")"$;
fmt := $""""g""""$;
printf(($g$,"keys: ", comma sep, key OF color map items, $l$));
fmt := color repr;
printf(($g$,"values: ", comma sep, value OF color map items, $l$))
) |
http://rosettacode.org/wiki/Anti-primes | Anti-primes | The anti-primes
(or highly composite numbers, sequence A002182 in the OEIS)
are the natural numbers with more factors than any smaller than itself.
Task
Generate and show here, the first twenty anti-primes.
Related tasks
Factors of an integer
Sieve of Eratosthenes
| #ALGOL_W | ALGOL W | begin
% find some anti-primes - numbers with more factors than the numbers %
% smaller than them %
% calculates the number of divisors of v %
integer procedure divisor_count( integer value v ) ; begin
integer total, n, p;
total := 1; n := v;
while not odd( n ) do begin
total := total + 1;
n := n div 2
end while_not_odd_n ;
p := 3;
while ( p * p ) <= n do begin
integer count;
count := 1;
while n rem p = 0 do begin
count := count + 1;
n := n div p
end while_n_rem_p_eq_0 ;
p := p + 2;
total := total * count
end while_p_x_p_le_n ;
if n > 1 then total := total * 2;
total
end divisor_count ;
begin
integer maxAntiPrime, antiPrimeCount, maxDivisors, n;
maxAntiPrime := 20;
n := maxDivisors := antiPrimeCount := 0;
while antiPrimeCount < maxAntiPrime do begin
integer divisors;
n := n + 1;
divisors := divisor_count( n );
if divisors > maxDivisors then begin
writeon( i_w := 1, s_w := 0, " ", n );
maxDivisors := divisors;
antiPrimeCount := antiPrimeCount + 1
end if_have_an_anti_prime
end while_antiPrimeCoiunt_lt_maxAntiPrime
end
end. |
http://rosettacode.org/wiki/Anti-primes | Anti-primes | The anti-primes
(or highly composite numbers, sequence A002182 in the OEIS)
are the natural numbers with more factors than any smaller than itself.
Task
Generate and show here, the first twenty anti-primes.
Related tasks
Factors of an integer
Sieve of Eratosthenes
| #AppleScript | AppleScript | on factorCount(n)
set counter to 0
set sqrt to n ^ 0.5
set limit to sqrt div 1
if (limit = sqrt) then
set counter to counter + 1
set limit to limit - 1
end if
repeat with i from limit to 1 by -1
if (n mod i is 0) then set counter to counter + 2
end repeat
return counter
end factorCount
on antiprimes(howMany)
set output to {}
set mostFactorsSoFar to 0
set n to 0
repeat until ((count output) = howMany)
set n to n + 1
tell (factorCount(n))
if (it > mostFactorsSoFar) then
set end of output to n
set mostFactorsSoFar to it
end if
end tell
end repeat
return output
end antiprimes
antiprimes(20) |
http://rosettacode.org/wiki/Atomic_updates | Atomic updates |
Task
Define a data type consisting of a fixed number of 'buckets', each containing a nonnegative integer value, which supports operations to:
get the current value of any bucket
remove a specified amount from one specified bucket and add it to another, preserving the total of all bucket values, and clamping the transferred amount to ensure the values remain non-negative
In order to exercise this data type, create one set of buckets, and start three concurrent tasks:
As often as possible, pick two buckets and make their values closer to equal.
As often as possible, pick two buckets and arbitrarily redistribute their values.
At whatever rate is convenient, display (by any means) the total value and, optionally, the individual values of each bucket.
The display task need not be explicit; use of e.g. a debugger or trace tool is acceptable provided it is simple to set up to provide the display.
This task is intended as an exercise in atomic operations. The sum of the bucket values must be preserved even if the two tasks attempt to perform transfers simultaneously, and a straightforward solution is to ensure that at any time, only one transfer is actually occurring — that the transfer operation is atomic.
| #Groovy | Groovy | class Buckets {
def cells = []
final n
Buckets(n, limit=1000, random=new Random()) {
this.n = n
(0..<n).each {
cells << random.nextInt(limit)
}
}
synchronized getAt(i) {
cells[i]
}
synchronized transfer(from, to, amount) {
assert from in (0..<n) && to in (0..<n)
def cappedAmt = [cells[from], amount].min()
cells[from] -= cappedAmt
cells[to] += cappedAmt
}
synchronized String toString() { cells.toString() }
}
def random = new Random()
def buckets = new Buckets(5)
def makeCloser = { i, j ->
synchronized(buckets) {
def targetDiff = (buckets[i]-buckets[j]).intdiv(2)
if (targetDiff < 0) {
buckets.transfer(j, i, -targetDiff)
} else {
buckets.transfer(i, j, targetDiff)
}
}
}
def randomize = { i, j ->
synchronized(buckets) {
def targetLimit = buckets[i] + buckets[j]
def targetI = random.nextInt(targetLimit + 1)
if (targetI < buckets[i]) {
buckets.transfer(i, j, buckets[i] - targetI)
} else {
buckets.transfer(j, i, targetI - buckets[i])
}
}
}
Thread.start {
def start = System.currentTimeMillis()
while (start + 10000 > System.currentTimeMillis()) {
def i = random.nextInt(buckets.n)
def j = random.nextInt(buckets.n)
makeCloser(i, j)
}
}
Thread.start {
def start = System.currentTimeMillis()
while (start + 10000 > System.currentTimeMillis()) {
def i = random.nextInt(buckets.n)
def j = random.nextInt(buckets.n)
randomize(i, j)
}
}
def start = System.currentTimeMillis()
while (start + 10000 > System.currentTimeMillis()) {
synchronized(buckets) {
def sum = buckets.cells.sum()
println "${new Date()}: checksum: ${sum} buckets: ${buckets}"
}
Thread.sleep(500)
} |
http://rosettacode.org/wiki/Assertions | Assertions | Assertions are a way of breaking out of code when there is an error or an unexpected input.
Some languages throw exceptions and some treat it as a break point.
Task
Show an assertion in your language by asserting that an integer variable is equal to 42.
| #Emacs_Lisp | Emacs Lisp | (require 'cl-lib)
(let ((x 41))
(cl-assert (= x 42) t "This shouldn't happen")) |
http://rosettacode.org/wiki/Assertions | Assertions | Assertions are a way of breaking out of code when there is an error or an unexpected input.
Some languages throw exceptions and some treat it as a break point.
Task
Show an assertion in your language by asserting that an integer variable is equal to 42.
| #Erlang | Erlang | 1> N = 42.
42
2> N = 43.
** exception error: no match of right hand side value 43
3> N = 42.
42
4> 44 = N.
** exception error: no match of right hand side value 42
5> 42 = N.
42 |
http://rosettacode.org/wiki/Apply_a_callback_to_an_array | Apply a callback to an array | Task
Take a combined set of elements and apply a function to each element.
| #ALGOL_W | ALGOL W | begin
procedure printSquare ( integer value x ) ; writeon( i_w := 1, s_w := 0, " ", x * x );
% applys f to each element of a from lb to ub (inclusive) %
procedure applyI ( procedure f; integer array a ( * ); integer value lb, ub ) ;
for i := lb until ub do f( a( i ) );
% test applyI %
begin
integer array a ( 1 :: 3 );
a( 1 ) := 1; a( 2 ) := 2; a( 3 ) := 3;
applyI( printSquare, a, 1, 3 )
end
end. |
http://rosettacode.org/wiki/Apply_a_callback_to_an_array | Apply a callback to an array | Task
Take a combined set of elements and apply a function to each element.
| #APL | APL | - 1 2 3
¯1 ¯2 ¯3
2 * 1 2 3 4
2 4 8 16
2 × ⍳4
2 4 6 8
3 * 3 3 ⍴ ⍳9
3 9 27
81 243 729
2187 6561 19683
|
http://rosettacode.org/wiki/Averages/Mode | Averages/Mode | Task[edit]
Write a program to find the mode value of a collection.
The case where the collection is empty may be ignored. Care must be taken to handle the case where the mode is non-unique.
If it is not appropriate or possible to support a general collection, use a vector (array), if possible. If it is not appropriate or possible to support an unspecified value type, use integers.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #MUMPS | MUMPS | MODE(X)
;X is assumed to be a list of numbers separated by "^"
;I is a loop index
;L is the length of X
;Y is a new array
;ML is the list of modes
;LOC is a placeholder to shorten the statement
Q:'$DATA(X) "No data"
Q:X="" "Empty Set"
NEW Y,I,L,LOC
SET L=$LENGTH(X,"^"),ML=""
FOR I=1:1:L SET LOC=+$P(X,"^",I),Y(LOC)=$S($DATA(Y(LOC)):Y(LOC)+1,1:1)
SET I="",I=$O(Y(I)),ML=I ;Prime the pump, rather than test for no data
FOR S I=$O(Y(I)) Q:I="" S ML=$SELECT(Y($P(ML,"^"))>Y(I):ML,Y($P(ML,"^"))<Y(I):I,Y($P(ML,"^"))=Y(I):ML_"^"_I)
QUIT ML |
http://rosettacode.org/wiki/Associative_array/Iteration | Associative array/Iteration | Show how to iterate over the key-value pairs of an associative array, and print each pair out.
Also show how to iterate just over the keys, or the values, if there is a separate way to do that in your language.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #C | C | using System;
using System.Collections.Generic;
namespace AssocArrays
{
class Program
{
static void Main(string[] args)
{
Dictionary<string,int> assocArray = new Dictionary<string,int>();
assocArray["Hello"] = 1;
assocArray.Add("World", 2);
assocArray["!"] = 3;
foreach (KeyValuePair<string, int> kvp in assocArray)
{
Console.WriteLine(kvp.Key + " : " + kvp.Value);
}
foreach (string key in assocArray.Keys)
{
Console.WriteLine(key);
}
foreach (int val in assocArray.Values)
{
Console.WriteLine(val.ToString());
}
}
}
}
|
http://rosettacode.org/wiki/Apply_a_digital_filter_(direct_form_II_transposed) | Apply a digital filter (direct form II transposed) | Digital filters are used to apply a mathematical operation to a sampled signal. One of the common formulations is the "direct form II transposed" which can represent both infinite impulse response (IIR) and finite impulse response (FIR) filters, as well as being more numerically stable than other forms. [1]
Task
Filter a signal using an order 3 low-pass Butterworth filter. The coefficients for the filter are a=[1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17] and b = [0.16666667, 0.5, 0.5, 0.16666667]
The signal that needs filtering is the following vector: [-0.917843918645, 0.141984778794, 1.20536903482, 0.190286794412, -0.662370894973, -1.00700480494, -0.404707073677 ,0.800482325044, 0.743500089861, 1.01090520172, 0.741527555207, 0.277841675195, 0.400833448236, -0.2085993586, -0.172842103641, -0.134316096293, 0.0259303398477, 0.490105989562, 0.549391221511, 0.9047198589]
| #Go | Go | package main
import "fmt"
type filter struct {
b, a []float64
}
func (f filter) filter(in []float64) []float64 {
out := make([]float64, len(in))
s := 1. / f.a[0]
for i := range in {
tmp := 0.
b := f.b
if i+1 < len(b) {
b = b[:i+1]
}
for j, bj := range b {
tmp += bj * in[i-j]
}
a := f.a[1:]
if i < len(a) {
a = a[:i]
}
for j, aj := range a {
tmp -= aj * out[i-j-1]
}
out[i] = tmp * s
}
return out
}
//Constants for a Butterworth filter (order 3, low pass)
var bwf = filter{
a: []float64{1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17},
b: []float64{0.16666667, 0.5, 0.5, 0.16666667},
}
var sig = []float64{
-0.917843918645, 0.141984778794, 1.20536903482, 0.190286794412,
-0.662370894973, -1.00700480494, -0.404707073677, 0.800482325044,
0.743500089861, 1.01090520172, 0.741527555207, 0.277841675195,
0.400833448236, -0.2085993586, -0.172842103641, -0.134316096293,
0.0259303398477, 0.490105989562, 0.549391221511, 0.9047198589,
}
func main() {
for _, v := range bwf.filter(sig) {
fmt.Printf("%9.6f\n", v)
}
} |
http://rosettacode.org/wiki/Averages/Arithmetic_mean | Averages/Arithmetic mean | Task[edit]
Write a program to find the mean (arithmetic average) of a numeric vector.
In case of a zero-length input, since the mean of an empty set of numbers is ill-defined, the program may choose to behave in any way it deems appropriate, though if the programming language has an established convention for conveying math errors or undefined values, it's preferable to follow it.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Chef | Chef | Mean.
Chef has no way to detect EOF, so rather than interpreting
some arbitrary number as meaning "end of input", this program
expects the first input to be the sample size. Pass in the samples
themselves as the other inputs. For example, if you wanted to
compute the mean of 10, 100, 47, you could pass in 3, 10, 100, and
47. To test the "zero-length vector" case, you need to pass in 0.
Ingredients.
0 g Sample Size
0 g Counter
0 g Current Sample
Method.
Take Sample Size from refrigerator.
Put Sample Size into mixing bowl.
Fold Counter into mixing bowl.
Put Current Sample into mixing bowl.
Loop Counter.
Take Current Sample from refrigerator.
Add Current Sample into mixing bowl.
Endloop Counter until looped.
If Sample Size.
Divide Sample Size into mixing bowl.
Put Counter into 2nd mixing bowl.
Fold Sample Size into 2nd mixing bowl.
Endif until ifed.
Pour contents of mixing bowl into baking dish.
Serves 1. |
http://rosettacode.org/wiki/Averages/Arithmetic_mean | Averages/Arithmetic mean | Task[edit]
Write a program to find the mean (arithmetic average) of a numeric vector.
In case of a zero-length input, since the mean of an empty set of numbers is ill-defined, the program may choose to behave in any way it deems appropriate, though if the programming language has an established convention for conveying math errors or undefined values, it's preferable to follow it.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Clojure | Clojure | (defn mean [sq]
(if (empty? sq)
0
(/ (reduce + sq) (count sq)))) |
http://rosettacode.org/wiki/Associative_array/Merging | Associative array/Merging | Task
Define two associative arrays, where one represents the following "base" data:
Key
Value
"name"
"Rocket Skates"
"price"
12.75
"color"
"yellow"
And the other represents "update" data:
Key
Value
"price"
15.25
"color"
"red"
"year"
1974
Merge these into a new associative array that contains every key found in either of the source ones. Each key should map to the value in the second (update) table if that exists, or else to the value in the first (base) table. If possible, do this in a way that does not mutate the original two associative arrays. Obviously this should be done in a way that would work for any data, not just the specific data given here, but in this example the result should be:
Key
Value
"name"
"Rocket Skates"
"price"
15.25
"color"
"red"
"year"
1974
| #Ol | Ol |
(define a1 {
'name "Rocket Skates"
'price 12.75
'color "yellow"
})
(define a2 {
'price 15.25
'color "red"
'year 1974
})
(print "a1: " a1)
(print "a2: " a2)
(define (collide a b) b) ; will use new key value
(print "merged a1 a2: " (ff-union collide a1 a2))
|
http://rosettacode.org/wiki/Associative_array/Merging | Associative array/Merging | Task
Define two associative arrays, where one represents the following "base" data:
Key
Value
"name"
"Rocket Skates"
"price"
12.75
"color"
"yellow"
And the other represents "update" data:
Key
Value
"price"
15.25
"color"
"red"
"year"
1974
Merge these into a new associative array that contains every key found in either of the source ones. Each key should map to the value in the second (update) table if that exists, or else to the value in the first (base) table. If possible, do this in a way that does not mutate the original two associative arrays. Obviously this should be done in a way that would work for any data, not just the specific data given here, but in this example the result should be:
Key
Value
"name"
"Rocket Skates"
"price"
15.25
"color"
"red"
"year"
1974
| #Perl | Perl | use strict;
use warnings;
my %base = ("name" => "Rocket Skates", "price" => 12.75, "color" => "yellow");
my %more = ("price" => 15.25, "color" => "red", "year" => 1974);
print "Update\n";
my %update = (%base, %more);
printf "%-7s %s\n", $_, $update{$_} for sort keys %update;
print "\nMerge\n";
my %merge;
$merge{$_} = [$base{$_}] for keys %base;
push @{$merge{$_}}, $more{$_} for keys %more;
printf "%-7s %s\n", $_, join ', ', @{$merge{$_}} for sort keys %merge; |
http://rosettacode.org/wiki/Average_loop_length | Average loop length | Let f be a uniformly-randomly chosen mapping from the numbers 1..N to the numbers 1..N (note: not necessarily a permutation of 1..N; the mapping could produce a number in more than one way or not at all). At some point, the sequence 1, f(1), f(f(1))... will contain a repetition, a number that occurring for the second time in the sequence.
Task
Write a program or a script that estimates, for each N, the average length until the first such repetition.
Also calculate this expected length using an analytical formula, and optionally compare the simulated result with the theoretical one.
This problem comes from the end of Donald Knuth's Christmas tree lecture 2011.
Example of expected output:
N average analytical (error)
=== ========= ============ =========
1 1.0000 1.0000 ( 0.00%)
2 1.4992 1.5000 ( 0.05%)
3 1.8784 1.8889 ( 0.56%)
4 2.2316 2.2188 ( 0.58%)
5 2.4982 2.5104 ( 0.49%)
6 2.7897 2.7747 ( 0.54%)
7 3.0153 3.0181 ( 0.09%)
8 3.2429 3.2450 ( 0.07%)
9 3.4536 3.4583 ( 0.14%)
10 3.6649 3.6602 ( 0.13%)
11 3.8091 3.8524 ( 1.12%)
12 3.9986 4.0361 ( 0.93%)
13 4.2074 4.2123 ( 0.12%)
14 4.3711 4.3820 ( 0.25%)
15 4.5275 4.5458 ( 0.40%)
16 4.6755 4.7043 ( 0.61%)
17 4.8877 4.8579 ( 0.61%)
18 4.9951 5.0071 ( 0.24%)
19 5.1312 5.1522 ( 0.41%)
20 5.2699 5.2936 ( 0.45%)
| #Phix | Phix | constant MAX = 20,
ITER = 1000000
function expected(integer n)
atom sum = 0
for i=1 to n do
sum += factorial(n) / power(n,i) / factorial(n-i)
end for
return sum
end function
function test(integer n)
integer count = 0, x, bits
for i=1 to ITER do
x = 1
bits = 0
while not and_bits(bits,x) do
count += 1
bits = or_bits(bits,x)
x = power(2,rand(n)-1)
end while
end for
return count/ITER
end function
atom av, ex
puts(1," n avg. exp. (error%)\n");
puts(1,"== ====== ====== ========\n");
for n=1 to MAX do
av = test(n)
ex = expected(n)
printf(1,"%2d %8.4f %8.4f (%5.3f%%)\n", {n,av,ex,abs(1-av/ex)*100})
end for
|
http://rosettacode.org/wiki/Averages/Simple_moving_average | Averages/Simple moving average | Computing the simple moving average of a series of numbers.
Task[edit]
Create a stateful function/class/instance that takes a period and returns a routine that takes a number as argument and returns a simple moving average of its arguments so far.
Description
A simple moving average is a method for computing an average of a stream of numbers by only averaging the last P numbers from the stream, where P is known as the period.
It can be implemented by calling an initialing routine with P as its argument, I(P), which should then return a routine that when called with individual, successive members of a stream of numbers, computes the mean of (up to), the last P of them, lets call this SMA().
The word stateful in the task description refers to the need for SMA() to remember certain information between calls to it:
The period, P
An ordered container of at least the last P numbers from each of its individual calls.
Stateful also means that successive calls to I(), the initializer, should return separate routines that do not share saved state so they could be used on two independent streams of data.
Pseudo-code for an implementation of SMA is:
function SMA(number: N):
stateful integer: P
stateful list: stream
number: average
stream.append_last(N)
if stream.length() > P:
# Only average the last P elements of the stream
stream.delete_first()
if stream.length() == 0:
average = 0
else:
average = sum( stream.values() ) / stream.length()
return average
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Ruby | Ruby | def simple_moving_average(size)
nums = []
sum = 0.0
lambda do |hello|
nums << hello
goodbye = nums.length > size ? nums.shift : 0
sum += hello - goodbye
sum / nums.length
end
end
ma3 = simple_moving_average(3)
ma5 = simple_moving_average(5)
(1.upto(5).to_a + 5.downto(1).to_a).each do |num|
printf "Next number = %d, SMA_3 = %.3f, SMA_5 = %.1f\n",
num, ma3.call(num), ma5.call(num)
end |
http://rosettacode.org/wiki/Attractive_numbers | Attractive numbers | A number is an attractive number if the number of its prime factors (whether distinct or not) is also prime.
Example
The number 20, whose prime decomposition is 2 × 2 × 5, is an attractive number because the number of its prime factors (3) is also prime.
Task
Show sequence items up to 120.
Reference
The OEIS entry: A063989: Numbers with a prime number of prime divisors.
| #Haskell | Haskell | import Data.Numbers.Primes
import Data.Bool (bool)
attractiveNumbers :: [Integer]
attractiveNumbers =
[1 ..] >>= (bool [] . return) <*> (isPrime . length . primeFactors)
main :: IO ()
main = print $ takeWhile (<= 120) attractiveNumbers |
http://rosettacode.org/wiki/Averages/Mean_time_of_day | Averages/Mean time of day | Task[edit]
A particular activity of bats occurs at these times of the day:
23:00:17, 23:40:20, 00:12:45, 00:17:19
Using the idea that there are twenty-four hours in a day,
which is analogous to there being 360 degrees in a circle,
map times of day to and from angles;
and using the ideas of Averages/Mean angle
compute and show the average time of the nocturnal activity
to an accuracy of one second of time.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Nim | Nim | import math, complex, strutils, sequtils
proc meanAngle(deg: openArray[float]): float =
var c: Complex[float]
for d in deg:
c += rect(1.0, degToRad(d))
radToDeg(phase(c / float(deg.len)))
proc meanTime(times: openArray[string]): string =
const day = 24 * 60 * 60
let
angles = times.map(proc(time: string): float =
let t = time.split(":")
(t[2].parseInt + t[1].parseInt * 60 + t[0].parseInt * 3600) * 360 / day)
ms = (angles.meanAngle * day / 360 + day) mod day
(h,m,s) = (ms.int div 3600, (ms.int mod 3600) div 60, ms.int mod 60)
align($h, 2, '0') & ":" & align($m, 2, '0') & ":" & align($s, 2, '0')
echo meanTime(["23:00:17", "23:40:20", "00:12:45", "00:17:19"]) |
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